1 Introduction and Background (Tuesday, January 11)


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.

Recall the definition of a left \(R{\hbox{-}}\)module. Some examples:

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*}

Recall the definitions of:

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Some examples:

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 direct product \(\prod_{k\in I} M_k \mathrel{\vcenter{:}}=\left\{{(m_k)_{k\in I} {~\mathrel{\Big\vert}~}m_k\in M_k }\right\}\), the set of all ordered sequences of elements from the \(M_k\), with addition defined pointwise. For the 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 irreducible representation is a simple \(R{\hbox{-}}\)module, and an indecomposable representation is an indecomposable \(R{\hbox{-}}\)module. An \(R{\hbox{-}}\)module is simple iff its only submodules are \(0, M\), and indecomposable iff \(M \not\cong M_1 \oplus M_2\) for any \(M_i\not\cong M\). Note that simple \(\implies\) indecomposable.

Note: is it possible for \(M \cong M \oplus M\)?

Some examples:

Toward homological algebra: free and projective modules. An \(R{\hbox{-}}\)module \(M\) is 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:

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Every \(M\in{\mathsf{R}{\hbox{-}}\mathsf{Mod}}\) is the image of a free \(R{\hbox{-}}\)module: let \(X\mathrel{\vcenter{:}}=\left\{{m_i}\right\}_{i\in I}\) 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:

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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 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.

2 Thursday, January 13

Last time:

Today: projective modules and their resolutions.

See Krull-Schmidt theorem.

Recall the definition of projective modules \(P\) and injective modules \(I\):

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Show that free implies projective using the universal properties, and conclude that every \(R{\hbox{-}}\)module has a projective cover.

Forming projective resolutions: take the minimal \(P_0 \xrightarrow[]{\delta_0} { \mathrel{\mkern-16mu}\rightarrow }\, M\to 0\) such that \(\Omega^1 \mathrel{\vcenter{:}}=\ker \delta_0\) has no projective summands. Continue in such a minimal way:

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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 complexity or 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 \(\dim V_A(M) = c_A(M)\).

Recall the definition of a SES \(\xi: 0\to A \xrightarrow{d_1} B \xrightarrow{d_2} C\to 0\) and show that TFAE:

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 \(\ker d_2 \cap\operatorname{im}s_r = \emptyset\).

It’s not necessarily true that if \(B \cong A \oplus C\) that \(\xi\) splits: consider

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Show that for \(P \in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}\), TFAE:

Show that \(\bigoplus_{i\in I} P_i\) is projective iff each \(P_i\) is projective.

Let \(Q\in{\mathsf{R}{\hbox{-}}\mathsf{Mod}}\) and show TFAE:

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.

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 \(0\to M\hookrightarrow Q\).

See Hungerford or Weibel. Prove it first for \(\mathsf{C} = {\mathsf{Z}{\hbox{-}}\mathsf{Mod}}\). The idea now is to apply \begin{align*} F({-}) \mathrel{\vcenter{:}}=\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

Show that for \(M\in{\mathsf{R}{\hbox{-}}\mathsf{Mod}}\) that \(\mathop{\mathrm{Hom}}_{\mathbb{Z}}(R, M) \cong M\).

Hint: try \(f\mapsto f(1)\).

Next week:

3 Tensor Products (Tuesday, January 18)

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.

The 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

The 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 \(\psi: M\otimes_R N\to P\):

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Uniqueness follows from the standard argument on universal properties:

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Existence: let \(\mathsf{Free}({-}): {\mathsf{Set}}\to {\mathsf{Ab}}{\mathsf{Grp}}\) and \(F\mathrel{\vcenter{:}}=\mathsf{Free}(M\times N)\), then set \(M\otimes_R N \mathrel{\vcenter{:}}= F/G\) where \(G\) is generated by

Then define the map as \begin{align*} \otimes: M\times N\to F \\ (x, y) &\mapsto x\otimes y \mathrel{\vcenter{:}}=(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:

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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*}

Note every \(z\in M\otimes_R N\) is a simple tensor of the form \(z=x\otimes y\)!

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

  1. \((A, B) \neq (C, D)\implies \mathop{\mathrm{Hom}}(A, B)\) and \(\mathop{\mathrm{Hom}}(C, D)\) are disjoint.
  2. Associativity of composition: \((h\circ g)\circ f = h\circ(g\circ f)\)
  3. Identities: \(\exists ! \operatorname{id}_A \in \mathop{\mathrm{Hom}}_{\mathsf{C}}(A, A)\) for all \(A\in \mathsf{C}\).

A subcategory \(\mathsf{D} \leq \mathsf{C}\) is a subclass of objects and morphisms, and is full if \(\mathop{\mathrm{Hom}}_{\mathsf{D}}(A, B) = \mathop{\mathrm{Hom}}_{\mathsf{C}}(A, B)\) for all objects in \(\mathsf{D}\).

Examples of categories:

Examples of fullness:

Recall the definition of covariant and contravariant functors, which requires that \(F(\operatorname{id}_A) = \operatorname{id}_{F(A)}\).

4 Thursday, January 20

RIP Brian Parshall and Fred Cohen… 😔

Recall the definition of a covariant functor. Some examples:

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?

Recall that contravariant functors reverse arrows. Functors with the same variance can be composed.

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 full if \(F_{AB}\) is injective for all \(A, B\in \mathsf{C}\), and faithful if \(F_{AB}\) is surjective for all \(A, B\).

A morphism of functors \(\eta: F\to G\) for \(F,G:\mathsf{C}\to \mathsf{D}\) is a natural transformation: a family of maps \(\eta_A\in \mathop{\mathrm{Hom}}_{\mathsf{D}}(FA, GA)\) satisfying the following naturality condition:

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If \(\eta_A\) is an isomorphism for all \(A\in \mathsf{C}\), then \(\eta\) is a natural isomorphism.

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 \(F\to G\).

Two categories \(\mathsf{C}, \mathsf{D}\) are isomorphic if there are functors \(F, G\) with \(F\circ G = \operatorname{id}_{\mathsf{D}}, G\circ F = \operatorname{id}_{\mathsf{C}}\) equal to the identities. They are equivalent if \(F\circ G, G\circ F\) are instead naturally isomorphic to the identity.

Some examples:

Producing inverse functors can be difficult, so we have the following:

Let \(F:\mathsf{C}\to \mathsf{D}\), then there exists an inverse inducing an equivalence iff

\(\implies\): 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*}

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 \(A \mathrel{\vcenter{:}}= GA'\in \mathsf{C}\) and use that \begin{align*} \mathop{\mathrm{Hom}}_{\mathsf{D}}(A', FGA') \mathrel{\vcenter{:}}=\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 \(FA \cong A'\).

#todo Missed a bit here so this doesn’t make sense as-is!

Let \(R\in \mathsf{Ring}\) and set \(S\mathrel{\vcenter{:}}=\operatorname{Mat}_{n\times n}(R)\), then \({\mathsf{R}{\hbox{-}}\mathsf{Mod}} \simeq{\mathsf{S}{\hbox{-}}\mathsf{Mod}}\).

5 Tuesday, January 25

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 \(F\circ G \cong \operatorname{id}\).

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

  1. \(F\) is faithful, i.e. \(\psi_F\) is injective,
  2. \(F\) is full, i.e. \(\psi_F\) is surjective,
  3. For any \(D\in \mathsf{D}\), there exists a \(C\in \mathsf{C}\) with \(F(C) \cong D\).

Let \(R\in \mathsf{Ring}\) and \(S=\operatorname{Mat}_{n\times n}(R)\), then \({\mathsf{R}{\hbox{-}}\mathsf{Mod}} \simeq{\mathsf{S}{\hbox{-}}\mathsf{Mod}}\).

Define a functor \(F:{\mathsf{R}{\hbox{-}}\mathsf{Mod}} \to {\mathsf{S}{\hbox{-}}\mathsf{Mod}}\) by \(F(M) \mathrel{\vcenter{:}}=\prod_{k\leq n} M\), 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.

\(F\) is fully faithful.

See also Jacobson Basic Algebra Part II p.31.

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 \(M\otimes(\oplus N_\alpha)\).

Define functors \(F,G{\mathsf{R}{\hbox{-}}\mathsf{Mod}} \to{\mathsf{{\mathbb{Z}}}{\hbox{-}}\mathsf{Mod}}\) by \(F({-}) \mathrel{\vcenter{:}}= M\otimes_R ({-})\) and \(G({-}) \mathrel{\vcenter{:}}=({-})\otimes_R N\) on objects, and on morphisms \(f:N\to N'\), set \(F(f) \mathrel{\vcenter{:}}=\operatorname{id}\otimes f\) and similarly for \(G\). Recall the definition of exactness, left-exactness, and right-exactness.

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.

Show that \(M\otimes_R({-})\) and \(({-})\otimes_R N\) are right exact for any \(M, N \in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}\).

Let \(0 \to A \xrightarrow{f} B \xrightarrow{g} C \to 0\) which maps to \(M\otimes A \xrightarrow{\operatorname{id}\otimes f} M\otimes B \xrightarrow{\operatorname{id}\otimes g} C\).

When is \(M\otimes_R ({-})\) exact?

6 Thursday, January 27

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.

\(M \mathrel{\vcenter{:}}=\bigoplus _{\alpha\in I} M_\alpha\) is flat iff \(M_\alpha\) is flat for all \(\alpha\in I\).

\begin{align*} M\otimes_R({-}) \mathrel{\vcenter{:}}=(\bigoplus M_\alpha)\otimes_R ({-}) \cong \bigoplus (M_\alpha \otimes_R ({-}) ) .\end{align*}

Show that projective \(\implies\) flat.

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.

Show that

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 \(0_{AB} \mathrel{\vcenter{:}}= 0_{B} \circ 0_A \in \mathop{\mathrm{Hom}}_{\mathsf{C}}(A, B)\).

7 Tuesday, February 01

A category \(\mathsf{C}\) is additive iff

A morphism: \(k:K\to A\) is monic iff whenever \(g_1, g_2: L\to K\), \(kg_1 = kg_2 \implies g_1 = g_2\):

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Define \(k\) to be epic by reversing the arrows.

Assume \(\mathsf{C}\) has a zero object. Then for \(f:A\to B\), the morphism \(k: K\to A\) is the kernel of \(f\) iff

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 \(G \xrightarrow{g'} \ker f \xhookrightarrow{k} A\).

For \(f: A\to B\), a morphism \(c: B\to C\) is a cokernel of \(f\) iff

For \(\mathsf{C} = {\mathsf{R}{\hbox{-}}\mathsf{Mod}}\) and \(f\in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}(A, B)\), set \(c: B\to B/\operatorname{im}f\).

Show that kernels are unique. Sketch:

\(\mathsf{C}\) is abelian iff \(\mathsf{C}\) is additive and

For \(f\in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}(A, B)\),

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Some notes:

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8 Thursday, February 03

8.1 Projective Resolutions and Chain Maps

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*}

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 \(\widehat{\alpha }= \widehat{\beta}\).

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*}


Let \(M\in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}\). A 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\mathrel{\vcenter{:}}={\varepsilon}} 0 .\end{align*}

In particular, \(d^2 = 0\), but this complex need not be exact. A projective resolution of \(M\) is an exact projective complex in the following sense:

Some projective resolutions:

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For \(\mu \in \mathsf{C}(M, M')\) and \(C \mathrel{\vcenter{:}}=({ {C}_{\scriptscriptstyle \bullet}} , d) \twoheadrightarrow M, C' \mathrel{\vcenter{:}}=({ {C}_{\scriptscriptstyle \bullet}} ', d')\twoheadrightarrow M'\), there is an induced chain map \(\alpha \in \mathsf{Ch}\mathsf{C}(C, C')\). Moreover, any other chain map \(\beta\) is chain homotopic to \(\alpha\).

Note that \(C\) can in fact be any projective complex over \(M\), not necessarily a resolution.

Using that \(C_0\) is projective, there is a lift of the following form:

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Now inductively, we want to construct the following lift:

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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:

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To see that any two such maps are chain homotopic, set \(\gamma \mathrel{\vcenter{:}}=\alpha - \beta\), 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:

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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:

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Thus follows from the fact that \(\operatorname{im}\gamma_n - s_{n-1} d_n \subseteq \ker d_n'\) and projectivity of \(C_n\).

Dually one can construct injective resolutions \(0 \to M \xhookrightarrow{\eta} { {D}_{\scriptscriptstyle \bullet}}\)

8.2 Derived Functors

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 \((L_n F)(M) \mathrel{\vcenter{:}}= H_n(F({ {C}_{\scriptscriptstyle \bullet}} ))\).

9 Tuesday, February 08

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 \mathrel{\vcenter{:}}= H_n F{ {C}_{\scriptscriptstyle \bullet}} .\end{align*}

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.

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Show that any two lifts \(\alpha, \alpha'\) induce the same map on homology.

Similarly, \({\mathbb{L}}F(M)\) does not depend on the choice of resolution:

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For \(0\to M' \to M\to M'' \to 0\) in \(\mathsf{C}\), a projective resolution is a collection of chain maps forming projective resolutions of each of the constituent modules:

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Show that such resolutions exist. This involves constructing \({\varepsilon}: C_0 \to M\):

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The claim is that \({\varepsilon}(x, x'') \mathrel{\vcenter{:}}=\gamma {\varepsilon}'(x') + {\varepsilon}^*(x'')\) works. To prove surjectivity, use the following:

Given a commutative diagram of the following form

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If \(g,h\) are mono (resp. epi, resp. iso) then \(f\) is mono (resp. epi, resp. iso).

The setup: