Note that we can also take complexes with arrows in the other direction. For $F$ a functor, we can rewrite these examples as $$\begin{align*} d_i^* \circ d_{i-1}^* = F(d_i) \circ F(d_{i-1}) = F(d_i \circ d_{i-1}) = F(0) = 0 ,\end{align*}$$ provided $F$ is nice enough and sends zero to zero. This follows from the fact that functors preserve composition. Even if the original complex is exact, the new one may not be, so we can define the following:

These will lead to $i$th derived functors, and category theory will be useful here. See appendix in Weibel. For a category $\mathcal{C}$ we’ll define

• $\mathrm{Obj}(\mathcal{C} )$ as the objects
• $\mathop{\mathrm{Hom}}_{\mathcal{C}}(A, B)$ a set of morphisms between them, where a more modern notation might be $\mathrm{Mor}(A, B)$.
• Morphisms compose: $A \xrightarrow{f} B \xrightarrow{g} C$ means that $g\circ f \in \mathop{\mathrm{Hom}}_{\mathcal{C}}(A, C)$
• Associativity
• Identity morphisms

See the appendix for diagrams defining zero objects and the zero map, which we’ll need to make sense of exactness. We’ll also needs notions of kernels and images, or potentially cokernels instead of images since they’re closely related.

In the examples, we had $\ker d_i \subseteq M_i$, but this need not be true since the objects in the category may not be sets. Such an example is the category of complexes of $R{\hbox{-}}$modules: $\operatorname{Cx}({\mathsf{R}{\hbox{-}}\mathsf{Mod}})$. In this setting, kernels will be subcomplexes but not subsets.

Taking cohomology yields the $i$th derived functors of $F$, for example $\operatorname{Ext}^i, \operatorname{Tor}_i$. Recall that functors can be covariant or contravariant. See section 1 for formulating simplicial and singular homology (from topology) in this language.

We can thus define a category $\mathrm{Ch}({\mathsf{R}{\hbox{-}}\mathsf{Mod}})$ where

• The objects are chain complexes,
• The morphisms are chain maps.

Each $H_n$ thus becomes a functor $\mathrm{Ch}({\mathsf{R}{\hbox{-}}\mathsf{Mod}}) \to {\mathsf{R}{\hbox{-}}\mathsf{Mod}}$ where $H_n(u) \mathrel{\vcenter{:}}= u_{*. n}$.

Recall that a chain complex is $C_{-}$ where $d^2 = 0$, and a map of chain complex is a ladder of commuting squares

Link to diagram

Recall that $u_n: Z_n(C) \to Z_n(D)$ and $u_n: B_n(C) \to B_n(D)$ preserves these submodules, so there are induced maps $u_{{-}, n}: H_n(D) \to H_n(D)$ where $H_n(C) \mathrel{\vcenter{:}}= Z_n(C) / B_nn-1(C)$. Moreover, taking $H_n({-})$ is a functor from $\mathsf{Ch}({\mathsf{R}{\hbox{-}}\mathsf{Mod}}) \to {\mathsf{R}{\hbox{-}}\mathsf{Mod}}$ for any fixed $n$ and on objects $C\mapsto H_n(C)$ and chain maps $u_{n} \to H_n(u) \mathrel{\vcenter{:}}= u_{*, n}$. Note the lower indices denote maps going down in degree.

Note that the usual notion of an isomorphism in the categorical sense might be too strong here.

There is a way to go back and forth bw chain complexes and cochain complexes: set $C_n \mathrel{\vcenter{:}}= C^{-n}$ and $d_n \mathrel{\vcenter{:}}= d^{-n}$. This yields $$\begin{align*} C^{-n} \xrightarrow{d^{-n}} C^{-n+1} \iff C_n \xrightarrow{d^n} C_{n-1} ,\end{align*}$$ and the notions of $d^2 = 0$ coincide.

See the book for classical applications:

• 1.1.3: Simplicial homology
• 1.1.5: Singular homology

Write $\mathsf{Ch}$ for $\mathsf{Ch}({\mathsf{R}{\hbox{-}}\mathsf{Mod}})$, then if $f,g: C\to D$ are chain maps then $f+g:C\to D$ can be defined as $(f+g)(x) = f(x) + g(x)$, since $D$ has an addition coming from its $R{\hbox{-}}$module structure. Thus the hom sets $\mathop{\mathrm{Hom}}_{\mathsf{Ch}}(C, D)$ becomes an abelian group. There is a distinguished zero object¹ $0$, defined as the chain complex with all zero objects and all zero maps. Note that we also have a zero map given by the composition $(C \to 0) \circ (0\to D)$.

Note that if the index set is finite, these notions coincide. By convention, finite direct products are written as direct sums.

These structures make $\mathsf{Ch}$ into an additive category. See appendix for definition: the homs are abelian groups where composition distributes over addition, existence of a zero object, and existence of finite products. Note that here we have arbitrary products.

This can be alternatively stated as saying the inclusion $i: B\to C$ given by $i_n: B_n \to C_n$ is a morphism of chain complexes. Recall that some squares need to commute, and this forces the condition on restrictions.

Suppose $f:B\to C$ is a chain map, then there exist induced maps on the levelwise kernels and cokernels, so we can form the kernel and cokernel complex:

Link to Diagram

Here $\ker f \leq B$ is a subcomplex, and $\operatorname{coker}f$ is a quotient complex of $C$. The chain map $i: \ker f\to B$ is a categorical kernel of $f$ in $\mathsf{Ch}$, and $\pi$ is similarly a cokernel. See appendix A 1.6. These constructions make $\mathsf{Ch}$ into an abelian category: roughly an additive category where every morphism has a kernel and a cokernel.

Consider a double complex:

Link to Diagram

All of the individual rows and columns are chain complexes, where $(d^h)^2 = 0$ and $(d^v)^2 = 0$, and the square anticommute: $d^v d^h + d^h d^v - 0$, so $d^v d^h = -d^h d^v$. This is almost a chain complex of chain complexes, i.e. an element of $\mathsf{Ch}(\mathsf{Ch}{\mathsf{R}{\hbox{-}}\mathsf{Mod}}))$. It’s useful here to consider lines parallel to the line $y=x$.

The squares anticommute, since the $d^v$ are not chain maps between the horizontal chain complexes. This can be fixed by changing every one out of four signs, defining $$\begin{align*} f_{*, q}: C_{*, q} \to C_{*, q-1} \\ f_{p, q} \mathrel{\vcenter{:}}=(-1)^p d^v_{p, q}: C_{p,q} \to C_{p, q-1} .\end{align*}$$

This yields a new double complex where the signs of each column alternate:

Now the squares commute and $f_{{-}, q}$ are chain maps, so this object is an element of $\mathsf{Ch}(\mathsf{Ch}{\mathsf{R}{\hbox{-}}\mathsf{Mod}})$.

Recall that products and coproducts of $R{\hbox{-}}$modules coincide when the indexing set is finite.

Some notes:

• $\operatorname{Tot}^{\oplus}(C) = \operatorname{Tot}^{\Pi}(C)$ when $C$ is bounded.

• The total complexes need not exist if $C$ is unbounded: one needs infinite direct products and infinite coproducts to exist in $\mathcal{C}$. A category admitting these is called complete or cocomplete.²

Note that $$\begin{align*} H_i(\tau_{\geq n} C) = \begin{cases} 0 & i < n \\ H_i(C) & i\geq n. \end{cases} .\end{align*}$$

We can make translation into a functor $[p]: \mathsf{Ch}\to \mathsf{Ch}$: given $f: C\to D$, define $f[p]: C[p] \to D[p]$ by $f[p]_n \mathrel{\vcenter{:}}= f_{n+p}$, and a similar definition for cochain complexes changing $p$ to $-p$.

Some terminology: in an abelian category $\mathcal{A}$ an example of an exact complex in $\mathsf{Ch}(\mathcal{A})$ is $$\begin{align*} \cdots \to 0 \to A \xrightarrow{f} B \xrightarrow{g} C \to 0 \to \cdots .\end{align*}$$

where exactness means $\ker = \operatorname{im}$ at each position, i.e. $\ker f = 0, \operatorname{im}f = \ker g, \operatorname{im}g = C$. We say $f$ is monic and $g$ epic.

As a special case, if $0\to A\to 0$ is exact then $A$ must be zero, since the image of the incoming map must be 0. This also happens when every other term is zero. If $0\to A \xrightarrow{f} B \to 0$, then $A \cong B$ since $f$ is both injective and surjective (say for $R{\hbox{-}}$modules).

We assumed that $\mathcal{A}= {\mathsf{R}{\hbox{-}}\mathsf{Mod}}$ here, so we could chase elements, but this happens to also be true in any abelian category $\mathcal{A}$ but by a different proof. The idea is to embed $\mathcal{A} \to {\mathsf{R}{\hbox{-}}\mathsf{Mod}}$ for some ring $R$, do the construction there, and pull the results back – but this doesn’t quite work! $\mathcal{A}$ can be too big. Instead, do this for the smallest subcategory $\mathcal{A}_0$ containing all of the modules and maps involved in the snake lemma. Then $\mathcal{A}_0$ is small enough to embed into ${\mathsf{R}{\hbox{-}}\mathsf{Mod}}$ by the Freyd-Mitchell Embedding Theorem.

Note that $d$ is natural, which means the following: there is a category $\mathcal{S}$ whose objects are SESs of chain complexes and whose maps are chain maps:

There is another full subcategory $\mathcal{L}$ of $\mathsf{Ch}$ whose objects are LESs of objects in the original abelian category, i.e. exact chain complexes. The claim is that the LES construction in the theorem defines a functor $\mathcal{S}\to \mathcal{L}$. We’ve seen how this maps objects, so what is the map on morphisms? Given a morphism as in the above diagram, there is an induced morphism:

Link to Diagram

The first two squares commute, and naturality means that the third square commutes as well.

It is sometimes useful to explicitly know how to compute snake lemma boundary elements. See the book for a recipe for computing ${{\partial}}(\xi)$:

• Lift $\xi$ to a cycle $c\in Z_n(C) \subseteq C_n$.
• Pull $c$ back to a preimage $b\in B_n$ by surjectivity.
• Apply the differential to get $d(b)\in Z_{n-1}(B)$, using that images are contained in kernels.
• Since this is in kernel of the outgoing map, it’s in the kernel of the incoming map and thus there exists an $a\in Z_{n-1}(A)$ such that $f(a) = db$
• So set $\delta(\xi) \mathrel{\vcenter{:}}=[a] \in H_{n-1}(A)$.

Why is naturality useful? Suppose $H_n(B) = 0$, you get isomorphisms, and this allows inductive arguments up the LES. The LES in homology is sometimes abbreviated as an exact triangle:

Here ${{\partial}}:H_*(C) \to H_*(A)[1]$ shifts degrees. Note that this motivates the idea of triangulated categories, which is important in modern research. See Weibel Ch.10, and exercise 1.4.5 for how to construct these as quotients of $\mathsf{Ch}$.

Assume for now that we’re in the situation of $R{\hbox{-}}$modules where $R$ is a field, i.e. vector spaces. The main fact/advantage here that is not generally true for $R{\hbox{-}}$modules: every subspace has a complement. Since $B_n \subseteq Z_n \subseteq C_n$, we can write $C_n = Z_n \oplus B_n'$ for every $n$, and $Z_n = B_n \oplus H_n$. This notation is suggestive, since $H_n \cong Z_n/B_n$ as a quotient of vector spaces. Substituting, we get $C_n = B_n \oplus H_n \oplus B_n'$. Consider the projection $C_n \to B_n$ by projecting onto the first factor. Identifying $B_n \mathrel{\vcenter{:}}=\operatorname{im}(C_{n+1} \to C_n) \cong C_{n+1}/Z_{n+1}$ by the 1st isomorphism theorem in the reverse direction. But this image is equal to $B_{n+1}'$, and we can embed this in $C_{n+1}$, so define $s_n: C_n \to C_{n+1}$ as the composition $$\begin{align*} s_n \mathrel{\vcenter{:}}=( C_n \xrightarrow{\mathop{\mathrm{proj}}} B_n = \operatorname{im}(C_{n+1} \to C_n) \xrightarrow{d_{n+1}^{-1}} C_{n+1}/Z_{n+1} \xrightarrow{\cong} B_{n+1}' \hookrightarrow C_{n+1} .\end{align*}$$

Note that when $C$ is split exact, we have

Link to Diagram

Note that we’ll skip mapping cylinders, since they don’t come up until the section on triangulated categories. The goal is to see how any two maps between homologies can be fit into a LES. This helps reduce questions about quasi-isomorphisms to questions about split exact complexes.

Given $f:B\to C$ we defined $\operatorname{cone}(f)_n \mathrel{\vcenter{:}}= B_{n-1} \oplus C_n$, which fits into a SES $$\begin{align*} 0 \to C \to \operatorname{cone}(f) \xrightarrow{\delta} B[-1] \to 0 \end{align*}$$ and thus yields a LES in cohomology.

Link to Diagram

So we can convert statements about quasi-isomorphisms of complexes into exactness of a single complex.

Setup: fix $M\in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}$, where $R$ is a ring with unit. Note that by an upcoming exercise, $\mathop{\mathrm{Hom}}_{R}(M, {-}): {\mathsf{Mod}{\hbox{-}}\mathsf{R}}\to {\mathsf{Ab}}$ is a left-exact functor, but not in general right-exact: given a SES $$\begin{align*} 0\to A \xrightarrow{f} B \xrightarrow{g} C\to 0 && \in \mathsf{Ch}({\mathsf{Mod}{\hbox{-}}\mathsf{R}}) ,\end{align*}$$ there is an exact sequence:

Link to Diagram

However, this is not generally surjective: not every $M\to C$ is given by composition with a morphism $M\to B$ (lifting). To create a LES here, one could use the cokernel construction, but we’d like to do this functorially by defining a sequence functors $F^n$ that extend this on on the right to form a LES:

Link to Diagram

It turns out such functors exist and are denoted $F^n({-}) \mathrel{\vcenter{:}}=\operatorname{Ext}_R^n(M, {-})$:

Link to Diagram

By convention, we set $\operatorname{Ext}_R^0({-}) \mathrel{\vcenter{:}}=\mathop{\mathrm{Hom}}_R(M, {-})$. This is an example of a general construction: right-derived functors of $\mathop{\mathrm{Hom}}_R(M, {-})$. More generally, if $\mathcal{A}$ is an abelian category (with a certain additional property) and $F: \mathcal{A} \to \mathcal{B}$ is a left-exact functor (where $\mathcal{B}$ is another abelian category) then we can define right-derived functors $R^n F: \mathcal{A} \to \mathcal{B}$. These send SESs in $\mathcal{A}$ to LESs in $\mathcal{B}$:

Link to Diagram

Similarly, if $F$ is right-exact instead, there are left-derived functors $L^n F$ which form a LES ending with 0 at the right:

Link to Diagram

Free modules are projective. Let $F = R^X$ be the free module on the set $X$. Then consider $\gamma(x)\in C$, by surjectivity these can be pulled back to some elements in $B$:

Link to Diagram

This follows from the universal property of free modules:

Link to Diagram

The definition of projective objects extends to any abelian category, not just $R{\hbox{-}}$modules.

The proof will only use that $P \xrightarrow{\epsilon} M$ is a chain complex of projective objects, i.e. $d^2 = 0$, and that $\epsilon \circ d_1^p = 0$. To make the notation more consistent, we’ll write $Z_{-1}(P) \mathrel{\vcenter{:}}= M$ and $Z_{-1}(Q) \mathrel{\vcenter{:}}= N$. Toward an induction, suppose that the $f_i$ have been constructed for $i\leq n$, so $f_{i-1} \circ d = d \circ f_i$.

All rings have 1 in this course!

Big question: what are injective modules really? These are pretty nonintuitive objects.

Over a PID, divisible is equivalent (?) to injective as a module.

As a consequence, ${\mathsf{Mod}{\hbox{-}}\mathsf{R}}$ has enough injectives for any ring $R$.

Note that $({-})^{\operatorname{op}}$ switches

• Monics and epis,
• Initial and terminal objects,
• Kernels and cokernels.

Moreover, $\mathcal{A}$ is abelian if and only if $\mathcal{A}^{\operatorname{op}}$ is abelian.

A quick look at why these maps are inverses: $$\begin{align*} \mu( \tau(f)) &= (\tau f)(m)(1_R) \\ &= f(m\cdot 1) \\ &= f(m) .\end{align*}$$

Conversely, $$\begin{align*} \tau(\mu(g))(m)(r) &= (\mu g)( mr) \\ &= g(mr)(1) \\ &= g(m\cdot r) && \text{since } g \in \operatorname{Mor}_{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}} \\ &= g(m)(r\cdot 1) && \text{by observation earlier} \\ &= g(m)(r) .\end{align*}$$

The $?$ functor in the lemma will be the forgetful functor applied to $M$, yielding an adjoint pair.

Last time: we had a lemma that for any $M\in {\mathsf{Mod}{\hbox{-}}\mathsf{R}}$ and $A\in {\mathsf{Ab}}$ there is an isomorphism $$\begin{align*} \mathop{\mathrm{Hom}}_{{\mathsf{Ab}}}( F(M), A) \cong \mathop{\mathrm{Hom}}_{{\mathsf{Mod}{\hbox{-}}\mathsf{R}}}(M, \mathop{\mathrm{Hom}}_{\mathsf{Ab}}(R, A) ) ,\end{align*}$$ where $F: {\mathsf{Mod}{\hbox{-}}\mathsf{R}}\to {\mathsf{Ab}}$ is the forgetful functor.

The lemma thus says that $\mathop{\mathrm{Hom}}_{{\mathsf{Ab}}}(R, {-}): {\mathsf{Ab}}\to {\mathsf{Mod}{\hbox{-}}\mathsf{R}}$ (using that $R\in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}$ is a left $R{\hbox{-}}$module) is right adjoint to the forgetful functor ${\mathsf{Mod}{\hbox{-}}\mathsf{R}}\to\ Ab$.

Recall that $F$ is additive if $\mathop{\mathrm{Hom}}_{\mathcal{B}}(B', B) \to \mathop{\mathrm{Hom}}_{\mathcal{A}}(FB', FB)$ is a morphism of abelian groups for all $B, B' \in \mathcal{B}$.

Goal: define left derived functors of a right exact functor $F$, with applications the bifunctor ${-}\otimes_R {-}$, which is right exact and covariant in each variable. We’re ultimately interested in Hom functors and Ext, but this is slightly more technical since it’s covariant in one slot and contravariant in the other, so focusing on this functor makes the theory slightly easier to develop. This can be fixed by switching $\mathcal{C}$ with $\mathcal{C}^{\operatorname{op}}$ once in a while. Everything for left derived functors will have a dual for right derived functors.

Let $\mathcal{A}, \mathcal{B}$ be abelian categories where $\mathcal{A}$ has enough projectives and $F: \mathcal{A} \to \mathcal{B}$ is a right exact functor (which implicitly assumes $F$ is additive). We want to define $L_i F: \mathcal{A} \to \mathcal{B}$ for $i\geq 0$.

Note that $P_2 \xrightarrow{d_2} P_1 \xrightarrow{d_1} P_0 \xrightarrow{\varepsilon} A\to 0$ is exact, and since $F$ is right exact, it can be shown that the following is a SES: $FP_1 \xrightarrow{Fd_1} FP_0 \xrightarrow{F \varepsilon} FA \to 0$. We can use this to compute the original homology, despite it not having any homology itself! From this, we can extra $L_0(A) \mathrel{\vcenter{:}}= FP_0 / (Fd_1)(FP_1) = FP_0 / \ker F(\varepsilon)$ using exactness at $FP_0$, and by the 1st isomorphism theorem this is isomorphic to the image $FA$ using surjectivity. So $L_0 F \cong F$.

Setup: Let $\mathcal{A}, \mathcal{B}$ and $F: {\mathcal{A}}\to {\mathcal{B}}$ where ${\mathcal{A}}$ has enough projectives. Let $P \xrightarrow{\varepsilon} A\in {\mathcal{A}}$ be a projective resolution, we want to define the left derived functors $L_i F(A) \mathrel{\vcenter{:}}= H_i(FP)$.

This is an interesting result, since it doesn’t depend on the functor! Short aside on $F{\hbox{-}}$acyclic objects – we don’t need something as strong as a projective resolution.

Note that projective implies $F{\hbox{-}}$acyclic for every $F$, but not conversely. For example, flat $R{\hbox{-}}$modules are acyclic for ${-}\otimes_R {-}$. In general, flat does not imply projective, although projective implies flat.

One can compute $L_iF(A) \cong H_i(FQ)$ for any $F{\hbox{-}}$acyclic resolution. For the $L_i F$ to be functors, we need to define them on maps!

We can now pick up the theorem from the end of last time:

See exercise 2.4.3 for “dimension shifting.” This is a useful tool for inductive arguments.

Last time: right-exact functors have left-derived functors where a SES induces a LES. The functors are natural with respect to the connecting morphisms in the sense that certain squares commute. Weibel refers to $\left\{{ L_i F }\right\}_{i\geq 0}$ as a homological $\delta{\hbox{-}}$functor, i.e. anything that takes SESs to LESs which are natural with respect to connecting morphism.

Here universal means that if $\left\{{ T_i }\right\} _{i\geq 0}$ is also a $\delta{\hbox{-}}$functor with a natural transformation (not necessarily an isomorphism) $\varphi_0: T_0 \to F$, then there exist unique morphism of $\delta{\hbox{-}}$functors $\left\{{ \varphi_i: T_i \to L_i F }\right\} _{i\geq 0}$, i.e. a family of natural transformations that commute with the respective $\delta$ maps coming from both the $T_i$ and the $L_i F$, which extend $\varphi_0$. This will be important later on when we try to show Ext and Tor are functors in either slot.

Today: right-derived functors of a left-exact functor. Luckily we can use some opposite category tricks which save us some work of re deriving everything.

Then $$\begin{align*} 0 \to FA \xrightarrow{F\varepsilon} FI^0 \xrightarrow{Fd^0} FI^1 \end{align*}$$ is exact, and $$\begin{align*} R^0 FA = \ker F(d^0) / \left\langle{ 0 }\right\rangle = \operatorname{im}F\varepsilon\cong FA ,\end{align*}$$ and so there is naturally an isomorphism $R^0 F \cong F$. Observe that $F$ yields a right-exact functor $F^{\operatorname{op}}: \mathcal{A}^{\operatorname{op}}\to \mathcal{B}^{\operatorname{op}}$, where we note that $F^{\operatorname{op}}(f^{\operatorname{op}}) = F(f)^{\operatorname{op}}$. Note that taking the opposite category sends injectives to projectives and so $\mathcal{A}^{\operatorname{op}}$ has enough projectives. This means that $L_i F^{\operatorname{op}}$ are defined using the projective resolution $I$, so we have $$\begin{align*} R^i F(A) = (L_i F^{\operatorname{op}})^{\operatorname{op}} .\end{align*}$$ Thus all results about left-derived functors translate to right-derived functors:

• $R_i F$ is independent of the choice of injective resolution, up to a natural isomorphism.
• If $A$ is injective, then $R^{i>0} F(A) = 0$.
• The collection $\left\{{ R^i F }\right\} _{i\geq 0 }$ forms a universal cohomological $\delta{\hbox{-}}$functor for $F$.
• An object $Q \in \mathcal{A}$ is $F{\hbox{-}}$acyclic if $R^{>0}F(Q) = 0$.
• $R^iF$ can be computed using $F{\hbox{-}}$acyclic objects instead of injective resolutions.

Exercises 2.5.1, 2.5.2 are important extensions of our existing characterizations of injectives and projectives in ${\mathsf{Mod}{\hbox{-}}\mathsf{R}}$. These upgrade the characterization involving $\mathop{\mathrm{Hom}}$ to one involving $\operatorname{Ext}$.³

Fix $B\in {\mathsf{Mod}{\hbox{-}}\mathsf{R}}$ and consider $G\mathrel{\vcenter{:}}=\mathop{\mathrm{Hom}}_{{\mathsf{Mod}{\hbox{-}}\mathsf{R}}}({-}, B): {\mathsf{Mod}{\hbox{-}}\mathsf{R}}\to {\mathsf{Ab}}$. Then $G$ is still left-exact, but is now contravariant. We can regard it as a covariant functor left-exact functor $G: {\mathsf{Mod}{\hbox{-}}\mathsf{R}}^{\operatorname{op}}\to {\mathsf{Ab}}$. So we define $R^i G(A)$ by an injective resolution of $A$ in $\mathcal{A}^{\operatorname{op}}$, and this is the same as a projective resolution of $A$ in $\mathcal{A}$. So apply $G$ and take cohomology. It turns out that $$\begin{align*} R^i \mathop{\mathrm{Hom}}_{{\mathsf{Mod}{\hbox{-}}\mathsf{R}}}({-}, B) \cong R^i \mathop{\mathrm{Hom}}_{{\mathsf{Mod}{\hbox{-}}\mathsf{R}}}(A, {-})(B) \mathrel{\vcenter{:}}=\operatorname{Ext}^i_{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}(A, B) ,\end{align*}$$ so we can use the same notation $\operatorname{Ext}^i_R({-}, B)$ for both cases.

Let

• $R, S \in \mathsf{Ring}$,
• $B\in (\mathsf{R}, \mathsf{S}){\hbox{-}}\mathsf{biMod}$,
• $C\in {\mathsf{S}{\hbox{-}}\mathsf{Mod}}$.

Then $\mathop{\mathrm{Hom}}_{S}(B, C)\in {\mathsf{Mod}{\hbox{-}}\mathsf{R}}$ in a natural way: given $f:B\to C$, define $(f\cdot r)(b) = f(rb)$.

We saw this structure earlier with $S={\mathbb{Z}}$, see p.41.

If $B$ is only a left $R{\hbox{-}}$module, we can always take $S = {\mathbb{Z}}$, which makes this into a functor $$\begin{align*} {-}\otimes_R B: {\mathsf{Mod}{\hbox{-}}\mathsf{R}}\to {\mathsf{Ab}} .\end{align*}$$ Since this is a right exact functor from a category with enough injectives, we can define left-derived functors.

Note that these are easier to work with, since they’re covariant in both variables.

We looked at $B\in (\mathsf{R}, \mathsf{S}){\hbox{-}}\mathsf{biMod}$ and showed ${-}\otimes_R B: {\mathsf{R}{\hbox{-}}\mathsf{Mod}} \to {\mathsf{S}{\hbox{-}}\mathsf{Mod}}$ is left adjoint to hom, and has left-derived functors $\operatorname{Tor}_n^R({-}, B) \mathrel{\vcenter{:}}= L_n({-}\otimes_R B)$. $$\begin{align*} \adjunction{{-}\otimes_R B}{ \mathop{\mathrm{Hom}}_{{S}}(B, {-}) }{\mathsf{R}{\hbox{-}}\mathsf{Mod}}{\mathsf{S}{\hbox{-}}\mathsf{Mod}} .\end{align*}$$

Note that $\operatorname{Tor}_0^R(A, B) \cong A\otimes_R B$.

$A\otimes_R {-}$ is also right exact, and it turns out that $$\begin{align*} L_n(A\otimes_R {-})(B) \cong L_n({-}\otimes_R B)(A) .\end{align*}$$ So unambiguously denote either of this left derived functors as $\operatorname{Tor}_n(A, B)$.

Both $\mathcal{A}, \mathcal{A}^{\mathcal{I}}$ are small, so we can consider functors between them.

Think about this last diagram: what happens when you mod out by larger modules?

We’ll prove this next time, note that $2\implies 1$ since coproducts are special cases of limits.

${\mathsf{Mod}{\hbox{-}}\mathsf{R}}$ satisfies (1), since direct sums of $R{\hbox{-}}$modules still have an $R{\hbox{-}}$module structure. Thus ${\mathsf{Mod}{\hbox{-}}\mathsf{R}}$ is cocomplete.

Note that this amounts to reversing arrows in the conditions of a colimit. Many of the results for colimits go through with arrows reversed. Examples: kernels, direct products. If $I$ is a poset, then limits are referred to as inverse limits, using $\varprojlim_{i\in I} A_i$.

Recall left adjoints are right-exact and have left-derived functors.

One can also show directly from the definition that $$\begin{align*} \operatorname{Tor}_*^R(A, \bigoplus_{i\in I} B_i) \cong \bigoplus_{i\in I} \operatorname{Tor}_*^R(A, B_i) .\end{align*}$$ This uses the fact that $P \otimes_R (\bigoplus_{i\in I} B_i) \cong \bigoplus_{i\in I} (P \otimes B_i)$.

We’ll skip the rest of this section, we (hopefully) won’t need filtered colimits.

Idea: their derived functors with either variable fixed will essentially be the same. We’ll start by showing that the two left-derived functors of ${-}\otimes_R {-}$ give the same results, and similarly for the two right-derived functors $\mathop{\mathrm{Hom}}_R({-}, {-})$. We’ll use double complexes!

Suppose we have two chain complexes $(P)_R \in \mathsf{Ch}( {\mathsf{Mod}{\hbox{-}}\mathsf{R}}), {}_R(Q) \in \mathsf{Ch}({\mathsf{R}{\hbox{-}}\mathsf{Mod}})$. Then there is a double complex where $i, j$ indexes rows and columns: $P \otimes_R Q = \left\{{ P_i \otimes_R Q_j }\right\}_{i, j}$, the tensor product double complex of $P$ and $Q$. We use the sign trick from 1.2.5:

• $d^h \mathrel{\vcenter{:}}= d^P \otimes\one$

• $d^v \mathrel{\vcenter{:}}=(-1)^i 1 \otimes d^Q$

Taking the direct sum totalization $\operatorname{Tor}^{ \oplus }(P \otimes_R Q )$ is the total tensor product chain complex of $P$ and $Q$. Note that this has a single differential! The big theorem from this section:

Note that this makes the right-hand side notation unambiguous.

It suffices to prove (1). Interchanging rows and columns by reflecting along the line $i=j$ interchanges the types showing up in (1) and (2), and doesn’t change the total complex. This similarly switches (3) and (4), so we have $1\implies 2$ and $4\implies 3$, so we’ll show that $1\implies 4$. Let $\tau_n C$ be the double complex obtained taking a good truncation of $C$ at level $n$: $$\begin{align*} (\tau_n C)_{ij} \mathrel{\vcenter{:}}= \begin{cases} C_{ij} & j > n \\ \ker( d^v: C_{i, n} \to C_{i, n-1} & j=n. \end{cases} .\end{align*}$$ Up to translation $\tau_n C$ is a 1st quadrant complex, and since we’re in case (4), we’re assuming the columns are exact. Now using (1), $\operatorname{Tot}^{\oplus }(\tau_n C)= \operatorname{Tot}^{\prod}(\tau_n C)$ since we now have a first quadrant complex and all diagonals are finite, and we can conclude both are exact. This implies that $\operatorname{Tot}^{\oplus }C$ is acyclic since every cycle in $\operatorname{Tot}^{\oplus }(C)$ is nonzero in only finitely many terms. Thus each such cycle is a cycle in $\operatorname{Tot}(\tau_n C)$ for some $n\ll 0$, and hence a boundary by the previous argument.

Note that this argument does not go through for the direct product, since then there may be infinitely many nonzero terms on any diagonal, and not every cycle would be represented after some finite truncation and shift.

This proves that ${-}\otimes_R{-}$ is balanced, i.e. taking the derived functors in either variable with the same pair $(A, B)$ results in the same thing. To prove a similar result for hom and ext, we want to consider $\mathop{\mathrm{Hom}}_R(A, {-})$ which requires injective resolutions, and $\mathop{\mathrm{Hom}}_R({-}, B)$ is contravariant and left-exact, so we take an injective resolution in $\mathcal{C}^{\operatorname{op}}$, i.e. a projective resolution in $\mathcal{C}$. So take a projective resolution $P\to A$ and an injective resolution $B\to I$ and make a first quadrant double complex $C_{i, j} \mathrel{\vcenter{:}}=\mathop{\mathrm{Hom}}(P_i, I^j)$ for $i, j\geq 0$. Define the differentials using the following sign convention:

Link to Diagram

Now applying a dual argument as the one for tor yields a “dual acyclic assembly lemma.”

We’ll skip the first 3 sections of chapter 3. It’s worth looking at 3.2 on tor and flatness. There’s a slightly circular statement that projective implies flat in the book, since we used this to show that certain rows were exact, so refer to a good algebra book for alternative proofs.

More generally, given $\xi$, let $\Theta(\xi) \mathrel{\vcenter{:}}={{\partial}}(\one_B) \in \operatorname{Ext}^1(A, B)$. Thus TFAE:

• $\xi$ is split
• $\one-B$ lifts to some $\sigma\in \mathop{\mathrm{Hom}}(X, B)$
• $\one_B \in \operatorname{im}f_* = \ker {{\partial}}$
• $\Theta(\xi) = 0$, even if $\operatorname{Ext}^1(A, B) \neq 0$.

Then $\Theta(\xi)$ is an obstruction to $\xi$ being split.

If $\xi'\sim \xi$ then ${{\partial}}'(\one_B) = {{\partial}}(\one_B)\in \operatorname{Ext}^1(A, B)$ by naturality of the connecting morphisms. So equivalent extensions have the same obstruction, i.e. $\Theta$ only depends only on the equivalence class $[\xi]$ of the SES.

Last time: we looked at group extensions. Given $\xi: 0\to B\to X \to A\to 0$, we had a canonical element in $\operatorname{Ext}^1(A, B)$, namely $\Theta(\xi) = \delta(\one_B)$. This only depends on the equivalence class of $\xi$.

Note the importance of the reversed directions after taking the Hom!

How can we upgrade this to a group homomorphism? One way is to pull back the group structure from the right-hand side to the left-hand side, but it turns out that Baer worked out an intrinsic group structure around 1934. We can construct the “smallest” extension such that $A$ is a quotient and $B$ is a submodule.

Next time we’ll check this by showing $\Theta(\phi) = \Theta(\xi) + \Theta(\xi')$.

Link to Diagram

Link to Diagram

We want to define $\xi' \oplus \xi''$, An important takeaway is that $\Theta$ can alternatively be defined as a map induced by the original boundary map coming from the SES, i.e. ${{\partial}}(\beta') = \Theta(\xi')$. This fits into the diagram as follows:

Link to Diagram

We define $$\begin{align*} \tilde X \mathrel{\vcenter{:}}=\left\{{ (x', x'') \in X' \times X'' {~\mathrel{\Big|}~}\pi'(x') = \pi''(x'') }\right\} \twoheadrightarrow Y ,\end{align*}$$ and note that we had a skew diagonal $\tilde\Delta\subseteq \tilde X$. This yields a YES $$\begin{align*} \phi: 0 \to B \to Y \to Y/B \cong A \to 0 .\end{align*}$$

One should check that this is well-defined since we’re using equivalence classes. There is a fast way to do both at once, i.e. showing $\Theta$ is well-defined and also a group morphism.

What about the 0 element for split SESs? Recall that additive functors preserve split exact sequences, since these are just in terms of sums of maps composing to the identity. Then applying the hom functor to the original SES produces another SES, which in particular has no Ext correction term.

Similarly, $\operatorname{Ext}^n(A, B)$ is identified with equivalence classes of longer sequences with $n+2$ terms, and an equivalence is a sequence of maps that result in commuting squares:

Link to Diagram

Note that if ${P}_{*} \to A\to 0$ is a projective resolution, then the comparison theorem yields maps and a commutative diagram

Link to Diagram

Then the dimension shifting theorem (Exc. 2.4.3) and its proof yields an exact sequence $$\begin{align*} \mathop{\mathrm{Hom}}(P_{n-1}, B) \to \mathop{\mathrm{Hom}}(M, B) \xrightarrow{{{\partial}}} \operatorname{Ext}^n(A, B) \to 0 ,\end{align*}$$ and the asserted bijection is then given by $\Theta(\xi) \mathrel{\vcenter{:}}={{\partial}}(\beta)$.

Note that the correction term vanishes if $R$ is a field.

In optimal situations the tor term vanishes, e.g. if either term is torsionfree (so no elements of finite order).

The following is a generalization for both.

Taking $R={\mathbb{Z}}$ makes $S_k(X)$ a free abelian group. If $M$ is any abelian group, we can define $H_n(X; M) \mathrel{\vcenter{:}}= H_n( {S}_{*}(X) \otimes_{\mathbb{Z}}M)$, the homology with coefficients in $M$. If no coefficients are specified, we write $H_n(X) \mathrel{\vcenter{:}}= H_n(X; {\mathbb{Z}})$. There is then a universal coefficient theorem in topology: $$\begin{align*} H_n(X; M) \cong \qty{ H_n(X) \otimes_{\mathbb{Z}}M} \oplus \operatorname{Tor}_1^{\mathbb{Z}}( H_{n-1}(X), M) .\end{align*}$$

Next week: group cohomology, spectral sequences next week. This will give us some objects to apply spectral sequences.

Let $X \in {\mathsf{Top}}$ and $S_k(X)$ be the free ${\mathbb{Z}}{\hbox{-}}$module on $\mathop{\mathrm{Hom}}_{\mathsf{Top}}( \Delta_k, X)$, which assemble into a chain complex $S(X)$. For $M\in {\mathsf{Ab}}$, we defined $H^n(X; M) \mathrel{\vcenter{:}}= H^n( \mathop{\mathrm{Hom}}(S(X), M))$ and write $H^n(X) \mathrel{\vcenter{:}}= H^n(X; {\mathbb{Z}})$. The universal coefficient theorem states $$\begin{align*} H^n(X; M) \cong \mathop{\mathrm{Hom}}_{\mathbb{Z}}(H_n(X), M) \oplus \operatorname{Ext}^1_{\mathbb{Z}}( H_{n-1}(X), M) .\end{align*}$$

Note that flat is weaker than projective for tensor products, but in an asymmetric situation, there’s nothing weaker than projective for the hom functors to be exact (since this is an iff).

There is an equality of categories ${\mathsf{G}{\hbox{-}}\mathsf{Mod}} = \mathsf{{\mathbb{Z}}G}{\hbox{-}}\mathsf{Mod}$. This is also the same as the functor category ${\mathsf{Ab}}^\mathcal{G}$ (a category of the form $\mathcal{A}^{\mathcal{I}}$) where $\mathcal{G}$ is the category with one object whose morphisms are the elements of $G$. In other words, ${\operatorname{Ob}}(\mathcal{G}) \mathrel{\vcenter{:}}=\left\{{ 1 }\right\}$ and $\mathop{\mathrm{Hom}}_{\mathcal{G}}(1, 1) = G$. Note that every morphism is invertible since $G$ is a group.

The right-hand side yields a $G{\hbox{-}}$module since $F(g)(a) = g.a$.

Any $G\in {\mathsf{Ab}}$ can be viewed as a trivial $G{\hbox{-}}$module in this way. This yields a functor $\operatorname{Triv}:{\mathsf{Ab}}\to \mathsf{G}{\hbox{-}}\mathsf{Mod}$. There is a distinguished trivial $G{\hbox{-}}$module, namely $A \mathrel{\vcenter{:}}={\mathbb{Z}}$ with the trivial $G{\hbox{-}}$action. There are two natural functors $\mathsf{G}{\hbox{-}}\mathsf{Mod}\to {\mathsf{Ab}}$:

• $A^G \mathrel{\vcenter{:}}=\left\{{ a\in A {~\mathrel{\Big|}~}g.a = a \forall g\in G }\right\}$, the invariant subgroup of $A$.
• $A_G \mathrel{\vcenter{:}}= A / \left\langle{ ga-a {~\mathrel{\Big|}~}g\in G, a\in A }\right\rangle$, where we take the $G{\hbox{-}}$module generated by the relation in the denominator, which are the coinvariants of $A$.

The exts here will classify extensions in the category of left ${\mathbb{Z}}{\hbox{-}}$modules. Note the switched order on the hom functor however!

See exercise 6.1.2 for $kG{\hbox{-}}$modules for $k\in \mathsf{Ring}$ arbitrary.

For $h\in G$, $$\begin{align*} hN = \sum_g hg = \sum_{g'\in G} g' = N ,\end{align*}$$ and so $N \in ({\mathbb{Z}}G)^g$. Similarly $Nh = N$ and so $Z({\mathbb{Z}}G)$ is in the center.

Note the two different $Z$s here!

Exercise 6.1.3 shows that $H^0(G; {\mathbb{Z}}G) = 0$ when $G$ is infinite, in which case $\mathcal{I} = \left\{{ a \in {\mathbb{Z}}G {~\mathrel{\Big|}~}N a = 0 }\right\}$ is the annihilator of the norm element. Next class we’ll start on spectral sequences.

Invented by John Leray, 1946 while a prisoner of war in Austria, as an algorithmic way to compute homology of chain complexes. Start with a first-quadrant double complex $\left\{{ E_{p, q} {~\mathrel{\Big|}~}p, q\geq 0 }\right\}$, say of $R{\hbox{-}}$modules. Let $T_n \mathrel{\vcenter{:}}=\bigoplus_{p}$ be the total complex (direct sum or product, since the diagonals are finite) where $d \mathrel{\vcenter{:}}= d^b + d^h$. Suppose one could compute the homology of each “piece” of the differential separately and independently. First forget $d^h$, and let this complex be $E_{p, q}^0$ (where the $0$ superscript denotes a “zeroth approximation”).

Link to Diagram

Now let $E^1{p, q} \mathrel{\vcenter{:}}= H_q(E_{p, q}^0)$ be the homology obtained from the vertical complexes, i.e. $E^1_{p, q} \mathrel{\vcenter{:}}=\ker d^v_{p, q} / \operatorname{im}d^v_{p, q-1}$. Recall that by convention we require anticommutativity, so $d^vd^h + d^h d^v = 0$, so this is not quite a complex of complexes. So these won’t quite give a chain map, but $d^vd^h = -d^h d^v$ is enough to induce well-defined maps on $E^1_{*, *}$ since they will preserve kernels and images. So $E^1$ has horizontal differentials $d^h: E^1_{*,*} \to E^1_{*-1, *}$:

Link to Diagram

We can now write $E_{p, q}^2$ for the horizontal homology $H_p(E^1_{*, q})$ at the $p, q$ spot. We’ve done the horizontal and vertical homology separately, how close is $\left\{{ E_{p, q}^2 {~\mathrel{\Big|}~}p+q = n }\right\}$ to giving us information about the total homology?

This yields differentials on $E^2$ on lines of slope $-1/2$ which move from the $n$th diagonal to the $n-1$st diagonal:

Link to Diagram

Link to Diagram

So we let $E^3$ be the homology, and it turns out there are differentials $d^3: E^3_{p, q} \to E^3_{p-3, q+2}$ which go from diagonal $n$ to $n-1$.

The term $E_{p, q}^{r+1}$ is a subquotient, i.e. a submodule of a quotient, of $E_{p, q}^r$, and hence inductively a subquotient of $E_{p, q}^a$ by transitivity of “being a subquotient.” The terms of total degree $n$ lie on a line of slope $-1$, and each differential $d^r_{p, q}$ decreases the total degree by 1.

There is a category of homology spectral sequences over a fixed abelian category $\mathcal{A}$. The objects consist of the above data of pages, differentials, and structure maps from the above definition The morphisms $f: E\to \tilde E$ are families of maps $$\begin{align*} f_{p, q}^r: E_{p, q}^r \to \tilde{E}^r_{p, q} \end{align*}$$ for all $r \geq\max\left\{{a, \tilde a}\right\}$ with $\tilde{d}^r f^r = f^r d^r$ such that $f_{p, q}^{r+1}$ is the map on homology induced by $f_{p, q}^r$.

Recall that we had

• $\left\{{ E_{p, q}^r {~\mathrel{\Big|}~}r\geq a, p,q\in {\mathbb{Z}}}\right\}$ for some $a$.
• $d_{p, q}^r: E_{p, q}^r \to E_{p-r, q+r-1}$ with $d^2=0$.
• $E_{p, q}^{r+1} \cong \ker d_{p, q}^r / \operatorname{im}d_{p+r,q-r+1}^r$.

For the rest of this course, we’ll restrict our attention to bounded spectral sequences.

We saw a case where the length of the filtration was 2, when we had $2$ columns. Recall that this only yields information up to extensions, since this only computes quotients.

We can form a similar definition for a cohomology spectral sequence. The conditions change slightly:

(2’) We have a decreasing filtration $$\begin{align*} H^n = F^s H^n \supseteq \cdots \supseteq F^p H^n \supseteq F^{p+1} H^n \supseteq \cdots \supseteq F^t H^n = 0 .\end{align*}$$ In this case we have $$\begin{align*} E_{\infty }^{p, q} \cong {F^p H^{p+q} \over F^{p+1} H^{p+q} } .\end{align*}$$ Then each $H_n$ will have a filtration of length $n+1$ by explicitly counting terms on the diagonal, so we obtain $$\begin{align*} 0 = F_{-1} H_n \subset F_0 H_n \subseteq \cdots \subseteq F_{n-1} H_n \subseteq F_n H_n = H_n .\end{align*}$$

Then $$\begin{align*} E_{0, n} &\cong F_0 H_n \hookrightarrow H_n\\ E_{p, n-p} &\cong {F_p H_n \over F_{p-1} H_n} \\ H_n \twoheadrightarrow E_{n, 0} &\cong {H_n \over F_{n-1} H_n} .\end{align*}$$

For a first quadrant cohomological spectral sequence, the edge maps are $$\begin{align*} E_{a}^{n, 0} \twoheadrightarrow E_{\infty }^{n, 0} \hookrightarrow H^n \\ H^n \twoheadrightarrow E_{\infty }^{0, n} \hookrightarrow E_{a}^{0, n} .\end{align*}$$

This implies that $E_{p, q}^r = E_{p, q}^{\infty }$ at this point. In this case, we can read off the single nonzero section:

Here we’ll have $$\begin{align*} E^{\infty }_{p, q} \cong {F_p H_n \over F_{p-1} H_n} \cong {H_n \over 0}\cong H_n .\end{align*}$$

A more common definition of a spectral sequence collapsing at $r$ is that for all $p, q$, the differentials $d_{p, q}^r = 0$. Note that this implies stabilization at $r$, but doesn’t allow for such a simple statement about the diagonals since they may intersect multiple nonzero objects.

Some things we’re skipping from the book, around the last part of 5.2:

• Definitions pertaining to unbounded spectral sequences.
• Weak convergence.
• Filtrations that are infinite in on or both filtrations.
• Filtrations that don’t limit to a union equal to $H_n$ or intersection to 0.
• Abutment, which is convergence when the filtration is not finite.

We’ll skip 5.3 on the Leray spectral sequence and jump to 5.4, constructing a spectral sequence.

The construction of the spectral sequence will show that $C$ and $\bigcup_p F_p C$ give rise to the same spectral sequence. So we will assume that all filtrations are exhaustive.

Note that this implies that each diagonal of total degree $n$ has only finitely many nonzero terms, so the spectral sequence will again be bounded. We’ll next show that this spectral sequence converges to $H_*(C)$.

Note that all elements on all pages are subquotients of $E^0$ elements, so they can only get smaller, and terms that become 0 on some page stay 0 for all remaining pages.

For ease of notation, we’ll suppress the subscript $q$ since it can always be recovered as $q = n-p$. Define the canonical quotients $$\begin{align*} \eta_p: F_p C \to F_p C / F_{p-1}C = E_p^0 .\end{align*}$$

Define $$\begin{align*} A^r_p \mathrel{\vcenter{:}}=\left\{{ c\in F_p C {~\mathrel{\Big|}~}d(c) \in F_{p-r}(C) }\right\} ,\end{align*}$$ which are elements of $F_p C$ which are cycles modulo $F_{p-r} C$, the approximate cycles. Note that any actual cycle is in all $A^r$. This differential takes things $r$ columns to the left, so we’ll want to define a differential that associates the following terms

Link to Diagram

Similarly, define $$\begin{align*} Z_p^r &\mathrel{\vcenter{:}}=\eta_p(A_p^r \subseteq E_p^0 \\ B_p^r &\mathrel{\vcenter{:}}=\eta_p(d A_{p+r-1}^{r-1}) \subseteq \eta_p(F_p C) \subseteq E_p^0 .\end{align*}$$

Some consequences:

$(1) \implies Z_p^0 = E_p^0$ (taking $r=0$ in the quotient map $\eta_p$).

$(2) \implies Z_p^{r+q} \subseteq Z_p^r$, since these are images of subgroups

$(3) \implies A_{p+r-1}^{r-1} \subseteq A_{p+r}^r$, replacing $p\mapsto p+r$ in the intersection formula. Then applying $d$ yields $B_p^r \subseteq D_p^{r+1}$.

$(1) \implies B_p^0 = \eta_p(d A_{p-1}^{-1}) \subseteq \eta_p(F_{p-1} C) = 0$, since this occurs in the denominator for $\eta_p$ and $d$ preserves filtration degree.

So the $Z_p$ get smaller and the $B_p$ get bigger. What happens in the middle?

Set $B_p^{\infty } \mathrel{\vcenter{:}}=\cup_{r\geq 1} B_p^r \subseteq Z_p^{\infty } \mathrel{\vcenter{:}}=\bigcap_{s\geq 1} Z_p^s$, which follows from a set theory exercise.

Combining and summarizing these results: for every $p\geq 0$, we have a tower of groups:

Link to Diagram

Note that using standard isomorphism theorems, we have $$\begin{align*} Z_p^r \cong {A_p^r \over A_p^r \cap F_{p-1}C C} \overset{(3)}{=} {A_p^r \over A_{p-1}^{r-1}} .\end{align*}$$ So set $$\begin{align*} E_p^r \mathrel{\vcenter{:}}= Z_p^r/B_p^r \cong {A_p^r + F_{p-1} C \over d A_{p+r-1}^{r-1} + F_{p-1}C } \cong {A_p^r \over d A_{p+r-q}^{r-1} + A_{p-1}^{r-1}} ,\end{align*}$$ making $E_p^r$ a quotient of $A_p^r$. Using a similar calculation, one can show $$\begin{align*} {Z_p^{r+1} \over B_p^r} \cong { A_p^{r+1} + A_{p-1}^{r-1} \over dA_{p+r-1}^{r-1} + A_{p-1}^{r-1} } .\end{align*}$$

There will be an induced differential on this quotient, which will follow from checking that the different preserves the numerator and denominator.

We have an increasing filtration $F_p C \subseteq F_{p+1}C$, where we defined $$\begin{align*} E_{p, q}^0 = { F_p C_{p+q} \over F_{p-1} C_{p+1} } && E_{p,q}^1 = H_{p+q} E_{p, *}^0 .\end{align*}$$

1. We have a map $$\begin{align*} \eta_p: F_p C \twoheadrightarrow{F_p C \over F_{p-1}C } = E_p^0 ,\end{align*}$$ where we’ve dropped the $q$ from notation.

2. $$\begin{align*} A_{p, q}^r = \left\{{ c \in C_p C {~\mathrel{\Big|}~}dc \in F_{p-1} C }\right\} ,\end{align*}$$ the eventual cycles. We defined $Z_p^r = \eta_p A_p^r$ and $B_p^r = \eta_p dA_{p+r-1}^{r-1}$, and wrote $A_p^r \cap F_{p-1}C = A_{p-1}^{r-1}$.

3. We had the chain of inclusions $$\begin{align*} 0 = B_p^r \subseteq \cdots \subseteq B_p^{\infty} \subset Z_p^{\infty } \subset \cdots \subseteq Z_p^1 = E_p^) .\end{align*}$$

4. We also have $E_p^r = Z_p^r/B_p^r = A_p^r / dA_{p+r-1}^{r-1} + A_{p-1}^{r-1}$

5. $Z_{p}^{r+1}/B_pr \cong {A_p^{r+1} +A_{p-1}^{r-1} \over dA_{p+r-1} ^{r-1} + A_{p-1}^{r-1}}$.

6. $dA_p^r \cap F_{p-r-1} C = dA_P^{r+1}$.

Obviously we have $$\begin{align*} d: A_p^r &\to A_{p-r}^{r} \\ d: A_{p-1}^r &\to dA_{p-1}^{r-1} ,\end{align*}$$ so $d$ induces a well-defined map $d_p^r: E_p^r \xrightarrow{} E_{p-r}^r$, which of course squares to zero, which goes $r$ columns to the left and decreases the total degree $n$ by 1 since the original $d$ did on $C_n$. This is what we need to set up a spectral sequence, since we now have pages and differentials, and it just remains to show that $E^{r+1} \cong H_*(E^r, d^r)$.

We’ll restrict our attention to bounded complexes.

A filtration $F$ on a chain complex $C$ induces a filtration on the homology $H_*C$, where $H_p H_n C = \operatorname{im}( H_n F_p C \to H_n C)$:

Link to Diagram

These inclusions induce a map from the homology of the subcomplex to the homology of the total complex.

If the filtration on $C$ is bounded, say $0 = F_s C_n \subseteq \cdots \subseteq F_t C_n = C_n$ for some $s<t$, then so is the induced filtration on $H_n C$. Also note that $F_t H_n = H_n$ and $F_s H_n = 0$.

Need to check next time that the $E^{\infty }_{p, q}$ terms give the proper quotients.

Last time: we’re trying to prove the classical convergence theorem in the bounded case. We have $$\begin{align*} E_{pq}^1 = H_{p+q}( F_p C/ F_{p-1} C ) \Rightarrow H_{p+q}C .\end{align*}$$ We’d like this converge, i.e. the $E^\infty$ page will be the sections of $H_{p+q}C$. Writing $C_n'\mathrel{\vcenter{:}}= F_p C_n$ for the filtered pieces, we have

Link to Diagram

Then the induced filtration on homology is $$\begin{align*} H_n' \mathrel{\vcenter{:}}={Z_n' \over B_n'} &\hookrightarrow H_n \mathrel{\vcenter{:}}={Z_n\over B_n} \\ z' + B_n' &\mapsto z' + B_n .\end{align*}$$

Consider two different filtrations of the total complex $\operatorname{Tot}(C)$ (either sum or product) of a double complex $C_{*, *}$. We know there is an spectral sequence associated to each and play them off of each other to get extra information about cohomology.

We can say something about the unbounded case. Suppose $C$ is 4th quadrant, then $F_{-1} \operatorname{Tot}(C) = 0$, so the first filtration ${}^I F$ is bounded below. The diagonals are infinite, so we take $\operatorname{Tot}(C) \mathrel{\vcenter{:}}=\operatorname{Tot}^{\oplus}(C)$. Every element of $(\operatorname{Tot}(C))_n$ lives in $\bigoplus _{p=0}^N C_{p, n-p}$ for some finite $N$ and the filtration is exhaustive, i.e. $\operatorname{Tot}^{\oplus}C = \bigcup_{p\geq 0} F_p \operatorname{Tot}^{\oplus}C$. A version of the classical convergence theorem will yield $$\begin{align*} {}^I E_{pq}^r \Rightarrow H_{p+q} \operatorname{Tot}^{\oplus}C .\end{align*}$$ However, this will not hold for $\operatorname{Tot}^{\Pi}$.

Next time: a second filtration and its spectral sequence, and how to play them off of each other.

Recall that we had two filtrations on a total complex: the first was fixing a vertical line and replacing everything to the right with zeros, which was given by $^{I}E_{p, q}^0 = F_p(\operatorname{Tot})/ F_{p-1}(\operatorname{Tot}) = C_{p, q}$. Taking homology with the vertical differentials yielded $^{I}E_{p, q}^1 = H_q^v(C_{p,*})$, and $^{I} E_{p, q}^2 = H_p^h H_q^v(C_{*, *})$. Applying the classical convergence theorem when this is 1st quadrant yields some spectral sequence with these as the pages which converges to $H_{p+q}(\operatorname{Tot}(C))$.

Note that the differential is $$\begin{align*} d^0: E^0_{p, q} &\to E^0_{p, q-1} \\ = d^h: C_{q, p} &\to C_{q-1, p} .\end{align*}$$

We similarly have ${}^II E_{p, q}^I = H_q^h(C_{*, p})$, again noting the switched indices, with differential $$\begin{align*} d^1: E^1_{p, q} &\to E_{p-1, q}^1 \\ =H^h(C_{q, p}) &\to H^h(C_{*, p-1}) \end{align*}$$ which comes from the original differential inducing a map on horizontal homology. Then ${}^{II} E^2_{p, q} = H_p^v H_q^h(C)$.

Note that transposing everything about the line $p=q$ interchanges filtrations $I$ and $II$, and thus the two spectral sequences ${}^{I}E_{p, q} \rightleftharpoons{}^{II} E_{q, p}$. Using that first quadrant sequences are canonically bounded, we can apply the classical convergence theorem to ${}^{II} E$ to obtain $$\begin{align*} {}^{II}E_{p, q}^2 \Rightarrow H_{p+q}( \operatorname{Tot}(C) ) .\end{align*}$$ Transposing sends $QIV$ to $QII$ and thus ${}^{II} E \Rightarrow H_{p+q}\operatorname{Tot}^{\oplus}(C)$. Note that this does not guarantee anything about $\operatorname{Tot}^{\Pi}(C)$.

In particular, if we have a $QI$ double complex, both filtrations converge to the homology of the total complex.

Our proof in 2.7 that $\operatorname{Tor}_*^R(A, B)$ could be computed either by a projective resolution ${P}_{*}\twoheadrightarrow A$ or a projective resolution ${Q}_{*}\twoheadrightarrow B$ was a disguised spectral sequence argument. So we’ll go recover it using the actual spectral sequence.

We have a $QI$ double complex $C$ given by $C_{p, q} \mathrel{\vcenter{:}}=(P\otimes Q)_{p, q} = P_p\otimes Q_q$, and we now have two spectral sequences converging to $H_*(\operatorname{Tot}(P\otimes Q))$. Taking the first filtration, we can write $$\begin{align*} H_q^v(\operatorname{Tot}(C)) = H_q(P_p \otimes Q_q) = P_p \otimes H_q(Q) .\end{align*}$$ Using that $P$ is an exact complex, and noting that we delete the augmentation when taking homology, we have $$\begin{align*} H_1^v(\operatorname{Tot}(C)) = \begin{cases} 0 & q>0 \\ P_p\otimes B & q=0. \end{cases} \end{align*}$$

Thus $$\begin{align*} E^2_{p, q} = \begin{cases} H_p^h(P_* \otimes B) & q=0 \\ 0 & 1>0, \end{cases} \end{align*}$$ meaning that this collapses at $E^2$ and we have $$\begin{align*} H_p (\operatorname{Tot}(P\otimes Q) ) \cong L_p({-}\otimes B)(A) \mathrel{\vcenter{:}}=\operatorname{Tor}^R_p(A, B) .\end{align*}$$

Now consider taking the second filtration, which yields $$\begin{align*} {}^{II} E_{p, q}^1 = H_q^h( P_q \otimes Q_p) = H_q(P_*) \otimes Q_p = \begin{cases} A_\otimes Q_p & q=0 \\ 0 & q>0. \end{cases} \end{align*}$$ The second pages comes from taking the vertical homology, so $$\begin{align*} {}^{II}E_{p, q}^2 = H_p^v H_q^h(P_q \otimes Q_p) = \begin{cases} H^v_p(A\otimes Q) & q=0 \\ 0 & q>0. \end{cases} ,\end{align*}$$ which is $L_p(A\otimes{-})(B)$ in $q=0$. Since ${}^{II}E_{p, q}^2 \Rightarrow H_{p+q}(\operatorname{Tot}(P\otimes Q)) = L_p({-}\otimes B)(A)$, and we thus have $$\begin{align*} L_p(A\otimes{-})(B) \cong L_p({-}\otimes B)(A) .\end{align*}$$

See the this section of Weibel for other applications in the exercises: the Kunneth formula, the Universal Coefficient Theorem, and the Acyclic Assembly Lemma.

We’d like to compute derived functors acting on chain complexes instead of just objects.

So we have the following situation

Link to Diagram

The situation in row $q$ will be:

Link to Diagram

Here when we take the homology of the complex along the rows $p$, we’ll obtain $$\begin{align*} H_q(P, d^h) = {Z_p(P, d^h)_q \over B_p(P, d^h)_q} ,\end{align*}$$ and since the induces maps preserve cycles and boundaries, we get induced maps on homology.

Exercise 5.7.1 shows that $P_{p, *} \xrightarrow{\varepsilon} A_p$ will be a projective resolution in $\mathcal{A}$ and so $Z_p(P, d^h)_* \to Z_p(A)$.

Last time: we talked about hypercohomology. We’re doing this so we can set up a Grothendieck spectral sequence.

The definition is set up so that $s^h + s^v: \operatorname{Tot}(D)_n \to \operatorname{Tot}(E)_{n+1}$ is a chain homotopy $\operatorname{Tot}^{\oplus}(D) \to \operatorname{Tot}^{\oplus}(E)$.

Exercises 5.7.2 and 5.7.3 show:

1. If $f:A\to B$ is a chain map and $P\to A, Q\to B$ are CE resolutions, then there is a map of double complexes $\tilde f: P\to Q$ lifting $f$.

2. If $f, g: A\to B$ are chain homotopic, then $\tilde f, \tilde g$ are chain homotopic in the sense just defined.

3. Any two CE resolutions $P, P'$ of $A$ are chain homotopy equivalent, as are $\operatorname{Tot}^{\oplus}(F(P))$ and $\operatorname{Tot}^{\oplus}(F(P'))$ for any additive functor $F$.

This last remark shouldn’t be too hard to believe: chain homotopies are defined in terms of addition.

Recall that chain homotopy yields a notion of equivalence, and chain homotopic maps induce the same map on homology. The same is true for double complexes. There is a lemma that shows a SES of double complexes induces a LES in homology.

There is a cohomology variant of this: everything dualizes to $R^i F(A)$ for a left exact functor $F: \mathcal{A}\to \mathcal{B}$ where $A\in \mathsf{Ch}(\mathcal{A})$, $\mathcal{A}$ has enough injectives, and $B$ is complete. Using a right CE resolution $I^{*, *}$ of injective objects in $A$ yields an upper half-plane complex with $A^{*} \to I^{*,0}$ such that the induces maps on cohomology are themselves injective resolutions of $B^p(A^{*})$ and $H^p(A^{*})$. In this case $$\begin{align*} R^i F(A^{*}) = H^i \operatorname{Tot}^{\Pi}F(I^{*, *}) .\end{align*}$$ We can prove dual version of all of the results about left hyper-derived functors, although there are some slight convergence issues to worry about due to the direct product.

Last time we talked about hypercohomology and hyper derived functors, and we proved that two spectra sequences converging to ${\mathbb{L}}_{p+q}F(A)$.

We’ll focus on the cohomological version, which gives a spectral sequence from a composition of functors. Let $\mathcal{A}, \mathcal{B}, \mathcal{C}$ be abelian categories with enough injectives, and let $G: \mathcal{A} \to \mathcal{B}$, $F: \mathcal{B} \to \mathcal{C}$ be left exact functors. By a previous result, $FG:\mathcal{A} \to \mathcal{C}$ is left exact, which follows from checking that it preserves 4-term exact sequences. Recall that $B \in \mathcal{B}$ is $F{\hbox{-}}$acyclic if $R^i F(B) = 0$ for all $i>0$.

We’re skipping the section on sheaf cohomology and 5.9, so we’ll move into chapter 6.

Let $H{~\trianglelefteq~}G$ and $A\in \mathsf{G}{\hbox{-}}\mathsf{Mod}$, then $A_H, A^H \in \mathsf{G/H}{\hbox{-}}\mathsf{Mod}$. The canonical projection $p: G\to G/H$ induces a forgetful functor $p^*: \mathsf{G/H}{\hbox{-}}\mathsf{Mod} \to {\mathsf{G}{\hbox{-}}\mathsf{Mod}}$ given by pullback. Note that $G/H{\hbox{-}}$modules are essentially $G{\hbox{-}}$modules where $H$ acts trivially, so this functor forgets the trivial $H$ action. Generally, this works a bit like the Frobenius map, which yields a representation that can be pulled back.

Note that we can identify the functors $$\begin{align*} ({-})^H, ({-})_H : \mathsf{G}{\hbox{-}}\mathsf{Mod} \to \mathsf{G/H}{\hbox{-}}\mathsf{Mod} ,\end{align*}$$ whose derived functors are group homology/cohomology. The idea will be that $G{\hbox{-}}$invariants can be written as a composition of other functors, and we can apply the Grothendieck spectral sequence construction.

We’re trying to prove the Lyndon-Hochschild-Serre spectral sequence for $H{~\trianglelefteq~}G$.

The exact sequence of low degree terms in the cohomological LHS spectral sequence are of the form

Link to Diagram

Note that these maps have particular name, inflation and restriction.

We thought of homology as a functor of the module $A$, but here we see it’s varying. Can this be thought of as a functor of the group instead?

Setup: let $\rho: H\to G$ be a group morphism, then recall that any $G{\hbox{-}}$module becomes an $H{\hbox{-}}$module by composition with $\rho$, which yields an exact functor $$\begin{align*} \rho^{\#}: {\mathsf{G}{\hbox{-}}\mathsf{Mod}} \to {\mathsf{H}{\hbox{-}}\mathsf{Mod}} .\end{align*}$$

Letting $A\in{\mathsf{G}{\hbox{-}}\mathsf{Mod}}$, set

• $T_n(A) \mathrel{\vcenter{:}}= H_n(G; A)$
• $T^n(A) \mathrel{\vcenter{:}}= H^n(G; A)$
• $S_n(A) \mathrel{\vcenter{:}}= H_n(G; \rho^{\#} A)$
• $S^n(A) \mathrel{\vcenter{:}}= H^n(G; \rho^{\#} A)$

A special case is when $H\leq G$ is a subgroup and $\rho: H\hookrightarrow G$ is the inclusion. Then we define a capital $\operatorname{Res}$ as $$\begin{align*} \rho^\sharp = \operatorname{Res}_H^G: {\mathsf{G}{\hbox{-}}\mathsf{Mod}} \to {\mathsf{H}{\hbox{-}}\mathsf{Mod}} ,\end{align*}$$ which is a restriction of the action to a subgroup and thus a type of forgetful functor.

Note that ${\mathbb{Z}G}$ is a free ${\mathbb{Z}}H{\hbox{-}}$module with basis being any set of coset representatives, thus any projective $G{\hbox{-}}$module restricts to a projective $H{\hbox{-}}$module, using the characterization of projective modules as direct summands of free modules.

Recall that $$\begin{align*} H_*(G; A) &\cong \operatorname{Tor}_*^{{\mathbb{Z}G}}({\mathbb{Z}}, A) \\ H_*(G; A) &\cong \operatorname{Ext}_{{\mathbb{Z}G}}^*({\mathbb{Z}}, A) .\end{align*}$$ We can compute both using a ${\mathbb{Z}G}{\hbox{-}}$projective resolution $P_* \to {\mathbb{Z}}$. This is also a ${\mathbb{Z}}H{\hbox{-}}$projective resolution, so we can use this to compute $H^*(H; {-})$ and $H_*(H; {-})$ as well.

When $*=0$, we can write $$\begin{align*} \operatorname{inf}: (A^H)^{G\over H} \to (A^H)^G \to A^G ,\end{align*}$$ and note that this is exactly the functor composition we needed to get the LHS spectral sequence. Similarly there is a LHS for homology, and an isomorphism $$\begin{align*} \operatorname{coinf}: A_G \to (A_H)_G \to (A_H)_{G\over H} .\end{align*}$$

When $A \in {\mathsf{H}{\hbox{-}}\mathsf{Mod}}^{{ \operatorname{Triv}}}$, $A_H\hookrightarrow A$ is the identity, so $A^H = A = A_H$. In this case $\operatorname{inf}= \operatorname{res}$ and $\operatorname{coinf}= \operatorname{cores}$.

Back to the LHS spectral sequence, the five-term exact sequence yields $$\begin{align*} 0 \to E_{2}^{1, 0} \to H^1(T) \to E_2^{0, 1} \xrightarrow{d_2} E_{2, 0} \to H^2(T) ,\end{align*}$$ which we can identify as $$\begin{align*} 0\to H^1\qty{{G\over H}; A^H} \xrightarrow{\operatorname{inf}} H^1(G; A) \xrightarrow{\operatorname{res}} H^1(H; A)^{G\over H} \xrightarrow{d_2} H^2\qty{ {G\over H}; A^H } \xrightarrow{\operatorname{inf}} H^2(G; A) .\end{align*}$$ There is a similar story in homology with coinflation and corestriction.

So this provides a way of computing homology on subgroups when the coefficients are in these induced/coinduced modules.

Recall that we had two ways of inducing an $H{\hbox{-}}$module up to a $G{\hbox{-}}$module for $H\leq G$ a subgroup. In this case, we can take cohomology with coefficients in any $B\in {\mathsf{{\mathbb{Z}}H}{\hbox{-}}\mathsf{Mod}}$. Shapiro’s lemma (or Frobenius Reciprocity) allowed compute homology and cohomology when the coefficients are in induced or coinduced modules:

$$\begin{align*} H_*(G; \operatorname{Ind}_H^G B) &\cong H_*(H; B) &&(1) \\ H^*(G; \operatorname{coInd}^G B) &\cong H^*(H; B) &&(2) .\end{align*}$$

Taking $A = {\mathbb{Z}}\in {\mathsf{{\mathbb{Z}G}}{\hbox{-}}\mathsf{Mod}}^{ \operatorname{Triv}}$, one gets result (2) in Shapiro’s lemma. This shows that $\operatorname{Ind}$ is left adjoint to $\operatorname{Res}$ and $\operatorname{coInd}$ is right adjoint to it, so these will have derived functors. A special case is when $H = \left\{{ 1 }\right\}$ is the trivial group, in which case any $H{\hbox{-}}$module $B$ is an abelian group such that $B^H = B = B_H$. So $({-})^H, ({-})_H$ are exact, and thus their higher derived functors are zero, i.e. $H_n(H, B) = 0 = H^n(H; B)$ for $n>0$. Moreover $$\begin{align*} H_n(G; {\mathbb{Z}G}\otimes_{\mathbb{Z}}B) \cong H^n( G, \mathop{\mathrm{Hom}}_{\mathbb{Z}}( {\mathbb{Z}}G, B)) \cong \begin{cases} B & n = 0 \\ 0 & n > 0. \end{cases} \end{align*}$$

Motivation and historical background: if $G$ is a Lie group, $G\in {\mathsf{Grp}}\cap{\mathsf{Mfd}}(C^\infty)$, i.e. the group operations are smooth maps. Usually these are real manifolds, they were introduced in the late 1800s by Sophus Lie who studied differential equations on such objects. Taking the tangent space at the identity, we write ${\mathfrak{g}}= T_e G$, which is a Lie Algebra. Lie showed that this is isomorphic to the vector space of left $G{\hbox{-}}$invariant vector fields (1st order differential operators) on $G$, which enjoys a bracket operation: $$\begin{align*} [X, Y](f) = X(Yf) - Y(Xf) && f\in C^{\infty } .\end{align*}$$ This turns out to again be a 1st order operator, despite looking like it might be 2nd order. This led to the study of abstract Lie algebras.

The product need not be associative, and $A$ need not have 1, so $A=0$ is an algebra.

Note that nilpotent implies solvable, since one can show by induction that ${\mathfrak{g}}^{(n)} \subseteq {\mathfrak{g}}^n$.

Next time: ${\mathfrak{g}}{\hbox{-}}$modules.

Last time: Lie algebras. Fix a cocommutative ring $k$, usually a field, then a Lie algebra ${\mathfrak{g}}$ over $k$ is a $k{\hbox{-}}$module with a bilinear product called the bracket such that

• $[xx] = 0$, so $[xy] = -[yx]$
• The Jacobi identity holds.

This yields a category $\mathsf{{\mathfrak{g}}}{\hbox{-}}\mathsf{Mod} \leq \mathsf{k}{\hbox{-}}\mathsf{Mod}$ which is a subcategory of $k{\hbox{-}}$modules, and this is in fact an abelian category. So we have notions of (co)kernels, injectives and projectives, etc. There is also a category $\mathsf{Mod}{\hbox{-}}\mathsf{{\mathfrak{g}}}$, but these can be sent to left ${\mathfrak{g}}{\hbox{-}}$modules by defining $x\cdot m \mathrel{\vcenter{:}}=-mx$ which makes ${\mathfrak{g}}$ anticommutative. Thus there is an equivalence of categories $$\begin{align*} \mathsf{{\mathfrak{g}}}{\hbox{-}}\mathsf{Mod} \xrightarrow{\sim} \mathsf{Mod}{\hbox{-}}\mathsf{{\mathfrak{g}}} ,\end{align*}$$ and so we usually just refer to left modules.

We’ll want to take homology and cohomology. There are some relevant functors:

• The trivial module functor: $$\begin{align*} { \operatorname{Triv}}: \mathsf{k}{\hbox{-}}\mathsf{Mod} \to \mathsf{{\mathfrak{g}}}{\hbox{-}}\mathsf{Mod} ,\end{align*}$$ which sends $M$ to itself, adding the structure of a trivial ${\mathfrak{g}}{\hbox{-}}$action.

• ${\mathfrak{g}}{\hbox{-}}$invariants: $$\begin{align*} ({-})^{\mathfrak{g}}: {\mathfrak{g}{\hbox{-}}\mathsf{Mod}}&\to {\mathsf{k}{\hbox{-}}\mathsf{Mod}}\\ M &\mapsto M^g \mathrel{\vcenter{:}}=\left\{{ x\in M {~\mathrel{\Big|}~}xm = 0 \forall \,\, x\in {\mathfrak{g}}}\right\} .\end{align*}$$

  • This yields the largest ${\mathfrak{g}}{\hbox{-}}$trivial submodule, and similarly $({-})^{\mathfrak{g}}$ is right-adjoint to ${ \operatorname{Triv}}$. $$\begin{align*} \adjunction{{ \operatorname{Triv}}}{({-})^{\mathfrak{g}}}{{\mathsf{k}{\hbox{-}}\mathsf{Mod}}}{{\mathfrak{g}{\hbox{-}}\mathsf{Mod}}} .\end{align*}$$
  • There is an isomorphism $$\begin{align*} \operatorname{ev}_1: \mathop{\mathrm{Hom}}_{\mathfrak{g}}(k, M) &\xrightarrow{\sim} M^{\mathfrak{g}}\\ f &\mapsto f(1_k) .\end{align*}$$ where $k$ is the trivial ${\mathfrak{g}}{\hbox{-}}$module.

• ${\mathfrak{g}}{\hbox{-}}$coinvariants: $$\begin{align*} ({-})_{\mathfrak{g}}: {\mathfrak{g}{\hbox{-}}\mathsf{Mod}}&\to {\mathsf{k}{\hbox{-}}\mathsf{Mod}}\\ M &\mapsto M/{\mathfrak{g}}M .\end{align*}$$

  • This is the largest ${\mathfrak{g}}{\hbox{-}}$trivial quotient of $M$, so this is left-adjoint to ${ \operatorname{Triv}}$: $$\begin{align*} \adjunction{({-})^{\mathfrak{g}}}{{ \operatorname{Triv}}}{{\mathfrak{g}{\hbox{-}}\mathsf{Mod}}}{{\mathsf{k}{\hbox{-}}\mathsf{Mod}}} .\end{align*}$$

We might expect this is related to some tensor product, but it may not be clear what ring one should tensor over.

Assume that ${\mathfrak{g}{\hbox{-}}\mathsf{Mod}}$ has enough projectives, which we’ll see is true in a later section by identifying this with a category ${\mathsf{R}{\hbox{-}}\mathsf{Mod}}$ of modules over a ring.

A better name might be the universal associative algebra. This plays an analogous role to the group algebra ${\mathbb{Z}G}$ of a group. We’ll assign an associative algebra ${\mathcal{U}(\mathfrak{g}) }$ to ${\mathfrak{g}}$, and there will be an equivalence of categories $$\begin{align*} \mathsf{{\mathfrak{g}}}{\hbox{-}}\mathsf{Mod} \xrightarrow{\sim} \mathsf{{\mathcal{U}(\mathfrak{g}) }}{\hbox{-}}\mathsf{Mod} ,\end{align*}$$ where we’ll know that the latter has enough projectives and injectives, allowing us to compute homology and cohomology with injective and projective resolutions.

Note that $T(M) \in {\mathsf{k}{\hbox{-}}\mathsf{Mod}}$ by extending the $k{\hbox{-}}$action over sums and tensor products in the obvious way, and in fact $T(M) \in {\mathsf{gr}\,}({\mathsf{Alg}_{/k} })$ where tensors in different degrees are juxtaposed. Explicitly, for $m\in M^{\otimes n}$ and $m' \in M^{\otimes n'}$, we write $m\otimes m' \in M^{\otimes(n+n')}$, which is what it means to be a graded algebra.

There is an inclusion map $$\begin{align*} M = M^{\otimes 1} \overset{\iota}\hookrightarrow T(M)_1 \hookrightarrow T(M) .\end{align*}$$ where $T(M)_j \mathrel{\vcenter{:}}=\bigoplus_{n\geq j} M^{\otimes n}$, and in fact $T(M)$ is generated as a $k{\hbox{-}}$algebra by $\iota(M)$. For example, for $m, m' \in M$, we have $\iota(m) \otimes\iota(m') \in T(M)_2$. This yields a functor $$\begin{align*} T: \mathsf{k}{\hbox{-}}\mathsf{Mod} \to {\mathsf{Alg}_{/k} }(\mathsf{Assoc}, \mathsf{Unital}) ,\end{align*}$$ as well as a forgetful functor $$\begin{align*} {\operatorname{Forget}}: {\mathsf{Alg}_{/k} }\to {\mathsf{k}{\hbox{-}}\mathsf{Mod}} .\end{align*}$$ The pair $(T, i)$ is a universal associative algebra in the following sense: if $M\in {\mathsf{k}{\hbox{-}}\mathsf{Mod}}$ and $A\in {\mathsf{Alg}_{/k} }(\mathsf{Assoc})$, then there is a $k{\hbox{-}}$module morphism $M\to {\operatorname{Forget}}(A)$ making the following diagram commute:

Link to Diagram

Note that the red portion of the diagram happens in ${\mathsf{k}{\hbox{-}}\mathsf{Mod}}$, while the blue portion is in ${\mathsf{Alg}_{/k} }$, so this allows lifting module morphisms to algebra morphisms. Commuting here means that $$\begin{align*} f(m_1) f(m_2) = \tilde f(m_1 m_2) \mathrel{\vcenter{:}}= f( \iota(m_1) \otimes\iota(m_2)) .\end{align*}$$ There is thus a natural isomorphism $$\begin{align*} \mathop{\mathrm{Hom}}_{{\mathsf{k}{\hbox{-}}\mathsf{Mod}}}(M, {\operatorname{Forget}}(A)) \xrightarrow{\sim} \mathop{\mathrm{Hom}}_{{\mathsf{Alg}_{/k} }}( T(M), A) .\end{align*}$$

Continuing section 7.3 on universal enveloping algebras.: Letting $k \in \mathsf{CRing}, {\mathfrak{g}}\in {\mathsf{Lie}{\hbox{-}}\mathsf{Alg}}_{/k}, M\in {\mathsf{k}{\hbox{-}}\mathsf{Mod}}$, we defined the tensor algebra $T(M) \mathrel{\vcenter{:}}= k \oplus \bigoplus_{i\geq 1} M^{\otimes n}\in {\mathsf{gr}\,}{\mathsf{Alg}_{/k} }(\mathsf{Assoc}, \mathsf{Unital})$ and noted that it was universal for maps from $M$ to $k{\hbox{-}}$algebras.

There is an injection $k\hookrightarrow{\mathcal{U}(\mathfrak{g}) }$, so ${\mathcal{U}(\mathfrak{g}) }$ is unital. The relations guarantee that there is a Lie algebra morphism $\iota: {\mathfrak{g}}\to {\mathcal{U}(\mathfrak{g}) }$. Note that we do not know if this is injective yet! Thus there is a functor $$\begin{align*} \mathcal{U}: {\mathsf{Lie}{\hbox{-}}\mathsf{Alg}}_{/k} \to {\mathsf{Alg}_{/k} } ,\end{align*}$$ and it turns out that this is adjoint to the $\operatorname{Lie}$ functor.

We’ll now set up an analog of the augmentation for group algebras, $\varepsilon: {\mathbb{Z}G}\to {\mathbb{Z}}$.

We can identify the coinvariants: $$\begin{align*} k \cong {\mathcal{U}(\mathfrak{g}) }/{\mathfrak{g}}= {\mathcal{U}(\mathfrak{g}) }/{\mathfrak{g}}\, {\mathcal{U}(\mathfrak{g}) }= {\mathcal{U}(\mathfrak{g}) }_{{\mathfrak{g}}} .\end{align*}$$

So Lie algebra (co)homology is just a special case of the usual Tor and Ext we’ve already looked at. We’ll next find a basis for ${\mathcal{U}(\mathfrak{g}) }$:

To at least see why these are a spanning set, suppose $\beta > \alpha$. We can commute elements: $$\begin{align*} x_{ \beta} x_{ \alpha} = x_{ \alpha} x_{ \beta} + [x_{ \beta} x_{ \alpha}] .\end{align*}$$ However, note that the commutator here has lower degree (here, the other factors are degree 2 and the commutator is degree 1). This decreases the number of misorders as well, so induction roughly works. The fact that these are linearly independent is harder and uses some actual representation theory.

Last time: the PBW theorem. Let ${\mathfrak{g}}\in {\mathsf{Lie}{\hbox{-}}\mathsf{Alg}}$ and free as a $k{\hbox{-}}$module with $k{\hbox{-}}$basis $\left\{{ x_ \alpha }\right\}_{\alpha\in A}$. Then ${\mathcal{U}(\mathfrak{g}) }$ has a $k{\hbox{-}}$basis $\left\{{ x_I }\right\}$ where $I = ( \alpha_1, \cdots, \alpha_p)$ is a finite increasing sequence from $A$

Recall that we have an augmentation ideal $\mathcal{I} {~\trianglelefteq~}{\mathcal{U}(\mathfrak{h}) }$ and a SES $$\begin{align*} 0 \to\mathcal{I}\to{\mathcal{U}(\mathfrak{g}) }\to k\to 0 .\end{align*}$$ Applying the functor $\mathop{\mathrm{Hom}}_{{\mathcal{U}(\mathfrak{g}) }}({-}, M)$ for a fixed $M\in {\mathfrak{g}{\hbox{-}}\mathsf{Mod}}$ yields a LES:

Link to Diagram

Here the red term vanishes since ${\mathcal{U}(\mathfrak{g}) }$ is free and this projective as a ${\mathfrak{g}}{\hbox{-}}$module. Note that for $n\geq 2$, we have the following situation:

Link to Diagram

Thus we get a degree shifting isomorphism $$\begin{align*} H^n({\mathfrak{g}}; M) \cong \operatorname{Ext}_{{\mathcal{U}(\mathfrak{g}) }}^{n-1}(\mathcal{I}, M) .\end{align*}$$

We thus have $$\begin{align*} H^1({\mathfrak{g}}; M) \cong { \mathop{\mathrm{Hom}}_{\mathcal{U}(\mathfrak{g}) }(\mathcal{I}, M) / \operatorname{im}\qty{ \mathop{\mathrm{Hom}}_{\mathcal{U}(\mathfrak{g}) }({\mathcal{U}(\mathfrak{g}) }, M) \cong M \to \mathop{\mathrm{Hom}}(\mathcal{I}, M) } } .\end{align*}$$ Next goal: to more concretely express all of the terms here as $M{\hbox{-}}$valued derivations on ${\mathfrak{g}}$.

The set of all such maps $\mathop{\mathrm{Der}}({\mathfrak{g}}, M) \leq_{{\mathsf{k}{\hbox{-}}\mathsf{Mod}}} \mathop{\mathrm{Hom}}_{{\mathsf{k}{\hbox{-}}\mathsf{Mod}}}({\mathfrak{g}}, M)$ is a $k{\hbox{-}}$submodule. A special case is taking $M \mathrel{\vcenter{:}}={\mathfrak{g}}$, regarded as a ${\mathfrak{g}}{\hbox{-}}$module using the adjoint representation. In fact, for any $k{\hbox{-}}$algebra (not necessarily associative), we get $$\begin{align*} D(ab) = (Da)\cdot b + a\cdot(Db) .\end{align*}$$ When $A \mathrel{\vcenter{:}}={\mathfrak{g}}\in {\mathsf{Lie}{\hbox{-}}\mathsf{Alg}}$ with the adjoint action, we obtain $$\begin{align*} D([xy]) &= [x, Dy] + [Dx, y] \\ &= [x, Dy] - [y, Dx] \\ &= x\cdot Dy - y\cdot (Dx) ,\end{align*}$$ recovering the previous definition.

Note that this is indeed a derivation: $$\begin{align*} D_m([xy]) ] [xy]\cdot m = x\cdot(y\cdot m) - y\cdot(x\cdot m) = x\cdot (D_m y) - y\cdot (D_m x) .\end{align*}$$ It also turns out that any inner derivation is of this form, bracketing against a fixed element.

Let ${\mathfrak{g}}\in{\mathsf{Lie}{\hbox{-}}\mathsf{Alg}}_{/k}$ and $M\in{\mathfrak{g}{\hbox{-}}\mathsf{Mod}}$, we were showing $$\begin{align*} H^1({\mathfrak{g}}; M) \cong { \mathop{\mathrm{Hom}}_{{\mathfrak{g}{\hbox{-}}\mathsf{Mod}}}( \mathcal{I}, M) \over \operatorname{im}\qty{ \mathop{\mathrm{Hom}}_{{\mathcal{U}(\mathfrak{g}) }}({\mathcal{U}(\mathfrak{g}) }, M) \to \mathop{\mathrm{Hom}}_{{\mathfrak{g}}}( \mathcal{I}, M) } } ,\end{align*}$$ where the source in the denominator is isomorphic to $M$, given by the map $\operatorname{ev}_1$. We found a map $$\begin{align*} \mathop{\mathrm{Hom}}_{\mathfrak{g}}(\mathcal{I}, M) \xrightarrow{\sim} \mathop{\mathrm{Der}}({\mathfrak{g}}, M) \\ \phi \mapsto (D_\phi: x\mapsto \phi(x)) .\end{align*}$$ We also defined inner derivations as those given by maps $D_m(x) \mathrel{\vcenter{:}}= mx$ for some $m\in M$.

If ${\mathfrak{h}}{~\trianglelefteq~}{\mathfrak{g}}$, there is a SES in ${\mathsf{Lie}{\hbox{-}}\mathsf{Alg}}$: $$\begin{align*} 0 \to {\mathfrak{h}}\to {\mathfrak{g}}\to {\mathfrak{g}}/{\mathfrak{h}}\to 0 .\end{align*}$$

This comes from a similar application of the Grothendieck spectral sequence. The exact sequences in low-degree terms are given as usual¹¹ and similar inflation and restriction maps appear here. This is useful because it allows computing homology of “smaller” algebras, which one might have control over by induction.

A computationally efficient way of compute Lie algebra cohomology using a projective resolution of the trivial ${\mathfrak{g}}{\hbox{-}}$module $k\in{\mathfrak{g}{\hbox{-}}\mathsf{Mod}}$, recalling that this involves acting by zero.¹²

We’re going to define a chain complex $$\begin{align*} V_*({\mathfrak{g}}) \xrightarrow[]{\varepsilon} { \mathrel{\mkern-16mu}\rightarrow }\, k ,\end{align*}$$ which will turn out to be supported in finitely many degrees when $\dim_k {\mathfrak{g}}< \infty$.

We’ll assume ${\mathfrak{g}}\in { {\mathsf{Lie}{\hbox{-}}\mathsf{Alg}}_{/k} }(\mathsf{Free})$, which happens e.g. if $k\in \mathsf{Field}$. Recall that the exterior algebra was a graded algebra defined as the quotient of the tensor algebra: $$\begin{align*} \bigwedge^* {\mathfrak{g}}\mathrel{\vcenter{:}}={ T({\mathfrak{g}}) \over \left\langle{x^{\otimes 2} {~\mathrel{\Big|}~}x\in {\mathfrak{g}} }\right\rangle} = \bigoplus _{p\geq 0} \bigwedge^p {\mathfrak{g}} .\end{align*}$$ We write $x_1\wedge x_2\wedge\cdots \wedge x_p$ for the image of ${\mathfrak{g}}\hookrightarrow\bigwedge^* {\mathfrak{g}}$ Note that this is a 2-sided homogeneous ideal, and since $x\wedge x = 0$ we have $x\wedge y = -y\wedge x$.

If $\left\{{ x_\alpha }\right\}$ is an ordered basis for ${\mathfrak{g}}$, then there is an ordered basis for $\bigwedge^p {\mathfrak{g}}$: $$\begin{align*} \left\{{ { {x_{\alpha}}_1, {x_{\alpha}}_2, \cdots, {x_{\alpha}}_{p}} {~\mathrel{\Big|}~}\alpha_1 < \cdots < \alpha_p }\right\} ,\end{align*}$$ where we note that the indices are strictly increasing like the sequences $I$ we had previously. One can always arrange this by commuting things to organize the sequence properly. We also have $\bigwedge^0 {\mathfrak{g}}\cong k$ with a basis of $1_k$, and $\bigwedge^1 {\mathfrak{g}}\cong {\mathfrak{g}}$. In particular, if $\dim {\mathfrak{g}}= n < \infty$, then $\bigwedge^p {\mathfrak{g}}= 0$ for all $p>n$, and in this case $\bigwedge^n {\mathfrak{g}}\cong k$.

We want this to be a projective resolution, so not just that $\ker \subset \operatorname{im}$, but rather we want exactness everywhere so $\ker = \operatorname{im}$. We’ll proceed by showing its homology vanishes.

Recall that ${\mathfrak{g}}$ was free over $k$ with an ordered basis $\left\{{ e_{\alpha} {~\mathrel{\Big|}~}\alpha\in \Omega }\right\}$. We defined $$\begin{align*} V_n({\mathfrak{g}}) \mathrel{\vcenter{:}}={\mathcal{U}(\mathfrak{g}) }\otimes_k \bigwedge^n {\mathfrak{g}} \end{align*}$$ with a differential $d = \theta_1 + \theta_2$. We claimed that $V_n({\mathfrak{g}}) \xrightarrow[]{\varepsilon} { \mathrel{\mkern-16mu}\rightarrow }\, k$ is a projective resolution, and we were showing that $V_*$ was an exact complex.

See section 4.5 on Koszul complex. We’ll do 7.8 next time.

Last time: $V_n({\mathfrak{g}}) \mathrel{\vcenter{:}}={\mathcal{U}(\mathfrak{g}) }\otimes_k \bigwedge^n {\mathfrak{g}} \xrightarrow[]{\varepsilon} { \mathrel{\mkern-16mu}\rightarrow }\, k$ is a projective resolution in ${\mathfrak{g}{\hbox{-}}\mathsf{Mod}}$. Note that we can introduce negative signs to easily interchange $\mathsf{{\mathfrak{g}}}{\hbox{-}}\mathsf{Mod}$ and $\mathsf{Mod}{\hbox{-}}\mathsf{{\mathfrak{g}}}$.

This is very concrete! Standard trick for exterior algebras: any $n{\hbox{-}}$cochain $f\in \mathop{\mathrm{Hom}}_k(\bigwedge^n {\mathfrak{g}}, M)$ can be viewed as an alternating $k{\hbox{-}}$multilinear function $f(x_1, \cdots, x_n): {\mathfrak{g}}\to M$. The cochain differential should increase degree, so we define $$\begin{align*} \Theta_1 f({ {x}_1, {x}_2, \cdots, {x}_{n}}) = \sum_{i=1}^{n+1} (-1)^{i+1} x_i \cdot f(x_1, \cdots, \widehat{x_i}, \cdots, x_n) + \sum_{i < j} (-1)^{i+1} f( [x_i x_j], x_1, \cdots, \widehat{x_i}, \cdots, \widehat{x_j}, \cdots, x_n) .\end{align*}$$ Note that the tor definition has the arguments switched compared to the original definition. This is to set up the tensor cancellation of $\cdots \otimes_{\mathcal{U}(\mathfrak{g}) }{\mathcal{U}(\mathfrak{g}) }\cdots$. Swapping factors and introducing signs makes this work for left ${\mathfrak{g}}{\hbox{-}}$modules.

Public service section since we won’t have a Lie algebras course next Fall. Semisimples: the most important and interesting classes of Lie algebras! These occur frequently and we can prove a lot about them. We’ll assume ${\mathfrak{g}}$ is a finite dimensional Lie algebra over a field $k$, where we’ll soon assume $\operatorname{ch}(k) = 0$.

Recall that ${\mathsf{Lie}{\hbox{-}}\mathsf{Alg}}^{\mathsf{Ab}}\approx {\mathsf{k}{\hbox{-}}\mathsf{Mod}}$ are vector spaces, and so if ${\mathfrak{g}}$ is abelian it automatically has a chain of ideals by just taking vector subspaces. These are closed under brackets since bracketing is zero. So if $\dim_k {\mathfrak{g}}\geq 2$, there are nontrivial ideals, so the abelian condition rules out all 1-dimensional Lie algebras – they’re all abelian by taking a generator, bracketing it with itself, and noting you get zero. So there’s only one 1-dimensional Lie algebra over any field $k$: the abelian one.

The derived algebra $[{\mathfrak{g}}{\mathfrak{g}}] {~\trianglelefteq~}{\mathfrak{g}}$ is a subalgebra and always an ideal, so if ${\mathfrak{g}}$ is simple then $[{\mathfrak{g}}{\mathfrak{g}}] = {\mathfrak{g}}$. So ${\mathfrak{gl}}_n({\mathbb{C}})$ is not simple, since $[{\mathfrak{gl}}_n {\mathfrak{gl}}_n] = {\mathfrak{sl}}_n$ by taking traces.

The vector space sum of any two solvable ideals is again a solvable ideal. Note that this works for products of solvable subgroups $N, H\leq G$ with $N$ normal. Use 2-out-of-3 property for solvable groups and quotient by $N$. By finite-dimensionality, we can find a maximal solvable ideal:

There shouldn’t be any solvable ideals in this quotient, otherwise you could lift. Next up, our most powerful tool for semisimple Lie algebras:

This is a symmetric bilinear form since traces don’t depend on the order of products. It has another nice property, ${\mathfrak{g}}{\hbox{-}}$invariance: $$\begin{align*} \kappa([xy], z] = \kappa(x, [yz]) .\end{align*}$$

These are like “orthogonal” ideals. So we can study semisimple Lie algebras by just studying simple Lie algebras.

Today: filling in some previous things, including proofs for Whitehead’s second lemma and Levi’s theorem.

Given such an extension, thinking of $M \subset E$, $M$ becomes a ${\mathfrak{g}}{\hbox{-}}$module in a natural way: given $m\in M$ and $x\in{\mathfrak{g}}$, choose $\tilde x\in E$ such that $\pi(\tilde x) = x$ and set $$\begin{align*} x\cdot m \mathrel{\vcenter{:}}=[\tilde x, m]_E \in M {~\trianglelefteq~}E ,\end{align*}$$ noting that $M$ is the kernel of a morphism and thus an ideal. Is this well-defined? If $\tilde x'\in \pi^{-1}(E)$, we have $\pi(\tilde x' - \tilde x) = 0$ which implies $\tilde x' -\tilde x\in \ker \pi = M$ by exactness. So we can write $\tilde x' = m' + \tilde x$ for some $m'\in M$, and since $M$ is abelian and its elements bracket to zero, we have $$\begin{align*} [\tilde x ', m] = [m' + \tilde x, m] = [\tilde x, m] .\end{align*}$$

The extension problem: given a ${\mathfrak{g}}{\hbox{-}}$module $M$ viewed as an abelian Lie algebra, how many (equivalence classes of) extensions of ${\mathfrak{g}}$ by $M$ are there for which the induced action above agrees with the given action? Here we view equivalence as existence of an isomorphism making the following diagram commute:

Link to Diagram

Write $\operatorname{Ext}_{{\mathsf{Lie}{\hbox{-}}\mathsf{Alg}}}({\mathfrak{g}}, M)$ for the set of equivalence classes of such extensions.

We can form semidirect products $M\rtimes{\mathfrak{g}}$ of Lie algebras in the following way: start with the $k{\hbox{-}}$module $M \times{\mathfrak{g}}$ with bracket $$\begin{align*} [(m, x), (n, y)] \mathrel{\vcenter{:}}=(x\cdot n - y\cdot m, [xy]) && m,n\in M,\,\, x,y\in{\mathfrak{g}} .\end{align*}$$ One checks that this is anticommutative and satisfies the Jacobi identity. This is a Lie algebra containing $M \times 0$ as an abelian ideal and $0 \times{\mathfrak{g}}$ as a subalgebra, which fits into a SES $$\begin{align*} 0 \to M \xrightarrow{\iota} M\rtimes{\mathfrak{g}}\xrightarrow{\pi} {\mathfrak{g}}\to 0 .\end{align*}$$ Moreover, the naturally induced action described previously agrees with this semidirect action. Identifying elements with their inclusions, we have $$\begin{align*} [(0, x), (m, 0)] = (x\cdot m - 0, [0, 0] ) = (x\cdot m, 0) .\end{align*}$$ Thus there is always at least one extension, called the split extension. There is a classification:

Note that the map $\pi$ makes $M$ into an $E{\hbox{-}}$module and makes $M$ into a trivial $M{\hbox{-}}$module. See Weibel for a functorial proof, using the same correspondence between $\operatorname{Ext}_R^1(A, B)$ and extensions of $A$ by $B$. Note that we have an algebra, an ideal, and its quotient, which is precisely the setup for the LHS spectral sequence for cohomology with coefficients in $M$. There was an associated 5-term exact sequence, which contains a classifying map $$\begin{align*} \mathop{\mathrm{Hom}}_{\mathfrak{g}{\hbox{-}}\mathsf{Mod}}(M, M) &\xrightarrow{d^2} \mathop{\mathrm{Hom}}_{{\mathfrak{g}{\hbox{-}}\mathsf{Mod}}}({\mathfrak{g}}, M) \\ \one_M &\mapsto d^2(\one_M) .\end{align*}$$ One checks that this only depends on the equivalence class of extensions, and turns out to be a bijection. Weibel’s proof uses some facts about free Lie algebras that we haven’t discussed yet, so we’ll instead do a slightly more down-to-earth proof from Knapp’s book using the Koszul complex.

One should check that choices differ by coboundaries, along with a few other things that we’re eliding.

Last time: we were proving the bijection between $H^2({\mathfrak{g}}; M)$ and extensions of ${\mathfrak{g}}$ by $M$ up to equivalence.

This was known much earlier for group cohomology: if $G \in {\mathsf{Grp}}, A\in {\mathsf{G}{\hbox{-}}\mathsf{Mod}}$, there is a bijection $$\begin{align*} \left\{{ 0\to A\to E \xrightarrow{\pi} G \to 1 }\right\} \mathrel{\vcenter{:}}= \left\{{ \text{Equivalence classes of extensions of $G$ by $A$ } }\right\} &\to H^2(G; A) ,\end{align*}$$ where $G$ may not be abelian, and one acts by conjugation instead. Analogy: bracketing is like the differential of conjugation.

Recall Whitehead’s Lemma 2¹⁵ for ${\mathfrak{g}}$ finite-dimensional and semisimple over $\operatorname{ch}(k) = 0$ and $M\in {\mathfrak{g}{\hbox{-}}\mathsf{Mod}}^{\operatorname{fd}}$, then $H^2({\mathfrak{g}}; M) = 0$.

Next time: Levi’s theorem.

Notational conventions:

• Finite direct products: $\bigoplus$
• Cohomological indexing: $C^i, {{\partial}}^i$
• Homological indexing: $C_i, {{\partial}}_i$
• Right-derived functors $R^iF$.
  • Come from left-exact functors
  • Require injective resolutions
  • Extend to the right: $0 \to F(A) \to F(B) \to F(C) \to L_1 F(A) \cdots$
• Left-derived functors $L_i F$.
  • Come from right-exact functors
  • Require projective resolutions
  • Extend to the left: $\cdots L_1F(C) \to F(A) \to F(B) \to F(C) \to 0$
• Colimits:
  • Examples: coproducts, direct limits, cokernels, initial objects, pushouts
  • Commute with left adjoints, i.e. $L(\mathop{\mathrm{colim}}\nolimits F_i) = \mathop{\mathrm{colim}}\nolimits LF_i$.
• Examples of limits:
  • Products, inverse limits, kernels, terminal objects, pullbacks
  • Commute with right adjoints. i.e. $R(\mathop{\mathrm{colim}}\nolimits F_i) = \mathop{\mathrm{colim}}\nolimits RF_i$.