\input{"preamble.tex"} \addbibresource{HomologicalAlgebra.bib} \let\Begin\begin \let\End\end \newcommand\wrapenv[1]{#1} \makeatletter \def\ScaleWidthIfNeeded{% \ifdim\Gin@nat@width>\linewidth \linewidth \else \Gin@nat@width \fi } \def\ScaleHeightIfNeeded{% \ifdim\Gin@nat@height>0.9\textheight 0.9\textheight \else \Gin@nat@width \fi } \makeatother \setkeys{Gin}{width=\ScaleWidthIfNeeded,height=\ScaleHeightIfNeeded,keepaspectratio}% \title{ \rule{\linewidth}{1pt} \\ \textbf{ Homological Algebra } \\ {\normalsize Lectures by Brian Boe. University of Georgia, Spring 2021} \\ \rule{\linewidth}{2pt} } \titlehead{ \begin{center} \includegraphics[width=\linewidth,height=0.45\textheight,keepaspectratio]{figures/cover.png} \end{center} \begin{minipage}{.35\linewidth} \begin{flushleft} \vspace{2em} {\fontsize{6pt}{2pt} \textit{Notes: These are notes live-tex'd from a graduate course in Homological Algebra taught by Brian Boe at the University of Georgia in Spring 2021. As such, any errors or inaccuracies are almost certainly my own. } } \\ \end{flushleft} \end{minipage} \hfill \begin{minipage}{.65\linewidth} \end{minipage} } \begin{document} \date{} \author{D. Zack Garza} \maketitle \begin{flushleft} \textit{D. Zack Garza} \\ \textit{University of Georgia} \\ \textit{\href{mailto: dzackgarza@gmail.com}{dzackgarza@gmail.com}} \\ {\tiny \textit{Last updated:} 2021-05-02 } \end{flushleft} \newpage % Note: addsec only in KomaScript \addsec{Table of Contents} \tableofcontents \newpage \hypertarget{wednesday-january-13}{% \section{Wednesday, January 13}\label{wednesday-january-13}} Reference: \begin{itemize} \item The course text is Weibel \autocite{weibel_2011}. \item See the many corrections/errata: \url{http://www.math.rutgers.edu/~weibel/Hbook-corrections.html} \item Sections we'll cover: \begin{itemize} \tightlist \item 1.1-1.5, \item 2.2-2.7, \item 3.4, \item 3.6, \item 6.1, \item 5.1-5.2, \item 5.4-5.8, \item 6.8, \item 6.7, \item 6.3, \item 7.1-7.5, \item 7.7-7.8, \item Appendix A (when needed) \end{itemize} \item Course Website: \url{https://uga.view.usg.edu/d2l/le/content/2218619/viewContent/33763436/View} \end{itemize} \hypertarget{overview}{% \subsection{Overview}\label{overview}} \begin{definition}[Exact complexes] A \textbf{complex} is given by \begin{align*} \cdots \xrightarrow{d_{i-1}} M_{i-1} \xrightarrow{d_i} M_i \xrightarrow{d_{i+1}}M_{i+1} \to \cdots .\end{align*} where $$M_i \in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}$$ and $$d_i \circ d_{i-1} = 0$$, which happens if and only if $$\operatorname{im}d_{i-1} \subseteq \ker d_i$$. If $$\operatorname{im}d_{i-1} = \ker d_i$$, this complex is \textbf{exact}. \end{definition} \begin{example}[?] We can apply a functor such as $$\otimes_R N$$ to get a new complex \begin{align*} \cdots \xrightarrow{d_{i-1} \otimes 1_N} M_{i-1} \otimes_R N \xrightarrow{d_i \otimes 1} M_i \otimes N \to M_{i+1} \xrightarrow{d_{i+1} \otimes 1} \cdots .\end{align*} \end{example} \begin{example}[?] Applying $$\mathop{\mathrm{Hom}}(N, {-})$$ similarly yields \begin{align*} \mathop{\mathrm{Hom}}_R(N, M_{i}) \xrightarrow{d_{i-1}^*} \mathop{\mathrm{Hom}}_R(N, M_{i+1}) ,\end{align*} where $$d_i^* = d_i \circ ({-})$$ is given by composition. \end{example} \begin{example}[?] Applying $$\mathop{\mathrm{Hom}}({-}, N)$$ yields \begin{align*} \mathop{\mathrm{Hom}}_R(M_i, N) \xrightarrow{d_{i}^*} \mathop{\mathrm{Hom}}_R(M_{i+1}, N) \end{align*} where $$d_i^* = ({-}) \circ d_i$$. \end{example} \begin{remark} Note that we can also take complexes with arrows in the other direction. For $$F$$ a functor, we can rewrite these examples as \begin{align*} d_i^* \circ d_{i-1}^* = F(d_i) \circ F(d_{i-1}) = F(d_i \circ d_{i-1}) = F(0) = 0 ,\end{align*} provided $$F$$ is nice enough and sends zero to zero. This follows from the fact that functors preserve composition. Even if the original complex is exact, the new one may not be, so we can define the following: \end{remark} \begin{definition}[Cohomology] \begin{align*} H^i(M^*) = \ker d_i^* / \operatorname{im}d_{i-1}^* .\end{align*} \end{definition} \begin{remark} These will lead to \textbf{$$i$$th derived functors}, and category theory will be useful here. See appendix in Weibel. For a category $$\mathcal{C}$$ we'll define \begin{itemize} \tightlist \item $$\mathrm{Obj}(\mathcal{C} )$$ as the objects \item $$\mathop{\mathrm{Hom}}_{\mathcal{C}}(A, B)$$ a set of morphisms between them, where a more modern notation might be $$\mathrm{Mor}(A, B)$$. \item Morphisms compose: $$A \xrightarrow{f} B \xrightarrow{g} C$$ means that $$g\circ f \in \mathop{\mathrm{Hom}}_{\mathcal{C}}(A, C)$$ \item Associativity \item Identity morphisms \end{itemize} See the appendix for diagrams defining zero objects and the zero map, which we'll need to make sense of exactness. We'll also needs notions of kernels and images, or potentially cokernels instead of images since they're closely related. \end{remark} \begin{remark} In the examples, we had $$\ker d_i \subseteq M_i$$, but this need not be true since the objects in the category may not be sets. Such an example is the category of complexes of $$R{\hbox{-}}$$modules: $$\operatorname{Cx}({\mathsf{R}{\hbox{-}}\mathsf{Mod}})$$. In this setting, kernels will be subcomplexes but not subsets. \end{remark} \begin{definition}[Functors] Recall that \textbf{functors} are functions'' between categories $$F: \mathcal{C}\to \mathcal{D}$$ such that \begin{itemize} \item Objects are sent to objects, \item Morphisms are sent to morphisms, so $$A \xrightarrow{f} B \leadsto F(A) \xrightarrow{F(f)} F(B)$$, \item $$F$$ respects composition and identities \end{itemize} \end{definition} \begin{example}[Hom] $$\mathop{\mathrm{Hom}}_R(N, {-}): {\mathsf{R}{\hbox{-}}\mathsf{Mod}}\to {\mathsf{Ab}}$$, noting that the hom set may not have an $$R{\hbox{-}}$$module structure. \end{example} \begin{remark} Taking cohomology yields the $$i$$th derived functors of $$F$$, for example $$\operatorname{Ext}^i, \operatorname{Tor}_i$$. Recall that functors can be \emph{covariant} or contravariant. See section 1 for formulating simplicial and singular homology (from topology) in this language. \end{remark} \hypertarget{chapter-1-chain-complexes}{% \subsection{Chapter 1: Chain Complexes}\label{chapter-1-chain-complexes}} \hypertarget{complexes-of-rhbox-modules}{% \subsubsection{\texorpdfstring{Complexes of $$R{\hbox{-}}$$modules}{Complexes of R\{\textbackslash hbox\{-\}\}modules}}\label{complexes-of-rhbox-modules}} \begin{definition}[Exactness] Let $$R$$ be a ring with 1 and define $${\mathsf{R}{\hbox{-}}\mathsf{Mod}}$$ to be the category of \emph{right} $$R{\hbox{-}}$$modules. $$A \xrightarrow{f} B \xrightarrow{g} C$$ is \textbf{exact} if and only if $$\ker g = \operatorname{im}f$$, and in particular $$g\circ f = 0$$. \end{definition} \begin{definition}[Chain Complex] A \textbf{chain complex} is \begin{align*} C_{-}\coloneqq(C_{-}, d_{-}) \coloneqq\qty{ \cdots \to C_{n+1} \xrightarrow{d_{n+1}} C_n \xrightarrow{d_n} C_{n-1} \to \cdots } \end{align*} for $$n \in {\mathbb{Z}}$$ such that $$d_n \circ d_{n+1} = 0$$. We drop the $$n$$ from the notation and write $$d^2 \coloneqq d\circ d = 0$$. \end{definition} \begin{definition}[Cycles and boundaries] \envlist \begin{itemize} \tightlist \item $$Z_n = Z_n(C_{-}) = \ker d_n$$ are referred to as \textbf{$$n{\hbox{-}}$$cycles}. \item $$B_n = B_n(C_{-}) = \operatorname{im}d_{n+1}$$ are the \textbf{$$n{\hbox{-}}$$boundaries}. \end{itemize} \end{definition} \begin{definition}[Homology of a chain complex] Note that if $$d^2 = 0$$ then $$B_n \leq Z_n \leq C_n$$. In this case, it makes sense to define the quotient module $$H^n(C_{-}) \coloneqq Z_n / B_n$$, the \textbf{$$n$$th homology} of $$C_{-}$$. \end{definition} \begin{definition}[Maps of chain complexes] A map $$u: C_{-}\to D_{-}$$ of chain complexes is a sequence of maps $$u_n: C_n \to D_n$$ such that all of the following squares commute: \begin{center} \begin{tikzcd} {\cdots} & {C_{n+1}} & {C_n} & {C_{n-1}} & {\cdots} \\ \\ {\cdots} & {D_{n+1}} & {D_n} & {D_{n-1}} & {\cdots} \arrow[from=1-1, to=1-2] \arrow[from=1-2, to=1-3] \arrow[from=1-3, to=1-4] \arrow[from=3-1, to=3-2] \arrow[from=3-2, to=3-3] \arrow[from=3-3, to=3-4] \arrow[from=3-4, to=3-5] \arrow[from=1-4, to=1-5] \arrow["{u_{n+1}}", from=1-2, to=3-2] \arrow["{u_n}", from=1-3, to=3-3] \arrow["{u_{n-1}}", from=1-4, to=3-4] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsMTAsWzEsMCwiQ197bisxfSJdLFsyLDAsIkNfbiJdLFszLDAsIkNfe24tMX0iXSxbMSwyLCJEX3tuKzF9Il0sWzIsMiwiRF9uIl0sWzMsMiwiRF97bi0xfSJdLFswLDAsIlxcYnVsbGV0Il0sWzAsMiwiXFxidWxsZXQiXSxbNCwyLCJcXGJ1bGxldCJdLFs0LDAsIlxcYnVsbGV0Il0sWzYsMF0sWzAsMV0sWzEsMl0sWzcsM10sWzMsNF0sWzQsNV0sWzUsOF0sWzIsOV0sWzAsMywidV97bisxfSIsMV0sWzEsNCwidV9uIiwxXSxbMiw1LCJ1X3tuLTF9IiwxXV0=}{Link to Diagram} \end{quote} \end{definition} \begin{remark} We can thus define a category $$\mathrm{Ch}({\mathsf{R}{\hbox{-}}\mathsf{Mod}})$$ where \begin{itemize} \tightlist \item The objects are chain complexes, \item The morphisms are chain maps. \end{itemize} \end{remark} \begin{exercise}[Weibel 1.1.2] A chain complex map $$u: C_{-}\to D_{-}$$ restricts to \begin{align*} u_n: Z_n(C_{-}) \to Z_n(D_{-}) \\ u_n: B_n(D_{-}) \to B_n(D_{-}) \end{align*} and thus induces a well-defined map $$u_{n, *}: H_n(C_{-}) \to H_n(D_{-})$$. \end{exercise} \begin{remark} Each $$H_n$$ thus becomes a functor $$\mathrm{Ch}({\mathsf{R}{\hbox{-}}\mathsf{Mod}}) \to {\mathsf{R}{\hbox{-}}\mathsf{Mod}}$$ where $$H_n(u) \coloneqq u_{*. n}$$. \end{remark} \hypertarget{friday-january-15}{% \section{Friday, January 15}\label{friday-january-15}} \hypertarget{review}{% \subsection{Review}\label{review}} \begin{quote} See assignment posted on ELC, due Wed Jan 27 \end{quote} \begin{remark} Recall that a chain complex is $$C_{-}$$ where $$d^2 = 0$$, and a map of chain complex is a ladder of commuting squares \begin{center} \begin{tikzcd} \cdots & {C_{n-1}} & {C_{n}} & {C_{n+1}} & \cdots \\ && {} \\ \cdots & {D_{n-1}} & {D_n} & {D_{n+1}} & \cdots \arrow["{u_{n-1}}", from=1-2, to=3-2] \arrow["{u_n}", from=1-3, to=3-3] \arrow["{u_{n+1}}", from=1-4, to=3-4] \arrow["{d_{n-1}}", from=1-2, to=1-3] \arrow["{d_n}", from=1-3, to=1-4] \arrow["{d_{n-1}}", from=3-2, to=3-3] \arrow["{d_n}"', from=3-3, to=3-4] \arrow[from=1-4, to=1-5] \arrow[from=3-4, to=3-5] \arrow[from=3-1, to=3-2] \arrow[from=1-1, to=1-2] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsMTEsWzEsMCwiQ197bi0xfSJdLFsyLDAsIkNfe259Il0sWzMsMCwiQ197bisxfSJdLFsyLDIsIkRfbiJdLFszLDIsIkRfe24rMX0iXSxbMSwyLCJEX3tuLTF9Il0sWzQsMCwiXFxidWxsZXQiXSxbNCwyLCJcXGJ1bGxldCJdLFswLDIsIlxcYnVsbGV0Il0sWzAsMCwiXFxidWxsZXQiXSxbMiwxXSxbMCw1LCJ1Il0sWzEsMywidV9uIl0sWzIsNCwidSJdLFswLDFdLFsxLDIsImRfbiJdLFs1LDNdLFszLDQsImRfbiIsMl0sWzIsNl0sWzQsN10sWzgsNV0sWzksMF1d}{Link to diagram} \end{quote} Recall that $$u_n: Z_n(C) \to Z_n(D)$$ and $$u_n: B_n(C) \to B_n(D)$$ preserves these submodules, so there are induced maps $$u_{{-}, n}: H_n(D) \to H_n(D)$$ where $$H_n(C) \coloneqq Z_n(C) / B_nn-1(C)$$. Moreover, taking $$H_n({-})$$ is a functor from $$\mathsf{Ch}({\mathsf{R}{\hbox{-}}\mathsf{Mod}}) \to {\mathsf{R}{\hbox{-}}\mathsf{Mod}}$$ for any fixed $$n$$ and on objects $$C\mapsto H_n(C)$$ and chain maps $$u_{n} \to H_n(u) \coloneqq u_{*, n}$$. Note the lower indices denote maps going down in degree. \end{remark} \hypertarget{cohomology}{% \subsection{Cohomology}\label{cohomology}} \begin{definition}[Quasi-isomorphism] A chain map $$u:C\to D$$ is a \textbf{quasi-isomorphism} if and only if the induced map $$u_{*, n}: H^n(C) \to H^n(D)$$ is an isomorphism of $$R{\hbox{-}}$$modules. \end{definition} \begin{remark} Note that the usual notion of an isomorphism in the categorical sense might be too strong here. \end{remark} \begin{definition}[Cohomology] A \textbf{cochain complex} is a complex of the form \begin{align*} \cdots \xrightarrow{d^{n-2}} C^{n-1} \xrightarrow{d^{n-1}} C^{n} \xrightarrow{d^{n}} C^{n+1} \cdots \end{align*} where $$d^n \circ d^{n-1} = 0$$. We similarly write $$Z^n(C) \coloneqq\ker d^n$$ and $$B^n(C) \coloneqq\operatorname{im}d^{n-1}$$ and write the $$R{\hbox{-}}$$module $$H^n(C) \coloneqq Z^n/B^n$$ for the $$n$$th \textbf{cohomology} of $$C$$. \end{definition} \begin{remark} There is a way to go back and forth bw chain complexes and cochain complexes: set $$C_n \coloneqq C^{-n}$$ and $$d_n \coloneqq d^{-n}$$. This yields \begin{align*} C^{-n} \xrightarrow{d^{-n}} C^{-n+1} \iff C_n \xrightarrow{d^n} C_{n-1} ,\end{align*} and the notions of $$d^2 = 0$$ coincide. \end{remark} \begin{definition}[Bounded complexes] A cochain complex $$C$$ is \textbf{bounded} if and only if there exists an $$a\leq b \in {\mathbb{Z}}$$ such that $$C_n \neq 0 \iff a\leq n \leq b$$. Similarly $$C^n$$ is bounded above if there is just a $$b$$, and \textbf{bounded below} for just an $$a$$. All of the same definitions are made for cochain complexes. \end{definition} \begin{remark} See the book for classical applications: \begin{itemize} \tightlist \item 1.1.3: Simplicial homology \item 1.1.5: Singular homology \end{itemize} \end{remark} \hypertarget{operations-on-chain-complexes}{% \subsection{Operations on Chain Complexes}\label{operations-on-chain-complexes}} \begin{remark} Write $$\mathsf{Ch}$$ for $$\mathsf{Ch}({\mathsf{R}{\hbox{-}}\mathsf{Mod}})$$, then if $$f,g: C\to D$$ are chain maps then $$f+g:C\to D$$ can be defined as $$(f+g)(x) = f(x) + g(x)$$, since $$D$$ has an addition coming from its $$R{\hbox{-}}$$module structure. Thus the hom sets $$\mathop{\mathrm{Hom}}_{\mathsf{Ch}}(C, D)$$ becomes an abelian group. There is a distinguished \textbf{zero object}\footnote{See appendix A 1.6 for initial and terminal objects. Note that $$\emptyset$$ is an initial but non-terminal object in $${\mathsf{Set}}$$, whereas zero objects are both.} $$0$$, defined as the chain complex with all zero objects and all zero maps. Note that we also have a zero map given by the composition $$(C \to 0) \circ (0\to D)$$. \end{remark} \begin{definition}[Products and Coproducts] If $$\left\{{A_ \alpha}\right\}$$ is a family of complexes, we can form two new complexes: \begin{itemize} \item The \textbf{product} $$\qty{ \prod_ \alpha A_ \alpha}_n \coloneqq\prod_ \alpha A _{\alpha, n}$$ with the differential \begin{align*} \qty{ \prod d_ \alpha}_n: \prod A _{\alpha, n} \xrightarrow{d _{\alpha, n}} \prod A _{\alpha, n-1} .\end{align*} \item The \textbf{coproduct} $$\qty{ \coprod _{\alpha} A _{\alpha}}_n \coloneqq\bigoplus _{\alpha} A _{\alpha, n}$$, i.e.~there are only finitely many nonzero entries, with exactly the same definition as above for the differential. \end{itemize} \end{definition} \begin{remark} Note that if the index set is finite, these notions coincide. By convention, finite direct products are written as direct sums. These structures make $$\mathsf{Ch}$$ into an \textbf{additive category}. See appendix for definition: the homs are abelian groups where composition distributes over addition, existence of a zero object, and existence of finite products. Note that here we have arbitrary products. \end{remark} \begin{definition}[Subcomplexes] We say $$B$$ is a \textbf{subcomplex} of $$C$$ if and only if \begin{itemize} \tightlist \item $$B_n \leq C_n \in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}$$ for all $$n$$, \item The differentials of $$B_n$$ are the restrictions of the differentials of $$C_n$$. \end{itemize} \end{definition} \begin{remark} This can be alternatively stated as saying the inclusion $$i: B\to C$$ given by $$i_n: B_n \to C_n$$ is a morphism of chain complexes. Recall that some squares need to commute, and this forces the condition on restrictions. \end{remark} \begin{definition}[Quotient Complexes] When $$B \leq C$$, we can form the quotient complex $$C/B$$ where \begin{align*} C_n/B_n \xrightarrow{\mkern 1.5mu\overline{\mkern-1.5mud_n\mkern-1.5mu}\mkern 1.5mu} C _{n-1} / B _{n-1} .\end{align*} Moreover there is a natural projection $$\pi: C\to C/B$$ which is a chain map. \end{definition} \begin{remark} Suppose $$f:B\to C$$ is a chain map, then there exist induced maps on the levelwise kernels and cokernels, so we can form the \textbf{kernel} and \textbf{cokernel} complex: \begin{center} \begin{tikzcd} \cdots && {\ker f_n} && {\ker f_{n-1}} && \cdots \\ \\ \cdots && {B_n} && {B_{n-1}} && \cdots \\ \\ \cdots && {C_n} && {C_{n-1}} && \cdots \\ &&& {} & {} \\ \cdots && {\operatorname{coker}f_n} && {\operatorname{coker}f_{n-1}} && \cdots \arrow["{d_n}", from=5-3, to=5-5] \arrow["{d_n}", from=3-3, to=3-5] \arrow["{\exists d_n}", dashed, from=1-3, to=1-5] \arrow["{\exists d_n}", dashed, from=7-3, to=7-5] \arrow["{i_{n}}"{description}, from=1-3, to=3-3] \arrow["{f_n}"{description}, from=3-3, to=5-3] \arrow["{\pi_n}"{description}, from=5-3, to=7-3] \arrow["{f_{n-1}}"{description}, from=3-5, to=5-5] \arrow["{\pi_{n-1}}"{description}, from=5-5, to=7-5] \arrow["{i_{n-1}}"{description}, from=1-5, to=3-5] \arrow[from=1-1, to=1-3] \arrow[from=3-1, to=3-3] \arrow[from=5-1, to=5-3] \arrow[from=7-1, to=7-3] \arrow[from=7-5, to=7-7] \arrow[from=5-5, to=5-7] \arrow[from=3-5, to=3-7] \arrow[from=1-5, to=1-7] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=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}{Link to Diagram} \end{quote} Here $$\ker f \leq B$$ is a subcomplex, and $$\operatorname{coker}f$$ is a quotient complex of $$C$$. The chain map $$i: \ker f\to B$$ is a categorical kernel of $$f$$ in $$\mathsf{Ch}$$, and $$\pi$$ is similarly a cokernel. See appendix A 1.6. These constructions make $$\mathsf{Ch}$$ into an \textbf{abelian category}: roughly an additive category where every morphism has a kernel and a cokernel. \end{remark} \hypertarget{chain-complex-of-chain-complexes-wednesday-january-20}{% \section{1.2: Chain Complex of Chain Complexes (Wednesday, January 20)}\label{chain-complex-of-chain-complexes-wednesday-january-20}} \begin{quote} See phone pic for missed first 10m \end{quote} \hypertarget{double-complexes}{% \subsection{Double Complexes}\label{double-complexes}} \begin{remark} Consider a double complex: \begin{center} \begin{tikzcd} &&&&&& {C_{p, \cdot}:} \\ &&&& \vdots && \vdots && \vdots \\ \\ && \cdots && {C_{p-1, q+1}} && {C_{p, q+1}} && {C_{p+1, q+1}} && \cdots \\ \\ {C_{\cdot, q}:} && \cdots && {C_{p-1, q}} && {C_{p, q}} && {C_{p+1, q}} && \cdots \\ \\ && \cdots && {C_{p-1, q+1}} && {C_{p, q+1}} && {C_{p+1, q+1}} && \cdots \\ \\ &&&& \vdots && \vdots && \vdots \arrow["{d_{p, q}^h}", from=6-7, to=6-5] \arrow["{d_{p, q}^v}", from=6-7, to=8-7] \arrow["{d_{p, q+1}^v}", from=4-7, to=6-7] \arrow["{d_{p+1, q+1}^v}", from=4-9, to=6-9] \arrow["{d_{p+1, q}^v}", from=6-9, to=8-9] \arrow["{d_{p-1, q}^v}", from=6-5, to=8-5] \arrow["{d_{p-1, q+1}^v}", from=4-5, to=6-5] \arrow[from=8-5, to=10-5] \arrow[from=8-7, to=10-7] \arrow[from=8-9, to=10-9] \arrow["{d_{p+1, q+1}^h}", from=8-9, to=8-7] \arrow["{d_{p+1, q}^h}", from=6-9, to=6-7] \arrow["{d_{p, q+1}^h}", from=8-7, to=8-5] \arrow["{d_{p+1, q+1}^h}"{description}, from=4-9, to=4-7] \arrow["{d_{p, q+1}^h}"{description}, from=4-7, to=4-5] \arrow[from=2-5, to=4-5] \arrow[from=2-7, to=4-7] \arrow[from=2-9, to=4-9] \arrow[from=4-5, to=4-3] \arrow[from=6-5, to=6-3] \arrow[from=8-5, to=8-3] \arrow[from=8-11, to=8-9] \arrow[from=6-11, to=6-9] \arrow[from=4-11, to=4-9] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=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}{Link to Diagram} \end{quote} All of the individual rows and columns are chain complexes, where $$(d^h)^2 = 0$$ and $$(d^v)^2 = 0$$, and the square anticommute: $$d^v d^h + d^h d^v - 0$$, so $$d^v d^h = -d^h d^v$$. This is almost a chain complex of chain complexes, i.e.~an element of $$\mathsf{Ch}(\mathsf{Ch}{\mathsf{R}{\hbox{-}}\mathsf{Mod}}))$$. It's useful here to consider lines parallel to the line $$y=x$$. \end{remark} \begin{definition}[Bounded Complexes] A double complex $$C_{{-}, {-}}$$ is \textbf{bounded} if and only if there are only finitely many nonzero terms along each constant diagonal $$p+q = n$$. \end{definition} \begin{example}[?] A \emph{first quadrant} double complex $$\left\{{C_{p, q}}\right\}_{p, q\geq 0}$$ is bounded: note that this can still have infinitely many terms, but each diagonal is finite because each will hit a coordinate axis. \end{example} \begin{remark}[The sign trick] The squares anticommute, since the $$d^v$$ are not chain maps between the horizontal chain complexes. This can be fixed by changing every one out of four signs, defining \begin{align*} f_{*, q}: C_{*, q} \to C_{*, q-1} \\ f_{p, q} \coloneqq(-1)^p d^v_{p, q}: C_{p,q} \to C_{p, q-1} .\end{align*} This yields a new double complex where the signs of each column alternate: \begin{center} \begin{tikzcd} {C_{0, q}} && {C_{1, q}} && {C_{2, q}} \\ \\ {C_{0, q-1}} && {C_{1, q-1}} && {C_{2, q-1}} \arrow["{d^v}", from=1-1, to=3-1] \arrow["{-d^v}", from=1-3, to=3-3] \arrow["{d^v}", from=1-5, to=3-5] \arrow["{d^h}"{description}, from=1-5, to=1-3] \arrow["{d^h}"{description}, from=1-3, to=1-1] \arrow["{d^h}"{description}, from=3-5, to=3-3] \arrow["{d^h}"{description}, from=3-3, to=3-1] \end{tikzcd} \end{center} Now the squares commute and $$f_{{-}, q}$$ are chain maps, so this object is an element of $$\mathsf{Ch}(\mathsf{Ch}{\mathsf{R}{\hbox{-}}\mathsf{Mod}})$$. \end{remark} \hypertarget{total-complexes}{% \subsection{Total Complexes}\label{total-complexes}} \begin{remark} Recall that products and coproducts of $$R{\hbox{-}}$$modules coincide when the indexing set is finite. \end{remark} \begin{definition}[Total Complexes] Given a double complex $$C_{{-}, {-}}$$, there are two ordinary chain complexes associated to it referred to as \textbf{total complexes}: \begin{align*} (\operatorname{Tot}^{\Pi}C)_n &\coloneqq\prod_{p+q = n} C_{p, q}\\ (\operatorname{Tot}^{\oplus}C)_n &\coloneqq\bigoplus_{p+q = n} C_{p, q} .\end{align*} Writing $$\operatorname{Tot}(C)$$ usually refers to the former. The differentials are given by \begin{align*} d_{p, q} = d^h + d^v: C_{p, q} \to C_{p-1, q} \oplus C_{p, q-1} ,\end{align*} where $$C_{p, q} \subseteq \operatorname{Tot}^{\oplus}(C)_n$$ and $$C_{p-1, q} \oplus C_{p, q-1} \subseteq \operatorname{Tot}^{\oplus}(C)_{n-1}$$. Then you extend this to a differential on the entire diagonal by defining $$d = \bigoplus_{p, q} d_{p, q}$$. \end{definition} \begin{exercise}[?] Check that $$d^2 = 0$$, using $$d^v d^h + d^h d^v = 0$$. \end{exercise} \begin{remark} Some notes: \begin{itemize} \item $$\operatorname{Tot}^{\oplus}(C) = \operatorname{Tot}^{\Pi}(C)$$ when $$C$$ is bounded. \item The total complexes need not exist if $$C$$ is unbounded: one needs infinite direct products and infinite coproducts to exist in $$\mathcal{C}$$. A category admitting these is called \textbf{complete} or \textbf{cocomplete}.\footnote{Recall that abelian categories are additive and only require \emph{finite} products/coproducts. A counterexample: categories of \emph{finite} abelian groups, where e.g.~you can't take infinite sums and stay within the category.} \end{itemize} \end{remark} \hypertarget{more-operations}{% \subsection{More Operations}\label{more-operations}} \begin{definition}[Truncation below] Fix $$n\in {\mathbb{Z}}$$, and define the \textbf{$$n$$th truncation} $$\tau_{\geq n}(C)$$ by \begin{align*} \tau_{\geq n}(C) = \begin{cases} 0 & i < n \\ Z_n & i= n \\ C_i & i > n . \end{cases} .\end{align*} Pictorially: \begin{center} \begin{tikzcd} \cdots & 0 & {Z_n} & {C_{n+1}} & {C_{n+2}} & \cdots \arrow[from=1-2, to=1-1] \arrow["{d_n}"', from=1-3, to=1-2] \arrow["{d_{n+1}}"', from=1-4, to=1-3] \arrow["{d_{n+2}}"', from=1-5, to=1-4] \arrow[from=1-6, to=1-5] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsNixbMCwwLCJcXGNkb3RzIl0sWzEsMCwiMCJdLFsyLDAsIlpfbiJdLFszLDAsIkNfe24rMX0iXSxbNCwwLCJDX3tuKzJ9Il0sWzUsMCwiXFxjZG90cyJdLFsxLDBdLFsyLDEsImRfbiIsMl0sWzMsMiwiZF97bisxfSIsMl0sWzQsMywiZF97bisyfSIsMl0sWzUsNF1d}{Link to diagram} \end{quote} This is sometimes call the \textbf{good truncation of $$C$$ below $$n$$}. \end{definition} \begin{remark} Note that \begin{align*} H_i(\tau_{\geq n} C) = \begin{cases} 0 & i < n \\ H_i(C) & i\geq n. \end{cases} .\end{align*} \end{remark} \begin{definition}[Truncation above] We define the quotient complex \begin{align*} \tau_{0} FA = 0\). \end{corollary} \begin{proof}[?] Use the projective resolution $$\cdots \to 0 \to A \xrightarrow{\one_A} A \to 0 \to \cdots$$. In this case $$H_{>0}(FP) = 0$$. \end{proof} \begin{remark} This is an interesting result, since it doesn't depend on the functor! Short aside on $$F{\hbox{-}}$$acyclic objects -- we don't need something as strong as a \emph{projective} resolution. \end{remark} \begin{definition}[$F\dash$acyclic objects] An object $$Q\in {\mathcal{A}}$$ is \textbf{$$F{\hbox{-}}$$acyclic} if $$L_{>0}FQ = 0$$. \end{definition} \begin{remark} Note that projective implies $$F{\hbox{-}}$$acyclic for every $$F$$, but not conversely. For example, flat $$R{\hbox{-}}$$modules are acyclic for $${-}\otimes_R {-}$$. In general, flat does not imply projective, although projective implies flat. \end{remark} \begin{definition}[$F\dash$acyclic resolutions] An \textbf{$$F{\hbox{-}}$$acyclic resolution} of $$A$$ is a left resolution $$Q\to A$$ for which every $$Q_i$$ is $$F{\hbox{-}}$$acyclic. \end{definition} \begin{remark} One can compute $$L_iF(A) \cong H_i(FQ)$$ for any $$F{\hbox{-}}$$acyclic resolution. For the $$L_i F$$ to be functors, we need to define them on maps! \end{remark} \begin{lemma}[?] If $$f: A\to A'$$, there is a natural associated morphism $$L_i F(f): L_iF(A) \to L_iF(A')$$. \end{lemma} \begin{proof}[?] Again use the comparison theorem: \begin{center} \begin{tikzcd} P && A && 0 \\ \\ {P'} && {A'} && 0 \arrow["f", from=1-3, to=3-3] \arrow[from=3-1, to=3-3] \arrow[from=1-1, to=1-3] \arrow[from=1-3, to=1-5] \arrow[from=3-3, to=3-5] \arrow["{\exists \tilde f}"', from=1-1, to=3-1] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsNixbMCwwLCJQIl0sWzIsMCwiQSJdLFsyLDIsIkEnIl0sWzAsMiwiUCciXSxbNCwwLCIwIl0sWzQsMiwiMCJdLFsxLDIsImYiXSxbMywyXSxbMCwxXSxbMSw0XSxbMiw1XSxbMCwzLCJcXGV4aXN0cyBcXHRpbGRlIGYiLDJdXQ==}{Link to Diagram} \end{quote} Then there is an induced map $$\tilde f_*: H_*(FP) \to H_*(FP')$$, noting that one first needs to apply $$F$$ to the above diagram. As before, this is independent of the lift using the same argument as before, using the additivity of $$F$$ and $$H_*$$ and the chain homotopy is pushed through $$F$$ appropriately. So set $$(L_i F)(f) \coloneqq(\tilde f_*)_i$$. \end{proof} \begin{remark} We can now pick up the theorem from the end of last time: \end{remark} \begin{theorem}[Left-derived functors are additive] $$L_iF: \mathcal{A}\to \mathcal{B}$$ are additive functors. \end{theorem} \begin{proof}[?] Done last time! \end{proof} \begin{theorem}[Existence of connecting maps for left-derived functors] Using the same assumptions as before, given a SES \begin{align*} 0 \to A' \to A \to A'' \to 0 \end{align*} there are natural connecting maps $$\delta$$ yielding a LES \begin{center} \begin{tikzcd} &&&& \cdots \\ \\ {L_iF(A')} && {L_iF(A)} && {L_iF(A'')} \\ \\ {L_{i-1}F(A')} && \cdots && \cdots \\ \\ {FA'} && FA && {FA''} && 0 \arrow["{\delta_i}"', from=3-5, to=5-1, out=0, in=180] \arrow[from=3-1, to=3-3] \arrow[from=3-3, to=3-5] \arrow["{\delta_{i+1}}"', from=1-5, to=3-1, out=0, in=180] \arrow[dashed, from=5-1, to=5-3] \arrow[from=7-1, to=7-3] \arrow[from=7-3, to=7-5] \arrow[from=7-5, to=7-7] \arrow[dashed, from=5-3, to=5-5] \arrow["{\delta_1}"', from=5-5, to=7-1, out=0, in=180] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsMTEsWzAsMiwiTF9pRihBJykiXSxbMiwyLCJMX2lGKEEpIl0sWzQsMiwiTF9pRihBJycpIl0sWzAsNCwiTF97aS0xfUYoQScpIl0sWzQsMCwiXFxjZG90cyJdLFsyLDQsIlxcY2RvdHMiXSxbMCw2LCJGQSciXSxbMiw2LCJGQSJdLFs0LDYsIkZBJyciXSxbNiw2LCIwIl0sWzQsNCwiXFxjZG90cyJdLFsyLDMsIlxcZGVsdGFfaSIsMl0sWzAsMV0sWzEsMl0sWzQsMCwiXFxkZWx0YV97aSsxfSIsMl0sWzMsNSwiIiwyLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV0sWzYsN10sWzcsOF0sWzgsOV0sWzUsMTAsIiIsMix7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFsxMCw2LCJcXGRlbHRhXzEiLDJdXQ==}{Link to Diagram} \end{quote} \end{theorem} \begin{proof}[?] Using the Horseshoe lemma, we can obtain the following map: \begin{center} \begin{tikzcd} &&& 0 \\ \\ {P'} &&& {A'} \\ \\ P &&& A && 0 \\ \\ {P''} &&& {A''} \\ \\ &&& 0 \arrow["\exists", dashed, from=5-1, to=5-4] \arrow[from=7-1, to=7-4] \arrow[from=7-4, to=9-4] \arrow[from=5-4, to=7-4] \arrow[from=5-4, to=5-6] \arrow[from=3-4, to=5-4] \arrow[from=3-1, to=3-4] \arrow[from=3-1, to=5-1] \arrow[from=5-1, to=7-1] \arrow[from=1-4, to=3-4] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsOSxbMCwyLCJQJyJdLFswLDQsIlAiXSxbMCw2LCJQJyciXSxbMyw2LCJBJyciXSxbMyw0LCJBIl0sWzMsMiwiQSciXSxbMywwLCIwIl0sWzMsOCwiMCJdLFs1LDQsIjAiXSxbMSw0LCJcXGV4aXN0cyIsMCx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFsyLDNdLFszLDddLFs0LDNdLFs0LDhdLFs1LDRdLFswLDVdLFswLDFdLFsxLDJdLFs2LDVdXQ==}{Link to Diagram} \end{quote} So we get a SES of complexes over $$\mathcal{A}$$, $$0 \to P' \to P \to P'' \to 0$$. One can use that $$P = P' \oplus P''$$, or alternatively that each $$P_n''$$ is a projective $$R{\hbox{-}}$$module, to see that there are splittings \begin{center} \begin{tikzcd} 0 && {P'} && P && {P''} && 0 \arrow[from=1-1, to=1-3] \arrow["f", from=1-3, to=1-5] \arrow["g", from=1-5, to=1-7] \arrow[from=1-7, to=1-9] \arrow["{g'}"{description}, curve={height=18pt}, dashed, from=1-7, to=1-5] \arrow["{f'}"{description}, curve={height=18pt}, dashed, from=1-5, to=1-3] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsNSxbMCwwLCIwIl0sWzIsMCwiUCciXSxbNCwwLCJQIl0sWzYsMCwiUCcnIl0sWzgsMCwiMCJdLFswLDFdLFsxLDIsImYiXSxbMiwzLCJnIl0sWzMsNF0sWzMsMiwiZyciLDEseyJjdXJ2ZSI6Mywic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV0sWzIsMSwiZiciLDEseyJjdXJ2ZSI6Mywic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV1d}{Link to Diagram} \end{quote} Note that this can be phrased in terms of $$g'g = \one, f'f = \one$$, or $$g'g + f'f = \one$$. Since $$F$$ is additive, it preserves all of these relations, particularly the ones that define being split exact. So additive functors preserve split exact sequences. Thus $$0 \to FP' \to FP \to FP'' \to 0$$ is still split exact, even though $$F$$ is only right exact. Now take homology and use the LES in homology to get the desired LES above, and $$\delta$$ is the connecting morphism that comes from the snake lemma. Proving naturality: we start with the following setup. \begin{center} \begin{tikzcd} 0 && {A'} && A && {A''} && 0 \\ \\ 0 && {B'} && B && {B''} && 0 \arrow["{g'}", from=1-3, to=3-3] \arrow["g", from=1-5, to=3-5] \arrow["{g''}", from=1-7, to=3-7] \arrow[from=1-1, to=1-3] \arrow[from=1-3, to=1-5] \arrow[from=1-5, to=1-7] \arrow[from=1-7, to=1-9] \arrow[from=3-1, to=3-3] \arrow[from=3-3, to=3-5] \arrow[from=3-5, to=3-7] \arrow[from=3-7, to=3-9] \end{tikzcd} \end{center} Naturality of $$\delta$$ will be showing that the following square commutes: \begin{center} \begin{tikzcd} L_{i+1}F(A'') \ar[r, "\delta"] \ar[d] & L_iF(A') \ar[d] \\ L_{i+1}F(B'') \ar[r, "\delta"] & L_iF(B') \end{tikzcd} \end{center} We now apply the horseshoe lemma several times: \begin{center} \begin{tikzcd} 0 && {P'} && \textcolor{rgb,255:red,214;green,92;blue,92}{P} && {P''} && 0 \\ \\ \\ & 0 && {A'} && A && {A''} && 0 \\ \\ & 0 && {B'} && B && {B''} && 0 \\ \\ \\ 0 && {Q'} && \textcolor{rgb,255:red,214;green,92;blue,92}{Q} && {Q''} && 0 \arrow["{g'}", from=4-4, to=6-4] \arrow["g", from=4-6, to=6-6] \arrow["{g''}", from=4-8, to=6-8] \arrow[from=4-2, to=4-4] \arrow[from=4-4, to=4-6] \arrow[from=4-6, to=4-8] \arrow[from=4-8, to=4-10] \arrow[from=6-2, to=6-4] \arrow[from=6-4, to=6-6] \arrow[from=6-6, to=6-8] \arrow[from=6-8, to=6-10] \arrow[from=1-7, to=4-8] \arrow[from=1-3, to=4-4] \arrow[color={rgb,255:red,214;green,92;blue,92}, dashed, from=1-5, to=4-6] \arrow[from=9-3, to=6-4] \arrow[from=9-7, to=6-8] \arrow[color={rgb,255:red,214;green,92;blue,92}, dashed, from=9-5, to=6-6] \arrow[from=9-3, to=9-5] \arrow[from=9-5, to=9-7] \arrow["{\exists G'}"{description}, curve={height=-6pt}, dotted, from=1-3, to=9-3] \arrow["{\exists G''}"{description}, curve={height=-12pt}, dotted, from=1-7, to=9-7] \arrow["{\exists G}"{description}, curve={height=6pt}, dashed, from=1-5, to=9-5] \arrow[from=1-3, to=1-5] \arrow[from=1-5, to=1-7] \end{tikzcd} \end{center} It turns out (details omitted see Weibel p.~46) that $$G$$ can be chosen such that we get a commutative diagram of chain complexes with exact rows: \begin{center} \begin{tikzcd} 0 && {P'} && P && {P''} && 0 \\ \\ 0 && {Q'} && Q && {Q''} && 0 \arrow[from=3-1, to=3-3] \arrow[from=3-3, to=3-5] \arrow[from=3-5, to=3-7] \arrow[from=3-7, to=3-9] \arrow[from=1-1, to=1-3] \arrow[from=1-3, to=1-5] \arrow[from=1-5, to=1-7] \arrow[from=1-7, to=1-9] \arrow["{G'}"{description}, from=1-3, to=3-3] \arrow["G"{description}, from=1-5, to=3-5] \arrow["{G''}"{description}, from=1-7, to=3-7] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsMTAsWzAsMCwiMCJdLFsyLDAsIlAnIl0sWzQsMCwiUCJdLFs2LDAsIlAnJyJdLFsyLDIsIlEnIl0sWzQsMiwiUSJdLFs2LDIsIlEnJyJdLFswLDIsIjAiXSxbOCwwLCIwIl0sWzgsMiwiMCJdLFs3LDRdLFs0LDVdLFs1LDZdLFs2LDldLFswLDFdLFsxLDJdLFsyLDNdLFszLDhdLFsxLDQsIkcnIiwxXSxbMiw1LCJHIiwxXSxbMyw2LCJHJyciLDFdXQ==}{Link to Diagram} \end{quote} We proved naturality of the connecting maps $${{\partial}}$$ in the corresponding LES in homology in general (see prop. 1.3.4). This translates to naturality of the maps $$\delta_i: L_{i} (A'') \to L_{i-1}(A')$$. \end{proof} \begin{remark} See exercise 2.4.3 for dimension shifting''. This is a useful tool for inductive arguments. \end{remark} \hypertarget{friday-february-12}{% \section{Friday, February 12}\label{friday-february-12}} \begin{remark} Last time: right-exact functors have left-derived functors where a SES induces a LES. The functors are \emph{natural} with respect to the connecting morphisms in the sense that certain squares commute. Weibel refers to $$\left\{{ L_i F }\right\}_{i\geq 0}$$ as a \textbf{homological $$\delta{\hbox{-}}$$functor}, i.e.~anything that takes SESs to LESs which are natural with respect to connecting morphism. \end{remark} \hypertarget{aside-natural-transformations}{% \subsection{Aside: Natural Transformations}\label{aside-natural-transformations}} \begin{definition}[Natural Transformation] Given functors $$F, G, \mathcal{C} \to \mathcal{D}$$, a \textbf{natural transformation} $$\eta: F \implies G$$ is the following data: \begin{itemize} \item For all $$C\in \mathcal{C}$$ there is a map $$F(C) \xrightarrow{\eta_C} G(C) \in \operatorname{Mor}(\mathcal{D})$$, sometimes referred to as $$\eta(C)$$. \item If $$C \xrightarrow{f} C' \in \operatorname{Mor}(\mathcal{C})$$, there is a diagram \end{itemize} \begin{center} \begin{tikzcd} FC && {FC'} \\ \\ GC && {GC'} \arrow["Gf", from=3-1, to=3-3] \arrow["Ff", from=1-1, to=1-3] \arrow["{\eta_C}"{description}, from=1-1, to=3-1] \arrow["{\eta_{C'}}"{description}, from=1-3, to=3-3] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJGQyJdLFsyLDAsIkZDJyJdLFswLDIsIkdDIl0sWzIsMiwiR0MnIl0sWzIsMywiR2YiXSxbMCwxLCJGZiJdLFswLDIsIlxcZXRhX0MiLDFdLFsxLDMsIlxcZXRhX3tDJ30iLDFdXQ==}{Link to Diagram} \end{quote} \begin{itemize} \tightlist \item $$\eta$$ is a \textbf{natural isomorphism} if all of the $$\eta_C$$ are isomorphisms, and we write $$F \cong C$$. \end{itemize} \end{definition} \begin{definition}[Equivalence of Categories] A functor $$F: \mathcal{C} \to \mathcal{D}$$ is an \textbf{equivalence of categories} if and only if there exists a $$G: \mathcal{D} \to \mathcal{C}$$ such that $$GF \cong \one_{\mathcal{C}}$$ and $$FG \cong \one_{\mathcal{D}}$$. \end{definition} \begin{example}[?] A category $$\mathcal{C}$$ is \textbf{small} if $${\operatorname{Ob}}(\mathcal{C})$$ is a set, then take $$\mathcal{C} \coloneqq\mathsf{Cat}$$ whose objects are categories and morphisms are functors. Note that in all categories, all collections of morphisms should be sets, and the small condition guarantees it. In this case, natural transformations $$\eta: F\to G$$ is an additional structure yielding morphisms of morphisms. These are called \textbf{2-morphisms}, and in this entire structure is a \textbf{2-category}, and our previous notion is referred to as a \textbf{1-category}. \end{example} \begin{theorem}[Left-derived functors of a right-exact functor form a universal $\delta\dash$functor] Assume $$\mathcal{A}, \mathcal{B}$$ are abelian and $$F:\mathcal{A} \to \mathcal{B}$$ is a right-exact additive functor where $$\mathcal{A}$$ has enough projectives. Then the family $$\left\{{ L_i F }\right\} _{i\geq 0}$$ is a \emph{universal $$\delta{\hbox{-}}$$functor} where $$L_0 F \cong F$$ is a natural isomorphism. \end{theorem} \begin{remark} Here \emph{universal} means that if $$\left\{{ T_i }\right\} _{i\geq 0}$$ is also a $$\delta{\hbox{-}}$$functor with a natural \emph{transformation} (not necessarily an isomorphism) $$\varphi_0: T_0 \to F$$, then there exist unique morphism of $$\delta{\hbox{-}}$$functors $$\left\{{ \varphi_i: T_i \to L_i F }\right\} _{i\geq 0}$$, i.e.~a family of natural transformations that commute with the respective $$\delta$$ maps coming from both the $$T_i$$ and the $$L_i F$$, which extend $$\varphi_0$$. This will be important later on when we try to show Ext and Tor are functors in either slot. \end{remark} \begin{proof}[?] Assume $$\left\{{ T_i }\right\} _{i\geq 0}$$ and $$\varphi_0$$ are given, and assume inductively that $$n>0$$ and we've defined $$\varphi_i: T_i \to F$$ for $$0\leq i< n$$ which commute with the $$\delta$$ maps. Step 1: given $$A\in \mathcal{A}$$, fix a reference exact sequence: pick a projective mapping onto $$A$$ and its kernel to obtain \begin{align*} 0 \to K \to P \to A \to 0 .\end{align*} Applying the functors $$T_i$$ and $$L_i F$$ yields \begin{center} \begin{tikzcd} && \textcolor{rgb,255:red,214;green,92;blue,214}{x} && k && 0 \\ && {T_nA} && {T_{n-1}K} && {T_{n-1}P} \\ \\ \\ \textcolor{rgb,255:red,214;green,92;blue,214}{L_n FP = 0} && {L_nFA} && {L_{n-1}FK} && {L_{n-1}FP} \\ && \textcolor{rgb,255:red,214;green,92;blue,214}{\exists ! y \coloneqq\phi_{n-1}(x)} && \ell && 0 \arrow["{\phi_{n-1}(K)}", from=2-5, to=5-5] \arrow["{\phi_{n-1}(P)}", from=2-7, to=5-7] \arrow["{\exists \phi_{n-1}(A)}", dashed, from=2-3, to=5-3] \arrow["\delta"{description}, from=2-3, to=2-5] \arrow[from=2-5, to=2-7] \arrow[from=5-1, to=5-3] \arrow["\delta"', hook, from=5-3, to=5-5] \arrow[from=5-5, to=5-7] \arrow[dotted, from=1-3, to=1-5] \arrow[dotted, from=1-5, to=1-7] \arrow[curve={height=18pt}, dotted, from=1-5, to=6-5] \arrow[curve={height=18pt}, dotted, from=1-7, to=6-7] \arrow[color={rgb,255:red,214;green,92;blue,214}, dotted, from=6-3, to=6-5] \arrow[dotted, from=6-5, to=6-7] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=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}{Link to Diagram} \end{quote} So define $$\varphi_{n}(A)(x) \coloneqq y$$, which makes the LHS square commute by construction. Note that $$L_n FP$$ vanishes (as do all its higher derived functors) since $$P$$ is projective. \begin{warnings} The map $$\varphi_n(A)$$ could depend on the choice of $$P$$! \end{warnings} We now want to show that $$\varphi_n$$ is a natural transformation. Supposing $$f:A' \to A$$, we need to show $$\varphi_n$$ commutes with $$f$$. \begin{center} \begin{tikzcd} 0 & {K'} & {P'} & A & 0 \\ \\ 0 & K & P & A & 0 \arrow["f", from=1-4, to=3-4] \arrow["{\exists g}", dashed, from=1-3, to=3-3] \arrow[from=3-1, to=3-2] \arrow[from=3-2, to=3-3] \arrow[from=3-3, to=3-4] \arrow[from=3-4, to=3-5] \arrow[from=1-1, to=1-2] \arrow[from=1-2, to=1-3] \arrow[from=1-3, to=1-4] \arrow[from=1-4, to=1-5] \arrow["{\exists h}", dashed, from=1-2, to=3-2] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsMTAsWzAsMCwiMCJdLFsxLDAsIksnIl0sWzIsMCwiUCciXSxbMywwLCJBIl0sWzMsMiwiQSJdLFsyLDIsIlAiXSxbMSwyLCJLIl0sWzAsMiwiMCJdLFs0LDIsIjAiXSxbNCwwLCIwIl0sWzMsNCwiZiJdLFsyLDUsIlxcZXhpc3RzIGciLDAseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XSxbNyw2XSxbNiw1XSxbNSw0XSxbNCw4XSxbMCwxXSxbMSwyXSxbMiwzXSxbMyw5XSxbMSw2LCJcXGV4aXN0cyBoIiwwLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV1d}{Link to Diagram} \end{quote} Since $$P'$$ is projective, we can lift $$f$$ to $$P'\to P$$, and then define $$h$$ to be the restriction of $$g$$ to $$K' \to K$$. \begin{center} \begin{tikzcd} {T_n A'} &&&& {T_nA} \\ & {T_{n-1}K'} && {T_{n-1}K} \\ \\ & {L_{n-1}FK'} && {L_{n-1}FK} \\ {L_n FA'} &&&& {L_nF(A)} \arrow["{T_nf}", from=1-1, to=1-5] \arrow["{L_nFf}", from=5-1, to=5-5] \arrow["{\phi_n(A)}", from=1-5, to=5-5] \arrow["{\phi_n(A')}", from=1-1, to=5-1] \arrow["{T_{n-1}h}", from=2-2, to=2-4] \arrow["{\delta'}"{description}, from=1-1, to=2-2] \arrow["\delta"{description}, from=1-5, to=2-4] \arrow["{L_{n-1}Fh}", from=4-2, to=4-4] \arrow["{\phi_{n-1}}", from=2-2, to=4-2] \arrow["{\phi_{n-1}}", from=2-4, to=4-4] \arrow["{\delta'}"{description}, from=5-1, to=4-2] \arrow["\delta"{description}, from=5-5, to=4-4] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsOCxbMCwwLCJUX24gQSciXSxbNCwwLCJUX25BIl0sWzAsNCwiTF9uIEZBJyJdLFs0LDQsIkxfbkYoQSkiXSxbMSwxLCJUX3tuLTF9SyciXSxbMywxLCJUX3tuLTF9SyJdLFsxLDMsIkxfe24tMX1GSyciXSxbMywzLCJMX3tuLTF9RksiXSxbMCwxLCJUX25mIl0sWzIsMywiTF9uRmYiXSxbMSwzLCJcXHBoaV9uKEEpIl0sWzAsMiwiXFxwaGlfbihBJykiXSxbNCw1LCJUX3tuLTF9aCJdLFswLDQsIlxcZGVsdGEnIiwxXSxbMSw1LCJcXGRlbHRhIiwxXSxbNiw3LCJMX3tuLTF9RmgiXSxbNCw2LCJcXHBoaV97bi0xfSJdLFs1LDcsIlxccGhpX3tuLTF9Il0sWzIsNiwiXFxkZWx0YSciLDFdLFszLDcsIlxcZGVsdGEiLDFdXQ==}{Link to Diagram} \end{quote} Note that all of the quadrilaterals here commute. The middle top and bottom come from naturality of $$T_*, L_*F$$ with respect to $$\delta$$, the RHS/LHS due to the construction of the $$\varphi_n$$, and $$\phi_{n-1}$$ is natural by the inductive hypothesis. Now in order to traverse $$T_nA' \to T_n A \to L_n F (A)$$, we can pass the path through one commuting square at a time to make it equal to $$T_nA' \to L_n FA' \to L_n FA$$, so the outer square commutes. We have \begin{align*} \delta \varphi_n(A) T_n F = \delta L_n Ff \varphi_n(A') ,\end{align*} and since $$\delta$$ is monic (using the previous vanishing due to projectivity), so we can cancel on the left and this yields the definition of naturality. \begin{corollary}[?] The definition of $$\varphi_n(A)$$ does not depend on the choice of $$P$$. Taking $$A' = A$$ in the previous argument with $$f = \one$$, suppose $$P'\neq P$$. Then $$T_n f = \one = L_n Ff$$ and setting $$\varphi_n'(A)$$ to be the map coming from $$P'$$, we get $$\varphi_n'(A) = \varphi_n(A)$$ using the following diagram: \begin{center} \begin{tikzcd} 0 & {K'} & {P'} & A & 0 \\ \\ 0 & K & P & A & 0 \arrow["\one", from=1-4, to=3-4] \arrow["{\exists g}", dashed, from=1-3, to=3-3] \arrow[from=3-1, to=3-2] \arrow[from=3-2, to=3-3] \arrow[from=3-3, to=3-4] \arrow[from=3-4, to=3-5] \arrow[from=1-1, to=1-2] \arrow[from=1-2, to=1-3] \arrow[from=1-3, to=1-4] \arrow[from=1-4, to=1-5] \arrow["{\exists h}", dashed, from=1-2, to=3-2] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsMTAsWzAsMCwiMCJdLFsxLDAsIksnIl0sWzIsMCwiUCciXSxbMywwLCJBIl0sWzMsMiwiQSJdLFsyLDIsIlAiXSxbMSwyLCJLIl0sWzAsMiwiMCJdLFs0LDIsIjAiXSxbNCwwLCIwIl0sWzMsNCwiXFxvbmUiXSxbMiw1LCJcXGV4aXN0cyBnIiwwLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV0sWzcsNl0sWzYsNV0sWzUsNF0sWzQsOF0sWzAsMV0sWzEsMl0sWzIsM10sWzMsOV0sWzEsNiwiXFxleGlzdHMgaCIsMCx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dXQ==}{Link to Diagram} \end{quote} \end{corollary} So the $$\varphi_n(A)$$ are uniquely defined. We now want to show that $$\varphi_n$$ commutes with the $$\delta_n$$ coming from an \emph{arbitrary} SES instead of a fixed reference SES. \begin{center} \begin{tikzcd} {T_n A} &&&&&&& {T_{n-1}C} \\ &&& {T_* \text{ a } \delta \text{ functor}} \\ {T_nA} &&&& {T_{n-1}K} \\ &&& {\text{reference}} &&& {\phi_{n-1} \text{natural}} \\ \\ {L_nFA} &&&& {L_{n-1}FA} \\ &&& {L_*F \text{ a } \delta \text{ functor}} \\ {L_n FA} &&&&&&& {L_{n-1}FC} \arrow["{\phi_n}", from=3-1, to=6-1] \arrow["{\delta_{(2)}}", from=3-1, to=3-5] \arrow["{\delta_{(1)}}", from=6-1, to=6-5] \arrow["{\phi_{n-1}}"', from=3-5, to=6-5] \arrow["{=}", from=1-1, to=3-1] \arrow["{=}", from=8-1, to=6-1] \arrow["{\delta_{(1)}}"', from=8-1, to=8-8] \arrow["{\delta_{(2)}}"', from=1-1, to=1-8] \arrow["{\phi_{n-1}}"{description}, from=1-8, to=8-8] \arrow["{T_{n-1}h}"{description}, from=3-5, to=1-8] \arrow["{L_{n-1}Fh}"{description}, from=6-5, to=8-8] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=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}{Link to Diagram} \end{quote} This diagram commutes for the reasons indicated, and commutativity of the outside square implies that $$\varphi_n$$ commutes with the $$\delta$$ coming from any SES. \begin{quote} See section 2.4 in Weibel. \end{quote} \end{proof} \hypertarget{monday-february-15}{% \section{Monday, February 15}\label{monday-february-15}} \hypertarget{right-derived-functors}{% \subsection{2.5: Right-Derived Functors}\label{right-derived-functors}} \begin{remark} Today: right-derived functors of a left-exact functor. Luckily we can use some opposite category tricks which save us some work of re deriving everything. \end{remark} \begin{definition}[Right Derived Functors] Let $$F: \mathcal{A} \to \mathcal{B}$$ be left-exact where $$\mathcal{A}$$ has enough injectives. Given $$A \in \mathcal{A}$$, fix an injective resolution $$0 \to A \xrightarrow{\varepsilon} I$$ and define \begin{align*} R^i \mathcal{F} \coloneqq H^i( FA ) && i \geq 0 .\end{align*} \end{definition} \begin{remark} Then \begin{align*} 0 \to FA \xrightarrow{F\varepsilon} FI^0 \xrightarrow{Fd^0} FI^1 \end{align*} is exact, and \begin{align*} R^0 FA = \ker F(d^0) / \left\langle{ 0 }\right\rangle = \operatorname{im}F\varepsilon\cong FA ,\end{align*} and so there is naturally an isomorphism $$R^0 F \cong F$$. Observe that $$F$$ yields a right-exact functor $$F^{\operatorname{op}}: \mathcal{A}^{\operatorname{op}}\to \mathcal{B}^{\operatorname{op}}$$, where we note that $$F^{\operatorname{op}}(f^{\operatorname{op}}) = F(f)^{\operatorname{op}}$$. Note that taking the opposite category sends injectives to projectives and so $$\mathcal{A}^{\operatorname{op}}$$ has enough projectives. This means that $$L_i F^{\operatorname{op}}$$ are defined using the projective resolution $$I$$, so we have \begin{align*} R^i F(A) = (L_i F^{\operatorname{op}})^{\operatorname{op}} .\end{align*} Thus all results about left-derived functors translate to right-derived functors: \begin{itemize} \tightlist \item $$R_i F$$ is independent of the choice of injective resolution, up to a natural isomorphism. \item If $$A$$ is injective, then $$R^{i>0} F(A) = 0$$. \item The collection $$\left\{{ R^i F }\right\} _{i\geq 0 }$$ forms a universal cohomological $$\delta{\hbox{-}}$$functor for $$F$$. \item An object $$Q \in \mathcal{A}$$ is \textbf{$$F{\hbox{-}}$$acyclic} if $$R^{>0}F(Q) = 0$$. \item $$R^iF$$ can be computed using $$F{\hbox{-}}$$acyclic objects instead of injective resolutions. \end{itemize} \end{remark} \begin{definition}[?] Fix a right $$R{\hbox{-}}$$module $$M \in {\mathsf{Mod}{\hbox{-}}\mathsf{R}}$$, then $$F \coloneqq\mathop{\mathrm{Hom}}_{{\mathsf{Mod}{\hbox{-}}\mathsf{R}}}(M, {-}): {\mathsf{Mod}{\hbox{-}}\mathsf{R}}\to {\mathsf{Ab}}$$ is a left-exact functor. Its right-derived functors are \textbf{ext functors} and denoted $$\operatorname{Ext}_{{\mathsf{Mod}{\hbox{-}}\mathsf{R}}}^i(M, {-})$$. \end{definition} \begin{example}[?] \begin{align*} \operatorname{Ext}_{{\mathsf{Mod}{\hbox{-}}\mathsf{R}}}^i(M, A) = (R^i F)(A) = [ R^i \mathop{\mathrm{Hom}}_{{\mathsf{Mod}{\hbox{-}}\mathsf{R}}}(M, {-}) ] (A) .\end{align*} \end{example} \begin{remark} Exercises 2.5.1, 2.5.2 are important extensions of our existing characterizations of injectives and projectives in $${\mathsf{Mod}{\hbox{-}}\mathsf{R}}$$. These upgrade the characterization involving $$\mathop{\mathrm{Hom}}$$ to one involving $$\operatorname{Ext}$$. \footnote{Note the typo in 2.5.1.3, it should say the following: $$B$$ is $$\mathop{\mathrm{Hom}}_{R}(A, {-})$$ is acyclic for all $$A$$.''} \end{remark} \begin{remark} Fix $$B\in {\mathsf{Mod}{\hbox{-}}\mathsf{R}}$$ and consider $$G\coloneqq\mathop{\mathrm{Hom}}_{{\mathsf{Mod}{\hbox{-}}\mathsf{R}}}({-}, B): {\mathsf{Mod}{\hbox{-}}\mathsf{R}}\to {\mathsf{Ab}}$$. Then $$G$$ is still left-exact, but is now \emph{contravariant}. We can regard it as a covariant functor left-exact functor $$G: {\mathsf{Mod}{\hbox{-}}\mathsf{R}}^{\operatorname{op}}\to {\mathsf{Ab}}$$. So we define $$R^i G(A)$$ by an injective resolution of $$A$$ in $$\mathcal{A}^{\operatorname{op}}$$, and this is the same as a projective resolution of $$A$$ in $$\mathcal{A}$$. So apply $$G$$ and take cohomology. It turns out that \begin{align*} R^i \mathop{\mathrm{Hom}}_{{\mathsf{Mod}{\hbox{-}}\mathsf{R}}}({-}, B) \cong R^i \mathop{\mathrm{Hom}}_{{\mathsf{Mod}{\hbox{-}}\mathsf{R}}}(A, {-})(B) \coloneqq\operatorname{Ext}^i_{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}(A, B) ,\end{align*} so we can use the same notation $$\operatorname{Ext}^i_R({-}, B)$$ for both cases. \end{remark} \hypertarget{adjoint-functors-and-leftright-exactness}{% \subsection{2.6: Adjoint Functors and Left/Right Exactness}\label{adjoint-functors-and-leftright-exactness}} \begin{slogan} $${-}$$ adjoints are $${-}^{\operatorname{op}}$$ exact, since $${-}$$ adjoints have $${-}{\hbox{-}}$$derived functors. \end{slogan} \begin{theorem}[Exactness of adjoint functors] Let \begin{align*} \adjunction{L}{R}{ \mathcal{A} } { \mathcal{B} } \end{align*} be an adjoint pair of functors. Then there exists a natural isomorphism \begin{align*} \tau_{AB}: \mathop{\mathrm{Hom}}_{\mathcal{B}}(LA, B) \xrightarrow{\sim} \mathop{\mathrm{Hom}}_{\mathcal{A}}(A, RB) \quad \forall A\in \mathcal{A}, B\in \mathcal{B} .\end{align*} Moreover, \begin{itemize} \tightlist \item $$L$$ is right exact, and \item $$B$$ is left exact. \end{itemize} \end{theorem} \begin{proposition}[1.6: Yoneda] A sequence \begin{align*} A \xrightarrow{\alpha} B \xrightarrow{\beta} C \end{align*} is exact in $$\mathcal{A}$$ if and only if for all $$M \in {\operatorname{Ob}}( \mathcal{A} )$$, the sequence \begin{align*} \mathop{\mathrm{Hom}}_{\mathcal{A}} (M, A) \xrightarrow{\alpha^* \coloneqq\alpha\circ {-}} \mathop{\mathrm{Hom}}_{\mathcal{A}} (M, B) \xrightarrow{\beta^* \coloneqq\beta \circ {-}} \mathop{\mathrm{Hom}}_{\mathcal{A}} (M, C) \end{align*} is exact. \end{proposition} \begin{proof}[?] \envlist \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \item Take $$M=A$$, then $$0 = \beta^* \alpha^*(\one_A) = \beta \alpha \one = \beta \alpha$$. Thus $$\operatorname{im}\alpha \subseteq \ker \beta$$. \item Take $$M = \ker \beta$$ and consider the inclusion $$\iota: \ker M \hookrightarrow B$$, then $$\beta^*(\iota) = \beta \iota = 0$$ and thus $$\iota \in \ker \beta^* = \operatorname{im}\alpha^*$$. So there exists $$\sigma\in \mathop{\mathrm{Hom}}( \ker \beta, A)$$ such that $$\iota = \alpha^* (\sigma) \coloneqq\alpha \sigma$$, and thus $$\ker \beta = \operatorname{im}\iota \subset \operatorname{im}\alpha$$. \end{enumerate} Thus $$\ker \beta= \operatorname{im}\alpha$$, yielding exactness of the bottom sequence. \end{proof} \begin{proof}[of theorem] We'll first prove that $$R$$ is left-exact. Take a SES in $$B$$, say \begin{align*} 0 \to B' \to B \to B'' \to 0 .\end{align*} Apply the left-exact covariant functor $$\mathop{\mathrm{Hom}}_{\mathcal{B}}(LA, {-})$$ followed by $$\tau$$: \begin{center} \begin{tikzcd} 0 && { \mathop{\mathrm{Hom}}_{\mathcal{B}} (LA, B') } && { \mathop{\mathrm{Hom}}_{\mathcal{B}} (LA, B) } && { \mathop{\mathrm{Hom}}_{\mathcal{B}} (LA, B'')} \\ \\ 0 && {\mathop{\mathrm{Hom}}_{\mathcal{B}} (A, RB')} && {\mathop{\mathrm{Hom}}_{\mathcal{B} }(A, RB)} && {\mathop{\mathrm{Hom}}_{\mathcal{B}} (A, RB'')} \arrow[from=1-1, to=1-3] \arrow[from=1-3, to=1-5] \arrow[from=1-5, to=1-7] \arrow[from=3-3, to=3-5] \arrow[from=3-5, to=3-7] \arrow["{\tau_{AB}}"{description}, squiggly, from=1-3, to=3-3] \arrow["{\tau_{AB}}"{description}, squiggly, from=1-5, to=3-5] \arrow["{\tau_{AB}}"{description}, squiggly, from=1-7, to=3-7] \arrow[dashed, from=3-1, to=3-3] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=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}{Link to Diagram} \end{quote} The bottom sequence is exact by naturality of $$\tau$$. Now applying the Yoneda lemma, we obtain an exact sequence \begin{align*} 0 \to \mathop{\mathrm{Hom}}_{\mathcal{A}}(A, RB') \to \mathop{\mathrm{Hom}}_{\mathcal{A}}(A, RB) \to \mathop{\mathrm{Hom}}_{\mathcal{A}}(A, RB'') .\end{align*} So $$R$$ is left exact. Now $$L^{\operatorname{op}}: \mathcal{A} \to \mathcal{B}$$ is right adjoint to $$R^{\operatorname{op}}$$, so $$L^{\operatorname{op}}$$ is left exact and thus $$L$$ is right exact. \end{proof} \hypertarget{tensor-product-functors-and-tor}{% \subsection{Tensor Product Functors and Tor}\label{tensor-product-functors-and-tor}} \begin{remark} Let \begin{itemize} \tightlist \item $$R, S \in \mathsf{Ring}$$, \item $$B\in (\mathsf{R}, \mathsf{S}){\hbox{-}}\mathsf{biMod}$$, \item $$C\in {\mathsf{S}{\hbox{-}}\mathsf{Mod}}$$. \end{itemize} Then $$\mathop{\mathrm{Hom}}_{S}(B, C)\in {\mathsf{Mod}{\hbox{-}}\mathsf{R}}$$ in a natural way: given $$f:B\to C$$, define $$(f\cdot r)(b) = f(rb)$$. \end{remark} \begin{exercise}[?] Check that this is a well-defined morphism of right $$S{\hbox{-}}$$modules. \end{exercise} \begin{remark} We saw this structure earlier with $$S={\mathbb{Z}}$$, see p.41. \end{remark} \begin{proposition}[Tensor-Hom adjunction] Fix $$R,S$$ and $${}_R B_S$$ as above. Then for every $$A \in {\mathsf{Mod}{\hbox{-}}\mathsf{R}}$$ and $$C\in {\mathsf{{\hbox{-}}}{\hbox{-}}\mathsf{Mod}}S$$ there is a natural isomorphism \begin{align*} \tau: \mathop{\mathrm{Hom}}_S( A\otimes_R B, C ) &\xrightarrow{\sim} \mathop{\mathrm{Hom}}_R(A, \mathop{\mathrm{Hom}}_S(B, C) ) \\ f &\mapsto g(a)(b) = f(a\otimes b) \\ f(a\otimes b) = g(a)(b) &\mapsfrom g .\end{align*} Note that the tensor product is a right $$S{\hbox{-}}$$module, and the hom on the right is a right $$R{\hbox{-}}$$module, so these expressions make sense. Here $$B$$ is fixed, so $$A$$ and $$C$$ are variables and this is a statement about bifunctors \begin{align*} {-}\otimes_R B: {\mathsf{Mod}{\hbox{-}}\mathsf{R}}\to \mathsf{Mod}{\hbox{-}}\mathsf{S} ,\end{align*} which is left adjoint to \begin{align*} \mathop{\mathrm{Hom}}_S(B, {-}): \mathsf{Mod}{\hbox{-}}\mathsf{S} \to {\mathsf{Mod}{\hbox{-}}\mathsf{R}} .\end{align*} So the former is a left adjoint and the latter is a right adjoint, so by the theorem, $${-}\otimes_R B$$ is right exact. \end{proposition} \begin{remark} If $$B$$ is only a left $$R{\hbox{-}}$$module, we can always take $$S = {\mathbb{Z}}$$, which makes this into a functor \begin{align*} {-}\otimes_R B: {\mathsf{Mod}{\hbox{-}}\mathsf{R}}\to {\mathsf{Ab}} .\end{align*} Since this is a right exact functor from a category with enough injectives, we can define left-derived functors. \end{remark} \begin{definition}[?] Let $$B\in (\mathsf{R}, \mathsf{S}){\hbox{-}}\mathsf{biMod}$$ and let \begin{align*} T({-}) \coloneqq{-}\otimes_R B: {\mathsf{Mod}{\hbox{-}}\mathsf{R}}\to \mathsf{Mod}{\hbox{-}}\mathsf{S} .\end{align*} Then define $$\operatorname{Tor}_n^R(A, B) \coloneqq L_n T(A)$$. \end{definition} \begin{remark} Note that these are easier to work with, since they're covariant in both variables. \end{remark} \hypertarget{friday-february-19}{% \section{Friday, February 19}\label{friday-february-19}} \begin{remark} We looked at $$B\in (\mathsf{R}, \mathsf{S}){\hbox{-}}\mathsf{biMod}$$ and showed $${-}\otimes_R B: {\mathsf{R}{\hbox{-}}\mathsf{Mod}} \to {\mathsf{S}{\hbox{-}}\mathsf{Mod}}$$ is left adjoint to hom, and has left-derived functors $$\operatorname{Tor}_n^R({-}, B) \coloneqq L_n({-}\otimes_R B)$$. \begin{align*} \adjunction{{-}\otimes_R B}{ \mathop{\mathrm{Hom}}_{{S}}(B, {-}) }{\mathsf{R}{\hbox{-}}\mathsf{Mod}}{\mathsf{S}{\hbox{-}}\mathsf{Mod}} .\end{align*} Note that $$\operatorname{Tor}_0^R(A, B) \cong A\otimes_R B$$. \end{remark} \begin{remark} $$A\otimes_R {-}$$ is also right exact, and it turns out that \begin{align*} L_n(A\otimes_R {-})(B) \cong L_n({-}\otimes_R B)(A) .\end{align*} So unambiguously denote either of this left derived functors as $$\operatorname{Tor}_n(A, B)$$. \end{remark} \hypertarget{limits-and-colimits}{% \subsection{Limits and Colimits}\label{limits-and-colimits}} \begin{definition}[Functor Category] Given categories $$\mathcal{I}, \mathcal{A}$$, define a \textbf{functor category} $$\mathcal{A}^{\mathcal{I}}$$ by \begin{itemize} \item $${\operatorname{Ob}}( \mathcal{A}^{\mathcal{I}} )$$: functors $$A: \mathcal{I} \to \mathcal{A}$$. \item $$\operatorname{Mor}(\mathcal{A} ^{\mathcal{I} })$$: natural transformations $$\eta:A\to B$$ between functors. \end{itemize} $$\mathcal{I}$$ is thought of as an index category, and we'll write $$A_i \coloneqq A(i) \in \mathcal{A}$$ for $$i\in \mathcal{I}$$. If $$\alpha:i \to j$$ is a morphism in $$I$$, then denote $$A(\alpha) \coloneqq\alpha_*$$, which is the morphism defined by the following: \begin{center} \begin{tikzcd} {A_i} &&& {A_j} \\ \\ {B_i} &&& {B_j} \arrow["{\alpha_*}", from=1-1, to=1-4] \arrow["{\alpha_*}", from=3-1, to=3-4] \arrow["{\eta_i}"{description}, from=1-1, to=3-1] \arrow["{\eta_j}"{description}, from=1-4, to=3-4] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJBX2kiXSxbMywwLCJBX2oiXSxbMCwyLCJCX2kiXSxbMywyLCJCX2oiXSxbMCwxLCJcXGFscGhhXyoiXSxbMiwzLCJcXGJldGFfKiJdLFswLDIsIlxcZXRhX2kiLDFdLFsxLDMsIlxcZXRhX2oiLDFdXQ==}{Link to Diagram} \end{quote} Composition is defined by $$A \xrightarrow{\eta} B \xrightarrow{\zeta} C$$ is given by $$(\zeta_\eta)_i = \zeta_i \circ \eta_i$$. We need the collection of morphisms to be sets, so we'll require $$\mathcal{I}$$ to be a \emph{small category} (i.e.~the class of objects forms a set). \end{definition} \begin{example}[Poset Category] Take $$(I, \leq)$$ a poset (which is reflexive, antisymmetric, transitive, but not every two elements are comparable), define a category by \begin{itemize} \item $${\operatorname{Ob}}(\mathcal{I}) = I$$ \item $${\left\lvert { \mathop{\mathrm{Hom}}_{\mathcal{I}}(i, j) } \right\rvert} \leq 1$$, and $$i\to j \iff i\leq j$$ \end{itemize} Note that if $$i\not\leq j$$, then $$\mathop{\mathrm{Hom}}_{\mathcal{I}}(i, j) = \emptyset$$. \end{example} \begin{remark} Both $$\mathcal{A}, \mathcal{A}^{\mathcal{I}}$$ are small, so we can consider functors between them. \end{remark} \begin{definition}[Diagonal Functor] The \textbf{diagonal functor} is defined as $$\Delta: \mathcal{A} \to \mathcal{A}^{\mathcal{I}}$$ where for $$B \in \mathcal{A}$$ the functor $$\Delta(B)$$ is the constant functor, i.e.~$$\Delta(B)_i = B$$ for all $$i\in \mathcal{I}$$. All morphism are sent to the identity, i.e.~$$i \xrightarrow{\alpha} j \xrightarrow{\Delta(B)} B \xrightarrow{\one_B} B$$. \end{definition} \todo[inline]{Work out how morphisms work here with respect to natural transformations.} \begin{definition}[Colimit] The \textbf{colimit} of a functor $$A: \mathcal{I} \to \mathcal{A}$$ is an object $$C\in \mathcal{A}$$ which we'll denote $$\mathop{\mathrm{colim}}\nolimits_{i\in \mathcal{I}} A_i$$, along with a natural transformation $$\eta:A\to \Delta(C)$$ which is universal among natural transformations of the form $$\theta: A\to \Delta(B)$$ for $$B\in \mathcal{A}$$. The unique map in the universal property is from $$C\to B$$, and we have the following situation: \begin{center} \begin{tikzcd} {\mathcal{I}} \\ i && {A_i} && C \\ \\ j && {A_j} && C \\ \\ && {A_i} \\ &&&& C && B \\ && {A_j} \arrow["\alpha", from=2-1, to=4-1] \arrow["{\alpha_*}", from=2-3, to=4-3] \arrow["{\eta_i}"', from=2-3, to=2-5] \arrow["{\eta_j}", from=4-3, to=4-5] \arrow[equals, from=2-5, to=4-5] \arrow["{\eta_i}"', from=6-3, to=7-5] \arrow["{\eta_j}", from=8-3, to=7-5] \arrow["{\alpha_*}", from=6-3, to=8-3] \arrow["{\theta_i}", curve={height=-12pt}, from=6-3, to=7-7] \arrow["{\theta_j}", curve={height=12pt}, from=8-3, to=7-7] \arrow["{\exists ! \gamma}", dashed, from=7-5, to=7-7] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=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}{Link to Diagram} \end{quote} \end{definition} \begin{example}[?] Let $$(I, \leq)$$ be a poset and take $$\mathcal{I}$$ its poset category. Then there are morphisms $$i\to j \iff i\leq j$$, and we have a diagram \begin{center} \begin{tikzcd} {A_i} \\ && C && D \\ {A_j} \\ {} \arrow["{\eta_i}", from=1-1, to=2-3] \arrow["{\eta_j}"', from=3-1, to=2-3] \arrow["{\theta_j}", curve={height=12pt}, from=3-1, to=2-5] \arrow["{\theta_i}"', curve={height=-12pt}, from=1-1, to=2-5] \arrow["{\exists ! \gamma}"{description}, dashed, from=2-3, to=2-5] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsNSxbMCwwLCJBX2kiXSxbMCwzXSxbMCwyLCJBX2oiXSxbMiwxLCJDIl0sWzQsMSwiRCJdLFswLDMsIlxcZXRhX2kiXSxbMiwzLCJcXGV0YV9qIiwyXSxbMiw0LCJcXHRoZXRhX2oiLDAseyJjdXJ2ZSI6Mn1dLFswLDQsIlxcdGhldGFfaSIsMix7ImN1cnZlIjotMn1dLFszLDQsIlxcZXhpc3RzICEgXFxnYW1tYSIsMSx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dXQ==}{Link to Diagram} \end{quote} This is the \textbf{direct limit}. Note that for a poset of category of subsets, this ends up being the union. \end{example} \begin{example}[?] Let $${\operatorname{Ob}}(\mathcal{I}) = \left\{{ 1, 2 }\right\}$$, and take two maps, one of which we'll label by 0'': \begin{center} \begin{tikzcd} 1 \\ \\ 2 \arrow[shift right=5, from=1-1, to=3-1] \arrow["0"{description}, shift left=5, from=1-1, to=3-1] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsMixbMCwwLCIxIl0sWzAsMiwiMiJdLFswLDEsIiIsMCx7Im9mZnNldCI6NX1dLFswLDEsIjAiLDEseyJvZmZzZXQiOi01fV1d}{Link to Diagram} \end{quote} Suppose now that $$\mathcal{A}$$ is an abelian category, and suppose we're given a morphism $$A_1 \xrightarrow{f} A_2$$ in $$\mathcal{A}$$. Define $$A\in \mathcal{A}^{\mathcal{I}}$$, and define a functor \begin{center} \begin{tikzcd} 1 && {A_1} \\ &&&&&& B \\ 2 && {A_2} \arrow[shift right=2, from=1-1, to=3-1] \arrow["0", shift left=2, from=1-1, to=3-1] \arrow["0", shift left=2, from=1-3, to=3-3] \arrow["f"', shift right=2, from=1-3, to=3-3] \arrow["{\theta_1}", curve={height=-12pt}, from=1-3, to=2-7] \arrow["{\theta_2}", curve={height=12pt}, from=3-3, to=2-7] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsNSxbMCwwLCIxIl0sWzAsMiwiMiJdLFsyLDAsIkFfMSJdLFsyLDIsIkFfMiJdLFs2LDEsIkIiXSxbMCwxLCIiLDAseyJvZmZzZXQiOjJ9XSxbMCwxLCIwIiwwLHsib2Zmc2V0IjotMn1dLFsyLDMsIjAiLDAseyJvZmZzZXQiOi0yfV0sWzIsMywiZiIsMix7Im9mZnNldCI6Mn1dLFsyLDQsIlxcdGhldGFfMSIsMCx7ImN1cnZlIjotMn1dLFszLDQsIlxcdGhldGFfMiIsMCx7ImN1cnZlIjoyfV1d}{Link to Diagram} \end{quote} By commutativity, \begin{itemize} \item $$\theta_2 \circ 0 = \theta_1 \implies \theta_1 = 0$$ \item $$\theta_2 \circ f = \theta_1 = 0$$. \end{itemize} So suppose there was a colimit $$C$$, then it'd fit into this diagram as follows: \begin{center} \begin{tikzcd} 1 && {A_1} \\ &&&& C && B \\ 2 && {A_2} \arrow[shift right=2, from=1-1, to=3-1] \arrow["0", shift left=2, from=1-1, to=3-1] \arrow["0", shift left=2, from=1-3, to=3-3] \arrow["f"', shift right=2, from=1-3, to=3-3] \arrow["{\theta_1}", curve={height=-12pt}, from=1-3, to=2-7] \arrow["{\theta_2}", curve={height=12pt}, from=3-3, to=2-7] \arrow["0", from=1-3, to=2-5] \arrow["p", from=3-3, to=2-5] \arrow[dashed, from=2-5, to=2-7] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsNixbMCwwLCIxIl0sWzAsMiwiMiJdLFsyLDAsIkFfMSJdLFsyLDIsIkFfMiJdLFs2LDEsIkIiXSxbNCwxLCJDIl0sWzAsMSwiIiwwLHsib2Zmc2V0IjoyfV0sWzAsMSwiMCIsMCx7Im9mZnNldCI6LTJ9XSxbMiwzLCIwIiwwLHsib2Zmc2V0IjotMn1dLFsyLDMsImYiLDIseyJvZmZzZXQiOjJ9XSxbMiw0LCJcXHRoZXRhXzEiLDAseyJjdXJ2ZSI6LTJ9XSxbMyw0LCJcXHRoZXRhXzIiLDAseyJjdXJ2ZSI6Mn1dLFsyLDUsIjAiXSxbMyw1LCJwIl0sWzUsNCwiIiwxLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV1d}{Link to Diagram} \end{quote} Note that $$C$$ is precisely the cokernel of $$f$$! \end{example} \begin{remark} Think about this last diagram: what happens when you mod out by larger modules? \end{remark} \begin{exercise}[Colimits always exist] Suppose $$I$$ is a discrete category, i.e.~$$\mathop{\mathrm{Hom}}(i, j) = \emptyset$$ unless $$i=j$$, in which case $$\mathop{\mathrm{Hom}}(i, i) = \left\{{ \one_i }\right\}$$. Supposing that $$A: I \to \mathcal{A}$$, show that $$\mathop{\mathrm{colim}}\nolimits_{i\in \mathcal{I}} = \coprod_{i} A_i$$. \end{exercise} \begin{definition}[?] A category $$\mathcal{A}$$ is \textbf{cocomplete} if every colimit $$\mathop{\mathrm{colim}}\nolimits_{i\in \mathcal{I}} A_i$$ exists for every $$A\in \mathcal{A}^{\mathcal{I}}$$ and all small categories $$\mathcal{I}$$. \end{definition} \begin{exercise}[Taking colimits defines a functor for cocomplete categories] Show that when $$\mathcal{A}$$ is cocomplete, $$\mathop{\mathrm{colim}}\nolimits: \mathcal{A}^{\mathcal{I}} \to \mathcal{A}$$ defines a functor. \end{exercise} \begin{exercise}[Weibel 2.6.4] Show that the functor $$\mathop{\mathrm{colim}}\nolimits$$ is left-adjoint to the diagonal functor $$\Delta$$, so there is an adjunction \begin{align*} \adjunction{\mathop{\mathrm{colim}}\nolimits}{\Delta}{\mathcal{A}^{\mathcal{I}} }{\mathcal{A} } .\end{align*} Thus when $$\mathcal{A}$$ is abelian and $$\mathop{\mathrm{colim}}\nolimits$$ exists, it is right-exact (since left-adjoints are always right-exact). Note that it's not exact in general. \end{exercise} \begin{proposition}[Cocomplete iff all coproducts exist] For any abelian category $$\mathcal{A}$$, the following are equivalent: \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \item $$\coprod A_i$$ exists in $$\mathcal{A}$$ for every set $$\left\{{ A_i }\right\}$$ of objects in $$\mathcal{A}$$ (\emph{set-indexed coproducts}). \item $$\mathcal{A}$$ is cocomplete. \end{enumerate} \end{proposition} \begin{remark} We'll prove this next time, note that $$2\implies 1$$ since coproducts are special cases of limits. \end{remark} \hypertarget{monday-february-22}{% \section{Monday, February 22}\label{monday-february-22}} \hypertarget{colimits-and-adjoints}{% \subsection{Colimits and Adjoints}\label{colimits-and-adjoints}} \begin{proposition}[Characterizations of cocomplete categories] Assume $$\mathcal{A}$$ is abelian so we have cokernels for maps. TFAE: \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \item $$\bigoplus A_i$$ exists in $$\mathcal{A}$$ for every set $$\left\{{A_i}\right\}$$ of objects in $$\mathcal{A}$$. \item $$\mathcal{A}$$ is cocomplete, i.e.~$$\mathop{\mathrm{colim}}\nolimits_{i\in I}A_i$$ exists for every functor $$\mathcal{I} \to \mathcal{A}$$ with $$\mathcal{I}$$ small. \end{enumerate} \end{proposition} \begin{proof}[?] Note that (1) is a special case of (2), so it suffices to show $$1\implies 2$$. Given a functor $$A: \mathcal{I} \to \mathcal{A}$$ and let $$f: \bigoplus _{\alpha i\to j} A_i \to \bigoplus_{i\in \mathcal{I}} A_i$$ where $$i,j \in \mathcal{I}$$. \begin{center} \begin{tikzcd} i && j && {\mathcal{I}} \\ \\ {A_i} && {A_j} && {\mathcal{A}} \arrow["\alpha", from=1-1, to=1-3] \arrow["{\alpha_*}", from=3-1, to=3-3] \arrow["A"{description}, squiggly, from=1-3, to=3-3] \arrow["A"{description}, squiggly, from=1-1, to=3-1] \arrow["A", squiggly, from=1-5, to=3-5] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsNixbMCwwLCJpIl0sWzIsMCwiaiJdLFswLDIsIkFfaSJdLFsyLDIsIkFfaiJdLFs0LDIsIlxcY2F0e0F9Il0sWzQsMCwiXFxjYXR7SX0iXSxbMCwxLCJcXGFscGhhIl0sWzIsMywiXFxhbHBoYV8qIl0sWzEsMywiQSIsMSx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6InNxdWlnZ2x5In19fV0sWzAsMiwiQSIsMSx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6InNxdWlnZ2x5In19fV0sWzUsNCwiQSIsMCx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6InNxdWlnZ2x5In19fV1d}{Link to Diagram} \end{quote} Then the map $$f( a_{i, \alpha}) = \alpha_*(a_i) - a_i \in A_j - A_i$$, so this is $$\alpha_* - \one$$. Let $$C\coloneqq\operatorname{coker}f \coloneqq\bigoplus_{i\in I} A_i / \operatorname{im}(f)$$, and we'll denote elements in this quotient with a bar. \begin{claim} $$C = \mathop{\mathrm{colim}}\nolimits_{i\in I} A_i$$ with \begin{align*} \eta_i: A_i &\to C \\ a_i &\mapsto \mkern 1.5mu\overline{\mkern-1.5mua_i\mkern-1.5mu}\mkern 1.5mu ,\end{align*} where we first embed $$A_i$$ into the direct sum and then take the quotient. \end{claim} \begin{exercise}[?] Use the universal property of cokernels in $$\mathcal{A}$$. Check that the following diagram commutes: \begin{center} \begin{tikzcd} {A_i} \\ && C \\ {A_j} \arrow["{\eta_i}", from=1-1, to=2-3] \arrow["{\eta_j}"', from=3-1, to=2-3] \arrow["{\alpha_*}"', from=1-1, to=3-1] \end{tikzcd} \end{center} This essentially follows from the fact that $$\mkern 1.5mu\overline{\mkern-1.5mu \alpha_*(a_i)\mkern-1.5mu}\mkern 1.5mu = \mkern 1.5mu\overline{\mkern-1.5mua_i\mkern-1.5mu}\mkern 1.5mu$$. \end{exercise} \end{proof} \begin{remark} $${\mathsf{Mod}{\hbox{-}}\mathsf{R}}$$ satisfies (1), since direct sums of $$R{\hbox{-}}$$modules still have an $$R{\hbox{-}}$$module structure. Thus $${\mathsf{Mod}{\hbox{-}}\mathsf{R}}$$ is cocomplete. \end{remark} \begin{definition}[Limits] The \textbf{limit} of a functor $$A:\mathcal{I} \to \mathcal{A}$$ is the colimit of the dual functor $$A^{\operatorname{op}}: I^{\operatorname{op}}\to \mathcal{A}^{\operatorname{op}}$$. \end{definition} \begin{remark} Note that this amounts to reversing arrows in the conditions of a colimit. Many of the results for colimits go through with arrows reversed. Examples: kernels, direct products. If $$I$$ is a poset, then limits are referred to as \textbf{inverse limits}, using $$\varprojlim_{i\in I} A_i$$. \end{remark} \begin{definition}[Complete Categories] $$\mathcal{A}$$ is \textbf{complete} if and only if $$\lim_{i\in I} A_i$$ exists whenever $$\mathcal{I}$$ is small and $$A: \mathcal{I} \to \mathcal{A}$$. \end{definition} \begin{theorem}[The Adjoint-Limit Theorem] Let $$\adjunction{L}{R}{ \mathcal{A} }{ \mathcal{ B} }$$ be an adjoint pair, where now $$\mathcal{A}, \mathcal{B}$$ are now arbitrary categories (not necessarily abelian). Then \begin{itemize} \item The \textbf{left adjoint} $$L$$ preserves \textbf{colimits} (direct sums, cokernels, etc). I.e. if $$A: \mathcal{I} \to \mathcal{A}$$ has a colimit, then so does $$(L \circ A ): \mathcal{I}\to \mathcal{B}$$, and $$L( \mathop{\mathrm{colim}}\nolimits A_i) = \mathop{\mathrm{colim}}\nolimits(LA_i)$$. \item The \textbf{right adjoint} $$R$$ preserves \textbf{limits} (direct products, kernels, etc). \end{itemize} \end{theorem} \begin{proof}[?] Not given in the book! See MacLane's \emph{Categories for the Working Mathematician}. \end{proof} \begin{remark} Recall left adjoints are right-exact and have left-derived functors. \end{remark} \begin{corollary}[?] If $$\mathcal{A}$$ is a cocomplete abelian category with enough projectives and $$\adjunction{F}{G}{ \mathcal{A} } { \mathcal{B} }$$. Then for every set-indexed collection of objects $$\left\{{ A_i }\right\}$$, \begin{align*} (L_* F)\qty{ \bigoplus_{i \in I } A_i } = \bigoplus _{i\in I} L_* F(A_i) ) ,\end{align*} so left-derived functors commute with direct sums. \end{corollary} \begin{proof}[?] Let $$P_i$$ be the projective resolution of $$A_i$$, so $$P_i \to A_i$$, then $$\bigoplus P_i \to \bigoplus A_i$$ is a projective resolution, and by definition \begin{align*} (L_* F) \qty{ \bigoplus A_i} &= H_*\qty{ F\qty{ \bigoplus P_i } } \\ &= H_* \qty{ \bigoplus FP_i } \quad \text{by the theorem} \\ &\cong \bigoplus H_*( FP_i) \text{homology commutes with $\oplus \in \mathsf{Ch}(\mathcal{A})$} \\ &= \bigoplus _i L_* F(A_i) .\end{align*} \end{proof} \begin{corollary}[?] For $$A_i \in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}, B\in {\mathsf{Mod}{\hbox{-}}\mathsf{R}}$$, \begin{align*} \operatorname{Tor}_*^R\qty{ \bigoplus_{i\in I} A_i, B } \cong \bigoplus_{i\in I} \operatorname{Tor}_*^R(A_i, B) .\end{align*} \end{corollary} \begin{proof}[of corollary] \begin{align*} \operatorname{Tor}_*^R({-}, B) = L_*F, && F \coloneqq({-}\otimes_R B) ,\end{align*} and $$F$$ is a left-adjoint by the tensor-hom adjunction. \end{proof} \begin{remark} One can also show directly from the definition that \begin{align*} \operatorname{Tor}_*^R(A, \bigoplus_{i\in I} B_i) \cong \bigoplus_{i\in I} \operatorname{Tor}_*^R(A, B_i) .\end{align*} This uses the fact that $$P \otimes_R (\bigoplus_{i\in I} B_i) \cong \bigoplus_{i\in I} (P \otimes B_i)$$. \end{remark} \begin{remark} We'll skip the rest of this section, we (hopefully) won't need filtered colimits. \end{remark} \hypertarget{balancing-tor-and-ext}{% \subsection{Balancing Tor and Ext}\label{balancing-tor-and-ext}} \begin{remark} Idea: their derived functors with either variable fixed will essentially be the same. We'll start by showing that the two left-derived functors of $${-}\otimes_R {-}$$ give the same results, and similarly for the two right-derived functors $$\mathop{\mathrm{Hom}}_R({-}, {-})$$. We'll use double complexes! \end{remark} \hypertarget{tensor-product-complexes}{% \subsubsection{Tensor Product Complexes}\label{tensor-product-complexes}} \begin{remark} Suppose we have two chain complexes $$(P)_R \in \mathsf{Ch}( {\mathsf{Mod}{\hbox{-}}\mathsf{R}}), {}_R(Q) \in \mathsf{Ch}({\mathsf{R}{\hbox{-}}\mathsf{Mod}})$$. Then there is a double complex where $$i, j$$ indexes rows and columns: $$P \otimes_R Q = \left\{{ P_i \otimes_R Q_j }\right\}_{i, j}$$, the \textbf{tensor product double complex} of $$P$$ and $$Q$$. We use the sign trick from 1.2.5: \begin{itemize} \item $$d^h \coloneqq d^P \otimes\one$$ \item $$d^v \coloneqq(-1)^i 1 \otimes d^Q$$ \end{itemize} Taking the direct sum totalization $$\operatorname{Tor}^{ \oplus }(P \otimes_R Q )$$ is the \textbf{total tensor product chain complex} of $$P$$ and $$Q$$. Note that this has a single differential! The big theorem from this section: \end{remark} \begin{theorem}[Tor is balanced] \begin{align*} L_n(A\otimes_R {-})(B) \cong L_n({-}\otimes_R B)(A) \coloneqq\operatorname{Tor}_n^R(A, B) .\end{align*} \end{theorem} \begin{remark} Note that this makes the right-hand side notation unambiguous. \end{remark} \begin{proof}[?] Choose projective resolutions $$P \xrightarrow{\varepsilon} A \in {\mathsf{Mod}{\hbox{-}}\mathsf{R}}$$ and $$Q \xrightarrow{\eta} B \in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}$$. We'll form 3 tensor product double complexes. \begin{itemize} \item $$P\otimes Q$$: A first quadrant double complex, since the projective resolutions have nonnegative indices. \item $$A\otimes Q$$, embedding $$A \hookrightarrow\mathsf{Ch}( \mathcal{A} )$$ as a complex concentrated in degree 0 (so one column) \item $$P \otimes B$$ (one row). \end{itemize} There are several maps of double complexes among these induced by $$\epsilon, \eta$$: \begin{center} \begin{tikzcd} && {} \\ && {} & \vdots && \vdots && \vdots \\ & {A \otimes Q_2} && {P_0 \otimes Q_2} && {P_1\otimes Q_2} && {P_2 \otimes Q_2} && \cdots \\ & {A \otimes Q_1} && {P_0 \otimes Q_1} && {P_1\otimes Q_1} && {P_2 \otimes Q_1} && \cdots \\ & {A \otimes Q_0} && {P_0 \otimes Q_0} && {P_1\otimes Q_0} && {P_2 \otimes Q_0} && \cdots \\ {} && {} &&&&&& {} \\ &&& {P_0 \otimes B} && {P_1 \otimes B} && {P_2 \otimes B} \\ && {} \arrow["{\epsilon \otimes 1}", from=3-4, to=3-2] \arrow["{\epsilon \otimes 1}", from=4-4, to=4-2] \arrow["{\epsilon \otimes 1}", from=5-4, to=5-2] \arrow["{1 \otimes\eta}"{description}, from=5-4, to=7-4] \arrow["{1 \otimes\eta}"{description}, from=5-6, to=7-6] \arrow["{1 \otimes\eta}"{description}, from=5-8, to=7-8] \arrow[from=2-4, to=3-4] \arrow["{d^P \otimes 1}"', from=3-6, to=3-4] \arrow[from=2-6, to=3-6] \arrow[from=2-8, to=3-8] \arrow["{d^P \otimes 1}"', from=3-8, to=3-6] \arrow["{d^P \otimes 1}", from=4-8, to=4-6] \arrow["{d^P \otimes 1}", from=5-8, to=5-6] \arrow["{d^P \otimes 1}", from=5-6, to=5-4] \arrow["{d^P \otimes 1}", from=4-6, to=4-4] \arrow["{1\otimes d^Q}", from=3-4, to=4-4] \arrow["{1\otimes d^Q}", from=4-4, to=5-4] \arrow["{1\otimes d^Q}", from=3-6, to=4-6] \arrow["{1\otimes d^Q}", from=4-6, to=5-6] \arrow["{1\otimes d^Q}", from=3-8, to=4-8] \arrow["{1\otimes d^Q}", from=4-8, to=5-8] \arrow[draw={rgb,255:red,214;green,92;blue,92}, dotted, no head, from=6-1, to=6-9] \arrow[draw={rgb,255:red,214;green,92;blue,92}, dotted, no head, from=1-3, to=8-3] \arrow[from=5-10, to=5-8] \arrow[from=4-10, to=4-8] \arrow[from=3-10, to=3-8] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=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}{Link to Diagram} \end{quote} We'll show there are two maps: \begin{align*} A \otimes Q = \operatorname{Tot}(A\otimes Q) \xleftarrow{\varepsilon\otimes\one} \operatorname{Tor}(P\otimes Q) \xrightarrow{1\otimes\eta} \operatorname{Tor}(P\otimes B) = P\otimes B ,\end{align*} using that totalizing a one-row or one-column complex is summing along diagonals where each has one term, yielding actual equality of the first and last terms respectively above. Moreover, we'll show these are \textbf{quasi-isomorphisms}, and so \begin{align*} L_*(A\otimes{-}) \xleftarrow{\varepsilon\otimes\one} H_*( \operatorname{Tor}(P\otimes Q) ) \xrightarrow{\one \otimes\eta} L_*( {-}\otimes B)(A) .\end{align*} We'll continue with the proof of this next time. \end{proof} \hypertarget{wednesday-february-24}{% \section{Wednesday, February 24}\label{wednesday-february-24}} \hypertarget{finishing-the-proof-of-balancing-tor}{% \subsection{Finishing the Proof of Balancing Tor}\label{finishing-the-proof-of-balancing-tor}} We were trying to prove that taking the left derived functors of the two slots in $$\operatorname{Tor}$$ yield the same thing. \begin{quote} See the diagram from last time! \end{quote} \begin{proof}[?] We'll need the following: \begin{claim} This induces a \emph{quasi-isomorphism} \begin{align*} P\otimes B \xleftarrow{1\otimes\eta} \operatorname{Tor}(P\otimes Q) \xrightarrow{\varepsilon\otimes\one} \operatorname{Tot}(A\otimes Q) = A\otimes Q ,\end{align*} i.e.~it is a morphism that induces an isomorphism on homology. \end{claim} Recall that by Corollary 1.5, a chain complex is a quasi-isomorphism if and only if the cone complex is acyclic/exact. In degree $$n$$ of the total complex, the $$n$$th piece is the $$n$$th diagonal and we have \begin{align*} (P_n \otimes Q_0) \oplus \cdots \oplus (P_0 \otimes Q_n) .\end{align*} where $$P_0 \xrightarrow{\varepsilon\otimes\one A \otimes Q_n}$$. Recall that for a map $$B_n \xrightarrow{f} C_n$$, the cone complex was given by \begin{center} \begin{tikzcd} {B_{n-1}} & \oplus & {C_n} \\ \\ {B_{n-2}} & \oplus & {C_{n-1}} \arrow["{d^C}", from=1-3, to=3-3] \arrow["{-d^B}", from=1-1, to=3-1] \arrow["{-f}"', from=1-1, to=3-3] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsNixbMCwwLCJCX3tuLTF9Il0sWzAsMiwiQl97bi0yfSJdLFsyLDAsIkNfbiJdLFsyLDIsIkNfe24tMX0iXSxbMSwyLCJcXG9wbHVzIl0sWzEsMCwiXFxvcGx1cyJdLFsyLDMsImReQyJdLFswLDEsIi1kXkIiXSxbMCwzLCItZiIsMl1d}{Link to Diagram} \end{quote} Writing one term out explicitly, we have \begin{center} \begin{tikzcd} {(P_{n-1}\otimes Q_0) \oplus \cdots \oplus (P_0 \otimes Q_{n-1}) \oplus Q_n} & \oplus && {A \otimes Q_n} \\ \\ {(P_{n-1}\otimes Q_0) \oplus \cdots \oplus (P_0 \otimes Q_{n-2}) \oplus Q_{n-1}} & \oplus && {A\otimes Q_{n-1}} \arrow["{-d^{\oplus}}", from=1-1, to=3-1] \arrow["{\one\otimes d}", from=1-4, to=3-4] \arrow["{-(\varepsilon\otimes\one)}"', from=1-1, to=3-4] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsNixbMCwwLCIoUF97bi0xfVxcdGVuc29yIFFfMCkgXFxvcGx1cyBcXGNkb3RzIFxcb3BsdXMgKFBfMCBcXHRlbnNvciBRX3tuLTF9KSBcXG9wbHVzIFFfbiJdLFswLDIsIihQX3tuLTF9XFx0ZW5zb3IgUV8wKSBcXG9wbHVzIFxcY2RvdHMgXFxvcGx1cyAoUF8wIFxcdGVuc29yIFFfe24tMn0pIFxcb3BsdXMgUV97bi0xfSJdLFszLDAsIkEgXFx0ZW5zb3IgUV9uIl0sWzMsMiwiQVxcdGVuc29yIFFfe24tMX0iXSxbMSwwLCJcXG9wbHVzIl0sWzEsMiwiXFxvcGx1cyJdLFswLDEsIi1kXntcXG9wbHVzfSJdLFsyLDMsIlxcb25lXFx0ZW5zb3IgZCJdLFswLDMsIi0oXFxlcHMgXFx0ZW5zb3IgXFxvbmUpIiwyXV0=}{Link to Diagram} \end{quote} Call this complex (2). \todo[inline]{Fix spacing} On the other hand, consider the double complex obtained from $$P\otimes Q$$ by adjoining the shifted complex $$(A\otimes Q)[1, 0]$$\footnote{The book may have the sign incorrect here.} in column $$i=-1$$. This has the effect of keeping the same complex but relabeling left-most column in degree 0'' into "degree $$-1$$. Note that this negatives the leftmost vertical differentials $$A\otimes Q_n \to A\otimes Q_{n-1}$$. Now call everything above the dotted line $$C$$. Consider $$\operatorname{Tot}(C)[-1]$$, which in degree $$n$$ is $$(\operatorname{Tot}(C))_{n-1}$$ and since this was an odd shift, negates all of the signs of differential. So in degree $$n$$, this explicitly looks like \begin{align*} n:\quad (P_{n-1} \otimes Q_0) \oplus \cdots \oplus (P_0 \otimes Q_{n-1}) \oplus (A\otimes Q_n)\\ \\ n:\quad (P_{n-1} \otimes Q_0) \oplus \cdots \oplus (P_0 \otimes Q_{n-1}) \oplus (A\otimes Q_n)\\ \\ .\end{align*} and we have \begin{center} \begin{tikzcd} {(P_{n-1}\otimes Q_0) \oplus \cdots } & {\oplus (P_0 \otimes Q_{n-1})} & \oplus & {A \otimes Q_n} \\ \\ {(P_{n-1}\otimes Q_0) \oplus \cdots \oplus (P_0 \otimes Q_{n-2})} && \oplus & {A\otimes Q_{n-1}} \arrow["{-d^{\oplus}}", from=1-1, to=3-1] \arrow["{\one\otimes d}", from=1-4, to=3-4] \arrow["{-(\varepsilon\otimes\one)}"', from=1-2, to=3-4] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsNyxbMCwwLCIoUF97bi0xfVxcdGVuc29yIFFfMCkgXFxvcGx1cyBcXGNkb3RzICJdLFswLDIsIihQX3tuLTF9XFx0ZW5zb3IgUV8wKSBcXG9wbHVzIFxcY2RvdHMgXFxvcGx1cyAoUF8wIFxcdGVuc29yIFFfe24tMn0pIl0sWzMsMCwiQSBcXHRlbnNvciBRX24iXSxbMywyLCJBXFx0ZW5zb3IgUV97bi0xfSJdLFsxLDAsIlxcb3BsdXMgKFBfMCBcXHRlbnNvciBRX3tuLTF9KSJdLFsyLDAsIlxcb3BsdXMiXSxbMiwyLCJcXG9wbHVzIl0sWzAsMSwiLWRee1xcb3BsdXN9Il0sWzIsMywiXFxvbmVcXHRlbnNvciBkIl0sWzQsMywiLShcXGVwcyBcXHRlbnNvciBcXG9uZSkiLDJdXQ==}{Link to Diagram} \end{quote} Calling this complex (3), we have (3) = (2), so it suffices to show (2) is exact, i.e.~$$\operatorname{Tot}(C)$$ is acyclic. This follow from the next result we'll prove, the \emph{acyclic assembly lemma}. Note that if $$Q_j$$ is projective, then it's an algebra fact that $${-}\otimes_R Q_j$$ is exact (not just right exact) since projective implies flat. This implies that the rows of $$C$$ are exact, since this is taking a project resolution (which is exact) and tensoring with a flat module. Using that $$C$$ is supported on the upper half-plane and has exact rows, by this part (3) of the acyclic assembly lemma, $$\operatorname{Tot}^{\oplus}(C)$$ will be acyclic. A similar argument will go through to show that $$\one\otimes\eta$$ is also a quasi-isomorphism by adjoining $$(P\otimes B)$$ as the $$-1$$st row and applying a version of the lemma for right half-plane complexes with exact columns. \end{proof} \hypertarget{acyclic-assembly-lemma}{% \subsection{Acyclic Assembly Lemma}\label{acyclic-assembly-lemma}} \begin{proposition}[Acyclic Assembly Lemma] Let $$C$$ be a double complex in $${\mathsf{Mod}{\hbox{-}}\mathsf{R}}$$, then \begin{itemize} \item $$\operatorname{Tot}^{\prod}(C)$$ is acyclic if either \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \tightlist \item $$C$$ is upper half-plane with exact columns, or \item $$C$$ is right half-plane with exact rows. \end{enumerate} \item $$\operatorname{Tot}^{\oplus }(C)$$ is acyclic if either \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \setcounter{enumi}{2} \tightlist \item $$C$$ is upper half-plane with exact rows\footnote{This is the part we used previously, and (4) is the one used for the other half of the argument.}, or \item $$C$$ is right half-plane with exact columns. \end{enumerate} \end{itemize} \end{proposition} \begin{remark} It suffices to prove (1). Interchanging rows and columns by reflecting along the line $$i=j$$ interchanges the types showing up in (1) and (2), and doesn't change the total complex. This similarly switches (3) and (4), so we have $$1\implies 2$$ and $$4\implies 3$$, so we'll show that $$1\implies 4$$. Let $$\tau_n C$$ be the double complex obtained taking a \emph{good truncation} of $$C$$ at level $$n$$: \begin{align*} (\tau_n C)_{ij} \coloneqq \begin{cases} C_{ij} & j > n \\ \ker( d^v: C_{i, n} \to C_{i, n-1} & j=n. \end{cases} .\end{align*} Up to translation $$\tau_n C$$ is a 1st quadrant complex, and since we're in case (4), we're assuming the columns are exact. Now using (1), $$\operatorname{Tot}^{\oplus }(\tau_n C)= \operatorname{Tot}^{\prod}(\tau_n C)$$ since we now have a first quadrant complex and all diagonals are finite, and we can conclude both are exact. This implies that $$\operatorname{Tot}^{\oplus }C$$ is acyclic since every cycle in $$\operatorname{Tot}^{\oplus }(C)$$ is nonzero in only finitely many terms. Thus each such cycle is a cycle in $$\operatorname{Tot}(\tau_n C)$$ for some $$n\ll 0$$, and hence a boundary by the previous argument. \end{remark} \begin{remark} Note that this argument does not go through for the direct product, since then there may be infinitely many nonzero terms on any diagonal, and not every cycle would be represented after some finite truncation and shift. \end{remark} \begin{proof}[of proposition] By translating $$C$$ left or right, it's enough to prove that $$H_0 \operatorname{Tot}^{\prod }C = 0$$. We can write \begin{align*} (\operatorname{Tot}^{\prod}C)_0 = \prod_{j \geq 0} C_{-j, j} \ni c\coloneqq (\cdots, c_{-j, j}, \cdots, c_{-2, 2}, c_{-1, 1}, c_{0, 0} ) ,\end{align*} letting the latter element by a 0-cycle. By inducting on $$j$$, we'll construct an element $$b$$ such that $$b_{-j, j+1} \in C_{-j, j+1} \subseteq (\operatorname{Tot}^{\prod} C)_1$$ such that \begin{align*} d^v( b_{-j, j+1}) + d^h( b_{-j+1, j}) = c_{-j, j} ,\end{align*} which will make $$c$$ a boundary. \end{proof} \hypertarget{friday-february-26}{% \section{Friday, February 26}\label{friday-february-26}} Today: trying to prove acyclic assembly lemma \begin{proof}[Of acyclic assembly lemma] We reduced to proving one case, where $$C$$ is a double complex upper half-plane with exact columns $$\implies \operatorname{Tot}^{\prod}(C)$$ is acyclic. It's enough to check in degree 0 by shifting. Fix a 0-cycle $$\mathbf{c} = (\cdots, c_{-j, j}, \cdots, c_{-2, 2}, c_{-1, 1}, c_{0, 0})$$. Find $$b \in \prod_{j\leq 0}C_{-j, j+1}$$\such that $$d(b) = c$$, so $$c_{-j, j} = d^v(b_{-j, j+1}) + d^h(b_{-j+1, j})$$. \begin{center} \begin{tikzcd} \textcolor{rgb,255:red,214;green,92;blue,92}{b_{-j, j+1}} \\ {c_{-j, j}} & {b_{-j+1, j}} \\ 0 && {c_{-j+1, j-1}} & {b_{-j+2, j-1}} \\ &&&& \ddots \\ &&&&& {c_{-2, 2}} & {b_{-1, 2}} \\ &&&&&& {c_{-1, 1}} & {b_{0, 1}} \\ &&&&&&& {c_{0, 0}} & \textcolor{rgb,255:red,92;green,214;blue,92}{b_{1,0} = 0} \\ \bullet &&&&&&&&&& \bullet \\ &&&&&&& 0 \arrow[dashed, equals, from=8-1, to=8-11] \arrow[from=7-8, to=9-8] \arrow["{d^v}", from=2-1, to=3-1] \arrow["{d^h}"', from=3-3, to=3-1] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=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}{Link to Diagram} \end{quote} Construct by induction on $$j$$: set $$b_{1, 0} = 0$$ and need $$c_{0, 0} = d^v(b_{0, 1})$$. Since $$d^vc_{0, 0} =0$$ and the columns are exact, we can lift this to some $$b_{0, 1}$$ such that $$d^v b_{0, 1} = c_{0, 0}$$. Inductively, we want $$d^v(b_{-j, j+1}) = c_{j, -j} - d^h(b_{-j+1, j})$$. Then \begin{align*} d^v( c_{j, -j} - d^h b_{-j+1, j} ) &= d^v c_{j, -j} + d^h d^v b_{-j+1, j} \\ &= d^v c_{j, -j} + d^h\qty{ c_{-j+1, j-1} - d^h b_{-j+2, j-1} } \\ &= d^v c_{j. -j} + d^h c_{-j+1, j-1} \\ &= 0 \text{ since } d^{\prod} = 0 .\end{align*} By exactness of column $$j$$, we can lift to $$b_{-j, j+1}$$, making $$c$$ a boundary. \end{proof} \begin{remark} This proves that $${-}\otimes_R{-}$$ is balanced, i.e.~taking the derived functors in either variable with the same pair $$(A, B)$$ results in the same thing. To prove a similar result for hom and ext, we want to consider $$\mathop{\mathrm{Hom}}_R(A, {-})$$ which requires injective resolutions, and $$\mathop{\mathrm{Hom}}_R({-}, B)$$ is contravariant and left-exact, so we take an injective resolution in $$\mathcal{C}^{\operatorname{op}}$$, i.e.~a projective resolution in $$\mathcal{C}$$. So take a projective resolution $$P\to A$$ and an injective resolution $$B\to I$$ and make a first quadrant double complex $$C_{i, j} \coloneqq\mathop{\mathrm{Hom}}(P_i, I^j)$$ for $$i, j\geq 0$$. Define the differentials using the following sign convention: \begin{center} \begin{tikzcd} \textcolor{rgb,255:red,92;green,92;blue,214}{(-1)^{i+j+1} d_I f(p)} & {\mathop{\mathrm{Hom}}(P_i, I^{j+1})} \\ & {\mathop{\mathrm{Hom}}(P_i, I^{j})} && {\mathop{\mathrm{Hom}}(P_{i+1}, I^{j})} \\ \textcolor{rgb,255:red,92;green,92;blue,214}{f(p)} &&& \textcolor{rgb,255:red,92;green,92;blue,214}{f(d^P p)} \arrow["{d^v}", from=2-2, to=1-2] \arrow["{d^h}"', from=2-2, to=2-4] \arrow[color={rgb,255:red,92;green,92;blue,214}, maps to, from=3-1, to=3-4] \arrow[color={rgb,255:red,92;green,92;blue,214}, maps to, from=3-1, to=1-1] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsNixbMSwwLCJcXEhvbShQX2ksIElee2orMX0pIl0sWzEsMSwiXFxIb20oUF9pLCBJXntqfSkiXSxbMywxLCJcXEhvbShQX3tpKzF9LCBJXntqfSkiXSxbMywyLCJmKGReUCBwKSIsWzI0MCw2MCw2MCwxXV0sWzAsMiwiZihwKSIsWzI0MCw2MCw2MCwxXV0sWzAsMCwiKC0xKV57aStqKzF9IGRfSSBmKHApIixbMjQwLDYwLDYwLDFdXSxbMSwwLCJkXnYiXSxbMSwyLCJkXmgiLDJdLFs0LDMsIiIsMix7ImNvbG91ciI6WzI0MCw2MCw2MF0sInN0eWxlIjp7InRhaWwiOnsibmFtZSI6Im1hcHMgdG8ifX19XSxbNCw1LCIiLDAseyJjb2xvdXIiOlsyNDAsNjAsNjBdLCJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJtYXBzIHRvIn19fV1d}{Link to Diagram} \end{quote} Now applying a dual argument as the one for tor yields a dual acyclic assembly lemma''. \end{remark} \begin{remark} We'll skip the first 3 sections of chapter 3. It's worth looking at 3.2 on tor and flatness. There's a slightly circular statement that projective implies flat in the book, since we used this to show that certain rows were exact, so refer to a good algebra book for alternative proofs. \end{remark} \hypertarget{operatornameext1-and-extensions}{% \subsection{\texorpdfstring{$$\operatorname{Ext}^1$$ and Extensions}{\textbackslash operatorname\{Ext\}\^{}1 and Extensions}}\label{operatornameext1-and-extensions}} \begin{definition}[Module Extensions] Let $$A, B\in {\mathsf{Mod}{\hbox{-}}\mathsf{R}}$$, then an \textbf{extension of $$A$$ by $$B$$} is a SES \begin{align*} \xi: 0 \to B\to X\to A\to 0 .\end{align*} \begin{figure} \centering \includegraphics{figures/image_2021-02-26-09-41-27.png} \caption{image\_2021-02-26-09-41-27} \end{figure} We say two extensions $$\xi, \xi'$$ are equivalent and write $$\xi \sim \xi'$$ iff \begin{center} \begin{tikzcd} 0 & B & X & A & 0 \\ \\ 0 & B & {X'} & A & 0 \arrow["\exists", dashed, from=1-3, to=3-3] \arrow[equals, from=1-4, to=3-4] \arrow[equals, from=1-2, to=3-2] \arrow[from=1-1, to=1-2] \arrow[from=1-2, to=1-3] \arrow[from=1-3, to=1-4] \arrow[from=1-4, to=1-5] \arrow[from=3-1, to=3-2] \arrow[from=3-3, to=3-4] \arrow[from=3-2, to=3-3] \arrow[from=3-4, to=3-5] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsMTAsWzAsMCwiMCJdLFsxLDAsIkIiXSxbMiwwLCJYIl0sWzMsMCwiQSJdLFs0LDAsIjAiXSxbMCwyLCIwIl0sWzIsMiwiWCciXSxbMSwyLCJCIl0sWzMsMiwiQSJdLFs0LDIsIjAiXSxbMiw2LCJcXGV4aXN0cyIsMCx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFszLDgsIiIsMCx7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XSxbMSw3LCIiLDAseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV0sWzAsMV0sWzEsMl0sWzIsM10sWzMsNF0sWzUsN10sWzYsOF0sWzcsNl0sWzgsOV1d}{Link to Diagram} \end{quote} An extension is \textbf{split} if and only if it is equivalent to \begin{align*} 0 \to B \overset{\iota}\hookrightarrow A \oplus B \to A \xrightarrow[]{\pi} { \mathrel{\mkern-16mu}\rightarrow }\, A \to 0 .\end{align*} \end{definition} \begin{warnings} Note that a SES as above is related to $$\operatorname{Ext}(A, B)$$, which reverses the order! \end{warnings} \begin{lemma}[?] If $$\operatorname{Ext}^1(A, B) = 0$$ then every extension of $$A$$ by $$B$$ is split. \end{lemma} \begin{warnings} There are lots of corrections needed to this proof in Weibel! \end{warnings} \begin{proof}[of lemma] Given an extension $$\xi$$, look at the LES associated to $$\mathop{\mathrm{Hom}}^*({-}, B)$$: \begin{center} \begin{tikzcd} && \cdots && {\operatorname{Ext}^1(A, B)} \\ \\ {\mathop{\mathrm{Hom}}(B, B)} && {\mathop{\mathrm{Hom}}(X, B)} && {\mathop{\mathrm{Hom}}(A, B)} \\ \textcolor{rgb,255:red,92;green,214;blue,92}{\one_B} && \textcolor{rgb,255:red,92;green,214;blue,92}{\sigma} \arrow[from=3-5, to=3-3] \arrow[from=3-3, to=3-1] \arrow[from=3-1, to=1-5, out=180, in=0] \arrow[from=1-5, to=1-3] \arrow[color={rgb,255:red,92;green,214;blue,92}, maps to, from=4-3, to=4-1] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsNyxbMiwwLCJcXGNkb3RzIl0sWzQsMCwiXFxFeHReMShBLCBCKSJdLFswLDIsIlxcSG9tKEIsIEIpIl0sWzIsMiwiXFxIb20oWCwgQikiXSxbNCwyLCJcXEhvbShBLCBCKSJdLFsyLDMsIlxcc2lnbWEiLFsxMjAsNjAsNjAsMV1dLFswLDMsIlxcb25lX0IiLFsxMjAsNjAsNjAsMV1dLFs0LDNdLFszLDJdLFsyLDFdLFsxLDBdLFs1LDYsIiIsMCx7ImNvbG91ciI6WzEyMCw2MCw2MF0sInN0eWxlIjp7InRhaWwiOnsibmFtZSI6Im1hcHMgdG8ifX19XV0=}{Link to Diagram} \end{quote} However, this gives a splitting: \begin{center} \begin{tikzcd} & \textcolor{rgb,255:red,92;green,214;blue,92}{B} \\ \\ 0 & B & X & A & 0 \\ \\ 0 & B & {X'} & A & 0 \arrow["\exists", dashed, from=3-3, to=5-3] \arrow[equals, from=3-4, to=5-4] \arrow[equals, from=3-2, to=5-2] \arrow[from=3-1, to=3-2] \arrow[from=3-2, to=3-3] \arrow[from=3-3, to=3-4] \arrow[from=3-4, to=3-5] \arrow[from=5-1, to=5-2] \arrow[from=5-3, to=5-4] \arrow[from=5-2, to=5-3] \arrow[from=5-4, to=5-5] \arrow["{\one_B}", color={rgb,255:red,92;green,214;blue,92}, from=3-2, to=1-2] \arrow["{\exists \sigma}", color={rgb,255:red,92;green,214;blue,92}, dashed, from=3-3, to=1-2] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=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}{Link to Diagram} \end{quote} Todo: label $$(X, B) \to (B, B)$$ as $$f_*$$. This is one of the many equivalent criteria for a SES of modules to be split. \end{proof} \begin{remark} More generally, given $$\xi$$, let $$\Theta(\xi) \coloneqq{{\partial}}(\one_B) \in \operatorname{Ext}^1(A, B)$$. Thus TFAE: \begin{itemize} \tightlist \item $$\xi$$ is split \item $$\one-B$$ lifts to some $$\sigma\in \mathop{\mathrm{Hom}}(X, B)$$ \item $$\one_B \in \operatorname{im}f_* = \ker {{\partial}}$$ \item $$\Theta(\xi) = 0$$, even if $$\operatorname{Ext}^1(A, B) \neq 0$$. \end{itemize} Then $$\Theta(\xi)$$ is an \emph{obstruction} to $$\xi$$ being split. \end{remark} \begin{remark} If $$\xi'\sim \xi$$ then $${{\partial}}'(\one_B) = {{\partial}}(\one_B)\in \operatorname{Ext}^1(A, B)$$ by naturality of the connecting morphisms. So equivalent extensions have the same obstruction, i.e.~$$\Theta$$ only depends only on the equivalence class $$[\xi]$$ of the SES. \end{remark} \begin{theorem}[Module extensions correspond to Ext groups] Given $$A, B\in {\mathsf{Mod}{\hbox{-}}\mathsf{R}}$$ (or an abelian category with enough projectives and injectives), there is a correspondence \begin{align*} \left\{{\substack{ 0 \to B \to X\to A \to 0 }}\right\}_{/\sim} \mathrel{\operatorname*{\rightleftharpoons}_{\Theta}^{\Psi}} \operatorname{Ext}^1(A, B) \end{align*} Note that this is a bijection of sets, but we'll upgrade it to a bijection of abelian groups. \end{theorem} \hypertarget{monday-march-01}{% \section{Monday, March 01}\label{monday-march-01}} \begin{remark} Last time: we looked at group extensions. Given $$\xi: 0\to B\to X \to A\to 0$$, we had a canonical element in $$\operatorname{Ext}^1(A, B)$$, namely $$\Theta(\xi) = \delta(\one_B)$$. This only depends on the equivalence class of $$\xi$$. \end{remark} \begin{theorem}[Module extensions biject with Ext groups] Given $$A, B\in {\mathsf{Mod}{\hbox{-}}\mathsf{R}}$$, there is a bijection \begin{align*} \left\{{\substack{ \text{Extensions ofA$by$B} }}\right\} \mathrel{\operatorname*{\rightleftharpoons}_{\Theta}^{\Phi}} \operatorname{Ext}_R^1(A, B) \end{align*} \end{theorem} \begin{proof}[?] \envlist \begin{claim} $$\Theta$$ is surjective. \end{claim} Fix a SES \begin{align*} 0 \to M \xrightarrow{j} P \xrightarrow{\pi} A \to 0 \end{align*} with $$P$$ projective, and take the LES resulting from applying $$\mathop{\mathrm{Hom}}({-}, B)$$: \begin{center} \begin{tikzcd} 0 \\ {\mathop{\mathrm{Hom}}(A, B)} & {\mathop{\mathrm{Hom}}(P, B)} & {\mathop{\mathrm{Hom}}(M, B)} \\ \\ {\operatorname{Ext}^1(A, B)} & {\operatorname{Ext}^1(P, B) = 0} \\ x \arrow[from=2-1, to=2-2] \arrow[from=2-2, to=2-3] \arrow["{{\partial}}", from=2-3, to=4-1, out=0, in=180] \arrow[from=4-1, to=4-2] \arrow[from=1-1, to=2-1] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsNyxbMCwxLCJcXEhvbShBLCBCKSJdLFsxLDEsIlxcSG9tKFAsIEIpIl0sWzIsMSwiXFxIb20oTSwgQikiXSxbMCwzLCJcXEV4dF4xKEEsIEIpIl0sWzEsMywiXFxFeHReMShQLCBCKSA9IDAiXSxbMCw0LCJ4Il0sWzAsMCwiMCJdLFswLDFdLFsxLDJdLFsyLDMsIlxcYmQiXSxbMyw0XSxbNiwwXV0=}{Link to Diagram} \end{quote} Letting $$x \in \operatorname{Ext}^1(A, B)$$ and choose $$\beta\in \mathop{\mathrm{Hom}}(M, B)$$ with $${{\partial}}\beta = x$$ using that $$P$$ is projective and thus $$\operatorname{Ext}^1(P, B)$$ vanishes. Now let $$X$$ be the \textbf{pushout} of $$j: M\to P$$ and $$\beta: M\to B$$. Note that we can apply the universal property of cokernels to get a map of the following form: \begin{center} \begin{tikzcd} M && {P\oplus B} && {X = \operatorname{coker}g} && 0 \\ \\ && A \arrow["{g = (j, -\beta)}", from=1-1, to=1-3] \arrow[from=1-3, to=1-5] \arrow["{\pi \oplus 0}", from=1-3, to=3-3] \arrow[from=1-5, to=1-7] \arrow["{\therefore 0}"', dotted, from=1-1, to=3-3] \arrow["{\exists ! \mu}", dashed, from=1-5, to=3-3] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsNSxbMCwwLCJNIl0sWzIsMCwiUFxcb3BsdXMgQiJdLFs0LDAsIlggPSBcXGNva2VyIGciXSxbMiwyLCJBIl0sWzYsMCwiMCJdLFswLDEsImcgPSAoaiwgLVxcYmV0YSkiXSxbMSwyXSxbMSwzLCJcXHBpIFxcb3BsdXMgMCJdLFsyLDRdLFswLDMsIlxcdGhlcmVmb3JlIDAiLDIseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkb3R0ZWQifX19XSxbMiwzLCJcXGV4aXN0cyAhIFxcbXUiLDAseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XV0=}{Link to Diagram} \end{quote} Taking the pushout yields a diagram: \begin{center} \begin{tikzcd} 0 && M && P && A && 0 \\ \\ 0 && B && X && A && 0 \arrow[from=1-1, to=1-3] \arrow["j"', from=1-3, to=1-5] \arrow["\pi"', from=1-5, to=1-7] \arrow[from=3-1, to=3-3] \arrow["\iota"', from=3-3, to=3-5] \arrow["\mu"', from=3-5, to=3-7] \arrow[from=1-1, to=3-1] \arrow["\beta"', from=1-3, to=3-3] \arrow["\sigma"{description}, from=1-5, to=3-5] \arrow[equals, from=1-7, to=3-7] \arrow[from=1-7, to=1-9] \arrow[from=3-7, to=3-9] \arrow[from=1-9, to=3-9] \arrow["\lrcorner"{anchor=center, pos=0.125, rotate=180}, draw=none, from=3-5, to=1-3] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsMTAsWzAsMCwiMCJdLFswLDIsIjAiXSxbMiwwLCJNIl0sWzIsMiwiQiJdLFs0LDIsIlgiXSxbNCwwLCJQIl0sWzYsMCwiQSJdLFs2LDIsIkEiXSxbOCwwLCIwIl0sWzgsMiwiMCJdLFswLDJdLFsyLDUsImoiLDJdLFs1LDYsIlxccGkiLDJdLFsxLDNdLFszLDQsIlxcaW90YSIsMl0sWzQsNywiXFxtdSIsMl0sWzAsMV0sWzIsMywiXFxiZXRhIiwyXSxbNSw0LCJcXHNpZ21hIiwxXSxbNiw3LCIiLDEseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV0sWzYsOF0sWzcsOV0sWzgsOV0sWzQsMiwiIiwyLHsic3R5bGUiOnsibmFtZSI6ImNvcm5lciJ9fV1d}{Link to Diagram} \end{quote} \begin{exercise}[?] Check that this diagram commutes and that the new row is exact. \end{exercise} Taking the LES for $$\mathop{\mathrm{Hom}}({-}, B)$$ yields \begin{center} \begin{tikzcd} &&&& \textcolor{rgb,255:red,92;green,92;blue,214}{\beta} && \textcolor{rgb,255:red,92;green,92;blue,214}{x} \\ \cdots && {\mathop{\mathrm{Hom}}(P, B)} && {\mathop{\mathrm{Hom}}(P, B)} && {\operatorname{Ext}^1(A, B)} \\ \\ \cdots && {\mathop{\mathrm{Hom}}(X, B)} && {\mathop{\mathrm{Hom}}(B, B)} && {\operatorname{Ext}^1(A, B)} \\ &&&& \textcolor{rgb,255:red,92;green,92;blue,214}{\one_B} && \textcolor{rgb,255:red,92;green,92;blue,214}{\Theta(\xi)} \arrow[from=2-1, to=2-3] \arrow[from=2-3, to=2-5] \arrow[from=4-1, to=4-3] \arrow[from=4-3, to=4-5] \arrow[from=2-5, to=2-7] \arrow[from=4-5, to=4-7] \arrow[no head, from=2-7, to=4-7] \arrow["{\beta_*}"{description}, from=4-5, to=2-5] \arrow["{\sigma_*}"', from=4-3, to=2-3] \arrow["{{\partial}}", color={rgb,255:red,92;green,92;blue,214}, maps to, from=5-5, to=5-7] \arrow[color={rgb,255:red,92;green,92;blue,214}, curve={height=-30pt}, maps to, from=1-5, to=5-5] \arrow[color={rgb,255:red,92;green,92;blue,214}, maps to, from=1-5, to=1-7] \arrow[color={rgb,255:red,92;green,92;blue,214}, curve={height=-30pt}, dashed, maps to, from=1-7, to=5-7] \end{tikzcd} \end{center} \begin{quote} $$(*)$$ \href{https://q.uiver.app/?q=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}{Link to Diagram} \end{quote} So we \begin{itemize} \tightlist \item Started with $$x$$ \item Took a reference SES \item Produce the cokernel \item Took a pushout and found $$\beta$$. \item Showed that $$\beta\mapsto x$$. \end{itemize} \todo[inline]{Review video: 9:28 AM!} This shows surjectivity, but depended on choice of $$\beta$$. \begin{claim} $$\Theta$$ is injective. \end{claim} Note that the previous construction there is a way to associate to $$x\in \operatorname{Ext}^1(A, B)$$ an extension of $$A$$ by $$B$$. To see that this gives a well-defined map $$\Psi$$, so $$\Psi(x) = [ \xi ]$$ as well, suppose $$\beta'\in \mathop{\mathrm{Hom}}(M, B)$$ is another lift of $$x$$. Note that although $$\operatorname{Ext}^1(P, B) =0$$, the fact that $$\ker {{\partial}}= \mathop{\mathrm{Hom}}(M, B) \neq 0$$, there are many such choices of lifts. Using exactness of diagram $$(*)$$, there exists an $$f\in \mathop{\mathrm{Hom}}(P, B)$$ such that $$\beta' = \beta + fj$$, recalling that $$j: M\to P$$. Now taking the pushout $$X'$$ of $$j$$ and $$\beta'$$, the maps $$i: B\to X$$ and $$\sigma + if: P\to X$$ induce an isomorphism $$X' \xrightarrow{\sim} X$$ and thus an equivalence $$\xi \xrightarrow{\sim} \xi'$$. \begin{exercise}[?] Check this isomorphism. \end{exercise} Moreover, given any extension $$\xi$$, we can fit it into a diagram of the following form: \begin{center} \begin{tikzcd} && 0 & M & P & A & 0 \\ \\ {\xi:} && 0 & B & X & A & 0 \arrow[equals, from=1-6, to=3-6] \arrow[from=1-3, to=1-4] \arrow[from=1-4, to=1-5] \arrow[from=1-5, to=1-6] \arrow[from=1-6, to=1-7] \arrow[from=3-3, to=3-4] \arrow[from=3-4, to=3-5] \arrow[from=3-5, to=3-6] \arrow[from=3-6, to=3-7] \arrow["{\exists \beta}"{description}, dashed, from=1-4, to=3-4] \arrow["{\exists \sigma}"{description}, dashed, from=1-5, to=3-5] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsMTEsWzIsMCwiMCJdLFszLDAsIk0iXSxbNCwwLCJQIl0sWzUsMCwiQSJdLFs2LDAsIjAiXSxbMiwyLCIwIl0sWzYsMiwiMCJdLFszLDIsIkIiXSxbNCwyLCJYIl0sWzUsMiwiQSJdLFswLDIsIlxceGk6Il0sWzMsOSwiIiwwLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dLFswLDFdLFsxLDJdLFsyLDNdLFszLDRdLFs1LDddLFs3LDhdLFs4LDldLFs5LDZdLFsxLDcsIlxcZXhpc3RzIFxcYmV0YSIsMSx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFsyLDgsIlxcZXhpc3RzIFxcc2lnbWEiLDEseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XV0=}{Link to Diagram} \end{quote} First we use projectivity of $$P$$ to get $$\sigma: P\to X$$. Then restricting $$\sigma$$ to the kernels of $$\pi, \mu$$ respectively makes $$\beta: M\to B$$, so this diagram commutes \begin{exercise}[?] Check that $$X$$ is the pushout of $$j$$ and $$\beta$$. \end{exercise} It follows that $$\Psi (\Theta(\xi)) = \xi$$ and thus $$\Theta$$ is injective, making it a bijection. \end{proof} \begin{remark} Note the importance of the reversed directions after taking the Hom! \end{remark} \begin{remark} How can we upgrade this to a group homomorphism? One way is to pull back the group structure from the right-hand side to the left-hand side, but it turns out that Baer worked out an intrinsic group structure around 1934. We can construct the smallest'' extension such that $$A$$ is a quotient and $$B$$ is a submodule. \end{remark} \begin{definition}[Baer Sum (1934)] Suppose we have two extensions of $$A$$ by $$B$$: \begin{align*} \xi: & 0\to B \xrightarrow{i} X \xrightarrow{\pi} A \to 0 \\ \xi': & 0\to B \xrightarrow{i'} X' \xrightarrow{\pi'} A \to 0 \\ .\end{align*} Let $$X''$$ be the \textbf{pullback} of $$\pi, \pi'$$, defined by \begin{align*} X'' \coloneqq\left\{{ (x, x') \in X \times X' {~\mathrel{\Big|}~}\pi(x) = \pi'(x') \in A }\right\} ,\end{align*} which identifies the two copies of $$A$$. This fits into a cartesian square \begin{center} \begin{tikzcd} {X''} && {X'} \\ \\ X && A \arrow["{\pi_1}"', from=1-1, to=3-1] \arrow["{\pi_2}", from=1-1, to=1-3] \arrow["{\pi'}"', from=1-3, to=3-3] \arrow["\pi", from=3-1, to=3-3] \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=3-3] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJYJyciXSxbMiwwLCJYJyJdLFsyLDIsIkEiXSxbMCwyLCJYIl0sWzAsMywiXFxwaV8xIiwyXSxbMCwxLCJcXHBpXzIiXSxbMSwyLCJcXHBpJyIsMl0sWzMsMiwiXFxwaSJdLFswLDIsIiIsMSx7InN0eWxlIjp7Im5hbWUiOiJjb3JuZXIifX1dXQ==}{Link to Diagram} \end{quote} Note that $$X''$$ contains 3 copies of $$B$$: \begin{itemize} \tightlist \item $$B \times 0$$, or really $$i(B) \times\left\{{ 0 }\right\} \subset X''$$ (using exactness). \item $$0\times B$$, i.e.~$$\left\{{ 0 }\right\} \times i'(B) \subseteq X''$$ (using exactness). \item $$\tilde\Delta= \left\{{ (-b, b) {~\mathrel{\Big|}~}b\in B }\right\}$$, the \textbf{skew diagonal}. One can check that $$\pi i (-b) = 0 = \pi' i' (b)$$. \end{itemize} Note that we're identifying $$B$$ with $$i(B), i'(B)$$. Set $$Y \coloneqq X'' / \tilde\Delta$$, then $$(b, 0) + (-b, b) = (0, b)$$ where $$(-b, b) \in \tilde \Delta$$, so $$B \times 0$$ and $$0 \times B$$ have the same image in $$Y$$, since \begin{align*} (B \times 0) \cap\tilde\Delta= \left\{{ (0, 0) }\right\} = (0 \times B) \cap\tilde\Delta .\end{align*} In fact this image in $$Y$$ is isomorphic to $$B$$, by construction of what we're quotienting out by. Denoting this subgroup of $$Y$$ by $$B$$, we get a SES \begin{align*} \phi: 0\to B \to Y \to Y/B \to 0 .\end{align*} What is $$Y/B$$? We can write this as \begin{align*} Y/B = { X'' / \tilde \Delta\over (0 \times B ) / \tilde\Delta} \cong {X'' \over (0 \times B) + \tilde\Delta} \cong {X'' / 0 \times B \over (\tilde\Delta+ (0 \times B) ) / (0 \times B)} .\end{align*} But the numerator is isomorphic to $$X$$ by $$\pi_1$$, and the denominator is isomorphic to $$B$$ by $$\pi_1$$. So $$\phi$$ is an extension of $$A$$ by $$B$$ called the \textbf{Baer sum} of $$\xi, \xi'$$. \end{definition} \begin{corollary}[?] The equivalence classes of extensions of $$A$$ by $$B$$ is an abelian group under Baer sums, where zero is the class of split extensions. Moreover, the map $$\Theta$$ from the previous theorem is an isomorphism of abelian groups. \end{corollary} \begin{remark} Next time we'll check this by showing $$\Theta(\phi) = \Theta(\xi) + \Theta(\xi')$$. \end{remark} \hypertarget{wednesday-march-03}{% \section{Wednesday, March 03}\label{wednesday-march-03}} \hypertarget{baer-sum-and-higher-exts}{% \subsection{Baer Sum and Higher Exts}\label{baer-sum-and-higher-exts}} Last time: Baer sum. \begin{remark} \begin{center} \begin{tikzcd} {\xi':} && 0 && B && {X'} && A && 0 \\ \\ {\text{Ref}:} && 0 && M && P && A && 0 \arrow[from=1-3, to=1-5] \arrow["{\iota'}", from=1-5, to=1-7] \arrow["{\sigma'}"', from=3-7, to=1-7] \arrow["\pi", from=3-7, to=3-9] \arrow["{\pi'}", from=1-7, to=1-9] \arrow[from=1-9, to=1-11] \arrow[from=3-9, to=3-11] \arrow["{\beta'}", from=3-5, to=1-5] \arrow["j", from=3-5, to=3-7] \arrow[from=3-3, to=3-5] \arrow[equals, from=3-9, to=1-9] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsMTIsWzQsMCwiQiJdLFs2LDAsIlgnIl0sWzgsMCwiQSJdLFsyLDAsIjAiXSxbNCwyLCJNIl0sWzYsMiwiUCJdLFs4LDIsIkEiXSxbMTAsMCwiMCJdLFsxMCwyLCIwIl0sWzIsMiwiMCJdLFswLDAsIlxceGknOiJdLFswLDIsIlxcdGV4dHtSZWZ9OiJdLFszLDBdLFswLDEsIlxcaW90YSciXSxbNSwxLCJcXHNpZ21hJyIsMl0sWzUsNiwiXFxwaSJdLFsxLDIsIlxccGknIl0sWzIsN10sWzYsOF0sWzQsMCwiXFxiZXRhJyJdLFs0LDUsImoiXSxbOSw0XSxbNiwyLCIiLDEseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV1d}{Link to Diagram} \end{quote} \begin{center} \begin{tikzcd} &&&& \textcolor{rgb,255:red,92;green,92;blue,214}{\one_B} && \textcolor{rgb,255:red,92;green,92;blue,214}{\Theta(\xi')} \\ \cdots && {\mathop{\mathrm{Hom}}(X', B)} && {\mathop{\mathrm{Hom}}(B, B)} && {\operatorname{Ext}^1_R(A, B)} \\ \\ \cdots && {\mathop{\mathrm{Hom}}(P, B)} && {\mathop{\mathrm{Hom}}(M, B)} && {\operatorname{Ext}^1_R(A, B)} \\ &&&& \textcolor{rgb,255:red,92;green,92;blue,214}{\beta'} && \textcolor{rgb,255:red,92;green,92;blue,214}{{{\partial}}(\beta') = \Theta(\xi')} \arrow["{\sigma'_*}", from=2-3, to=4-3] \arrow["{\beta'_*}", from=2-5, to=4-5] \arrow[from=4-1, to=4-3] \arrow[from=2-1, to=2-3] \arrow[from=2-3, to=2-5] \arrow[from=2-5, to=2-7] \arrow[from=4-3, to=4-5] \arrow[from=4-5, to=4-7] \arrow[equals, from=2-7, to=4-7] \arrow[color={rgb,255:red,92;green,92;blue,214}, maps to, from=5-5, to=5-7] \arrow[color={rgb,255:red,92;green,92;blue,214}, maps to, from=1-5, to=1-7] \arrow[color={rgb,255:red,92;green,92;blue,214}, curve={height=-30pt}, maps to, from=1-5, to=5-5] \arrow[color={rgb,255:red,92;green,92;blue,214}, curve={height=-30pt}, maps to, from=1-7, to=5-7] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=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}{Link to Diagram} \end{quote} We want to define $$\xi' \oplus \xi''$$, An important takeaway is that $$\Theta$$ can alternatively be defined as a map induced by the original boundary map coming from the SES, i.e.~$${{\partial}}(\beta') = \Theta(\xi')$$. This fits into the diagram as follows: \begin{center} \begin{tikzcd} {\xi':} && 0 && B && {X'} && A && 0 \\ \\ {\text{Ref}:} && 0 && M && P && A && 0 \\ \\ {\xi'':} && 0 && B && {X''} && A && 0 \arrow[from=1-3, to=1-5] \arrow["{\iota'}", from=1-5, to=1-7] \arrow["{\sigma'}", from=3-7, to=1-7] \arrow["\pi", from=3-7, to=3-9] \arrow["{\pi'}", from=1-7, to=1-9] \arrow[from=1-9, to=1-11] \arrow[from=3-9, to=3-11] \arrow["{\beta'}", from=3-5, to=1-5] \arrow["j", from=3-5, to=3-7] \arrow[from=3-3, to=3-5] \arrow[equals, from=3-9, to=1-9] \arrow[from=5-3, to=5-5] \arrow["{\iota''}", from=5-5, to=5-7] \arrow["{\pi''}", from=5-7, to=5-9] \arrow[from=5-9, to=5-11] \arrow["{\beta''}"', from=3-5, to=5-5] \arrow["{\sigma''}"', from=3-7, to=5-7] \arrow[equals, from=3-9, to=5-9] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=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}{Link to Diagram} \end{quote} We define \begin{align*} \tilde X \coloneqq\left\{{ (x', x'') \in X' \times X'' {~\mathrel{\Big|}~}\pi'(x') = \pi''(x'') }\right\} \twoheadrightarrow Y ,\end{align*} and note that we had a skew diagonal $$\tilde\Delta\subseteq \tilde X$$. This yields a YES \begin{align*} \phi: 0 \to B \to Y \to Y/B \cong A \to 0 .\end{align*} \end{remark} \begin{corollary}[?] The set of equivalence classes of extensions of $$A$$ by $$B$$ is an abelian group under the Baer sum, where \begin{align*} [\xi] \oplus [\xi'] \coloneqq[\varphi] ,\end{align*} where the identity element $$0$$ is the class of split extensions. The map $$\Theta$$ is an isomorphism of abelian groups. \end{corollary} \begin{remark} One should check that this is well-defined since we're using equivalence classes. There is a fast way to do both at once, i.e.~showing $$\Theta$$ is well-defined and also a group morphism. \end{remark} \begin{proof}[?] We'll show that \begin{align*} \Theta(\varphi) = \Theta(\xi) + \Theta(\xi'') \in \operatorname{Ext}^1_R(A, B) ,\end{align*} which will make it a group isomorphism since $$\Theta$$ was already a set bijection. Considering commutativity in the 3-row diagram, we can get a well-defined map \begin{align*} \sigma\coloneqq\sigma' \oplus \sigma'': P \to \tilde{X} .\end{align*} So let $$\mkern 1.5mu\overline{\mkern-1.5mu \sigma\mkern-1.5mu}\mkern 1.5mu: P\to Y$$ be the induced map. The restriction of $$\mkern 1.5mu\overline{\mkern-1.5mu \sigma\mkern-1.5mu}\mkern 1.5mu$$ to $$M$$ is induced by the map \begin{align*} \beta' + \beta'': M\to (B \times 0) + (0 \times B) \subseteq \tilde X .\end{align*} These both map to $$B$$ in $$Y$$ under the SES $$0\to B\to Y\to Y/B\to 0$$. This gives a commutative diagram \begin{center} \begin{tikzcd} 0 && M && P && A && 0 \\ \\ 0 && B && Y && A && 0 \arrow[equals, from=1-7, to=3-7] \arrow[from=1-1, to=1-3] \arrow[from=1-3, to=1-5] \arrow[from=1-5, to=1-7] \arrow[from=3-1, to=3-3] \arrow[from=3-3, to=3-5] \arrow[from=3-5, to=3-7] \arrow[from=3-7, to=3-9] \arrow[from=1-7, to=1-9] \arrow["{\beta'+\beta''}"', from=1-3, to=3-3] \arrow["{\mkern 1.5mu\overline{\mkern-1.5mu\sigma\mkern-1.5mu}\mkern 1.5mu}"', from=1-5, to=3-5] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsMTAsWzAsMCwiMCJdLFsyLDAsIk0iXSxbNCwwLCJQIl0sWzYsMCwiQSJdLFswLDIsIjAiXSxbMiwyLCJCIl0sWzQsMiwiWSJdLFs2LDIsIkEiXSxbOCwyLCIwIl0sWzgsMCwiMCJdLFszLDcsIiIsMCx7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XSxbMCwxXSxbMSwyXSxbMiwzXSxbNCw1XSxbNSw2XSxbNiw3XSxbNyw4XSxbMyw5XSxbMSw1LCJcXGJldGEnK1xcYmV0YScnIiwyXSxbMiw2LCJcXGJhcntcXHNpZ21hfSIsMl1d}{Link to Diagram} \end{quote} We then have $$\Theta(\varphi) = {{\partial}}( \beta' + \beta'') = {{\partial}}(\beta') + {{\partial}}(\beta'')$$ using that $${{\partial}}\in \operatorname{Mor}({\mathsf{R}{\hbox{-}}\mathsf{Mod}})$$. But this is equal to $$\Theta(\xi') + \Theta(\xi'')$$, which is what we wanted to show. \end{proof} \begin{remark} What about the 0 element for split SESs? Recall that additive functors preserve split exact sequences, since these are just in terms of sums of maps composing to the identity. Then applying the hom functor to the original SES produces another SES, which in particular has no Ext correction term. \end{remark} \begin{remark} Similarly, $$\operatorname{Ext}^n(A, B)$$ is identified with equivalence classes of longer sequences with $$n+2$$ terms, and an equivalence is a sequence of maps that result in commuting squares: \begin{center} \begin{tikzcd} {\xi:} && 0 & B & {X_n} & \cdots & {X_1} & A & 0 \\ \\ {\xi':} && 0 & B & {X_n'} & \cdots & {X_1'} & A & 0 \arrow[from=1-3, to=1-4] \arrow[from=1-4, to=1-5] \arrow[from=1-5, to=1-6] \arrow[from=1-6, to=1-7] \arrow[from=1-7, to=1-8] \arrow[from=1-8, to=1-9] \arrow[equals, from=1-4, to=3-4] \arrow[equals, from=1-8, to=3-8] \arrow[dashed, from=1-5, to=3-5] \arrow[dashed, from=1-7, to=3-7] \arrow[dashed, from=1-6, to=3-6] \arrow[from=3-3, to=3-4] \arrow[from=3-4, to=3-5] \arrow[from=3-5, to=3-6] \arrow[from=3-6, to=3-7] \arrow[from=3-7, to=3-8] \arrow[from=3-8, to=3-9] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=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}{Link to Diagram} \end{quote} Note that if $${P}_{*} \to A\to 0$$ is a projective resolution, then the comparison theorem yields maps and a commutative diagram \begin{center} \begin{tikzcd} {\phi:} && 0 & M & {P_{n-1}} & \cdots & {P_0} & A & 0 \\ \\ {\xi':} && 0 & B & {X_n'} & \cdots & {X_1'} & A & 0 \arrow[from=1-3, to=1-4] \arrow[from=1-4, to=1-5] \arrow[from=1-5, to=1-6] \arrow[from=1-6, to=1-7] \arrow[from=1-7, to=1-8] \arrow[from=1-8, to=1-9] \arrow["{\exists \beta}", dashed, from=1-4, to=3-4] \arrow["{\one_A}", equals, from=1-8, to=3-8] \arrow["\exists", dashed, from=1-5, to=3-5] \arrow["\exists", dashed, from=1-7, to=3-7] \arrow[dashed, from=1-6, to=3-6] \arrow[from=3-3, to=3-4] \arrow[from=3-4, to=3-5] \arrow[from=3-5, to=3-6] \arrow[from=3-6, to=3-7] \arrow[from=3-7, to=3-8] \arrow[from=3-8, to=3-9] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=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}{Link to Diagram} \end{quote} Then the dimension shifting theorem (Exc. 2.4.3) and its proof yields an exact sequence \begin{align*} \mathop{\mathrm{Hom}}(P_{n-1}, B) \to \mathop{\mathrm{Hom}}(M, B) \xrightarrow{{{\partial}}} \operatorname{Ext}^n(A, B) \to 0 ,\end{align*} and the asserted bijection is then given by $$\Theta(\xi) \coloneqq{{\partial}}(\beta)$$. \end{remark} \hypertarget{kunneth-and-universal-coefficient-theorems}{% \subsection{3.6: Kunneth and Universal Coefficient Theorems}\label{kunneth-and-universal-coefficient-theorems}} \begin{observation} If $$R$$ is a field $$F$$ then $$\operatorname{Tor}_n^F(A, B) = 0$$ for all $$n>0$$, i.e.~every module over a field is a complex space, hence free, hence projective, hence flat, and so $$A\otimes_F {-}$$ is exact. \end{observation} \begin{question} If $${P}_{*} \in \mathsf{Ch}({\mathsf{Mod}{\hbox{-}}\mathsf{R}})$$ is a complex of of right $$R{\hbox{-}}$$modules and $$M \in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}$$ is a left $$R{\hbox{-}}$$module, how is the homology of $${P}_{*}$$ and that of $${P}_{*} \otimes_R M$$ related? \end{question} \begin{lemma}[?] Given a 5-term exact sequence \begin{align*} A_1 \xrightarrow{\alpha} A_2 \xrightarrow{f} B \xrightarrow{g} C_1 \xrightarrow{\gamma} C_2 ,\end{align*} there is a corresponding SES \begin{center} \begin{tikzcd} 0 & A & B & C & 0 \\ & {\substack{ A_2/\ker f = A_2/\operatorname{im}\alpha \\ = \operatorname{coker}\alpha} } && {\operatorname{im}g = \ker f} \arrow[from=1-1, to=1-2] \arrow["{\mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu}", from=1-2, to=1-3] \arrow["g", from=1-3, to=1-4] \arrow[from=1-4, to=1-5] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsNyxbMCwwLCIwIl0sWzEsMCwiQSJdLFsyLDAsIkIiXSxbMywwLCJDIl0sWzQsMCwiMCJdLFszLDEsIlxcaW0gZyA9IFxca2VyIGYiXSxbMSwxLCJBXzIvXFxrZXIgZiA9IEFfMi9cXGltXFxhbHBoYSA9IFxcY29rZXIgXFxhbHBoYSJdLFswLDFdLFsxLDIsIlxcYmFyIGYiXSxbMiwzLCJnIl0sWzMsNF1d}{Link to Diagram} \end{quote} In particular, we can always take $$A = \operatorname{coker}\alpha$$ and $$C = \ker \gamma$$ in any abelian category. \end{lemma} \begin{theorem}[The Kunneth Formula] Let $${P}_{*}\in \mathsf{Ch}({\mathsf{Mod}{\hbox{-}}\mathsf{R}})$$ be a chain complex of flat right $$R{\hbox{-}}$$modules such that each boundary module $$dP_n$$ is again flat. Then for every $$M\in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}$$ and all $$N$$, there is an exact sequence \begin{center} \begin{tikzcd} 0 && {H_n({P}_{*})\otimes_R M} && {H_n({P}_{*} \otimes_R M)} && {\operatorname{Tor}^1_R(H_{n-1}({P}_{*}), M)} && 0 \arrow[from=1-7, to=1-9] \arrow[from=1-5, to=1-7] \arrow[from=1-3, to=1-5] \arrow[from=1-1, to=1-3] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsNSxbMCwwLCIwIl0sWzIsMCwiSF9uKFxcdmVjdG9yIFApXFx0ZW5zb3JfUiBNIl0sWzQsMCwiSF9uKFxcdmVjdG9yIFAgXFx0ZW5zb3JfUiBNKSJdLFs2LDAsIlxcVG9yXjFfUihIX3tuLTF9KFxcdmVjdG9yIFApLCBNKSJdLFs4LDAsIjAiXSxbMyw0XSxbMiwzXSxbMSwyXSxbMCwxXV0=}{Link to Diagram} \end{quote} \end{theorem} \begin{remark} Note that the correction term vanishes if $$R$$ is a field. \end{remark} \begin{proof}[?] Let $$Z_n \coloneqq Z_n({P}_{*})$$, there there is a SES \begin{align*} 0 \to Z_n \to P_n \xrightarrow{d} dP_n \to 0 .\end{align*} Since $$P_n, dP_n$$ are flat by assumption, by Exc. 3.2.2, $$Z_n$$ is also flat. Taking the LES from applying $${-}\otimes_R M$$, noting that $$M$$ is arbitrary yields \begin{center} \begin{tikzcd} &&&& 0 \\ {Z_n\otimes_R M} && {P_n\otimes_R M} && {dP_n\otimes_R M} \\ \\ && \cdots && {\operatorname{Tor}_1(dP_n, M)} \arrow[from=4-5, to=2-1, out=0, in=180] \arrow[from=2-1, to=2-3] \arrow[from=2-3, to=2-5] \arrow[from=4-3, to=4-5] \arrow[from=2-5, to=1-5] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsNixbNCwxLCJkUF9uXFx0ZW5zb3JfUiBNIl0sWzIsMSwiUF9uXFx0ZW5zb3JfUiBNIl0sWzAsMSwiWl9uXFx0ZW5zb3JfUiBNIl0sWzQsMywiXFxUb3JfMShkUF9uLCBNKSJdLFsyLDMsIlxcY2RvdHMiXSxbNCwwLCIwIl0sWzMsMl0sWzIsMV0sWzEsMF0sWzQsM10sWzAsNV1d}{Link to Diagram} \end{quote} Here $$\operatorname{Tor}_1(dP_n, M)=0$$ since $$dP_n$$ is flat, noting that one could also apply $$\operatorname{Tor}(dP_n, {-})$$ to get a similar LES. So this lifts to a SES of complexes \begin{align*} 0 \to {Z}_{*}\otimes M \to {P}_{*}\otimes M \to {dP}_{*}\otimes M \to 0 ,\end{align*} where we can consider $$d\otimes\one$$ in the middle. We'll pick this up next time! \end{proof} \hypertarget{friday-march-05}{% \section{Friday, March 05}\label{friday-march-05}} \begin{quote} See first 10m \end{quote} \begin{observation} For a SES \begin{align*} A_1 \xrightarrow{\alpha} A_2 \xrightarrow{f} B \xrightarrow{g} C_1 \xrightarrow{\gamma} C_2 ,\end{align*} one can obtain an exact sequence \begin{align*} 0\to \operatorname{coker}\alpha \xrightarrow{\mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu} B \xrightarrow{g} \ker \gamma \to 0 .\end{align*} \end{observation} \begin{observation} For a SES \begin{align*} 0 \to Y \xrightarrow{i} Z \xrightarrow{\pi} {Z\over Y} \to 0 \end{align*} there is an induced exact sequence \end{observation} Some missed stuff here. \begin{proof}[of Kunneth Formula (continued)] Note that \begin{align*} 0\to {Z}_{*} \otimes M \to {P}_{*}\otimes M \to d{P}_{*}\otimes M\to 0 ,\end{align*} where the differentials for the end terms are zero, and the homology will recover the original complex. \begin{center} \begin{tikzcd} &&&& {} && {H_{n+1}(dP \otimes M)= dP \otimes M} \\ \\ {} && {H_{n}(Z \otimes M) = Z\otimes M} && {H_{n}(P \otimes M)} && {H_{n}(dP \otimes M) = dP \otimes M} \\ \\ && {H_{n-1}(Z \otimes M) = Z_{n-1}\otimes M} \arrow[from=1-7, to=3-3, in=180, out=0] \arrow[from=3-3, to=3-5] \arrow[from=3-5, to=3-7] \arrow[from=3-7, to=5-3, in=180, out=0] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsNyxbMCwyXSxbMiwyLCJIX3tufShaIFxcdGVuc29yIE0pID0gWlxcdGVuc29yIE0iXSxbNCwyLCJIX3tufShQIFxcdGVuc29yIE0pIl0sWzQsMF0sWzYsMCwiSF97bisxfShkUCBcXHRlbnNvciBNKT0gZFAgXFx0ZW5zb3IgTSJdLFs2LDIsIkhfe259KGRQIFxcdGVuc29yIE0pID0gZFAgXFx0ZW5zb3IgTSJdLFsyLDQsIkhfe24tMX0oWiBcXHRlbnNvciBNKSA9IFpfe24tMX1cXHRlbnNvciBNIl0sWzQsMV0sWzEsMl0sWzIsNV0sWzUsNl1d}{Link to Diagram} \end{quote} By using the explicit formula for $${{\partial}}$$, it turns out that $${{\partial}}= (dP_{i+1} \overset{i}\hookrightarrow Z) \otimes\one M$$. By observation one, we get a SES \begin{align*} 0 \to {Z_n\otimes M \over dP_{n+1} \otimes M } \to H_n(P\otimes M) \to \ker i( \otimes\one_M) \to 0 .\end{align*} By observation 1, the first term equals $$H_n({P}_{*})\otimes M$$. From this, we get a flat resolution of $$H_{n-1}(P)$$: \begin{center} \begin{tikzcd} \textcolor{rgb,255:red,214;green,92;blue,92}{\deg:} & \textcolor{rgb,255:red,214;green,92;blue,92}{2} & \textcolor{rgb,255:red,214;green,92;blue,92}{1} & \textcolor{rgb,255:red,214;green,92;blue,92}{0} \\ 0 & 0 & {dP_n} & {Z_{n-1}} & {H_{n-1}(P)} & 0 \arrow[from=2-2, to=2-3] \arrow[from=2-3, to=2-4] \arrow[from=2-4, to=2-5] \arrow[from=2-5, to=2-6] \arrow[from=2-1, to=2-2] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsMTAsWzAsMSwiMCJdLFsxLDEsIjAiXSxbMSwwLCIyIixbMCw2MCw2MCwxXV0sWzIsMCwiMSIsWzAsNjAsNjAsMV1dLFszLDAsIjAiLFswLDYwLDYwLDFdXSxbMiwxLCJkUF9uIl0sWzMsMSwiWl97bi0xfSJdLFs0LDEsIkhfe24tMX0oUCkiXSxbNSwxLCIwIl0sWzAsMCwiXFxkZWc6IixbMCw2MCw2MCwxXV0sWzEsNV0sWzUsNl0sWzYsN10sWzcsOF0sWzAsMV1d}{Link to Diagram} \end{quote} So we can use this to compute $$\operatorname{Tor}(H_{n-1}(P), M)$$ by taking homology: \begin{center} \begin{tikzcd} \textcolor{rgb,255:red,214;green,92;blue,92}{\deg} & \textcolor{rgb,255:red,214;green,92;blue,92}{2} & \textcolor{rgb,255:red,214;green,92;blue,92}{1} & \textcolor{rgb,255:red,214;green,92;blue,92}{0} \\ 0 & 0 & {dP_n \otimes M} & {Z_{n-1}\otimes M} & 0 \arrow[from=2-1, to=2-2] \arrow[from=2-2, to=2-3] \arrow["{i\otimes\one}", from=2-3, to=2-4] \arrow[from=2-4, to=2-5] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsOSxbMCwxLCIwIl0sWzEsMSwiMCJdLFsyLDEsImRQX24gXFx0ZW5zb3IgTSJdLFszLDEsIlpfe24tMX1cXHRlbnNvciBNIl0sWzQsMSwiMCJdLFszLDAsIjAiLFswLDYwLDYwLDFdXSxbMiwwLCIxIixbMCw2MCw2MCwxXV0sWzEsMCwiMiIsWzAsNjAsNjAsMV1dLFswLDAsIlxcZGVnIixbMCw2MCw2MCwxXV0sWzAsMV0sWzEsMl0sWzIsMywiaVxcdGVuc29yIFxcMSJdLFszLDRdXQ==}{Link to Diagram} \end{quote} Thus \begin{align*} \ker(i\otimes\one_M) = \operatorname{Tor}_1( H_{n-1}(P), M) \cong \ker (dP_m \xrightarrow{{{\partial}}} Z_{n-1} \otimes M) .\end{align*} \end{proof} \begin{theorem}[Universal Coefficient Theorem] Let $${P}_{*}$$ be a chain complex of free abelian groups. For every abelian groups $$M$$ and every $$n$$, the Kunneth sequence splits non-canonically as \begin{align*} H_n({P}_{*} \otimes M) \cong \qty{ H_n( {P}_{*} )\otimes M } \oplus \operatorname{Tor}_1^{{\mathbb{Z}}}(H_{n-1}(P), M) .\end{align*} \end{theorem} \begin{remark} In optimal situations the tor term vanishes, e.g.~if either term is torsionfree (so no elements of finite order). \end{remark} \begin{fact} Every subgroup of a free abelian group is free (hence projective, hence flat). \end{fact} \begin{proof}[?] Since $$dP_n \leq dP_{n-1}$$, we can conclude $$dP_n$$ is free. Thus the following SES splits: \begin{align*} 0\to Z_n \to P_n \xrightarrow{d} dP_n \to 0 .\end{align*} So any lift of the identity map on $$dP_n$$ gives an isomorphic copy of the last term in the middle term, yielding $$P_n \cong Z_n \oplus dP_n$$. Now tensoring with $$M$$ and using that it distributes over direct sums yields \begin{align*} P_n \otimes M \cong (Z_n \otimes M) \oplus (dP_n \otimes M) .\end{align*} The left-hand side contains a copy of $$\ker(d_n \otimes\one: P_n \otimes M \to P_{n-1} \otimes M)$$, which itself contains a copy of $$Z_n\otimes M$$. So by a linear algebra exercise, we have $$\ker(d_n \otimes\one) \cong (Z_n \otimes M) \oplus A$$ for some unknown $$A$$, and since $$dP_{n+1} \otimes M = \operatorname{im}(d_{n+1}\otimes\one)$$ is contained in the first term, we can use the partial exactness of tensoring to preserve quotients and obtain \begin{align*} H_n(P\otimes M) = \qty{ H_n(P) \otimes M} \oplus C' \end{align*} for some $$C'$$. Now applying the Kunneth formula we find that $$C' = \operatorname{Tor}^{\mathbb{Z}}_1( H_{n-1}(P), M)$$, yielding the claimed direct sum. \end{proof} \begin{remark} The following is a generalization for both. \end{remark} \begin{theorem}[Kunneth formula for complexes] Let $$P, Q \in \mathsf{Ch}({\mathsf{R}{\hbox{-}}\mathsf{Mod}})$$ be complexes, then \begin{align*} P\otimes Q \coloneqq\operatorname{Tot}^{\oplus}(P\otimes Q)_n \coloneqq\bigoplus_{p+q = n} P_p \otimes Q_q \end{align*} with differential\footnote{Recall that the squares would commute if we took the usual differentials, so we use a sign trick to get $$d^2=0$$.} \begin{align*} d(a\otimes b) = (da)\otimes b + (-1)^pa \otimes(db) .\end{align*} If $$P_n, dP_n$$ are flat for all $$n$$, then there exists a SES \begin{align*} 0 \to \bigoplus_{p+q=n} H_p(P)\otimes H_q(Q) \to H_n(P\otimes Q) \to \bigoplus_{p+q=n-1} \operatorname{Tor}^R_1(H_p(P), H_q(Q) ) \to 0 .\end{align*} \end{theorem} \begin{proof}[?] Omitted here, but uses same ideas as the previous proofs. Hint: take $$Q$$ to have $$M$$ in degree 0. \end{proof} \hypertarget{applications-to-topology}{% \subsection{Applications to Topology}\label{applications-to-topology}} \begin{definition}[Simplicial Homology] See some applications in section 1 of Weibel, e.g.~simplicial and singular homology. The setup: $$X\in {\mathsf{Top}}, R\in \mathsf{Ring}$$ unital, and for $$k\geq 0$$ let $$S_k = S_k(X)$$ be the free $$R{\hbox{-}}$$module on $$\mathop{\mathrm{Hom}}_{\mathsf{Top}}( \Delta_k, X)$$ where $$\Delta_k$$ is the standard simplex By ordering the vertices, this induces an ordering on the faces by taking lexicographic ordering. Then the restriction of a map $$\Delta_k \to X$$ to the $$i$$th face of $$\Delta_k$$ gives a map $$\Delta_{k-1} \to X$$, which induces an $$R{\hbox{-}}$$module morphism $${{\partial}}_i: S_k \to S_{k-1}$$ By summing these we can define $$d \coloneqq\sum_{i=0}^k (-1)^i {{\partial}}_i: S_k\to S_{k-1}$$ and it turns out that $$d^2 = 0$$. So we can define a complex \begin{align*} \cdots \to S_2 \xrightarrow{d} \to S_1\to S_0 \to 0 \in \mathsf{Ch}({\mathsf{R}{\hbox{-}}\mathsf{Mod}}) .\end{align*} Taking it homology yields the \textbf{simplicial homology} of the complex $$H_n(X; R) \coloneqq H_n({S}_{*}(X) )$$. \end{definition} \begin{remark} Taking $$R={\mathbb{Z}}$$ makes $$S_k(X)$$ a free abelian group. If $$M$$ is any abelian group, we can define $$H_n(X; M) \coloneqq H_n( {S}_{*}(X) \otimes_{\mathbb{Z}}M)$$, the homology with \textbf{coefficients} in $$M$$. If no coefficients are specified, we write $$H_n(X) \coloneqq H_n(X; {\mathbb{Z}})$$. There is then a universal coefficient theorem in topology: \begin{align*} H_n(X; M) \cong \qty{ H_n(X) \otimes_{\mathbb{Z}}M} \oplus \operatorname{Tor}_1^{\mathbb{Z}}( H_{n-1}(X), M) .\end{align*} \end{remark} \begin{remark} Next week: group cohomology, spectral sequences next week. This will give us some objects to apply spectral sequences. \end{remark} \hypertarget{monday-march-08}{% \section{Monday, March 08}\label{monday-march-08}} \hypertarget{universal-coefficients-theorem}{% \subsection{3.6: Universal Coefficients Theorem}\label{universal-coefficients-theorem}} \begin{remark} Let $$X \in {\mathsf{Top}}$$ and $$S_k(X)$$ be the free $${\mathbb{Z}}{\hbox{-}}$$module on $$\mathop{\mathrm{Hom}}_{\mathsf{Top}}( \Delta_k, X)$$, which assemble into a chain complex $$S(X)$$. For $$M\in {\mathsf{Ab}}$$, we defined $$H^n(X; M) \coloneqq H^n( \mathop{\mathrm{Hom}}(S(X), M))$$ and write $$H^n(X) \coloneqq H^n(X; {\mathbb{Z}})$$. The universal coefficient theorem states \begin{align*} H^n(X; M) \cong \mathop{\mathrm{Hom}}_{\mathbb{Z}}(H_n(X), M) \oplus \operatorname{Ext}^1_{\mathbb{Z}}( H_{n-1}(X), M) .\end{align*} \end{remark} \begin{warnings} Note that this is homology on the RHS, not cohomology! \end{warnings} \begin{theorem}[Universal Coefficients Theorem for Cohomology] Let $${P}_{*}$$ be a chain complex of projective $$R{\hbox{-}}$$modules. Assume $$dP_n$$ is also projective for all $$n$$. For $$M\in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}$$, there is a split SES \begin{align*} 0 \to \operatorname{Ext}_R^1(H_{n-1}(P), M) \to H^n( \mathop{\mathrm{Hom}}_R( {P}_{*}, M)) \to \mathop{\mathrm{Hom}}_R (H_n(P), M) \to 0 .\end{align*} \end{theorem} \todo[inline]{Ask about naturality!} \begin{proof}[Sketch] As in the last lecture with free abelian groups, since the $$dP_n$$ are projective we can split $$P_n \cong Z_n \oplus dP_n$$ since $$Z_n = \ker d$$. Applying homs, since it's an additive functor this yields a new split exact sequence \begin{align*} 0 \to \mathop{\mathrm{Hom}}(dP_n, M) \to \mathop{\mathrm{Hom}}(P_n, M) \to \mathop{\mathrm{Hom}}(Z_n, M) \to 0 .\end{align*} Now running the proof for the original Kunneth formula and replacing tensor products to homs, these assemble into a split exact sequence of complexes and this yields the desired SES. Using the strategy of the proof of the UCF for free abelian groups to see that the sequence splits (although non-canonically). \end{proof} \begin{remark} Note that flat is weaker than projective for tensor products, but in an asymmetric situation, there's nothing weaker than projective for the hom functors to be exact (since this is an iff). \end{remark} \hypertarget{ch.-6-group-homology-and-cohomology}{% \subsection{Ch. 6: Group Homology and Cohomology}\label{ch.-6-group-homology-and-cohomology}} \hypertarget{definitions-and-properties}{% \subsubsection{Definitions and Properties}\label{definitions-and-properties}} \begin{definition}[Modules of Groups] Let $$G\in {\mathsf{Grp}}$$ be any group, finite or infinite, and let $$A\in \mathsf{G}{\hbox{-}}\mathsf{Mod}$$ be a left $$G{\hbox{-}}$$module, i.e.~an abelian group on which $$G$$ acts by additive maps on the left, written $$g.a$$ or $$ga$$ for $$g\in G, a\in A$$. Here \emph{additive} means that $$g.(a_1 + a_2) = g.a_1 + g.a_2$$. Note that this implies $$g.0 = 0, -g.a = -(g.a), g_1 (g_2 . a) = (g_1 g_2).a, 1_G.a = a$$. Writing $$\mathop{\mathrm{End}}_R(A) \coloneqq\mathop{\mathrm{Hom}}_R(A, A)$$, we have a group morphism \begin{align*} G &\to \mathop{\mathrm{End}}_{\mathbb{Z}}(A) \\ g &\mapsto g.({-}) .\end{align*} \end{definition} \begin{definition}[Equivariant Maps] If $$B \in \mathsf{G}{\hbox{-}}\mathsf{Mod}$$ is another left $$G{\hbox{-}}$$module, then \begin{align*} \mathop{\mathrm{Hom}}_G(A, B) = \left\{{ f\in \mathop{\mathrm{Hom}}_{\mathbb{Z}}(A, B) {~\mathrel{\Big|}~}f(g.a) = g(f(a)) \quad \forall a\in A, \forall g\in G }\right\} ,\end{align*} which are \textbf{$$G{\hbox{-}}$$equivariant maps}. \end{definition} \begin{definition}[Integral Group Ring] We define \begin{align*} {\mathbb{Z}}G\coloneqq\left\{{ \sum_{i=1}^N m_i g_i {~\mathrel{\Big|}~}m_i\in {\mathbb{Z}}, g_i\in G, n\in {\mathbb{N}}}\right\} .\end{align*} We can equip this with a ring structure using $$(mg)(m' g') = mm' gg'$$ and extending $${\mathbb{Z}}{\hbox{-}}$$linearly. \end{definition} \begin{remark} There is an equality of categories $${\mathsf{G}{\hbox{-}}\mathsf{Mod}} = \mathsf{{\mathbb{Z}}G}{\hbox{-}}\mathsf{Mod}$$. This is also the same as the functor category $${\mathsf{Ab}}^\mathcal{G}$$ (a category of the form $$\mathcal{A}^{\mathcal{I}}$$) where $$\mathcal{G}$$ is the category with one object whose morphisms are the elements of $$G$$. In other words, $${\operatorname{Ob}}(\mathcal{G}) \coloneqq\left\{{ 1 }\right\}$$ and $$\mathop{\mathrm{Hom}}_{\mathcal{G}}(1, 1) = G$$. Note that every morphism is invertible since $$G$$ is a group. \begin{figure} \centering \includegraphics{figures/image_2021-03-08-09-36-58.png} \caption{image\_2021-03-08-09-36-58} \end{figure} The right-hand side yields a $$G{\hbox{-}}$$module since $$F(g)(a) = g.a$$. \end{remark} \begin{definition}[Trivial modules] An object $$A\in \mathsf{G}{\hbox{-}}\mathsf{Mod}$$ is a \textbf{trivial} module if and only if $$g.a = a$$ for all $$g\in G$$. \end{definition} \begin{remark} Any $$G\in {\mathsf{Ab}}$$ can be viewed as a trivial $$G{\hbox{-}}$$module in this way. This yields a functor $$\operatorname{Triv}:{\mathsf{Ab}}\to \mathsf{G}{\hbox{-}}\mathsf{Mod}$$. There is a distinguished trivial $$G{\hbox{-}}$$module, namely $$A \coloneqq{\mathbb{Z}}$$ with the trivial $$G{\hbox{-}}$$action. There are two natural functors $$\mathsf{G}{\hbox{-}}\mathsf{Mod}\to {\mathsf{Ab}}$$: \begin{itemize} \tightlist \item $$A^G \coloneqq\left\{{ a\in A {~\mathrel{\Big|}~}g.a = a \forall g\in G }\right\}$$, the \textbf{invariant subgroup} of $$A$$. \item $$A_G \coloneqq A / \left\langle{ ga-a {~\mathrel{\Big|}~}g\in G, a\in A }\right\rangle$$, where we take the $$G{\hbox{-}}$$module generated by the relation in the denominator, which are the \textbf{coinvariants} of $$A$$. \end{itemize} \end{remark} \begin{exercise}[6.1.1] \envlist \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \item $$A^G$$ is the maximal trivial submodule of $$A$$, so the functor $$({-})^G$$ is right-adjoint to $$\operatorname{Triv}$$. These should both be easy checks! So this is left-exact and has right-derived functors (similar to ext). \item $$A_G$$ is the largest $$G{\hbox{-}}$$trivial quotient of $$A$$, and $$({-})_G$$ is left-adjoint to $$\operatorname{Triv}$$. Thus it is right-exact and has left-derived functors (similar to tor). \end{enumerate} \end{exercise} \begin{lemma}[?] Let $$A\in \mathsf{G}{\hbox{-}}\mathsf{Mod}$$ and $${\mathbb{Z}}$$ be the trivial $$G{\hbox{-}}$$module. Then \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \tightlist \item $$A_G \cong {\mathbb{Z}}\otimes_{{\mathbb{Z}}G} A$$, and \item $$A^G \cong \mathop{\mathrm{Hom}}_G({\mathbb{Z}}, A)$$ (\textbf{important!!}) \end{enumerate} \end{lemma} \begin{warnings} Number 2 above is important to remember! \end{warnings} \begin{proof}[of 1] Viewing $${\mathbb{Z}}= _{{\mathbb{Z}}} {\mathbb{Z}}_{{\mathbb{Z}}G} \in (\mathsf{{\mathbb{Z}}}, \mathsf{{\mathbb{Z}}G}){\hbox{-}}\mathsf{biMod}$$ with the trivial structure, recall\footnote{See Weibel p.~41.} that we have a functor \begin{align*} \mathop{\mathrm{Hom}}_({\mathbb{Z}}, {-}): \mathsf{{\mathbb{Z}}}{\hbox{-}}\mathsf{Mod} &\to \mathsf{{\mathbb{Z}}G}{\hbox{-}}\mathsf{Mod}\\ \end{align*} where $$\mathop{\mathrm{Hom}}_{\mathbb{Z}}({\mathbb{Z}}, A)$$ has an action $$(g.f)(x) \coloneqq f(x. g)$$ for $$x\in {\mathbb{Z}}g\in G$$. Since $$x.g = x$$ for all $$x, g$$, we have $$g.f = f$$ and thus $$\mathop{\mathrm{Hom}}_{\mathbb{Z}}({\mathbb{Z}}, A)$$ is a trivial $$G{\hbox{-}}$$module, and there is an isomorphism in $${\mathsf{Ab}}$$: \begin{align*} \mathop{\mathrm{Hom}}_{\mathbb{Z}}({\mathbb{Z}}, A) &\underset{{\mathsf{Ab}}}{\xrightarrow{\sim}} A \\ f &\mapsto f(a) .\end{align*} Thus $$\mathop{\mathrm{Hom}}_{\mathbb{Z}}({\mathbb{Z}}, {-}) \cong \operatorname{Triv}({-})$$. By prop 2.6.3, the functor $${\mathbb{Z}}\otimes_{{\mathbb{Z}}G} ({-})$$ is left-adjoint to $$\mathop{\mathrm{Hom}}_{\mathbb{Z}}( _{{\mathbb{Z}}} {\mathbb{Z}}_{{\mathbb{Z}}G}, {-})$$. Now applying exercise 6.1.1 part 2, $$({-})_G \cong \operatorname{Triv}({-})$$. Since left-derived functors are universal $$\delta{\hbox{-}}$$functors, we have a natural isomorphism $$({-})_G \cong {\mathbb{Z}}\otimes_{{\mathbb{Z}}G} ({-})$$ since they're both left-adjoint to the same functor. \end{proof} \begin{proof}[of 2 ] Taking $$f(1)$$, we have $$A^G \cong \mathop{\mathrm{Hom}}_{\mathbb{Z}}({\mathbb{Z}}, A^G)$$. Using the adjoint property from exercise 6.1.1 part 1, this is isomorphic to $$\mathop{\mathrm{Hom}}_G( \operatorname{Triv}({\mathbb{Z}}), A)$$. Thus $$({-})^G \cong \mathop{\mathrm{Hom}}_G({\mathbb{Z}}, {-})$$. \end{proof} \begin{remark} The exts here will classify extensions in the category of left $${\mathbb{Z}}{\hbox{-}}$$modules. Note the switched order on the hom functor however! \end{remark} \hypertarget{ch.-6-group-homology-and-cohomology-wednesday-march-10}{% \section{Ch. 6: Group Homology and Cohomology (Wednesday, March 10)}\label{ch.-6-group-homology-and-cohomology-wednesday-march-10}} \begin{lemma}[?] Last time: started setting up group homology. For $$G$$ a group and $$A\in \mathsf{G}{\hbox{-}}\mathsf{Mod}$$, we think of $${\mathbb{Z}}$$ as a trivial $$G{\hbox{-}}$$module and \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \item $$A_G \cong {\mathbb{Z}}\otimes_{{\mathbb{Z}}G} A$$, the $$G{\hbox{-}}$$coinvariants. \item $$A^G \cong \mathop{\mathrm{Hom}}_{{\mathbb{Z}}G}( {\mathbb{Z}}, A)$$. the $$G{\hbox{-}}$$invariants, this is the largest $$G{\hbox{-}}$$trivial submodule of $$A$$ \end{enumerate} \end{lemma} \begin{definition}[?] For $$A\in \mathsf{G}{\hbox{-}}\mathsf{Mod}$$, \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \item $$H_*(G; A) \coloneqq L_*({-}))G (A)$$ are the \textbf{homology groups of $$G$$ with coefficients in $$A$$}. It is isomorphic to $$\operatorname{Tor}_*^{{\mathbb{Z}}G}({\mathbb{Z}}, A)$$ by (1) in the lemma above. In particular, $$H_0(G; A) \cong A_G$$. \item $$H^*(G; A) \coloneqq R^*({-})^G(A)$$ is the \textbf{cohomology of $$G$$ with coefficients in $$A$$}. It is isomorphic to $$\operatorname{Ext}^*_{{\mathbb{Z}}G}({\mathbb{Z}}, A)$$ by (2) in the lemma. In particular, $$H^0(G; A) \cong A^G$$. \end{enumerate} \end{definition} \todo[inline]{Ask about contructing resolutions: take any "augmentation" map and iterate kernels? Different resolution lengths?} \begin{example}[?] For $$G = \left\{{ 1 }\right\}$$, for any $$A\in \mathsf{G}{\hbox{-}}\mathsf{Mod}$$ we have $$A^G = A = A_G$$. Forgetful functors are usually exact, and in this case $$({-})^G, ({-})_G: \mathsf{G}{\hbox{-}}\mathsf{Mod} \to {\mathsf{Ab}}$$ is really a forgetful functor and thus exact. Here $$H_n(G; A) = 0 = H^n(G; A)$$ for $$n>0$$. \end{example} \begin{example}[?] Let $$G$$ be infinite cyclic, which we'll write multiplicatively to prevent the notation from conflicting with the addition on $${\mathbb{Z}}G$$, so $$G\coloneqq T = \left\langle{ t }\right\rangle= \left\{{ t^n {~\mathrel{\Big|}~}n\in {\mathbb{Z}}}\right\}$$. Then $${\mathbb{Z}}G = {\mathbb{Z}}[t, t ^{-1} ]$$ are integral Laurent polynomials, since we're taking integer linear combinations of various $$t^n$$. Computing $$H_*(T, A) \cong \operatorname{Tor}_*^{{\mathbb{Z}}T}({\mathbb{Z}}, A)$$ and $$H^*(T; A) \cong \operatorname{Ext}^*_{{\mathbb{Z}}T}({\mathbb{Z}}, A)$$ using a projective resolution of $${\mathbb{Z}}$$ as a $${\mathbb{Z}}T{\hbox{-}}$$module, since the first slot Ext requires an injective resolution in the opposite category. It suffices to take a free resolution: \begin{align*} \cdots \to P_2 \to P_1 \to P_0 \to {\mathbb{Z}}\to 0 \coloneqq \cdots \to 0\to {\mathbb{Z}}T \xrightarrow{\times (t-1)} {\mathbb{Z}}T \xrightarrow{\operatorname{ev}_1} {\mathbb{Z}}\to 0 .\end{align*} Note that the resolution ends here because the multiplication $$\times(t-1)$$ is injective on polynomials rings. Thus $$H_{>\geq 2}(T; A) = H^{\geq 2}(T; A) = 0$$. The zeroth terms are invariants/coinvariants. For $$\operatorname{Tor}$$, we apply $${-}\otimes_{{\mathbb{Z}}T} A$$ to this resolution to obtain \begin{align*} 0\to FP_1 \to FP_0 \to 0 &\coloneqq 0 \to {\mathbb{Z}}T \otimes_{{\mathbb{Z}}T} A \xrightarrow{(t-1) \otimes\one} {\mathbb{Z}}T \otimes_{{\mathbb{Z}}T} A \to 0\\ &= 0 \to A \xrightarrow{(t-1) \otimes\one} A \to 0 .\end{align*} One can check that \begin{itemize} \tightlist \item $$\ker (t-1) \otimes\one = A^T = H_1(T; A)$$ is equal to the invariants and \item $$\operatorname{coker}(t-1) \otimes\one = A_T = H_0(T; A)$$ is equal to the coinvariants. \end{itemize} The second fact had to be true, but the first is surprising! For $$\operatorname{Ext}^*$$, we apply the contravariant $$\mathop{\mathrm{Hom}}_{{\mathbb{Z}}T}({-}, A)$$ to obtain \begin{align*} 0 \to \mathop{\mathrm{Hom}}_{{\mathbb{Z}}T}({\mathbb{Z}}T, A) \xrightarrow{{-}\circ (t-1)} \mathop{\mathrm{Hom}}_{{\mathbb{Z}}T}({\mathbb{Z}}T, A) \to 0 .\end{align*} One checks \begin{itemize} \tightlist \item $$\operatorname{coker}( {-}\circ (t-1)) = A_T = H^1(T; A)$$ (surprising!) and \item $$\ker( {-}\circ (t-1)) = A^T = H^0(T; A)$$ \end{itemize} \end{example} \begin{remark} See exercise 6.1.2 for $$kG{\hbox{-}}$$modules for $$k\in \mathsf{Ring}$$ arbitrary. \end{remark} \begin{question} What can we say about $$H_0$$ and $$H^0$$ for more general groups? \end{question} \hypertarget{h_0-for-groups}{% \subsection{\texorpdfstring{$$H_0$$ for Groups}{H\_0 for Groups}}\label{h_0-for-groups}} \begin{definition}[Augmentation Maps] Define the \textbf{augmentation map} \begin{align*} \varepsilon: {\mathbb{Z}}G\to {\mathbb{Z}}\\ \sum n_i g_i &\mapsto \sum n_i ,\end{align*} which is a ring morphism. Define $$\mathcal{I} \coloneqq\ker \varepsilon$$ to be the \textbf{augmentation ideal}. \end{definition} \begin{observation} There is a basis of $${\mathbb{Z}}G$$ as a $${\mathbb{Z}}{\hbox{-}}$$module given by \begin{align*} \mathcal{B}\coloneqq B_1 \cup B_2 \coloneqq\left\{{ 1 }\right\} \cup\left\{{ g-1 {~\mathrel{\Big|}~}1\neq g\in G }\right\} .\end{align*} Note that $$\varepsilon(g-1) = 0$$, so $$\mathcal{I}$$ is a free $${\mathbb{Z}}{\hbox{-}}$$module with basis $$B_2$$. Here the kernel should be expected to have codimension 1! We also have $${\mathbb{Z}}G/ \mathcal{I} \cong {\mathbb{Z}}$$ as rings, where the left-hand side is a $$G{\hbox{-}}$$module. Letting $$\mkern 1.5mu\overline{\mkern-1.5mu{-}\mkern-1.5mu}\mkern 1.5mu$$ denote coset/equivalence class representatives, we have \begin{align*} g \mkern 1.5mu\overline{\mkern-1.5mu1\mkern-1.5mu}\mkern 1.5mu = \mkern 1.5mu\overline{\mkern-1.5mug1\mkern-1.5mu}\mkern 1.5mu = \mkern 1.5mu\overline{\mkern-1.5mug\mkern-1.5mu}\mkern 1.5mu = \mkern 1.5mu\overline{\mkern-1.5mu1\mkern-1.5mu}\mkern 1.5mu ,\end{align*} and so the action $$G \curvearrowright{\mathbb{Z}}G/ \mathcal{I}$$ is trivial. \begin{fact} For $$R$$ a ring and $$\mathcal{I} {~\trianglelefteq~}R$$ a (left? right?) ideal and $$M\in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}$$, \begin{align*} R/I \otimes_R M \cong M/IM .\end{align*} \end{fact} So for any $$A\in \mathsf{G}{\hbox{-}}\mathsf{Mod}$$ we have \begin{align*} H_0(G; A) &= A_G \\ &\cong {\mathbb{Z}}\otimes_{{\mathbb{Z}}G} A \\ &= \operatorname{Tor}_0^{{\mathbb{Z}}G}({\mathbb{Z}}; A) \\ &= {\mathbb{Z}}G/\mathcal{I} \otimes_{{\mathbb{Z}}G} A \\ &\cong A/ \mathcal{I} A .\end{align*} \end{observation} \begin{example}[?] \envlist \begin{itemize} \item $$H_0(G; {\mathbb{Z}}) \cong {\mathbb{Z}}/ \mathcal{I} {\mathbb{Z}}\cong {\mathbb{Z}}$$, where $$\mathcal{I} {\mathbb{Z}}= 0$$ since $${\mathbb{Z}}$$ is the trivial $$G{\hbox{-}}$$module and $$(g-1)a = ga-1a=a-a=0$$. \item $$H_0(G; {\mathbb{Z}}G) \cong {\mathbb{Z}}G/ \mathcal{I} \cong {\mathbb{Z}}$$. \item $$H_0(G; \mathcal{I} ) \cong \mathcal{I} / \mathcal{I}^2$$. \end{itemize} \end{example} \begin{example}[?] Noting that $$A = {\mathbb{Z}}G$$ is projective in $$\mathsf{{\mathbb{Z}}G}{\hbox{-}}\mathsf{Mod}$$, so $$H_n(G; {\mathbb{Z}}G) = 0$$ for $$n>0$$, using that this was a version of $$\operatorname{Tor}$$ and projective implies flat. \end{example} \hypertarget{h0-for-groups}{% \subsection{\texorpdfstring{$$H^0$$ for Groups}{H\^{}0 for Groups}}\label{h0-for-groups}} \begin{definition}[Norm Element] Let $$G$$ be a finite group, then the \textbf{norm element} is defined by \begin{align*} N = \sum_{g\in G} g\in {\mathbb{Z}}G .\end{align*} \end{definition} \begin{remark} For $$h\in G$$, \begin{align*} hN = \sum_g hg = \sum_{g'\in G} g' = N ,\end{align*} and so $$N \in ({\mathbb{Z}}G)^g$$. Similarly $$Nh = N$$ and so $$Z({\mathbb{Z}}G)$$ is in the center. \begin{quote} Note the two different $$Z$$s here! \end{quote} \end{remark} \begin{lemma}[?] Let $$G$$ be finite, then \begin{align*} H^0(G; {\mathbb{Z}}G) = ({\mathbb{Z}}G)^G = {\mathbb{Z}}N ,\end{align*} which is a two-sided ideal of $${\mathbb{Z}}G$$ that is isomorphic to $${\mathbb{Z}}$$. \end{lemma} \begin{proof}[?] The inclusion $${\mathbb{Z}}N \subseteq ({\mathbb{Z}}G)^G$$ is clear from the previous remark, so it remains to show the other inclusion. Suppose \begin{align*} a\in \sum_{g\in G} n_g g \in ({\mathbb{Z}}G)^G .\end{align*} Then for all $$h\in G$$, we have \begin{align*} a = ha = \sum n_g h_g .\end{align*} Now note that the $$g$$ are a free $${\mathbb{Z}}{\hbox{-}}$$basis for $${\mathbb{Z}}G$$, so we can equate coefficients of $$h$$ to find that $$n_h = n_1$$. Since $$h$$ was arbitrary, we have $$a = n_1 N \in {\mathbb{Z}}N$$. \end{proof} \begin{remark} Exercise 6.1.3 shows that $$H^0(G; {\mathbb{Z}}G) = 0$$ when $$G$$ is infinite, in which case $$\mathcal{I} = \left\{{ a \in {\mathbb{Z}}G {~\mathrel{\Big|}~}N a = 0 }\right\}$$ is the annihilator of the norm element. Next class we'll start on spectral sequences. \end{remark} \hypertarget{spectral-sequences-monday-march-15}{% \section{Spectral Sequences (Monday, March 15)}\label{spectral-sequences-monday-march-15}} \hypertarget{motivation}{% \subsection{Motivation}\label{motivation}} \begin{remark} Invented by John Leray, 1946 while a prisoner of war in Austria, as an algorithmic way to compute homology of chain complexes. Start with a first-quadrant double complex $$\left\{{ E_{p, q} {~\mathrel{\Big|}~}p, q\geq 0 }\right\}$$, say of $$R{\hbox{-}}$$modules. Let $$T_n \coloneqq\bigoplus_{p}$$ be the total complex (direct sum or product, since the diagonals are finite) where $$d \coloneqq d^b + d^h$$. Suppose one could compute the homology of each piece'' of the differential separately and independently. First forget $$d^h$$, and let this complex be $$E_{p, q}^0$$ (where the $$0$$ superscript denotes a zeroth approximation''). \begin{center} \begin{tikzcd} q \\ \\ 3 & \bullet & \bullet & \bullet \\ 2 & \bullet & \bullet & \bullet \\ 1 & \bullet & \bullet & \bullet \\ 0 & 1 & 2 & 3 &&& p \arrow[from=3-4, to=4-4] \arrow[from=4-4, to=5-4] \arrow[from=3-3, to=4-3] \arrow[from=4-3, to=5-3] \arrow[from=3-2, to=4-2] \arrow[from=4-2, to=5-2] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=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}{Link to Diagram} \end{quote} Now let $$E^1{p, q} \coloneqq H_q(E_{p, q}^0)$$ be the homology obtained from the vertical complexes, i.e.~$$E^1_{p, q} \coloneqq\ker d^v_{p, q} / \operatorname{im}d^v_{p, q-1}$$. Recall that by convention we require anticommutativity, so $$d^vd^h + d^h d^v = 0$$, so this is not quite a complex of complexes. So these won't quite give a chain map, but $$d^vd^h = -d^h d^v$$ is enough to induce well-defined maps on $$E^1_{*, *}$$ since they will preserve kernels and images. So $$E^1$$ has horizontal differentials $$d^h: E^1_{*,*} \to E^1_{*-1, *}$$: \begin{center} \begin{tikzcd} {E^1_{p, q}:} & q \\ \\ & 3 & \bullet & \bullet & \bullet \\ & 2 & \bullet & \bullet & \bullet \\ & 1 & \bullet & \bullet & \bullet \\ & 0 & 1 & 2 & 3 &&& p \arrow[from=5-5, to=5-4] \arrow[from=4-5, to=4-4] \arrow[from=3-5, to=3-4] \arrow[from=3-4, to=3-3] \arrow[from=4-4, to=4-3] \arrow[from=5-4, to=5-3] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=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}{Link to Diagram} \end{quote} We can now write $$E_{p, q}^2$$ for the horizontal homology $$H_p(E^1_{*, q})$$ at the $$p, q$$ spot. We've done the horizontal and vertical homology separately, how close is $$\left\{{ E_{p, q}^2 {~\mathrel{\Big|}~}p+q = n }\right\}$$ to giving us information about the total homology? \end{remark} \begin{exercise}[5.1.1] If $$E^0_{*,*}$$ consists of only two columns $$p$$ and $$p-1$$, then there is a SES \begin{align*} 0 \to E_{p-1, q+1}^2 \to H_{p+q}(T) \to E_{p, q}^2 \to 0 .\end{align*} \begin{center} \begin{tikzcd} n \\ & \vdots \\ {q+1} && \bullet \\ {n-p=q} &&& \bullet \\ &&&& \vdots \\ && {p-1} & p && n \arrow[from=2-2, to=3-3] \arrow[from=3-3, to=4-4] \arrow[from=4-4, to=5-5] \arrow[from=1-1, to=2-2] \arrow[from=5-5, to=6-6] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsMTAsWzIsNSwicC0xIl0sWzMsNSwicCJdLFswLDMsIm4tcD1xIl0sWzAsMiwicSsxIl0sWzIsMiwiXFxidWxsZXQiXSxbMywzLCJcXGJ1bGxldCJdLFswLDAsIm4iXSxbNSw1LCJuIl0sWzEsMSwiXFx2ZG90cyJdLFs0LDQsIlxcdmRvdHMiXSxbOCw0XSxbNCw1XSxbNSw5XSxbNiw4XSxbOSw3XV0=}{Link to Diagram} \end{quote} \begin{figure} \centering \includegraphics{figures/image_2021-03-15-09-29-09.png} \caption{image\_2021-03-15-09-29-09} \end{figure} So in general, $$H_*(T)$$ is determined up to extensions. \end{exercise} \begin{exercise}[5.1.2] We view $$E^2_{*, *}$$ as a 2nd order approximation to $$H_*( {T}_{*} )$$. We've used both differentials, so how do we continue? There are well-defined maps $$d_{p, q}^2: E_{p, q}^2 \to E^{2}_{p-2, q+1}$$ such that $$d^2_{*,*} \circ d^2_{*, *} = 0$$ (noting that these are superscripts, not squaring). \end{exercise} \begin{remark} This yields differentials on $$E^2$$ on lines of slope $$-1/2$$ which move from the $$n$$th diagonal to the $$n-1$$st diagonal: \begin{center} \begin{tikzcd} & {} & {} \\ {q+1} && \bullet \\ q &&&& \bullet \\ \\ && {p-2} && p & {} & {} \arrow["n"{pos=0}, color={rgb,255:red,92;green,214;blue,214}, dashed, no head, from=1-3, to=5-7] \arrow["{n-1}"{pos=0}, color={rgb,255:red,92;green,214;blue,214}, dashed, no head, from=1-2, to=5-6] \arrow[from=3-5, to=2-3] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsMTAsWzIsNCwicC0yIl0sWzQsNCwicCJdLFswLDIsInEiXSxbMCwxLCJxKzEiXSxbMiwxLCJcXGJ1bGxldCJdLFs0LDIsIlxcYnVsbGV0Il0sWzIsMF0sWzYsNF0sWzEsMF0sWzUsNF0sWzYsNywibiIsMCx7ImxhYmVsX3Bvc2l0aW9uIjowLCJjb2xvdXIiOlsxODAsNjAsNjBdLCJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifSwiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX0sWzE4MCw2MCw2MCwxXV0sWzgsOSwibi0xIiwwLHsibGFiZWxfcG9zaXRpb24iOjAsImNvbG91ciI6WzE4MCw2MCw2MF0sInN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9LCJoZWFkIjp7Im5hbWUiOiJub25lIn19fSxbMTgwLDYwLDYwLDFdXSxbNSw0XV0=}{Link to Diagram} \end{quote} \begin{center} \begin{tikzcd} \bullet \\ & \bullet & {(p, q)} \\ && \bullet \arrow["{d^0}", from=2-3, to=3-3] \arrow["{d^2}"', from=2-3, to=1-1] \arrow["{d^1}", from=2-3, to=2-2] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsNCxbMiwxLCIocCwgcSkiXSxbMiwyLCJcXGJ1bGxldCJdLFswLDAsIlxcYnVsbGV0Il0sWzEsMSwiXFxidWxsZXQiXSxbMCwxLCJkXjAiXSxbMCwyLCJkXjIiLDJdLFswLDMsImReMSJdXQ==}{Link to Diagram} \end{quote} So we let $$E^3$$ be the homology, and it turns out there are differentials $$d^3: E^3_{p, q} \to E^3_{p-3, q+2}$$ which go from diagonal $$n$$ to $$n-1$$. \end{remark} \hypertarget{setup}{% \subsection{Setup}\label{setup}} \begin{definition}[Homology Spectral Sequences] A \textbf{homology spectral sequence} starting with $$E^a$$ for $$a\in {\mathbb{Z}}$$ in an abelian category $$\mathcal{A}$$ consists of the following data: \begin{enumerate} \def\labelenumi{\alph{enumi}.} \item Pages: For all $$r\geq a$$ and all $$p, q\in {\mathbb{Z}}$$, a family $$\left\{{ E_{p, q}^r }\right\}$$ of objects in $$\mathcal{A}$$ (some of which my be zero), where typically $$a=1, 2$$. \item Differentials: A family of maps $$\left\{{ d_{p, q}^r: E_{p, q}^r \to E_{p-r, q+r-1}^r }\right\}$$ with $$d^r \circ d^r =0$$ of slope $$-\frac{r-1}{r}$$ in that lattice $$E_{*, *}^r$$ the form chain complexes. We take the convention that the differentials go to the left: \end{enumerate} \begin{center} \begin{tikzcd} \bullet \\ \\ && \bullet \\ \\ &&&& \bullet \arrow["{d^r}", from=5-5, to=3-3] \arrow["{d^r}", from=3-3, to=1-1] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsMyxbNCw0LCJcXGJ1bGxldCJdLFsyLDIsIlxcYnVsbGV0Il0sWzAsMCwiXFxidWxsZXQiXSxbMCwxLCJkXnIiXSxbMSwyLCJkXnIiXV0=}{Link to Diagram} \end{quote} \begin{enumerate} \def\labelenumi{\alph{enumi}.} \setcounter{enumi}{2} \tightlist \item Structure Maps: Isomorphisms $$E_{p, q}^{r+1} \cong \ker d_{p, q}^r / \operatorname{im}d_{p+r, q-r+1}^r$$. \end{enumerate} We denote $$E^r_{*,*}$$ to be the \textbf{$$r$$th page} of the sequence, and the \textbf{total degree} of an entry $$E_{p, q}^r$$ is $$p+q$$. \end{definition} \begin{remark} The term $$E_{p, q}^{r+1}$$ is a \emph{subquotient}, i.e.~a submodule of a quotient, of $$E_{p, q}^r$$, and hence inductively a subquotient of $$E_{p, q}^a$$ by transitivity of being a subquotient''. The terms of total degree $$n$$ lie on a line of slope $$-1$$, and each differential $$d^r_{p, q}$$ decreases the total degree by 1. \end{remark} \begin{remark} There is a category of homology spectral sequences over a fixed abelian category $$\mathcal{A}$$. The objects consist of the above data of pages, differentials, and structure maps from the above definition The morphisms $$f: E\to \tilde E$$ are families of maps \begin{align*} f_{p, q}^r: E_{p, q}^r \to \tilde{E}^r_{p, q} \end{align*} for all $$r \geq\max\left\{{a, \tilde a}\right\}$$ with $$\tilde{d}^r f^r = f^r d^r$$ such that $$f_{p, q}^{r+1}$$ is the map on homology induced by $$f_{p, q}^r$$. \end{remark} \begin{definition}[Cohomology Spectral Sequence] A \textbf{cohomology} spectral sequence is defined dually: we'll write this as $$E_r^{p, q}, d_r^{p, q}$$, where the differentials go down and to the right, and increase the total degree by 1: \begin{align*} d_r^{p, q}: E_r^{p, q} \to E_{r}^{p+r, q-r+1} .\end{align*} \begin{center} \begin{tikzcd} q & \bullet \\ \\ {q-r+1} &&&&& \bullet \\ & p &&&& {p+r} \arrow["{d_r^{p, q}}", from=1-2, to=3-6] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsNixbMSwwLCJcXGJ1bGxldCJdLFs1LDIsIlxcYnVsbGV0Il0sWzEsMywicCJdLFs1LDMsInArciJdLFswLDAsInEiXSxbMCwyLCJxLXIrMSJdLFswLDEsImRfcl57cCwgcX0iXV0=}{Link to Diagram} \end{quote} There is similarly a category of these. \end{definition} \begin{lemma}[Mapping Lemma] Let $$f:E\to \tilde{E}$$ be a morphism of spectral sequences (homology or cohomology) such that for some fixed $$r$$, the map $$f^r: E_{p, q}^r\to \tilde{E}_{p, q}^r$$ is an isomorphism for all $$p, q$$. Then all $$f^s_{p, q}$$ are isomorphisms for all $$s\geq r$$ and all $$p, q$$. \end{lemma} \begin{proof}[of the mapping lemma] There is a commutative diagram with exact rows: \begin{center} \begin{tikzcd} 0 && {B_{p, q}^r} && {Z_{p, q}^r} && {E_{p, q}^{r+1}} && 0 && 0 \\ \\ 0 && {\tilde B_{p, q}^r} && {\tilde Z_{p, q}^r} && {\tilde E_{p, q}^r} && 0 && 0 \arrow[from=1-1, to=1-3] \arrow[from=1-3, to=1-5] \arrow[from=1-5, to=1-7] \arrow[from=1-7, to=1-9] \arrow[from=3-1, to=3-3] \arrow[from=3-3, to=3-5] \arrow[from=3-5, to=3-7] \arrow[from=3-7, to=3-9] \arrow["{f_{p ,q}^r (\sim)}"', from=1-5, to=3-5] \arrow["{f_{p ,q}^r (\sim)}"', from=1-3, to=3-3] \arrow["{f_{p, q}^{r+1}}"', from=1-7, to=3-7] \arrow[from=1-9, to=1-11] \arrow[from=3-9, to=3-11] \arrow[no head, from=1-11, to=3-11] \arrow[no head, from=1-9, to=3-9] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=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}{Link to Diagram} \end{quote} Extending the right-hand side as indicate, we can apply the Five Lemma to conclude that $$f_{p, q}^{r+1}$$ is an isomorphism. Now do induction on $$r$$. \end{proof} \hypertarget{wednesday-march-17}{% \section{Wednesday, March 17}\label{wednesday-march-17}} \hypertarget{spectral-sequences}{% \subsection{5.2: Spectral Sequences}\label{spectral-sequences}} \begin{remark} Recall that we had \begin{itemize} \tightlist \item $$\left\{{ E_{p, q}^r {~\mathrel{\Big|}~}r\geq a, p,q\in {\mathbb{Z}}}\right\}$$ for some $$a$$. \item $$d_{p, q}^r: E_{p, q}^r \to E_{p-r, q+r-1}$$ with $$d^2=0$$. \item $$E_{p, q}^{r+1} \cong \ker d_{p, q}^r / \operatorname{im}d_{p+r,q-r+1}^r$$. \end{itemize} \end{remark} \begin{example}[First quadrant spectral sequences] A \textbf{first quadrant} (homology) spectral sequence is one with $$E_{p, q}^r = 0$$ for $$p, q<0$$. Note that for a fixed $$p, q$$, there is an $$r \gg 0$$ such that the differential entering and leaving $$E_{p, q}^r$$ will be zero. The domain will be in quadrant 2 and the codomain in quadrant 4. In this case $$E_{p, q}^r \cong E_{p, q}^{r+1}$$ and we call this stable'' module $$E_{p, q}^{\infty }$$. Note that $$r=r(p, q)$$ can generally depend on $$p, q$$. \end{example} \begin{definition}[Bounded] We say a spectral sequence is \textbf{bounded} if there are only finitely many nonzero terms of total degree $$n$$. If so, there exists some uniform $$r_0$$ such that for $$r\geq r_0$$, we have $$E^{r}_{p, q} \cong E_{p, q}^{r+1} \cong E_{p, q}^{\infty }$$. \todo[inline]{See video for image.} \end{definition} \begin{remark} For the rest of this course, we'll restrict our attention to bounded spectral sequences. \end{remark} \begin{definition}[Convergence of a homology spectral sequences] A bounded spectral sequences $$E$$ \textbf{converges} to $$H_*$$ if we are given \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \item A family of objects $$\left\{{ H_n }\right\}_{n\in {\mathbb{Z}}}$$ \item For each $$n$$, a finite (here increasing) filtration \begin{align*} 0 = F_s H_n \subseteq \cdots \subseteq F_{p-1} H_n \subseteq F_p H_n \subseteq \cdots \subseteq F_t H_n = H_n \end{align*} where each $$F_i H_n$$ is a subobject of $$H_n$$ \item Isomorphisms \begin{align*} E_{p, q}^{\infty } \cong { F_p H_{p +q} \over F_{p-1} H_{p+q}} ,\end{align*} or equivalently \begin{align*} E_{p, n-p}^{\infty } \cong { F_p H_n \over F_{p-1} H_n} ,\end{align*} which are the $$t-s$$ \textbf{successive quotients} (or \textbf{sections}) of the filtration, which depend on $$n$$. We refer to $$t-s$$ as the \textbf{length} of the filtration \end{enumerate} In this case we write \begin{align*} E_{p, q}^a \Rightarrow H_{p+q} ,\end{align*} thinking of $$a\to \infty$$. \end{definition} \begin{remark} We saw a case where the length of the filtration was 2, when we had $$2$$ columns. Recall that this only yields information up to extensions, since this only computes quotients. \end{remark} \begin{remark} We can form a similar definition for a cohomology spectral sequence. The conditions change slightly: (2') We have a \emph{decreasing} filtration \begin{align*} H^n = F^s H^n \supseteq \cdots \supseteq F^p H^n \supseteq F^{p+1} H^n \supseteq \cdots \supseteq F^t H^n = 0 .\end{align*} In this case we have \begin{align*} E_{\infty }^{p, q} \cong {F^p H^{p+q} \over F^{p+1} H^{p+q} } .\end{align*} Then each $$H_n$$ will have a filtration of length $$n+1$$ by explicitly counting terms on the diagonal, so we obtain \begin{align*} 0 = F_{-1} H_n \subset F_0 H_n \subseteq \cdots \subseteq F_{n-1} H_n \subseteq F_n H_n = H_n .\end{align*} Then \begin{align*} E_{0, n} &\cong F_0 H_n \hookrightarrow H_n\\ E_{p, n-p} &\cong {F_p H_n \over F_{p-1} H_n} \\ H_n \twoheadrightarrow E_{n, 0} &\cong {H_n \over F_{n-1} H_n} .\end{align*} \todo[inline]{See video for remarks!} \end{remark} \begin{definition}[Edge maps] Assume that $$a\geq 1$$. Provided $$a\geq 1$$, note that $$E_{0, n}^r$$ is a quotient of $$E_{0, n}^a$$ for all $$r$$, since the outgoing (?) differentials are all zero. Similarly, $$E_{n, 0}^r$$ is a subobject of $$E_{n, 0}^a$$ for all $$r$$. We thus have maps \begin{align*} E_{0, n}^a \twoheadrightarrow E_{0, n}^{\infty } \hookrightarrow H_n \\ H_n \twoheadrightarrow E_{n, 0}^{\infty } \hookrightarrow E_{n, 0}^a .\end{align*} These compositions are referred to as the \textbf{edge maps}. \begin{figure} \centering \includegraphics{figures/EdgeMaps.png} \caption{Edges of a spectral sequence} \end{figure} \end{definition} \begin{remark} For a first quadrant \emph{cohomological} spectral sequence, the edge maps are \begin{align*} E_{a}^{n, 0} \twoheadrightarrow E_{\infty }^{n, 0} \hookrightarrow H^n \\ H^n \twoheadrightarrow E_{\infty }^{0, n} \hookrightarrow E_{a}^{0, n} .\end{align*} \end{remark} \begin{definition}[Collapsing of a spectral sequence] A spectral sequence $$E$$ \textbf{collapses} at $$E^r$$ if there is exactly one nonzero row (or column) in $$E_{*, *}^r$$. \end{definition} \begin{remark} This implies that $$E_{p, q}^r = E_{p, q}^{\infty }$$ at this point. In this case, we can read off the single nonzero section: \begin{figure} \centering \includegraphics{figures/image_2021-03-17-09-55-34.png} \caption{image\_2021-03-17-09-55-34} \end{figure} Here we'll have \begin{align*} E^{\infty }_{p, q} \cong {F_p H_n \over F_{p-1} H_n} \cong {H_n \over 0}\cong H_n .\end{align*} \end{remark} \begin{remark} A more common definition of a spectral sequence \textbf{collapsing at $$r$$} is that for all $$p, q$$, the differentials $$d_{p, q}^r = 0$$. Note that this implies stabilization at $$r$$, but doesn't allow for such a simple statement about the diagonals since they may intersect multiple nonzero objects. \end{remark} \begin{remark} Some things we're skipping from the book, around the last part of 5.2: \begin{itemize} \tightlist \item Definitions pertaining to unbounded spectral sequences. \item Weak convergence. \item Filtrations that are infinite in on or both filtrations. \item Filtrations that don't limit to a union equal to $$H_n$$ or intersection to 0. \item Abutment, which is convergence when the filtration is not finite. \end{itemize} We'll skip 5.3 on the Leray spectral sequence and jump to 5.4, constructing a spectral sequence. \end{remark} \hypertarget{friday-march-19}{% \section{Friday, March 19}\label{friday-march-19}} \hypertarget{spectral-sequence-of-a-filtration}{% \subsection{Spectral Sequence of a Filtration}\label{spectral-sequence-of-a-filtration}} \begin{definition}[?] A \textbf{filtration} of a chain complex $$C$$ is an ordered family of subcomplexes \begin{align*} F\coloneqq& \cdots \subseteq F_{p-1}C \subseteq F_p C \subseteq \cdots \subseteq C && p\in {\mathbb{Z}} \end{align*} such that there are commutative diagrams \begin{center} \begin{tikzcd} {F_pC_n} && {C_n} \\ \\ {F_pC_{n-1}} && {C_{n-1}} \arrow["d", from=1-1, to=3-1] \arrow["d", from=1-3, to=3-3] \arrow[hook, from=1-1, to=1-3] \arrow[hook, from=3-1, to=3-3] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJGX3BDX24iXSxbMiwwLCJDX24iXSxbMCwyLCJGX3BDX3tuLTF9Il0sWzIsMiwiQ197bi0xfSJdLFswLDIsImQiXSxbMSwzLCJkIl0sWzAsMSwiIiwxLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbMiwzLCIiLDEseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dXQ==}{Link to Diagram} \end{quote} A filtration is \textbf{exhaustive} if $$\bigcup_{p\in {\mathbb{Z}}} F_p C_n = C_n$$ for all $$n$$. \end{definition} \begin{remark} The construction of the spectral sequence will show that $$C$$ and $$\bigcup_p F_p C$$ give rise to the same spectral sequence. So we will assume that all filtrations are exhaustive. \end{remark} \begin{theorem}[Construction of the spectral sequence of a filtration] A filtration $$F$$ of $$C\in \mathsf{Ch}({\mathsf{R}{\hbox{-}}\mathsf{Mod}})$$ determines a spectral sequence starting with \begin{align*} E_{p, q}^0 { F_p C_{p+q} \over F_{p-1} C_{p+q} } && E_{p ,q}^1 = H_{p+q}(E^0_{p, *}) .\end{align*} Since $$d$$ preserves numerators and denominators, we get well-defined differentials $$\mkern 1.5mu\overline{\mkern-1.5mud\mkern-1.5mu}\mkern 1.5mu$$ on the quotients: \begin{center} \begin{tikzcd} &&&&& \textcolor{rgb,255:red,92;green,92;blue,214}{E_{p-1, q+1}^0} \\ && {F_{p-1}C_{p+q+1}} & {} & {F_{p}C_{p+q+1}} && \textcolor{rgb,255:red,92;green,92;blue,214}{E_{p, q+1}^0} \\ &&&&&&& \ddots \\ \textcolor{rgb,255:red,92;green,92;blue,214}{F_{p-2}C_{p+q}} && \textcolor{rgb,255:red,92;green,92;blue,214}{F_{p-1}C_{p+q}} && {F_{p}C_{p+q}} && {E_{p, q}^0} \\ \\ && {F_{p-1}C_{p+q-1}} && {F_{p}C_{p+q-1}} && {E_{p, q-1}^0} \arrow[from=2-3, to=2-5] \arrow[from=2-5, to=2-7] \arrow[from=4-3, to=4-5] \arrow[from=4-5, to=4-7] \arrow[from=6-3, to=6-5] \arrow[from=6-5, to=6-7] \arrow["{\mkern 1.5mu\overline{\mkern-1.5mud\mkern-1.5mu}\mkern 1.5mu}"', from=2-7, to=4-7] \arrow["{\mkern 1.5mu\overline{\mkern-1.5mud\mkern-1.5mu}\mkern 1.5mu}"', from=4-7, to=6-7] \arrow["d"', from=2-5, to=4-5] \arrow["d"', from=4-5, to=6-5] \arrow["d"', from=2-3, to=4-3] \arrow["d"', from=4-3, to=6-3] \arrow[color={rgb,255:red,92;green,92;blue,214}, from=4-1, to=4-3] \arrow[color={rgb,255:red,92;green,92;blue,214}, from=1-6, to=2-7] \arrow[from=2-7, to=3-8] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=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}{Link to Diagram} \end{quote} Taking vertical homology of the $$E^0$$ terms on the right yields $$E_{p, q}^1$$. Note that the blue terms contribute to the same diagonal $$p+q=n$$. \end{theorem} \begin{definition}[Bounded Filtrations] A filtration $$F$$ on a chain complex $$C$$ is \textbf{bounded} if for each $$n$$ there are sn, p-1\geq n). \end{cases} \end{align*} So \(E becomes a first quadrant spectral sequence. \end{definition} \begin{remark} Note that all elements on all pages are subquotients of $$E^0$$ elements, so they can only get smaller, and terms that become 0 on some page stay 0 for all remaining pages. \end{remark} \hypertarget{construction-of-the-spectral-sequence-of-a-filtration}{% \subsection{Construction of the Spectral Sequence of a Filtration}\label{construction-of-the-spectral-sequence-of-a-filtration}} \begin{remark} For ease of notation, we'll suppress the subscript $$q$$ since it can always be recovered as $$q = n-p$$. Define the canonical quotients \begin{align*} \eta_p: F_p C \to F_p C / F_{p-1}C = E_p^0 .\end{align*} Define \begin{align*} A^r_p \coloneqq\left\{{ c\in F_p C {~\mathrel{\Big|}~}d(c) \in F_{p-r}(C) }\right\} ,\end{align*} which are elements of $$F_p C$$ which are cycles modulo $$F_{p-r} C$$, the \textbf{approximate cycles}. Note that any actual cycle is in all $$A^r$$. This differential takes things $$r$$ columns to the left, so we'll want to define a differential that associates the following terms \begin{center} \begin{tikzcd} &&& {F_{p-1}C_{n+1}} & {} & {F_{p}C_{n+1}} \\ \\ &&& \textcolor{rgb,255:red,153;green,92;blue,214}{F_{p-1}C_{n}} && {F_{p}C_{n}} & c \\ \\ \textcolor{rgb,255:red,153;green,92;blue,214}{F_{p-r}C} & \cdots && {F_{p-1}C_{n-1}} && {F_{p}C_{n-1}} & dc \arrow[hook, from=1-4, to=1-6] \arrow[hook, from=3-4, to=3-6] \arrow[hook, from=5-4, to=5-6] \arrow["d"', from=1-6, to=3-6] \arrow["d"', from=3-6, to=5-6] \arrow["d"', from=1-4, to=3-4] \arrow["d"', from=3-4, to=5-4] \arrow[hook, from=5-1, to=5-2] \arrow[maps to, from=3-7, to=5-7] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=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}{Link to Diagram} \end{quote} Similarly, define \begin{align*} Z_p^r &\coloneqq\eta_p(A_p^r \subseteq E_p^0 \\ B_p^r &\coloneqq\eta_p(d A_{p+r-1}^{r-1}) \subseteq \eta_p(F_p C) \subseteq E_p^0 .\end{align*} \end{remark} \begin{observation} Some key observations: \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \item $$F_p C = A_p^0 = A_p^{-1} = A_p^{-2} = \cdots$$ \item $$A_p^{r+1} \subseteq A_p^r$$ \item $$A_p^r \cap F_{p-1} C = A_{p-1}^{r-1}$$. \end{enumerate} \end{observation} \begin{exercise}[?] Work through these facts using the diagram above. \end{exercise} \begin{remark} Some consequences: $$(1) \implies Z_p^0 = E_p^0$$ (taking $$r=0$$ in the quotient map $$\eta_p$$). $$(2) \implies Z_p^{r+q} \subseteq Z_p^r$$, since these are images of subgroups $$(3) \implies A_{p+r-1}^{r-1} \subseteq A_{p+r}^r$$, replacing $$p\mapsto p+r$$ in the intersection formula. Then applying $$d$$ yields $$B_p^r \subseteq D_p^{r+1}$$. $$(1) \implies B_p^0 = \eta_p(d A_{p-1}^{-1}) \subseteq \eta_p(F_{p-1} C) = 0$$, since this occurs in the denominator for $$\eta_p$$ and $$d$$ preserves filtration degree. So the $$Z_p$$ get smaller and the $$B_p$$ get bigger. What happens in the middle? \end{remark} \begin{proposition}[All boundaries are contained in all cycles in a spectral sequence] $$B_p^r \subseteq Z_p^s$$ for all $$r, s\geq 0$$. \end{proposition} \begin{proof}[?] A sequence of implications: \begin{align*} B_p^r \ni x = \eta_p(dc) \text{ for some }c &\implies d(dc) = 0 \in F_{p-s}C \, \forall s \\ &\implies dc \in A_p^s \\ &\implies \eta_p(dc) \in Z_p^s .\end{align*} \end{proof} \begin{remark} Set $$B_p^{\infty } \coloneqq\cup_{r\geq 1} B_p^r \subseteq Z_p^{\infty } \coloneqq\bigcap_{s\geq 1} Z_p^s$$, which follows from a set theory exercise. \end{remark} \begin{remark} Combining and summarizing these results: for every $$p\geq 0$$, we have a tower of groups: \begin{center} \begin{tikzcd} {0 = B_p^0} & {B_p^1} & \cdots & {B_p^r} & \cdots & {B_p^\infty} & {Z_p^{\infty}} & \cdots & {Z_p^{r}} & \cdots & {Z_p^{1}} & {Z_p^{0} = E_p^0} \arrow[hook, from=1-1, to=1-2] \arrow[hook, from=1-2, to=1-3] \arrow[hook, from=1-3, to=1-4] \arrow[hook, from=1-4, to=1-5] \arrow[hook, from=1-5, to=1-6] \arrow[hook, from=1-6, to=1-7] \arrow[hook, from=1-7, to=1-8] \arrow[hook, from=1-8, to=1-9] \arrow[hook, from=1-9, to=1-10] \arrow[hook, from=1-10, to=1-11] \arrow[hook, from=1-11, to=1-12] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=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}{Link to Diagram} \end{quote} \end{remark} \begin{remark} Note that using standard isomorphism theorems, we have \begin{align*} Z_p^r \cong {A_p^r \over A_p^r \cap F_{p-1}C C} \overset{(3)}{=} {A_p^r \over A_{p-1}^{r-1}} .\end{align*} So set \begin{align*} E_p^r \coloneqq Z_p^r/B_p^r \cong {A_p^r + F_{p-1} C \over d A_{p+r-1}^{r-1} + F_{p-1}C } \cong {A_p^r \over d A_{p+r-q}^{r-1} + A_{p-1}^{r-1}} ,\end{align*} making $$E_p^r$$ a quotient of $$A_p^r$$. Using a similar calculation, one can show \begin{align*} {Z_p^{r+1} \over B_p^r} \cong { A_p^{r+1} + A_{p-1}^{r-1} \over dA_{p+r-1}^{r-1} + A_{p-1}^{r-1} } .\end{align*} \end{remark} \begin{remark} There will be an induced differential on this quotient, which will follow from checking that the different preserves the numerator and denominator. \end{remark} \hypertarget{monday-march-22}{% \section{Monday, March 22}\label{monday-march-22}} \hypertarget{spectral-sequence-of-a-filtration-1}{% \subsection{5.4: Spectral Sequence of a Filtration}\label{spectral-sequence-of-a-filtration-1}} \begin{remark} We have an increasing filtration $$F_p C \subseteq F_{p+1}C$$, where we defined \begin{align*} E_{p, q}^0 = { F_p C_{p+q} \over F_{p-1} C_{p+1} } && E_{p,q}^1 = H_{p+q} E_{p, *}^0 .\end{align*} \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \item We have a map \begin{align*} \eta_p: F_p C \twoheadrightarrow{F_p C \over F_{p-1}C } = E_p^0 ,\end{align*} where we've dropped the $$q$$ from notation. \item \begin{align*} A_{p, q}^r = \left\{{ c \in C_p C {~\mathrel{\Big|}~}dc \in F_{p-1} C }\right\} ,\end{align*} the eventual cycles. We defined $$Z_p^r = \eta_p A_p^r$$ and $$B_p^r = \eta_p dA_{p+r-1}^{r-1}$$, and wrote $$A_p^r \cap F_{p-1}C = A_{p-1}^{r-1}$$. \item We had the chain of inclusions \begin{align*} 0 = B_p^r \subseteq \cdots \subseteq B_p^{\infty} \subset Z_p^{\infty } \subset \cdots \subseteq Z_p^1 = E_p^) .\end{align*} \item We also have $$E_p^r = Z_p^r/B_p^r = A_p^r / dA_{p+r-1}^{r-1} + A_{p-1}^{r-1}$$ \item $$Z_{p}^{r+1}/B_pr \cong {A_p^{r+1} +A_{p-1}^{r-1} \over dA_{p+r-1} ^{r-1} + A_{p-1}^{r-1}}$$. \item $$dA_p^r \cap F_{p-r-1} C = dA_P^{r+1}$$. \end{enumerate} \todo[inline]{See video for missed spoken details!} Obviously we have \begin{align*} d: A_p^r &\to A_{p-r}^{r} \\ d: A_{p-1}^r &\to dA_{p-1}^{r-1} ,\end{align*} so $$d$$ induces a well-defined map $$d_p^r: E_p^r \xrightarrow{} E_{p-r}^r$$, which of course squares to zero, which goes $$r$$ columns to the left and decreases the total degree $$n$$ by 1 since the original $$d$$ did on $$C_n$$. This is what we need to set up a spectral sequence, since we now have pages and differentials, and it just remains to show that $$E^{r+1} \cong H_*(E^r, d^r)$$. \end{remark} \begin{lemma}[?] $$d$$ determines isomorphisms $$Z_{p}^r/Z_p^{r+1} \xrightarrow{\sim} B_{p-r}^{r+1} / B_{p-r}^r$$. \end{lemma} \begin{proof}[?] Unwind definitions! Note that we have $$B_{p-r}^{r+1} = \eta_{p-r} dA_p^r$$, using that the lower index on $$B$$ and upper index on $$A$$ should sum to the lower index on $$A$$. This is equal to $$dA_p^r / dA_p^r \cap F_{p-r-1} C$$, where the latter term is $$\ker\eta_{p-r}$$ and $$B_{p-r}^r = \eta_{p-r} dA_{p-1}^{r-1}$$. This yields \begin{align*} {B^{r+1}_{p-r} \over B_{p-r}^r} \cong { dA_{p}^r \over dA_{p-1}^{r-1} + (dA_p^r \cap F_{p-r-1} C) } .\end{align*} Similarly, \begin{align*} {Z_p^r \over Z_p^{r+1}} \coloneqq{ \eta_p A_p^r \over \eta_p A_p^{r+1} } \cong {A_p^r \over A_p^{r+1} + (A_p^r \cap F_{p-1} C )} \overset{(3)}{\cong} {A_p^r \over A_p^{r+1} + A_{p-1}^{r-1} } .\end{align*} Now applying the map induced by $$d: A_p^r \to F_{p-r}C$$ to this quotient, we have $$\ker { \left.{{d}} \right|_{{A_p^r}} } \subseteq A_p^{r+1}$$. These go down $$r$$ steps, but everything in the kernel goes down as far as you'd like! So $$d$$ kills one of the denominator terms, and thus induces an injective map on the quotient. Thus $${Z_p^r \over Z_p^{r+1}} \xrightarrow{\sim} {dA_p^r \over dA_p^{r+1} + dA_{p-1}^{r-1} }$$, which is exactly the previous expression with the order switched, so this is isomorphic to $$B_{p-r}^{r+1} / B_{p-r}^r$$. \end{proof} \begin{proposition}[Ther+1$st page is the homology of the$rth page] \begin{align*} { \ker d_p^r \over \operatorname{im}d_{p+r}^r } \cong E_p^{r+1} \coloneqq{Z_p^{r+1} \over B_p^{r+1} } .\end{align*} \end{proposition} \begin{proof}[?] Recall that $$d_p^r: E_p^r \to E_{p-r}^r$$ and by (4), $$E_p^r \cong {A_p^r \over dA_{p+r-1}^{r-1} + A_{p-1}^{r-1}}$$. Substituting $$p \mapsfrom p-r$$, we have \begin{align*} \ker d_p^r = { \left\{{ z\in A_p^r {~\mathrel{\Big|}~}dz \in dA_{p-1}^{r-1} + A_{p-r-1}^{r-1} }\right\} \over dA_{p+r-1}^{r-1} + A_{p-1}^{r-1} } = { A_{p-1}^{r-1} + A_{p}^{r+1} \over dA_{p+r-1}^{r-1} + A_{p-1}^{r-1} } \overset{(5)}{\cong} {Z_p^{r+1} \over B_p^r} && \text{which is } (6) .\end{align*} Here we've used that $$x\in F_p C\implies dx \in F_{p-r-1} C \implies dx\in A^{?}_{p-r-1}$$. What is the image of $$d_p^r$$ in general? Note that later we can replace $$p\mapsfrom p+r$$. By the 1st isomorphism theorem, we have \begin{align*} d_p^r: E_p^r = Z_p^r / B_p^r \xrightarrow{\sim} {Z_p^r / B_p^r \over Z_p^{r+1} / B_p^r} \xrightarrow{\sim} {Z_p^r \over Z_p^{r+1}} \xrightarrow{d} {B_{p-r}^{r+1} \over B_{p-r}^r} \hookrightarrow{Z_{p-r}^r \over B_{p-r}^r} = E_{p-r}^r ,\end{align*} where we've applied the lemma from last time, and we've used the fact that in the last map, all of the $$B$$ are contained in all of the $$Z$$, so we can choose any superscript we want. These are all isomorphisms up until the last part, so \begin{align*} \operatorname{im}d_p^r \cong B_{p-r}^{r+1} / B_{p-r}^{r+1} .\end{align*} . Replacing $$p\mapsfrom p+r$$, we get a 7th fact \begin{fact}[7] \begin{align*} \operatorname{im}d_{p+r}^r \cong B_{p}^{r+1} / B_{p}^{r+1} .\end{align*} \end{fact} Now combining (6) and (7), we have \begin{align*} {\ker d_p^r \over \operatorname{im}d_{p+r}^{r} } \xrightarrow{\sim} {Z_p^{r+1} / B_p^r \over B_{p}^{r+1} / B_p^r } \cong {Z_p^{r+1} \over B_p^{r+1}} = E_p^{r+1} .\end{align*} \end{proof} \hypertarget{convergence-of-the-spectral-sequence-of-a-filtration}{% \subsection{5.5: Convergence of the Spectral Sequence of a Filtration}\label{convergence-of-the-spectral-sequence-of-a-filtration}} \begin{remark} We'll restrict our attention to bounded complexes. \end{remark} \begin{remark} A filtration $$F$$ on a chain complex $$C$$ induces a filtration on the homology $$H_*C$$, where $$H_p H_n C = \operatorname{im}( H_n F_p C \to H_n C)$$: \begin{center} \begin{tikzcd} {F_p C_{n+1}} && {C_{n+1}} \\ \\ {F_p C_{n}} && {C_{n}} \\ \\ {F_p C_{n-1}} && {C_{n-1}} \arrow[hook, from=1-1, to=1-3] \arrow[hook, from=3-1, to=3-3] \arrow[hook, from=5-1, to=5-3] \arrow["d"{description}, from=1-1, to=3-1] \arrow["d"{description}, from=3-1, to=5-1] \arrow["d"{description}, from=1-3, to=3-3] \arrow["d"{description}, from=3-3, to=5-3] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsNixbMCwwLCJGX3AgQ197bisxfSJdLFsyLDAsIkNfe24rMX0iXSxbMiwyLCJDX3tufSJdLFsyLDQsIkNfe24tMX0iXSxbMCwyLCJGX3AgQ197bn0iXSxbMCw0LCJGX3AgQ197bi0xfSJdLFswLDEsIiIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzQsMiwiIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbNSwzLCIiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFswLDQsImQiLDFdLFs0LDUsImQiLDFdLFsxLDIsImQiLDFdLFsyLDMsImQiLDFdXQ==}{Link to Diagram} \end{quote} \todo[inline]{See video for missed details.} These inclusions induce a map from the homology of the subcomplex to the homology of the total complex. \end{remark} \begin{remark} If the filtration on $$C$$ is bounded, say $$0 = F_s C_n \subseteq \cdots \subseteq F_t C_n = C_n$$ for some $$s0} dA_{p+r}^r \subseteq \ker \eta_p = F_{p-1} C$$. \end{proof} \hypertarget{applications-two-spectral-sequences-of-a-double-complex}{% \subsection{Applications: Two Spectral Sequences of a Double Complex}\label{applications-two-spectral-sequences-of-a-double-complex}} \begin{remark} Consider two different filtrations of the total complex $$\operatorname{Tot}(C)$$ (either sum or product) of a double complex $$C_{*, *}$$. We know there is an spectral sequence associated to each and play them off of each other to get extra information about cohomology. \end{remark} \begin{definition}[Filtration I: by columns (of a double complex)] Let $${}^IF_n \operatorname{Tot}(C)$$ be the total subcomplex obtain by applying truncation functors: \begin{align*} \qty{{}^I \tau_{\leq n} C}_{p, q} \coloneqq \begin{cases} C_{p, q} & p \leq n \\ 0 & p > n. \end{cases} .\end{align*} \begin{center} \begin{tikzcd} \ddots & \vdots & \vdots & {} & \vdots & \vdots \\ \cdots & \bullet & \bullet && 0 & 0 & \cdots \\ \cdots & \bullet & \bullet && 0 & 0 & \cdots \\ \cdots & \bullet & \bullet && 0 & 0 & \cdots \\ \bullet & \vdots & \vdots & {} & \vdots & \vdots & \bullet \\ && n & \bullet & {n+1} \arrow[no head, from=5-1, to=5-7] \arrow[no head, from=1-4, to=6-4] \arrow[from=2-3, to=2-2] \arrow[from=2-3, to=3-3] \arrow[from=3-3, to=3-2] \arrow[from=3-3, to=4-3] \arrow[from=4-3, to=4-2] \arrow[color={rgb,255:red,92;green,92;blue,214}, dotted, no head, from=1-1, to=5-5] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=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}{Link to Diagram} \end{quote} We still have $$d = d^v + d^h: {}^IF_n \to {}^I F_n$$. By the construction theorem, there is a spectral sequence $$\left\{{ {}^I E_{p,q}^r}\right\}$$ starting with $${}^I E_{p, q}^0 = C_{p, q}$$ and \begin{align*} {}^I E_{p, q}^0 = { F_p \operatorname{Tot}(C)_{p+q} \over F_{p-1} \operatorname{Tot}(C)_{p+q}} .\end{align*} \begin{center} \begin{tikzcd} & \bullet &&&& 0 \\ && \bullet &&& 0 \\ &&& {\bullet F_{p-1} } && 0 \\ q &&&& \textcolor{rgb,255:red,92;green,92;blue,214}{\bullet (F_p) = C_{p, q}} & 0 \\ &&&&& 0 \\ &&& {} & {p-1} & p \arrow[color={rgb,255:red,92;green,92;blue,214}, dashed, no head, from=1-2, to=5-6] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsMTMsWzUsMCwiMCJdLFs1LDEsIjAiXSxbNSwyLCIwIl0sWzUsMywiMCJdLFs1LDQsIjAiXSxbNCwzLCJcXGJ1bGxldCAoRl9wKSA9IENfe3AsIHF9IixbMjQwLDYwLDYwLDFdXSxbMywyLCJcXGJ1bGxldCBGX3twLTF9ICJdLFsyLDEsIlxcYnVsbGV0Il0sWzMsNV0sWzQsNSwicC0xIl0sWzUsNSwicCJdLFsxLDAsIlxcYnVsbGV0Il0sWzAsMywicSJdLFsxMSw0LCIiLDAseyJjb2xvdXIiOlsyNDAsNjAsNjBdLCJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifSwiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dXQ==}{Link to Diagram} \end{quote} Recall that $$d_p^r: E_p^r \to E_{p-r}^{r}$$ (going $$r$$ columns to the left, where we've suppressed $$q$$) is the map induced from $$d: \operatorname{Tot}(C)_n \to \operatorname{Tot}(C)_{n-1}$$. So for $$r=0$$, we have $$d_{p, q}^0: E_{p, q}^0 \to E_{p, q-1}^0$$. But the left-hand side is $$C_{p, q}$$ and the right-hand side is $$C_{p, q-1}$$, so it's perhaps not surprising that this coincides with the original $$d^v$$ from $$C_{*, *}$$. Thus $${}^IE_{pq}^1= H_q^v(C_{p, *})$$ by taking homology in the vertical direction. For the differential, we want $$d_{pq}^1: E_{pq}^1\to E_{p-1, q}^1$$, and these will just be the maps induced on the vertical homology by $$d^h$$. So we write $${}^I E_{p, q}^2 = H_p^h H_q^v (C_{**})$$. If $$C$$ is a first quadrant complex, the filtration is canonically bounded since $$F_{-1} \operatorname{Tot}(C) = 0$$ and $$F_n \operatorname{Tot}(C)_n = \operatorname{Tot}(C)_n$$. So we get the spectral sequence that we started constructing in section 5.1, and we now know it converges to $$H_* \operatorname{Tot}(C)$$ by the classical convergence theorem. So \begin{align*} {}^I E_{p, q}^2 = H_p^h H_q^v(C) \Rightarrow H_{p+q} \operatorname{Tot}(C) .\end{align*} \end{definition} \begin{remark} We can say something about the unbounded case. Suppose $$C$$ is 4th quadrant, then $$F_{-1} \operatorname{Tot}(C) = 0$$, so the first filtration $${}^I F$$ is bounded below. The diagonals are infinite, so we take $$\operatorname{Tot}(C) \coloneqq\operatorname{Tot}^{\oplus}(C)$$. Every element of $$(\operatorname{Tot}(C))_n$$ lives in $$\bigoplus _{p=0}^N C_{p, n-p}$$ for some finite $$N$$ and the filtration is exhaustive, i.e.~$$\operatorname{Tot}^{\oplus}C = \bigcup_{p\geq 0} F_p \operatorname{Tot}^{\oplus}C$$. A version of the classical convergence theorem will yield \begin{align*} {}^I E_{pq}^r \Rightarrow H_{p+q} \operatorname{Tot}^{\oplus}C .\end{align*} However, this will not hold for $$\operatorname{Tot}^{\Pi}$$. \end{remark} \begin{remark} Next time: a second filtration and its spectral sequence, and how to play them off of each other. \end{remark} \hypertarget{friday-march-26}{% \section{Friday, March 26}\label{friday-march-26}} \hypertarget{two-spectral-sequences-on-total-complexes}{% \subsection{5.6: Two Spectral Sequences on Total Complexes}\label{two-spectral-sequences-on-total-complexes}} \begin{remark} Recall that we had two filtrations on a total complex: the first was fixing a vertical line and replacing everything to the right with zeros, which was given by $$^{I}E_{p, q}^0 = F_p(\operatorname{Tot})/ F_{p-1}(\operatorname{Tot}) = C_{p, q}$$. Taking homology with the vertical differentials yielded $$^{I}E_{p, q}^1 = H_q^v(C_{p,*})$$, and $$^{I} E_{p, q}^2 = H_p^h H_q^v(C_{*, *})$$. Applying the classical convergence theorem when this is 1st quadrant yields some spectral sequence with these as the pages which converges to $$H_{p+q}(\operatorname{Tot}(C))$$. \end{remark} \begin{definition}[The second filtration] We'll define a filtration by rows: let $$^{II}F_n \operatorname{Tot}(C)$$ be the total complex of the double complex \begin{align*} ({}^{II} \tau_{\leq n} C)_{p, q} &= \begin{cases} C_{p, q} & p, q\leq \\ 0 & p, q > n. \end{cases} \end{align*} This is the complex gotten by replacing everything below the $$n$$th row with zeros. We define the 0th page \begin{align*} {}^{II} E^{0}_{p, q} = { {}^{II} F_p \operatorname{Tot}(C)_{p+q} \over {}^{II} F_{p-1} \operatorname{Tot}(C)_{p+q} } = C_{q, p} ,\end{align*} which follows from the fact that we are modding out a full diagonal by a diagonal with one fewer elements: \begin{center} \begin{tikzcd} & \ddots & \vdots & \vdots & \vdots & \vdots & \ddots \\ & \cdots & 0 & 0 & 0 & 0 & \cdots \\ p &&&&&& {} & {} \\ & \cdots & \textcolor{rgb,255:red,214;green,92;blue,92}{\bullet} & \bullet & \bullet & \bullet & \cdots \\ & \cdots & \bullet & \textcolor{rgb,255:red,92;green,92;blue,214}{\bullet} & \bullet & \bullet & \cdots \\ & \cdots & \bullet & \bullet & \bullet & \bullet & \cdots \\ & \cdots & \bullet & \bullet & \bullet & \bullet & \ddots \\ & \ddots & \vdots & \vdots & \vdots & \ddots & \ddots \\ && q \arrow["{F_p}"', shift right=3, color={rgb,255:red,214;green,92;blue,92}, squiggly, no head, from=7-6, to=4-3] \arrow["{F_{p-1}}", shift left=3, color={rgb,255:red,92;green,92;blue,214}, squiggly, no head, from=7-6, to=5-4] \arrow[dashed, no head, from=3-1, to=3-8] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=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}{Link to Diagram} \end{quote} \end{definition} \begin{warnings} Note the switched order! \end{warnings} \begin{remark} Note that the differential is \begin{align*} d^0: E^0_{p, q} &\to E^0_{p, q-1} \\ = d^h: C_{q, p} &\to C_{q-1, p} .\end{align*} We similarly have $${}^II E_{p, q}^I = H_q^h(C_{*, p})$$, again noting the switched indices, with differential \begin{align*} d^1: E^1_{p, q} &\to E_{p-1, q}^1 \\ =H^h(C_{q, p}) &\to H^h(C_{*, p-1}) \end{align*} which comes from the original differential inducing a map on horizontal homology. Then $${}^{II} E^2_{p, q} = H_p^v H_q^h(C)$$. \end{remark} \begin{remark} Note that transposing everything about the line $$p=q$$ interchanges filtrations $$I$$ and $$II$$, and thus the two spectral sequences $${}^{I}E_{p, q} \rightleftharpoons{}^{II} E_{q, p}$$. Using that first quadrant sequences are canonically bounded, we can apply the classical convergence theorem to $${}^{II} E$$ to obtain \begin{align*} {}^{II}E_{p, q}^2 \Rightarrow H_{p+q}( \operatorname{Tot}(C) ) .\end{align*} Transposing sends $$QIV$$ to $$QII$$ and thus $${}^{II} E \Rightarrow H_{p+q}\operatorname{Tot}^{\oplus}(C)$$. Note that this does not guarantee anything about $$\operatorname{Tot}^{\Pi}(C)$$. \end{remark} \begin{remark} In particular, if we have a $$QI$$ double complex, both filtrations converge to the homology of the total complex. \end{remark} \hypertarget{application-balancing-tor}{% \subsection{Application: Balancing Tor}\label{application-balancing-tor}} \begin{remark} Our proof in 2.7 that $$\operatorname{Tor}_*^R(A, B)$$ could be computed either by a projective resolution $${P}_{*}\twoheadrightarrow A$$ or a projective resolution $${Q}_{*}\twoheadrightarrow B$$ was a disguised spectral sequence argument. So we'll go recover it using the actual spectral sequence. \end{remark} \begin{remark} We have a $$QI$$ double complex $$C$$ given by $$C_{p, q} \coloneqq(P\otimes Q)_{p, q} = P_p\otimes Q_q$$, and we now have two spectral sequences converging to $$H_*(\operatorname{Tot}(P\otimes Q))$$. Taking the first filtration, we can write \begin{align*} H_q^v(\operatorname{Tot}(C)) = H_q(P_p \otimes Q_q) = P_p \otimes H_q(Q) .\end{align*} Using that $$P$$ is an exact complex, and noting that we delete the augmentation when taking homology, we have \begin{align*} H_1^v(\operatorname{Tot}(C)) = \begin{cases} 0 & q>0 \\ P_p\otimes B & q=0. \end{cases} \end{align*} Thus \begin{align*} E^2_{p, q} = \begin{cases} H_p^h(P_* \otimes B) & q=0 \\ 0 & 1>0, \end{cases} \end{align*} meaning that this collapses at $$E^2$$ and we have \begin{align*} H_p (\operatorname{Tot}(P\otimes Q) ) \cong L_p({-}\otimes B)(A) \coloneqq\operatorname{Tor}^R_p(A, B) .\end{align*} Now consider taking the second filtration, which yields \begin{align*} {}^{II} E_{p, q}^1 = H_q^h( P_q \otimes Q_p) = H_q(P_*) \otimes Q_p = \begin{cases} A_\otimes Q_p & q=0 \\ 0 & q>0. \end{cases} \end{align*} The second pages comes from taking the vertical homology, so \begin{align*} {}^{II}E_{p, q}^2 = H_p^v H_q^h(P_q \otimes Q_p) = \begin{cases} H^v_p(A\otimes Q) & q=0 \\ 0 & q>0. \end{cases} ,\end{align*} which is $$L_p(A\otimes{-})(B)$$ in $$q=0$$. Since $${}^{II}E_{p, q}^2 \Rightarrow H_{p+q}(\operatorname{Tot}(P\otimes Q)) = L_p({-}\otimes B)(A)$$, and we thus have \begin{align*} L_p(A\otimes{-})(B) \cong L_p({-}\otimes B)(A) .\end{align*} \end{remark} \begin{remark} See the this section of Weibel for other applications in the exercises: the Kunneth formula, the Universal Coefficient Theorem, and the Acyclic Assembly Lemma. \end{remark} \hypertarget{hypercohomology}{% \subsection{Hypercohomology}\label{hypercohomology}} \begin{remark} We'd like to compute derived functors acting on chain complexes instead of just objects. \end{remark} \begin{definition}[Cartan-Eilenberg Resolutions] Let $$\mathcal{A}$$ be an abelian category with enough projectives and let $${A}_{*} \in \mathsf{Ch}(\mathcal{A})$$. A (left) \textbf{Cartan-Eilenberg resolution} (a CE resolution) $$P_{*, *}$$ of $$A_*$$ is an upper half-plane complex (so $$P_{p, q} = 0$$ when $$q<0$$) of projective objects and an augmentation chain map $$P_{*, 0} \xrightarrow{\varepsilon} A_*$$ such that \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \item If $$A_p=0$$ then the entire column $$P_{p, *}$$ is zero. \item The augmentation induces maps on boundaries and in homology which are projective resolutions in $$\mathcal{A}$$: \begin{align*} B_p(P, d^h) &\xrightarrow{B_p(\varepsilon)} B_p(A) \\ H_p(P, d^h) &\xrightarrow{H_p(\varepsilon)} H_p(A) .\end{align*} \end{enumerate} \end{definition} \begin{remark} So we have the following situation \begin{center} \begin{tikzcd} {q:} & \cdots && {P_{p+1, q}} && {P_{p, q}} && {P_{p-1, q}} && \cdots \\ &&& \vdots && \vdots && \vdots \\ & \cdots && {P_{p+1, 1}} && {P_{p, 1}} & {} & {P_{p-1, 1}} && \cdots \\ \\ & \cdots && {P_{p+1, 0}} && {P_{p, 0}} && {P_{p-1, 0}} && \cdots \\ {} &&&&&&&&&& {} \\ & \cdots && {A_{p-1}} && {A_{p}} && {A_{p-1}} && \cdots \arrow[from=3-8, to=3-6] \arrow[from=7-8, to=7-6] \arrow[from=7-6, to=7-4] \arrow[from=5-8, to=5-6] \arrow[from=5-6, to=5-4] \arrow[from=3-6, to=3-4] \arrow[from=3-4, to=5-4] \arrow[from=5-4, to=7-4] \arrow[from=3-6, to=5-6] \arrow[from=5-6, to=7-6] \arrow[from=3-8, to=5-8] \arrow[from=5-8, to=7-8] \arrow[from=3-10, to=3-8] \arrow[from=5-10, to=5-8] \arrow[from=7-10, to=7-8] \arrow[from=7-4, to=7-2] \arrow[from=5-4, to=5-2] \arrow[from=3-4, to=3-2] \arrow[from=1-10, to=1-8] \arrow[from=1-8, to=1-6] \arrow[from=1-6, to=1-4] \arrow[from=1-4, to=1-2] \arrow[color={rgb,255:red,92;green,92;blue,214}, dotted, no head, from=6-1, to=6-11] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=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}{Link to Diagram} \end{quote} The situation in row $$q$$ will be: \begin{center} \begin{tikzcd} \cdots && {P_{p+1, q}} && {P_{p, q}} && {P_{p-1, q}} && \cdots \\ \\ &&&& {Z_p(P, d^h)} \\ &&&&&& {H_p(P, d^h)_q} \\ &&&& {B_p(P, d^h)} \arrow[from=1-9, to=1-7] \arrow[from=1-7, to=1-5] \arrow[from=1-5, to=1-3] \arrow[from=1-3, to=1-1] \arrow[hook, from=3-5, to=1-5] \arrow[hook, from=5-5, to=3-5] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsOCxbNiwwLCJQX3twLTEsIHF9Il0sWzQsMCwiUF97cCwgcX0iXSxbMiwwLCJQX3twKzEsIHF9Il0sWzgsMCwiXFxjZG90cyJdLFswLDAsIlxcY2RvdHMiXSxbNCwyLCJaX3AoUCwgZF5oKSJdLFs0LDQsIkJfcChQLCBkXmgpIl0sWzYsMywiSF9wKFAsIGReaClfcSJdLFszLDBdLFswLDFdLFsxLDJdLFsyLDRdLFs1LDEsIiIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzYsNSwiIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XV0=}{Link to Diagram} \end{quote} Here when we take the homology of the complex along the rows $$p$$, we'll obtain \begin{align*} H_q(P, d^h) = {Z_p(P, d^h)_q \over B_p(P, d^h)_q} ,\end{align*} and since the induces maps preserve cycles and boundaries, we get induced maps on homology. Exercise 5.7.1 shows that $$P_{p, *} \xrightarrow{\varepsilon} A_p$$ will be a projective resolution in $$\mathcal{A}$$ and so $$Z_p(P, d^h)_* \to Z_p(A)$$. \end{remark} \begin{lemma}[?] Every $$A_*$$ has a CE resolution $$P_{*, *} \xrightarrow{\varepsilon} A$$. \end{lemma} \begin{proof}[?] Choose a levelwise resolution and use the horseshoe lemma: \begin{center} \begin{tikzcd} 0 && {B_p(A)} && {Z_p(A)} && {H_p(A)} && 0 \\ \\ 0 && {P^B_{p, *}} && \textcolor{rgb,255:red,92;green,92;blue,214}{P^Z_{p, *}} && {P^H_{p, *}} && 0 \arrow[from=1-1, to=1-3] \arrow[from=1-3, to=1-5] \arrow[from=1-5, to=1-7] \arrow[from=1-7, to=1-9] \arrow[from=3-3, to=1-3] \arrow[from=3-7, to=1-7] \arrow[from=3-1, to=3-3] \arrow[draw={rgb,255:red,92;green,92;blue,214}, dashed, from=3-3, to=3-5] \arrow[draw={rgb,255:red,92;green,92;blue,214}, dashed, from=3-5, to=3-7] \arrow[from=3-7, to=3-9] \arrow[draw={rgb,255:red,92;green,92;blue,214}, dashed, from=3-5, to=1-5] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=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}{Link to Diagram} \end{quote} Recall that this involved a direct sum construction. Now do a similar thing for the following SES: \begin{center} \begin{tikzcd} 0 && {Z_p(A)} && {A_p} && {B_{p-1}(A)} && 0 \\ \\ 0 && {P^Z_{p, *}} && \textcolor{rgb,255:red,92;green,92;blue,214}{P^A_{p, *}} && {P^B_{p-1, *}} && 0 \arrow[from=1-1, to=1-3] \arrow[from=1-3, to=1-5] \arrow["{d_p}", from=1-5, to=1-7] \arrow[from=1-7, to=1-9] \arrow[from=3-3, to=1-3] \arrow[from=3-7, to=1-7] \arrow[from=3-1, to=3-3] \arrow[draw={rgb,255:red,92;green,92;blue,214}, dashed, from=3-3, to=3-5] \arrow["{\tilde{d_p}}", draw={rgb,255:red,92;green,92;blue,214}, dashed, from=3-5, to=3-7] \arrow[from=3-7, to=3-9] \arrow[draw={rgb,255:red,92;green,92;blue,214}, dashed, from=3-5, to=1-5] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsMTAsWzAsMCwiMCJdLFsyLDAsIlpfcChBKSJdLFs0LDAsIkFfcCJdLFs2LDAsIkJfe3AtMX0oQSkiXSxbOCwwLCIwIl0sWzYsMiwiUF5CX3twLTEsICp9Il0sWzQsMiwiUF5BX3twLCAqfSIsWzI0MCw2MCw2MCwxXV0sWzIsMiwiUF5aX3twLCAqfSJdLFswLDIsIjAiXSxbOCwyLCIwIl0sWzAsMV0sWzEsMl0sWzIsMywiZF9wIl0sWzMsNF0sWzcsMV0sWzUsM10sWzgsN10sWzcsNiwiIiwxLHsiY29sb3VyIjpbMjQwLDYwLDYwXSwic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV0sWzYsNSwiXFx0aWxkZXtkX3B9IiwwLHsiY29sb3VyIjpbMjQwLDYwLDYwXSwic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV0sWzUsOV0sWzYsMiwiIiwxLHsiY29sb3VyIjpbMjQwLDYwLDYwXSwic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV1d}{Link to Diagram} \end{quote} We use the fact that we have the two side resolutions from the previous step. So set $$P_{p, q} \coloneqq P_{p, q}^A$$ assembled into a double complex using the sign trick: $$d^v \coloneqq(-1)^p d$$ where we used the differential $$d$$ from $$P_{p, *}^A$$. We can now define \begin{align*} d^h: P^A_{p+1, *} \xrightarrow{\tilde d_{p+1} } P_{p, *}^B \hookrightarrow P_{p, *}^Z \hookrightarrow P_{p, *}^A .\end{align*} One then checks that $$B_p(\varepsilon)$$ and $$H_p(\varepsilon)$$ are indeed projective resolutions. \end{proof} \hypertarget{monday-march-29}{% \section{Monday, March 29}\label{monday-march-29}} \hypertarget{maps-of-double-complexes}{% \subsection{Maps of Double Complexes}\label{maps-of-double-complexes}} \begin{remark} Last time: we talked about hypercohomology. We're doing this so we can set up a Grothendieck spectral sequence. \end{remark} \begin{definition}[Chain homotopies of double complexes] Let $$f, g:D\to E$$ be two maps between double complexes. A \textbf{chain homotopy} from $$f$$ to $$g$$ consists of $$s_{p, q}^h: D_{p, q} \to E_{p+1, q}$$ and $$s_{p, q}^v: D_{p, q} \to E_{p, q+1}$$ for all $$p, q$$ satisfying the following conditions: \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \tightlist \item All of the possible maps $$D_{p, q} \to E_{p, q}$$ summed should be equal to $$g-f$$, i.e.~$$g-f = (d^h s^h + s^h d^h) + (d^v s^v + s^v d^v)$$: \end{enumerate} \begin{center} \begin{tikzcd} &&&& {E_{p, q+1}} \\ \\ &&&& {E_{p, q}} && {E_{p+1, q}} \\ \\ {D_{p-1, q}} && {D_{p, q}} \\ \\ && {D_{p, q-1}} \arrow["{d^v}"{description}, draw={rgb,255:red,16;green,178;blue,32}, from=1-5, to=3-5] \arrow["{d^v}"{description}, draw={rgb,255:red,16;green,178;blue,32}, from=5-3, to=7-3] \arrow["{s_{p-1, q}^h}"{description, pos=0.2}, color={rgb,255:red,149;green,68;blue,55}, dashed, from=5-1, to=3-5] \arrow["{s^v_{p, q-1}}"{description, pos=0.9}, color={rgb,255:red,16;green,178;blue,32}, dashed, from=7-3, to=3-5] \arrow["{d^h}"{description}, draw={rgb,255:red,149;green,68;blue,55}, from=5-3, to=5-1] \arrow["{d^h}"{description}, draw={rgb,255:red,149;green,68;blue,55}, from=3-7, to=3-5] \arrow["{s_{p, q}^h}"{description, pos=0.2}, color={rgb,255:red,149;green,68;blue,55}, dashed, from=5-3, to=3-7] \arrow["{s^v_{p, q}}"{description, pos=0.9}, color={rgb,255:red,16;green,178;blue,32}, dashed, from=5-3, to=1-5] \arrow["{g-f}"{description}, curve={height=-30pt}, from=5-3, to=3-5] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=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}{Link to Diagram} \end{quote} \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \setcounter{enumi}{1} \tightlist \item The two rectangles below should be zero, i.e.~$$s^v d^h + d^h s^v = 0 = s^h d^v + d^v s^h$$: \end{enumerate} \begin{center} \begin{tikzcd} && {E_{p-1, q+1}} && {E_{p, q+1}} \\ \\ &&&& {E_{p, q}} && {E_{p+1, q}} \\ \\ {D_{p-1, q}} && {D_{p, q}} &&&& {E_{p+1, q-1}} \\ \\ && {D_{p, q-1}} \arrow["{d^v}", from=1-5, to=3-5] \arrow["{d^v}", color={rgb,255:red,171;green,43;blue,43}, from=5-3, to=7-3] \arrow["{d^h}"', color={rgb,255:red,38;green,151;blue,38}, from=5-3, to=5-1] \arrow["{d^h}"', from=3-7, to=3-5] \arrow["{g-f}"{description}, from=5-3, to=3-5] \arrow["{d^v}"{description}, color={rgb,255:red,171;green,43;blue,43}, from=3-7, to=5-7] \arrow["{s^h_{p, q-1}}"', color={rgb,255:red,171;green,43;blue,43}, from=7-3, to=5-7] \arrow["{s^h_{p, q}}"', color={rgb,255:red,171;green,43;blue,43}, from=5-3, to=3-7] \arrow["{s^v_{p-1, q}}", color={rgb,255:red,38;green,151;blue,38}, from=5-1, to=1-3] \arrow["{d^h}"{description}, color={rgb,255:red,38;green,151;blue,38}, from=1-5, to=1-3] \arrow["{s^v_{p, q}}", color={rgb,255:red,38;green,151;blue,38}, from=5-3, to=1-5] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=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}{Link to Diagram} \end{quote} \end{definition} \begin{remark} The definition is set up so that $$s^h + s^v: \operatorname{Tot}(D)_n \to \operatorname{Tot}(E)_{n+1}$$ is a chain homotopy $$\operatorname{Tot}^{\oplus}(D) \to \operatorname{Tot}^{\oplus}(E)$$. \end{remark} \begin{remark} Exercises 5.7.2 and 5.7.3 show: \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \item If $$f:A\to B$$ is a chain map and $$P\to A, Q\to B$$ are CE resolutions, then there is a map of double complexes $$\tilde f: P\to Q$$ lifting $$f$$. \item If $$f, g: A\to B$$ are chain homotopic, then $$\tilde f, \tilde g$$ are chain homotopic in the sense just defined. \item Any two CE resolutions $$P, P'$$ of $$A$$ are chain homotopy equivalent, as are $$\operatorname{Tot}^{\oplus}(F(P))$$ and $$\operatorname{Tot}^{\oplus}(F(P'))$$ for any additive functor $$F$$. \end{enumerate} \end{remark} \begin{remark} This last remark shouldn't be too hard to believe: chain homotopies are defined in terms of addition. \end{remark} \hypertarget{hypercohomology-1}{% \subsection{Hypercohomology}\label{hypercohomology-1}} \begin{definition}[Hyper Left-Derived Functors] Let $$F : \mathcal{A} \to \mathcal{B}$$ be a right-exact functor where $$\mathcal{A}$$ has enough projectives and $$\mathcal{B}$$ is cocomplete (closed under direct sums/coproducts). If $$A \in \mathsf{Ch}(\mathcal{A})$$ is a chain complex and $$P\to A$$ a CE resolution, define \begin{align*} {\mathbb{L}}_i F(A) \coloneqq H_i \operatorname{Tot}^{\oplus}F(P): \mathsf{Ch}(\mathcal{A}) \to \mathcal{B} .\end{align*} If $$f:A\to B$$ is a chain map in $$\mathsf{Ch}(\mathcal{A})$$ and $$\tilde f: P\to Q$$ where $$P, Q$$ are CE resolutions of $$A, B$$ resp., define $$L_iF(f)$$ to be the map \begin{align*} H_i \operatorname{Tot}(F\tilde f) \to {\mathbb{L}}_i F(B) .\end{align*} This yields a functor \begin{align*} {\mathbb{L}}_i F: \mathsf{Ch}(\mathcal{A}) \to \mathcal{B} ,\end{align*} the \textbf{hyper left-derived functor} of $$F$$. \end{definition} \begin{remark} Recall that chain homotopy yields a notion of equivalence, and chain homotopic maps induce the same map on homology. The same is true for double complexes. There is a lemma that shows a SES of double complexes induces a LES in homology. \end{remark} \begin{proposition}[Convergence of spectral sequences and filtration comparison] \envlist \begin{enumerate} \def\labelenumi{\alph{enumi}.} \item There is always a convergent spectral sequence \begin{align*} {}^{II} E^2_{p, q} (L_p F)(H_q(A)) \Rightarrow{\mathbb{L}}_{p+q} F(A) .\end{align*} \item If $$A$$ is bounded below complex, so there exists a $$p_0$$ such that $$A_p=0$$ for $$p< p_0$$, then there is another spectral sequence \begin{align*} {}^{I} E_{p, q}^2 = H_p L_q F(A) \Rightarrow{\mathbb{L}}_{p+q} F(A) .\end{align*} \end{enumerate} \end{proposition} \begin{proof}[of (a)] These are the spectral sequences associated to the upper half-plane double complex $$FP_{*, *}$$. Recall that $${}^{II} E^2_{p, q} = H_p^v H_q^h (FP) \Rightarrow H_{p+q} \operatorname{Tot}^{\oplus}FP \coloneqq{\mathbb{L}}_{p+q} F(A)$$. The filtration by rows is exhaustive since we are taking the direct sum, so any cycle or boundary is supported in some finite row. So what we want to show is that \begin{align*} {}^{II}E_{p, q}^2 (L_p F)(H_q A) = H_p^v H_q^h (FP) .\end{align*} The main claim is the following: $$H_q^h(FP) = F H_q^h(P)$$. Fix a row $$p$$ of the double complex so we can drop $$p$$ and $$h$$ from the notation. We have the following situation: \begin{center} \begin{tikzcd} \cdots && {P_{q-1}} && {P_{q}} && {P_{q+1}} && \cdots \\ \\ &&&& {Z_q} \\ \\ &&&& {B_q} \arrow[hook, from=5-5, to=3-5] \arrow[hook, from=3-5, to=1-5] \arrow["d"', from=1-9, to=1-7] \arrow["d"', from=1-7, to=1-5] \arrow["d"', from=1-5, to=1-3] \arrow["d"', from=1-3, to=1-1] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsNyxbMiwwLCJQX3txLTF9Il0sWzQsMCwiUF97cX0iXSxbNiwwLCJQX3txKzF9Il0sWzQsMiwiWl9xIl0sWzQsNCwiQl9xIl0sWzgsMCwiXFxjZG90cyJdLFswLDAsIlxcY2RvdHMiXSxbNCwzLCIiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFszLDEsIiIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzUsMiwiZCIsMl0sWzIsMSwiZCIsMl0sWzEsMCwiZCIsMl0sWzAsNiwiZCIsMl1d}{Link to Diagram} \end{quote} We have a SES \begin{align*} 0 \to B_q \to Z_q \to H_q \to 0 ,\end{align*} which induces a LES \begin{center} \begin{tikzcd} \cdots && {L_2FH_q} && {L_1FH_q} \\ \\ {FB_q} && {FZ_q} && {FH_q} && 0 \arrow[from=3-1, to=3-3] \arrow[from=3-3, to=3-5] \arrow[from=3-5, to=3-7] \arrow[from=1-5, to=3-1, out=0, in=180] \arrow[from=1-3, to=1-5] \arrow[from=1-1, to=1-3] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsNyxbNiwyLCIwIl0sWzQsMiwiRkhfcSJdLFsyLDIsIkZaX3EiXSxbMCwyLCJGQl9xIl0sWzQsMCwiTF8xRkhfcSJdLFsyLDAsIkxfMkZIX3EiXSxbMCwwLCJcXGNkb3RzIl0sWzMsMl0sWzIsMV0sWzEsMF0sWzQsM10sWzUsNF0sWzYsNV1d}{Link to Diagram} \end{quote} We have $$L_1 FH_q = 0$$, since in the CE resolution we assume that $$H_q(P, d^h)$$ is projective. The second SES we have is \begin{align*} 0 \to Z_q \to P_q \xrightarrow{d} B_{q-1} \end{align*} inducing the LES \begin{center} \begin{tikzcd} \cdots && {L_2FP_q} && {L_1FB_{q-1}} \\ \\ {FZ_q} && {FP_q} && {FB_{q-1}} && 0 \arrow[from=3-1, to=3-3] \arrow[from=3-3, to=3-5] \arrow[from=3-5, to=3-7] \arrow[from=1-5, to=3-1, in=180, out=0] \arrow[from=1-3, to=1-5] \arrow[from=1-1, to=1-3] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsNyxbNiwyLCIwIl0sWzQsMiwiRkJfe3EtMX0iXSxbMiwyLCJGUF9xIl0sWzAsMiwiRlpfcSJdLFs0LDAsIkxfMUZCX3txLTF9Il0sWzIsMCwiTF8yRlBfcSJdLFswLDAsIlxcY2RvdHMiXSxbMywyXSxbMiwxXSxbMSwwXSxbNCwzXSxbNSw0XSxbNiw1XV0=}{Link to Diagram} \end{quote} Here $$L_i F B_{q-1} = 0$$ since $$B_{p-q}(P, d^h)$$ was projective. Putting these together, we have \begin{align*} H_{q}(FP) = { \ker Fd : FP_q \to FP_{q-1} \over \operatorname{im}Fd : FP_{q+1} \to FP_{q} } \cong {FZ_q \over FB_q} \cong FH_q(P_{*, *}) .\end{align*} Now what is its vertical homology? The map $$H_q(P_{*, *}) \to H_q(A)$$ is a projective resolution, so apply $$F$$ to the source -- it's no longer exact, and you get $$FH_q(P)$$ from above, and taking homology yields the left-derived functors applied to the source. Thus \begin{align*} H_p^v FH_q^h(P) = L_p F( H_q (A)) ,\end{align*} and the left-hand side is equal to $$H_p^v H_q^h (FP)$$. \end{proof} \begin{exercise}[Prove (b)] Prove part (b) of the proposition. \end{exercise} \begin{remark} There is a cohomology variant of this: everything dualizes to $$R^i F(A)$$ for a left exact functor $$F: \mathcal{A}\to \mathcal{B}$$ where $$A\in \mathsf{Ch}(\mathcal{A})$$, $$\mathcal{A}$$ has enough injectives, and $$B$$ is complete. Using a right CE resolution $$I^{*, *}$$ of injective objects in $$A$$ yields an upper half-plane complex with $$A^{*} \to I^{*,0}$$ such that the induces maps on cohomology are themselves injective resolutions of $$B^p(A^{*})$$ and $$H^p(A^{*})$$. In this case \begin{align*} R^i F(A^{*}) = H^i \operatorname{Tot}^{\Pi}F(I^{*, *}) .\end{align*} We can prove dual version of all of the results about left hyper-derived functors, although there are some slight convergence issues to worry about due to the direct product. \end{remark} \hypertarget{wednesday-march-31}{% \section{Wednesday, March 31}\label{wednesday-march-31}} \begin{remark} Last time we talked about hypercohomology and hyper derived functors, and we proved that two spectra sequences converging to $${\mathbb{L}}_{p+q}F(A)$$. \end{remark} \hypertarget{grothendieck-spectral-sequences}{% \subsection{Grothendieck Spectral Sequences}\label{grothendieck-spectral-sequences}} \begin{remark} We'll focus on the cohomological version, which gives a spectral sequence from a composition of functors. Let $$\mathcal{A}, \mathcal{B}, \mathcal{C}$$ be abelian categories with enough injectives, and let $$G: \mathcal{A} \to \mathcal{B}$$, $$F: \mathcal{B} \to \mathcal{C}$$ be left exact functors. By a previous result, $$FG:\mathcal{A} \to \mathcal{C}$$ is left exact, which follows from checking that it preserves 4-term exact sequences. Recall that $$B \in \mathcal{B}$$ is $$F{\hbox{-}}$$acyclic if $$R^i F(B) = 0$$ for all $$i>0$$. \end{remark} \begin{theorem}[Grothendieck Spectral Sequence] Assume the above setup, and that $$G$$ sends injectives in $$\mathcal{A}$$ to $$F{\hbox{-}}$$acyclic objects in $$\mathcal{B}$$. Then there is a convergent QI spectral sequence for each $$A \in \mathcal{A}$$: \begin{align*} E_2^{p, q} = (R^p F)(R^q G)(A) \Rightarrow R^{p+q}(FG)(A) .\end{align*} The edge maps are the natural maps \begin{align*} (R^p F)(GA) &\to R^p(FG)(A) \\ R^q (FG)(A) &\to F( R^qG(A)) .\end{align*} The exact sequences of the low-degree terms are \begin{align*} 0 \to (R^jF)(GA) \to R^j(FG)(A) \to F(R^j G(A)) \to (R^j F)(GA) \to R^j(FG)(A) .\end{align*} \end{theorem} \begin{proof}[?] Choose an injective resolution $$A\hookrightarrow I$$ in $$\mathcal{A}$$ and apply $$G$$ to form the cochain complex $$G(I)\in \mathcal{B}$$. Using a first quadrant CE resolution of $$G(I)$$, form the hyper right-derived functors $${\mathbb{R}}^i F(G(I))$$. We have the two spectral sequences that converge to this, since the complex is bounded below: \begin{align*} {}^I E_1^{p, q} = H^p R^q F(GI) \Rightarrow({\mathbb{R}}^{p+q} F)(GI) .\end{align*} By hypothesis $$I^p$$ is injective in $$\mathcal{A}$$, and thus $$G(I^p)$$ is $$F{\hbox{-}}$$acyclic in $$\mathcal{B}$$, so this spectral sequence collapses onto the horizontal axis at the 2nd page. So $$({\mathbb{R}}^p F)(GI) = H^p(FG(I))$$, which is by definition $$R^p(FG)(A)$$, and this holds for all $$p>0$$. This follows because only one term survives on each diagonal, and the associated graded is just to those terms, so it lifts to just being the actual homology. The second spectral sequence converges to the same thing, and so by reindexing the previous limiting term $$p\mapsto p+q$$, we can write \begin{align*} {}^{II} E_2^{p, q} = (R^p F)(H^q(GI)) \Rightarrow R^{p+q} (FG)(A) .\end{align*} But this is $$(R^p F)(R^q G)(A)$$ by definition. By example 5.2.6, the edge maps from the $$p{\hbox{-}}$$axis are \begin{align*} E_2^{p, 0} \to E_{\infty }^{p, 0} \hookrightarrow H^p ,\end{align*} and composing these yields $$(R^p F)(GA) \to R^p(FG)(A)$$. We also have $$H^q \twoheadrightarrow E_{\infty }^{p, 0} \hookrightarrow E_2^{0, q}$$. \end{proof} \begin{remark} We're skipping the section on sheaf cohomology and 5.9, so we'll move into chapter 6. \end{remark} \hypertarget{the-lyndon-hochschild-serre-spectral-sequence}{% \subsection{6.8: The Lyndon-Hochschild-Serre Spectral Sequence}\label{the-lyndon-hochschild-serre-spectral-sequence}} \begin{remark} Let $$H{~\trianglelefteq~}G$$ and $$A\in \mathsf{G}{\hbox{-}}\mathsf{Mod}$$, then $$A_H, A^H \in \mathsf{G/H}{\hbox{-}}\mathsf{Mod}$$. The canonical projection $$p: G\to G/H$$ induces a forgetful functor $$p^*: \mathsf{G/H}{\hbox{-}}\mathsf{Mod} \to {\mathsf{G}{\hbox{-}}\mathsf{Mod}}$$ given by pullback. Note that $$G/H{\hbox{-}}$$modules are essentially $$G{\hbox{-}}$$modules where $$H$$ acts trivially, so this functor forgets the trivial $$H$$ action. Generally, this works a bit like the Frobenius map, which yields a representation that can be pulled back. \end{remark} \begin{lemma}[?] The invariant functor $$({-})_H$$ has a left adjoint and the coinvariant functor $$({-})^H$$ has a right adjoint. \end{lemma} \begin{proof}[?] A $$G/H{\hbox{-}}$$module is a $$G{\hbox{-}}$$module with a trivial $$H$$ action, so both $$A_H, A^H$$ are $$G/H{\hbox{-}}$$modules. One needs to check that although $$H$$ preserves these submodules, so does $$G$$. The universal property of $$A^H \hookrightarrow A$$ as the largest trivial submodule and $$A\to A_H$$ as the largest trivial quotient imply that there are natural isomorphisms: for $$A\in \mathsf{G}{\hbox{-}}\mathsf{Mod}$$ and $$B\in \mathsf{G/H}{\hbox{-}}\mathsf{Mod}$$, \begin{align*} \mathop{\mathrm{Hom}}_G(p^* B, A) &\xrightarrow{\sim} \mathop{\mathrm{Hom}}_{G/H}(B, A^H) \\ f &\mapsto f \end{align*} which is well-defined since $$f(b) = f(hb) = hf(b) = f(b)$$, putting $$f(b) \in A^H$$. We also have \begin{align*} \mathop{\mathrm{Hom}}_G(A, p^\sharp B) &\xrightarrow{\sim} \mathop{\mathrm{Hom}}_{G/H}(A_H, B) \\ (\tilde f: A \xrightarrow{\pi} A_H \xrightarrow{f} B ) &\mapsfrom f ,\end{align*} and these give the required adjunction. \end{proof} \begin{theorem}[Lyndon-Hochschild-Serre Spectral Sequence] Let $$H{~\trianglelefteq~}G$$ for $$A\in \mathsf{G}{\hbox{-}}\mathsf{Mod}$$, then there are two QI spectral sequences: \begin{align*} E_{p, q}^2 &= H_p (G/H, H_q(H, A)) \\ E_2^{p, q} &= H^p(G/H, H^q(H, A)) .\end{align*} \end{theorem} \begin{remark} Note that we can identify the functors \begin{align*} ({-})^H, ({-})_H : \mathsf{G}{\hbox{-}}\mathsf{Mod} \to \mathsf{G/H}{\hbox{-}}\mathsf{Mod} ,\end{align*} whose derived functors are group homology/cohomology. The idea will be that $$G{\hbox{-}}$$invariants can be written as a composition of other functors, and we can apply the Grothendieck spectral sequence construction. \end{remark} \hypertarget{friday-april-02}{% \section{Friday, April 02}\label{friday-april-02}} \hypertarget{review-the-lyndon-hochschild-serre-spectral-sequence}{% \subsection{Review: The Lyndon-Hochschild-Serre Spectral Sequence}\label{review-the-lyndon-hochschild-serre-spectral-sequence}} \begin{remark} We're trying to prove the Lyndon-Hochschild-Serre spectral sequence for $$H{~\trianglelefteq~}G$$. \end{remark} \begin{lemma}[?] Let $$H{~\trianglelefteq~}G$$ and $$A\in\mathsf{G}{\hbox{-}}\mathsf{Mod}$$ with \begin{align*} \rho: G\to { G \over H} .\end{align*} Then $$A_H, A^H$$ are in $$\mathsf{{G \over H}}{\hbox{-}}\mathsf{Mod}$$ and $$({-})^H$$ (respectively $$({-})_H$$) are right (respectively left) adjoin to \begin{align*} \phi^\#: \mathsf{G \over H}{\hbox{-}}\mathsf{Mod} \to \mathsf{G}{\hbox{-}}\mathsf{Mod} .\end{align*} \end{lemma} \begin{theorem}[Lyndon-Hochschild-Serre Spectral Sequence] Let $$H{~\trianglelefteq~}G$$ and $$A\in {\mathsf{G}{\hbox{-}}\mathsf{Mod}}$$, then there exist two $$Q1$$ spectral sequences: \begin{align*} E_{p, q}^2 &= H_p\qty{ {G \over H}, H_q(H;A)} \Rightarrow H_{p+q}(G; A) \\ E^{p, q}_2 &= H^p\qty{ { G \over H}, H_q(H;A)} \Rightarrow H^{p+q}(G; A) .\end{align*} \end{theorem} \begin{proof}[?] We want to write this as a composition of functors: \begin{center} \begin{tikzcd} {{\mathsf{G}{\hbox{-}}\mathsf{Mod}}} && {{\mathsf{G/H}{\hbox{-}}\mathsf{Mod}}} && {} \\ \\ && {\mathsf{Ab}} \arrow["{({-})^H}", from=1-1, to=1-3] \arrow["{({-})^{G\over H}}", from=1-3, to=3-3] \arrow["{({-})^G}"', dashed, from=1-1, to=3-3] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJcXG1vZHN7R30iXSxbMiwwLCJcXG1vZHN7Ry9IfSJdLFsyLDIsIlxcQWIiXSxbNCwwXSxbMCwxLCIoXFx3YWl0KV5IIl0sWzEsMiwiKFxcd2FpdClee0dcXG92ZXIgSH0iXSxbMCwyLCIoXFx3YWl0KV5HIiwyLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV1d}{Link to Diagram} \end{quote} We can write \begin{align*} (A^H)^{G/H} &= \left\{{ a\in A {~\mathrel{\Big|}~}ha = a \forall h\in H }\right\} \\ &= \left\{{ a\in A^H {~\mathrel{\Big|}~}\mkern 1.5mu\overline{\mkern-1.5mug\mkern-1.5mu}\mkern 1.5mu a =a \forall \mkern 1.5mu\overline{\mkern-1.5mug\mkern-1.5mu}\mkern 1.5mu\in G/H}\right\} \\ &= \left\{{ a\in A {~\mathrel{\Big|}~}\alpha= a \forall g\in G }\right\} \\ &= A^G .\end{align*} By the lemma, $$({-})^H$$ is right adjoint to $$\rho^{\#}$$, which is exact. By prop 2.3.10, it sends injectives to injectives, and injectives are $$F{\hbox{-}}$$acyclic for $$F({-}) = ({-})^{G \over H}$$. So this is a valid setup for the Grothendieck spectral sequence. \end{proof} \hypertarget{application-bootstrapping-homology-of-cyclic-groups}{% \subsection{Application: Bootstrapping Homology of Cyclic Groups}\label{application-bootstrapping-homology-of-cyclic-groups}} \begin{example}[?] Let $$C_m$$ be cyclic of order $$m$$, and suppose we have the results from section 6.2: \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \item If $$m$$ is odd, \begin{align*} H_q(C_m; {\mathbb{Z}}) = \begin{cases} {\mathbb{Z}}& q=0 \\ {\mathbb{Z}}/m & q \text{ odd} \\ 0 & q\text{ even}. \end{cases} \end{align*} \item If $$H\leq Z(G)$$ and $$A$$ is a trivial $$G{\hbox{-}}$$module, then $$G/H \curvearrowright H_*(H; A)$$ trivially as well. \footnote{Note that this can be phrased in terms of the image of the functor lying in trivial modules.} \item If $$A$$ is a trivial $$C_2{\hbox{-}}$$module and let $$\times 2:A\to A$$ be multiplication, then \begin{align*} H_p(C_2; A) = \begin{cases} A & p = 0 \\ \operatorname{coker}(\times 2) = A/2A & p \text{ odd} \\ \ker(\times s) = \left\{{ a\in A {~\mathrel{\Big|}~}2a = 0 }\right\} & p \text{ even}. \end{cases} \end{align*} Note that the previous fact was a special case of multiplication by $$m$$. \end{enumerate} Using the SES \begin{align*} 0 \to C_m \to C_{2m} \to C_2 \to 0 ,\end{align*} we can use the LHS spectral sequence to compute \begin{align*} E_{p, q}^2 = H_p( C_2; H_q(C_m; {\mathbb{Z}})) \Rightarrow H_{p+q}(C_{2m}; {\mathbb{Z}}) .\end{align*} Let $$A = H_q(C_m; {\mathbb{Z}})$$, then by fact (2) we'll get a trivial $$C_2{\hbox{-}}$$module, and we can then use fact (3). \begin{itemize} \item For $$q=0$$ we have \begin{align*} E_{p, 0}^2 &= H_p(C_2; {\mathbb{Z}}) \\ &= \begin{cases} {\mathbb{Z}}& p=0 \\ {\mathbb{Z}}/2 & p \text{ odd} \\ 0 & p \text{ even} \end{cases} && \text{by (3)} .\end{align*} \item For $$p=0$$ we have \begin{align*} E_{0, q}^2 &= H_q(C_m; {\mathbb{Z}}) \\ &= \begin{cases} {\mathbb{Z}}& p=0 \\ {\mathbb{Z}}/m & p \text{ odd} \\ 0 & p \text{ even}. \end{cases} \end{align*} \item For $$q>0$$ odd and $$p>0$$ odd, note that $${\mathbb{Z}}/m \xrightarrow{\times 2} {\mathbb{Z}}/m$$ is a bijection for odd $$m$$, so \begin{align*} E_{p, q}^2 = H_p(C_2; {\mathbb{Z}}/m) = 0 && \text{since } { {\mathbb{Z}}/m \over 2{\mathbb{Z}}/m} = 0 .\end{align*} \item For $$q>0$$ odd and $$p>0$$ even, \begin{align*} E_{p, q}^2 = H_p(C_2; {\mathbb{Z}}/m) = 0 .\end{align*} \item For $$q>0$$ even and $$p>0$$, \begin{align*} H_q(C_m; {\mathbb{Z}}) = 0 \implies E_{p, q}^2 = 0 .\end{align*} \end{itemize} Thus the $$E_2$$ page of the LHS spectral sequence looks like the following, where there is only one possible nontrivial differential which is forced to be zero: \begin{center} \begin{tikzcd} q \\ & \bullet \\ && \vdots \\ 5 && {{\mathbb{Z}}/m} & \bullet &&&&&& {E_2} \\ 4 && \bullet & \bullet & \bullet \\ 3 && {{\mathbb{Z}}/m} & \bullet & \bullet & \bullet \\ 2 && \bullet & \bullet & \bullet & \bullet & \bullet \\ 1 && {{\mathbb{Z}}/m} & \bullet & \bullet & \bullet & \bullet & \bullet \\ 0 && {\mathbb{Z}}& {{\mathbb{Z}}/2} & \bullet & {{\mathbb{Z}}/2} & \bullet & {{\mathbb{Z}}/2} & \bullet & \cdots \\ \bullet &&&&&&&&&&& \bullet \\ & \bullet & 0 & 1 & 2 & 3 & 4 & 5 &&&& p \arrow[dashed, no head, from=10-1, to=10-12] \arrow["{d=0}"', color={rgb,255:red,92;green,92;blue,214}, from=9-5, to=8-3] \arrow[dashed, no head, from=2-2, to=11-2] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=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}{Link to Diagram} \end{quote} Note that each diagonal only has (at most) two nonzero terms along the axes, and so we'll get a 2-term filtration. Recall that in general we get $$\left\{{ F_i H_n }\right\}_{i=1}^n$$ where $$F_{\leq -1} H_n =0$$ and $$F_{\geq n}H_n = H_n$$. Here $$E_{0, n}^{\infty }$$ comes from $$F_{-1}, F_0$$ and $$E_{n, 0}^{\infty }$$ comes from $$F_{n-1}, F_n$$. So we have \begin{align*} H_0(C_{2m}; {\mathbb{Z}}) &= {\mathbb{Z}}\\ H_n(C_{2m}; {\mathbb{Z}}) &= 0 \text{ forneven} .\end{align*} For $$n$$ odd, we get a SES \begin{align*} 0 \to {\mathbb{Z}}/m \to H_n(C_{2m}; {\mathbb{Z}}) \to {\mathbb{Z}}/2 \to 0 .\end{align*} Letting $$B\in {\mathsf{Ab}}$$ be the middle term, its order is $$2m$$, the product of the two outer elements. By Cauchy's theorem, since $$2\bigm|\# B$$, there is an element $$y\in B$$ of order 2. So send the generator of $${\mathbb{Z}}/2$$ to $$y$$ to form the splitting. Thus \begin{align*} B\cong {\mathbb{Z}}/m \oplus {\mathbb{Z}}/2 \cong {\mathbb{Z}}/m \times {\mathbb{Z}}/2 \cong {\mathbb{Z}}/2m ,\end{align*} where we've now used the $$\gcd(2, m) = 1$$. So \begin{align*} H_n(C_{2m}; {\mathbb{Z}}) = \begin{cases} {\mathbb{Z}}& n=0 \\ {\mathbb{Z}}/2m & n\text{ even} \\ 0 & n \text{ odd}. \end{cases} \end{align*} \end{example} \begin{question} Can you get the group homology of any cyclic group this way? Similar formulas likely hold, see section 6.2. \end{question} \hypertarget{restriction-and-inflation}{% \subsection{Restriction and Inflation}\label{restriction-and-inflation}} \begin{remark} The exact sequence of low degree terms in the cohomological LHS spectral sequence are of the form \begin{center} \begin{tikzcd} 0 && {H^1(G/H; A^H)} && {H^1(G; A)} && {H^1(H; A)} \\ \\ && {H^2(G/H; A^H)} && {H^2(G; A)} \arrow[from=1-1, to=1-3] \arrow["{\text{inflation}}", from=1-3, to=1-5] \arrow["{\text{restriction}}", from=1-5, to=1-7] \arrow["{d_2}"', from=1-7, to=3-3, out=0, in=180] \arrow["{\text{inflation}}", from=3-3, to=3-5] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsNixbMCwwLCIwIl0sWzIsMCwiSF4xKEcvSDsgQV5IKSJdLFs0LDAsIkheMShHOyBBKSJdLFs2LDAsIkheMShIOyBBKSJdLFsyLDIsIkheMihHL0g7IEFeSCkiXSxbNCwyLCJIXjIoRzsgQSkiXSxbMCwxXSxbMSwyLCJcXHRleHR7aW5mbGF0aW9ufSJdLFsyLDMsIlxcdGV4dHtyZXN0cmljdGlvbn0iXSxbMyw0LCJkXzIiLDJdLFs0LDUsIlxcdGV4dHtpbmZsYXRpb259Il1d}{Link to Diagram} \end{quote} Note that these maps have particular name, \textbf{inflation} and \textbf{restriction}. \end{remark} \begin{remark} We thought of homology as a functor of the module $$A$$, but here we see it's varying. Can this be thought of as a functor of the group instead? Setup: let $$\rho: H\to G$$ be a group morphism, then recall that any $$G{\hbox{-}}$$module becomes an $$H{\hbox{-}}$$module by composition with $$\rho$$, which yields an exact functor \begin{align*} \rho^{\#}: {\mathsf{G}{\hbox{-}}\mathsf{Mod}} \to {\mathsf{H}{\hbox{-}}\mathsf{Mod}} .\end{align*} Letting $$A\in{\mathsf{G}{\hbox{-}}\mathsf{Mod}}$$, set \begin{itemize} \tightlist \item $$T_n(A) \coloneqq H_n(G; A)$$ \item $$T^n(A) \coloneqq H^n(G; A)$$ \item $$S_n(A) \coloneqq H_n(G; \rho^{\#} A)$$ \item $$S^n(A) \coloneqq H^n(G; \rho^{\#} A)$$ \end{itemize} \end{remark} \hypertarget{monday-april-05}{% \section{Monday, April 05}\label{monday-april-05}} \hypertarget{restriction-and-inflation-1}{% \subsection{Restriction and Inflation}\label{restriction-and-inflation-1}} \begin{definition}[Restriction and Corestriction] Let $$\rho: H\to G$$ be a group morphism, this induces an exact functor $$\rho^\sharp: {\mathsf{G}{\hbox{-}}\mathsf{Mod}} \to {\mathsf{H}{\hbox{-}}\mathsf{Mod}}$$. We define \begin{itemize} \item $$T_n(A) \coloneqq H_n(G; A)$$ \item $$T^n(A) \coloneqq H^n(G; A)$$ \item $$S_n(A) \coloneqq H_n(\rho^\sharp G; A)$$ \item $$S^n(A) \coloneqq H^n(\rho^\sharp G; A)$$ \end{itemize} These are all functors $${\mathsf{G}{\hbox{-}}\mathsf{Mod}} \to {\mathsf{{\mathbb{Z}}}{\hbox{-}}\mathsf{Mod}}$$. As in section 2.1, $$H_n$$ defines a homological $$\delta{\hbox{-}}$$functor, and since $$\rho^\sharp$$ is exact, $$T_n, S_n$$ are homological $$\delta{\hbox{-}}$$functors as well. We have a map \begin{align*} A^G &\hookrightarrow(\rho^\sharp A)^H \\ T^0 A&\to S^0 A .\end{align*} and similarly \begin{align*} (\rho^\sharp A)_H &\to A_G\\ S_0 A &\to T_0 A .\end{align*} These maps on the 0th terms extend to morphisms of $$\delta{\hbox{-}}$$functors. There thus exist two maps \begin{align*} \operatorname{res}_H^G H^*(G; A) \to H^*(G; \rho^\sharp A) && \text{restriction} \\ \operatorname{cores}_H^G H_*(G; \rho^\sharp A) \to H_*(G; A) && \text{corestriction} .\end{align*} \end{definition} \begin{remark} A special case is when $$H\leq G$$ is a subgroup and $$\rho: H\hookrightarrow G$$ is the inclusion. Then we define a capital $$\operatorname{Res}$$ as \begin{align*} \rho^\sharp = \operatorname{Res}_H^G: {\mathsf{G}{\hbox{-}}\mathsf{Mod}} \to {\mathsf{H}{\hbox{-}}\mathsf{Mod}} ,\end{align*} which is a restriction of the action to a subgroup and thus a type of forgetful functor. \end{remark} \begin{remark} Note that $${\mathbb{Z}G}$$ is a free $${\mathbb{Z}}H{\hbox{-}}$$module with basis being any set of coset representatives, thus any projective $$G{\hbox{-}}$$module restricts to a projective $$H{\hbox{-}}$$module, using the characterization of projective modules as direct summands of free modules. \end{remark} \begin{remark} Recall that \begin{align*} H_*(G; A) &\cong \operatorname{Tor}_*^{{\mathbb{Z}G}}({\mathbb{Z}}, A) \\ H_*(G; A) &\cong \operatorname{Ext}_{{\mathbb{Z}G}}^*({\mathbb{Z}}, A) .\end{align*} We can compute both using a $${\mathbb{Z}G}{\hbox{-}}$$projective resolution $$P_* \to {\mathbb{Z}}$$. This is also a $${\mathbb{Z}}H{\hbox{-}}$$projective resolution, so we can use this to compute $$H^*(H; {-})$$ and $$H_*(H; {-})$$ as well. \end{remark} \begin{fact} \envlist \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \item There's a natural chain map induced by the forgetful functor: \begin{align*} \beta: \mathop{\mathrm{Hom}}_G(P_*, A) \to \mathop{\mathrm{Hom}}_H(P^*, A) .\end{align*} \item There is an induced map \begin{align*} H^*(\beta): \operatorname{Ext}_G^*({\mathbb{Z}}, A) \to \operatorname{Ext}_H^*({\mathbb{Z}}, A) ,\end{align*} which is equal to the map \begin{align*} \operatorname{res}_H^G: H^*(G; A) \to H^*(H; A) ,\end{align*} giving a way to calculate $$\operatorname{res}$$ from something just coming from restriction of functions. \item There is a chain map \begin{align*} \alpha: P_* \otimes_{{\mathbb{Z}H}} A &\to P_* \otimes{{\mathbb{Z}H}} P_* \otimes_{{\mathbb{Z}G}} A \\ p\otimes a &\mapsto p\otimes a ,\end{align*} which induces \begin{align*} H( \alpha): \operatorname{Tor}_*^H({\mathbb{Z}}, A) \to \operatorname{Tor}_*^G({\mathbb{Z}}, A) \end{align*} which is equal to \begin{align*} \operatorname{cores}_H^G: H_*(H; A) \to H_*(G; A) .\end{align*} So this can be computed from tensor products. \end{enumerate} \end{fact} \begin{definition}[Inflation and Coinflation] Now consider quotient groups instead: assume $$H{~\trianglelefteq~}G$$ and let $$\rho:G\to G/H$$. By precomposing with $$\rho$$, we get a map $$\rho^\sharp: {\mathsf{G\over H}{\hbox{-}}\mathsf{Mod}}\to{\mathsf{G}{\hbox{-}}\mathsf{Mod}}$$. Given a $$G{\hbox{-}}$$module, taking $$H$$ invariants yields a $$G/H{\hbox{-}}$$module, so $$H^*(G/H; A^H) \in {\mathsf{G\over H}{\hbox{-}}\mathsf{Mod}}$$. We form the following composition: \begin{center} \begin{tikzcd} {H^*\qty{{G\over H}; A^H}} && {H^*(G; A^H)} &&&& {H^*(G; A)} \arrow["{H^*(G; {-})(A^H \hookrightarrow A)}", from=1-3, to=1-7] \arrow["\operatorname{res}", from=1-1, to=1-3] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsMyxbMCwwLCJIXipcXHF0eXt7R1xcb3ZlciBIfTsgQV5IfSJdLFsyLDAsIkheKihHOyBBXkgpIl0sWzYsMCwiSF4qKEc7IEEpIl0sWzEsMiwiSF4qKEc7IFxcd2FpdCkoQV5IIFxcaW5qZWN0cyBBKSJdLFswLDEsIlxccmVzIl1d}{Link to Diagram} \end{quote} We'll refer to this as \textbf{inflation}. We similarly define \textbf{coinflation} as the following composition: \begin{center} \begin{tikzcd} {H_*(G; A)} &&& {H(G; A_H)} && {H_*\qty{{G\over H}, A_H}} \arrow["\operatorname{cores}", from=1-4, to=1-6] \arrow["{H_*(G; {-})(A \twoheadrightarrow A_H)}", from=1-1, to=1-4] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsMyxbMCwwLCJIXyooRzsgQSkiXSxbMywwLCJIKEc7IEFfSCkiXSxbNSwwLCJIXypcXHF0eXt7R1xcb3ZlciBIfSwgQV9IfSJdLFsxLDIsIlxcY29yZXMiXSxbMCwxLCJIXyooRzsgXFx3YWl0KShBIFxcc3VyamVjdHMgQV9IKSJdXQ==}{Link to Diagram} \end{quote} \end{definition} \begin{remark} When $$*=0$$, we can write \begin{align*} \operatorname{inf}: (A^H)^{G\over H} \to (A^H)^G \to A^G ,\end{align*} and note that this is exactly the functor composition we needed to get the LHS spectral sequence. Similarly there is a LHS for homology, and an isomorphism \begin{align*} \operatorname{coinf}: A_G \to (A_H)_G \to (A_H)_{G\over H} .\end{align*} \end{remark} \begin{remark} When $$A \in {\mathsf{H}{\hbox{-}}\mathsf{Mod}}^{{ \operatorname{Triv}}}$$, $$A_H\hookrightarrow A$$ is the identity, so $$A^H = A = A_H$$. In this case $$\operatorname{inf}= \operatorname{res}$$ and $$\operatorname{coinf}= \operatorname{cores}$$. \end{remark} \begin{remark} Back to the LHS spectral sequence, the five-term exact sequence yields \begin{align*} 0 \to E_{2}^{1, 0} \to H^1(T) \to E_2^{0, 1} \xrightarrow{d_2} E_{2, 0} \to H^2(T) ,\end{align*} which we can identify as \begin{align*} 0\to H^1\qty{{G\over H}; A^H} \xrightarrow{\operatorname{inf}} H^1(G; A) \xrightarrow{\operatorname{res}} H^1(H; A)^{G\over H} \xrightarrow{d_2} H^2\qty{ {G\over H}; A^H } \xrightarrow{\operatorname{inf}} H^2(G; A) .\end{align*} There is a similar story in homology with coinflation and corestriction. \end{remark} \hypertarget{shapiros-lemma-inducedcoinduced-modules}{% \subsection{Shapiro's Lemma, Induced/Coinduced Modules}\label{shapiros-lemma-inducedcoinduced-modules}} \begin{definition}[Induced and Coinduced Modules] Let $$H\leq G$$ and $$B\in {\mathsf{{\mathbb{Z}}H}{\hbox{-}}\mathsf{Mod}}$$. Define the \textbf{induced $$G{\hbox{-}}$$module} (or tensor-induced $$G{\hbox{-}}$$module) \begin{align*} \operatorname{Ind}_H^G(B) \coloneqq{\mathbb{Z}G}\otimes_{{\mathbb{Z}H}} B \in {\mathsf{{\mathbb{Z}}G}{\hbox{-}}\mathsf{Mod}} .\end{align*} This is a $${\mathbb{Z}G}{\hbox{-}}$$module with an action on the first tensor factor. Similarly define the \textbf{coinduced} or \textbf{hom-induced $$G{\hbox{-}}$$module}. \begin{align*} \operatorname{coInd}_H^G(B) \coloneqq\mathop{\mathrm{Hom}}_{H}({\mathbb{Z}G}, B) \in {\mathsf{{\mathbb{Z}}G}{\hbox{-}}\mathsf{Mod}} .\end{align*} Here the action is $$(g.f)(g') \coloneqq f(gg')$$. \end{definition} \begin{lemma}[Shapiro's Lemma (Frobenius Reciprocity)] \begin{align*} H_*(G; \operatorname{Ind}_H^G B) &\cong H_*(H; B) &&(1) \\ H^*(G; \operatorname{coInd}^G B) &\cong H^*(H; B) &&(2) .\end{align*} \end{lemma} \begin{remark} So this provides a way of computing homology on subgroups when the coefficients are in these induced/coinduced modules. \end{remark} \hypertarget{wednesday-april-07}{% \section{Wednesday, April 07}\label{wednesday-april-07}} \hypertarget{shapiros-lemma-coinduced-modules-cont}{% \subsection{6.3: Shapiro's Lemma, (co)Induced Modules (cont)}\label{shapiros-lemma-coinduced-modules-cont}} \begin{remark} Recall that we had two ways of inducing an $$H{\hbox{-}}$$module up to a $$G{\hbox{-}}$$module for $$H\leq G$$ a subgroup. In this case, we can take cohomology with coefficients in any $$B\in {\mathsf{{\mathbb{Z}}H}{\hbox{-}}\mathsf{Mod}}$$. Shapiro's lemma (or Frobenius Reciprocity) allowed compute homology and cohomology when the coefficients are in induced or coinduced modules: \begin{align*} H_*(G; \operatorname{Ind}_H^G B) &\cong H_*(H; B) &&(1) \\ H^*(G; \operatorname{coInd}^G B) &\cong H^*(H; B) &&(2) .\end{align*} \end{remark} \begin{proof}[of Shapiro's lemma] Let $$P_* \to {\mathbb{Z}}$$ be a right $${\mathbb{Z}G}{\hbox{-}}$$projective resolution of $${\mathbb{Z}}$$. Since $${\mathbb{Z}G}$$ is a free $${\mathbb{Z}}H$$ module, these are still projective over $${\mathbb{Z}}H$$. Then take \begin{align*} P_* \otimes_{{\mathbb{Z}G}} ({\mathbb{Z}G}\otimes_{{\mathbb{Z}}H} B) \cong P_* \otimes_{{\mathbb{Z}}H} B .\end{align*} The homology of the left-hand side computes $$\operatorname{Tor}_*^{{\mathbb{Z}G}}({\mathbb{Z}}, \operatorname{Ind}_H^G B)$$. On the other hand, we can consider $$P_*$$ to be a projective resolution in $${\mathbb{Z}}H$$ and thus the homology of the right-hand side is $$\operatorname{Tor}_*^{{\mathbb{Z}}H}({\mathbb{Z}}, B)$$, which is $$H_*(H; B)$$. For (2), use the tensor-hom adjunction. \footnote{See proposition 2.6.3 in Weibel.} \end{proof} \begin{theorem}[Adjoints of Restriction are Induction and Coinduction] For $$H\leq G, A\in {\mathsf{{\mathbb{Z}G}}{\hbox{-}}\mathsf{Mod}}, B\in {\mathsf{{\mathbb{Z}H}}{\hbox{-}}\mathsf{Mod}}$$, \begin{align*} \operatorname{Ext}_G^*(\operatorname{Ind}_h^G B, A) \cong \operatorname{Ext}_H^*(B, \operatorname{Res}_H^G A) && (1) \\ \operatorname{Ext}_G^*(A, \operatorname{coInd}_H^G B) \cong \operatorname{Ext}_H^*(\operatorname{Res}_H^G A, B) && (1) .\end{align*} \end{theorem} \begin{remark} Taking $$A = {\mathbb{Z}}\in {\mathsf{{\mathbb{Z}G}}{\hbox{-}}\mathsf{Mod}}^{ \operatorname{Triv}}$$, one gets result (2) in Shapiro's lemma. This shows that $$\operatorname{Ind}$$ is left adjoint to $$\operatorname{Res}$$ and $$\operatorname{coInd}$$ is right adjoint to it, so these will have derived functors. A special case is when $$H = \left\{{ 1 }\right\}$$ is the trivial group, in which case any $$H{\hbox{-}}$$module $$B$$ is an abelian group such that $$B^H = B = B_H$$. So $$({-})^H, ({-})_H$$ are exact, and thus their higher derived functors are zero, i.e.~$$H_n(H, B) = 0 = H^n(H; B)$$ for $$n>0$$. Moreover \begin{align*} H_n(G; {\mathbb{Z}G}\otimes_{\mathbb{Z}}B) \cong H^n( G, \mathop{\mathrm{Hom}}_{\mathbb{Z}}( {\mathbb{Z}}G, B)) \cong \begin{cases} B & n = 0 \\ 0 & n > 0. \end{cases} \end{align*} \end{remark} \begin{lemma}[?] If the index $$[G: H]$$ (i.e.~the number of left or right cosets) is finite, then \begin{align*} \operatorname{Ind}_H^G B \cong \operatorname{coInd}_H^G B && \in {\mathsf{G}{\hbox{-}}\mathsf{Mod}} .\end{align*} \end{lemma} \begin{proof}[?] Let $$X$$ be a set of left coset representatives for $$G/H$$, where we'll take the convention that left cosets are of the form $$gH$$. Then $$X$$ is a free $$\mathsf{Mod}{\hbox{-}}\mathsf{{\mathbb{Z}H}}{\hbox{-}}$$basis of $${\mathbb{Z}G}$$, so \begin{align*} \operatorname{Ind}_H^G B \cong {\mathbb{Z}G}\otimes_{{\mathbb{Z}H}} B \cong \bigoplus_{x\in X} x\otimes B &&\in {\mathsf{{\mathbb{Z}}}{\hbox{-}}\mathsf{Mod}} .\end{align*} How does $$g\curvearrowright x\otimes b$$ for $$g\in G$$? We have $$gx\in yH$$ for some $$y\in X$$, so for some $$h\in H$$ we have \begin{align*} gx = gh .\end{align*} We can then compute \begin{align*} g(x\otimes b) &= gx \otimes b\\ &= yh \otimes b\\ &= y \otimes hb .\end{align*} since $$h\in {\mathbb{Z}}H$$. Now $$X^{-1}\coloneqq\left\{{ x^{-1}{~\mathrel{\Big|}~}x\in X }\right\}$$ is a set of coset representatives for $${}_{H}\mkern-.5mu\backslash\mkern-2mu^{G}$$ and hence a left $${\mathbb{Z}H}{\hbox{-}}$$basis for $${\mathbb{Z}G}$$. We can thus write \begin{align*} \operatorname{coInd}_H^G B &\cong \mathop{\mathrm{Hom}}_{{\mathbb{Z}}H}({\mathbb{Z}G}, B) \\ &\cong \mathop{\mathrm{Hom}}_{{\mathbb{Z}H}} \qty{ \bigoplus_{x\in X} {\mathbb{Z}H}x^{-1}, B } \\ &= \prod_{x\in X} \mathop{\mathrm{Hom}}_{{\mathbb{Z}H}}({\mathbb{Z}H}x^{-1}, B) && \text{by exc. A.1.4}\\ &= \prod_{x\in X} \pi_x(A) ,\end{align*} where each term is a copy of $$A$$. This follows because we can specify such a module hom by specifying the image of a basis. So here for $$b\in B$$, $$\pi_x(B)$$ for a fixed $$x$$ is the $$H{\hbox{-}}$$module morphism $${\mathbb{Z}G}\to B$$ where $$x^{-1}\mapsto b$$ and $$z^{-1}\mapsto 0$$ for $$z\neq x$$. How does $$G$$ act on these homs? Using equation (??) \todo[inline]{Sort out which equation this was!} we have \begin{align*} y^{-1}g = hx^{-1} ,\end{align*} and thus \begin{align*} ( g\cdot \pi_x(b))(y^{-1}) &= (\pi_x(b))( y^{-1}g) \\ &= (\pi_x(b)) (hx^{-1}) \\ &= h(\pi_x(b))(x^{-1}) \\ &= hb ,\end{align*} and $$y^{-1}$$ is the only one that lights up for the $$G{\hbox{-}}$$action, i.e.~$$(g\cdot \pi_x(b))(z^{-1}) =0$$ for $$y\neq z$$, and thus \begin{align*} g\cdot \pi_x(b) = \pi_y(hb) .\end{align*} Thus we have a $$G{\hbox{-}}$$module map \begin{align*} \operatorname{Ind}_H^G B & \xrightarrow{\sim} \operatorname{coInd}_H^G B \\ \bigoplus_{x\in X} x\otimes B & \xrightarrow{\sim} \bigoplus_{x\in X} \pi_x B \\ x\otimes B &\mapsto \pi_x(b) ,\end{align*} which is an isomorphism since \begin{align*} g\cdot(x\otimes b) = y\otimes hb \mapsto \pi_y(hb) = g\cdot \pi_x(b) .\end{align*} \end{proof} \begin{corollary}[?] If $$G$$ is a finite group, then for any $$A\in {\mathsf{G}{\hbox{-}}\mathsf{Mod}}$$, \begin{align*} H^{>0}(G; {\mathbb{Z}G}\otimes_{\mathbb{Z}}A) = 0 .\end{align*} \end{corollary} \begin{proof}[?] We think of $$A$$ as a module for the trivial subgroup, and so \begin{align*} H^n(G; {\mathbb{Z}G}\otimes_{\mathbb{Z}}A) &\cong H^n(G, \operatorname{Ind}_1^G A) \\ &\cong H^n(G; \operatorname{coInd}_1^G A) && \text{by the lemma} \\ &= H^n(1; A) && \text{by Shapiro's lemma} \\ &= 0 ,\end{align*} for $$n>0$$, since these are the higher derived functors of taking fixed points, and everything is fixed by 1. \end{proof} \hypertarget{lie-algebra-cohomology}{% \subsection{Lie Algebra (Co)homology}\label{lie-algebra-cohomology}} \begin{remark} Motivation and historical background: if $$G$$ is a Lie group, $$G\in {\mathsf{Grp}}\cap{\mathsf{Mfd}}(C^\infty)$$, i.e.~the group operations are smooth maps. Usually these are real manifolds, they were introduced in the late 1800s by Sophus Lie who studied differential equations on such objects. Taking the tangent space at the identity, we write $${\mathfrak{g}}= T_e G$$, which is a \textbf{Lie Algebra}. Lie showed that this is isomorphic to the vector space of left $$G{\hbox{-}}$$invariant vector fields (1st order differential operators) on $$G$$, which enjoys a \textbf{bracket} operation: \begin{align*} [X, Y](f) = X(Yf) - Y(Xf) && f\in C^{\infty } .\end{align*} This turns out to again be a 1st order operator, despite looking like it might be 2nd order. This led to the study of abstract Lie algebras. \end{remark} \hypertarget{section-7.1-lie-algebras-friday-april-09}{% \section{Section 7.1: Lie Algebras (Friday, April 09)}\label{section-7.1-lie-algebras-friday-april-09}} \hypertarget{definitions}{% \subsection{Definitions}\label{definitions}} \begin{definition}[k\dashalgebras] Let $$k \in \mathsf{CRing}$$, e.g.~a field. An \textbf{algebra} over $$k$$ is a $$k{\hbox{-}}$$module with a bilinear product $$A^{\otimes 2} \to A$$. \end{definition} \begin{remark} The product need not be associative, and $$A$$ need not have 1, so $$A=0$$ is an algebra. \end{remark} \begin{definition}[Lie Algebra Definitions] A \textbf{Lie algebra} $${\mathfrak{g}}$$ is a $$k{\hbox{-}}$$algebra whose product (denoted $$[{-}, {-}]$$) is called the \textbf{Lie bracket}, which satisfies \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \tightlist \item $$[xx] = 0$$ for all $$x\in{\mathfrak{g}}$$, and skew-symmetry: $$[xy] = -[yx]$$ for all $$x,y\in {\mathfrak{g}}$$. \item The Jacobi identity: \end{enumerate} \begin{align*} [x [yz]] + [y[zx]] + [z[xy]] = 0 \iff [x[yz]] = [[xy]z] = [y[xz]] ,\end{align*} so the product behaves like a derivation. \begin{itemize} \tightlist \item There is an \textbf{adjoint} map $$\operatorname{ad}_x \coloneqq[{-}, x]: {\mathfrak{g}}{\circlearrowleft}$$. \item A \textbf{(2-sided) ideal} $${\mathfrak{h}}{~\trianglelefteq~}{\mathfrak{g}}$$ is a $$k{\hbox{-}}$$submodule absorbing under the bracket, so $$[xh] \in {\mathfrak{h}}\forall x\in {\mathfrak{g}}, h\in {\mathfrak{h}}$$ In particular, $${\mathfrak{h}}\leq {\mathfrak{g}}$$ is a subalgebra. \item A \textbf{morphism} $$\rho: {\mathfrak{g}}\to {\mathfrak{g}}'$$ of Lie algebras is a $$k{\hbox{-}}$$module map which preserves the bracket, so $$[\rho(x) \rho(y)] = \rho([xy])$$, so we get a category $${\mathsf{Lie}{\hbox{-}}\mathsf{Alg}}_{/k}$$. \item If $${\mathfrak{h}}{~\trianglelefteq~}{\mathfrak{g}}$$, there is a \textbf{quotient} Lie algebra $${\mathfrak{g}}/{\mathfrak{h}}$$ consisting of additive coset $$x+ {\mathfrak{h}}$$, and a SES \begin{align*} 0\to {\mathfrak{h}}\to {\mathfrak{g}}\to {\mathfrak{g}}/{\mathfrak{h}}\to 0 .\end{align*} \item A Lie algebra $${\mathfrak{g}}$$ is \textbf{abelian} if $$[xy] = 0$$ for all $$x,y\in {\mathfrak{g}}$$. Any $$k{\hbox{-}}$$module can be made into an abelian Lie algebra by setting $$[xy] \coloneqq 0$$. \item The \textbf{derived subalgebra} of $${\mathfrak{g}}$$ is $$[{\mathfrak{g}}{\mathfrak{g}}] {~\trianglelefteq~}{\mathfrak{g}}$$, the $$k{\hbox{-}}$$submodule of $${\mathfrak{g}}$$ generated by all brackets $$[xy]$$. The largest abelian quotient of $${\mathfrak{g}}$$ is given by $${\mathfrak{g}}/ [{\mathfrak{g}}{\mathfrak{g}}]$$. \end{itemize} \end{definition} \hypertarget{examples}{% \subsection{Examples}\label{examples}} \begin{example}[?] Let $$A$$ be any associative $$k{\hbox{-}}$$algebra, not necessarily with 1, and let $${\mathfrak{g}}\coloneqq\operatorname{Lie}(A)$$ be the same $$k{\hbox{-}}$$module with a bracket defined as $$[xy] \coloneqq xy-yx$$. One can check that this satisfies the Jacobi identity. So there is a functor \begin{align*} \operatorname{Lie}: {\mathsf{Alg}_{/k} }(\mathsf{Assoc}) \to {\mathsf{Lie}{\hbox{-}}\mathsf{Alg}}_{/k} .\end{align*} In particular, for $$A \in {\mathsf{Alg}_{/k} }(\mathsf{Assoc})$$ (e.g.~$$A=k$$), the ring $$\operatorname{Mat}(m\times m; A) \in {\mathsf{Alg}_{/k} }(\mathsf{Assoc})$$ and can be mapped into Lie Algebras. We write \begin{align*} {\mathfrak{gl}}_m(A) \coloneqq\operatorname{Lie}(\operatorname{Mat}(m\times m; A)) \cong \operatorname{Lie}(\mathop{\mathrm{End}}_k(A^m)) ,\end{align*} and often omit notation to write $${\mathfrak{gl}}_m \coloneqq{\mathfrak{gl}}_m(k)$$ where $$[xy] \coloneqq xy-yx$$ as the \textbf{general linear Lie algebra} over $$A$$. \end{example} \begin{example}[Important special cases] Let $$A \in {\mathsf{Alg}_{/k} }(\mathsf{Assoc}, \mathsf{Comm})$$ be an associative commutative $$k{\hbox{-}}$$algebra, then \begin{itemize} \item $${\mathfrak{sl}}_m(A)$$ is the \textbf{special linear Lie algebra}, which consists of all trace zero matrices in $${\mathfrak{gl}}_m(A)$$. \item $${\mathfrak{o}}_m(A)$$ is the \textbf{orthogonal algebra} of all skew-symmetric matrices, i.e.~$$x^t = -x$$. \item $${\mathfrak{t}}_m(A)$$ is the \textbf{upper triangular matrices}, so $$x_{ij} = 0$$ if $$i>j$$. \item $${\mathfrak{n}}_m(A)$$ are the \textbf{strictly upper triangular matrices}. \item $${\mathfrak{d}}(A)$$ are the \textbf{diagonal matrices}, so $$x_{ij} = 0$$ if $$i\neq j$$.\footnote{Note that this is referred to as $${\mathfrak{h}}$$ or sometimes $${\mathfrak{t}}$$, since it's the torus.} \end{itemize} \end{example} \begin{definition}[Derivation Algebras] Let $$A \in {\mathsf{Alg}_{/k} }$$, not necessarily associative. A \textbf{derivation} $$D$$ of $$A$$ (or from $$A$$ to $$A$$) is a $$k{\hbox{-}}$$module endomorphism of $$A$$ satisfying the \textbf{Leibniz rule}: \begin{align*} D(ab) = (Da)b + a(Db) && \forall a, b\in A .\end{align*} We write $$\mathop{\mathrm{Der}}(A) \leq \mathop{\mathrm{End}}_k(A)$$ as the $$k{\hbox{-}}$$submodule of all derivations. One can check that $$[D_1, D_2]$$ is again a derivation for derivations $$D_i$$, so $$\mathop{\mathrm{Der}}(A) \in {\mathsf{Lie}{\hbox{-}}\mathsf{Alg}}$$ called the \textbf{derivation algebra} of $$A$$. \end{definition} \begin{definition}[Nilpotent Algebras] Let $${\mathfrak{g}}\in {\mathsf{Lie}{\hbox{-}}\mathsf{Alg}}$$, and define a decreasing sequence of ideals \begin{align*} {\mathfrak{g}}^0 &\coloneqq{\mathfrak{g}}, \quad {\mathfrak{g}}^1 \coloneqq[{\mathfrak{g}}{\mathfrak{g}}], \quad \cdots {\mathfrak{g}}^n \coloneqq[{\mathfrak{g}}^{n-1} {\mathfrak{g}}] .\end{align*} This yields the \textbf{lower central series} \begin{align*} {\mathfrak{g}}^0 \supseteq {\mathfrak{g}}^1 \supseteq \cdots \supseteq {\mathfrak{g}}^n \supseteq \cdots ,\end{align*} and $${\mathfrak{g}}$$ is said to be \textbf{nilpotent} if $${\mathfrak{g}}^n = 0$$ for some $$n$$. \end{definition} \begin{example}[?] For $${\mathfrak{g}}\coloneqq{\mathfrak{n}}_m(A)$$ the strictly upper triangular matrices, we have $$x\in {\mathfrak{g}}^n \iff x_{ij}=0$$ unless $$j \geq i + (n+1)$$. So we get $$n+1$$ diagonals of all zeros: \begin{align*} \begin{bmatrix} 0 & 0 & \cdot & \cdot & \cdot \\ \vdots & 0 & 0 &\cdot & \cdot \\ \vdots & \ddots & 0 & 0 & \cdot \\ \vdots & \vdots& \ddots & 0 & 0 \\ 0 & \cdots & \cdots & \ddots & 0 \end{bmatrix} \end{align*} \end{example} \begin{definition}[Solvable Algebras] Define \begin{align*} {\mathfrak{g}}^{(0)} \coloneqq{\mathfrak{g}}, \quad {\mathfrak{g}}^{(1)} \coloneqq[{\mathfrak{g}}^{(0)} {\mathfrak{g}}^{(0)}], \quad {\mathfrak{g}}^{(n+1)} \coloneqq[{\mathfrak{g}}^{(n)} {\mathfrak{g}}^{(n)}] ,\end{align*} which yields a decreasing sequence of ideals, the \textbf{derived series}, \begin{align*} {\mathfrak{g}}^{(0)} \supseteq {\mathfrak{g}}^{(0)} \supseteq \cdots \supseteq {\mathfrak{g}}^{(0)} \supseteq \cdots ,\end{align*} $${\mathfrak{g}}$$ is \textbf{solvable} if $${\mathfrak{g}}^{(n)} = 0$$ for some $$n$$. \end{definition} \begin{remark} Note that nilpotent implies solvable, since one can show by induction that $${\mathfrak{g}}^{(n)} \subseteq {\mathfrak{g}}^n$$. \end{remark} \begin{example}[?] For $${\mathfrak{g}}= {\mathfrak{t}}_m(A)$$ for $$A$$ commutative, the diagonal of the product is the product along the diagonals, so \begin{itemize} \tightlist \item $${\mathfrak{g}}^{(1)}$$ are matrices with zeros on the diagonal, \item $${\mathfrak{g}}^{(2)}$$ are matrices with zeros on 2 diagonals, \item $${\mathfrak{g}}^{(3)}$$ are matrices with zeros on 4 diagonals, \end{itemize} and so on, so $${\mathfrak{g}}$$ is solvable. On the other hand, taking brackets with one diagonal of zeros doesn't introduce new zero diagonals, and $${\mathfrak{g}}^2 = {\mathfrak{g}}^1$$. So $${\mathfrak{g}}$$ is not nilpotent, provided $$m\geq 2$$ \end{example} \begin{remark} Next time: $${\mathfrak{g}}{\hbox{-}}$$modules. \end{remark} \hypertarget{monday-april-12}{% \section{Monday, April 12}\label{monday-april-12}} \hypertarget{lie-algebra-homology}{% \subsection{Lie Algebra Homology}\label{lie-algebra-homology}} \begin{remark} Last time: Lie algebras. Fix a cocommutative ring $$k$$, usually a field, then a Lie algebra $${\mathfrak{g}}$$ over $$k$$ is a $$k{\hbox{-}}$$module with a bilinear product called the bracket such that \begin{itemize} \tightlist \item $$[xx] = 0$$, so $$[xy] = -[yx]$$ \item The Jacobi identity holds. \end{itemize} \end{remark} \begin{definition}[Modules over Lie algebras] A left $${\mathfrak{g}}{\hbox{-}}$$module $$M$$ is a $$k{\hbox{-}}$$module with a $$k{\hbox{-}}$$bilinear product \begin{align*} \cdot: {\mathfrak{g}}\otimes_k M &\to M \\ x\otimes m &\mapsto x\cdot m \end{align*} which is compatible with the bracket in the following sense: $$[xy]m = x(ym) - y(xm) \quad \forall x,y\in {\mathfrak{g}}, m\in M \label{eq:assoc_formula_lie_algebra} ,$$ i.e.~there is a Lie algebra morphism $${\mathfrak{g}}\to {\mathfrak{gl}}(M) \coloneqq\operatorname{Lie}( \mathop{\mathrm{End}}_k(M))$$, the Lie algebra of the endomorphism algebra. \end{definition} \begin{example}[Algebra Commutators] For $$A \in {\mathsf{Alg}_{/k} }(\mathsf{Assoc})$$ and $${\mathfrak{g}}\in \operatorname{Lie}(A)$$, then any $$M\in \mathsf{A}{\hbox{-}}\mathsf{Mod}$$ (so the action is associative) can be made into an $$M' \in \mathsf{{\mathfrak{g}}}{\hbox{-}}\mathsf{Mod}$$ by the formula \cref{eq:assoc_formula_lie_algebra}. \end{example} \begin{example}[Adjoint Representations] Any Lie algebra $${\mathfrak{g}}$$ is a module over itself by the \textbf{adjoint representation}, where $$\operatorname{ad}_x({-}) \coloneqq[x, {-}]$$. \end{example} \begin{example}[Trivial Modules] Any $$M\in\mathsf{k}{\hbox{-}}\mathsf{Mod}$$ becomes a trivial $${\mathfrak{g}}{\hbox{-}}$$module by defining $$xm = 0$$ for all $$x\in {\mathfrak{g}}, m\in M$$. Note that this is acting by zero instead of the identity: this is motivated from Lie algebras obtained from Lie groups by taking tangent spaces at the identity. A trivial group action on the elements would be the identity, but then taking its derivative acting on tangent vectors to curves would be zero. \hfill\break There is a \emph{unique} trivial $${\mathfrak{g}}{\hbox{-}}$$module, namely $$k$$ with this trivial action. \end{example} \begin{definition}[Morphisms of Lie algebra modules] A morphism $$M \xrightarrow{f} N$$ of $${\mathfrak{g}}{\hbox{-}}$$modules is a morphism of $$k{\hbox{-}}$$modules commuting with the module action, so $$f(xm) = x(fm)$$ for $$x\in {\mathfrak{g}}, m\in M$$. This yields $$\mathop{\mathrm{Hom}}_{\mathfrak{g}}(M, N) \leq \mathop{\mathrm{Hom}}_k(M, N)$$ as a $$k{\hbox{-}}$$submodule. \end{definition} \begin{remark} This yields a category $$\mathsf{{\mathfrak{g}}}{\hbox{-}}\mathsf{Mod} \leq \mathsf{k}{\hbox{-}}\mathsf{Mod}$$ which is a subcategory of $$k{\hbox{-}}$$modules, and this is in fact an abelian category. So we have notions of (co)kernels, injectives and projectives, etc. There is also a category $$\mathsf{Mod}{\hbox{-}}\mathsf{{\mathfrak{g}}}$$, but these can be sent to left $${\mathfrak{g}}{\hbox{-}}$$modules by defining $$x\cdot m \coloneqq-mx$$ which makes $${\mathfrak{g}}$$ anticommutative. Thus there is an equivalence of categories \begin{align*} \mathsf{{\mathfrak{g}}}{\hbox{-}}\mathsf{Mod} \xrightarrow{\sim} \mathsf{Mod}{\hbox{-}}\mathsf{{\mathfrak{g}}} ,\end{align*} and so we usually just refer to left modules. \end{remark} \begin{remark} We'll want to take homology and cohomology. There are some relevant functors: \begin{itemize} \item The trivial module functor: \begin{align*} { \operatorname{Triv}}: \mathsf{k}{\hbox{-}}\mathsf{Mod} \to \mathsf{{\mathfrak{g}}}{\hbox{-}}\mathsf{Mod} ,\end{align*} which sends $$M$$ to itself, adding the structure of a trivial $${\mathfrak{g}}{\hbox{-}}$$action. \item $${\mathfrak{g}}{\hbox{-}}$$invariants: \begin{align*} ({-})^{\mathfrak{g}}: {\mathfrak{g}{\hbox{-}}\mathsf{Mod}}&\to {\mathsf{k}{\hbox{-}}\mathsf{Mod}}\\ M &\mapsto M^g \coloneqq\left\{{ x\in M {~\mathrel{\Big|}~}xm = 0 \forall \,\, x\in {\mathfrak{g}}}\right\} .\end{align*} \begin{itemize} \tightlist \item This yields the largest $${\mathfrak{g}}{\hbox{-}}$$trivial submodule, and similarly $$({-})^{\mathfrak{g}}$$ is right-adjoint to $${ \operatorname{Triv}}$$. \begin{align*} \adjunction{{ \operatorname{Triv}}}{({-})^{\mathfrak{g}}}{{\mathsf{k}{\hbox{-}}\mathsf{Mod}}}{{\mathfrak{g}{\hbox{-}}\mathsf{Mod}}} .\end{align*} \item There is an isomorphism \begin{align*} \operatorname{ev}_1: \mathop{\mathrm{Hom}}_{\mathfrak{g}}(k, M) &\xrightarrow{\sim} M^{\mathfrak{g}}\\ f &\mapsto f(1_k) .\end{align*} where $$k$$ is the trivial $${\mathfrak{g}}{\hbox{-}}$$module. \end{itemize} \item $${\mathfrak{g}}{\hbox{-}}$$coinvariants: \begin{align*} ({-})_{\mathfrak{g}}: {\mathfrak{g}{\hbox{-}}\mathsf{Mod}}&\to {\mathsf{k}{\hbox{-}}\mathsf{Mod}}\\ M &\mapsto M/{\mathfrak{g}}M .\end{align*} \begin{itemize} \tightlist \item This is the largest $${\mathfrak{g}}{\hbox{-}}$$trivial \emph{quotient} of $$M$$, so this is left-adjoint to $${ \operatorname{Triv}}$$: \begin{align*} \adjunction{({-})^{\mathfrak{g}}}{{ \operatorname{Triv}}}{{\mathfrak{g}{\hbox{-}}\mathsf{Mod}}}{{\mathsf{k}{\hbox{-}}\mathsf{Mod}}} .\end{align*} \end{itemize} \end{itemize} \begin{quote} We might expect this is related to some tensor product, but it may not be clear what ring one should tensor over. \end{quote} \end{remark} \begin{remark} Assume that $${\mathfrak{g}{\hbox{-}}\mathsf{Mod}}$$ has enough projectives, which we'll see is true in a later section by identifying this with a category $${\mathsf{R}{\hbox{-}}\mathsf{Mod}}$$ of modules over a ring. \end{remark} \begin{definition}[Cohomology of Lie algebras] Define the \textbf{(co)homology of $${\mathfrak{g}}$$ with coefficients in $$M$$} as \begin{align*} H_n({\mathfrak{g}}; M) &\coloneqq{\mathbb{L}}({-})_{\mathfrak{g}}(M) \\ H^n({\mathfrak{g}}; M) &\coloneqq{\mathbb{R}}({-})^{\mathfrak{g}}(M) .\end{align*} \end{definition} \begin{example}[?] If $${\mathfrak{g}}= \left\{{ 0 }\right\}$$, then $$M^{\mathfrak{g}}= M = M_{\mathfrak{g}}$$ and these functors are exact (and are essentially the identity) and thus their higher derived functors are zero. So $$H^n(0; M) = 0 = H_n(0; M)$$. \end{example} \hypertarget{the-universal-enveloping-algebra}{% \subsection{The Universal Enveloping Algebra}\label{the-universal-enveloping-algebra}} \begin{remark} A better name might be the universal \emph{associative} algebra. This plays an analogous role to the group algebra $${\mathbb{Z}G}$$ of a group. We'll assign an associative algebra $${\mathcal{U}(\mathfrak{g}) }$$ to $${\mathfrak{g}}$$, and there will be an equivalence of categories \begin{align*} \mathsf{{\mathfrak{g}}}{\hbox{-}}\mathsf{Mod} \xrightarrow{\sim} \mathsf{{\mathcal{U}(\mathfrak{g}) }}{\hbox{-}}\mathsf{Mod} ,\end{align*} where we'll know that the latter has enough projectives and injectives, allowing us to compute homology and cohomology with injective and projective resolutions. \end{remark} \begin{definition}[Tensor Algebra] For $$k \in \mathsf{CRing}$$ and $$M\in {\mathsf{k}{\hbox{-}}\mathsf{Mod}}$$, and \textbf{tensor algebra} is defined as \begin{align*} T(M) \coloneqq\bigoplus_{i\geq 0} M^{\otimes_k n} \coloneqq k \otimes\bigoplus _{n\geq 1} M^{\otimes_k n} .\end{align*} \end{definition} \begin{remark} Note that $$T(M) \in {\mathsf{k}{\hbox{-}}\mathsf{Mod}}$$ by extending the $$k{\hbox{-}}$$action over sums and tensor products in the obvious way, and in fact $$T(M) \in {\mathsf{gr}\,}({\mathsf{Alg}_{/k} })$$ where tensors in different degrees are juxtaposed. Explicitly, for $$m\in M^{\otimes n}$$ and $$m' \in M^{\otimes n'}$$, we write $$m\otimes m' \in M^{\otimes(n+n')}$$, which is what it means to be a \emph{graded} algebra. \end{remark} \begin{remark} There is an inclusion map \begin{align*} M = M^{\otimes 1} \overset{\iota}\hookrightarrow T(M)_1 \hookrightarrow T(M) .\end{align*} where $$T(M)_j \coloneqq\bigoplus_{n\geq j} M^{\otimes n}$$, and in fact $$T(M)$$ is generated as a $$k{\hbox{-}}$$algebra by $$\iota(M)$$. For example, for $$m, m' \in M$$, we have $$\iota(m) \otimes\iota(m') \in T(M)_2$$. This yields a functor \begin{align*} T: \mathsf{k}{\hbox{-}}\mathsf{Mod} \to {\mathsf{Alg}_{/k} }(\mathsf{Assoc}, \mathsf{Unital}) ,\end{align*} as well as a forgetful functor \begin{align*} {\operatorname{Forget}}: {\mathsf{Alg}_{/k} }\to {\mathsf{k}{\hbox{-}}\mathsf{Mod}} .\end{align*} The pair $$(T, i)$$ is a \textbf{universal} associative algebra in the following sense: if $$M\in {\mathsf{k}{\hbox{-}}\mathsf{Mod}}$$ and $$A\in {\mathsf{Alg}_{/k} }(\mathsf{Assoc})$$, then there is a $$k{\hbox{-}}$$module morphism $$M\to {\operatorname{Forget}}(A)$$ making the following diagram commute: \begin{center} \begin{tikzcd} \textcolor{rgb,255:red,214;green,92;blue,92}{M} && \textcolor{rgb,255:red,92;green,92;blue,214}{T(M)} \\ \\ && \textcolor{rgb,255:red,92;green,92;blue,214}{A} \arrow["f", color={rgb,255:red,214;green,92;blue,92}, from=1-1, to=3-3] \arrow["{\exists ! \tilde f \in {\mathsf{Alg}_{/k} }}", color={rgb,255:red,92;green,92;blue,214}, dashed, from=1-3, to=3-3] \arrow["\iota", from=1-1, to=1-3] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsMyxbMCwwLCJNIixbMCw2MCw2MCwxXV0sWzIsMiwiQSIsWzI0MCw2MCw2MCwxXV0sWzIsMCwiVChNKSIsWzI0MCw2MCw2MCwxXV0sWzAsMSwiZiIsMCx7ImNvbG91ciI6WzAsNjAsNjBdfSxbMCw2MCw2MCwxXV0sWzIsMSwiXFxleGlzdHMgISBcXHRpbGRlIGYgXFxpbiBcXGthbGciLDAseyJjb2xvdXIiOlsyNDAsNjAsNjBdLCJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19LFsyNDAsNjAsNjAsMV1dLFswLDIsIlxcaW90YSJdXQ==}{Link to Diagram} \end{quote} Note that the red portion of the diagram happens in $${\mathsf{k}{\hbox{-}}\mathsf{Mod}}$$, while the blue portion is in $${\mathsf{Alg}_{/k} }$$, so this allows lifting module morphisms to algebra morphisms. Commuting here means that \begin{align*} f(m_1) f(m_2) = \tilde f(m_1 m_2) \coloneqq f( \iota(m_1) \otimes\iota(m_2)) .\end{align*} There is thus a natural isomorphism \begin{align*} \mathop{\mathrm{Hom}}_{{\mathsf{k}{\hbox{-}}\mathsf{Mod}}}(M, {\operatorname{Forget}}(A)) \xrightarrow{\sim} \mathop{\mathrm{Hom}}_{{\mathsf{Alg}_{/k} }}( T(M), A) .\end{align*} \end{remark} \hypertarget{universal-enveloping-algebras-wednesday-april-14}{% \section{Universal Enveloping Algebras (Wednesday, April 14)}\label{universal-enveloping-algebras-wednesday-april-14}} \begin{remark} Continuing section 7.3 on universal enveloping algebras.: Letting $$k \in \mathsf{CRing}, {\mathfrak{g}}\in {\mathsf{Lie}{\hbox{-}}\mathsf{Alg}}_{/k}, M\in {\mathsf{k}{\hbox{-}}\mathsf{Mod}}$$, we defined the tensor algebra $$T(M) \coloneqq k \oplus \bigoplus_{i\geq 1} M^{\otimes n}\in {\mathsf{gr}\,}{\mathsf{Alg}_{/k} }(\mathsf{Assoc}, \mathsf{Unital})$$ and noted that it was universal for maps from $$M$$ to $$k{\hbox{-}}$$algebras. \end{remark} \begin{definition}[Universal Enveloping Algebra] Let $${\mathfrak{g}}\in{\mathsf{Lie}{\hbox{-}}\mathsf{Alg}}_{/k}$$, then define the \textbf{universal enveloping algebra} of $${\mathfrak{g}}$$ as \begin{align*} {\mathcal{U}(\mathfrak{g}) }\coloneqq{T({\mathfrak{g}}) \over \left\langle{ xy -yx - [xy] {~\mathrel{\Big|}~}x,y\in {\mathfrak{g}}}\right\rangle } .\end{align*} \end{definition} \begin{remark} There is an injection $$k\hookrightarrow{\mathcal{U}(\mathfrak{g}) }$$, so $${\mathcal{U}(\mathfrak{g}) }$$ is unital. The relations guarantee that there is a Lie algebra morphism $$\iota: {\mathfrak{g}}\to {\mathcal{U}(\mathfrak{g}) }$$. Note that we do not know if this is injective yet! Thus there is a functor \begin{align*} \mathcal{U}: {\mathsf{Lie}{\hbox{-}}\mathsf{Alg}}_{/k} \to {\mathsf{Alg}_{/k} } ,\end{align*} and it turns out that this is adjoint to the $$\operatorname{Lie}$$ functor. \end{remark} \begin{fact} There is an adjunction \begin{align*} \adjunction{ \mathcal{U}}{\operatorname{Lie}}{{\mathsf{Lie}{\hbox{-}}\mathsf{Alg}}}{{\mathsf{Alg}_{/k} }} .\end{align*} Thus for every $$f:{\mathfrak{g}}\to \operatorname{Lie}(A)$$ for $$A \in {\mathsf{Alg}_{/k} }(\mathsf{Assoc})$$, we have a commuting diagram \begin{center} \begin{tikzcd} \textcolor{rgb,255:red,214;green,92;blue,92}{{\mathfrak{g}}} && \textcolor{rgb,255:red,92;green,92;blue,214}{{\mathcal{U}(\mathfrak{g}) }} \\ \\ && \textcolor{rgb,255:red,92;green,92;blue,214}{A} \arrow["{\exists f \in {\mathsf{Alg}_{/k} }}", color={rgb,255:red,92;green,92;blue,214}, dashed, from=1-3, to=3-3] \arrow["{f\in {\mathsf{Lie}{\hbox{-}}\mathsf{Alg}}}"', color={rgb,255:red,214;green,92;blue,92}, from=1-1, to=3-3] \arrow["\iota", from=1-1, to=1-3] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsMyxbMCwwLCJcXGxpZWciLFswLDYwLDYwLDFdXSxbMiwyLCJBIixbMjQwLDYwLDYwLDFdXSxbMiwwLCJcXFVnIixbMjQwLDYwLDYwLDFdXSxbMiwxLCJcXGV4aXN0cyBmIFxcaW4gXFxrYWxnIiwwLHsiY29sb3VyIjpbMjQwLDYwLDYwXSwic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fSxbMjQwLDYwLDYwLDFdXSxbMCwxLCJmXFxpbiBcXGxpZWFsZyIsMix7ImNvbG91ciI6WzAsNjAsNjBdfSxbMCw2MCw2MCwxXV0sWzAsMiwiXFxpb3RhIl1d}{Link to Diagram} \end{quote} Thus there is a natural isomorphism \begin{align*} \mathop{\mathrm{Hom}}_{{\mathsf{Lie}{\hbox{-}}\mathsf{Alg}}}({\mathfrak{g}}, \operatorname{Lie}(A)) \xrightarrow{\sim} \mathop{\mathrm{Hom}}_{{\mathsf{Alg}_{/k} }}({\mathcal{U}(\mathfrak{g}) }, A) .\end{align*} \end{fact} \begin{theorem}[?] There is an equivalence of categories \begin{align*} {\mathfrak{g}{\hbox{-}}\mathsf{Mod}}\xrightarrow{\sim} \mathsf{{\mathcal{U}(\mathfrak{g}) }}{\hbox{-}}\mathsf{Mod} ,\end{align*} where we use the fact that $${\mathcal{U}(\mathfrak{g}) }$$ has an underlying ring structure. \hfill\break Concretely, if $$M\in {\mathfrak{g}{\hbox{-}}\mathsf{Mod}}$$ and $$\prod x_i \in {\mathcal{U}(\mathfrak{g}) }$$, then setting $$(x_1 \cdots x_n)m = x_1(\cdots x_n m)$$ (and similarly for every $$i$$) for $$m\in M$$ makes $$m$$ into a $${\mathcal{U}(\mathfrak{g}) }{\hbox{-}}$$module. Conversely, if $$M\in \mathsf{{\mathcal{U}(\mathfrak{g}) }}{\hbox{-}}\mathsf{Mod}$$, we can set $$xm \coloneqq\iota(x) m$$ for $$x\in {\mathfrak{g}}$$ to make $$M$$ into a $${\mathfrak{g}}{\hbox{-}}$$module. \end{theorem} \begin{proof}[?] Let $$M\in {\mathsf{k}{\hbox{-}}\mathsf{Mod}}$$ and set $$E \coloneqq\mathop{\mathrm{End}}_k(M) \in {\mathsf{Alg}_{/k} }$$. Note that a $${\mathfrak{g}}{\hbox{-}}$$module is a $$k{\hbox{-}}$$module $$M$$ with a morphism of Lie algebras $${\mathfrak{g}}\to \operatorname{Lie}(E)$$. Using the adjunction, we can map such a morphism to $$\tilde f: {\mathcal{U}(\mathfrak{g}) }\to E$$, and by definition a $${\mathcal{U}(\mathfrak{g}) }{\hbox{-}}$$module is a $$k{\hbox{-}}$$module $$M$$ with a $$k{\hbox{-}}$$algebra morphism $${\mathcal{U}(\mathfrak{g}) }\to \mathop{\mathrm{End}}_k(M) = E$$. \end{proof} \begin{corollary}[?] The category $${\mathfrak{g}{\hbox{-}}\mathsf{Mod}}$$ has enough projectives and injectives. \end{corollary} \begin{remark} We'll now set up an analog of the augmentation for group algebras, $$\varepsilon: {\mathbb{Z}G}\to {\mathbb{Z}}$$. \end{remark} \begin{definition}[Augmentation Ideal for Lie Algebras] There is a unique surjective morphism $$\varepsilon\in {\mathsf{Alg}_{/k} }({\mathcal{U}(\mathfrak{g}) }, k)$$ where $$\varepsilon\circ \iota({\mathfrak{g}}) =0$$. The kernel $$I \coloneqq\ker\varepsilon$$ is defined as the \textbf{augmentation ideal}, and is a two-sided ideal of $${\mathcal{U}(\mathfrak{g}) }$$ generated by $$\iota({\mathfrak{g}})$$ and write $${\mathfrak{g}}\,{\mathcal{U}(\mathfrak{g}) }= {\mathcal{U}(\mathfrak{g}) }{\mathfrak{g}}$$, i.e.~those elements which contain at least one tensor factor. \end{definition} \begin{remark} We can identify the coinvariants: \begin{align*} k \cong {\mathcal{U}(\mathfrak{g}) }/{\mathfrak{g}}= {\mathcal{U}(\mathfrak{g}) }/{\mathfrak{g}}\, {\mathcal{U}(\mathfrak{g}) }= {\mathcal{U}(\mathfrak{g}) }_{{\mathfrak{g}}} .\end{align*} \end{remark} \begin{corollary}[?] \envlist \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \tightlist \item $$H_*({\mathfrak{g}}; M) \cong \operatorname{Tor}_*^{{\mathcal{U}(\mathfrak{g}) }}(k, M)$$, \item $$H^*({\mathfrak{g}}; M) \cong \operatorname{Ext}^*_{{\mathcal{U}(\mathfrak{g}) }}(k, M)$$, \end{enumerate} \end{corollary} \begin{proof}[?] To show that two derived functors are isomorphic, it's enough to show that their underlying functors (the degree 0 parts) are isomorphic. Starting with (2), we observed that $$M^g \cong \mathop{\mathrm{Hom}}_{{\mathfrak{g}}}(k, M) \cong \mathop{\mathrm{Hom}}_{{\mathcal{U}(\mathfrak{g}) }}(k, M)$$. ~ For (1), we can write \begin{align*} k \otimes_{{\mathcal{U}(\mathfrak{g}) }} M \cong \qty{{\mathcal{U}(\mathfrak{g}) }\over \mathcal{I} } \otimes_{{\mathcal{U}(\mathfrak{g}) }} M \cong M/ \mathcal{I} M \cong M/ {\mathfrak{g}}M = M_{\mathfrak{g}} ,\end{align*} so $$k\otimes_{{\mathcal{U}(\mathfrak{g}) }}({-}) = ({-})_{{\mathfrak{g}}}$$. \end{proof} \begin{remark} So Lie algebra (co)homology is just a special case of the usual Tor and Ext we've already looked at. We'll next find a basis for $${\mathcal{U}(\mathfrak{g}) }$$: \end{remark} \begin{theorem}[PoincarÃ©-Birkhoff-Witt (PBW) Theorem] Let $${\mathfrak{g}}$$ be free in $${\mathsf{k}{\hbox{-}}\mathsf{Mod}}$$ and fix a $$k{\hbox{-}}$$basis, so $${\mathfrak{g}}\in {\mathsf{Vect}}_{/k}$$. Note that this makes $$\iota: {\mathfrak{g}}\hookrightarrow{\mathcal{U}(\mathfrak{g}) }$$ an injection. Let $$\left\{{ x_{ \alpha} }\right\}_{ \alpha\in A}$$ be a fixed totally ordered $$k{\hbox{-}}$$basis for $${\mathfrak{g}}$$. If $$I = (\alpha_1, \cdots, \alpha_p) \in A^p$$, we'll write monomials as $$x_I \coloneqq x_{ \alpha_1} \cdots x_{ \alpha_p} \in {\mathcal{U}(\mathfrak{g}) }$$, where we'll suppress writing $$\iota(x_{\alpha_j})$$. We'll say $$I$$ is (weakly) increasing if $$\alpha_1 \leq \cdots \leq \alpha_p \in A$$. Noting that the empty sequence $$\emptyset \in A^0$$ is increasing, set $$x_\emptyset \coloneqq 1 \in {\mathcal{U}(\mathfrak{g}) }$$, and if $$I = ( \alpha ) \in A^1$$ is a single index, then we'll write $$x_{ \alpha} \in {\mathfrak{g}}$$ and $$x_{( \alpha) } \in {\mathcal{U}(\mathfrak{g}) }$$. \hfill\break Then if $${\mathfrak{g}}\in {\mathsf{Lie}{\hbox{-}}\mathsf{Alg}}_{/k}$$ is a free $$k{\hbox{-}}$$module, a $$k{\hbox{-}}$$basis for $${\mathcal{U}(\mathfrak{g}) }$$ is given by the monomials $$x_I$$ as $$I$$ ranges over finite increasing sequences from $$A$$. \end{theorem} \begin{proof}[?] Omitted. \end{proof} \begin{remark} To at least see why these are a spanning set, suppose $$\beta > \alpha$$. We can commute elements: \begin{align*} x_{ \beta} x_{ \alpha} = x_{ \alpha} x_{ \beta} + [x_{ \beta} x_{ \alpha}] .\end{align*} However, note that the commutator here has lower degree (here, the other factors are degree 2 and the commutator is degree 1). This decreases the number of misorders as well, so induction roughly works. The fact that these are linearly independent is harder and uses some actual representation theory. \end{remark} \hypertarget{friday-april-16}{% \section{Friday, April 16}\label{friday-april-16}} \hypertarget{the-enveloping-algebra-continued}{% \subsection{The Enveloping Algebra (Continued)}\label{the-enveloping-algebra-continued}} \begin{remark} Last time: the PBW theorem. Let $${\mathfrak{g}}\in {\mathsf{Lie}{\hbox{-}}\mathsf{Alg}}$$ and free as a $$k{\hbox{-}}$$module with $$k{\hbox{-}}$$basis $$\left\{{ x_ \alpha }\right\}_{\alpha\in A}$$. Then $${\mathcal{U}(\mathfrak{g}) }$$ has a $$k{\hbox{-}}$$basis $$\left\{{ x_I }\right\}$$ where $$I = ( \alpha_1, \cdots, \alpha_p)$$ is a finite increasing sequence from $$A$$ \end{remark} \begin{example}[?] If $$k$$ is a field and $$\dim_k {\mathfrak{g}}$$ is finite with basis $$\left\{{ x_1, \cdots, x_n }\right\}$$. Take $$I = (1,\cdots, 1, 2\cdots, 2, n\cdots, n)$$ where each $$i$$ occurs $$a_i$$ times. Then a basis for $${\mathcal{U}(\mathfrak{g}) }$$ is $$\left\{{ x_1^{a_1} x_2^{a_2} \cdots x_n^{a_n} {~\mathrel{\Big|}~}a_i \geq 0 }\right\}$$. \end{example} \begin{corollary}[?] The map $$\iota: {\mathfrak{g}}\to {\mathcal{U}(\mathfrak{g}) }$$ is injective, so we can identify $$\iota({\mathfrak{g}})$$ with $${\mathfrak{g}}$$. \end{corollary} \begin{proof}[?] The elements $$x_{(\alpha)} \coloneqq\iota(x_{ \alpha} ) \in {\mathcal{U}(\mathfrak{g}) }$$ are $$k{\hbox{-}}$$linearly independent. \end{proof} \begin{corollary}[?] If $${\mathfrak{h}}\leq {\mathfrak{g}}$$ is a subalgebra and $$k$$ is a field, then $${\mathcal{U}(\mathfrak{g}) }$$ is free as a $$\mathcal{U}({\mathfrak{h}}){\hbox{-}}$$module. \end{corollary} \begin{proof}[?] Choose an ordered basis for $${\mathfrak{h}}$$ first and then extend this to an ordered basis for $${\mathfrak{g}}$$ -- that one can do this is a fact from linear algebra. Then the $$x_I$$ where $$I = ( \alpha_1, \cdots, \alpha_p )$$ is increasing and no $$x_{\alpha_i} \in {\mathfrak{h}}$$ will be a basis for $${\mathcal{U}(\mathfrak{g}) }$$ over $${\mathcal{U}(\mathfrak{h}) }$$. \end{proof} \begin{example}[?] If $$\dim_k {\mathfrak{g}}< \infty$$ and $$\left\{{ x_1, \cdots, x_k }\right\}$$ is a basis for $${\mathfrak{h}}$$ and $$\left\{{ x_1,\cdots, x_k, x_{k+1} \cdots, x_n }\right\}$$ is a basis for $${\mathfrak{g}}$$, then the PBW basis is given by $$\left\{{ x_1^{a_1} \cdots x_k^{a_k} x_{k+1}^{a_{k+1}} \cdots x_n^{a_n} {~\mathrel{\Big|}~}a_i \geq 0 }\right\}$$. Then $$\left\{{ x_{k+1]^{a_{k+1}} \cdots x_n ^{a_n} }}\right\}$$ form a free left $${\mathcal{U}(\mathfrak{h}) }{\hbox{-}}$$module basis for $${\mathcal{U}(\mathfrak{g}) }$$. \end{example} \begin{exercise}[?] Some suggested exercises: \begin{itemize} \tightlist \item 7.3.4 \item 7.3.6 \item 7.3.7 for working with $${\mathcal{U}(\mathfrak{h}) }$$ as a Hopf algebra. \item 7.3.9 for representations of Lie algebras in characteristic $$p$$. \end{itemize} \end{exercise} \hypertarget{h1-for-lie-algebras-weibel-7.4}{% \subsection{\texorpdfstring{$$H^1$$ for Lie Algebras (Weibel 7.4)}{H\^{}1 for Lie Algebras (Weibel 7.4)}}\label{h1-for-lie-algebras-weibel-7.4}} \begin{remark} Recall that we have an augmentation ideal $$\mathcal{I} {~\trianglelefteq~}{\mathcal{U}(\mathfrak{h}) }$$ and a SES \begin{align*} 0 \to\mathcal{I}\to{\mathcal{U}(\mathfrak{g}) }\to k\to 0 .\end{align*} Applying the functor $$\mathop{\mathrm{Hom}}_{{\mathcal{U}(\mathfrak{g}) }}({-}, M)$$ for a fixed $$M\in {\mathfrak{g}{\hbox{-}}\mathsf{Mod}}$$ yields a LES: \begin{center} \begin{tikzcd} 0 && {\mathop{\mathrm{Hom}}_{{\mathcal{U}(\mathfrak{g}) }}(k, M)} && {\mathop{\mathrm{Hom}}_{{\mathcal{U}(\mathfrak{g}) }}({\mathcal{U}(\mathfrak{g}) }, M)} && {\mathop{\mathrm{Hom}}_{{\mathcal{U}(\mathfrak{g}) }}(\mathcal{I}, M)} \\ \\ && {\operatorname{Ext}^1_{{\mathcal{U}(\mathfrak{g}) }}(k, M) = H^1({\mathfrak{g}}; M)} && \textcolor{rgb,255:red,214;green,92;blue,92}{\operatorname{Ext}^1_{{\mathcal{U}(\mathfrak{g}) }}({\mathcal{U}(\mathfrak{g}) }, M)=0} && \cdots \arrow["\delta", from=1-7, to=3-3] \arrow[from=1-1, to=1-3] \arrow[from=1-3, to=1-5] \arrow[from=1-5, to=1-7] \arrow[from=3-3, to=3-5] \arrow[from=3-5, to=3-7] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsNyxbMCwwLCIwIl0sWzIsMCwiXFxIb21fe1xcVWd9KGssIE0pIl0sWzQsMCwiXFxIb21fe1xcVWd9KFxcVWcsIE0pIl0sWzYsMCwiXFxIb21fe1xcVWd9KFxcbWF0aGNhbHtJfSwgTSkiXSxbMiwyLCJcXEV4dF4xX3tcXFVnfShrLCBNKSA9IEheMShcXGxpZWc7IE0pIl0sWzQsMiwiXFxFeHReMV97XFxVZ30oXFxVZywgTSk9MCIsWzAsNjAsNjAsMV1dLFs2LDIsIlxcY2RvdHMiXSxbMyw0LCJcXGRlbHRhIl0sWzAsMV0sWzEsMl0sWzIsM10sWzQsNV0sWzUsNl1d}{Link to Diagram} \end{quote} Here the red term vanishes since $${\mathcal{U}(\mathfrak{g}) }$$ is free and this projective as a $${\mathfrak{g}}{\hbox{-}}$$module. Note that for $$n\geq 2$$, we have the following situation: \begin{center} \begin{tikzcd} \cdots && \textcolor{rgb,255:red,214;green,92;blue,92}{\operatorname{Ext}^{n-1}_{{\mathcal{U}(\mathfrak{g}) }}({\mathcal{U}(\mathfrak{g}) }, M)=0} && {\operatorname{Ext}_{{\mathcal{U}(\mathfrak{g}) }}^{n-1}(\mathcal{I}, M)} \\ \\ {\operatorname{Ext}^1_{{\mathcal{U}(\mathfrak{g}) }}(k, M) = H^1({\mathfrak{g}}; M)} && \textcolor{rgb,255:red,214;green,92;blue,92}{\operatorname{Ext}^{n}_{{\mathcal{U}(\mathfrak{g}) }}({\mathcal{U}(\mathfrak{g}) }, M)=0} && \cdots \arrow["{\delta \quad \cong}", color={rgb,255:red,92;green,92;blue,214}, from=1-5, to=3-1] \arrow[from=1-1, to=1-3] \arrow[from=1-3, to=1-5] \arrow[from=3-1, to=3-3] \arrow[from=3-3, to=3-5] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsNixbMCwwLCJcXGNkb3RzIl0sWzIsMCwiXFxFeHRee24tMX1fe1xcVWd9KFxcVWcsIE0pPTAiLFswLDYwLDYwLDFdXSxbNCwwLCJcXEV4dF97XFxVZ31ee24tMX0oXFxtYXRoY2Fse0l9LCBNKSJdLFswLDIsIlxcRXh0XjFfe1xcVWd9KGssIE0pID0gSF4xKFxcbGllZzsgTSkiXSxbMiwyLCJcXEV4dF57bn1fe1xcVWd9KFxcVWcsIE0pPTAiLFswLDYwLDYwLDFdXSxbNCwyLCJcXGNkb3RzIl0sWzIsMywiXFxkZWx0YSBcXHF1YWQgXFxjb25nIiwwLHsiY29sb3VyIjpbMjQwLDYwLDYwXX0sWzI0MCw2MCw2MCwxXV0sWzAsMV0sWzEsMl0sWzMsNF0sWzQsNV1d}{Link to Diagram} \end{quote} Thus we get a degree shifting isomorphism \begin{align*} H^n({\mathfrak{g}}; M) \cong \operatorname{Ext}_{{\mathcal{U}(\mathfrak{g}) }}^{n-1}(\mathcal{I}, M) .\end{align*} \end{remark} \begin{remark} We thus have \begin{align*} H^1({\mathfrak{g}}; M) \cong { \mathop{\mathrm{Hom}}_{\mathcal{U}(\mathfrak{g}) }(\mathcal{I}, M) / \operatorname{im}\qty{ \mathop{\mathrm{Hom}}_{\mathcal{U}(\mathfrak{g}) }({\mathcal{U}(\mathfrak{g}) }, M) \cong M \to \mathop{\mathrm{Hom}}(\mathcal{I}, M) } } .\end{align*} Next goal: to more concretely express all of the terms here as $$M{\hbox{-}}$$valued derivations on $${\mathfrak{g}}$$. \end{remark} \begin{definition}[Derivations of an algebra] Let $$M\in {\mathfrak{g}{\hbox{-}}\mathsf{Mod}}$$, then a \textbf{derivation from $${\mathfrak{g}}$$ into $$M$$} is a $$k{\hbox{-}}$$linear map $$D:{\mathfrak{g}}\to M$$ satisfying the Leibniz rule: \begin{align*} D([xy]) = x\cdot (Dy) - y\cdot (Dx) && x,y\in {\mathfrak{g}} .\end{align*} \end{definition} \begin{remark} The set of all such maps $$\mathop{\mathrm{Der}}({\mathfrak{g}}, M) \leq_{{\mathsf{k}{\hbox{-}}\mathsf{Mod}}} \mathop{\mathrm{Hom}}_{{\mathsf{k}{\hbox{-}}\mathsf{Mod}}}({\mathfrak{g}}, M)$$ is a $$k{\hbox{-}}$$submodule. A special case is taking $$M \coloneqq{\mathfrak{g}}$$, regarded as a $${\mathfrak{g}}{\hbox{-}}$$module using the adjoint representation. In fact, for any $$k{\hbox{-}}$$algebra (not necessarily associative), we get \begin{align*} D(ab) = (Da)\cdot b + a\cdot(Db) .\end{align*} When $$A \coloneqq{\mathfrak{g}}\in {\mathsf{Lie}{\hbox{-}}\mathsf{Alg}}$$ with the adjoint action, we obtain \begin{align*} D([xy]) &= [x, Dy] + [Dx, y] \\ &= [x, Dy] - [y, Dx] \\ &= x\cdot Dy - y\cdot (Dx) ,\end{align*} recovering the previous definition. \end{remark} \begin{definition}[Inner Derivations] For $$M\in {\mathfrak{g}{\hbox{-}}\mathsf{Mod}}$$, fix an $$m\in M$$. We then define \begin{align*} D_m: {\mathfrak{g}}&\to M \\ x &\mapsto x\cdot m .\end{align*} Any derivation of this form is said to be an \textbf{inner derivation}, and this yields a $$k{\hbox{-}}$$submodule \begin{align*} {\operatorname{Inn}}({\mathfrak{g}}, M) \leq_{{\mathsf{k}{\hbox{-}}\mathsf{Mod}}} \mathop{\mathrm{Der}}({\mathfrak{g}}, M) .\end{align*} \end{definition} \begin{remark} Note that this is indeed a derivation: \begin{align*} D_m([xy]) ] [xy]\cdot m = x\cdot(y\cdot m) - y\cdot(x\cdot m) = x\cdot (D_m y) - y\cdot (D_m x) .\end{align*} It also turns out that any inner derivation is of this form, bracketing against a fixed element. \end{remark} \begin{proposition}[?] \begin{align*} \mathop{\mathrm{Hom}}_{\mathfrak{g}{\hbox{-}}\mathsf{Mod}}(\mathcal{I}, M) \cong \mathop{\mathrm{Der}}({\mathfrak{g}}, M) .\end{align*} \end{proposition} \begin{proof}[?] \begin{claim} There exists such a map. \end{claim} Say $$\varphi\in \mathop{\mathrm{Hom}}_{\mathfrak{g}}(\mathcal{I}, M)$$ and set \begin{align*} D_{\varphi}:{\mathfrak{g}}&\to M \\ x &\mapsto \phi(x) .\end{align*} Then $$D_{ \varphi}$$ is a derivation, so we have \begin{align*} D_{ \varphi}([xy]) &\coloneqq\phi([xy]) \\ &= \phi(xy-yx) \\ &= x \varphi(y) - y \varphi(x) && \text{since\phi$is${\mathfrak{g}}{\hbox{-}}linear} \\ &= x D_{\varphi}(y) - y D_{\varphi}(x) .\end{align*} \begin{claim} This map is a natural isomorphism, in the sense that it doesn't depend on any choices: \begin{align*} \mathop{\mathrm{Hom}}_{\mathfrak{g}}(\mathcal{I}, M) &\to \mathop{\mathrm{Der}}({\mathfrak{g}}, M) \\ \varphi&\mapsto D_{ \varphi} .\end{align*} \end{claim} \begin{proof}[of surjectivity] Recall that we can write $$\mathcal{I} = {\mathcal{U}(\mathfrak{g}) }\,{\mathfrak{g}}$$, so the following product map is a surjection: \begin{align*} \theta: {\mathcal{U}(\mathfrak{g}) }\otimes_k {\mathfrak{g}}&\twoheadrightarrow{\mathcal{U}(\mathfrak{g}) }\,{\mathfrak{g}}= \mathcal{I} \\ x\otimes y &\mapsto xy .\end{align*} One checks that the kernel is given by \begin{align*} \ker(\theta) = \left\{{ u \otimes[xy] - \qty{ux\otimes y - uy\otimes x} {~\mathrel{\Big|}~}x,y\in {\mathfrak{g}}, u\in {\mathcal{U}(\mathfrak{g}) }}\right\} .\end{align*} Now given $$D \in \mathop{\mathrm{Der}}({\mathfrak{g}}, M)$$, consider the map \begin{align*} f: {\mathcal{U}(\mathfrak{g}) }\otimes_k {\mathfrak{g}}&\to M \\ f(u\otimes x) &= u\cdot Dx .\end{align*} One can compute the following, using that $$D$$ is a derivation: \begin{align*} f\qty{ u\otimes[xy] - ux\otimes y - uy\otimes x } &= u D([xy]) - (ux)\cdot D(y) + (uy) \cdot D(x) \\ &= u (x\cdot Dy - y\cdot Dx) - u\cdot(x\cdot Dy) + u\cdot(y\cdot Dx) \\ &= 0 .\end{align*} So $$f$$ induces a well-defined morphism of $${\mathfrak{g}}{\hbox{-}}$$modules, and descends to a map \begin{align*} \phi: {\mathcal{U}(\mathfrak{g}) }\,{\mathfrak{g}}= \mathcal{I} &\to M ,\end{align*} which is clearly also a morphism of $${\mathfrak{g}}{\hbox{-}}$$modules. So $$\varphi\in \mathop{\mathrm{Hom}}_{\mathfrak{g}}(\mathcal{I}, M)$$ and $$D_{\varphi}(x) = \varphi(x) = \varphi(1\cdot x) = f(1\cdot x) = 1\cdot Dx = Dx$$, and so $$D = D_{ \varphi}$$. \end{proof} \todo[inline]{Might as well find-and-replace "map" with "morphism"!} \begin{proof}[of injectivity] Suppose over $$D$$ that we have $$D_{\psi}$$ for some $$\psi \in \mathop{\mathrm{Hom}}_{\mathfrak{g}}(\mathcal{I}, M)$$. We then have \begin{align*} \phi(ux) = uD(x) = u \Psi(x) = \Psi(ux) && \forall u\in {\mathcal{U}(\mathfrak{g}) }, x\in {\mathfrak{g}} .\end{align*} Since $$\mathcal{I} = {\mathcal{U}(\mathfrak{g}) }\, {\mathfrak{g}}$$ and $$\phi = \Psi$$, yielding a 1-to-1 map. \end{proof} \end{proof} \hypertarget{lie-algebra-cohomology-monday-april-19}{% \section{Lie Algebra Cohomology (Monday, April 19)}\label{lie-algebra-cohomology-monday-april-19}} \hypertarget{identification-of-h1-as-derivations}{% \subsection{\texorpdfstring{Identification of $$H^1$$ as Derivations}{Identification of H\^{}1 as Derivations}}\label{identification-of-h1-as-derivations}} \begin{remark} Let $${\mathfrak{g}}\in{\mathsf{Lie}{\hbox{-}}\mathsf{Alg}}_{/k}$$ and $$M\in{\mathfrak{g}{\hbox{-}}\mathsf{Mod}}$$, we were showing \begin{align*} H^1({\mathfrak{g}}; M) \cong { \mathop{\mathrm{Hom}}_{{\mathfrak{g}{\hbox{-}}\mathsf{Mod}}}( \mathcal{I}, M) \over \operatorname{im}\qty{ \mathop{\mathrm{Hom}}_{{\mathcal{U}(\mathfrak{g}) }}({\mathcal{U}(\mathfrak{g}) }, M) \to \mathop{\mathrm{Hom}}_{{\mathfrak{g}}}( \mathcal{I}, M) } } ,\end{align*} where the source in the denominator is isomorphic to $$M$$, given by the map $$\operatorname{ev}_1$$. We found a map \begin{align*} \mathop{\mathrm{Hom}}_{\mathfrak{g}}(\mathcal{I}, M) \xrightarrow{\sim} \mathop{\mathrm{Der}}({\mathfrak{g}}, M) \\ \phi \mapsto (D_\phi: x\mapsto \phi(x)) .\end{align*} We also defined inner derivations as those given by maps $$D_m(x) \coloneqq mx$$ for some $$m\in M$$. \end{remark} \begin{theorem}[?] \begin{align*} H^1({\mathfrak{g}}; M) \cong {\mathop{\mathrm{Der}}({\mathfrak{g}}, M) \over {\operatorname{Inn}}({\mathfrak{g}}, M) } .\end{align*} \end{theorem} \begin{proof}[?] In the formula, we already know that the numerator is isomorphic to $$\mathop{\mathrm{Der}}({\mathfrak{g}}, M)$$, so it remains to look at the denominator. The map appearing there is restriction to $$\mathcal{I}$$, i.e.~$$\phi \mapsto { \left.{{\phi}} \right|_{{\mathcal{I}}} }$$. The associated derivation is given by \begin{align*} D_\phi(x) = \phi(x) = \phi(x\cdot 1) = x \phi(1) = xm \coloneqq D_m(x) ,\end{align*} and so $$D_\phi = D_m$$. Conversely, given an $$m$$, we get a derivation $$D_m$$, and thus the image is precisely all inner derivations. \end{proof} \hypertarget{lhs-spectral-sequences}{% \subsection{LHS Spectral Sequences}\label{lhs-spectral-sequences}} \begin{remark} If $${\mathfrak{h}}{~\trianglelefteq~}{\mathfrak{g}}$$, there is a SES in $${\mathsf{Lie}{\hbox{-}}\mathsf{Alg}}$$: \begin{align*} 0 \to {\mathfrak{h}}\to {\mathfrak{g}}\to {\mathfrak{g}}/{\mathfrak{h}}\to 0 .\end{align*} \end{remark} \begin{theorem}[LHS Spectral Sequence] Let $${\mathfrak{h}}{~\trianglelefteq~}{\mathfrak{g}}$$ and $$M\in{\mathfrak{g}{\hbox{-}}\mathsf{Mod}}$$, then there are first quadrant spectral sequences \begin{align*} E_{p, q}^2 &= H_p( {\mathfrak{g}}/{\mathfrak{h}}; H_q({\mathfrak{h}}; M ) ) \Rightarrow H_{p+q}({\mathfrak{g}}; M) \\ E_{p, q}^2 &= H^p( {\mathfrak{g}}/{\mathfrak{h}}; H_q({\mathfrak{h}}; M)) \Rightarrow H^{p+q}({\mathfrak{g}}; M) .\end{align*} \end{theorem} \begin{remark} This comes from a similar application of the Grothendieck spectral sequence. The exact sequences in low-degree terms are given as usual\footnote{See Weibel p.233.} and similar inflation and restriction maps appear here. This is useful because it allows computing homology of smaller'' algebras, which one might have control over by induction. \end{remark} \hypertarget{chevalley-eilenberg-koszul-complex}{% \subsubsection{7.7: Chevalley-Eilenberg (Koszul) Complex}\label{chevalley-eilenberg-koszul-complex}} \begin{remark} A computationally efficient way of compute Lie algebra cohomology using a projective resolution of the trivial $${\mathfrak{g}}{\hbox{-}}$$module $$k\in{\mathfrak{g}{\hbox{-}}\mathsf{Mod}}$$, recalling that this involves acting by zero. \footnote{See VIGRE project at UGA: programmed this resolution in GAP to compute Lie algebra cohomology!} We're going to define a chain complex \begin{align*} V_*({\mathfrak{g}}) \xrightarrow[]{\varepsilon} { \mathrel{\mkern-16mu}\rightarrow }\, k ,\end{align*} which will turn out to be supported in finitely many degrees when $$\dim_k {\mathfrak{g}}< \infty$$. \end{remark} \begin{remark} We'll assume $${\mathfrak{g}}\in { {\mathsf{Lie}{\hbox{-}}\mathsf{Alg}}_{/k} }(\mathsf{Free})$$, which happens e.g.~if $$k\in \mathsf{Field}$$. Recall that the exterior algebra was a graded algebra defined as the quotient of the tensor algebra: \begin{align*} \bigwedge^* {\mathfrak{g}}\coloneqq{ T({\mathfrak{g}}) \over \left\langle{x^{\otimes 2} {~\mathrel{\Big|}~}x\in {\mathfrak{g}} }\right\rangle} = \bigoplus _{p\geq 0} \bigwedge^p {\mathfrak{g}} .\end{align*} We write $$x_1\wedge x_2\wedge\cdots \wedge x_p$$ for the image of $${\mathfrak{g}}\hookrightarrow\bigwedge^* {\mathfrak{g}}$$ Note that this is a 2-sided homogeneous ideal, and since $$x\wedge x = 0$$ we have $$x\wedge y = -y\wedge x$$. \end{remark} \begin{remark} If $$\left\{{ x_\alpha }\right\}$$ is an ordered basis for $${\mathfrak{g}}$$, then there is an ordered basis for $$\bigwedge^p {\mathfrak{g}}$$: \begin{align*} \left\{{ { {x_{\alpha}}_1, {x_{\alpha}}_2, \cdots, {x_{\alpha}}_{p}} {~\mathrel{\Big|}~}\alpha_1 < \cdots < \alpha_p }\right\} ,\end{align*} where we note that the indices are strictly increasing like the sequences $$I$$ we had previously. One can always arrange this by commuting things to organize the sequence properly. We also have $$\bigwedge^0 {\mathfrak{g}}\cong k$$ with a basis of $$1_k$$, and $$\bigwedge^1 {\mathfrak{g}}\cong {\mathfrak{g}}$$. In particular, if $$\dim {\mathfrak{g}}= n < \infty$$, then $$\bigwedge^p {\mathfrak{g}}= 0$$ for all $$p>n$$, and in this case $$\bigwedge^n {\mathfrak{g}}\cong k$$. \end{remark} \begin{definition}[The Chevalley-Eilenberg (or Koszul) Complex] Define \begin{align*} V_p({\mathfrak{g}}) \coloneqq{\mathcal{U}(\mathfrak{g}) }\otimes_k \bigwedge^p{\mathfrak{g}} ,\end{align*} where the maps are given below. \end{definition} \begin{fact} $$V_p({\mathfrak{g}})$$ is free in $$\mathsf{{\mathcal{U}(\mathfrak{g}) }}{\hbox{-}}\mathsf{Mod}$$, since we've constructed a free basis, and so in particular it is projective. The maps in the complex are given by the following: \begin{center} \begin{tikzcd} &&&&&& {\mathcal{I} = \ker \varepsilon= {\mathcal{U}(\mathfrak{g}) }\,{\mathfrak{g}}} \\ \\ &&&&&&&& {\mathcal{U}(\mathfrak{g}) }\\ \\ {V_p({\mathfrak{g}})} && {V_{p-1}({\mathfrak{g}})} && \cdots && {V_1({\mathfrak{g}})} && {V_0({\mathfrak{g}})} && k \\ \\ {{\mathcal{U}(\mathfrak{g}) }\otimes_k \bigwedge^p {\mathfrak{g}}} && {{\mathcal{U}(\mathfrak{g}) }\otimes_k \bigwedge^{p-1} {\mathfrak{g}}} &&&& {{\mathcal{U}(\mathfrak{g}) }\otimes_k {\mathfrak{g}}} && {\mathcal{U}(\mathfrak{g}) }&&&& 0 \\ {u \otimes\bigwedge_j x_{i_j}} && {\theta_1 + \theta_2} &&&& {u\otimes x} && ux \arrow["\cong", no head, from=3-9, to=5-9] \arrow["{\exists \varepsilon}", dashed, from=5-9, to=5-11] \arrow["\varepsilon", from=3-9, to=5-11] \arrow[hook, from=1-7, to=3-9] \arrow["{d_p}", from=5-1, to=5-3] \arrow[from=5-3, to=5-5] \arrow[from=5-5, to=5-7] \arrow["{d_1}", from=5-7, to=5-9] \arrow[maps to, from=8-7, to=8-9] \arrow[no head, from=7-7, to=5-7] \arrow[no head, from=7-9, to=5-9] \arrow[maps to, from=8-1, to=8-3] \arrow["{d_p}", from=7-1, to=7-3] \arrow[no head, from=7-3, to=5-3] \arrow[no head, from=7-1, to=5-1] \arrow["{d_1}", from=7-7, to=7-9] \arrow[from=5-11, to=7-13] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=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}{Link to Diagram} \end{quote} Here we define \begin{align*} \theta_1 &\coloneqq\sum_{i=1}^p ux_i \otimes x_1 \wedge x_2 \wedge \cdots \wedge \widehat{x_i} \wedge \cdots \wedge x_p \\ \theta_2 &\coloneqq \sum_{i < j}^p (-1)^{i+j} u\otimes[x_i x_j] \otimes x_1 \wedge x_2 \wedge \cdots \wedge \widehat{x_i} \wedge \cdots \wedge \widehat{x_j} \wedge \cdots \wedge x_p ,\end{align*} where the hat denotes omitting a term. Note that $$\operatorname{im}d_1 = {\mathcal{U}(\mathfrak{g}) }\,{\mathfrak{g}}= \mathcal{I} = \ker \varepsilon$$, so we get exactness at the first position, and exercise 7.7.1 shows that $$d^2 = 0$$. \end{fact} \begin{example}[?] For $$p=2$$, we have \begin{align*} d(u\otimes x \otimes y) = \qty{ux\otimes y - uy\otimes x} + \qty{- u \otimes[xy] } .\end{align*} \end{example} \begin{remark} We want this to be a projective resolution, so not just that $$\ker \subset \operatorname{im}$$, but rather we want exactness everywhere so $$\ker = \operatorname{im}$$. We'll proceed by showing its homology vanishes. \end{remark} \begin{theorem}[Koszul Resolution] The Koszul complex $$V_*({\mathfrak{g}}) \xrightarrow[]{\varepsilon} { \mathrel{\mkern-16mu}\rightarrow }\, k\,$$ is a projective resolution in $${\mathfrak{g}{\hbox{-}}\mathsf{Mod}}$$. \end{theorem} \begin{proof}[of theorem] Choose an ordered basis $$\left\{{ e_{ \alpha}}\right\}_{\alpha\in \Omega}$$, where $$\Omega$$ some totally ordered index set, for $${\mathfrak{g}}$$ over $$k$$. By the PBW theorem, $$V_n \coloneqq V_n({\mathfrak{g}})$$ has a free $$k{\hbox{-}}$$basis given by $$\label{basis_elts_pbw_koszul} e_I \otimes\qty{e_{\alpha_1} \otimes\cdots e_{\alpha_n}} .$$ for $$I = [ \beta_1, \cdots, \beta_m ]$$ some weakly increasing sequence from $$\Omega$$. This gives a filtration, so we're heading toward using the spectral sequence of a filtered complex. The filtered pieces are given by $$F_p V_n$$ defined as the $$k{\hbox{-}}$$module generated by elements of the form given in \cref{basis_elts_pbw_koszul} where $$m+n \leq p$$. Looking at the formula for $$d$$, we will get a differential \begin{align*} d_n F_p V_n \to F_p V_{n-1} .\end{align*} \end{proof} \hypertarget{exactness-of-the-chevalley-eilenberg-resolution-wednesday-april-21}{% \section{Exactness of the Chevalley-Eilenberg Resolution (Wednesday, April 21)}\label{exactness-of-the-chevalley-eilenberg-resolution-wednesday-april-21}} \begin{remark} Recall that $${\mathfrak{g}}$$ was free over $$k$$ with an ordered basis $$\left\{{ e_{\alpha} {~\mathrel{\Big|}~}\alpha\in \Omega }\right\}$$. We defined \begin{align*} V_n({\mathfrak{g}}) \coloneqq{\mathcal{U}(\mathfrak{g}) }\otimes_k \bigwedge^n {\mathfrak{g}} \end{align*} with a differential $$d = \theta_1 + \theta_2$$. We claimed that $$V_n({\mathfrak{g}}) \xrightarrow[]{\varepsilon} { \mathrel{\mkern-16mu}\rightarrow }\, k$$ is a projective resolution, and we were showing that $$V_*$$ was an exact complex. \end{remark} \begin{proof}[of theorem, continued] We define a filtration \begin{align*} F_p V_n \coloneqq k \left\langle{ e_I \otimes e_{ \alpha_1} \wedge \cdots \wedge e_{\alpha_n} {~\mathrel{\Big|}~}I = [{ {\beta}_1, {\beta}_2, \cdots, {\beta}_{m}}], \alpha_1 \leq \cdots \leq \alpha_n,\, m+n\leq p }\right\rangle .\end{align*} Note that $$d: F_p V_n \to F_p V_{n-1}$$, and in fact $$\theta_2$$ maps into $$F_{p-1} V_{n-1}$$. We'll focus on $$\theta_1$$ for simplicity. It lands in the same complex since we can rearrange elements in the sum defining the differential to express everything in terms of the given basis, where every expression will be of length one less. The commutation relation was \begin{align*} e_{ \beta} e_{\alpha} = e_{\alpha} e_{\beta} + [e_{\beta} e_{ \alpha} ] ,\end{align*} where the left-hand side is degree 2, and the right-hand side is a degree 2 term plus a degree 1 term. Moreover $$d$$ preserves the filtration: when rearranging, the degree $$u$$ term in $$\theta_1$$ will decrease to $$m-1$$, the expression following it may increase to $$n+1$$, and $$(m-1) + (n+1) = m+n$$. So $$F_p V_*$$ is a subcomplex of $$V_*$$, and we have \begin{align*} 0 = F_{-1} V_* \subseteq F_0 V_* \subseteq \cdots \subseteq F_p V_* \subseteq V_* = \bigcup_{p\geq 0} F_p V_* ,\end{align*} which is not a finite filtration, but is bounded below and exhaustive. So by the canonical convergence theorem (Weibel 5.5.1), there is a convergent spectral sequence \begin{align*} E_{p, q}^0 \coloneqq{F_p V_{p+q} \over F_{p-1} V_{p+q} } \Rightarrow H_{p+q}(V_*({\mathfrak{g}})) .\end{align*} We have $$E_{p, q}^0 = 0$$ unless \begin{itemize} \item $$p\geq 0$$, since the exterior algebra is only graded in positive degrees. \item $$p+q = n \geq 0$$ \item $$q\leq 0$$, which requires some explanation. We have $$m+n\leq p$$, and so if $$p+q=n \leq p-m$$ for $$m\geq 0$$, $$q\leq -m \leq 0$$. \end{itemize} So this is a 4th quadrant spectral sequence that is supported above the line $$y=-x$$. Recall that $$E_{p, q}^1 = H_q^v(E_{p, *}^0)$$. \begin{center} \begin{tikzcd} \bullet && \bullet \\ \\ \bullet &&& \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet \\ &&&& \bullet & \bullet & \bullet & \bullet & \bullet & \bullet \\ &&&&& \bullet & \bullet & \bullet & \bullet & \bullet \\ &&&&&& \bullet & \bullet & \bullet & \bullet \\ &&&&&&& \bullet & \bullet & \bullet \\ &&&&&&&& \bullet \\ && \bullet \arrow[from=3-1, to=3-11] \arrow[from=1-3, to=9-3] \arrow[from=1-1, to=8-9] \arrow[from=3-5, to=4-5] \arrow[from=3-6, to=4-6] \arrow[from=4-6, to=5-6] \arrow["{d^0}", from=3-7, to=4-7] \arrow[from=4-7, to=5-7] \arrow[from=5-7, to=6-7] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=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}{Link to Diagram} \end{quote} Note that \begin{align*} E_{\infty}^0 \coloneqq F_0 V_0 / F_{-1} V_0 = k\otimes_k k \cong k ,\end{align*} since we take expressions with length zero in each factor defining $$F_p V_n$$. Moreover this position is already stable provided the $$E_{p, 0}^{1} = 0$$ for all $$p$$, the first and third quadrants are all zeros, and thus all differentials will be trivial from $$E^1$$ onward. \begin{claim} For $$p>0$$, $$E_{p, *}^0$$ is exact, and thus the spectral sequence collapses at $$E^1$$. \end{claim} Note that turning the page yields \begin{align*} E_{p, q}^1 = \begin{cases} k & (p, q) = (0, 0) \\ 0 & \text{else}. \end{cases} \end{align*} Thus $$H_n(V_*({\mathfrak{g}})) = k$$ in $$n=0$$ and zero elsewhere, which proves the result. \begin{proof}[Sketch] For $$q\gg 0$$, define $$A_q \coloneqq k \left\langle{ e_I {~\mathrel{\Big|}~}I = [{ {\beta}_1, {\beta}_2, \cdots, {\beta}_{q}}] \text{ increasing} }\right\rangle \subseteq {\mathcal{U}(\mathfrak{g}) }$$. So $$A_q$$ is the $$q$$th graded piece of the standard increasing filtration by degree, \begin{align*} k = U_0 \subset U_1 \subset \cdots \subseteq {\mathcal{U}(\mathfrak{g}) } .\end{align*} Note that this is a section standard filtration of $${\mathcal{U}(\mathfrak{g}) }$$ by degree with respect to the PBW basis\footnote{See exercise 7.3.6.}. We have $$A_q \cong F_q V_0 / F_{q-1} V_0$$ and \begin{align*} E_{p, q}^0 = {F_p V_{p+q} \over F_{p-1} V_{p+q} } \cong A_{-q} \otimes_k \bigwedge^{p+q}{\mathfrak{g}} .\end{align*} The negative sign is introduced since this is nonzero precisely when $$-p\leq q \leq 0$$ so $$q$$ is negative and $$-q$$ is positive. Now using the definition of $$d: V_n \to V_{n-1}$$, $$d^0$$ is vertical and \begin{align*} d^0: E_{p, q}^0 &\to E_{p, q-1}^0 \quad\quad n = p+q \\ \\ \cong d^0: A_{-1} \otimes_k \bigwedge^n{\mathfrak{g}}&\to A_{-q+1}\otimes_k \bigwedge^{n-1} {\mathfrak{g}} \quad\quad n = p+q .\end{align*} Recalling how $$d^0$$ was defined, note that we're modding out by lower order terms and thus brackets get killed when we commute elements to order them. \hfill\break By Weibel 7.3.6, $$A \coloneqq\bigoplus_{q\geq 0} A_q$$ is in fact a graded algebra, and $$A \cong {\mathsf{gr}\,}{\mathcal{U}(\mathfrak{g}) }$$, the associated graded of $${\mathcal{U}(\mathfrak{g}) }$$. This turns out to be a polynomial ring on the indeterminates $$\mathbf{x} = \left\{{ e_{ \alpha } }\right\}_{\alpha\in \Omega}$$, i.e.~$$A\cong k[\mathbf{x}]$$. In Weibel section 4.5, Weibel studies the \emph{Koszul} complex and the map $$A \otimes_k \bigwedge^* {\mathfrak{g}}\to A$$. By comparing the formula for $$d$$ between these two complexes, one observes that the Koszul complex differentials are equal to the $$d^0$$ here. So we have an equality of complexes \begin{align*} A \otimes_k \bigwedge^* {\mathfrak{g}}= \bigoplus _{p\geq 0} E_{p, *}^0 .\end{align*} Weibel section 4.5 shows that when $$A\in \mathsf{CRing}$$ with no zero divisors, e.g.~a polynomial ring, then \begin{align*} H_n \qty{ A \otimes_k \bigwedge^* {\mathfrak{g}}} = \begin{cases} k & n=0 \\ 0 & \text{else}. \end{cases} \end{align*} On the other hand, we have \begin{align*} H_n \qty{ A \otimes_k \bigwedge^* {\mathfrak{g}}} &= \bigoplus H_{n-p}^v( E_{p, *}^0 ) \hspace{4em} p+q=n \implies q=n-p \\ &= \bigoplus _{p\geq 0} E_{p, n-p}^1 .\end{align*} But we've already shown that $$E_{0, 0}^1 = k$$, so all of the other $$E^1$$ terms must be zero. \end{proof} \end{proof} \begin{remark} See section 4.5 on Koszul complex. We'll do 7.8 next time. \end{remark} \hypertarget{friday-april-23}{% \section{Friday, April 23}\label{friday-april-23}} \hypertarget{applications-chevalley-eilenberg-complex}{% \subsection{Applications Chevalley-Eilenberg Complex}\label{applications-chevalley-eilenberg-complex}} \begin{remark} Last time: $$V_n({\mathfrak{g}}) \coloneqq{\mathcal{U}(\mathfrak{g}) }\otimes_k \bigwedge^n {\mathfrak{g}} \xrightarrow[]{\varepsilon} { \mathrel{\mkern-16mu}\rightarrow }\, k$$ is a projective resolution in $${\mathfrak{g}{\hbox{-}}\mathsf{Mod}}$$. Note that we can introduce negative signs to easily interchange $$\mathsf{{\mathfrak{g}}}{\hbox{-}}\mathsf{Mod}$$ and $$\mathsf{Mod}{\hbox{-}}\mathsf{{\mathfrak{g}}}$$. \end{remark} \begin{corollary}[Chevalley-Eilenberg] Let $$M\in \mathsf{Mod}{\hbox{-}}\mathsf{{\mathfrak{g}}}$$, then \begin{align*} H_*({\mathfrak{g}}; M) \cong \operatorname{Tor}_*^{{\mathcal{U}(\mathfrak{g}) }}(M, k) \end{align*} is the homology of the following complex: \begin{align*} M\otimes_{\mathcal{U}(\mathfrak{g}) }V_*({\mathfrak{g}}) \coloneqq M \otimes_{\mathcal{U}(\mathfrak{g}) }{\mathcal{U}(\mathfrak{g}) }\otimes_k \bigwedge^* {\mathfrak{g}} \cong M\otimes_k \bigwedge^* {\mathfrak{g}} ,\end{align*} where we have a concrete differential $$d$$ on $$\bigwedge^* {\mathfrak{g}}$$ and we can define $$\partial \coloneqq\one \otimes d$$. If $$M\in \mathsf{{\mathfrak{g}}}{\hbox{-}}\mathsf{Mod})$$ (which is more convenient for cohomology), then \begin{align*} H^*({\mathfrak{g}}; M) \cong \operatorname{Ext}_{\mathcal{U}(\mathfrak{g}) }(k, M) \end{align*} is the cohomology of the cochain complex \begin{align*} \mathop{\mathrm{Hom}}_{\mathcal{U}(\mathfrak{g}) }(V_*({\mathfrak{g}}), M) \coloneqq\mathop{\mathrm{Hom}}_{{\mathcal{U}(\mathfrak{g}) }}({\mathcal{U}(\mathfrak{g}) }\otimes_k \bigwedge^* {\mathfrak{g}}, M) \cong \mathop{\mathrm{Hom}}_k(\bigwedge^*{\mathfrak{g}}, M) .\end{align*} \end{corollary} \begin{remark} This is very concrete! Standard trick for exterior algebras: any $$n{\hbox{-}}$$cochain $$f\in \mathop{\mathrm{Hom}}_k(\bigwedge^n {\mathfrak{g}}, M)$$ can be viewed as an alternating $$k{\hbox{-}}$$multilinear function $$f(x_1, \cdots, x_n): {\mathfrak{g}}\to M$$. The cochain differential should increase degree, so we define \begin{align*} \Theta_1 f({ {x}_1, {x}_2, \cdots, {x}_{n}}) = \sum_{i=1}^{n+1} (-1)^{i+1} x_i \cdot f(x_1, \cdots, \widehat{x_i}, \cdots, x_n) + \sum_{i < j} (-1)^{i+1} f( [x_i x_j], x_1, \cdots, \widehat{x_i}, \cdots, \widehat{x_j}, \cdots, x_n) .\end{align*} Note that the tor definition has the arguments switched compared to the original definition. This is to set up the tensor cancellation of $$\cdots \otimes_{\mathcal{U}(\mathfrak{g}) }{\mathcal{U}(\mathfrak{g}) }\cdots$$. Swapping factors and introducing signs makes this work for left $${\mathfrak{g}}{\hbox{-}}$$modules. \end{remark} \begin{corollary}[?] If $$k$$ is a field and $$\dim_k {\mathfrak{g}}= n$$, then for any $$M \in {\mathfrak{g}{\hbox{-}}\mathsf{Mod}}$$, \begin{align*} H^i({\mathfrak{g}}; M) = 0 = H_i({\mathfrak{g}}; M) && \forall i\geq n+1 .\end{align*} \end{corollary} \begin{proof}[?] This follows from the fact that $$\bigwedge^{\geq n+1} {\mathfrak{g}}=0$$. \end{proof} \begin{example}[?] Take $${\mathfrak{g}}= {\mathfrak{sl}}_2({\mathbb{C}})$$, then $$\dim_{\mathbb{C}}{\mathfrak{g}}= 3$$ (4 dimensions and one linear condition). Then $$H^i({\mathfrak{g}}; m) = 0$$ for all $$i > 3$$. \end{example} \hypertarget{brief-intro-to-semisimple-lie-algebras-weibel-7.8}{% \subsection{Brief Intro to Semisimple Lie Algebras (Weibel 7.8)}\label{brief-intro-to-semisimple-lie-algebras-weibel-7.8}} \begin{remark} Public service section since we won't have a Lie algebras course next Fall. Semisimples: the most important and interesting classes of Lie algebras! These occur frequently and we can prove a lot about them. We'll assume $${\mathfrak{g}}$$ is a finite dimensional Lie algebra over a field $$k$$, where we'll soon assume $$\operatorname{ch}(k) = 0$$. \end{remark} \begin{definition}[Simple Lie Algebras] A Lie algebra $${\mathfrak{g}}$$ is \textbf{simple} if it has no ideals other than $$0$$ and $${\mathfrak{g}}$$ and $$[{\mathfrak{g}}{\mathfrak{g}}] \neq 0$$ (i.e.~$${\mathfrak{g}}$$ is not abelian). \end{definition} \begin{remark} Recall that $${\mathsf{Lie}{\hbox{-}}\mathsf{Alg}}^{\mathsf{Ab}}\approx {\mathsf{k}{\hbox{-}}\mathsf{Mod}}$$ are vector spaces, and so if $${\mathfrak{g}}$$ is abelian it automatically has a chain of ideals by just taking vector subspaces. These are closed under brackets since bracketing is zero. So if $$\dim_k {\mathfrak{g}}\geq 2$$, there are nontrivial ideals, so the abelian condition rules out all 1-dimensional Lie algebras -- they're all abelian by taking a generator, bracketing it with itself, and noting you get zero. So there's only one 1-dimensional Lie algebra over any field $$k$$: the abelian one. \end{remark} \begin{remark} The derived algebra $$[{\mathfrak{g}}{\mathfrak{g}}] {~\trianglelefteq~}{\mathfrak{g}}$$ is a subalgebra and always an ideal, so if $${\mathfrak{g}}$$ is simple then $$[{\mathfrak{g}}{\mathfrak{g}}] = {\mathfrak{g}}$$. So $${\mathfrak{gl}}_n({\mathbb{C}})$$ is not simple, since $$[{\mathfrak{gl}}_n {\mathfrak{gl}}_n] = {\mathfrak{sl}}_n$$ by taking traces. \end{remark} \begin{remark} The vector space sum of any two solvable ideals is again a solvable ideal. Note that this works for products of solvable subgroups $$N, H\leq G$$ with $$N$$ normal. Use 2-out-of-3 property for solvable groups and quotient by $$N$$. By finite-dimensionality, we can find a maximal solvable ideal: \end{remark} \begin{definition}[?] For $$\dim_k {\mathfrak{g}}< \infty$$, define the \textbf{radical} to be $$\mathop{\mathrm{rad}}{\mathfrak{g}}\coloneqq\sum I_j$$ be the sum of all solvable ideals $$I_j{~\trianglelefteq~}{\mathfrak{g}}$$. We say $${\mathfrak{g}}$$ is \textbf{semisimple} if $$\mathop{\mathrm{rad}}{\mathfrak{g}}= 0$$. \end{definition} \begin{lemma}[?] Simple implies semisimple. \end{lemma} \begin{lemma}[?] $${\mathfrak{g}}/ \mathop{\mathrm{rad}}{\mathfrak{g}}$$ is always semisimple. \end{lemma} \begin{remark} There shouldn't be any solvable ideals in this quotient, otherwise you could lift. Next up, our most powerful tool for semisimple Lie algebras: \end{remark} \begin{definition}[?] Recall that for $$x\in{\mathfrak{g}}$$ we can define $$\operatorname{ad}_x \in \mathop{\mathrm{End}}_k({\mathfrak{g}})$$ where $$\operatorname{ad}_x(y) \coloneqq[x, y]$$. It has a well-defined (and basis-independent) trace, so define the \textbf{Killing form}\footnote{Named for a mathematician \emph{named} Killing.}: \begin{align*} \kappa(x, y) \coloneqq\operatorname{Tr}(\operatorname{ad}_x \circ \operatorname{ad}_y) \in k && x,y\in{\mathfrak{g}} .\end{align*} \end{definition} \begin{remark} This is a symmetric bilinear form since traces don't depend on the order of products. It has another nice property, $${\mathfrak{g}}{\hbox{-}}$$invariance: \begin{align*} \kappa([xy], z] = \kappa(x, [yz]) .\end{align*} \end{remark} \begin{proposition}[Cartan's Criterion] Let $$\operatorname{ch}(k) = 0$$ and $$\dim_k {\mathfrak{g}}< \infty$$. Then $${\mathfrak{g}}$$ is semisimple $$\iff \kappa$$ is nondegenerate. \end{proposition} \begin{proof}[?] Omitted, see Humphreys. \end{proof} \begin{theorem}[?] Let $$\operatorname{ch}(k) = 0$$, then $${\mathfrak{g}}$$ is semisimple $$\iff {\mathfrak{g}}\cong \bigoplus_{i=1}^r {\mathfrak{g}}_i$$ as a direct sum/product of simple ideals, so $$[{\mathfrak{g}}_i {\mathfrak{g}}_j] = 0$$ for $$i\neq j$$ and $$[{\mathfrak{g}}_i {\mathfrak{g}}_i ] = {\mathfrak{g}}_i$$. In particular, every ideal of $${\mathfrak{g}}$$ is a sum of sum of certain $${\mathfrak{g}}_i$$'s, and $${\mathfrak{g}}= [{\mathfrak{g}}{\mathfrak{g}}]$$. \end{theorem} \begin{remark} These are like orthogonal'' ideals. So we can study semisimple Lie algebras by just studying simple Lie algebras. \end{remark} \begin{observation} Reminder: if $$M\in {\mathfrak{g}{\hbox{-}}\mathsf{Mod}}({ \operatorname{Triv}})$$, then any derivation $$D \in \mathop{\mathrm{Der}}({\mathfrak{g}}, M)$$ satisfies $$D([xy]) = 0$$ for all $$x,y\in {\mathfrak{g}}$$. This follows from expanding the Leibniz rule and using trivial modules act by zero. There is an isomorphism \begin{align*} \mathop{\mathrm{Der}}({\mathfrak{g}}, M) \cong \mathop{\mathrm{Hom}}_{\mathsf{k}{\hbox{-}}\mathsf{Mod}}({\mathfrak{g}}^{\operatorname{ab}}, M) && {\mathfrak{g}}^{\operatorname{ab}}\coloneqq{\mathfrak{g}}/ [{\mathfrak{g}}{\mathfrak{g}}] .\end{align*} Recall that $$H^1$$ is related to derivations. \end{observation} \begin{corollary}[?] Let $${\mathfrak{g}}\in {\mathfrak{g}{\hbox{-}}\mathsf{Mod}}({\operatorname{ss}})$$ with $$\dim_k {\mathfrak{g}}< \infty$$, then \begin{align*} H^1({\mathfrak{g}};k ) = 0 = H_1({\mathfrak{g}}; k) .\end{align*} \end{corollary} \begin{proof}[?] Since $$[{\mathfrak{g}}{\mathfrak{g}}] = {\mathfrak{g}}$$, we have $${\mathfrak{g}}^{\operatorname{ab}}= 0$$. By Weibel theorem 7.4.1, one can check that $$H_1({\mathfrak{g}}; k) \cong {\mathfrak{g}}^{\operatorname{ab}}= 0$$. We also had $$\mathop{\mathrm{Der}}({\mathfrak{g}}, k) \twoheadrightarrow H^1({\mathfrak{g}}; k)$$ (it was outer derivations), the left-hand side is isomorphic to $$\mathop{\mathrm{Hom}}_k({\mathfrak{g}}^{\operatorname{ab}}; k)$$. \end{proof} \begin{theorem}[?] Let $${\mathfrak{g}}\in {\mathsf{Lie}{\hbox{-}}\mathsf{Alg}}({\operatorname{ss}})$$ with $$\dim_k {\mathfrak{g}}< \infty$$ and $$\operatorname{ch}(k) = 0$$. Then if $$k\neq M$$ is a simple $${\mathfrak{g}}{\hbox{-}}$$module (where simple means no proper nontrivial $${\mathfrak{g}}{\hbox{-}}$$invariant submodules), then \begin{align*} H^i({\mathfrak{g}}; M) = 0 = H_i({\mathfrak{g}}; M) .\end{align*} \end{theorem} \begin{proof}[?] Omitted. This uses the Casimir operator for $$M$$, which is in the center $$Z({\mathcal{U}(\mathfrak{g}) })$$. \end{proof} \hypertarget{section-7.6-wednesday-april-28}{% \section{Section 7.6 (Wednesday, April 28)}\label{section-7.6-wednesday-april-28}} \begin{remark} Today: filling in some previous things, including proofs for Whitehead's second lemma and Levi's theorem. \end{remark} \begin{definition}[?] Let $${\mathfrak{g}}\in{\mathsf{Lie}{\hbox{-}}\mathsf{Alg}}_{/k}$$ for $$k\in \mathsf{CRing}$$ and let $$M\in {\mathsf{k}{\hbox{-}}\mathsf{Mod}}$$ viewed as a trivial $${\mathfrak{g}}{\hbox{-}}$$ module. An \textbf{extension of $${\mathfrak{g}}$$ by $$M$$} is a SES in $${\mathfrak{g}{\hbox{-}}\mathsf{Mod}}$$ of the following form: \begin{align*} 0 \to M \xrightarrow{\iota} E \xrightarrow{\pi} {\mathfrak{g}}\to 0 .\end{align*} \end{definition} \begin{remark} Given such an extension, thinking of $$M \subset E$$, $$M$$ becomes a $${\mathfrak{g}}{\hbox{-}}$$module in a natural way: given $$m\in M$$ and $$x\in{\mathfrak{g}}$$, choose $$\tilde x\in E$$ such that $$\pi(\tilde x) = x$$ and set \begin{align*} x\cdot m \coloneqq[\tilde x, m]_E \in M {~\trianglelefteq~}E ,\end{align*} noting that $$M$$ is the kernel of a morphism and thus an ideal. Is this well-defined? If $$\tilde x'\in \pi^{-1}(E)$$, we have $$\pi(\tilde x' - \tilde x) = 0$$ which implies $$\tilde x' -\tilde x\in \ker \pi = M$$ by exactness. So we can write $$\tilde x' = m' + \tilde x$$ for some $$m'\in M$$, and since $$M$$ is abelian and its elements bracket to zero, we have \begin{align*} [\tilde x ', m] = [m' + \tilde x, m] = [\tilde x, m] .\end{align*} \end{remark} \begin{remark} The extension problem: given a $${\mathfrak{g}}{\hbox{-}}$$module $$M$$ viewed as an abelian Lie algebra, how many (equivalence classes of) extensions of $${\mathfrak{g}}$$ by $$M$$ are there for which the induced action above agrees with the given action? Here we view equivalence as existence of an isomorphism making the following diagram commute: \begin{center} \begin{tikzcd} 0 && M && E && {\mathfrak{g}}&& 0 \\ \\ &&&& {E'} \arrow["\sim", dashed, from=1-5, to=3-5] \arrow["{\iota'}"', from=1-3, to=3-5] \arrow[from=1-1, to=1-3] \arrow["\iota", from=1-3, to=1-5] \arrow["\pi", from=1-5, to=1-7] \arrow[from=1-7, to=1-9] \arrow["{\pi'}"', from=3-5, to=1-7] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsNixbMCwwLCIwIl0sWzIsMCwiTSJdLFs0LDAsIkUiXSxbNiwwLCJcXGxpZWciXSxbOCwwLCIwIl0sWzQsMiwiRSciXSxbMiw1LCJcXHNpbSIsMCx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFsxLDUsIlxcaW90YSciLDJdLFswLDFdLFsxLDIsIlxcaW90YSJdLFsyLDMsIlxccGkiXSxbMyw0XSxbNSwzLCJcXHBpJyIsMl1d}{Link to Diagram} \end{quote} Write $$\operatorname{Ext}_{{\mathsf{Lie}{\hbox{-}}\mathsf{Alg}}}({\mathfrak{g}}, M)$$ for the set of equivalence classes of such extensions. \end{remark} \begin{remark} We can form semidirect products $$M\rtimes{\mathfrak{g}}$$ of Lie algebras in the following way: start with the $$k{\hbox{-}}$$module $$M \times{\mathfrak{g}}$$ with bracket \begin{align*} [(m, x), (n, y)] \coloneqq(x\cdot n - y\cdot m, [xy]) && m,n\in M,\,\, x,y\in{\mathfrak{g}} .\end{align*} One checks that this is anticommutative and satisfies the Jacobi identity. This is a Lie algebra containing $$M \times 0$$ as an abelian ideal and $$0 \times{\mathfrak{g}}$$ as a subalgebra, which fits into a SES \begin{align*} 0 \to M \xrightarrow{\iota} M\rtimes{\mathfrak{g}}\xrightarrow{\pi} {\mathfrak{g}}\to 0 .\end{align*} Moreover, the naturally induced action described previously agrees with this semidirect action. Identifying elements with their inclusions, we have \begin{align*} [(0, x), (m, 0)] = (x\cdot m - 0, [0, 0] ) = (x\cdot m, 0) .\end{align*} Thus there is always at least one extension, called the \textbf{split extension}. There is a classification: \end{remark} \begin{theorem}[Classification of Extensions] Let $$M\in {\mathfrak{g}{\hbox{-}}\mathsf{Mod}}$$, then there is a bijection of sets \begin{align*} \operatorname{Ext}({\mathfrak{g}}, M) &\rightleftharpoons H^2({\mathfrak{g}}; M) \end{align*} \end{theorem} \begin{remark} Note that the map $$\pi$$ makes $$M$$ into an $$E{\hbox{-}}$$module and makes $$M$$ into a trivial $$M{\hbox{-}}$$module. See Weibel for a functorial proof, using the same correspondence between $$\operatorname{Ext}_R^1(A, B)$$ and extensions of $$A$$ by $$B$$. Note that we have an algebra, an ideal, and its quotient, which is precisely the setup for the LHS spectral sequence for cohomology with coefficients in $$M$$. There was an associated 5-term exact sequence, which contains a \textbf{classifying map} \begin{align*} \mathop{\mathrm{Hom}}_{\mathfrak{g}{\hbox{-}}\mathsf{Mod}}(M, M) &\xrightarrow{d^2} \mathop{\mathrm{Hom}}_{{\mathfrak{g}{\hbox{-}}\mathsf{Mod}}}({\mathfrak{g}}, M) \\ \one_M &\mapsto d^2(\one_M) .\end{align*} One checks that this only depends on the equivalence class of extensions, and turns out to be a bijection. Weibel's proof uses some facts about free Lie algebras that we haven't discussed yet, so we'll instead do a slightly more down-to-earth proof from Knapp's book using the Koszul complex. \end{remark} \begin{proof}[of classification theorem] We'll need to assume $$k\in \mathsf{Field}$$. Choose a splitting of the following SES as a $$k{\hbox{-}}$$vector space: \begin{center} \begin{tikzcd} 0 && M && E && {\mathfrak{g}}&& 0 \arrow[from=1-1, to=1-3] \arrow["\iota", from=1-3, to=1-5] \arrow["\pi", shift left=2, from=1-5, to=1-7] \arrow[from=1-7, to=1-9] \arrow["j", shift left=3, dotted, from=1-7, to=1-5] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsNSxbMCwwLCIwIl0sWzIsMCwiTSJdLFs0LDAsIkUiXSxbNiwwLCJcXGxpZWciXSxbOCwwLCIwIl0sWzAsMV0sWzEsMiwiXFxpb3RhIl0sWzIsMywiXFxwaSIsMCx7Im9mZnNldCI6LTJ9XSxbMyw0XSxbMywyLCJqIiwwLHsib2Zmc2V0IjotMywic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZG90dGVkIn19fV1d}{Link to Diagram} \end{quote} So here $$\pi \circ j = \one_{\mathfrak{g}}$$. Note that we can use $$j(x)$$ for our $$\tilde x$$. From section 7.7, we can characterize $$H^2({\mathfrak{g}}; M)$$ is a subquotient of $$\mathop{\mathrm{Hom}}_k\qty{ \bigwedge^2 {\mathfrak{g}}, M}$$, recalling that we canceled a $${\mathcal{U}(\mathfrak{g}) }$$ when taking the resolution \begin{align*} {\mathcal{U}(\mathfrak{g}) }\otimes\bigwedge^* {\mathfrak{g}} \xrightarrow[]{\varepsilon} { \mathrel{\mkern-16mu}\rightarrow }\, k \end{align*} and applying $$\mathop{\mathrm{Hom}}_k({-}, M)$$. Specifically, it is $$\ker \delta / \operatorname{im}\delta$$ for the coboundary $$\delta$$ from corollary 7.7.3. Recall that we define a hom from an $$n$$th piece of an exterior algebra is equivalence to an alternating $$n{\hbox{-}}$$argument function, and define $$w\in \mathop{\mathrm{Hom}}_k\qty{ \bigwedge^2 {\mathfrak{g}}, M }$$ by \begin{align*} w(x, y) = [jx, hy]_E - j\qty{ [xy]_M } \in E ,\end{align*} where we'll omit parentheses and bracket subscripts immediately. We want to detect if this is in $$M$$, so use that $$M = \ker \pi$$ and check \begin{align*} \pi ([jx, jy] - j[xy]) &= [\pi j x, \pi j y] - \pi j[xy] \\ &= [xy] - [xy] \\ &= 0 ,\end{align*} and so $$w(x, y) \in M$$ as needed. We now want to compute $$\delta w$$ to compute the action $$x\cdot m \coloneqq[\tilde x, m]_M$$, so take $$\tilde x \coloneqq j(x)$$. Use that $$\delta$$ has graded degree $$+1$$, so \begin{align*} \delta w(x,y,z) &= x\cdot w(y,z) - y\cdot w(x, z) + z\cdot w(x, y) \\ &\quad -w([xy], z) + w([xz], y) - w([yz], x) \\ \\ &= [jx, [jy, jz]] - [jz, j[yz]] \\ &\quad - [jy, [jz, jz]] + [jy, j[xz]] \\ &\quad + [jz, [jx, jy]] - [jz, j[xy]] \\ &\quad - [j[xy], jz] + j [[xy], z] \\ &\quad + [j[xz], jy] - j [[xz], y] \\ &\quad - [j[yz], jx] + j [[yz], x] .\end{align*} There is a lot of cancellation here! Use the Jacobi identity for terms in red, and sign rules to cancel the rest: \begin{figure} \centering \includegraphics{figures/image_2021-04-28-10-01-22.png} \caption{image\_2021-04-28-10-01-22} \end{figure} So $$w\in \ker \delta$$. \begin{quote} To be continued. \end{quote} \end{proof} \begin{remark} One should check that choices differ by coboundaries, along with a few other things that we're eliding. \end{remark} \hypertarget{friday-april-30}{% \section{Friday, April 30}\label{friday-april-30}} \hypertarget{proof-continued}{% \subsection{Proof Continued}\label{proof-continued}} \begin{remark} Last time: we were proving the bijection between $$H^2({\mathfrak{g}}; M)$$ and extensions of $${\mathfrak{g}}$$ by $$M$$ up to equivalence. \end{remark} \begin{proof}[of the classification theorem, continued] We chose a vector space splitting $${\mathfrak{g}}\xrightarrow{j} E$$ and used the Cartan-Eilenberg resolution to construct a 2-cocycle $$w\in \mathop{\mathrm{Hom}}_{{\mathsf{k}{\hbox{-}}\mathsf{Mod}}}(\bigwedge^2 {\mathfrak{g}}, M)$$ given by \begin{align*} w(x, y) \coloneqq[jx, jy] - j[x, y] && x,y{\mathfrak{g}} ,\end{align*} and we saw that $$d(w) = 0$$. Say we change $$j$$ to $$j': {\mathfrak{g}}\to E$$ to $$j': {\mathfrak{g}}\to E$$ with $$\pi j' = \one_{{\mathfrak{g}}}$$, and let $$w'$$ be the corresponding 2-cocycle. Letting $$\alpha \coloneqq j-j'$$, then $$\pi \circ \alpha = 0$$ by linearity and so $$\alpha : {\mathfrak{g}}\to \ker \pi = M$$ and thus $$\alpha\in\mathop{\mathrm{Hom}}_{{\mathsf{k}{\hbox{-}}\mathsf{Mod}}}(\bigwedge^1 {\mathfrak{g}}, M)$$. We then have \begin{align*} \delta \alpha (x, y) &= x \alpha(y) - y \alpha(x) - \alpha([xy]) \\ &= [j'x, j'y - jy] - [jy, j'x - jx]-j'[xy] + j[xy] \\ &= \qty{ [j'x, j'y] - j'[xy] } - [j'x, jy] + [j'x, jy] - \qty{[jx, jy] - j[xy] } \\ &= \qty{ [j'x, j'y] - j'[xy] } - \qty{[jx, jy] - j[xy] } \\ &= w'(x, y) - w(x, y) ,\end{align*} so $$\delta\alpha = w' -w$$. So their difference is a coboundary, yielding $$w = w' \in H^2({\mathfrak{g}}, M)$$, making this construction independent of the choice of $$j$$. \begin{exercise}[?] Show that equivalent extensions also lead to the same element in $$H^2$$. \end{exercise} This yields a well-defined map \begin{align*} \left\{{ \text{Extensions of{\mathfrak{g}}$by$M} }\right\} &\to H^2({\mathfrak{g}}; M) \\ (0\to M\to E\to {\mathfrak{g}}\to 0) &\mapsto w .\end{align*} Conversely, given a 2-cocycle $$\tilde w\in\mathop{\mathrm{Hom}}_{{\mathsf{k}{\hbox{-}}\mathsf{Mod}}}(\bigwedge^2 {\mathfrak{g}}, M)$$ with $$M$$ abelian, define \begin{itemize} \tightlist \item $$E \coloneqq M \oplus {\mathfrak{g}}\in {\mathsf{Vect}}_k$$ \item A bracket using the following rules (identifying $$M, {\mathfrak{g}}$$ with their images in $$E$$): \begin{itemize} \tightlist \item $$[m, n]_E \coloneqq 0$$ \item $$[x, m]_E \coloneqq x\cdot m = -[m, x]_E$$ \begin{itemize} \tightlist \item Note that this guarantees that the actions agree. \end{itemize} \item $$[x, y]_E \coloneqq\tilde w(x, y) + [x, y]_{\mathfrak{g}}$$ \end{itemize} \end{itemize} One can check that the last definition is anticommutative since $$w$$ was alternating, and further that this makes $$E$$ into a Lie algebra that fits into a SES of the desired form with canonical maps $$i, \pi, j$$. The cocycle $$w$$ coming from this extension is given by \begin{align*} w(x, y) &= [x, y]_E - [x, y]_{\mathfrak{g}}\\ &= \tilde w(x, y) + [x, y]_{\mathfrak{g}}- [x, y]_{{\mathfrak{g}}} \\ &= \tilde w(x, y) \end{align*} where here $$j$$ is a direct sum inclusions that we'll suppress. So $$H^2 \to \operatorname{Ext}/\sim \to H^2$$ is the identity. One can similarly check that $$\operatorname{Ext}/\sim \to H^2 \to \operatorname{Ext}/\sim$$ is also the identity, since it produces an equivalent extension. So this defines a bijection of sets. \end{proof} \begin{remark} This was known much earlier for group cohomology: if $$G \in {\mathsf{Grp}}, A\in {\mathsf{G}{\hbox{-}}\mathsf{Mod}}$$, there is a bijection \begin{align*} \left\{{ 0\to A\to E \xrightarrow{\pi} G \to 1 }\right\} \coloneqq \left\{{ \text{Equivalence classes of extensions ofG$by$A} }\right\} &\to H^2(G; A) ,\end{align*} where $$G$$ may not be abelian, and one acts by conjugation instead. Analogy: bracketing is like the differential of conjugation. \end{remark} \hypertarget{proof-backlog-from-monday}{% \subsection{Proof Backlog from Monday}\label{proof-backlog-from-monday}} \begin{remark} Recall Whitehead's Lemma 2\footnote{Weibel corollary 7.8.12} for $${\mathfrak{g}}$$ finite-dimensional and semisimple over $$\operatorname{ch}(k) = 0$$ and $$M\in {\mathfrak{g}{\hbox{-}}\mathsf{Mod}}^{\operatorname{fd}}$$, then $$H^2({\mathfrak{g}}; M) = 0$$. \end{remark} \begin{proof}[?] By Weyl's theorem, $$M$$ is a direct sum of simple $${\mathfrak{g}}{\hbox{-}}$$modules and $$H^*$$ commutes with direct sums, so it suffices to show this when $$M$$ is simple. By Weibel theorem 7.8.9 (structure of semisimple Lie algebras using the Casimir operator) we have $$H^n({\mathfrak{g}}; M) = 0$$ for $$M\neq k$$ and for all $$n$$, so we reduce to showing this for $$M=k$$. By the classification theorem, we need to show that every extension of the following form splits: \begin{align*} 0\to k\to E \xrightarrow{\pi} {\mathfrak{g}}\to 0 ,\end{align*} where we view $$k \in {\mathsf{Lie}{\hbox{-}}\mathsf{Alg}}^{\operatorname{ab}}$$. We proceed in an unanticipated way by reducing Lie algebra maps to $${\mathfrak{g}}{\hbox{-}}$$module maps.\\ First note that $$k \subset Z{\mathfrak{g}}$$, since $$E \cong k \oplus {\mathfrak{g}}\in {\mathsf{Vect}}_k$$, so there is an embedding $${\mathfrak{g}}\hookrightarrow E$$ where say $$x\mapsto \tilde x$$. For $$c\in k$$ and $$x\in {\mathfrak{g}}$$, we have $$[\tilde x, c] \coloneqq x\cdot c = 0$$ since $$k \in {\mathsf{Lie}{\hbox{-}}\mathsf{Alg}}^{\operatorname{ab}}$$, and by linearity this will show that $$k$$ commutes with everything. We now make $$E$$ into a $${\mathfrak{g}}{\hbox{-}}$$module by defining $$x\cdot e \coloneqq[\tilde x, e]$$ for $$x\in {\mathfrak{g}}, e\in E$$. If $$\tilde x'$$ is another other representative in $$E$$ of $$x$$, then noting that $$k \in \ker \pi$$ we can write $$\tilde x' = [\tilde x + c, e] = [\tilde x, e]$$ using that $$c\in Z(E)$$. This action makes $$\pi$$ into a $${\mathfrak{g}}{\hbox{-}}$$module map, and we have \begin{align*} \pi(x\cdot e) &\coloneqq\pi( [\tilde x, e] ) \\ &= [\pi(\tilde x), \pi(e) ] && \pi \in {\mathsf{Lie}{\hbox{-}}\mathsf{Alg}}({\mathfrak{g}}, E) \\ &\coloneqq[x, \pi(e)] \\ &\coloneqq x\cdot \pi(e) \in {\mathfrak{g}} ,\end{align*} since this is acting via the adjoint action. By Weyl's theorem, both $$E$$ and $${\mathfrak{g}}$$ decompose into direct sums of simple $${\mathfrak{g}}{\hbox{-}}$$modules. Using that $$j$$ is injective and a $${\mathfrak{g}}{\hbox{-}}$$module map, it must send simple submodules of $${\mathfrak{g}}$$ to simple submodules of $$E$$, using that maps to (from?) simple modules are either zero or isomorphisms and a dimension count. One can check (easily!) that there is a $${\mathfrak{g}}{\hbox{-}}$$module map $$\sigma: {\mathfrak{g}}\hookrightarrow E$$ such that $$E \cong K \oplus \sigma({\mathfrak{g}}) \in {\mathfrak{g}{\hbox{-}}\mathsf{Mod}}$$. So choose $$\tilde x \coloneqq\sigma(x)$$, then $$\sigma \in {\mathsf{Lie}{\hbox{-}}\mathsf{Alg}}({\mathfrak{g}}, E)$$, and so \begin{align*} \sigma( [x, y] ) &\coloneqq x \cdot \sigma(y) \\ &= [\tilde x, \sigma(y) ] \\ &= [\sigma(x), \sigma(y)] ,\end{align*} making $$\sigma({\mathfrak{g}}) \leq E$$ a Lie-subalgebra. Since $$\sigma({\mathfrak{g}}) \cong {\mathfrak{g}}$$, this is precisely a semidirect product and we obtain $$E \cong k \rtimes{\mathfrak{g}}$$, and the sequence splits as desired. \end{proof} \begin{remark} Next time: Levi's theorem. \end{remark} \hypertarget{appendix-extra-definitions}{% \section{Appendix: Extra Definitions}\label{appendix-extra-definitions}} \hypertarget{extra-references}{% \section{Extra References}\label{extra-references}} \begin{itemize} \tightlist \item \url{https://www.math.wisc.edu/~csimpson6/notes/2020_spring_homological_algebra/notes.pdf} \end{itemize} \hypertarget{useful-facts}{% \section{Useful Facts}\label{useful-facts}} \begin{definition}[Acyclic] A chain complex $$C$$ is \textbf{acyclic} if and only if $$H_*(C) = 0$$. \end{definition} \begin{proposition}[Algebra Facts] \envlist \begin{itemize} \tightlist \item Free $$\implies$$ projective $$\implies$$ flat $$\implies$$ torsionfree (for finitely-generated $$R{\hbox{-}}$$modules) \begin{itemize} \tightlist \item Over $$R$$ a PID: free $$\iff$$ torsionfree \end{itemize} \end{itemize} \end{proposition} \begin{remark} Notational conventions: \begin{itemize} \tightlist \item Finite direct products: $$\bigoplus$$ \item Cohomological indexing: $$C^i, {{\partial}}^i$$ \item Homological indexing: $$C_i, {{\partial}}_i$$ \item Right-derived functors $$R^iF$$. \begin{itemize} \tightlist \item Come from left-exact functors \item Require \emph{injective} resolutions \item Extend to the right: $$0 \to F(A) \to F(B) \to F(C) \to L_1 F(A) \cdots$$ \end{itemize} \item Left-derived functors $$L_i F$$. \begin{itemize} \tightlist \item Come from right-exact functors \item Require \emph{projective} resolutions \item Extend to the left: $$\cdots L_1F(C) \to F(A) \to F(B) \to F(C) \to 0$$ \end{itemize} \item Colimits: \begin{itemize} \tightlist \item Examples: coproducts, direct limits, cokernels, initial objects, pushouts \item Commute with left adjoints, i.e.~$$L(\mathop{\mathrm{colim}}\nolimits F_i) = \mathop{\mathrm{colim}}\nolimits LF_i$$. \end{itemize} \item Examples of limits: \begin{itemize} \tightlist \item Products, inverse limits, kernels, terminal objects, pullbacks \item Commute with right adjoints. i.e.~$$R(\mathop{\mathrm{colim}}\nolimits F_i) = \mathop{\mathrm{colim}}\nolimits RF_i$$. \end{itemize} \end{itemize} \end{remark} \hypertarget{hom-and-ext}{% \subsection{Hom and Ext}\label{hom-and-ext}} \begin{proposition}[Basic properties of Hom] \envlist \begin{itemize} \tightlist \item $$\mathop{\mathrm{Hom}}_R(A, {-})$$ is: \begin{itemize} \tightlist \item Covariant \item Left-exact \item Is a functor that sends $$f:X\to Y$$ to $$f_*: \mathop{\mathrm{Hom}}(A, X) \to \mathop{\mathrm{Hom}}(A, Y)$$ given by $$f_*(h) = f\circ h$$. \item Has right-derived functors $$\operatorname{Ext}^i_R(A, B) \coloneqq R^i \mathop{\mathrm{Hom}}_R(A, {-})(B)$$ computed using \emph{injective} resolutions. \end{itemize} \item $$\mathop{\mathrm{Hom}}_R({-}, B)$$ is: \begin{itemize} \tightlist \item Contravariant \item Right-exact \item Is a functor that sends $$f:X\to Y$$ to $$f^*: \mathop{\mathrm{Hom}}(Y, B) \to \mathop{\mathrm{Hom}}(X, B)$$ given by $$f^*(h) = h\circ f$$. \item Has left-derived functors $$\operatorname{Ext}^i_R(A, B) \coloneqq L_i \mathop{\mathrm{Hom}}_R({-}, B)(A)$$ computed using \emph{projective} resolutions. \end{itemize} \item For $$N \in (\mathsf{R}, \mathsf{S'}){\hbox{-}}\mathsf{biMod}$$ and $$M\in (\mathsf{R}, \mathsf{S}){\hbox{-}}\mathsf{biMod}$$, $$\mathop{\mathrm{Hom}}_R(M, N) \in (\mathsf{S}, \mathsf{S'}){\hbox{-}}\mathsf{biMod}$$. \begin{itemize} \tightlist \item Mnemonic: the slots of $$\mathop{\mathrm{Hom}}_R$$ use up a left $$R{\hbox{-}}$$action. In the first slot, the right $$S{\hbox{-}}$$action on $$M$$ becomes a left $$S{\hbox{-}}$$action on Hom. In the second slot, the right $$S'{\hbox{-}}$$action on $$N$$ becomes a right $$S'{\hbox{-}}$$action on Hom. \end{itemize} \end{itemize} \end{proposition} \begin{proposition}[Basic Properties of Ext] \envlist \begin{itemize} \tightlist \item $$\operatorname{Ext}^{>1}(A, B) = 0$$ for any $$A$$ projective or $$B$$ injective. \end{itemize} \end{proposition} \begin{fact} A maps $$A \xrightarrow{f} B$$ in $${\mathsf{R}{\hbox{-}}\mathsf{Mod}}$$ is injective if and only if $$f(a) = 0_B \implies a = 0_A$$. Monomorphisms are injective maps in $${\mathsf{R}{\hbox{-}}\mathsf{Mod}}$$. \end{fact} \begin{proposition}[Recipe for computing\Ext_R^i] Write $$F({-}) \coloneqq\mathop{\mathrm{Hom}}_R(A, {-})$$. This is left-exact and thus has right-derived functors $$\operatorname{Ext}^i_R(A, B) \coloneqq R^iF(B)$$. To compute this: \begin{itemize} \item Take an \emph{injective} resolution: \begin{align*} 1 \to B \xrightarrow{\varepsilon} I^0 \xrightarrow{d^0} I^1 \xrightarrow{d^1} \cdots .\end{align*} \item Remove the augmentation $$\varepsilon$$ and just keep the complex \begin{align*} I^{-}\coloneqq\qty{ 1 \xrightarrow{d^{-1}} I^0 \xrightarrow{d^0} I^1 \xrightarrow{d^1} \cdots } .\end{align*} \item Apply $$F({-})$$ to get a new (and usually \textbf{not exact}) complex \begin{align*} F(I)^{-}\coloneqq\qty{ 1 \xrightarrow{{{\partial}}^{-1}} F(I^0) \xrightarrow{{{\partial}}^0} F(I^1) \xrightarrow{{{\partial}}^1} \cdots } ,\end{align*} where $${{\partial}}^i \coloneqq F(d^i)$$. \item Take homology, i.e.~kernels mod images: \begin{align*} R^iF(B) \coloneqq{ \ker d^i \over \operatorname{im}d^{i-1}} .\end{align*} \end{itemize} Note that $$R^0 F(B) \cong F(B)$$ canonically: \begin{itemize} \item This is defined as $$\ker {{\partial}}^0 / \operatorname{im}{{\partial}}^{-1} = \ker {{\partial}}^0 / 1 = \ker {{\partial}}^0$$. \item Use the fact that $$F({-})$$ is left exact and apply it to the \emph{augmented} complex to obtain \begin{align*} 1 \to F(B) \xrightarrow{F(\varepsilon)} F(I^0) \xrightarrow{{{\partial}}^0} F(I^1) \xrightarrow{{{\partial}}^1} \cdots .\end{align*} \item By exactness, there is an isomorphism $$\ker {{\partial}}^0 \cong F(B)$$. \end{itemize} \end{proposition} \begin{proposition}[Computing\Hom_\ZZ(\ZZ, \ZZ/n)\$] $$\phi: \mathop{\mathrm{Hom}}_{{\mathbb{Z}}}({\mathbb{Z}}, {\mathbb{Z}}/n) \xrightarrow{\sim} {\mathbb{Z}}/n$$, where $$\phi(g) \coloneqq g(1)$$. \begin{itemize} \item That this is an isomorphism follows from \item Surjectivity: for each $$\ell \in {\mathbb{Z}}/n$$ define a map \begin{align*} \psi_y: {\mathbb{Z}}&\to {\mathbb{Z}}/n \\ 1 &\mapsto [\ell]_n .\end{align*} \item Injectivity: if $$g(1) = [0]_n$$, then \begin{align*} g(x) = xg(1) = x[0]_n = [0]_n .\end{align*} \item $${\mathbb{Z}}{\hbox{-}}$$module morphism: \begin{align*} \phi(gf) \coloneqq\phi(g\circ f) \coloneqq(g\circ f)(1) = g(f(1)) = f(1)g(1) = \phi(g)\phi(f) ,\end{align*} where we've used the fact that $${\mathbb{Z}}/n$$ is commutative. \end{itemize} \end{proposition} \begin{proposition}[Common Hom Groups] \begin{itemize} \tightlist \item $$\mathop{\mathrm{Hom}}_{\mathbb{Z}}({\mathbb{Z}}/m, {\mathbb{Z}}) = 0$$. \item $$\mathop{\mathrm{Hom}}_{\mathbb{Z}}({\mathbb{Z}}/m, {\mathbb{Z}}/n) = {\mathbb{Z}}/d$$. \item $$\mathop{\mathrm{Hom}}_{\mathbb{Z}}({\mathbb{Q}}, {\mathbb{Q}}) = {\mathbb{Q}}$$. \end{itemize} \end{proposition} \begin{proposition}[Common Ext Groups] \begin{itemize} \tightlist \item $$\operatorname{Ext}_{\mathbb{Z}}({\mathbb{Z}}/m, G) \cong G/mG$$ \begin{itemize} \tightlist \item Use $$1 \to {\mathbb{Z}}\xrightarrow{\times m} {\mathbb{Z}}\xrightarrow{} {\mathbb{Z}}/m \to 1$$ and apply $$\mathop{\mathrm{Hom}}_{\mathbb{Z}}({-}, {\mathbb{Z}})$$. \end{itemize} \item $$\operatorname{Ext}_{\mathbb{Z}}({\mathbb{Z}}/m, {\mathbb{Z}}/n) = {\mathbb{Z}}/d$$. \end{itemize} \end{proposition} \begin{slogan} \envlist \begin{itemize} \item In $${\mathsf{Ab}}$$, direct colimits commute with finite limits. Inverse limits do not generally commute with finite colimits. \item Left adjoints are right-exact with left-derived functors. Right adjoints are left-exact with right-derived functors. \item Left adjoints commute with colimits: $$L( \mathop{\mathrm{colim}}\nolimits F) = \mathop{\mathrm{colim}}\nolimits(L\circ F)$$ \end{itemize} \end{slogan} \begin{proposition}[Characterizations of Splittings] TFAE in $${\mathsf{R}{\hbox{-}}\mathsf{Mod}}$$: \begin{itemize} \tightlist \item A SES $$0\to A\to B \to C\to 0$$ is split. \item ? \end{itemize} \end{proposition} \hypertarget{tensor-and-tor}{% \subsection{Tensor and Tor}\label{tensor-and-tor}} \begin{proposition}[Basic Properties of the Tensor Product] \envlist \begin{itemize} \tightlist \item $$A\otimes_R {-}$$ is: \begin{itemize} \tightlist \item Covariant \item Right-exact \item Left-exact \item Has left-derived functors $$\operatorname{Ext}^i_R(A, B) \coloneqq L_i \mathop{\mathrm{Hom}}_R({-}, B)(A)$$ computed using \emph{projective} resolutions. \end{itemize} \item $${-}\otimes_R B$$ is: \begin{itemize} \tightlist \item Covariant \item Right-exact \item Has left-derived functors $$\operatorname{Ext}^i_R(A, B) \coloneqq L_i \mathop{\mathrm{Hom}}_R({-}, B)(A)$$ computed using \emph{projective} resolutions. \end{itemize} \item Tensor commutes with colimits: $$(\mathop{\mathrm{colim}}\nolimits A_i)\otimes_R M = \mathop{\mathrm{colim}}\nolimits(A_i \otimes_R M)$$. \end{itemize} \end{proposition} \begin{proposition}[Basic Properties of Tor] \envlist \begin{itemize} \tightlist \item $$\operatorname{Tor}_n^R(A, B) = 0$$ for either $$A$$ or $$B$$ flat. \end{itemize} \end{proposition} \begin{fact} The most useful SES for proofs here: \begin{align*} 0 \to {\mathbb{Z}}\xrightarrow{n} {\mathbb{Z}}\xrightarrow{\pi} {\mathbb{Z}}/n \to 0 .\end{align*} \end{fact} \begin{proposition}[Common Tensor Products] \envlist \begin{itemize} \tightlist \item $${\mathbb{Z}}/n \otimes_{\mathbb{Z}}G \cong G/nG$$ \item $${\mathbb{Z}}/n \otimes_{\mathbb{Z}}{\mathbb{Z}}/m \cong {\mathbb{Z}}/d$$. \item $${\mathbb{Q}}\otimes_{\mathbb{Z}}{\mathbb{Z}}/n \cong 0$$. \end{itemize} \end{proposition} \begin{proposition}[Common Tor Groups] \begin{itemize} \tightlist \item $$\operatorname{Tor}^{\mathbb{Z}}_1({\mathbb{Z}}/n, G) \cong \left\{{ h\in H {~\mathrel{\Big|}~}nh = e }\right\}$$ \item $$\operatorname{Tor}^{\mathbb{Z}}_1({\mathbb{Z}}/n, {\mathbb{Q}}) \cong 0$$. \item $$\operatorname{Tor}^{\mathbb{Z}}_1({\mathbb{Z}}/n, {\mathbb{Z}}/m) \cong {\mathbb{Z}}/d$$. \end{itemize} \end{proposition} \hypertarget{universal-properties}{% \subsection{Universal Properties}\label{universal-properties}} \begin{proposition}[Universal Property of the Quotient for Groups] If $$f: G\to K$$ and $$H{~\trianglelefteq~}G$$ (so that $$G/H$$ is defined), then the map $$f$$ descends to the quotient if and only if $$H \subseteq \ker(f)$$. \end{proposition} \begin{proposition}[Kernels as pullbacks and cokernels as pushouts] The kernel $$\ker f$$ of a morphism $$f$$ can be characterized as a cartesian square, and the cokernel $$\operatorname{coker}f$$ as a cocartesian square: \begin{center} \begin{tikzcd} K \\ & {\ker f} && \textcolor{rgb,255:red,92;green,92;blue,214}{A} && 0 \\ \\ & 0 && \textcolor{rgb,255:red,92;green,92;blue,214}{B} && {\operatorname{coker}f} \\ &&&&&& C \arrow[dotted, from=2-6, to=4-6] \arrow[from=2-4, to=2-6] \arrow["f"', color={rgb,255:red,92;green,92;blue,214}, from=2-4, to=4-4] \arrow["0"', dotted, from=4-4, to=4-6] \arrow["\lrcorner"{anchor=center, pos=0.125, rotate=180}, draw=none, from=4-6, to=2-4] \arrow["{\exists !}"', dashed, from=4-6, to=5-7] \arrow[curve={height=12pt}, from=4-4, to=5-7] \arrow[curve={height=-12pt}, from=2-6, to=5-7] \arrow[dotted, from=2-2, to=2-4] \arrow[from=4-2, to=4-4] \arrow[dotted, from=2-2, to=4-2] \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=2-2, to=4-4] \arrow[curve={height=-12pt}, from=1-1, to=2-4] \arrow[curve={height=12pt}, from=1-1, to=4-2] \arrow["{\exists !}"', dashed, from=1-1, to=2-2] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsOCxbMywxLCJBIixbMjQwLDYwLDYwLDFdXSxbNSwxLCIwIl0sWzMsMywiQiIsWzI0MCw2MCw2MCwxXV0sWzUsMywiXFxjb2tlciBmIl0sWzYsNCwiQyJdLFsxLDEsIlxca2VyIGYiXSxbMSwzLCIwIl0sWzAsMCwiSyJdLFsxLDMsIiIsMCx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRvdHRlZCJ9fX1dLFswLDFdLFswLDIsImYiLDIseyJjb2xvdXIiOlsyNDAsNjAsNjBdfSxbMjQwLDYwLDYwLDFdXSxbMiwzLCIwIiwyLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZG90dGVkIn19fV0sWzMsMCwiIiwwLHsic3R5bGUiOnsibmFtZSI6ImNvcm5lciJ9fV0sWzMsNCwiXFxleGlzdHMgISIsMix7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFsyLDQsIiIsMCx7ImN1cnZlIjoyfV0sWzEsNCwiIiwwLHsiY3VydmUiOi0yfV0sWzUsMCwiIiwwLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZG90dGVkIn19fV0sWzYsMl0sWzUsNiwiIiwxLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZG90dGVkIn19fV0sWzUsMiwiIiwxLHsic3R5bGUiOnsibmFtZSI6ImNvcm5lciJ9fV0sWzcsMCwiIiwxLHsiY3VydmUiOi0yfV0sWzcsNiwiIiwxLHsiY3VydmUiOjJ9XSxbNyw1LCJcXGV4aXN0cyAhIiwyLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV1d}{Link to Diagram} \end{quote} \end{proposition} \hypertarget{adjunctions}{% \subsection{Adjunctions}\label{adjunctions}} \begin{definition}[Adjoints] \todo[inline]{todo} \end{definition} \begin{proposition}[Tensor-Hom Adjunction] For a fixed $$M\in (\mathsf{R}, \mathsf{S}){\hbox{-}}\mathsf{biMod}$$, there is an adjunction \begin{align*} \adjunction{ {-}\otimes_R M }{\mathop{\mathrm{Hom}}_S(M, {-})}{ \mathsf{Mod}{\hbox{-}}\mathsf{R} } { \mathsf{Mod}{\hbox{-}}\mathsf{S} } ,\end{align*} so for $$Y \in (\mathsf{A}, \mathsf{R}){\hbox{-}}\mathsf{biMod}$$ and $$Z \in (\mathsf{B}, \mathsf{S}){\hbox{-}}\mathsf{biMod}$$, there is a (natural) isomorphism in $$(\mathsf{B}, \mathsf{A}){\hbox{-}}\mathsf{biMod}$$: \begin{align*} \mathop{\mathrm{Hom}}_S(X \otimes_R M, Z) \xrightarrow{\sim} \mathop{\mathrm{Hom}}_R( X, \mathop{\mathrm{Hom}}_S(M, Z) ) .\end{align*} \end{proposition} \begin{proposition}[Forgetful Adjunctions] Let $$F: {\mathsf{R}{\hbox{-}}\mathsf{Mod}} \to {\mathsf{{\mathbb{Z}}}{\hbox{-}}\mathsf{Mod}}$$ be the forgetful functor, then there are adjunctions \begin{align*} \adjunction{F}{ \mathop{\mathrm{Hom}}_{\mathbb{Z}}(R, {-})} {{\mathsf{R}{\hbox{-}}\mathsf{Mod}} } {{\mathsf{{\mathbb{Z}}}{\hbox{-}}\mathsf{Mod}} } \\ \\ \adjunction{R\otimes_{\mathbb{Z}}{-}}{F}{ {\mathsf{{\mathbb{Z}}}{\hbox{-}}\mathsf{Mod}} }{ {\mathsf{R}{\hbox{-}}\mathsf{Mod}} } .\end{align*} \end{proposition} \addsec{ToDos} \listoftodos[List of Todos] \cleardoublepage % Hook into amsthm environments to list them. \addsec{Definitions} \renewcommand{\listtheoremname}{} \listoftheorems[ignoreall,show={definition}, numwidth=3.5em] \cleardoublepage \addsec{Theorems} \renewcommand{\listtheoremname}{} \listoftheorems[ignoreall,show={theorem,proposition}, numwidth=3.5em] \cleardoublepage \addsec{Exercises} \renewcommand{\listtheoremname}{} \listoftheorems[ignoreall,show={exercise}, numwidth=3.5em] \cleardoublepage \addsec{Figures} \listoffigures \cleardoublepage \printbibliography[title=Bibliography] \end{document}