\input{"preamble.tex"}
\addbibresource{CohomologyRepTheory.bib}
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\title{
\rule{\linewidth}{1pt} \\
\textbf{
Cohomology in Representation Theory
}
\\ {\normalsize Lectures by Dan Nakano. University of Georgia,
Spring 2022} \\
\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 Cohomology in Representation Theory taught by
Dan Nakano at the University of Georgia in Spring 2022. 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{}
\maketitle
\begin{flushleft}
\textit{D. Zack Garza} \\
\textit{University of Georgia} \\
\textit{\href{mailto: dzackgarza@gmail.com}{dzackgarza@gmail.com}} \\
{\tiny \textit{Last updated:} 2022-05-29 }
\end{flushleft}
\newpage
% Note: addsec only in KomaScript
\addsec{Table of Contents}
\tableofcontents
\newpage
\hypertarget{introduction-and-background-tuesday-january-11}{%
\section{Introduction and Background (Tuesday, January
11)}\label{introduction-and-background-tuesday-january-11}}
\begin{remark}
References: \autocite{jacobson_2009}.
\end{remark}
\begin{remark}
Idea: study representation by studying associated geometric objects, and
use homological methods to bridge the two. The representation theory
side will mostly be rings/modules, and the geometric side will involve
algebraic geometry and commutative algebra. Throughout the course, all
rings will be unital and all actions on the left.
\end{remark}
\begin{example}[of categories of modules]
Recall the definition of a left \(R{\hbox{-}}\)module. Some examples:
\begin{itemize}
\tightlist
\item
\(k\in \mathsf{Field}\implies {\mathsf{k}{\hbox{-}}\mathsf{Mod}} = { \mathsf{Vect} }_k\)
\item
\(R={\mathbb{Z}}\implies {\mathsf{{\mathbb{Z}}}{\hbox{-}}\mathsf{Mod}} = {\mathsf{Ab}}{\mathsf{Grp}}\).
\item
\(A\in{\mathsf{Alg}}_{/ {k}}\), which is a ring \((A, +, \cdot)\)
where \((A, +, .)\) (using scalar multiplication) is a vector space.
\begin{itemize}
\tightlist
\item
E.g. \(\operatorname{Mat}(n\times n, {\mathbb{C}})\).
\item
E.g. for \(G\) a finite group, the group algebra \(kG\) for
\(k\in \mathsf{Field}\).
\item
E.g. \(U({\mathfrak{g}})\) for
\({\mathfrak{g}}\in \mathsf{Lie}{\mathsf{Alg}}\) or a super algebra.
\end{itemize}
\end{itemize}
\end{example}
\begin{remark}
Connecting this to representation theory: for
\(A\in {\mathsf{Alg}}_{/ {k}}\) and
\(M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}\), a representation of \(A\)
is a morphism of algebras \(A \xrightarrow{\rho} {\mathfrak{gl}}_n(k)\),
the algebra of all \(n\times n\) matrices (not necessarily invertible).
Note that for groups, one instead asks for maps
\(kG\to \operatorname{GL}_n\), the invertible matrices. There is a
correspondence between
\({\mathsf{A}{\hbox{-}}\mathsf{Mod}} \rightleftharpoons{\mathsf{Rep}}(A)\):
given \(M\), one can define the action as
\begin{align*}
\rho: A &\to \mathop{\mathrm{End}}_k(M) \\
\rho(a)(m) &= a.m
.\end{align*}
\end{remark}
\begin{remark}
Recall the definitions of:
\begin{itemize}
\tightlist
\item
Morphisms of \(R{\hbox{-}}\)modules:
\(f(r.m_1 + m_2) = r.f(m_1) + f(m_2)\)
\item
Submodules: \(N\leq M \iff r.n \in N\) and \(N\) is closed under
\(+\).
\item
Quotient modules: \(M/N = \left\{{m + N}\right\}\).
\item
The fundamental homomorphism theorem: for \(M \xrightarrow{f} N\),
there is an induced \(\psi: M/\ker f\to N\) where
\(M/\ker f\cong \operatorname{im}f\).
\end{itemize}
\begin{center}
\begin{tikzcd}
M && N & \textcolor{rgb,255:red,92;green,92;blue,214}{f(m)} \\
\\
{M/\ker f} \\
\textcolor{rgb,255:red,92;green,92;blue,214}{m + \ker f}
\arrow["\eta"', from=1-1, to=3-1]
\arrow["f", from=1-1, to=1-3]
\arrow["{\exists \psi}"', dashed, from=3-1, to=1-3]
\arrow[color={rgb,255:red,92;green,92;blue,214}, maps to, from=4-1, to=1-4]
\end{tikzcd}
\end{center}
\begin{quote}
\href{https://q.uiver.app/?q=WzAsNSxbMCwwLCJNIl0sWzIsMCwiTiJdLFswLDIsIk0vXFxrZXIgZiJdLFswLDMsIm0gKyBcXGtlciBmIixbMjQwLDYwLDYwLDFdXSxbMywwLCJmKG0pIixbMjQwLDYwLDYwLDFdXSxbMCwyLCJcXGV0YSIsMl0sWzAsMSwiZiJdLFsyLDEsIlxcZXhpc3RzIFxccHNpIiwyLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV0sWzMsNCwiIiwyLHsiY29sb3VyIjpbMjQwLDYwLDYwXSwic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoibWFwcyB0byJ9fX1dXQ==}{Link
to Diagram}
\end{quote}
\begin{itemize}
\tightlist
\item
The fundamental SES
\begin{align*}
0\to \ker f \xhookrightarrow{g} M \xrightarrow{f} \operatorname{im}f \to 0
,\end{align*}
where one generally needs \(\operatorname{im}g = \ker f\) for
exactness.
\begin{itemize}
\tightlist
\item
More generally, need monomorphisms, epimorphisms.
\end{itemize}
\end{itemize}
\end{remark}
\begin{example}[?]
Some examples:
\begin{itemize}
\tightlist
\item
\(f:{\mathbb{Z}}\to {\mathbb{Z}}\) where \(f(m) \coloneqq 4m\) yields
\(0\to {\mathbb{Z}}\xrightarrow{f} {\mathbb{Z}}\to {\mathbb{Z}}/4\to 0\)
in \({\mathsf{{\mathbb{Z}}}{\hbox{-}}\mathsf{Mod}}\).
\item
In \({\mathsf{{\mathbb{C}}}{\hbox{-}}\mathsf{Mod}}\), one can take
\(0\to {\mathbb{C}}\xrightarrow{\Delta: x\mapsto (x, x)} {\mathbb{C}}{ {}^{ \scriptscriptstyle\times^{2} } } \to {\mathbb{C}}\to 0\).
\end{itemize}
\end{example}
\begin{remark}
Direct sums, products, and indecomposables. Let \(I\) be an index set
and \(\left\{{M_k}\right\}_{k\in I}\) \(R{\hbox{-}}\)modules to define
the \textbf{direct product}
\(\prod_{k\in I} M_k \coloneqq\left\{{(m_k)_{k\in I} {~\mathrel{\Big\vert}~}m_k\in M_k }\right\}\),
the set of all ordered sequences of elements from the \(M_k\), with
addition defined pointwise. For the \textbf{direct sum}
\(\bigoplus _{k\in I} M_k\) to be those sequences with only finitely
many nonzero components. For internal direct sums, if \(M = M_1 + M_2\)
then \(M \cong M_1 \oplus M_2\) iff \(M \cap M_2 = 0\). An
\textbf{irreducible representation} is a simple \(R{\hbox{-}}\)module,
and an \textbf{indecomposable representation} is an indecomposable
\(R{\hbox{-}}\)module. An \(R{\hbox{-}}\)module is \textbf{simple} iff
its only submodules are \(0, M\), and \textbf{indecomposable} iff
\(M \not\cong M_1 \oplus M_2\) for any \(M_i\not\cong M\). Note that
simple \(\implies\) indecomposable.
\begin{quote}
Note: is it possible for \(M \cong M \oplus M\)?
\end{quote}
\end{remark}
\begin{example}[?]
Some examples:
\begin{itemize}
\item
Simple objects in \({\mathsf{k}{\hbox{-}}\mathsf{Mod}}\) are
isomorphic to \(k\), and indecomposables are also isomorphic to \(k\)
if we restrict to finite dimensional modules.
\item
Simple objects in \({\mathsf{{\mathbb{Z}}}{\hbox{-}}\mathsf{Mod}}\)
are cyclic groups of prime order, \(C_p\). Indecomposables are
\({\mathbb{Z}}, C_{p^k}\), using the classification theorem to rule
out composites.
\item
For \(A\in {\mathsf{Alg}}^{\mathrm{fd}}_{/ {k}}\), the simple objects
in \({\mathsf{A}{\hbox{-}}\mathsf{Mod}}\) are hard to determine in
general. The same goes for indecomposables, and is undecidable in many
cases (equivalent to the word problem in finite groups).
\begin{quote}
See \textbf{finite}, \textbf{tame}, and \textbf{wild} representation
types.
\end{quote}
\end{itemize}
\end{example}
\begin{remark}
Toward homological algebra: free and projective modules. An
\(R{\hbox{-}}\)module \(M\) is \textbf{free} iff
\(M\cong \bigoplus_{i\in I} R_i\) for some indexing set where
\(R_i \cong R\) as a left \(R{\hbox{-}}\)module. Equivalently, \(M\) has
a linearly independent spanning set, or there exists an \(X\) and a
unique \(\phi\) such that the following diagram commutes:
\begin{center}
\begin{tikzcd}
M \\
\\
X && N
\arrow["{\mathsf{Set}}", from=3-1, to=3-3]
\arrow["{\iota\in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}}", hook', from=3-1, to=1-1]
\arrow["{\exists ! \phi \in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}}", dashed, from=1-1, to=3-3]
\end{tikzcd}
\end{center}
\begin{quote}
\href{https://q.uiver.app/?q=WzAsMyxbMCwyLCJYIl0sWzAsMCwiTSJdLFsyLDIsIk4iXSxbMCwyLCJcXFNldCJdLFswLDEsIlxcaW90YVxcaW4gXFxtb2Rze1J9IiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJib3R0b20ifX19XSxbMSwyLCJcXGV4aXN0cyAhIFxccGhpIFxcaW4gXFxtb2Rze1J9IiwwLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV1d}{Link
to Diagram}
\end{quote}
Every \(M\in{\mathsf{R}{\hbox{-}}\mathsf{Mod}}\) is the image of a free
\(R{\hbox{-}}\)module: let \(X\coloneqq\left\{{m_i}\right\}_{i\in I}\)
generate \(M\), so \(X\hookrightarrow M\) by inclusion. Define
\(X \to \bigoplus \bigoplus_{i\in I} R_i\) sending
\(m_i \to (0,\cdots, 1, \cdots, 0)\) with a 1 in the \(i\)th position,
then since \(X\) is a generating set this will lift to a surjection
\(\bigoplus _i R_i\to M\). We can use this to define a free resolution:
\begin{center}
\begin{tikzcd}
{\ker \delta_1} \\
\cdots & \textcolor{rgb,255:red,92;green,92;blue,214}{\exists F_1} && {F_0} && M && 0 \\
&& \textcolor{rgb,255:red,214;green,92;blue,92}{\ker \delta_0} \\
& \textcolor{rgb,255:red,214;green,92;blue,92}{0} && \textcolor{rgb,255:red,92;green,92;blue,214}{0}
\arrow[from=4-2, to=3-3]
\arrow[color={rgb,255:red,214;green,92;blue,92}, no head, from=3-3, to=4-2]
\arrow["{\delta_0}", from=2-4, to=2-6]
\arrow[from=2-6, to=2-8]
\arrow[color={rgb,255:red,214;green,92;blue,92}, hook, from=3-3, to=2-4]
\arrow[color={rgb,255:red,92;green,92;blue,214}, two heads, from=2-2, to=3-3]
\arrow[color={rgb,255:red,92;green,92;blue,214}, from=3-3, to=4-4]
\arrow["{\exists\delta_1}", color={rgb,255:red,92;green,92;blue,214}, dashed, from=2-2, to=2-4]
\arrow[hook, from=1-1, to=2-2]
\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}
Let \(A\in {\mathsf{Alg}}^{\mathrm{fd}}_{/ {k}}\) and
\(F \cong \bigoplus A\) be free, and suppose \(e\in A\) is idempotent,
so \(e^2 = e\) -- these are useful because they can split algebras up.
There is a \emph{Pierce decomposition} of \(1\) given by
\(1 = e + (1-e)\). Noting that \(1-e\) is also idempotent, there is a
decomposition \(A \cong Ae \oplus A(1-e)\). Since \(Ae\) is direct
summand of \(A\) which is free, this yields a way to construct
projective modules.
\end{remark}
\hypertarget{thursday-january-13}{%
\section{Thursday, January 13}\label{thursday-january-13}}
\begin{remark}
Last time:
\begin{itemize}
\tightlist
\item
\(R{\hbox{-}}\)modules and their morphisms
\item
Free resolutions \(F \twoheadrightarrow R\).
\end{itemize}
Today: projective modules and their resolutions.
\begin{quote}
See Krull-Schmidt theorem.
\end{quote}
\end{remark}
\begin{remark}
Recall the definition of projective modules \(P\) and injective modules
\(I\):
\begin{center}
\begin{tikzcd}
&&&&&&& \textcolor{rgb,255:red,92;green,92;blue,214}{P} \\
\\
{\forall \xi:} & 0 && A && B && C && 0 \\
\\
&&& \textcolor{rgb,255:red,92;green,92;blue,214}{I}
\arrow[from=3-2, to=3-4]
\arrow[from=3-4, to=3-6]
\arrow[from=3-6, to=3-8]
\arrow[from=3-8, to=3-10]
\arrow[from=1-8, to=3-8]
\arrow["\exists", dashed, from=1-8, to=3-6]
\arrow["\exists", dashed, from=3-6, to=5-4]
\arrow[from=3-4, to=5-4]
\end{tikzcd}
\end{center}
\begin{quote}
\href{https://q.uiver.app/?q=WzAsOCxbMSwyLCIwIl0sWzMsMiwiQSJdLFs1LDIsIkIiXSxbNywyLCJDIl0sWzksMiwiMCJdLFs3LDAsIlAiLFsyNDAsNjAsNjAsMV1dLFszLDQsIkkiLFsyNDAsNjAsNjAsMV1dLFswLDIsIlxcZm9yYWxsOiJdLFswLDFdLFsxLDJdLFsyLDNdLFszLDRdLFs1LDNdLFs1LDIsIlxcZXhpc3QiLDAseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XSxbMiw2LCJcXGV4aXN0IiwwLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV0sWzEsNl1d}{Link
to Diagram}
\end{quote}
\end{remark}
\begin{exercise}[?]
Show that free implies projective using the universal properties, and
conclude that every \(R{\hbox{-}}\)module has a projective cover.
\end{exercise}
\begin{remark}
Forming projective resolutions: take the minimal
\(P_0 \xrightarrow[]{\delta_0} { \mathrel{\mkern-16mu}\rightarrow }\, M\to 0\)
such that \(\Omega^1 \coloneqq\ker \delta_0\) has no projective
summands. Continue in such a minimal way:
\begin{center}
\begin{tikzcd}
& 0 && 0 \\
& {\Omega^2} && {\Omega^1} \\
\\
\cdots && {P_1} && {P_0} && M && 0
\arrow[two heads, from=4-5, to=4-7]
\arrow[from=4-7, to=4-9]
\arrow[hook, two heads, from=2-4, to=4-5]
\arrow[two heads, from=4-3, to=2-4]
\arrow["\exists", dashed, from=4-3, to=4-5]
\arrow[dashed, from=4-1, to=4-3]
\arrow[hook, no head, from=2-2, to=4-3]
\arrow[from=1-2, to=2-2]
\arrow[from=1-4, to=2-4]
\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}
For modules \(M\) over an algebra \(A\), if \(\dim_k(M)\) is finite,
then each \(P_i\) can be chosen to be finite dimensional. Otherwise,
define a \textbf{complexity} or \textbf{rate of growth}
\(s c_A(M) \geq 0\) such that \(\dim P_n \leq C n^{s-1}\) for some
constant \(C\). A theorem we'll prove is that \(s\) is finite when
\(A = k G\) for every finite dimensional \(G{\hbox{-}}\)module. When
\(A = kG\), this is a numerical invariant but has a nice geometric
interpretation in terms of support varieties \(V_A(M)\), an affine
algebraic variety where \(\dim V_A(M) = c_A(M)\).
\end{remark}
\begin{exercise}[?]
Recall the definition of a SES
\(\xi: 0\to A \xrightarrow{d_1} B \xrightarrow{d_2} C\to 0\) and show
that TFAE:
\begin{itemize}
\tightlist
\item
\(\xi\) splits
\item
\(\xi\) admits a right section \(s_r: C\to B\)
\item
\(\xi\) admits a left section \(s_\ell B\to A\)
\end{itemize}
\begin{quote}
Hint: for the right section, show that \(s_r\) is injective. Get that
\(\operatorname{im}f + \operatorname{im}h \subseteq M_2\), use exactness
to write \(\operatorname{im}d_1 = \ker d_2\) and show that
\(\ker d_2 \cap\operatorname{im}s_r = \emptyset\).
\end{quote}
\end{exercise}
\begin{warnings}
It's not necessarily true that if \(B \cong A \oplus C\) that \(\xi\)
splits: consider
\begin{center}
\begin{tikzcd}
0 && {C_2} && {C_4} && {C_2} && 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]
\end{tikzcd}
\end{center}
\begin{quote}
\href{https://q.uiver.app/?q=WzAsNSxbMCwwLCIwIl0sWzIsMCwiQ18yIl0sWzQsMCwiQ180Il0sWzYsMCwiQ18yIl0sWzgsMCwiMCJdLFswLDFdLFsxLDJdLFsyLDNdLFszLDRdXQ==}{Link
to Diagram}
\end{quote}
\end{warnings}
\begin{exercise}[?]
Show that for \(P \in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}\), TFAE:
\begin{itemize}
\tightlist
\item
\(P\) is projective.
\item
Every SES \(\xi: 0\to A\to B\to P \to 0\) splits.
\item
There exists a free module \(F\) such that \(F = P \oplus K\).
\end{itemize}
\end{exercise}
\begin{exercise}[?]
Show that \(\bigoplus_{i\in I} P_i\) is projective iff each \(P_i\) is
projective.
\end{exercise}
\begin{example}[?]
\begin{itemize}
\item
If \(R=k\in \mathsf{Field}\), then every
\(M\in {\mathsf{k}{\hbox{-}}\mathsf{Mod}}\) is free and thus
projective since \(M \cong \bigoplus_{i\in I} k\) with \(k\) free in
\({\mathsf{k}{\hbox{-}}\mathsf{Mod}}\).
\item
If \(R={\mathbb{Z}}\), let
\(P\in {\mathsf{{\mathbb{Z}}}{\hbox{-}}\mathsf{Mod}}\) be projective
and \(F\) free and consider \(0\to K\to F\to P\to 0\). Since
\(F\cong P \oplus K\), \(P\) is a submodule of \(F\), making \(P\)
free since \({\mathbb{Z}}\) is a PID. So projective implies free.
\item
Not every \(M\in {\mathsf{{\mathbb{Z}}}{\hbox{-}}\mathsf{Mod}}\) is
projective: take
\(C_6\in {\mathsf{{\mathbb{Z}}}{\hbox{-}}\mathsf{Mod}}\), then
\(C_6 \cong C_2 \oplus C_3\) so \(C_2, C_3\) are projective in
\({\mathsf{C_6}{\hbox{-}}\mathsf{Mod}}\) but not free here.
\end{itemize}
\end{example}
\begin{exercise}[?]
Let \(Q\in{\mathsf{R}{\hbox{-}}\mathsf{Mod}}\) and show TFAE:
\begin{itemize}
\tightlist
\item
\(Q\) is injective
\item
Every SES \(\xi: 0\to Q\to B\to C\to 0\) splits.
\end{itemize}
\end{exercise}
\begin{exercise}[?]
Show that \(\prod_{i\in I}Q_i\) is injective iff each \(Q_i\) is
injective. Note that one needs to use direct products instead of direct
sums here.
\end{exercise}
\begin{theorem}[?]
The category \({\mathsf{R}{\hbox{-}}\mathsf{Mod}}\) has enough
injectives, i.e.~for every \(M\in{\mathsf{R}{\hbox{-}}\mathsf{Mod}}\)
there is an injective \(Q\) and a SES \(0\to M\hookrightarrow Q\).
\end{theorem}
\begin{proof}[Sketch]
See Hungerford or Weibel. Prove it first for
\(\mathsf{C} = {\mathsf{Z}{\hbox{-}}\mathsf{Mod}}\). The idea now is to
apply
\begin{align*}
F({-}) \coloneqq\mathop{\mathrm{Hom}}_{\mathbb{Z}}(R,{-}): ({{\mathbb{Z}}}, {{\mathbb{Z}}}){\hbox{-}}\mathsf{biMod} &\to ({R}, {{\mathbb{Z}}}){\hbox{-}}\mathsf{biMod}
,\end{align*}
the left-exact contravariant hom. Using that
\(R\in ({R}, {R}){\hbox{-}}\mathsf{biMod}\hookrightarrow({{\mathbb{Z}}}, {R}){\hbox{-}}\mathsf{biMod}\),
one can use the right action \(R\) on itself to define a left action on
\(\mathop{\mathrm{Hom}}_{\mathbb{Z}}(R, M)\). Then check that
\begin{itemize}
\tightlist
\item
\(f\) is left exact
\item
\(f\) sends injectives to injectives.
\item
If \(R\in{\mathsf{{\mathbb{Z}}}{\hbox{-}}\mathsf{Mod}}\) has an
\(R{\hbox{-}}\)module structure, then \(F(R)\) is again an
\(R{\hbox{-}}\)module.
\end{itemize}
\end{proof}
\begin{exercise}[?]
Show that for \(M\in{\mathsf{R}{\hbox{-}}\mathsf{Mod}}\) that
\(\mathop{\mathrm{Hom}}_{\mathbb{Z}}(R, M) \cong M\).
\begin{quote}
Hint: try \(f\mapsto f(1)\).
\end{quote}
\end{exercise}
\begin{remark}
Next week:
\begin{itemize}
\tightlist
\item
Tensor products
\item
Categories
\item
Tensor and Hom
\end{itemize}
\end{remark}
\hypertarget{tensor-products-tuesday-january-18}{%
\section{Tensor Products (Tuesday, January
18)}\label{tensor-products-tuesday-january-18}}
\begin{remark}
Setup: \(R\in \mathsf{Ring}, M_R \in \mathsf{Mod}{\hbox{-}}\mathsf{R}\),
and \({}_R N \in \mathsf{R}{\hbox{-}}\mathsf{Mod}\). Note that \(R\) is
not necessarily commutative. The goal is to define \(M\otimes_R N\) as
an abelian group.
\end{remark}
\begin{definition}[The Tensor Product]
The \textbf{balanced product} of \(M\) and \(N\) is a
\(P \in {\mathsf{Ab}}{\mathsf{Grp}}\) with a map \(f: M\times N\to P\)
such that
\begin{itemize}
\tightlist
\item
\(f(x+x', y) = f(x, y) + f(x', y)\)
\item
\(f(x, y+y') = f(x,y) + f(x, y')\)
\item
\(f(ax, y) = f(x, ay)\).
\end{itemize}
The \textbf{tensor product} \((M\otimes_R N, \otimes)\) of \(M\) and
\(N\) is the initial balanced product, i.e.~if \(P\) is a balanced
product with \(M\times N \xrightarrow{f} P\) then there is a unique map
\(\psi: M\otimes_R N\to P\):
\begin{center}
\begin{tikzcd}
& {M\otimes_R N} \\
\\
{M\times N} && P
\arrow["\otimes", from=3-1, to=1-2]
\arrow["{\exists !\psi}", dashed, from=1-2, to=3-3]
\arrow["f"', from=3-1, to=3-3]
\end{tikzcd}
\end{center}
\begin{quote}
\href{https://q.uiver.app/?q=WzAsMyxbMCwyLCJNXFx0aW1lcyBOIl0sWzIsMiwiUCJdLFsxLDAsIk1cXHRlbnNvcl9SIE4iXSxbMCwyLCJcXHRlbnNvciJdLFsyLDEsIlxcZXhpc3RzICFcXHBzaSIsMCx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFswLDEsImYiLDJdXQ==}{Link
to Diagram}
\end{quote}
Uniqueness follows from the standard argument on universal properties:
\begin{center}
\begin{tikzcd}
&&& {(M\otimes N)_1} \\
\\
{M\times N} &&& {(M\otimes N)_2} \\
\\
&&& {(M\otimes N)_1}
\arrow["{\otimes_2}"{description}, from=3-1, to=3-4]
\arrow["{\otimes_1}"{description}, from=3-1, to=5-4]
\arrow["{\exists \psi_{12}}"{description}, curve={height=-18pt}, dashed, from=5-4, to=3-4]
\arrow["{\exists \psi_{21}}"{description}, curve={height=-18pt}, dashed, from=3-4, to=1-4]
\arrow["\operatorname{id}"', curve={height=30pt}, from=5-4, to=1-4]
\arrow["{\otimes_1}"{description}, from=3-1, to=1-4]
\end{tikzcd}
\end{center}
\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwyLCJNXFx0aW1lcyBOIl0sWzMsNCwiKE1cXHRlbnNvciBOKV8xIl0sWzMsMiwiKE1cXHRlbnNvciBOKV8yIl0sWzMsMCwiKE1cXHRlbnNvciBOKV8xIl0sWzAsMiwiXFx0ZW5zb3JfMiIsMV0sWzAsMSwiXFx0ZW5zb3JfMSIsMV0sWzEsMiwiXFxleGlzdHMgXFxwc2lfezEyfSIsMSx7ImN1cnZlIjotMywic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV0sWzIsMywiXFxleGlzdHMgXFxwc2lfezIxfSIsMSx7ImN1cnZlIjotMywic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV0sWzEsMywiXFxpZCIsMix7ImN1cnZlIjo1fV0sWzAsMywiXFx0ZW5zb3JfMSIsMV1d}{Link
to Diagram}
\end{quote}
Existence: let
\(\mathsf{Free}({-}): {\mathsf{Set}}\to {\mathsf{Ab}}{\mathsf{Grp}}\)
and \(F\coloneqq\mathsf{Free}(M\times N)\), then set
\(M\otimes_R N \coloneqq F/G\) where \(G\) is generated by
\begin{itemize}
\tightlist
\item
\((x+x', y) - \qty{ (x, y) + (x', y) }\)
\item
\((x, y+y') - \qty{ (x, y) + (x, y') }\)
\item
\((ax, y) - (x, ay)\).
\end{itemize}
Then define the map as
\begin{align*}
\otimes: M\times N\to F \\
(x, y) &\mapsto x\otimes y \coloneqq(x, y) + G
.\end{align*}
Why it satisfies the universal property: use the universal property of
free groups to get a map to \(F\) and check that the following diagram
commutes:
\begin{center}
\begin{tikzcd}
{M\times N} && F && {M\otimes_R N \coloneqq F/G} \\
\\
&& P
\arrow[from=1-1, to=3-3]
\arrow["\otimes"', from=1-1, to=1-3]
\arrow["{({-})/G}"', from=1-3, to=1-5]
\arrow["{\exists }"', dashed, from=1-3, to=3-3]
\arrow["{\exists \psi}", from=1-5, to=3-3]
\end{tikzcd}
\end{center}
\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJNXFx0aW1lcyBOIl0sWzIsMCwiRiJdLFs0LDAsIk1cXHRlbnNvcl9SIE4gXFxkYSBGL0ciXSxbMiwyLCJQIl0sWzAsM10sWzAsMSwiXFx0ZW5zb3IiLDJdLFsxLDIsIihcXHdhaXQpL0ciLDJdLFsxLDMsIlxcZXhpc3RzICIsMix7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFsyLDMsIlxcZXhpc3RzICBcXHBzaSJdXQ==}{Link
to Diagram}
\end{quote}
Morphisms: for \(f:M\to M'\) and \(g: N\to N'\), form
\begin{align*}
f\otimes g: M\otimes N &\to M'\otimes N' \\
x\otimes y &\mapsto f(x) \otimes g(y)
.\end{align*}
\end{definition}
\begin{warnings}
Note every \(z\in M\otimes_R N\) is a simple tensor of the form
\(z=x\otimes y\)!
\end{warnings}
\begin{example}[?]
\begin{itemize}
\item
For \(R=k\in \mathsf{Field}\),
\(M\otimes_k N \in ({k}, {k}){\hbox{-}}\mathsf{biMod}\). If
\(M = \left\langle{m_i}\right\rangle\) and
\(N = \left\langle{n_j}\right\rangle\), then
\(M\otimes_k n = \left\langle{m_i\otimes n_j}\right\rangle\) and
\(\dim_k M\otimes_k N = \dim_k M \cdot \dim_k N\).
\item
For \(A\in {\mathsf{Ab}}{\mathsf{Grp}}\),
\(A\otimes_{\mathbb{Z}}{\mathbb{Z}}\cong A\) since
\(x\otimes y = xy\otimes 1\).
\item
\(M\coloneqq C_p\otimes_{\mathbb{Z}}{\mathbb{Q}}= 0\). It suffices to
check on simple tensors:
\begin{align*}
x\otimes y
&= x\otimes{p\over p} y \\
&= x\otimes p\qty{1\over p} y \\
&= px\otimes\qty{1\over p} y \\
&= 0\otimes{1\over p}y \\
&= 0
.\end{align*}
\item
More generally, if \(A\in {\mathsf{Ab}}{\mathsf{Grp}}\) is torsion
then \(A\otimes_{\mathbb{Z}}{\mathbb{Q}}= 0\).
\end{itemize}
\end{example}
\begin{definition}[Categories]
A category \(\mathsf{C}\) is a class of objects \(A\in \mathsf{C}\) and
for any pair \((A, B)\), a set of morphism
\(\mathop{\mathrm{Hom}}_{\mathsf{C}}(A, B)\) such that
\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
\((A, B) \neq (C, D)\implies \mathop{\mathrm{Hom}}(A, B)\) and
\(\mathop{\mathrm{Hom}}(C, D)\) are disjoint.
\item
Associativity of composition: \((h\circ g)\circ f = h\circ(g\circ f)\)
\item
Identities:
\(\exists ! \operatorname{id}_A \in \mathop{\mathrm{Hom}}_{\mathsf{C}}(A, A)\)
for all \(A\in \mathsf{C}\).
\end{enumerate}
A \textbf{subcategory} \(\mathsf{D} \leq \mathsf{C}\) is a subclass of
objects and morphisms, and is \textbf{full} if
\(\mathop{\mathrm{Hom}}_{\mathsf{D}}(A, B) = \mathop{\mathrm{Hom}}_{\mathsf{C}}(A, B)\)
for all objects in \(\mathsf{D}\).
\end{definition}
\begin{example}[?]
Examples of categories:
\begin{itemize}
\tightlist
\item
\(\mathsf{C} = {\mathsf{Set}}\),
\item
\(\mathsf{C} = {\mathsf{Grp}}\),
\item
\(\mathsf{C} = {\mathsf{R}{\hbox{-}}\mathsf{Mod}}\),
\item
\(\mathsf{C} = {\mathsf{Top}}\) with continuous maps.
\end{itemize}
\end{example}
\begin{example}[?]
Examples of fullness:
\begin{itemize}
\tightlist
\item
\({\mathsf{Grp}}\leq {\mathsf{Set}}\) is not a full subcategory, since
not all set morphisms are group morphisms.
\item
\({\mathsf{Ab}}{\mathsf{Grp}}\leq {\mathsf{Grp}}\) is a full
subcategory.
\end{itemize}
\end{example}
\begin{remark}
Recall the definition of covariant and contravariant functors, which
requires that \(F(\operatorname{id}_A) = \operatorname{id}_{F(A)}\).
\end{remark}
\hypertarget{thursday-january-20}{%
\section{Thursday, January 20}\label{thursday-january-20}}
\begin{remark}
RIP Brian Parshall and Fred Cohen\ldots{} 😔
\end{remark}
\begin{remark}
Recall the definition of a covariant functor. Some examples:
\begin{itemize}
\tightlist
\item
\(F(R) = U(R) = R^{\times}= {\mathbb{G}}_m(R)\), the group of units of
\(R\).
\item
The forgetful functor \({\mathsf{Grp}}\to {\mathsf{Set}}\).
\item
\(\mathop{\mathrm{Hom}}_{\mathbb{Z}}(R, {-})\) for
\(R\in ({{\mathbb{Z}}}, {R}){\hbox{-}}\mathsf{biMod}\) is a functor
\(\mathsf{{\mathbb{Z}}}{\hbox{-}}\mathsf{Mod}\to \mathsf{R}{\hbox{-}}\mathsf{Mod}\).
\end{itemize}
\end{remark}
\begin{exercise}[?]
Formulate \(\mathop{\mathrm{Hom}}_{\mathbb{Z}}({-}, {-})\) in terms of
functors between bimodule categories. How does this ``use up an action''
in the way \({-}\otimes_{\mathbb{Z}}{-}\) does?
\end{exercise}
\begin{remark}
Recall that contravariant functors reverse arrows. Functors with the
same variance can be composed.
\end{remark}
\begin{definition}[Full and Faithful Functors]
Let \(F: \mathsf{C}\to \mathsf{D}\) and consider the set map
\begin{align*}
F_{AB}: \mathop{\mathrm{Hom}}(A, B) &\to \mathop{\mathrm{Hom}}(FA, FB) \\
f &\mapsto F(f)
.\end{align*}
We say \(F\) is \textbf{full} if \(F_{AB}\) is injective for all
\(A, B\in \mathsf{C}\), and \textbf{faithful} if \(F_{AB}\) is
surjective for all \(A, B\).
\end{definition}
\begin{definition}[Natural Transformations]
A morphism of functors \(\eta: F\to G\) for
\(F,G:\mathsf{C}\to \mathsf{D}\) is a \textbf{natural transformation}: a
family of maps \(\eta_A\in \mathop{\mathrm{Hom}}_{\mathsf{D}}(FA, GA)\)
satisfying the following naturality condition:
\begin{center}
\begin{tikzcd}
A &&& FA && GA \\
&&&&&& {\in \mathsf{D}} \\
B &&& FB && GB
\arrow["{\eta_A}", from=1-4, to=1-6]
\arrow["{G(f)}", from=1-6, to=3-6]
\arrow[""{name=0, anchor=center, inner sep=0}, "{F(f)}", from=1-4, to=3-4]
\arrow["{\eta_B}"', from=3-4, to=3-6]
\arrow[""{name=1, anchor=center, inner sep=0}, "{f \in \mathsf{C}}"', from=1-1, to=3-1]
\arrow[shorten <=19pt, shorten >=19pt, Rightarrow, from=1, to=0]
\end{tikzcd}
\end{center}
\begin{quote}
\href{https://q.uiver.app/?q=WzAsNyxbMywwLCJGQSJdLFszLDIsIkZCIl0sWzUsMCwiR0EiXSxbNSwyLCJHQiJdLFswLDAsIkEiXSxbMCwyLCJCIl0sWzYsMSwiXFxpbiBcXGNhdCBEIl0sWzAsMiwiXFxldGFfQSJdLFsyLDMsIkcoZikiXSxbMCwxLCJGKGYpIl0sWzEsMywiXFxldGFfQiIsMl0sWzQsNSwiZiBcXGluIFxcY2F0e0N9IiwyXSxbMTEsOSwiIiwwLHsic2hvcnRlbiI6eyJzb3VyY2UiOjIwLCJ0YXJnZXQiOjIwfX1dXQ==}{Link
to Diagram}
\end{quote}
If \(\eta_A\) is an isomorphism for all \(A\in \mathsf{C}\), then
\(\eta\) is a \textbf{natural isomorphism}.
\end{definition}
\begin{exercise}[?]
For
\(\mathsf{C}, \mathsf{D} = { \mathsf{Vect} }^{{\mathrm{fd}}}_{/ {k}}\)
finite-dimensional vector spaces, take \(F = \operatorname{id}\) and
\(G({-}) = ({-}) {}^{ \vee } {}^{ \vee }\). Note that
\(\mathop{\mathrm{Hom}}(FV, GV) \cong \mathop{\mathrm{Hom}}(V, V {}^{ \vee } {}^{ \vee }) \cong \mathop{\mathrm{Hom}}(V, V)\),
so set \(\eta_V\) to be the image of \(\operatorname{id}_V\) under this
chain of isomorphisms. Show that
\(\left\{{\eta_V }\right\}_{V\in \mathsf{C}}\) assemble to a natural
transformation \(F\to G\).
\end{exercise}
\begin{definition}[Isomorphisms and Equivalences of categories]
Two categories \(\mathsf{C}, \mathsf{D}\) are \textbf{isomorphic} if
there are functors \(F, G\) with
\(F\circ G = \operatorname{id}_{\mathsf{D}}, G\circ F = \operatorname{id}_{\mathsf{C}}\)
\emph{equal} to the identities. They are \textbf{equivalent} if
\(F\circ G, G\circ F\) are instead \emph{naturally isomorphic} to the
identity.
\end{definition}
\begin{example}[?]
Some examples:
\begin{itemize}
\item
\(\mathsf{C} = {\mathsf{Ab}}{\mathsf{Grp}}\) and
\(\mathsf{D} = {\mathsf{{\mathbb{Z}}}{\hbox{-}}\mathsf{Mod}}\) by
taking \(G:\mathsf{D}\to \mathsf{C}\) the forgetful functor, and for
\(F\), using the same underlying set and defining the
\({\mathbb{Z}}{\hbox{-}}\)module structure by
\(n\cdot m \coloneqq m + m + \cdots + m\).
\item
\(\mathsf{C}={\mathsf{R}{\hbox{-}}\mathsf{Mod}}\) and
\(\mathsf{D} = {\mathsf{\operatorname{Mat}_{n\times n}(R)}{\hbox{-}}\mathsf{Mod}}\).
For \({\mathsf{k}{\hbox{-}}\mathsf{Mod}}\), the simple objects are
\(k\), but for
\({\mathsf{\operatorname{Mat}_{n\times n}(R)}{\hbox{-}}\mathsf{Mod}}\),
the simple objects are \(k^n\), so these categories are not
isomorphic. However, it turns out that they are equivalent.
\end{itemize}
Producing inverse functors can be difficult, so we have the following:
\end{example}
\begin{proposition}[A useful criterion for equivalence of categories]
Let \(F:\mathsf{C}\to \mathsf{D}\), then there exists an inverse
inducing an \emph{equivalence} iff
\begin{itemize}
\tightlist
\item
\(F\) is fully faithful,
\item
Surjectivity on objects: for every \(A'\in \mathsf{D}\), there exists
an \(A\in \mathsf{C}\) such that \(F(A) \cong A'\).
\end{itemize}
\end{proposition}
\begin{proof}[?]
\(\implies\): Suppose \(F, G\) induce an equivalence
\(\mathsf{C} \simeq\mathsf{D}\), so
\(F\circ G\simeq\operatorname{id}_{\mathsf{D}}\) and
\(G\circ F \simeq\operatorname{id}_{\mathsf{C}}\). To show \(f\to F(f)\)
is injective, check that
\begin{align*}
F(f) &= F(g) \\
\implies GF(f) &= GF(g) \\
\operatorname{id}(f) &= \operatorname{id}(g) \\
\implies f= g
.\end{align*}
\end{proof}
\begin{exercise}[?]
Show surjectivity.
A hint:
Let \(A'\in \mathsf{D}\) with
\(FG \simeq\operatorname{id}_{\mathsf{D}}\) and
\(\eta_{A'} \in \mathop{\mathrm{Hom}}_{\mathsf{D}}(A', FGA')\) is an
iso. Set \(A \coloneqq GA'\in \mathsf{C}\) and use that
\begin{align*}
\mathop{\mathrm{Hom}}_{\mathsf{D}}(A', FGA') \coloneqq\mathop{\mathrm{Hom}}(A', FA)
,\end{align*}
So if there is an isomorphism in \(\mathop{\mathrm{Hom}}(A', FA)\),
there exists an isomorphism in \(\mathop{\mathrm{Hom}}(FA, A')\) and
thus \(FA \cong A'\).
\begin{quote}
\#todo Missed a bit here so this doesn't make sense as-is!
\end{quote}
\end{exercise}
\begin{proposition}[?]
Let \(R\in \mathsf{Ring}\) and set
\(S\coloneqq\operatorname{Mat}_{n\times n}(R)\), then
\({\mathsf{R}{\hbox{-}}\mathsf{Mod}} \simeq{\mathsf{S}{\hbox{-}}\mathsf{Mod}}\).
\end{proposition}
\hypertarget{tuesday-january-25}{%
\section{Tuesday, January 25}\label{tuesday-january-25}}
\begin{remark}
Recall isomorphisms \(\mathsf{C} \cong \mathsf{D}\) of categories, so
\(F\circ G = \operatorname{id}\), vs equivalences of categories
\(\mathsf{C} \simeq\mathsf{D}\) so \(F\circ G \cong \operatorname{id}\).
\end{remark}
\begin{theorem}[?]
For \(F:\mathsf{C} \to \mathsf{D}\) and \(G:\mathsf{D}\to \mathsf{C}\)
and write
\(\psi_F: \mathop{\mathrm{Hom}}_{\mathsf{C}}(A, B) \to \mathop{\mathrm{Hom}}_{\mathsf{D}}(F(A), F(B))\).
This pair induces an equivalence iff
\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
\(F\) is faithful, i.e.~\(\psi_F\) is injective,
\item
\(F\) is full, i.e.~\(\psi_F\) is surjective,
\item
For any \(D\in \mathsf{D}\), there exists a \(C\in \mathsf{C}\) with
\(F(C) \cong D\).
\end{enumerate}
\end{theorem}
\begin{proposition}[?]
Let \(R\in \mathsf{Ring}\) and \(S=\operatorname{Mat}_{n\times n}(R)\),
then
\({\mathsf{R}{\hbox{-}}\mathsf{Mod}} \simeq{\mathsf{S}{\hbox{-}}\mathsf{Mod}}\).
\end{proposition}
\begin{proof}[?]
Define a functor
\(F:{\mathsf{R}{\hbox{-}}\mathsf{Mod}} \to {\mathsf{S}{\hbox{-}}\mathsf{Mod}}\)
by \(F(M) \coloneqq\prod_{k\leq n} M\), regarding this as a column
vector and letting \(S\) act by matrix multiplication. On morphisms,
define \(F(f)(\mathbf{x}) = {\left[ {f(x_1), \cdots, f(x_n)} \right]}\)
for \(\mathbf{x} \in \prod M\). Then
\(F(\operatorname{id}) = \operatorname{id}\), and (exercise) \(F(f)\) is
a morphism of \(S{\hbox{-}}\)modules and composes correctly:
\begin{align*}
F(g\circ f)(\mathbf{x}) = {\left[ {gf(x_1), \cdots, gf(x_n)} \right]} = F(g){\left[ {f(x_1), \cdots, f(x_n) } \right]} = \qty{ F(g)\circ F(f) } \mathbf{x}
.\end{align*}
So this defines a functor.
\begin{claim}
\(F\) is fully faithful.
\end{claim}
\begin{itemize}
\item
Faithfulness: if \(F(f_1) = F(f_2)\), then \(f_1(x_j) = f_2(x_j)\) for
all \(j\), making \(f_1=f_2\).
\item
Fullness: let \(g\in \mathop{\mathrm{Hom}}_S(M^n, N^n)\) for
\(M, N \in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}\) and \(e_{ij}\) be the
elementary matrix with a 1 only in the \(i, j\) position. Check that
\(e_{11} M^n = \left\{{{\left[ {x,0,\cdots} \right]} {~\mathrel{\Big\vert}~}x\in M}\right\}\),
\(e_{11} N^n = \left\{{{\left[ {y,0,\cdots} \right]}{~\mathrel{\Big\vert}~}y\in N}\right\}\),
and \(\operatorname{diag}(x)\) be a matrix with only copies of \(x\)
on the diagonal. Then
\(g(e_{11} M^n) \subseteq e_{11} g(M^n) \subseteq e_{11}N^n\) and
\(g{\left[ {x, 0, \cdots} \right]} = {\left[ {y, 0, \cdots} \right]}\).
Define \(f:M\to N\) by \(f(x) = y\), then on one hand,
\begin{align*}
g(\operatorname{diag}(a) {\left[ {x, 0,\cdots} \right]}) = g{\left[ {ax, 0, \cdots} \right]} = {\left[ {f(ax), 0, \cdots} \right]}
,\end{align*}
but since \(g\) is a morphism of \(S{\hbox{-}}\)modules, this also
equals
\(\operatorname{diag}(a)\cdot g{\left[ {x,0,\cdots} \right]} = {\left[ {ay,0,\cdots} \right]}\).
Then \(f(ax) = ay = af(x)\), so \(f\) is a morphism of
\(R{\hbox{-}}\)modules.
Note that
\(e_{j1} \mathbf{x} = {\left[ {0, \cdots, x,\cdots 0} \right]}\) with
\(x\) in the \(j\)th position. Check that
\(g(e_{j1}\mathbf{x}) = g{\left[ {0, \cdots, x, \cdots, 0} \right]}\).
The LHS is
\begin{align*}
e_{j1} g(\mathbf{x}) = e_{j1}{\left[ {f(x), 0, \cdots} \right]} = {\left[ { 0,\cdots, f(x), \cdots, 0} \right]}
\end{align*}
with \(f(x)\) in the \(j\)th position. Hence
\(g(\mathbf{x}) = {\left[ {f(x_1), \cdots, f(x_n)} \right]}\), making
\(F\) full.
\end{itemize}
\begin{quote}
See also Jacobson \emph{Basic Algebra Part II} p.31.
\end{quote}
\end{proof}
\begin{exercise}[Tensors commute with direct sums]
Show that
\begin{align*}
\qty{ \bigoplus _{\alpha \in I} M_\alpha } \otimes_R N
&\cong \bigoplus _{\alpha\in I} \qty{M_\alpha \otimes_R N}
,\end{align*}
and similarly for \(M\otimes(\oplus N_\alpha)\).
\end{exercise}
\begin{remark}
Define functors
\(F,G{\mathsf{R}{\hbox{-}}\mathsf{Mod}} \to{\mathsf{{\mathbb{Z}}}{\hbox{-}}\mathsf{Mod}}\)
by \(F({-}) \coloneqq M\otimes_R ({-})\) and
\(G({-}) \coloneqq({-})\otimes_R N\) on objects, and on morphisms
\(f:N\to N'\), set \(F(f) \coloneqq\operatorname{id}\otimes f\) and
similarly for \(G\). Recall the definition of exactness, left-exactness,
and right-exactness.
\end{remark}
\begin{example}[Tensoring may not be left exact]
Consider
\begin{align*}
\xi: 0\to p{\mathbb{Z}}\xrightarrow{f} {\mathbb{Z}}\xrightarrow{g} {\mathbb{Z}}/p{\mathbb{Z}}\to 0
\end{align*}
and apply \(({-})\otimes_{\mathbb{Z}}{\mathbb{Z}}/p{\mathbb{Z}}\). Use
that \(p{\mathbb{Z}}\cong {\mathbb{Z}}\) in
\({\mathsf{{\mathbb{Z}}}{\hbox{-}}\mathsf{Mod}}\) to get
\begin{align*}
F(\xi): C_p \xrightarrow{f\otimes\operatorname{id}} C_p \xrightarrow{g\otimes\operatorname{id}} C_p
,\end{align*}
and
\begin{align*}
(f\otimes\operatorname{id})(px\otimes y) = px\otimes y = x\otimes py = 0
,\end{align*}
using that \(f\) is the inclusion.
\end{example}
\begin{exercise}[?]
Show that \(M\otimes_R({-})\) and \(({-})\otimes_R N\) are right exact
for any \(M, N \in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}\).
\end{exercise}
\begin{solution}
Let \(0 \to A \xrightarrow{f} B \xrightarrow{g} C \to 0\) which maps to
\(M\otimes A \xrightarrow{\operatorname{id}\otimes f} M\otimes B \xrightarrow{\operatorname{id}\otimes g} C\).
\begin{itemize}
\item
Show \(\operatorname{id}\otimes g\) is surjective: write
\(m\in M\otimes C\) as \(m=\sum x_i\otimes y_j\), pull back the
\(y_j\) via \(g\) to get \(z_j\) with \(g(z_j) = y_j\). Then
\begin{align*}
(\operatorname{id}\otimes g)(\sum x_i \otimes z_J) = \sum x_i\otimes g(z_j) = \sum x_i \otimes y_j
.\end{align*}
\item
Exactness,
\(\operatorname{im}(\operatorname{id}\otimes f) = \ker (\operatorname{id}\otimes g)\):
Use that \(gf=0\) by exactness of the original sequence, and
\((\operatorname{id}\otimes g)\circ (\operatorname{id}\otimes f) = \operatorname{id}\otimes(g\circ f) = 0\),
so
\(\operatorname{im}(\operatorname{id}\otimes f) \subseteq \ker(\operatorname{id}\otimes g)\).
\begin{itemize}
\tightlist
\item
For the reverse containment, use that
\(\operatorname{id}\otimes g: M\otimes B\to M\otimes C\) and define
a map
\begin{align*}
\Gamma: {M\otimes B \over \operatorname{im}(\operatorname{id}\otimes f)} \to M\otimes C \\
m\otimes n + \operatorname{im}(\operatorname{id}\otimes f)&\mapsto m\otimes g(n)
.\end{align*}
Then \(\phi\) is an isomorphism iff
\(\operatorname{im}(\operatorname{id}\otimes f) = \ker (\operatorname{id}\otimes g)\).
Define
\begin{align*}
\Psi: M\times C &\to {M\otimes B\over \operatorname{im}(\operatorname{id}\otimes f)} \\
(x, y) &\mapsto x \otimes z + \operatorname{im}(\operatorname{id}\otimes f)
,\end{align*}
where \(g(z) = y\), so \(z\) is a lift of \(y\).
\end{itemize}
Why is this well-defined? Check \(g(z_1) = y = g(z_2)\) implies
\(z_1 -z_2\in \ker g = \operatorname{im}f\), so write
\(f(y) = z_1-z_2\) for some \(y\). Then
\(x\otimes z_1 + \operatorname{im}f = x\otimes z_2 + \operatorname{im}f\).
Why does this factor through the tensor product? Check that \(\Psi\)
is a balanced product, this yields
\(\mkern 1.5mu\overline{\mkern-1.5mu\Psi\mkern-1.5mu}\mkern 1.5mu: M\otimes C\to {M\otimes B\over \operatorname{im}(\operatorname{id}\otimes f)}\).
Now check that
\(\mkern 1.5mu\overline{\mkern-1.5mu\Psi\mkern-1.5mu}\mkern 1.5mu, \Gamma\)
are mutually inverse:
\begin{align*}
\Gamma\Psi(x\otimes y)
&= \Gamma(x\otimes z + \operatorname{im}(\operatorname{id}\otimes f)) = x\otimes g(z) = x\otimes y \\
\Psi\Gamma(x\otimes z + \operatorname{im}(\operatorname{id}\otimes f))
&= (x\otimes g(z) ) = x\otimes z + \operatorname{im}f
.\end{align*}
\end{itemize}
\end{solution}
\begin{question}
When is \(M\otimes_R ({-})\) exact?
\end{question}
\hypertarget{thursday-january-27}{%
\section{Thursday, January 27}\label{thursday-january-27}}
\begin{remark}
Recall that \(M\in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}\) is flat iff for
every \(N, N'\) and
\(f\in \mathop{\mathrm{Hom}}_{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}(N, N')\),
the induced map
\begin{align*}
\operatorname{id}_M\otimes f: M\otimes_R N \to M\otimes_R N'
\end{align*}
is a monomorphism. Equivalently, \(M\otimes_R ({-})\) is left exact and
thus exact.
\end{remark}
\begin{proposition}[?]
\(M \coloneqq\bigoplus _{\alpha\in I} M_\alpha\) is flat iff
\(M_\alpha\) is flat for all \(\alpha\in I\).
\end{proposition}
\begin{proof}[?]
\begin{align*}
M\otimes_R({-}) \coloneqq(\bigoplus M_\alpha)\otimes_R ({-}) \cong \bigoplus (M_\alpha \otimes_R ({-}) )
.\end{align*}
\end{proof}
\begin{exercise}[?]
Show that projective \(\implies\) flat.
\end{exercise}
\begin{exercise}[?]
Prove that the hom functors
\(\mathop{\mathrm{Hom}}_{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}(M, {-}), \mathop{\mathrm{Hom}}_{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}({-}, M): {\mathsf{R}{\hbox{-}}\mathsf{Mod}}\to {\mathbb{Z}{\hbox{-}}\mathsf{Mod}}\)
are left exact.
\end{exercise}
\begin{exercise}[?]
Show that
\begin{itemize}
\tightlist
\item
\(P\) is projective iff
\(\mathop{\mathrm{Hom}}_{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}(P, {-})\)
is exact
\item
\(I\) is projective iff
\(\mathop{\mathrm{Hom}}_{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}({-}, I)\)
is exact
\end{itemize}
\end{exercise}
\begin{remark}
An object \(Z\in \mathsf{C}\) is a zero object iff
\(\mathop{\mathrm{Hom}}_{\mathsf{C}}(A, Z), \mathop{\mathrm{Hom}}_{\mathsf{C}}(Z, A)\)
are singletons for all \(A\in \mathsf{C}\). Write this as
\(0_A \in \mathop{\mathrm{Hom}}_{\mathsf{C}}(A, Z)\). If \(\mathsf{C}\)
has a zero object, define the zero morphism as
\(0_{AB} \coloneqq 0_{B} \circ 0_A \in \mathop{\mathrm{Hom}}_{\mathsf{C}}(A, B)\).
\end{remark}
\hypertarget{tuesday-february-01}{%
\section{Tuesday, February 01}\label{tuesday-february-01}}
\begin{definition}[Additive categories]
A category \(\mathsf{C}\) is \textbf{additive} iff
\begin{itemize}
\tightlist
\item
\(\mathsf{C}\) has zero object
\item
There exists a binary operation
\(+: \mathop{\mathrm{Hom}}(A, B){ {}^{ \scriptscriptstyle\times^{2} } }\to \mathop{\mathrm{Hom}}(A, B)\)
for all \(A, B\in \mathsf{C}\) making \(\mathop{\mathrm{Hom}}(A ,B)\)
an abelian group.
\item
Distributivity with respect to composition:
\((g_1 + g_2)f = g_1f + g_2 f\)
\item
For any collection \(\left\{{A_1,\cdots, A_n}\right\}\), there exists
an object \(A\), projections \(p_j: A\to A_j\) with sections
\(i_k: A_k\to A\) with \(p_j i_j = \operatorname{id}_A\),
\(p_j i_k = 0\) for \(j\neq k\), and
\(\sum i_j p_j = \operatorname{id}_A\).
\end{itemize}
\end{definition}
\begin{definition}[Monomorphisms and epimorphisms]
A morphism: \(k:K\to A\) is \textbf{monic} iff whenever
\(g_1, g_2: L\to K\), \(kg_1 = kg_2 \implies g_1 = g_2\):
\begin{center}
\begin{tikzcd}
L && K && A
\arrow["k", from=1-3, to=1-5]
\arrow["{g_1}", shift left=3, from=1-1, to=1-3]
\arrow["{g_2}", shift right=3, from=1-1, to=1-3]
\end{tikzcd}
\end{center}
\begin{quote}
\href{https://q.uiver.app/?q=WzAsMyxbMCwwLCJMIl0sWzIsMCwiSyJdLFs0LDAsIkEiXSxbMSwyLCJrIl0sWzAsMSwiZ18xIiwwLHsib2Zmc2V0IjotM31dLFswLDEsImdfMiIsMCx7Im9mZnNldCI6M31dXQ==}{Link
to Diagram}
\end{quote}
Define \(k\) to be \textbf{epic} by reversing the arrows.
\end{definition}
\begin{definition}[Kernel]
Assume \(\mathsf{C}\) has a zero object. Then for \(f:A\to B\), the
\emph{morphism} \(k: K\to A\) is the \textbf{kernel} of \(f\) iff
\begin{itemize}
\tightlist
\item
\(k\) is monic
\item
\(fk=0\)
\item
For any \(g:G\to A\) with \(fg=0\), there exists a \(g'\) with
\(g=kg'\).
\end{itemize}
\end{definition}
\begin{example}[?]
For \(f\in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}(A, B)\), take
\(k: \ker f\hookrightarrow A\). If \(g\in \mathsf{C}(G, A)\) with
\(f(g(x)) = 0\) for all \(x\in G\), then
\(\operatorname{im}g \subseteq \ker f\) and we can factor \(g\) as
\(G \xrightarrow{g'} \ker f \xhookrightarrow{k} A\).
\end{example}
\begin{definition}[Cokernel]
For \(f: A\to B\), a morphism \(c: B\to C\) is a \textbf{cokernel of
\(f\)} iff
\begin{itemize}
\tightlist
\item
\(c\) is epic,
\item
\(cf=0\)
\item
For any \(h\in \mathsf{C}(B, H)\) with \(hf=0\), there is a lift
\(h': C\to G\) with \(h=h'c\).
\end{itemize}
\end{definition}
\begin{example}[?]
For \(\mathsf{C} = {\mathsf{R}{\hbox{-}}\mathsf{Mod}}\) and
\(f\in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}(A, B)\), set
\(c: B\to B/\operatorname{im}f\).
\end{example}
\begin{exercise}[?]
Show that kernels are unique. Sketch:
\begin{itemize}
\tightlist
\item
Set \(k:K\to A\), \(k': K'\to A\).
\item
Factor \(k=k' u_1\) and \(k' = ku_2\).
\item
Then
\(k\operatorname{id}= k(u_2 u_1) \implies \operatorname{id}= u_2 u_1\),
similarly \(u_1u_2=\operatorname{id}\).
\end{itemize}
\end{exercise}
\begin{definition}[Abelian categories]
\(\mathsf{C}\) is \textbf{abelian} iff \(\mathsf{C}\) is additive and
\begin{itemize}
\tightlist
\item
A5: Every morphism admits kernels and cokernels.
\item
A6: Every monic is the kernel of its cokernel, and every epic is the
cokernel of its kernel.
\item
A7: Every morphism \(f\) factors as \(f=me\) with \(m\) monic and
\(e\) epic.
\end{itemize}
\end{definition}
\begin{example}[?]
For \(f\in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}(A, B)\),
\begin{itemize}
\tightlist
\item
A5: Take \(k: \ker f\hookrightarrow A\) and
\(c: B\twoheadrightarrow B/\operatorname{im}f\)
\item
A6: For \(m: A\hookrightarrow B\) monic, consider the composition
\(A\hookrightarrow B \xrightarrow{\operatorname{coker}m} B/A\) and
check \(A\cong \ker(\operatorname{coker}m)\).
\item
A7: Use the 1st isomorphism theorem:
\end{itemize}
\begin{center}
\begin{tikzcd}
A &&&& B \\
\\
& {A/\ker f} && {\operatorname{im}f}
\arrow["f", from=1-1, to=1-5]
\arrow["i"', two heads, from=1-1, to=3-2]
\arrow["{\text{1st iso}}"', from=3-2, to=3-4]
\arrow["m"', hook, from=3-4, to=1-5]
\end{tikzcd}
\end{center}
\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJBIl0sWzQsMCwiQiJdLFszLDIsIlxcaW0gZiJdLFsxLDIsIkEvXFxrZXIgZiJdLFswLDEsImYiXSxbMCwzLCJpIiwyLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoiZXBpIn19fV0sWzMsMiwiXFx0ZXh0ezFzdCBpc299IiwyXSxbMiwxLCJtIiwyLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XV0=}{Link
to Diagram}
\end{quote}
\end{example}
\begin{remark}
Some notes:
\begin{itemize}
\tightlist
\item
Recall the definition the category of chain complexes
\(\mathsf{Ch}(\mathsf{C})\) over an abelian category:
\(d_i d_{i+1} = 0\), so
\(\operatorname{im}d_i \subseteq \ker d_{i+1}\).
\item
Every exact sequence is an acyclic complex.
\item
\(\mathsf{C}\hookrightarrow\mathsf{Ch}(\mathsf{C})\) by
\(M\mapsto \cdots \to 0 \to M \to 0 \to \cdots\). Note that this isn't
an acyclic complex.
\item
Morphisms between complexes: chain maps, just levelwise maps forming
commutative squares, i.e.~maps commuting with the differentials.
\item
\(\mathsf{Ch}(\mathsf{C})\) is additive: given
\(\alpha_\bullet, \beta_\bullet\in \mathsf{Ch}\mathsf{C}( (A, d), (B, \delta) )\),
check that
\((\alpha_{i-1} + \beta_{i-1})d_i = \delta_i (\alpha_i + \beta_i)\).
\item
There are direct sums: \((A \oplus B)_i \coloneqq A_i \oplus B_i\)
with \(d \coloneqq d_A + d_B\).
\item
Define cycles as
\(Z_i \coloneqq\ker\qty{ C_i \xrightarrow{d_i} C_{i-1}}\) for
\(C_\bullet \in \mathsf{Ch}(\mathsf{C})\), and boundaries
\(B_i \coloneqq\operatorname{im}\qty{C_{i+1} \xrightarrow{d_{i+1}} C_i} \subseteq \ker d_i\).
\item
Define \(H_i(C_\bullet )\coloneqq Z_i/B_i\).
\item
Show that chain morphisms induce morphisms on homology:
\begin{itemize}
\tightlist
\item
Let \(\alpha\in \mathsf{Ch}(\mathsf{C})(C, C')\), then
\(\alpha_i(Z_i) \subseteq Z_i'\).
\item
Check \(d_2(a_i(Z_i)) = a_{i-1} d_i(Z_i) = 0\).
\item
Factor
\(Z_i \xrightarrow{\alpha_i} Z_i' \twoheadrightarrow Z_i'/B_i'\).
\item
Show that \(x\in B_i\) maps lands in \(B_i'\) to get well-defined
map on \(H_i\).
\item
Use \(\alpha(B_i) \subseteq Z_i'\), so pull back \(x\in B_i\) to
\(y\in C_{i+1}\).
\item
Check \(d_{i+1}(y) = x\), so \(\alpha(d_{i+1}(y)) = \alpha(x)\).
\item
The LHS is \(d_{i+1}'(\alpha_{i+1}(y))\), so
\(\alpha_i(x) in \operatorname{im}d_{i+1}' = B_{i+1}'\)
\end{itemize}
\item
Chain homotopies: for
\(\alpha, \beta\in \mathsf{Ch}(\mathsf{C})(C, C')\), write
\(\alpha \simeq\beta\) iff there exists
\(\left\{{s_i: C_i \to C_{i+1}' }\right\}\) with
\(\alpha_i - \beta_i = d_{i+1}' s_i + s_{i-1} d_i\).
\end{itemize}
\begin{center}
\begin{tikzcd}
\cdots && {C_{i+1}} && {C_{i}} && {C_{i-1}} && \cdots \\
\\
\cdots && {C_{i+1}'} && {C_{i}'} && {C_{i-1}'} && \cdots
\arrow[from=1-1, to=1-3]
\arrow["{d_{i+1}}", from=1-3, to=1-5]
\arrow["{d_{i+1}}", color={rgb,255:red,92;green,92;blue,214}, from=1-5, to=1-7]
\arrow[from=1-7, to=1-9]
\arrow[from=3-1, to=3-3]
\arrow["{d_{i}'}", from=3-5, to=3-7]
\arrow["{d_{i+1}'}", color={rgb,255:red,214;green,92;blue,92}, from=3-3, to=3-5]
\arrow[from=3-7, to=3-9]
\arrow[from=1-3, to=3-3]
\arrow["{\alpha_i-\beta_i}"{description}, color={rgb,255:red,92;green,92;blue,214}, from=1-5, to=3-5]
\arrow[from=1-7, to=3-7]
\arrow["{s_i}"{description}, color={rgb,255:red,214;green,92;blue,92}, from=1-5, to=3-3]
\arrow["{s_{i-1}}"{description}, color={rgb,255:red,214;green,153;blue,92}, from=1-7, to=3-5]
\end{tikzcd}
\end{center}
\begin{quote}
\href{https://q.uiver.app/?q=WzAsMTAsWzIsMCwiQ197aSsxfSJdLFs0LDAsIkNfe2l9Il0sWzYsMCwiQ197aS0xfSJdLFsyLDIsIkNfe2krMX0nIl0sWzQsMiwiQ197aX0nIl0sWzYsMiwiQ197aS0xfSciXSxbMCwwLCJcXGNkb3RzIl0sWzgsMCwiXFxjZG90cyJdLFswLDIsIlxcY2RvdHMiXSxbOCwyLCJcXGNkb3RzIl0sWzYsMF0sWzAsMSwiZF97aSsxfSJdLFsxLDIsImRfe2krMX0iLDAseyJjb2xvdXIiOlsyNDAsNjAsNjBdfSxbMjQwLDYwLDYwLDFdXSxbMiw3XSxbOCwzXSxbNCw1LCJkX3tpfSciXSxbMyw0LCJkX3tpKzF9JyIsMCx7ImNvbG91ciI6WzAsNjAsNjBdfSxbMCw2MCw2MCwxXV0sWzUsOV0sWzAsM10sWzEsNCwiXFxhbHBoYV9pLVxcYmV0YV9pIiwxLHsiY29sb3VyIjpbMjQwLDYwLDYwXX0sWzI0MCw2MCw2MCwxXV0sWzIsNV0sWzEsMywic19pIiwxLHsiY29sb3VyIjpbMCw2MCw2MF19LFswLDYwLDYwLDFdXSxbMiw0LCJzX3tpLTF9IiwxLHsiY29sb3VyIjpbMzAsNjAsNjBdfSxbMzAsNjAsNjAsMV1dXQ==}{Link
to Diagram}
\end{quote}
\end{remark}
\hypertarget{thursday-february-03}{%
\section{Thursday, February 03}\label{thursday-february-03}}
\hypertarget{projective-resolutions-and-chain-maps}{%
\subsection{Projective Resolutions and Chain
Maps}\label{projective-resolutions-and-chain-maps}}
\begin{remark}
Also check that \(\simeq\) is an equivalence relation, i.e.~it is
symmetric, transitive, and reflexive. For transitivity: given
\begin{align*}
\alpha_i - \beta_i &= d_{i+1}' s_i +s_{i-1} d_i \\
\beta_i - \gamma_i &= d_{i+1}' t_{i} + t_{i-1} d_i
,\end{align*}
one can write
\begin{align*}
\alpha_i - \gamma_i &= d_{i+1}'(s_i + t_i) + (s_{i-1} + t_{i-1} ) d_i
.\end{align*}
\end{remark}
\begin{theorem}[?]
Let \(\alpha, \beta \in \mathsf{Ch}\mathsf{C}(A, B)\) with induced maps
\(\widehat{\alpha}, \widehat{\beta }\in \mathsf{Ch}\mathsf{C}(H^* A, H^* B)\)
on homology. If \(\alpha \simeq\beta\), then
\(\widehat{\alpha }= \widehat{\beta}\).
\end{theorem}
\begin{proof}[?]
A computation:
\begin{align*}
\widehat{\alpha}_{1}(&\left.z_{1}+B_{i}\right)=\alpha_{1}\left(z_{i}\right)+B_{i}^{\prime} \\
&=\beta_{i}\left(z_{i}\right)+\delta_{i+1}^{\prime} s_{1}\left(z_{i}\right)+s_{i-1}^{\prime \prime} \delta_{i}\left(z_{i}\right) + B_i'\\
&=\beta_{i}\left(z_{i}\right)+B_{i}^{\prime} \\
&=\widehat{\beta}_{i}\left(z_{i}+B_{i}\right)
\end{align*}
\end{proof}
\begin{remark}
Roadmap:
\begin{itemize}
\tightlist
\item
Homological algebra
\item
Commutative rings
\item
Support theory
\item
Tensor triangular geometry
\end{itemize}
\end{remark}
\begin{definition}[?]
Let \(M\in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}\). A \textbf{projective
complex} for \(M\) is a chain complex
\((C_i, d_i)_{i\in {\mathbb{Z}}}\), indexed homologically:
\begin{align*}
\cdots \to C_2 \xrightarrow{d_2} C_1 \xrightarrow{d_1} C_0 \xrightarrow{d_0\coloneqq{\varepsilon}} 0
.\end{align*}
In particular, \(d^2 = 0\), but this complex need not be exact. A
\textbf{projective resolution} of \(M\) is an \emph{exact} projective
complex in the following sense:
\begin{itemize}
\tightlist
\item
\(H_{k\geq 1}({ {C}_{\scriptscriptstyle \bullet}} ) = 0\)
\item
\(H_0({ {C}_{\scriptscriptstyle \bullet}} ) = C_0/d(C_1) = C_0/\ker {\varepsilon}\cong M\).
\end{itemize}
\end{definition}
\begin{example}[?]
Some projective resolutions:
\begin{itemize}
\tightlist
\item
For \(M\in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}\), projective
resolutions exist since we can find covers by free modules:
\end{itemize}
\begin{center}
\begin{tikzcd}
\cdots & {F_2} & {F_1} & {F_0} & M & 0 \\
&& {\ker d_1} & {\ker {\varepsilon}}
\arrow[from=1-4, to=2-4]
\arrow[from=2-4, to=1-5]
\arrow[from=1-3, to=2-3]
\arrow[from=2-3, to=1-4]
\arrow[from=1-2, to=1-3]
\arrow["{d_1}", from=1-3, to=1-4]
\arrow["{\varepsilon}", two heads, from=1-4, to=1-5]
\arrow[from=1-5, to=1-6]
\arrow[from=1-1, to=1-2]
\end{tikzcd}
\end{center}
\begin{quote}
\href{https://q.uiver.app/?q=WzAsOCxbMCwwLCJcXGNkb3RzIl0sWzEsMCwiRl8yIl0sWzIsMCwiRl8xIl0sWzMsMCwiRl8wIl0sWzQsMCwiTSJdLFs1LDAsIjAiXSxbMywxLCJcXGtlciBcXGVwcyJdLFsyLDEsIlxca2VyIGRfMSJdLFszLDZdLFs2LDRdLFsyLDddLFs3LDNdLFsxLDJdLFsyLDMsImRfMSJdLFszLDQsIlxcZXBzIiwwLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoiZXBpIn19fV0sWzQsNV0sWzAsMV1d}{Link
to Diagram}
\end{quote}
\begin{itemize}
\tightlist
\item
For \(M\in {\mathsf{Z}{\hbox{-}}\mathsf{Mod}}\), every module has a
2-stage resolution:
\end{itemize}
\begin{center}
\begin{tikzcd}
0 & {\ker {\varepsilon}\cong {\mathbb{Z}}{ {}^{ \scriptscriptstyle\oplus^{m} } }} & {{\mathbb{Z}}{ {}^{ \scriptscriptstyle\oplus^{n} } }} & M & 0
\arrow[from=1-4, to=1-5]
\arrow["{\varepsilon}", two heads, from=1-3, to=1-4]
\arrow[from=1-2, to=1-3]
\arrow[from=1-1, to=1-2]
\end{tikzcd}
\end{center}
\begin{quote}
\href{https://q.uiver.app/?q=WzAsNSxbNCwwLCIwIl0sWzMsMCwiTSJdLFsyLDAsIlxcWlpcXHN1bXBvd2Vye259Il0sWzEsMCwiXFxrZXIgXFxlcHMgXFxjb25nIFxcWlpcXHN1bXBvd2Vye219Il0sWzAsMCwiMCJdLFsxLDBdLFsyLDEsIlxcZXBzIiwwLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoiZXBpIn19fV0sWzMsMl0sWzQsM11d}{Link
to Diagram}
\end{quote}
\end{example}
\begin{theorem}[?]
For \(\mu \in \mathsf{C}(M, M')\) and
\(C \coloneqq({ {C}_{\scriptscriptstyle \bullet}} , d) \twoheadrightarrow M, C' \coloneqq({ {C}_{\scriptscriptstyle \bullet}} ', d')\twoheadrightarrow M'\),
there is an induced chain map
\(\alpha \in \mathsf{Ch}\mathsf{C}(C, C')\). Moreover, any other chain
map \(\beta\) is chain homotopic to \(\alpha\).
\begin{quote}
Note that \(C\) can in fact be any projective complex over \(M\), not
necessarily a resolution.
\end{quote}
\end{theorem}
\begin{proof}[?]
Using that \(C_0\) is projective, there is a lift of the following form:
\begin{center}
\begin{tikzcd}
{C_0} && M \\
\\
{C_0'} && {M'}
\arrow["\mu", from=1-3, to=3-3]
\arrow["{\varepsilon}"', two heads, from=3-1, to=3-3]
\arrow["{\varepsilon}", two heads, from=1-1, to=1-3]
\arrow["{\exists \alpha_0}"', dashed, from=1-1, to=3-1]
\end{tikzcd}
\end{center}
\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwyLCJDXzAnIl0sWzIsMiwiTSciXSxbMiwwLCJNIl0sWzAsMCwiQ18wIl0sWzIsMSwiXFxtdSJdLFswLDEsIlxcZXBzIiwyLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoiZXBpIn19fV0sWzMsMiwiXFxlcHMiLDAseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJlcGkifX19XSxbMywwLCJcXGV4aXN0cyBcXGFscGhhXzAiLDIseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XV0=}{Link
to Diagram}
\end{quote}
Now inductively, we want to construct the following lift:
\begin{center}
\begin{tikzcd}
{C_n} && {C_{n-1}} && {C_{n-2}} \\
\\
{C_{n}'} && {C_{n-1}'} && {C_{n-2}'} \\
& {\operatorname{im}d_n' = \ker d_{n-1}'}
\arrow["{d_n}", from=1-1, to=1-3]
\arrow["{d_{n-1}}", from=1-3, to=1-5]
\arrow["{\alpha_{n-2}}", from=1-5, to=3-5]
\arrow["{d_{n-1}'}"', from=3-3, to=3-5]
\arrow["{d_n'}"', from=3-1, to=3-3]
\arrow["{\alpha_{n-1}}", from=1-3, to=3-3]
\arrow["\exists"', dashed, from=1-1, to=3-1]
\arrow[two heads, from=3-1, to=4-2]
\arrow[hook, from=4-2, to=3-3]
\end{tikzcd}
\end{center}
\begin{quote}
\href{https://q.uiver.app/?q=WzAsNyxbMCwwLCJDX24iXSxbMiwwLCJDX3tuLTF9Il0sWzQsMCwiQ197bi0yfSJdLFs0LDIsIkNfe24tMn0nIl0sWzIsMiwiQ197bi0xfSciXSxbMCwyLCJDX3tufSciXSxbMSwzLCJcXGltIGRfbicgPSBcXGtlciBkX3tuLTF9JyJdLFswLDEsImRfbiJdLFsxLDIsImRfe24tMX0iXSxbMiwzLCJcXGFscGhhX3tuLTJ9Il0sWzQsMywiZF97bi0xfSciLDJdLFs1LDQsImRfbiciLDJdLFsxLDQsIlxcYWxwaGFfe24tMX0iXSxbMCw1LCJcXGV4aXN0cyIsMix7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFs1LDYsIiIsMCx7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6ImVwaSJ9fX1dLFs2LDQsIiIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV1d}{Link
to Diagram}
\end{quote}
STS \(\operatorname{im}\alpha_{n-1} d_n \subseteq \ker d_{n-1}'\), which
follows from
\begin{align*}
d_{n-1}' \alpha_{n-1} d_n(x) = \alpha_{n-1} d_{n-1} d_n(x)
.\end{align*}
So there is a map \(C_n \to \operatorname{im}d_n'\), and using
projectivity produces the desired lift by the same argument as in the
case case:
\begin{center}
\begin{tikzcd}
{C_n} && {C_{n-1}} && {C_{n-2}} \\
\\
{C_{n}'} && {C_{n-1}'} && {C_{n-2}'} \\
& {\operatorname{im}d_n' = \ker d_{n-1}'}
\arrow["{d_n}", from=1-1, to=1-3]
\arrow["{d_{n-1}}", from=1-3, to=1-5]
\arrow["{\alpha_{n-2}}", from=1-5, to=3-5]
\arrow["{d_{n-1}'}"', from=3-3, to=3-5]
\arrow["{d_n'}"'{pos=0.4}, from=3-1, to=3-3]
\arrow["{\alpha_{n-1}}", from=1-3, to=3-3]
\arrow[""{name=0, anchor=center, inner sep=0}, "{\exists \text{ by projectivity}}"', dashed, from=1-1, to=3-1]
\arrow[two heads, from=3-1, to=4-2]
\arrow[hook, from=4-2, to=3-3]
\arrow[""{name=1, anchor=center, inner sep=0}, "\exists"{description}, curve={height=-18pt}, dashed, from=1-1, to=4-2]
\arrow[shorten <=8pt, shorten >=8pt, Rightarrow, from=1, to=0]
\end{tikzcd}
\end{center}
\begin{quote}
\href{https://q.uiver.app/?q=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}{Link
to Diagram}
\end{quote}
To see that any two such maps are chain homotopic, set
\(\gamma \coloneqq\alpha - \beta\), then
\begin{align*}
{\varepsilon}'( \gamma_0) = {\varepsilon}'( \alpha_i - \beta_i) = \mu{\varepsilon}- \mu {\varepsilon}=0
,\end{align*}
and
\begin{align*}
d_n'(\gamma_n)
&- d_n'( \alpha_n - \beta_n) \\
&= d_n' \alpha_n - d_n' \beta_n \\
&= \alpha_{n-1} d_n - \beta_{n-1} d_n \\
&= \gamma_{n-1} d_n
,\end{align*}
so \(\gamma\) yields a well-defined chain map.
We'll now construct the chain homotopy inductively. There is a lift
\(s_0\) of the following form:
\begin{center}
\begin{tikzcd}
&& {C_0} \\
\\
{C_1'} && {C'_0} & {M'} & 0 \\
& {\operatorname{im}d_1'}
\arrow["{d_1'}"', two heads, from=3-1, to=4-2]
\arrow[hook, from=4-2, to=3-3]
\arrow["{\gamma_0}"', from=1-3, to=3-3]
\arrow["{\exists s_0}"', dashed, from=1-3, to=3-1]
\arrow[from=3-1, to=3-3]
\arrow["{{\varepsilon}'}", two heads, from=3-3, to=3-4]
\arrow[from=3-4, to=3-5]
\end{tikzcd}
\end{center}
\begin{quote}
\href{https://q.uiver.app/?q=WzAsNixbMywyLCJNJyJdLFs0LDIsIjAiXSxbMiwyLCJDJ18wIl0sWzEsMywiXFxpbSBkXzEnIl0sWzIsMCwiQ18wIl0sWzAsMiwiQ18xJyJdLFs1LDMsImRfMSciLDIseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJlcGkifX19XSxbMywyLCIiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFs0LDIsIlxcZ2FtbWFfMCIsMl0sWzQsNSwiXFxleGlzdHMgc18wIiwyLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV0sWzUsMl0sWzIsMCwiXFxlcHMnIiwwLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoiZXBpIn19fV0sWzAsMV1d}{Link
to Diagram}
\end{quote}
This follows because \(\operatorname{im}d_1' = \ker {\varepsilon}'\) and
\({\varepsilon}' \gamma_0 = 0\) by the previous calculation.
Assuming all \(s_{i\leq n-1}\) are constructed, set
\(\gamma_i = d_{i+1}' s_i + s_{i-1} d_i\). Setting
\(\gamma_n - s_{n-1}d_n: C_n \to C_n'\), then
\begin{align*}
d_n'( \gamma_n - s_{n-1} d_n)
&= d_n' \gamma_n - d_n' s_{n-1} d_n \\
&= \gamma_{n-1} d_n - d_n' s_{n-1} d_n \\
&= (\gamma_{n-1} - d_n' s_{n-1})d_n \\
&= s_{n-2} d_{n-1} d_n \\
&= 0
,\end{align*}
using \(d^2 = 0\). Now there is a lift \(s_n\) of the following form:
\begin{center}
\begin{tikzcd}
&& {C_n} \\
\\
{C_{n+1}'} && {C_n'} && {C_{n-1}} \\
& {\operatorname{im}d_{n+1} = \ker d_n'}
\arrow["{\gamma_n - s_{n-1} d_N}", from=1-3, to=3-3]
\arrow["{d_n'}", from=3-3, to=3-5]
\arrow[from=3-1, to=3-3]
\arrow["{d_{n+1}}"', from=3-1, to=4-2]
\arrow[from=4-2, to=3-3]
\arrow["{s_n}"', dashed, from=1-3, to=3-1]
\end{tikzcd}
\end{center}
\begin{quote}
\href{https://q.uiver.app/?q=WzAsNSxbMiwwLCJDX24iXSxbMiwyLCJDX24nIl0sWzQsMiwiQ197bi0xfSJdLFswLDIsIkNfe24rMX0nIl0sWzEsMywiXFxpbSBkX3tuKzF9ID0gXFxrZXIgZF9uJyJdLFswLDEsIlxcZ2FtbWFfbiAtIHNfe24tMX0gZF9OIl0sWzEsMiwiZF9uJyJdLFszLDFdLFszLDQsImRfe24rMX0iLDJdLFs0LDFdLFswLDMsInNfbiIsMix7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dXQ==}{Link
to Diagram}
\end{quote}
Thus follows from the fact that
\(\operatorname{im}\gamma_n - s_{n-1} d_n \subseteq \ker d_n'\) and
projectivity of \(C_n\).
\end{proof}
\begin{remark}
Dually one can construct injective resolutions
\(0 \to M \xhookrightarrow{\eta} { {D}_{\scriptscriptstyle \bullet}}\)
\end{remark}
\hypertarget{derived-functors}{%
\subsection{Derived Functors}\label{derived-functors}}
\begin{remark}
Setup:
\(F: {\mathsf{R}{\hbox{-}}\mathsf{Mod}}\to {\mathbb{Z}{\hbox{-}}\mathsf{Mod}}\)
is an additive covariant functor, e.g.~\(({-}) \otimes_R N\) or
\(M\otimes_R({-})\), and
\({ {C}_{\scriptscriptstyle \bullet}} \xrightarrow[]{{\varepsilon}} { \mathrel{\mkern-16mu}\rightarrow }\, M\)
a complex over \(M\). We define the left-derived functors as
\((L_n F)(M) \coloneqq H_n(F({ {C}_{\scriptscriptstyle \bullet}} ))\).
\end{remark}
\hypertarget{tuesday-february-08}{%
\section{Tuesday, February 08}\label{tuesday-february-08}}
\begin{remark}
Defining derived functors: for \(F\) an additive functor and
\(M\in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}\), take a projective
resolution and apply \(F\):
\begin{align*}
\cdots \to C_2 \xrightarrow{d_2} C_1 \xrightarrow{d_1} C_0 \xrightarrow{{\varepsilon}= d_0} M \to 0 \leadsto F(C_2) \xrightarrow{Fd_2} F(C_1) \xrightarrow{Fd_1} \cdots
,\end{align*}
so \({ {C}_{\scriptscriptstyle \bullet}} \rightrightarrows F\).
Define the left-derived functor
\begin{align*}
{\mathbb{L}}F M \coloneqq H_n F{ {C}_{\scriptscriptstyle \bullet}}
.\end{align*}
\end{remark}
\begin{remark}
Any \(\mu \in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}(M, M')\) induces a
chain map
\(\widehat{\alpha }\in \mathsf{Ch}{\mathsf{R}{\hbox{-}}\mathsf{Mod}}(H_* F{ {C}_{\scriptscriptstyle \bullet}} , H_* F{ {C}_{\scriptscriptstyle \bullet}} ' )\),
where \(\alpha\) is any lift of \(\mu\) to their resolutions.
\begin{center}
\begin{tikzcd}
{{ {C}_{\scriptscriptstyle \bullet}} } && M \\
\\
{{ {C}_{\scriptscriptstyle \bullet}} '} && {M'}
\arrow["{\varepsilon}", Rightarrow, from=1-1, to=1-3]
\arrow["\mu", from=1-3, to=3-3]
\arrow["{{\varepsilon}'}"', Rightarrow, from=3-1, to=3-3]
\arrow["\alpha"', from=1-1, to=3-1]
\end{tikzcd}
\end{center}
\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMiwwLCJNIl0sWzIsMiwiTSciXSxbMCwyLCJcXGNvbXBsZXh7Q30nIl0sWzAsMCwiXFxjb21wbGV4e0N9Il0sWzMsMCwiXFxlcHMiLDAseyJsZXZlbCI6Mn1dLFswLDEsIlxcbXUiXSxbMiwxLCJcXGVwcyciLDIseyJsZXZlbCI6Mn1dLFszLDIsIlxcYWxwaGEiLDJdXQ==}{Link
to Diagram}
\end{quote}
\end{remark}
\begin{exercise}[?]
Show that any two lifts \(\alpha, \alpha'\) induce the same map on
homology.
\end{exercise}
\begin{remark}
Similarly, \({\mathbb{L}}F(M)\) does not depend on the choice of
resolution:
\begin{center}
\begin{tikzcd}
{{ {C}_{\scriptscriptstyle \bullet}} } && M &&&& {F{ {C}_{\scriptscriptstyle \bullet}} } && {F(M)} \\
\\
{{ {C}_{\scriptscriptstyle \bullet}} '} && M & \leadsto &&& {F{ {C}_{\scriptscriptstyle \bullet}} '} && {F(M)} \\
\\
{{ {C}_{\scriptscriptstyle \bullet}} } && M &&&& {F{ {C}_{\scriptscriptstyle \bullet}} } && {F(M)}
\arrow["{\operatorname{id}_M}", from=1-3, to=3-3]
\arrow["{\operatorname{id}_M}", from=3-3, to=5-3]
\arrow["\alpha", from=1-1, to=3-1]
\arrow["\beta", from=3-1, to=5-1]
\arrow["{\varepsilon}", from=5-1, to=5-3]
\arrow["{\varepsilon}", from=3-1, to=3-3]
\arrow["{\varepsilon}", from=1-1, to=1-3]
\arrow[from=1-9, to=3-9]
\arrow[from=3-9, to=5-9]
\arrow[from=5-7, to=5-9]
\arrow[from=3-7, to=3-9]
\arrow[from=1-7, to=1-9]
\arrow["{F(\alpha)}", from=1-7, to=3-7]
\arrow["{F(\beta)}"', from=3-7, to=5-7]
\arrow["{\operatorname{id}_{{ {C}_{\scriptscriptstyle \bullet}} }}"', curve={height=30pt}, from=1-1, to=5-1]
\arrow["{\therefore \operatorname{id}_{F { {C}_{\scriptscriptstyle \bullet}} }}"', curve={height=30pt}, dashed, 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}
\end{remark}
\begin{definition}[Projective resolution of a SES]
For \(0\to M' \to M\to M'' \to 0\) in \(\mathsf{C}\), a
\textbf{projective resolution} is a collection of chain maps forming
projective resolutions of each of the constituent modules:
\begin{center}
\begin{tikzcd}
0 && {{ {C}_{\scriptscriptstyle \bullet}} '} && {{ {C}_{\scriptscriptstyle \bullet}} } && {{ {C}_{\scriptscriptstyle \bullet}} ''} && 0 \\
\\
0 && {M'} && M && {M''} && 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=1-7, to=1-9]
\arrow[from=1-5, to=1-7]
\arrow[from=1-3, to=1-5]
\arrow[from=1-1, to=1-3]
\arrow[Rightarrow, from=1-3, to=3-3]
\arrow[Rightarrow, from=1-5, to=3-5]
\arrow[Rightarrow, from=1-7, to=3-7]
\end{tikzcd}
\end{center}
\begin{quote}
\href{https://q.uiver.app/?q=WzAsMTAsWzAsMiwiMCJdLFsyLDIsIk0nIl0sWzQsMiwiTSJdLFs2LDIsIk0nJyJdLFs4LDIsIjAiXSxbMCwwLCIwIl0sWzIsMCwiXFxjb21wbGV4e0N9JyJdLFs0LDAsIlxcY29tcGxleHtDfSJdLFs2LDAsIlxcY29tcGxleHtDfScnIl0sWzgsMCwiMCJdLFswLDFdLFsxLDJdLFsyLDNdLFszLDRdLFs4LDldLFs3LDhdLFs2LDddLFs1LDZdLFs2LDEsIiIsMSx7ImxldmVsIjoyfV0sWzcsMiwiIiwxLHsibGV2ZWwiOjJ9XSxbOCwzLCIiLDEseyJsZXZlbCI6Mn1dXQ==}{Link
to Diagram}
\end{quote}
\end{definition}
\begin{exercise}[?]
Show that such resolutions exist. This involves constructing
\({\varepsilon}: C_0 \to M\):
\begin{center}
\begin{tikzcd}
0 && {C_0'} && {C \cong C_0' \oplus C_0''} && {C_0''} && 0 \\
\\
0 && {M'} && M && {M''} && 0
\arrow[from=3-1, to=3-3]
\arrow["\gamma", hook, from=3-3, to=3-5]
\arrow["\sigma", two heads, from=3-5, to=3-7]
\arrow[from=3-7, to=3-9]
\arrow[from=1-7, to=1-9]
\arrow["{p_0}", two heads, from=1-5, to=1-7]
\arrow["{\iota_0}", hook, from=1-3, to=1-5]
\arrow[from=1-1, to=1-3]
\arrow["{\varepsilon}"', from=1-3, to=3-3]
\arrow["{\therefore \exists {\varepsilon}}"', dashed, from=1-5, to=3-5]
\arrow["{{\varepsilon}''}"', two heads, from=1-7, to=3-7]
\arrow["{\exists {\varepsilon}^*}"', dashed, from=1-7, to=3-5]
\end{tikzcd}
\end{center}
\begin{quote}
\href{https://q.uiver.app/?q=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}{Link
to Diagram}
\end{quote}
The claim is that
\({\varepsilon}(x, x'') \coloneqq\gamma {\varepsilon}'(x') + {\varepsilon}^*(x'')\)
works. To prove surjectivity, use the following:
\end{exercise}
\begin{proposition}[Short Five Lemma]
Given a commutative diagram of the following form
\begin{center}
\begin{tikzcd}
0 && A && B && C && 0 \\
\\
0 && {A'} && {B'} && {C'} && 0
\arrow[from=3-1, to=3-3]
\arrow["s", from=3-3, to=3-5]
\arrow["t", from=3-5, to=3-7]
\arrow[from=3-7, to=3-9]
\arrow[from=1-1, to=1-3]
\arrow["p", from=1-3, to=1-5]
\arrow["q", from=1-5, to=1-7]
\arrow[from=1-7, to=1-9]
\arrow["h"', from=1-7, to=3-7]
\arrow["g"', from=1-5, to=3-5]
\arrow["f"', from=1-3, to=3-3]
\end{tikzcd}
\end{center}
\begin{quote}
\href{https://q.uiver.app/?q=WzAsMTAsWzAsMCwiMCJdLFsyLDAsIkEiXSxbNCwwLCJCIl0sWzYsMCwiQyJdLFs4LDAsIjAiXSxbMCwyLCIwIl0sWzIsMiwiQSciXSxbNCwyLCJCJyJdLFs2LDIsIkMnIl0sWzgsMiwiMCJdLFs1LDZdLFs2LDcsInMiXSxbNyw4LCJ0Il0sWzgsOV0sWzAsMV0sWzEsMiwicCJdLFsyLDMsInEiXSxbMyw0XSxbMyw4LCJoIiwyXSxbMiw3LCJnIiwyXSxbMSw2LCJmIiwyXV0=}{Link
to Diagram}
\end{quote}
If \(g,h\) are mono (resp. epi, resp. iso) then \(f\) is mono (resp.
epi, resp. iso).
\end{proposition}
\begin{proof}[of surjectivity, alternative by diagram chase]
\envlist
\begin{itemize}
\item
Let \(x\in M\)
\item
Set \(y=\sigma(x)\)
\item
Find \(z\in C_0\) such that \({\varepsilon}'' p_0 (z) = y\).
\item
Consider \({\varepsilon}(z) - x\) and apply \(\sigma\):
\begin{align*}
\sigma({\varepsilon}(z) - x)
&= \sigma {\varepsilon}(x) - \sigma(x) \\
&= {\varepsilon}'' p_0(x) - \sigma(x) \\
&= y-y \\
&= 0
.\end{align*}
\item
So \({\varepsilon}(z) - x\in \ker \sigma = \operatorname{im}\gamma\)
\item
Pull back to \(w\in C_0'\) such that
\(\gamma {\varepsilon}'(w) = {\varepsilon}(z) - x\)
\item
Check
\({\varepsilon}i_0 (w) = \gamma {\varepsilon}'(w) = {\varepsilon}(z) - x\),
so \({\varepsilon}(i_0(w) - z) = -x\).
\end{itemize}
\end{proof}
\begin{proof}[of existence]
The setup:
\begin{center}
\begin{tikzcd}
&& 0 && 0 && 0 \\
\\
0 && {\ker {\varepsilon}'} && {\ker {\varepsilon}} && {\ker {\varepsilon}''} && 0 \\
\\
0 && {C_0'} && {C_0} && {C_0''} && 0 \\
\\
0 && {M'} && M && {M''} && 0
\arrow[from=7-1, to=7-3]
\arrow["\gamma", hook, from=7-3, to=7-5]
\arrow["\sigma", two heads, from=7-5, to=7-7]
\arrow[from=7-7, to=7-9]
\arrow[from=5-7, to=5-9]
\arrow["{p_0}", two heads, from=5-5, to=5-7]
\arrow["{\iota_0}", hook, from=5-3, to=5-5]
\arrow[from=5-1, to=5-3]
\arrow["{{\varepsilon}'}"', from=5-3, to=7-3]
\arrow["{\therefore \exists {\varepsilon}}"', dashed, from=5-5, to=7-5]
\arrow["{{\varepsilon}''}"', two heads, from=5-7, to=7-7]
\arrow["{\exists {\varepsilon}^*}"', dashed, from=5-7, to=7-5]
\arrow[from=3-1, to=3-3]
\arrow["f", hook, from=3-3, to=3-5]
\arrow["g", from=3-5, to=3-7]
\arrow[from=3-7, to=3-9]
\arrow[hook, from=3-7, to=5-7]
\arrow[hook, from=3-5, to=5-5]
\arrow[hook, from=3-3, to=5-3]
\arrow[from=1-3, to=3-3]
\arrow[from=1-5, to=3-5]
\arrow[from=1-7, to=3-7]
\end{tikzcd}
\end{center}
\begin{quote}
\href{https://q.uiver.app/?q=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}{Link
to Diagram}
\end{quote}
This is exact and commutative by a diagram chase:
\begin{itemize}
\tightlist
\item
\(f = i \circ \downarrow_{\ker {\varepsilon}'}\) shows
\(g(\ker {\varepsilon}) \subseteq \ker {\varepsilon}''\)
\item
\(g = p \circ \downarrow_{\ker {\varepsilon}}\) shows
\(f(\ker {\varepsilon}') \subseteq \ker {\varepsilon}\).
\end{itemize}
To show exactness along the top line:
\begin{itemize}
\tightlist
\item
\(f\) is injective, since it's the restriction of an injective map.
\item
\(g\) is surjective:
\begin{itemize}
\tightlist
\item
Let \(x\in \ker {\varepsilon}''\), so \({\varepsilon}''(x) = 0\).
\item
\(\exists y\in C_0\) with \(p_0(y) = x\) by surjectivity of \(p_0\).
\item
Check \({\varepsilon}''(p_0(y)) = {\varepsilon}(x) = 0\) in \(M''\),
so \(\sigma{\varepsilon}(y) = 0\)
\item
Thus \({\varepsilon}(y)\in \ker \sigma = \operatorname{im}\gamma\)
\item
By surjectivity there exists \(w \in C_0'\) such that
\(\gamma( {\varepsilon}'(w)) = {\varepsilon}(y)\).
\item
Use commutativity to verify
\begin{align*}
{\varepsilon}(i_0(w) - y)
&= {\varepsilon}(i_0(w)) - {\varepsilon}(y) \\
&= \gamma{\varepsilon}'(w) - {\varepsilon}(y) \\
&= {\varepsilon}(y) - {\varepsilon}(y) \\
&= 0
.\end{align*}
\item
Then
\begin{align*}
g(i_0(w) - y)
&= p_0(i_0 (w)) - g(y) \\
&= -g(y) \\
&= -p_0(y) \\
&= -x
.\end{align*}
\end{itemize}
\item
Exactness at the middle, i.e.~\(\operatorname{im}f = \ker g\):
\begin{itemize}
\tightlist
\item
\(\operatorname{im}f \subseteq \ker g\) by exactness of the second
row, so it STS \(\ker g \subseteq \operatorname{im}f\).
\item
Let \(y\in \ker g\), then by commutativity
\(y\in \ker p_0 = \operatorname{im}i_0\). Note that
\(y\in \ker {\varepsilon}\) by definition.
\item
Write \(y = i_0(x)\) for some \(x\in C_0'\)
\item
Note
\(\gamma {\varepsilon}' (x) = {\varepsilon}i_0(x) = {\varepsilon}(y) = 0\)
since \(y\in \ker {\varepsilon}\).
\item
Since \(\gamma'\) is mono, \({\varepsilon}'(x) = 0\), so
\(y = i_0(x) = f(x)\).
\end{itemize}
\end{itemize}
\end{proof}
\begin{proposition}[?]
For
\(F: {\mathsf{R}{\hbox{-}}\mathsf{Mod}}\to{\mathbb{Z}{\hbox{-}}\mathsf{Mod}}\)
additive and a SES
\begin{align*}
\xi: 0\to M' \xrightarrow{f} M \xrightarrow{g} M'' \to 0
,\end{align*}
note that there are morphisms
\begin{align*}
{\mathbb{L}}F M'' \to {\mathbb{L}}F M\to {\mathbb{L}}FM'
.\end{align*}
There is a connecting morphism
\begin{align*}
\Delta: {\mathbb{L}}F M'' \to \Sigma^{-1} {\mathbb{L}}F M'
,\end{align*}
which in components looks like
\begin{center}
\begin{tikzcd}
0 && {{\mathbb{L}}_0 F(M'')} && {{\mathbb{L}}_0 F(M)} && {{\mathbb{L}}_0 F(M')} \\
\\
&& {{\mathbb{L}}_1 F(M'')} && {{\mathbb{L}}_1 F(M)} && {{\mathbb{L}}_1 F(M')} \\
\\
&& {{\mathbb{L}}_2 F(M'')} && \cdots
\arrow[from=1-3, to=1-1]
\arrow[from=1-5, to=1-3]
\arrow[from=1-7, to=1-5]
\arrow[from=3-3, to=1-7]
\arrow[from=3-5, to=3-3]
\arrow[from=3-7, to=3-5]
\arrow[from=5-3, to=3-7]
\arrow[from=5-5, to=5-3]
\end{tikzcd}
\end{center}
\begin{quote}
\href{https://q.uiver.app/?q=WzAsOSxbMCwwLCIwIl0sWzIsMCwiXFxMTF8wIEYoTScnKSJdLFs0LDAsIlxcTExfMCBGKE0pIl0sWzYsMCwiXFxMTF8wIEYoTScpIl0sWzIsMiwiXFxMTF8xIEYoTScnKSJdLFs0LDIsIlxcTExfMSBGKE0pIl0sWzYsMiwiXFxMTF8xIEYoTScpIl0sWzIsNCwiXFxMTF8yIEYoTScnKSJdLFs0LDQsIlxcY2RvdHMiXSxbMSwwXSxbMiwxXSxbMywyXSxbNCwzXSxbNSw0XSxbNiw1XSxbNyw2XSxbOCw3XV0=}{Link
to Diagram}
\end{quote}
\end{proposition}
\hypertarget{thursday-february-10}{%
\section{Thursday, February 10}\label{thursday-february-10}}
Missed! Please send me notes. :)
\hypertarget{tuesday-february-15}{%
\section{Tuesday, February 15}\label{tuesday-february-15}}
\hypertarget{tuesday-february-22}{%
\section{Tuesday, February 22}\label{tuesday-february-22}}
\hypertarget{prime-ideals}{%
\subsection{Prime Ideals}\label{prime-ideals}}
\begin{remark}
Plan: commutative ring theory, aiming toward tensor triangular geometry.
\end{remark}
\begin{remark}
\envlist
\begin{itemize}
\tightlist
\item
Recall the definition of prime ideals.
\item
Show \({\mathfrak{p}}\in \operatorname{Spec}R \iff R/{\mathfrak{p}}\)
is an integral domain.
\item
Recall
\({\mathfrak{m}}\in \operatorname{mSpec}R \iff R/{\mathfrak{m}}\) is a
field.
\item
Recall the definition of a monoid
\item
Note that \(R\setminus{\mathfrak{p}}\ni 1\) and
\(R\setminus{\mathfrak{p}}\) is a submonoid of \((R, \cdot)\).
\item
Examples of primes:
\begin{itemize}
\tightlist
\item
\(\left\langle{p}\right\rangle \in \operatorname{Spec}R\) and if
\(p\neq 0\) then
\(\left\langle{p}\right\rangle \in \operatorname{mSpec}R\).
\item
\(R = k[x]\) is a PID and
\(\left\langle{f}\right\rangle \in \operatorname{Spec}R \iff f\) is
irreducible.
\end{itemize}
\item
Recall the set of nilpotent elements and the nilradical
\({\sqrt{0_{R}} }\).
\begin{itemize}
\tightlist
\item
Show \({\sqrt{0_{R}} } \in \operatorname{Id}(R)\).
\item
Show that \(R_{ \text{red} }\coloneqq R/{\sqrt{0_{R}} }\) is reduced
(no nonzero nilpotents).
\end{itemize}
\end{itemize}
\end{remark}
\begin{lemma}[Prime Avoidance]
Let \(A, I_j \in \operatorname{Id}(R)\) where at most two of the \(I_j\)
are not prime and \(A \subseteq \displaystyle\bigcup_j I_j\). Then
\(A \subseteq I_j\) for some \(j\).
\end{lemma}
\begin{proof}[of lemma]
The case \(n=1\) is clear. For \(n>1\), if
\(A \subseteq \tilde I_k \coloneqq I_1 \cup I_2 \cup\cdots \widehat{I}_k \cup\cdots \cup I_n\)
then the result holds by the IH. So suppose
\(A \not\subseteq \tilde I_k\) and pick some \(a_k \not\in \tilde I_k\).
Since \(A \subseteq \displaystyle\bigcup I_j\), we must have
\(a_k\in I_k\).
Case 1: \(n=2\). If \(a_1 + a_2\in A\) with
\(a_1 \in I_1 \setminus I_2\) and \(a_2\in I_2\setminus I_1\), then
\(a_1 + a_2\not\in I_1 \cup I_2\) -- otherwise
\(a_1 + a_2 \in I_1 \implies a_2\in I_1\), and similarly if
\(a_1 + a_2\in I_2\). So \(A \subseteq I_1 \cup I_2\).
Case 2: \(n>2\). At least one \(I_j\) is prime, without loss of
generality \(I_1\). However,
\(a_1 + a_2a_3\cdots a_n\in A \setminus\displaystyle\bigcup_{j\geq 1} I_j\).
Since \(a_j\in I_j\), we have \(a_2\cdots a_n \in I_j\), contradicting
\(a_1\not\in I_j\) for \(j\neq 1\).
\end{proof}
\begin{proposition}[?]
Let \(S\leq (R, \cdot)\) be a submonoid and
\(P\in \operatorname{Id}(R)\) proper with \(P \cap S = \emptyset\) and
\(P\) is maximal with respect to this property, so if \(P' \supseteq P\)
and \(P' \cap S = \emptyset\) then \(P' = P\). Then
\(P\in \operatorname{Spec}R\) is prime.
\end{proposition}
\begin{proof}[?]
By contrapositive, we'll show \(a,b\not\in P \implies ab\not\in P\). If
\(a,b\not\in P\), then \(P \subsetneq aR + P, bR + P\) is a proper
subset. By maximality, \((aR + P) \cap S \neq \emptyset\) and
\((bR + P) \cap S \neq \emptyset\). Pick \(s_1, s_2\in S\) with
\(s_1 = x_1 a + p_1, s_2 = x_2 b + p_2\). Then \(s_1 s_2\in S\) and thus
\begin{align*}
s_1 s_2 = x_1x_2 ab + x_1 ap_2 + x_2 b p_1 + p_1 p_2\in x_1x_2 ab + P + P + P
,\end{align*}
hence \(ab\not\in P\) -- otherwise \(S \cap P \neq \emptyset\).
\(\contradiction\)
\end{proof}
\begin{proposition}[?]
Let \(S \leq R\) be a monoid and let \(I \in \operatorname{Id}(R)\) with
\(I \cap S = \emptyset\). Then there exists some
\(p\in \operatorname{Spec}R\) such that
\begin{itemize}
\tightlist
\item
\(I \subseteq p\)
\item
\(p \cap S = \emptyset\)
\end{itemize}
\end{proposition}
\begin{proof}[?]
Set
\(B = \left\{{I' \supseteq I {~\mathrel{\Big\vert}~}I' \cap S = \emptyset}\right\}\),
then \(B \neq \emptyset\). Apply Zorn's lemma to get a maximal element
\(p\), which is prime by the previous proposition.
\end{proof}
\begin{theorem}[Krull]
\begin{align*}
{\sqrt{0_{R}} } = \cap_{p\in \operatorname{Spec}R} p
.\end{align*}
\end{theorem}
\begin{exercise}[?]
Prove this!
\end{exercise}
\hypertarget{localization}{%
\subsection{Localization}\label{localization}}
\begin{remark}
Recall the definition of \({\mathbb{Q}}\) as
\({\mathbb{Z}}{ \left[ { \scriptstyle \frac{1}{S} } \right] }\) where
\(S = {\mathbb{Z}}\setminus\left\{{0}\right\}\) using the arithmetic of
fractions. More generally, for \(D\) an integral domain, there is a
field of fractions \(F\) with \(D \hookrightarrow F\) satisfying a
universal property and thus uniqueness. Recall the definition of
localization and the universal property: if \(\eta: R\to R'\) with
\(\eta(S) \subseteq (R')^{\times}\) then
\(\exists \tilde\eta: R \left[ { \scriptstyle { {S}^{-1}} } \right] \to R'\).
\end{remark}
\begin{remark}
Next time:
\begin{itemize}
\tightlist
\item
Existence of \(R \left[ { \scriptstyle { {S}^{-1}} } \right]\)
\item
Localization for \({\mathsf{R}{\hbox{-}}\mathsf{Mod}}\).
\item
Localization using tensor products.
\end{itemize}
\end{remark}
\hypertarget{tuesday-march-01}{%
\section{Tuesday, March 01}\label{tuesday-march-01}}
\begin{remark}
Recall the definition of the localization of an
\(R\in \mathsf{CRing}^{\operatorname{unital}}\) at a submonoid
\(S \leq (M, \cdot)\), written
\(R \left[ { \scriptstyle { {S}^{-1}} } \right]\). Similarly for
\(M\in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}\), one can form
\(M \left[ { \scriptstyle { {S}^{-1}} } \right]\), and
\(({-}) \left[ { \scriptstyle { {S}^{-1}} } \right]\) is a functor where
the induced map on \(M \xrightarrow{f} N\) is
\(f_S(m/s) \coloneqq f(m)/s\).
\end{remark}
\begin{proposition}[?]
For \(I\in \operatorname{Id}(R)\), let
\(j(I) \coloneqq\left\{{a\in R{~\mathrel{\Big\vert}~}a/s\in I \text{ for some } s\in S}\right\}\)
which is again an ideal in \(R\). Then
\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
\(j(I)_S = I\),
\item
\(I_S = R_S \iff I\) contains an element of \(S\).
\end{enumerate}
\end{proposition}
\begin{proof}[of 2]
\(\impliedby\): \(I_S \subseteq R_S\) is clear. Let \(x/t\in R_S\) and
\(s\in I \cap S\), then \({sx\over st} = {x\over t}\in I_S\).
\(\implies\): Write \(1=i/s\) to produce \(t\in s\) with \(t(s-i) = 0\).
Then \(z=ts \in S\) and \(z=it\in I\) so \(z \in I \cap S\).
\end{proof}
\begin{proposition}[?]
Let \(P\in \operatorname{Spec}R\) with \(S \cap p = \emptyset\), then
\(j(P_S) = P\).
\end{proposition}
\begin{proof}[?]
\(\supseteq\): Clear.
\(\subseteq\): Let \(a\in j(P_S)\), so \(a/s=p/t\) for
\(s,t\in S, p\in P\) and \(\exists u\in S\) such that
\(u(at-sp)=0\in P\), so \(uat - usp\in P\) where \(usp\in P\). Thus
\(uat\in P \implies a(ut)\in P\implies a\in P\), since \(ut\in S\) and
\(ut\not\in P\).
\end{proof}
\begin{proposition}[?]
There is an order-preserving correspondence
\begin{align*}
\left\{{p\in \operatorname{Spec}R {~\mathrel{\Big\vert}~}p \cap S = \emptyset}\right\} &\rightleftharpoons\operatorname{Spec}R \left[ { \scriptstyle { {S}^{-1}} } \right] \\
P &\mapsto P \left[ { \scriptstyle { {S}^{-1}} } \right] \\
j(P') &\mapsfrom P'
.\end{align*}
\end{proposition}
\begin{proof}[?]
We need to show
\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
\(P \left[ { \scriptstyle { {S}^{-1}} } \right] \in \operatorname{Spec}R \left[ { \scriptstyle { {S}^{-1}} } \right]\)
is actually prime.
\item
If
\(P'\in \operatorname{Spec}R \left[ { \scriptstyle { {S}^{-1}} } \right]\)
then \(j(P')\in \operatorname{Spec}R\) with
\(j(P') \cap S = \emptyset\).
\end{enumerate}
For one:
\begin{align*}
{x\over t}, {y\over t} \in P_S
&\implies {xy\over st} \in P_S \\
&\implies xy \in j(P_S) = P \\
&\implies x\in P \text{ or } y\in P \\
&\implies x/s\in P \text{ or } y/s\in P
.\end{align*}
For two:
\begin{align*}
xy\in j(P')
&\implies {xy\over s}\in P' \\
&\implies {x\over 1}{y\over s}\in P' \\
&\implies {x\over 1}\in P' \text{ or } {y\over s}\in P' \\
&\implies {x}\in P' \text{ or } {y}\in P' \\
.\end{align*}
If \(x\in j(P') \cap S\) then \({x\over t}\in P'\) so
\({t\over x}{x\over t}\in P'\). \(\contradiction\)
One can then check that these two maps compose to the identity.
\end{proof}
\begin{exercise}[?]
Show that if \(p\in \operatorname{Spec}R\) then
\(R_p \in \mathsf{Loc}\mathsf{Ring}\) is local. Use that the image of
\(p\) in \(R_p\) is \(P_p = R_p\setminus R_p^{\times}\), making it
maximal and unique.
\end{exercise}
\begin{exercise}[?]
Show that
\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
\(M=0 \iff M_S = 0\) for all \(S\),
\item
\(M=0 \iff M_p = 0\, \forall p\in \operatorname{mSpec}R\),
\item
\(M=0 \iff M_p = 0\, \forall p\in \operatorname{Spec}R\), noting that
this is a stronger condition than maximal.
\end{enumerate}
For (2), use that \(\operatorname{Ann}_R(x)\) is a proper ideal and thus
contained in a maximal, and show by contradiction that
\(x/1\neq 0\in M_p\).
\end{exercise}
\begin{exercise}[?]
Show that if \(f\in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}(M, N)\) then
\begin{itemize}
\tightlist
\item
\(f\) injective (resp. surjective) \(\implies f_S\) injective (resp.
surjective)
\item
If \(f_p\) is injective for all \(p\in \operatorname{Spec}R\), then
\(f\) is injective (resp. surjective)
\item
If \(M\) is flat then \(M_S\) is flat
\item
If \(M_p\) is flat for all \(p\) then \(M\) is flat.
\end{itemize}
\end{exercise}
\begin{remark}
Recall that for \(A \subseteq R\),
\(V(A) \coloneqq\left\{{p\in \operatorname{Spec}R{~\mathrel{\Big\vert}~}p\supseteq A}\right\}\).
Letting \(I(A)\) be the ideal generated by \(A\), then check that
\(V(I(A)) = V(A)\) and \(V(I) = V(\sqrt I)\).
\end{remark}
\begin{exercise}[?]
Check that defining closed sets as
\(\left\{{V(A) {~\mathrel{\Big\vert}~}A \subseteq R}\right\}\) forms the
basis for a topology on \(\operatorname{Spec}R\), and
\(V(p) \cap V(q) = V(pq)\).
\end{exercise}
\begin{remark}
Next time: generic points, idempotents, irreducible sets.
\end{remark}
\hypertarget{tuesday-march-15}{%
\section{Tuesday, March 15}\label{tuesday-march-15}}
\begin{quote}
See \url{https://www.math.ucla.edu/~balmer/Pubfile/TTG.pdf}
\end{quote}
\begin{remark}
Recall that
\(V(B) \coloneqq\left\{{p\in \operatorname{Spec}R {~\mathrel{\Big\vert}~}p\supseteq B}\right\}\)
are the closed sets for the Zariski topology, and
\(V(B) = V(\left\langle{B}\right\rangle)\). Write
\(I(A) = \displaystyle\bigcap_{p\in A} p\) for the vanishing ideal of
\(A\), and note
\(V(I(A)) = { \operatorname{cl}} _{\operatorname{Spec}R} A\). Recall
\(\sqrt{J} = \displaystyle\bigcap_{p\supseteq J} = \left\{{x\in R {~\mathrel{\Big\vert}~}\exists n\, \text{ such that } x^n \in J}\right\}\),
so \(\sqrt{0}\) is the nilradical, i.e.~all nilpotent elements. An ideal
\(J\) is radical iff \(\sqrt J = J\).
\end{remark}
\begin{theorem}[?]
For \(X=\operatorname{Spec}R\), \(I(V(J)) = \sqrt{J}\), and there is a
bijection between closed subsets of \(X\) and radical ideals in \(R\).
\end{theorem}
\begin{proof}[?]
\begin{align*}
I(V(J)) = \displaystyle\bigcap_{p\in V(J)} p = \displaystyle\bigcap_{p\supseteq J} p = \sqrt{J}
,\end{align*}
and
\begin{align*}
J \xrightarrow{V} V(J) \xrightarrow{I} I(V(J)) = \sqrt{J} = J
.\end{align*}
\end{proof}
\begin{remark}
Recall that \(X\) is \textbf{reducible} iff \(X= X_1 \cup X_2\) with
\(X_i\) nonempty proper and closed.
\end{remark}
\begin{theorem}[?]
For \(R\in \mathsf{CRing}\), a closed subset \(A \subseteq X\) is
irreducible iff \(I(A)\) is a prime ideal.
\end{theorem}
\begin{proof}[?]
\(\implies\): Suppose \(A\) is irreducible, let
\(fg\in I(A) = \displaystyle\bigcap_{p\in A} p\). Then
\(fg\in p\implies f\in [\) without loss of generality for all
\(p\in A\), and \(A = (A \cap V(f)) \cup(A \cap V(g))\) so
\(A \subseteq V(f)\) or \(A \subseteq V(g)\). Thus
\(f\in \sqrt{\left\langle{f}\right\rangle} = I(V(f)) \subseteq I(A)\)
(similarly for \(g\)).
\(\impliedby\): Suppose \(I(A)\) is a prime ideal and
\(A = A_1 \cup A_2\) with \(A_j\) closed, so \(I(A) \subseteq I(A_j)\).
Then
\begin{align*}
I(A) = I(A_1 \cup A_2) = I(A_1) \cap I(A_2)
.\end{align*}
If \(I(A_j) \subsetneq I(A)\) are proper containments, then one reaches
a contradiction: if \(x\in I(A_1)\) and \(y\in I(A_2)\), use that
\(xy\in I(A)\) to conclude \(x\in I(A)\) or \(y\in I(A)\).
\end{proof}
\begin{theorem}[?]
Let \(X\in {\mathsf{Top}}\); TFAE:
\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
\(X\) is irreducible.
\item
Any two open nonempty sets intersect.
\item
Any nonempty open is dense in \(X\).
\end{enumerate}
\end{theorem}
\begin{proposition}[?]
\envlist
\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
Any irreducible subset of \(X\) is entirely contained in a single
irreducible component.
\item
Any space is a union of its irreducible components.
\end{enumerate}
\end{proposition}
\begin{remark}
\begin{itemize}
\tightlist
\item
A space is Noetherian iff any descending chain of closed sets
stabilizes, and if \(R\) is a Noetherian ring then
\(X=\operatorname{Spec}R\) is a Noetherian space. Note that the
converse may not hold in general!
\item
A Noetherian space has a unique decomposition into irreducibles.
\item
Any irreducible component is the closure of a point.
\item
Any nonempty irreducible closed subset
\(A \subseteq \operatorname{Spec}R\) contains a unique generic point
\(p = I(A)\).
\end{itemize}
\end{remark}
\begin{remark}
Coming up:
\begin{itemize}
\tightlist
\item
Group cohomology, the Hopf algebra structure on \(kG\)
\item
Cohomology using minimal resolutions
\item
\(R = H^0(G; k) = \operatorname{Ext} _{kG}^0(k, k)\) which is a
Noetherian ring
\item
Use minimal resolutions to define \(c_{kG}(M)\), the rate of growth of
a minimal projective resolution of \(M\) (1977)
\item
Support varieties:
\(R\coloneqq\operatorname{Ext} ^i_{kG}(k,k)\curvearrowright\tilde M\coloneqq\operatorname{Ext} ^0_{kG}(M, M)\),
let \(J = \operatorname{Ann}_R(\tilde M)\) and
\(V_G(M) = \operatorname{Spec}(R/J)\).
\item
An equality of numerical invariants: \(c_{kG}(M) = \dim V_G(M)\).
\item
Paul Balmer's tensor triangular geometry.
\end{itemize}
\end{remark}
\hypertarget{tuesday-march-22}{%
\section{Tuesday, March 22}\label{tuesday-march-22}}
\hypertarget{hilbert-serre}{%
\subsection{Hilbert-Serre}\label{hilbert-serre}}
\begin{remark}
Setup:
\(V\in {\mathsf{gr}\,}_{\mathbb{Z}}{\mathsf{k}{\hbox{-}}\mathsf{Mod}}\)
a graded vector space, so \(V = \bigoplus _{r\geq 0} V_r\) with
\(\dim_k V_r < \infty\). Define the \textbf{Poincare series}
\begin{align*}
p(V, t) = \sum_{r\geq 0} \dim V_r t^r
.\end{align*}
\end{remark}
\begin{theorem}[Hilbert-Serre]
Let \(R\in {\mathsf{gr}\,}_{\mathbb{Z}}\mathsf{CRing}\) be of finite
type over \(A_0\) for \(A\in {{k}{\hbox{-}}\mathsf{Alg}}\) and suppose
\(R\) is finitely generated over \(A_0\) by homogeneous elements of
degrees \(k_1,\cdots, k_s\). Supposing
\(V\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\),
\begin{align*}
p(V, t) = {f(t) \over \prod_{1\leq j\leq s} 1-t^{k_j} }, \qquad f(t) \in {\mathbb{Z}}[t]
.\end{align*}
\end{theorem}
\begin{proposition}[?]
Suppose that
\begin{align*}
p(V, t) = {f(t) \over \prod_{1\leq j\leq s} 1-t^{k_j} } = \sum_{r\geq 0} a_r t^r, \qquad f(t) \in {\mathbb{Z}}[t], a_r\in {\mathbb{Z}}_{\geq 0}
.\end{align*}
Let \(\gamma\) be the order of the pole of \(p(t)\) at \(t=1\). Then
\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
There exists \(K > 0\) such that \(a_n \leq K n^{\gamma-1}\) for
\(n\geq 0\)
\item
There does \emph{not} exist \(k > 0\) such that
\(a_n \leq k n^{\gamma - 2}\).
\end{enumerate}
\end{proposition}
\begin{definition}[?]
Let \(V\) be a graded vector space of finite type over \(k\). The
\textbf{rate of growth} \(\gamma(V)\) of \(V\) is the smallest
\(\gamma\) such that \(\dim V_n \leq C n^{\gamma-1}\) for all
\(n\geq 0\) for some constant \(C\).
\end{definition}
\begin{remark}
Compare this to the complexity \(C_G(M) = \gamma(P_0)\) where
\(P^0 \rightrightarrows M\) is a minimal projective resolution.
\end{remark}
\hypertarget{finite-generation-of-cohomology}{%
\subsection{Finite Generation of
Cohomology}\label{finite-generation-of-cohomology}}
\begin{remark}
Fix \(G \in {\mathsf{Fin}}{\mathsf{Grp}}\). Recall that
\({ {H}^{\scriptscriptstyle \bullet}} (G; k) { {\operatorname{Ext} }^{\scriptscriptstyle \bullet}} _{G}(k, k)\)
has an algebra structure given by concatenation of LESs:
\begin{center}
\begin{tikzcd}
{\xi_M:} & 0 & k & {M_1} & \cdots & {M_n} & k & 0 & {\in \operatorname{Ext} ^n_G(k, k)} \\
\\
{\xi_N:} & 0 & k & {N_1} & \cdots & {N_m} & k & 0 & {\in \operatorname{Ext} ^m_G(k, k)}
\arrow[from=1-2, to=1-3]
\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["{\xi_M \cdot \xi_N}"{description}, color={rgb,255:red,92;green,92;blue,214}, dashed, from=1-7, to=3-3]
\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=3-2, to=3-3]
\end{tikzcd}
\end{center}
\begin{quote}
\href{https://q.uiver.app/?q=WzAsMTgsWzEsMCwiMCJdLFsyLDAsImsiXSxbMywwLCJNXzEiXSxbNCwwLCJcXGNkb3RzIl0sWzUsMCwiTV9uIl0sWzYsMCwiayJdLFs3LDAsIjAiXSxbMiwyLCJrIl0sWzYsMiwiayJdLFszLDIsIk5fMSJdLFs1LDIsIk5fbSJdLFs0LDIsIlxcY2RvdHMiXSxbMSwyLCIwIl0sWzcsMiwiMCJdLFs4LDAsIlxcaW4gXFxFeHRebl9HKGssIGspIl0sWzgsMiwiXFxpbiBcXEV4dF5tX0coaywgaykiXSxbMCwwLCJcXHhpX006Il0sWzAsMiwiXFx4aV9OOiJdLFswLDFdLFsxLDJdLFsyLDNdLFszLDRdLFs0LDVdLFs1LDZdLFs1LDcsIlxceGlfTSBcXGNkb3QgXFx4aV9OIiwxLHsiY29sb3VyIjpbMjQwLDYwLDYwXSwic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fSxbMjQwLDYwLDYwLDFdXSxbNyw5XSxbOSwxMV0sWzExLDEwXSxbMTAsOF0sWzgsMTNdLFsxMiw3XV0=}{Link
to Diagram}
\end{quote}
Recall that
\(\operatorname{Ext} ^n_{G}(k, k) = \mathop{\mathrm{Hom}}_{kG}(P_n, k)\),
providing the additive structure. Moreover,
\(\operatorname{Ext} _{kG}(M, M)\) is a ring, and if
\(N\in {\mathsf{kG}{\hbox{-}}\mathsf{Mod}}\), then
\(\operatorname{Ext} _{kG}{N, M} \in {\mathsf{\operatorname{Ext} _{kG}(M, M)}{\hbox{-}}\mathsf{Mod}}\).
Similarly
\(\operatorname{Ext} ^0_{kG}(N, M) \in {\mathsf{ { {\operatorname{Ext} }^{\scriptscriptstyle \bullet}} (k, k)}{\hbox{-}}\mathsf{Mod}}\)
by tensoring LESs.
\end{remark}
\begin{remark}
There is a coproduct
\begin{align*}
kG &\xrightarrow{\Delta} kG \otimes_k kG \\
g &\mapsto g\otimes g
.\end{align*}
There is a cup product:
\begin{center}
\begin{tikzcd}
{\bigoplus _{s+t=m} \operatorname{Ext} _{kG}^s(k, N) \otimes_k \operatorname{Ext} ^t_{kG}(k, M)} && {\operatorname{Ext} ^{m}_{kG{ {}^{ \scriptstyle\otimes_{k}^{2} } }}(k\otimes_k N, k\otimes_k M) } \\
\\
&& {\operatorname{Ext} _{kG}^m(N, M)}
\arrow["\cong", tail reversed, from=1-1, to=1-3]
\arrow[from=1-3, to=3-3]
\arrow["{(a, b)\mapsto a\smile b}"', from=1-1, to=3-3]
\end{tikzcd}
\end{center}
\begin{quote}
\href{https://q.uiver.app/?q=WzAsMyxbMCwwLCJcXGJpZ29wbHVzIF97cyt0PW19IFxcRXh0X3trR31ecyhrLCBOKSBcXHRlbnNvcl9rIFxcRXh0XnRfe2tHfShrLCBNKSJdLFsyLDAsIlxcRXh0XnttfV97a0dcXHRlbnNvcnBvd2VyIGsgMn0oa1xcdGVuc29yX2sgTiwga1xcdGVuc29yX2sgTSkgIl0sWzIsMiwiXFxFeHRfe2tHfV5tKE4sIE0pIl0sWzAsMSwiXFxjb25nIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiYXJyb3doZWFkIn19fV0sWzEsMl0sWzAsMiwiKGEsIGIpXFxtYXBzdG8gYVxcY3VwcHJvZCBiIiwyXV0=}{Link
to Diagram}
\end{quote}
It is a theorem that this coincides with the Yoneda product.
\end{remark}
\begin{theorem}[?]
\envlist
\begin{itemize}
\tightlist
\item
\(H^0(G, k)\) is a graded commutative ring, so
\(xy = (-1)^{{\left\lvert {x} \right\rvert} {\left\lvert {y} \right\rvert}} yx\)
\item
The even part
\({ {H}^{\scriptscriptstyle \bullet}} ^{\text{even}}(G; k)\) is a
(usual) commutative ring.
\end{itemize}
\end{theorem}
\begin{theorem}[Evans-Venkov, 61]
\envlist
\begin{itemize}
\tightlist
\item
\(H^0(G; k)\) is a finitely generated in \({\mathsf{Alg}_{/k} }\)
\item
If \(M\in {\mathsf{kG}{\hbox{-}}\mathsf{Mod}}\) then
\(\operatorname{Ext} ^0_{kG}(k, M) \in {\mathsf{ { {H}^{\scriptscriptstyle \bullet}} (G; k) }{\hbox{-}}\mathsf{Mod}}\).
\end{itemize}
\end{theorem}
\begin{remark}
Quillen described
\(\operatorname{mSpec} { {H}^{\scriptscriptstyle \bullet}} (G, k)^{ \text{red} }\)
in the 70s. Idea: look at \(E \hookrightarrow G\) the elementary abelian
subgroups, so \(E \cong C_p{ {}^{ \scriptscriptstyle\times^{m} } }\)
where \(p = \operatorname{ch}k\), and consider
\(V_G(k) = \displaystyle\bigcup_{E\leq G} V_E(k)/\sim\) the union of all
elementary abelian subgroups, where
\(V_G(k) \coloneqq\operatorname{mSpec} { {H}^{\scriptscriptstyle \bullet}} ^{}(G; k)^{ \text{red} }\).
Note that in characteristic zero, this is semisimple and only \(H^0=k\)
survives.
\end{remark}
\begin{example}[?]
\envlist
\begin{itemize}
\item
For \(A = C_p\) with \(\operatorname{ch}k = p > 0\), then
\begin{align*}
R \coloneqq H^0(C_p; k) \cong
\begin{cases}
k[x,y]/\left\langle{y^2}\right\rangle, {\left\lvert {x} \right\rvert} = 2, {\left\lvert {y} \right\rvert} = 1 & p\geq 3
\\
k[x], {\left\lvert {x} \right\rvert} = 1 & p = 2.
\end{cases}, \qquad \operatorname{mSpec}R \cong {\mathbb{A}}^1_{/ {k}}
.\end{align*}
\item
Dan's favorite: \(A = u({\mathfrak{g}})\) for
\({\mathfrak{g}}= {\mathfrak{sl}}_2\) with
\(\operatorname{ch}k = p \geq 3\) for \(u\) the \emph{small enveloping
algebra}. Friedlander-Parshall show
\(\operatorname{mSpec}R = k[{\mathcal{N}}]\) for
\({\mathcal{N}}\coloneqq\left\{{M { \begin{bmatrix} {a} & {b} \\ {c} & {-a} \end{bmatrix} } {~\mathrel{\Big\vert}~}M\text{ is nilpotent}}\right\}\).
This can be presented as
\begin{align*}
k[{\mathcal{N}}] \cong k[x,y,z] / \left\langle{z^2 + xy}\right\rangle, {\left\lvert {x} \right\rvert}, {\left\lvert {y} \right\rvert}, {\left\lvert {z} \right\rvert} = 2
,\end{align*}
and we'll see how finite generation is used in this setting.
\end{itemize}
\end{example}
\hypertarget{tuesday-march-29}{%
\section{Tuesday, March 29}\label{tuesday-march-29}}
\begin{remark}
Setup: for \(G \in {\mathsf{Fin}}{\mathsf{Grp}}, k\in \mathsf{Field}\)
with \(\operatorname{ch}k = p \divides {\sharp}G\). For
\(M\in {\mathsf{kG}{\hbox{-}}\mathsf{Mod}}\), we associate
\(V_G(M) \subseteq \operatorname{mSpec}(R)\) for
\(R\coloneqq H^0(G; k)\). There is a ring morphism
\(\Phi_M: R\to \operatorname{Ext} ^0_{kG}(M, M)\), we set
\(I_G(M) = \left\{{x\in R {~\mathrel{\Big\vert}~}\Phi_M(x) = 0}\right\}\)
and define the support variety as
\(V_G(M) = \operatorname{mSpec}(R/I_G(M))\).
\end{remark}
\begin{example}[?]
Let \(G = C_p{ {}^{ \scriptscriptstyle\times^{n} } }\), then
\begin{itemize}
\tightlist
\item
\(H^2(G; k) = k[x_1, \cdots, x_{n}]\) for
\(\operatorname{ch}k = p \geq 3\).
\item
\(\operatorname{mSpec}R = {\mathbb{A}}^n \supseteq V_E(M)\)
\end{itemize}
\end{example}
\hypertarget{rank-varieties}{%
\subsection{Rank Varieties}\label{rank-varieties}}
\begin{definition}[Rank varieties]
For
\(kG = k[z_1,\cdots, z_n]/\left\langle{z_1^p,\cdots, z_n^p}\right\rangle\),
let \(x_{\mathbf{a}} \coloneqq\sum a_i z_i\) for \(a_i\in k\). Define
the \textbf{rank variety}
\begin{align*}
V_E^r(M) = \left\{{\mathbf{a} {~\mathrel{\Big\vert}~}\mathop{\mathrm{Res}}^{kG}_{ \left\langle{x_{\mathbf{a}}}\right\rangle } \text{ is not free} }\right\} \cup\left\{{0}\right\}
.\end{align*}
\end{definition}
\begin{theorem}[Carlson]
\begin{align*}
V_E(M) \cong V_E^r(M)
.\end{align*}
\end{theorem}
\begin{remark}
Note that
\(\operatorname{Ext} ^0(M, M)\curvearrowright\operatorname{Ext} ^0(M', M)\)
by splicing, so we can define
\(I_G(M', M) \coloneqq\operatorname{Ann}_R \operatorname{Ext} _{kG}^1(M', M)\)
and the \textbf{relative support} variety
\(V_G(M', M) = \operatorname{mSpec}(R/ I_G(M', M))\). This recovers the
previous notion by \(V_G(M, M) = V_G(M)\).
\end{remark}
\begin{remark}
Since \(I_G(M', M) \supseteq I_G(M) + I_G(M')\),
\begin{align*}
V_G(M', M) \subseteq V_G(M) \cap V_G(M')
,\end{align*}
which relates relative support varieties to the usual support varieties.
\end{remark}
\begin{remark}
If \(0\to A\to B\to C\to 0\) is a SES, there is a LES in
\(\operatorname{Ext} _{kG}\) and by considering annihilators we have
\begin{align*}
I_G(A, M)\cdot I_G(B, M) \subseteq I_G(C, M) \implies V_G(C, M) \subseteq V_G(A, M)\cup V_G(C, M)
.\end{align*}
\end{remark}
\begin{proposition}[?]
Let \(M\in \mathsf{kG}{\hbox{-}}\mathsf{Mod}\), then
\begin{align*}
V_G(M) \subseteq \displaystyle\bigcup_{S\leq M \text{ simple}} V_G(S, M)
.\end{align*}
\end{proposition}
\begin{proof}[?]
Take the SES \(0\to S_1 \to M \to M/S_1\to 0\), then
\(V_G(M) = V_G(M, M) \subseteq V_G(S_1, M) \cup V_G(M/S_1, M)\).
Continuing this way yields a union of \(V(T, M)\) over all composition
factors \(T\). Conversely, by the intersection formula above, this union
is contained in \(V_G(M)\), so these are all equal.
\end{proof}
\begin{theorem}[?]
Let \(M \in {\mathsf{kG}{\hbox{-}}\mathsf{Mod}}\), then
\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
\(c_G(M) = \dim V_G(M)\)
\item
\(V_G(M) = \left\{{0}\right\}\) (as a conical varieties) iff \(M\) is
projective.
\end{enumerate}
\end{theorem}
\begin{proof}[?]
Note (2) follows from (1), since complexity zero modules are precisely
projectives. Consider \(\Phi_M: R\to \operatorname{Ext} ^0_{kG}(M, M)\),
which induces
\(R/I_G(M) \hookrightarrow\operatorname{Ext} _{kG}^0(M, M)\) which is
finitely generated over \(R/I_{G}(M)\). A computation shows
\begin{align*}
c_G(M)
&= \gamma(\operatorname{Ext} _{kG}^0(M, M)) \\
&= \gamma( R/I_G(M) ) \\
&= \operatorname{krulldim}(R/I_G(M)) \\
&= \dim V_G(M)
.\end{align*}
\end{proof}
\begin{remark}
Consider a LES
\(0\to M\to M_1\to \cdots \to M_n \to M\to 0 \in \operatorname{Ext} _{kG}^n(M, M)\).
Apply \(\Omega^n({-})\), which arises from projective covers
\({ {P}^{\scriptscriptstyle \bullet}} \rightrightarrows M\) and
truncating to get
\(0\to \Omega^n \to P^{n-1}\to \cdots \to P_0 \to M\to 0\). Similarly
define \(\Omega^{-n}\) in terms of injective resolutions. There is an
isomorphism
\(\operatorname{Ext} _{kG}^n(M, M) \cong \operatorname{Ext} _{kG}^n(\Omega^s M, \Omega^s M)\)
which is compatible with the \(R\) action. Thus
\(V_G(M) \cong V_G (\Omega^s M)\) for any \(s\). Since \(kG\) is a Hopf
algebra, dualizing yields
\(\operatorname{Ext} _{kG}^n(M, M) \cong \operatorname{Ext} _{kG}^n(M {}^{ \vee }, M {}^{ \vee })\)
and thus \(V_G(M) \cong V_G(M {}^{ \vee })\).
\end{remark}
\hypertarget{properties-of-support-varieties}{%
\subsection{Properties of support
varieties}\label{properties-of-support-varieties}}
\begin{proposition}[?]
\begin{align*}
V_G(M_1 \bigoplus M_2) \cong V_G(M_1)\cup V_G(M_2)
.\end{align*}
\end{proposition}
\begin{proof}[?]
Distribute:
\begin{align*}
\operatorname{Ext} _{kG}^0(M_1 \oplus M_2, M_1 \oplus M_2)
& \cong
\operatorname{Ext} _{kG}^0(M_1, M_1) \oplus
\operatorname{Ext} _{kG}^0(M_1, M_2) \oplus
\operatorname{Ext} _{kG}^0(M_2, M_1) \oplus
\operatorname{Ext} _{kG}^0(M_3, M_2)
.\end{align*}
Now \(I_G(M_1 \bigoplus M_2) \subseteq I_G(M_1) \oplus I_G(M_2)\), so
\(V_G(M_1) \cup V_G(M_2) \subseteq V_G(M_1 \oplus M_2)\). Applying the 2
out of 3 property,
\(V_G(M_1 \oplus M_2) \subseteq V_G(M_1) \cup V_G(M_2)\) since there is
a SES \(0\to M_1 \to M_1 \oplus M_2 \to M_2\to 0\).
\end{proof}
\begin{theorem}[Tensor product property]
Let \(M, N\in {\mathsf{kG}{\hbox{-}}\mathsf{Mod}}\), then
\begin{align*}
V_G(M\otimes_k N) = V_G(M) \cap V_G(N)
.\end{align*}
\end{theorem}
\begin{remark}
Conjectured by Carlson, proved by Arvrunin-Scott (82). Prove for
elementary abelians, piece together using the Quillen stratification.
\end{remark}
\begin{theorem}[Carlson]
Let \(X = \operatorname{mSpec}R\), which is a conical variety, and let
\(W \subseteq X\) be a closed conical subvariety (e.g.~a line through
the origin). Then there exists an
\(M\in {\mathsf{kG}{\hbox{-}}\mathsf{Mod}}\) such that \(V_G(M) = W\).
\end{theorem}
\begin{remark}
Take \(\zeta: \Omega^n k \to k\), so \(\zeta\in R/I_G(M)\), and define
certain \(L_\zeta\) modules and set \(Z(\zeta) \coloneqq V_G(L_\zeta)\).
\end{remark}
\begin{theorem}[Carlson]
Let \(M \in {\mathsf{kG}{\hbox{-}}\mathsf{Mod}}\) be indecomposable.
Then the projectivization \(\mathop{\mathrm{Proj}}V_G(M)\) is connected.
\end{theorem}
\hypertarget{supports-using-primes}{%
\subsection{Supports using primes}\label{supports-using-primes}}
\begin{remark}
As before, set \(R = H^{\text{even}}(G; k), X= \operatorname{Spec}R\),
and now define
\begin{align*}
V_G(M) = \left\{{p\in X{~\mathrel{\Big\vert}~}\operatorname{Ext} _{kG}^0(M, M)_p \neq 0}\right\}
.\end{align*}
All of the theorems mentioned today go through with this new definition.
\end{remark}
\begin{exercise}[?]
Let
\(I_G(M) = \operatorname{Ann}_R \operatorname{Ext} _{kG}^0(M, M) {~\trianglelefteq~}R\),
and show
\begin{align*}
V_G(M) = \left\{{p\in X{~\mathrel{\Big\vert}~}p\supseteq I_G(M) }\right\} = V(I_G(M))
\end{align*}
is a closed set.
\end{exercise}
\begin{remark}
Let \({\mathfrak{g}}\in \mathsf{Lie}{\mathsf{Alg}}_{/ {k}}\) with
\(\operatorname{ch}k = p > 0\),
e.g.~\({\mathfrak{g}}= {\mathfrak{gl}}_n(k)\). Then there is a \(p\)th
power operation \(x^{{\left\lceil p \right\rceil}} = x\cdot x\cdots x\).
The pair \(({\mathfrak{g}}, {\left\lceil p \right\rceil})\) forms a
restricted Lie algebra. Consider the enveloping algebra
\(U({\mathfrak{g}})\), and define
\begin{align*}
u({\mathfrak{g}}) \coloneqq U({\mathfrak{g}})/ \left\langle{x^p - x{ {}^{ \scriptstyle\otimes_{k}^{p} } } {~\mathrel{\Big\vert}~}x\in {\mathfrak{g}}}\right\rangle
,\end{align*}
which is a finite-dimensional Hopf algebra:
\begin{itemize}
\tightlist
\item
The counit is \({\varepsilon}(g) = 0\) for \(g\in {\mathfrak{g}}\)
\item
The antipode is \(\theta(g) = -g\)
\item
The comultiplication is \(\Delta(g) = g\otimes 1 + 1\otimes g\).
\end{itemize}
The dimension is given by
\(\dim u({\mathfrak{g}}) = p^{\dim {\mathfrak{g}}}\).
\end{remark}
\hypertarget{tuesday-april-05}{%
\section{Tuesday, April 05}\label{tuesday-april-05}}
\hypertarget{lie-theory}{%
\subsection{Lie Theory}\label{lie-theory}}
\begin{remark}
Setup:
\(k = { \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu }\),
\(\operatorname{ch}k = p > 0\), \({\mathfrak{g}}\) a restricted Lie
algebra (e.g.~\({\mathfrak{g}}= \mathsf{Lie}(G)\) for
\(G\in{\mathsf{Aff}}{\mathsf{Alg}}{\mathsf{Grp}}_{/ {k}}\)). Write
\(A^{{\left\lceil p \right\rceil} } = AA\cdots A\) and set
\(A = u({\mathfrak{g}}) = U({\mathfrak{g}})/ J\) where
\(J = \left\langle{x{ {}^{ \scriptstyle\otimes_{k}^{p} } } - x^{{\left\lceil p \right\rceil}}}\right\rangle\)
which is an ideal generated by central elements. Note that \(A\) is a
finite-dimensional Hopf algebra.
Proved last time: \(H^0(A; k) \in {\mathsf{Alg}_{/k} }^{\mathrm{fg}}\),
using a spectral sequence argument. From the spectral sequence, there is
a finite morphism
\begin{align*}
\Phi: S({\mathfrak{g}}^+)^{(1)} \to H^0(A; k)
,\end{align*}
making \(H^0(A; k)\) an integral extension of \(\operatorname{im}\Phi\).
This induces a map
\begin{align*}
\Phi: \operatorname{mSpec}H^0(A; k) \hookrightarrow{\mathfrak{g}}
.\end{align*}
\end{remark}
\begin{theorem}[Jantzen]
\begin{align*}
\operatorname{mSpec}H^0(A; k) \cong {\mathcal{N}}_p \coloneqq\left\{{x\in {\mathfrak{g}}{~\mathrel{\Big\vert}~}x^{{\left\lceil p \right\rceil}}}\right\}
.\end{align*}
\end{theorem}
\begin{example}[?]
For \({\mathfrak{g}}= {\mathfrak{gl}}_n\),
\({\mathcal{N}}_p \leq {\mathcal{N}}\) is a subvariety of the nilpotent
cone. Moreover \({\mathcal{N}}_p\) is stable under
\(G = \operatorname{GL}_n\), and there are only finitely many orbits.
There is a decomposition into finitely many irreducible orbit closures
\begin{align*}
{\mathcal{N}}_p = \displaystyle\bigcup_i \mkern 1.5mu\overline{\mkern-1.5muGx_i\mkern-1.5mu}\mkern 1.5mu
.\end{align*}
This corresponds to Jordan decompositions with blocks of size at most
\(p\).
\end{example}
\begin{remark}
Using spectral sequences one can show that if
\(M, N \in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}\) then
\(\operatorname{Ext} ^0_A(M, N)\) is finitely-generated as a module over
\(R\coloneqq H^0(A; k)\). So one can define support varieties
\(V_{{\mathfrak{g}}}(M) = \operatorname{mSpec}R/J_M\) where
\(I_M = \operatorname{Ann}_R \operatorname{Ext} ^0_A(M, M)\). Some
facts:
\begin{itemize}
\tightlist
\item
\(V_{{\mathfrak{g}}}(M) \subseteq {\mathcal{N}}_p \subseteq {\mathfrak{g}}\)
\item
If \(M\) is a \(G{\hbox{-}}\)module in addition to being a
\({\mathfrak{g}}{\hbox{-}}\)module, then \(V_G(M)\) is a
\(G{\hbox{-}}\)stable closed subvariety of \({\mathcal{N}}_p\).
\end{itemize}
\end{remark}
\begin{theorem}[Friedlander-Parshall (Inventiones 86)]
Given \(M\in {\mathsf{u({\mathfrak{g}})}{\hbox{-}}\mathsf{Mod}}\),
\begin{align*}
V_{{\mathfrak{g}}}(M) \cong \left\{{x\in {\mathfrak{g}}{~\mathrel{\Big\vert}~}x^{[p]} = 0, M \downarrow_{U(\left\langle{x}\right\rangle)} \text{ is not free over } u(\left\langle{x}\right\rangle) \leq u({\mathfrak{g}}) }\right\} \cup\left\{{0}\right\}
,\end{align*}
which is similar to the rank variety for finite groups, concretely
realize the support variety.
\end{theorem}
\begin{remark}
Here \(\left\langle{x}\right\rangle = kx\) is a 1-dimensional Lie
algebra, and if \(x^{[p]} = 0\) then
\(u(\left\langle{x}\right\rangle) = k[x] / \left\langle{x^p}\right\rangle\)
is a PID. We know how to classify modules over a PID: there are only
finitely many indecomposable such modules.
\end{remark}
\hypertarget{reductive-algebraic-groups}{%
\subsection{Reductive algebraic
groups}\label{reductive-algebraic-groups}}
\begin{example}[?]
For type \(A_n \sim \operatorname{GL}_{n+1}\),
\(\alpha_0 = \tilde \alpha_n = \sum_{1\leq i \leq n} \alpha_i\) and
\(h=n+1\). For \({\mathsf{G}}_2\),
\(\tilde \alpha_n = 3\alpha_1 + 2\alpha_2\) and \(h=6\).
\end{example}
\begin{fact}
If \(p\geq h\) then \({\mathcal{N}}_p({\mathfrak{g}}) = {\mathcal{N}}\).
\end{fact}
\begin{definition}[Good and bad primes]
A prime is \emph{bad} if it divides any coefficient of the highest
weight. By type:
\begin{longtable}[]{@{}ll@{}}
\toprule
Type & Bad primes \\
\midrule
\endhead
\(A_n\) & None \\
\(B_n\) & 2 \\
\(C_n\) & 2 \\
\(D_n\) & 2 \\
\(E_6\) & 2,3 \\
\(E_7\) & 2,3 \\
\(E_8\) & 2,3,5 \\
\(F_4\) & 2,3 \\
\(G_2\) & 2,3 \\
\bottomrule
\end{longtable}
\end{definition}
\begin{theorem}[Carlson-Lin-Nakano-Parshall (good primes), UGA VIGRE (bad primes)]
\({\mathcal{N}}_p = \mkern 1.5mu\overline{\mkern-1.5mu{\mathcal{O}}\mkern-1.5mu}\mkern 1.5mu\)
is an orbit closure, where \({\mathcal{O}}\) is a \(G{\hbox{-}}\)orbit
in \({\mathcal{N}}\). Hence \({\mathcal{N}}_p({\mathfrak{g}})\) is an
irreducible variety.
\end{theorem}
\begin{remark}
Let \(X = X(T)\) be the weight lattice and let \(\lambda \in X\), then
\begin{align*}
\Phi_\lambda \coloneqq\left\{{ \alpha\in \Phi {~\mathrel{\Big\vert}~}{\left\langle {\lambda + \rho},~{\alpha {}^{ \vee }} \right\rangle} \in p{\mathbb{Z}}}\right\}
.\end{align*}
Under the action of the affine Weyl group, this is empty when
\(\lambda\) is on a wall (non-regular) and otherwise contains some roots
for regular weights. When \(p\) is a good prime, there exists a
\(w\in W\) with \(w(\Phi_\lambda) = \Phi_J\) for \(J \subseteq \Delta\)
a subsystem of simple roots. In this case, there is a \textbf{Levi
decomposition}
\begin{align*}
{\mathfrak{g}}= u_J \oplus \ell_J \oplus u_J^+
.\end{align*}
\end{remark}
\begin{remark}
On Levis: consider type \(A_5 \sim \operatorname{GL}_6\) with simple
roots \(\alpha_i\).
\includegraphics{figures/2022-04-05_10-34-18.png}
\end{remark}
\begin{remark}
Consider induced/costandard modules
\(H^0( \lambda) = \operatorname{Ind}_B^G \lambda = \nabla(\lambda)\),
which are nonzero only when \(\lambda \in X_+\) is a dominant weight.
Their characters are given by Weyl's character formula, and their duals
are essentially \emph{Weyl modules} which admit Weyl filtrations. What
are their support varieties?
\end{remark}
\begin{theorem}[Nakano-Parshall-Vella, 2008]
Let \(\lambda\in X_+\) and let \(p\) be a good prime, and let \(w\in W\)
such that \(w(\Phi_\lambda ) = \Phi_J\) for \(J \subseteq \Delta\). Then
\begin{align*}
V_{{\mathfrak{g}}} H^0( \lambda) = G\cdot u_J = \mkern 1.5mu\overline{\mkern-1.5mu{\mathcal{O}}\mkern-1.5mu}\mkern 1.5mu
\end{align*}
is the closure of a ``Richardson orbit''.
\end{theorem}
\begin{remark}
\envlist
\begin{itemize}
\tightlist
\item
This theorem was conjectured by Jantzen in 87, proved for type \(A\).
\item
For bad primes, \(H^0(\lambda)\) is computed in one of seven VIGRE
papers (2007). These still yield orbit closures that are irreducible,
but need not be Richardson orbits.
\end{itemize}
Natural progression: what about tilting modules (good filtrations with
costandard sections and good + Weyl filtrations)? We're aiming for the
Humphreys conjecture.
\end{remark}
\begin{remark}
Let \(T( \lambda)\) be a tilting module for \(\lambda \in X_+\). A
conjecture of Humphreys: \(V_{{\mathfrak{g}}} T( \lambda)\) arises from
considering 2-sided cells of the affine Weyl group, which biject with
nilpotent orbits.
\end{remark}
\begin{example}[?]
In type \(A_2\):
\includegraphics{figures/2022-04-05_10-39-31.png}
There are three nilpotent orbits corresponding to Jordan blocks of type
\(X\alpha_1: (1,0)\) and \(X_\mathrm{reg}: (1,1)\) in
\({\mathfrak{gl}}_3\). Three cases:
\begin{itemize}
\tightlist
\item
\(V_{{\mathfrak{g}}} T( \lambda) = {\mathcal{N}}= \mkern 1.5mu\overline{\mkern-1.5muG X_\mathrm{reg}\mkern-1.5mu}\mkern 1.5mu\)
\item
\(V_{{\mathfrak{g}}} T( \lambda) = \mkern 1.5mu\overline{\mkern-1.5muG X_{ \alpha_1}\mkern-1.5mu}\mkern 1.5mu\)
\item
\(V_{{\mathfrak{g}}} T( \lambda) = \left\{{0}\right\}\)
\end{itemize}
\includegraphics{figures/2022-04-05_10-44-03.png}
\end{example}
\begin{remark}
The computation of \(V_G T( \lambda)\) is still open. Some recent work:
\begin{itemize}
\tightlist
\item
\(p=2, A_n\): done by B. Cooper,
\item
\(p > n+1, A_n\) by W. Hardesty,
\item
\(p \gg 1\), Achar, Hardesty, Riche.
\end{itemize}
\end{remark}
\begin{remark}
What about simple \(G{\hbox{-}}\)modules? Recall
\(L(\lambda) = \mathop{\mathrm{Soc}}_G \nabla( \lambda) \subseteq \nabla( \lambda)\)
-- computing \(V_G L( \lambda)\) is open.
\end{remark}
\begin{theorem}[Drupieski-N-Parshall]
Let \(p > h\) and \(w( \Phi_ \lambda) = \Phi_J\), then
\begin{align*}
V_{u_q({\mathfrak{g}})} L( \lambda) = G u_J
,\end{align*}
i.e.~the support varieties in the quantum case are known. This uses that
the Lusztig character formula is know for \(u_q( {\mathfrak{g}})\).
\end{theorem}
\hypertarget{tuesday-april-12}{%
\section{Tuesday, April 12}\label{tuesday-april-12}}
\hypertarget{tensor-triangular-geometry}{%
\subsection{Tensor triangular
geometry}\label{tensor-triangular-geometry}}
\begin{remark}
Last time: tensor categories and triangulated categories. Idea due to
Balmer: treat categories like rings.
\end{remark}
\begin{definition}[Tensor triangulated categories]
A \textbf{tensor triangulated category} (TTC) is a triple
\((K, \otimes, 1)\) where
\begin{itemize}
\tightlist
\item
\(K\) is a triangulated category
\item
\((K, \otimes)\) is a symmetric monoidal category
\item
\(1\) is a unit, so
\(X\otimes 1 { \, \xrightarrow{\sim}\, }X { \, \xrightarrow{\sim}\, }1\otimes X\)
for all \(X\) in \(K\).
\end{itemize}
\end{definition}
\begin{remark}
We'll have notions of ideals, thick ideals, and prime ideals in \(K\).
Define \(\operatorname{Spc}K\) to be the set of prime ideals with the
following topology: for a collection
\(C \subseteq \operatorname{Spec}K\), define
\(Z(C) = \left\{{p\in \operatorname{Spc}K {~\mathrel{\Big\vert}~}C \cap p = \emptyset}\right\}\).
Note that there is a universal categorical construction of
\(\operatorname{Spc}K\) which we won't discuss here.
\end{remark}
\begin{remark}
TTC philosophy: let \(K\) be a compactly generated TTC with a generating
set \(K^c\). Note that \(K\) can include ``infinitely generated''
objects, while \(K^c\) should thought of as ``finite-dimensional''
objects. Problems:
\begin{itemize}
\tightlist
\item
What is the homeomorphism type of \(\operatorname{Spc}K^c\)?
\item
What are the thick ideals in \(K^c\)?
\end{itemize}
Although not all objects can be classified, there is a classification of
thick tensor ideals. Idea: use the algebraic topology philosophy of
passing to infinitely generated objects to simplify classification.
\end{remark}
\begin{remark}
We'll need a candidate space
\(X\cong_{\mathsf{Top}}\operatorname{Spc}(K^c)\), e.g.~a Zariski space:
Noetherian, and every irreducible contains a generic point. We'll also
need an assignment \(V: K^c\leadsto X_{{ \operatorname{cl}} }\) (the
closed sets of \(X\)) satisfying certain properties, which is called a
\emph{support datum}. For \(I\) a thick tensor ideal, define
\begin{align*}
\Gamma(I) \coloneqq\displaystyle\bigcup_{M\in I} V(M) \in X_{\mathrm{sp}}
,\end{align*}
a union of closed sets which is called \emph{specialization closed}.
Conversely, for \(W\) a specialized closed set, define a thick tensor
ideal
\begin{align*}
\Theta(W) \coloneqq\left\{{M\in K^c {~\mathrel{\Big\vert}~}V(M) \subseteq W}\right\}
.\end{align*}
One can check that a tensor product property holds: if \(M\in K^c\) and
\(N\in \Theta(W)\), check
\(V(M\otimes N) = V(M) \cap V(N) \subseteq W\). Under suitable
conditions, a deep result is that
\(\Gamma \circ \Theta = \operatorname{id}\) and
\(\Theta \circ \Gamma = \operatorname{id}\). This yields a bijection
\begin{align*}
\left\{{\substack{
\text{Thick tensor ideals of } K^c
}}\right\}
&\rightleftharpoons
\left\{{\substack{
\text{Specialization closed sets of } X
}}\right\} \\
I &\mapsto \Gamma(I) \\
\Theta(W) &\mapsfrom W
\end{align*}
\end{remark}
\begin{remark}
Define
\begin{align*}
f: X\to \operatorname{Spc}K^c \\
x &\mapsto P_x \coloneqq\left\{{M \in K^c {~\mathrel{\Big\vert}~}x\not\int V(M)}\right\}
.\end{align*}
This is a prime ideal: if \(M\otimes N\in P_x\), then
\(x\not \in V(M\otimes N) = V(M) \cap V(N)\), so \(M\in P_x\) or
\(N\in P_x\).
\end{remark}
\hypertarget{zariski-spaces}{%
\subsection{Zariski spaces}\label{zariski-spaces}}
\begin{definition}[Zariski spaces]
A space \(X\in {\mathsf{Top}}\) is a \textbf{Zariski space} iff
\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
\(X\) is a Noetherian space, and
\item
Every irreducible closed set has a unique generic point.
\end{enumerate}
Note that since \(X\) is Noetherian, it admits a decomposition into
irreducible components
\(X = \displaystyle\bigcup_{1\leq i \leq t} W_i\).
\end{definition}
\begin{example}[?]
The basic examples:
\begin{itemize}
\tightlist
\item
For \(R\) a unital Noetherian commutative ring,
\(X = \operatorname{Spec}R\) is Zariski.
\item
For \(R\) a graded unital Noetherian ring, taking homogeneous prime
ideals \(\mathop{\mathrm{Proj}}R\).
\item
For \(G\in {\mathsf{Aff}}{\mathsf{Alg}}{\mathsf{Grp}}\) with
\(G\curvearrowright R\) a graded ring by automorphisms (permuting the
graded pieces), the stack \(X \coloneqq\mathop{\mathrm{Proj}}_G(R)\)
(which is not Proj of the fixed points) is the set of
\(G{\hbox{-}}\)invariant homogeneous prime ideals. There's a map
\(\rho: \mathop{\mathrm{Proj}}R\to \mathop{\mathrm{Proj}}_G R\) where
\(P\mapsto \cap_{g\in G} gP\) which gives
\(\mathop{\mathrm{Proj}}_G R\) the quotient topology:
\(W\in \mathop{\mathrm{Proj}}_G R\) is closed iff \(\rho\in R\) is
close din \(\mathop{\mathrm{Proj}}R\). This topologizes orbit
closures.
\end{itemize}
\end{example}
\begin{remark}
Notation:
\begin{itemize}
\tightlist
\item
\({\mathcal{X}}= 2^X\) for the powerset of \(X\),
\item
\({\mathcal{X}}_{{ \operatorname{cl}} }\) the closed sets,
\item
\({\mathcal{X}}_{{\mathrm{irr}}}\) the irreducible closed sets,
\item
\({\mathcal{X}}_{\mathrm{sp} }\) the specialization-closed sets.
\end{itemize}
\end{remark}
\hypertarget{support-data}{%
\subsection{Support data}\label{support-data}}
\begin{remark}
Recall
\begin{itemize}
\tightlist
\item
\(M = {\mathsf{kG}{\hbox{-}}\mathsf{Mod}}\)
\item
\(R = H^{\text{even}}(G; k)\)
\item
\(V_G(M) = \left\{{p\in \mathop{\mathrm{Proj}}R {~\mathrel{\Big\vert}~}\operatorname{Ext} _{kG}(M, M)_p\neq 0 }\right\}\).
\end{itemize}
Note that \(V_G(P) = \emptyset\) for any projective and
\(V_G(k) = \emptyset\). In general, we'll similarly want
\(V_G(0) = \emptyset\) and \(V_G(1) = X\).
\end{remark}
\begin{definition}[Support data]
A \textbf{support datum} is an assignment \(V: K \to {\mathcal{X}}\)
such that
\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
\(V(0) = \emptyset\) and \(V(1) = X\).
\item
\(V\qty{\bigoplus _{i\in I} M_i = \displaystyle\bigcup_{i\in I} V(M_i) }\)
\item
\(V(\Sigma M) = V(M)\) (similar to \(\Omega\))
\item
For any distinguished triangle
\(M\to N\to Q\to \Sigma M, V(N) \subseteq V(M) \cup V(Q)\).
\item
\(V(M\otimes N) = V(M) \cap V(N)\).
\end{enumerate}
We'll need two more properties for the Balmer classification:
\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\setcounter{enumi}{5}
\tightlist
\item
Faithfulness: \(V(M) = \emptyset \iff M \cong 0\).
\item
Realization: for any \(W\in {\mathcal{X}}_{{ \operatorname{cl}} }\)
there exists a compact \(M\in K^c\) with \(V(M) = W\).
\end{enumerate}
\end{definition}
\begin{remark}
Note that (6) holds for group cohomology, and (7) is Carlson's
realization theorem.
\end{remark}
\begin{lemma}[?]
Let \(K\) be a TTC which is closed under set-indexed coproducts and let
\(V:K\to {\mathcal{X}}\) be a support datum. Let \(C\) be a collection
of objects in \(K\) and suppose \(W \subseteq X\) with
\(V(M) \subseteq W\) for all \(M\in C\). Then \(V(M) \subseteq W\) for
all \(M\) in \(\mathsf{Loc}(C)\).
\end{lemma}
\begin{proof}[?]
Note that \(\mathsf{Loc}(C)\) is closed under
\begin{itemize}
\tightlist
\item
Applying \(\Sigma\) or \(\Sigma^{-1}\),
\item
2-out-of-3: if two objects in a distinguished triangle are in
\(\mathsf{Loc}(C)\), the third is in \(\mathsf{Loc}(C)\),
\item
Taking direct summands,
\item
Taking set-indexed coproducts.
\end{itemize}
These follow directly from the properties of support data and properties
of \(\mathsf{Loc}(C)\).
\end{proof}
\hypertarget{extension-of-support-data}{%
\subsection{Extension of support data}\label{extension-of-support-data}}
\begin{remark}
Let \(X\) be a Zariski space and let \(K\supseteq K^c\) be a compactly
generated TTC. Let \(V: K^c\to {\mathcal{X}}_{{ \operatorname{cl}} }\)
be a support data on compact objects, we then seek an \emph{extension}:
a support datum \({\mathcal{V}}\) on \(K\) forming a commutative
diagram:
\begin{center}
\begin{tikzcd}
K && {\mathcal{X}}\\
\\
{K^c} && {{\mathcal{X}}_{{ \operatorname{cl}} }}
\arrow[hook, from=3-1, to=1-1]
\arrow[hook, from=3-3, to=1-3]
\arrow["V", from=3-1, to=3-3]
\arrow["{\mathcal{V}}", from=1-1, to=1-3]
\end{tikzcd}
\end{center}
\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwyLCJLXmMiXSxbMCwwLCJLIl0sWzIsMCwiXFxtY3giXSxbMiwyLCJcXG1jeF97XFxjbH0iXSxbMCwxLCIiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFszLDIsIiIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzAsMywiViJdLFsxLDIsIlxcbWN2Il1d}{Link
to Diagram}
\end{quote}
\end{remark}
\begin{definition}[?]
Let \(K\) be a compactly generated TTC and
\(V: K^c\to {\mathcal{X}}_{{ \operatorname{cl}} }\) be a support datum.
Then \({\mathcal{V}}: K\to {\mathcal{X}}\) \textbf{extends} \(V\) iff
\begin{itemize}
\tightlist
\item
\({\mathcal{V}}\) satisfies properties (1) -- (5) above,
\item
\(V(M) = {\mathcal{V}}(M)\)for any \(M\in K^c\).
\item
If \(V\) is faithful then \({\mathcal{V}}\) is faithful.
\end{itemize}
\end{definition}
\begin{remark}
We'll need Hopkins' theorem to analyze such extensions.
\end{remark}
\hypertarget{tuesday-april-19}{%
\section{Tuesday, April 19}\label{tuesday-april-19}}
\hypertarget{hopkins-theorem}{%
\subsection{Hopkins' Theorem}\label{hopkins-theorem}}
\begin{remark}
Let \(\mathsf{K}\) be a compactly generated tensor triangulated category
with \(\mathsf{K}^c\) a subcategory of compact objects. Goal: classify
\(\operatorname{Spc}\mathsf{K}^c\). A candidate for its homeomorphism
type: we'll build a Zariski space \(X\) and a homeomorphism
\(\operatorname{Spc}\mathsf{K}^c \to X\). We'll use support data
\(\mathbf{V}: \mathsf{K}^c\to {\mathcal{X}}_{{ \operatorname{cl}} }\)
which satisfies the faithfulness and realization properties. We'll
extend this to \(\mathcal{V}: \mathsf{K} \to {\mathcal{X}}\). So we need
\begin{itemize}
\tightlist
\item
A Zariski space \(X\),
\item
Support data \(\mathbf V\),
\item
An extension \({\mathcal{V}}\).
\end{itemize}
\end{remark}
\hypertarget{localization-functors}{%
\subsection{Localization functors}\label{localization-functors}}
\begin{remark}
Let \(\mathsf{C} \leq \mathsf{K}\) be a thick subcategory for
\(\mathsf{K}\in {\mathsf{triang}}\mathsf{Cat}\). A mysterious sequence:
\begin{align*}
\Gamma_c(M) \to M \to L_c(M)
.\end{align*}
Suppose \(W\in {\mathcal{X}}_{{\mathrm{irr}}}\) is nonempty and let
\(Z = \left\{{x\in X{~\mathrel{\Big\vert}~}w\not\subseteq { \operatorname{cl}} _X\left\{{x}\right\}}\right\}\).
Define a functor \(\nabla_W = \Gamma_{I_W} L_{I_Z}\) and
\({\mathcal{V}}(M) \coloneqq\left\{{x\in X {~\mathrel{\Big\vert}~}\nabla_{\left\{{x}\right\}} (M) = 0}\right\}\).
\end{remark}
\begin{theorem}[Hopkins-Neeman]
Let \(\mathsf{K}\) be a compactly generated tensor triangulated
category, \(X\) a Zariski space, and
\({\mathcal{X}}_{{ \operatorname{cl}} }\) the closed sets. Given a
compact object \(M\in \mathsf{K}^c\), let
\(\left\langle{M}\right\rangle_{\mathsf{K} ^c}\) be the thick tensor
ideal in \(\mathsf{K}^c\) generated by \(M\). Let
\(\mathbf V: \mathsf{K}^c\to {\mathcal{X}}_{{ \operatorname{cl}} }\) be
support data satisfying the faithfulness condition and suppose
\({\mathcal{V}}: \mathsf{K}\to {\mathcal{X}}\) is an extension. Set
\(W = \mathbf V(M)\) and
\(I_W = \left\{{N\in \mathsf{K}^c {~\mathrel{\Big\vert}~}V(N) \subseteq W}\right\}\).
Then
\begin{align*}
I_W = \left\langle{M}\right\rangle_{\mathsf{K}^c}
,\end{align*}
i.e.~this is generated by a single object.
\end{theorem}
\begin{proof}[?]
Let \(I \coloneqq I_W\) and
\(I' \coloneqq\left\langle{M}\right\rangle_{\mathsf{K}^c}\).
\(I' \subseteq I\): If \(N\in I'\), then \(N\) is obtained by taking
direct sums, direct summands, distinguished triangles, shifts, etc.
These all preserve support containment, so \(\mathbf V(N) \subseteq W\)
and \(N\in I = I_W\).
\(I \subseteq I'\): Let \(N\in {\mathsf{K^c}}\). Apply the functorial
triangle \(\Gamma_{I'} \to \operatorname{id}\to L_{I'}\) to
\(\Gamma_I(N)\) to obtain
\begin{align*}
\Gamma_{I'} \Gamma_I N\to \Gamma_I(N) \to L_{I'} \Gamma_I N
.\end{align*}
From above, \(I' \subseteq I\) so the first term is in
\(\mathsf{Loc}(I)\). Since the second term is as well, the 2-out-of-3
property guarantees that the third term satisfies
\(L_{I'} \Gamma_I N \in\mathsf{Loc}(I)\). By the lemma,
\(V(L_{I'} \Gamma_I N) \subseteq W\). There are no nonzero maps
\(I' \to VL_{I'}\Gamma_I N\), therefore for \(S\in {\mathsf{K^c}}\),
noting that \(S\otimes M \in I'\),
\begin{align*}
0
= \mathop{\mathrm{Hom}}_{\mathsf{K}}(S\otimes M, L_{I'} \Gamma_I M)
= \mathop{\mathrm{Hom}}_{\mathsf{K}}(S, M {}^{ \vee }\otimes L_{I'} \Gamma_I N)
,\end{align*}
and since \(S\) is an arbitrary compact object, this forces
\(M {}^{ \vee }\otimes L_{I'} \Gamma_I N = 0\). By faithfulness, and the
tensor product property,
\begin{align*}
\emptyset
&= {\mathcal{V}}(M {}^{ \vee }\otimes L_{I'} \Gamma_I N)\\
&= {\mathcal{V}}(M {}^{ \vee }) \cap{\mathcal{V}}(L_{I'}\Gamma_I N)\\
&= \mathbf{V}(M) \cap{\mathcal{V}}(L_{I'} \Gamma_I N)\\
&= W \cap{\mathcal{V}}(L_{I'} \Gamma_I N)\\
&= {\mathcal{V}}(L_{I'} \Gamma_I N)
,\end{align*}
so by faithfulness (again) \(L_{I'} \Gamma_I N = 0\). Thus by the
localization triangle, \(\Gamma_{I'} \Gamma_I N \cong \Gamma_I N\).
Now specialize to \(N\in I\); the localization triangle yields
\begin{align*}
\Gamma_I N \to N \xrightarrow{0} L_I(N) \implies \Gamma_I N \cong N
.\end{align*}
Now replacing \(I\) with \(I'\) yields \(\Gamma_{I'} N \cong N\) since
\(L_{I'} N \cong L_{I'} \Gamma_I N \cong 0\) by the previous part. Thus
\(N\in \mathsf{Loc}(I')\) by applying a result of Neeman, implying
\(N\in I'\) and \(I \subseteq I'\).
\end{proof}
\begin{remark}
Many different takes on classification of thick tensor ideals:
\begin{itemize}
\tightlist
\item
Benson, Carlson, Rickard at UGA in the late 90s, for finite group
representations (now extended).
\item
Benson, Iyengar, Krause: axiomatic approach and description of
supports.
\item
Dell'Ambrogio
\item
Boe, Kujawa, Nakano
\end{itemize}
\end{remark}
\begin{theorem}[?]
Let
\begin{itemize}
\tightlist
\item
\(\mathsf{K}\) be a compactly generated tensor triangulated category,
\item
\(X\) be a Zariski space,
\item
\(\mathbf{V}: {\mathsf{K^c}}\to {\mathcal{X}}_{{ \operatorname{cl}} }\)
be a support datum satisfying both the faithfulness \emph{and}
realization properties,
\item
\({\mathcal{V}}: \mathsf{K}\to C\) be an extension of \(\mathbf{V}\).
\end{itemize}
Let \(\operatorname{Id}({\mathsf{K^c}})\) be the set of thick tensor
ideals in \({\mathsf{K^c}}\), then there is a bijection
\begin{align*}
\operatorname{Id}({\mathsf{K^c}})
&\rightleftharpoons
{\mathcal{X}}_{\mathrm{sp} } \\
I &\mapsto \Gamma(I) \coloneqq\displaystyle\bigcup_{M\in I} \mathbf{V}(I) \\
\Theta(W) = I_W \coloneqq\left\{{N\in {\mathsf{K^c}}{~\mathrel{\Big\vert}~}\mathbf V(N) \subseteq W}\right\} &\mapsfrom W
.\end{align*}
\end{theorem}
\begin{exercise}[?]
Show that \(I_W\in \operatorname{Id}({\mathsf{K^c}})\) is in fact a
thick tensor ideal.
\end{exercise}
\begin{proof}[?]
\(\Gamma \circ \Theta = \operatorname{id}\): Check that
\begin{align*}
\Gamma\Theta W = \Gamma(I_W) = \displaystyle\bigcup_{M\in I_W} \mathbf{V}(M) \subseteq W
.\end{align*}
For the reverse inclusion, let \(W = \displaystyle\bigcup_{j\in W} W_j\)
where \(W_j\in {\mathcal{X}}_{{ \operatorname{cl}} }\). By the
realization property, there exist \(N_j \in {\mathsf{K^c}}\) such that
\(\mathbf{V}(N_j) = W_j\), so \(N_j\in I_W\). Now
\(W \subseteq \displaystyle\bigcup_{M\in I_W} \mathbf{V}(M)\), so
\(W = \displaystyle\bigcup_{M\in I_W} \mathbf{V}(M)\).
\begin{center}\rule{0.5\linewidth}{0.5pt}\end{center}
\(\Theta \circ \Gamma = \operatorname{id}\): For
\(I\in \operatorname{Id}({\mathsf{K^c}})\), set
\(W \coloneqq\Gamma(I) = \displaystyle\bigcup_{M\in I} \mathbf{V}(M)\),
then
\begin{align*}
\Theta\Gamma I = \Theta(W) = I_W \supseteq I
.\end{align*}
For the reverse inclusion \(I_W \subseteq I\): let \(N\in I_W\). Since
\(X\) is a Zariski space, \(X\) is Noetherian and there is an
irreducible component decomposition
\(V(N) = \displaystyle\bigcup_i W_i\) with each \(W_i\) irreducible with
a unique generic point, so
\(W_i = { \operatorname{cl}} _{W_i} \left\{{x_i}\right\}\). Since each
\(W_i \subseteq W\), each
\(x_i\in W = \displaystyle\bigcup\mathbf{V}(M)\), so there exist \(M_i\)
with \(x_i \in \mathbf{V}(M_i)\). Since supports are closed,
\(W_i = { \operatorname{cl}} _{W_i}\left\{{x_i}\right\} \subseteq \mathbf{V}(M_i)\).
Setting \(M\coloneqq\bigoplus _i M_i\in I\) yields
\(V(N) \subseteq \displaystyle\bigcup V(M_i) = V(M) \subseteq W\).
\begin{claim}
\begin{align*}
N \in \left\langle{M}\right\rangle_{{\mathsf{K^c}}}
.\end{align*}
\end{claim}
Proving the claim will complete the proof, since \(I\) is a thick ideal
containing \(M\), so
\(\left\langle{M}\right\rangle_{{\mathsf{K^c}}} \subseteq I\) and
\(N\in I\).
\begin{proof}[of claim]
By Hopkins' theorem,
\(\left\langle{M}\right\rangle_{{\mathsf{K^c}}} = I_Z\) where
\(Z = \mathbf{V}(M)\). Since \(V(N) \subseteq V(M) = Z\), we have
\(N\in I_Z = \left\langle{M}\right\rangle_{{\mathsf{K^c}}}\).
\end{proof}
\end{proof}
\begin{remark}
Next time:
\begin{itemize}
\tightlist
\item
Showing
\(\operatorname{Spc}{\mathsf{K^c}}\underset{{\mathsf{Top}}}{\cong} X\)
\item
Examples: \({\mathsf{kG}{\hbox{-}}\mathsf{stMod}}\),
\({\mathsf{u({\mathfrak{g}})}{\hbox{-}}\mathsf{stMod}}\), and
\({\mathbb{D}}{\mathsf{R}{\hbox{-}}\mathsf{Mod}}\).
\end{itemize}
\end{remark}
\hypertarget{thursday-april-21}{%
\section{Thursday, April 21}\label{thursday-april-21}}
\hypertarget{classification-theorem}{%
\subsection{Classification theorem}\label{classification-theorem}}
\begin{theorem}[?]
Let \(\mathsf{K}\) be a compactly generated tensor-triangulated category
and let \(X\) be a Zariski space. Suppose that
\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
\(\mathbf{V}: \mathsf{K}^c\to {\mathcal{X}}_{{ \operatorname{cl}} }\)
is a support datum,
\item
\(\mathbf{V}\) satisfies the faithfulness property,
\item
\({\mathcal{V}}: \mathsf{K}\to {\mathcal{X}}\) extends \(\mathbf{V}\).
\end{enumerate}
Then there exists a bijective correspondence
\begin{align*}
\adjunction{\Gamma}{\Theta}{\operatorname{Id}(\mathsf{K}^c) }{{\mathcal{X}}_{\mathrm{sp}} }
\end{align*}
where \(\Gamma(I) \coloneqq\displaystyle\bigcup_{M\in I} \mathbf{V}(M)\)
and
\(\Theta(W) \coloneqq\left\{{N\in\mathsf{K}^c{~\mathrel{\Big\vert}~}\mathbf{V}(N) \subseteq W}\right\}\).
\end{theorem}
\begin{remark}
This relies on Hopkins' theorem.
\end{remark}
\hypertarget{balmer-spectrum}{%
\subsection{Balmer spectrum}\label{balmer-spectrum}}
\begin{theorem}[?]
Let \(\mathsf{K}\) and \(X\) be as in the previous theorem, satisfying
the same assumptions. Then there exists a homeomorphism
\(f: X\to \operatorname{Spc}\mathsf{K}^c\).
\end{theorem}
\begin{proof}[?]
Since
\(\mathbf{V}: \mathsf{K}^c\to {\mathcal{X}}_{{ \operatorname{cl}} }\) is
a support datum, Balmer shows there exists a continuous map
\begin{align*}
f: X &\to \operatorname{Spc}\mathsf{K}^c \\
x &\mapsto P_x \coloneqq\left\{{M{~\mathrel{\Big\vert}~}x\not\in\mathbf{V}(M) }\right\}
.\end{align*}
Note that \(P_x\) is a prime ideal:
\begin{align*}
M\otimes N\in P_x
&\implies x\not\in\mathbf{V}(M\otimes N) \\
&\implies x\not\in\mathbf{V}(M) \cap\mathbf{V}(N) \\
&\implies x\not\in \mathbf{V}(M) \text{ or } x\not\in \mathbf{V}(N) \\
&\implies M\in P_x \text{ or } N\in P_x
.\end{align*}
Applying the classification theorem, this yields a bijection.
\end{proof}
\begin{remark}
Examples of classification:
For
\(G\in{\mathsf{Fin}}{\mathsf{Grp}}, \operatorname{ch}k = p\divides {\sharp}G\),
take \(\mathsf{K} = {\mathsf{kG}{\hbox{-}}\mathsf{stMod}}\),
\(R = H^{\mathrm{even}}(G; k)\), and
\(X = \mathop{\mathrm{Proj}}R = \mathop{\mathrm{Proj}}(\operatorname{Spec}R)\).
Checking that this satisfies the 4 properties in the theorem:
\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
For \(M\in \mathsf{K}^c\), we take
\(\mathbf{V}(M) = \left\{{p\in X{~\mathrel{\Big\vert}~} { {\operatorname{Ext} }^{\scriptscriptstyle \bullet}} _{kG}(M, M) \left[ { \scriptstyle { {p}^{-1}} } \right] \neq 0 }\right\}\).
This yields a support datum.
\item
The tensor product property holds because
\(\mathbf{V}_E(M) = \mathbf{V}_E^r(M)\) (the rank variety), and we
showed that \(\mathbf{V}\) satisfies faithfulness and (Carlson)
realization properties.
\item
We can use localization functors to define
\({\mathcal{V}}: \mathsf{K}\to {\mathcal{X}}\) which satisfies the
same support data properties. For this to be an extension, one should
check that
\end{enumerate}
\begin{itemize}
\tightlist
\item
\(\mathbf{V}(M) = {\mathcal{V}}(M)\) for every compact
\(M\in \mathsf{K}^c\).
\item
\(\mathbf{V}(M\otimes N) = {\mathcal{V}}(M) \cap{\mathcal{V}}(N)\) for
all \(M,N\in \mathsf{K}\)
\item
If \({\mathcal{V}}(M)\) is empty then \(M = 0\).
\end{itemize}
\end{remark}
\begin{remark}
To prove these properties, Benson-Carlson-Rickard start with \(E\)
elementary abelian, so
\(E = \left\langle{x_1,\cdots, x_n}\right\rangle \cong C_p{ {}^{ \scriptscriptstyle\times^{n} } }\)
with \(o(x_i) = p\) for all \(i\). Set \(y_i = x_i-1 \in kE\), so
\(y_i^p=0\), and define cyclic subgroups
\(\mathbf{\alpha }= {\left[ {\alpha_1,\cdots, \alpha_n} \right]} \in L^n\)
where \(L/k\) is a field of large transcendence degree. Define
\(y_{\mathbf{\alpha}} \coloneqq\sum_{1\leq i\leq n} \alpha_i y_i\) and
define a rank variety
\begin{align*}
{\mathcal{V}}_E^r(M) = \left\{{ \mathbf{\alpha }\in L^n {~\mathrel{\Big\vert}~}L\otimes_k M \downarrow_{\left\langle{y_{\mathbf{\alpha}}}\right\rangle} \text{ is not free } }\right\}\cup\left\{{0}\right\}
.\end{align*}
\end{remark}
\begin{theorem}[?]
Let \(E\) be as above and suppose \(\operatorname{trdeg}(L/k) \geq n\).
Then if \(M\in \mathsf{K}\),
\({\mathcal{V}}_E(M) \cong {\mathcal{V}}_E^r(M)\), and the three
properties for (3) above hold for \(E\).
\end{theorem}
\begin{theorem}[?]
Let \(A = kG\) for \(G\) a finite group scheme, and let
\(R = H^{\mathrm{even}}(G; k)\) and \(X = \mathop{\mathrm{Proj}}(R)\).
Then
\begin{itemize}
\item
There is a bijective correspondence
\begin{align*}
\adjunction{\Gamma}{\Theta}{{\mathsf{kG}{\hbox{-}}\mathsf{stMod}}}{{\mathcal{X}}_{\mathrm{sp}}}
.\end{align*}
\item
\(\operatorname{Spc}({\mathsf{kG}{\hbox{-}}\mathsf{stMod}}) \underset{{\mathsf{Top}}}{\cong} X\).
\end{itemize}
\end{theorem}
\begin{remark}
Some remarks:
\begin{itemize}
\item
This theorem is an indication of why cohomology is central in
understanding the tensor structure of representation categories. If
\(G\in {\mathsf{Fin}}{\mathsf{Grp}}{\mathsf{Sch}}_{/ {k}}\) then the
coordinate ring \(k[G]\) is a commutative Hopf algebra, so
\(A = kG = k[G] {}^{ \vee }\) is a finite dimensional cocommutative
Hopf algebra. So there is an equivalence of categories between
\({\mathsf{Rep}}G\) and \({\mathsf{Rep}}A\) for \(A\) such a Hopf
algebra. By a result of Friedlander-Suslin, \(R\) is finitely
generated.
\item
The realization of \(\mathbf{V}\) and \({\mathcal{V}}\) for a general
group scheme involve so-called \emph{\(\pi{\hbox{-}}\)points}
developed be Friedlander-Pevtsovz and the construction of explicit
rank varieties.
\end{itemize}
\end{remark}
\begin{remark}
A special case: let \({\mathfrak{g}}= \mathsf{Lie}G\) for
\(G\in{\mathsf{Alg}}{\mathsf{Grp}}_{/ {k}}\) reductive and \(k\)
positive characteristic. Let \(A = u({\mathfrak{g}})\), which is a
finite-dimensional cocommutative Hopf algebra. If \(p > h\) for \(h\)
the Coxeter number,
\begin{align*}
{\mathcal{N}}_p = \left\{{x\in {\mathfrak{g}}{~\mathrel{\Big\vert}~}x^{[p]} = 0 }\right\} = {\mathcal{N}}, \text{ the nilpotent cone}
,\end{align*}
\(R = H^{\mathrm{even}}(u({\mathfrak{g}}); k) = k[{\mathcal{N}}]\), and
\(X = \mathop{\mathrm{Proj}}(k[{\mathcal{N}}])\), then applying the
theorem,
\begin{itemize}
\item
There is a correspondence
\begin{align*}
\adjunction{}{}{{\mathsf{u({\mathfrak{g}})}{\hbox{-}}\mathsf{stMod}}}{{\mathcal{X}}_{\mathrm{sp}}}
.\end{align*}
\item
There is a homeomorphism
\begin{align*}
\operatorname{Spc}\qty{ {\mathsf{u({\mathfrak{g}})}{\hbox{-}}\mathsf{stMod}} } \underset{{\mathsf{Top}}}{\cong} \mathop{\mathrm{Proj}}(k[{\mathcal{N}}])
.\end{align*}
\end{itemize}
\end{remark}
\begin{theorem}[Arkhipov-Bezrukavikov-Ginzburg]
Let \(\tilde {\mathcal{N}}\to {\mathcal{N}}\) be the Springer
resolution. There is an equivalence of derived categories
\begin{align*}
{\mathbb{D}}^b {\mathsf{ u_\zeta({\mathfrak{g}})_0}{\hbox{-}}\mathsf{Mod}} { \, \xrightarrow{\sim}\, }{\mathbb{D}}^b {\mathsf{Coh}}^{G\times {\mathbb{C}}^{\times}} k[\tilde{\mathcal{N}}] { \, \xrightarrow{\sim}\, }{\mathbb{D}}^b \mathsf{Perv}({\Omega}{\operatorname{Gr}})
.\end{align*}
where \(\mathsf{Perv}({-})\) is the category of perverse sheaves and
\({\Omega}{\operatorname{Gr}}\) is the loop Grassmannian.
\end{theorem}
\begin{remark}
For \(M\) a \(u_\zeta({\mathfrak{g}}){\hbox{-}}\)module and
\(R = H^{\mathrm{even}}(u_\zeta({\mathfrak{g}}); M) = {\mathbb{C}}[{\mathcal{N}}] \cong {\mathbb{C}}[\tilde {\mathcal{N}}]\).
There is an action of \(R\) on
\({ {H}^{\scriptscriptstyle \bullet}} (u_\zeta({\mathfrak{g}}); M)\).
Next time: examples for Lie superalgebras and Thomason's reconstruction
theorem for rings.
\end{remark}
\hypertarget{tuesday-april-26}{%
\section{Tuesday, April 26}\label{tuesday-april-26}}
\begin{quote}
See Boe-Kujawa-Nakano, Adv. Math 2017.
\end{quote}
\begin{remark}
Setup: \(\mathsf{K}^c \leq \mathsf{K}\in {\mathsf{TTC}}\), \(X\) a
Zariski space,
\(V:\mathsf{K}^c\to {\mathcal{X}}_{{ \operatorname{cl}} }\) with an
extension \({\mathcal{V}}:\mathsf{K}\to {\mathcal{X}}\). Let
\({\mathfrak{g}}= {\mathfrak{g}}_{0} \oplus {\mathfrak{g}}_1\) be a Lie
superalgebra with a \(C_2\) grading over \(k= {\mathbb{C}}\) where
\({\mathfrak{g}}_0\curvearrowright{\mathfrak{g}}_1\),
e.g.~\({\mathfrak{gl}}_{m, n} = {\mathfrak{gl}}_m \times {\mathfrak{gl}}_n\)
with matrices
\({ \begin{bmatrix} {{\mathfrak{g}}_0} & {{\mathfrak{g}}_1} \\ {{\mathfrak{g}}_1} & {{\mathfrak{g}}_0} \end{bmatrix} }\)
with the bracket action. Write \(\mathsf{Lie}G_0 = {\mathfrak{g}}_0\),
and note that \(G_0\) is reductive. Let
\({\mathcal{F}}({\mathfrak{g}}, {\mathfrak{g}}_0)\) be the category of
finite-dimensional \({\mathfrak{g}}{\hbox{-}}\)supermodules which are
completely reducible over \({\mathfrak{g}}_0\). Take
\(\mathsf{K}^c = {\mathsf{{\mathcal{F}}({\mathfrak{g}},{\mathfrak{g}}_0)}{\hbox{-}}\mathsf{stMod}} \leq \mathsf{K} = {\mathsf{C({\mathfrak{g}}, {\mathfrak{g}}_0)}{\hbox{-}}\mathsf{stMod}}\),
where for \(C\) we drop the finite-dimensional condition.
Set
\(R = H^0({\mathfrak{g}}_1, {\mathfrak{g}}_0; {\mathbb{C}}) = \operatorname{Ext} ({\mathbb{C}},{\mathbb{C}}) \cong S({\mathfrak{g}}_0 {}^{ \vee })^{G_0}\).
By a theorem of Hilbert, \(\operatorname{Ext} (M, M)\) is finitely
generated over \(R\). Write
\(V_{{\mathfrak{g}},{\mathfrak{g}}_0}(M) = \mathrm{sp}ec R/J_M\) -- for
Kac modules
\(K(\lambda) = U({\mathfrak{g}}) \otimes_{U(P^0)} L_0(\lambda)\),
\(V = 0\) but not every \(K(\lambda)\) is projective.
\end{remark}
\begin{remark}
Idea: use detecting subalgebras. For
\({\mathfrak{g}}= {\mathfrak{gl}}_{n,n}\), let \(f_1\) be the ``torus'':
\includegraphics{figures/2022-04-26_09-58-21.png}
Then define \(f_0 = [f_1, f_1]\).
\end{remark}
\begin{remark}
Let \(X = N\mathop{\mathrm{Proj}}(S^*(f_1 {}^{ \vee }))\) where
\(S^*(f_1 {}^{ \vee }) \cong \operatorname{Ext} _{f_1, f_0}({\mathbb{C}}, {\mathbb{C}}) = R'\)
and \(N = { N }_{G_0}(f_1)\), which is a reductive algebraic group.
Define a support datum by
\(\mathbf{V}(M) = \left\{{p\in X{~\mathrel{\Big\vert}~}\operatorname{Ext} _{f, f_0}(M,M)_p = 0}\right\}\).
The goal is to construct \({\mathcal{V}}: K\to {\mathcal{X}}\) using
localization functors -- one needs to show the tensor product formula,
and the faithfulness and realization properties, which follows from
Dede's lemma. It turns out that
\(f_1\cong {\mathfrak{sl}}(1,1){ {}^{ \scriptscriptstyle\times^{m} } }\)
and it suffices to define the rank variety on \(f_1\). Define
\begin{align*}
V_{f_1}^{\operatorname{rank}}(M) = \left\{{\mkern 1.5mu\overline{\mkern-1.5mux\mkern-1.5mu}\mkern 1.5mu = \tilde K\otimes_{\mathbb{Q}}f_1 {~\mathrel{\Big\vert}~}K\otimes_{\mathbb{C}}M\downarrow{\left\langle{\mkern 1.5mu\overline{\mkern-1.5mux\mkern-1.5mu}\mkern 1.5mu}\right\rangle} \text{ is not projective} }\right\}
\end{align*}
where \(\tilde K\supseteq{\mathbb{C}}\) is an extension with
\(\operatorname{trdeg}_{\mathbb{C}}\tilde K \geq \dim f_1\). A theorem
shows \({\mathcal{V}}(M) = V_{f_1}^{\operatorname{rank}}(M)\) for
\(M\in K\). This yields a classification for \({\mathfrak{gl}}_{m, n}\)
of thick tensor ideals in \(K^c\) in terms of
\({\mathcal{X}}_{\mathrm{sp}}\).
\end{remark}
\begin{remark}
What is the classification of other Lie superalgebras? This is an open
problem.
\end{remark}
\hypertarget{noncommutative-theory}{%
\subsection{Noncommutative theory}\label{noncommutative-theory}}
\begin{remark}
How does one extend this theory to noncommutative TTCs? See
Nakano-Vashaw-Yakomov, to appear in Amer J. Math.
\end{remark}
\begin{remark}
Let \(K\) be a compactly generated monoidal triangulated category, not
necessarily symmetric. One approaches this via noncommutative ring
theory, where e.g.~even the definition of prime ideals differs. We'll
only consider 2-sided ideals.
\end{remark}
\begin{definition}[(Noncommutative) prime ideals]
A thick triangulated subcategory \(P\) is a \textbf{completely prime}
ideal iff \(M\otimes N\in P\implies M\in P\) or \(N\in P\). The ideal
\(P\) is \textbf{prime} iff
\(I\otimes J\subseteq P \implies I \subseteq P\) or \(J \subseteq P\),
where \(I,J\) are themselves ideals. Define \(\mathrm{sp}c K\) to be
prime ideals and \(\mathrm{CP}\operatorname{Spc}K\) to be completely
prime ideals.
\end{definition}
\begin{example}[?]
Let \(A\in \mathsf{Hopf}{\mathsf{Alg}_{/k} }^{{\mathrm{fd}}}\) where the
coproduct \(\Delta: A\to A{ {}^{ \scriptstyle\otimes_{k}^{2} } }\) is
not necessarily commutative, e.g.~in the setting of quantum groups. Some
remarks:
\begin{itemize}
\tightlist
\item
Note that \(M\otimes N \not\cong N\otimes M\) in general.
\item
Here \(\mathrm{sp}c K^c\) is not known, but there is a conjectural
answer.
\item
In general \(\mathrm{sp}c K^c\not\cong \mathop{\mathrm{Proj}}R\) for
\(R = H(A; k)\).
\item
\(R\) is not known to be finitely-generated. Etingof-Ostrik conjecture
this in the setting of finite tensor categories.
\item
The definition of prime ideals is due to Buan-Krause-Solberg in 2007.
\item
A weird example: there are nilpotents where \(M\neq 0\) (is not
projective) but \(M{ {}^{ \scriptstyle\otimes_{k}^{2} } } = 0\) (is
projective).
\item
Being a prime ideal \(P\) is equivalent to
\(A\otimes C\otimes B \in P\) for all \(C\) \(\implies A\in P\) or
\(B\in P\).
\end{itemize}
\end{example}
\begin{definition}[(Noncommutative) support data]
Let \(K\) be a monoidal triangulated category, \(X\) a Zariski space,
and \({\mathcal{X}}= 2^X\) the subsets of \(X\). A map
\(\sigma: K\to{\mathcal{X}}\) is a \textbf{weak support datum} iff
\begin{itemize}
\tightlist
\item
\(\sigma(0) = \emptyset\) and \(\sigma(\one) = X\)
\item
\(\sigma(A\otimes B) = \sigma(A) \cup\sigma(B)\)
\item
\(\sigma(\Sigma A) = \sigma(A)\)
\item
If \(A\to B\to C\) is exact then
\(\sigma(A) \subseteq \sigma(B) \cup\sigma(C)\).
\end{itemize}
Set \(\Phi_\sigma(I) \coloneqq\displaystyle\bigcup_{M\in I} \sigma(I)\);
Then \(\sigma\) is a \textbf{support datum} if additionally
\begin{itemize}
\tightlist
\item
\(\displaystyle\bigcup_{C\in K} \sigma(A\otimes C\otimes B)= \sigma(A) \cap\sigma(B)\)
and
\item
\(\Phi_\sigma(I\otimes J) = \Phi_\sigma(I) \cap\Phi_\sigma(J)\).
\end{itemize}
\end{definition}
\begin{remark}
Next time:
\begin{itemize}
\tightlist
\item
Classification theorems
\item
The NVY conjecture for finite-dimensional Hopf algebras.
\item
Tensor product theorems.
\item
Examples of applications.
\end{itemize}
\end{remark}
\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
\newpage
\printbibliography[title=Bibliography]
\end{document}