1 Introduction and Background (Tuesday, January 11)

References: [1].

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*}

Roadmap:

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:

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This is exact and commutative by a diagram chase:

To show exactness along the top line:

For \(F: {\mathsf{R}{\hbox{-}}\mathsf{Mod}}\to{\mathbb{Z}{\hbox{-}}\mathsf{Mod}}\) additive and a SES \begin{align*} \xi: 0\to M' \xrightarrow{f} M \xrightarrow{g} M'' \to 0 ,\end{align*}

note that there are morphisms \begin{align*} {\mathbb{L}}F M'' \to {\mathbb{L}}F M\to {\mathbb{L}}FM' .\end{align*}

There is a connecting morphism \begin{align*} \Delta: {\mathbb{L}}F M'' \to \Sigma^{-1} {\mathbb{L}}F M' ,\end{align*} which in components looks like

Link to Diagram

10 Thursday, February 10

Missed! Please send me notes. :)

11 Tuesday, February 15

12 Tuesday, February 22

12.1 Prime Ideals

Plan: commutative ring theory, aiming toward tensor triangular geometry.

Let \(A, I_j \in \operatorname{Id}(R)\) where at most two of the \(I_j\) are not prime and \(A \subseteq \displaystyle\bigcup_j I_j\). Then \(A \subseteq I_j\) for some \(j\).

The case \(n=1\) is clear. For \(n>1\), if \(A \subseteq \tilde I_k \mathrel{\vcenter{:}}= I_1 \cup I_2 \cup\cdots \widehat{I}_k \cup\cdots \cup I_n\) then the result holds by the IH. So suppose \(A \not\subseteq \tilde I_k\) and pick some \(a_k \not\in \tilde I_k\). Since \(A \subseteq \displaystyle\bigcup I_j\), we must have \(a_k\in I_k\).

Case 1: \(n=2\). If \(a_1 + a_2\in A\) with \(a_1 \in I_1 \setminus I_2\) and \(a_2\in I_2\setminus I_1\), then \(a_1 + a_2\not\in I_1 \cup I_2\) – otherwise \(a_1 + a_2 \in I_1 \implies a_2\in I_1\), and similarly if \(a_1 + a_2\in I_2\). So \(A \subseteq I_1 \cup I_2\).

Case 2: \(n>2\). At least one \(I_j\) is prime, without loss of generality \(I_1\). However, \(a_1 + a_2a_3\cdots a_n\in A \setminus\displaystyle\bigcup_{j\geq 1} I_j\). Since \(a_j\in I_j\), we have \(a_2\cdots a_n \in I_j\), contradicting \(a_1\not\in I_j\) for \(j\neq 1\).

Let \(S\leq (R, \cdot)\) be a submonoid and \(P\in \operatorname{Id}(R)\) proper with \(P \cap S = \emptyset\) and \(P\) is maximal with respect to this property, so if \(P' \supseteq P\) and \(P' \cap S = \emptyset\) then \(P' = P\). Then \(P\in \operatorname{Spec}R\) is prime.

By contrapositive, we’ll show \(a,b\not\in P \implies ab\not\in P\). If \(a,b\not\in P\), then \(P \subsetneq aR + P, bR + P\) is a proper subset. By maximality, \((aR + P) \cap S \neq \emptyset\) and \((bR + P) \cap S \neq \emptyset\). Pick \(s_1, s_2\in S\) with \(s_1 = x_1 a + p_1, s_2 = x_2 b + p_2\). Then \(s_1 s_2\in S\) and thus \begin{align*} s_1 s_2 = x_1x_2 ab + x_1 ap_2 + x_2 b p_1 + p_1 p_2\in x_1x_2 ab + P + P + P ,\end{align*} hence \(ab\not\in P\) – otherwise \(S \cap P \neq \emptyset\). \(\contradiction\)

Let \(S \leq R\) be a monoid and let \(I \in \operatorname{Id}(R)\) with \(I \cap S = \emptyset\). Then there exists some \(p\in \operatorname{Spec}R\) such that

Set \(B = \left\{{I' \supseteq I {~\mathrel{\Big\vert}~}I' \cap S = \emptyset}\right\}\), then \(B \neq \emptyset\). Apply Zorn’s lemma to get a maximal element \(p\), which is prime by the previous proposition.

\begin{align*} {\sqrt{0_{R}} } = \cap_{p\in \operatorname{Spec}R} p .\end{align*}

Prove this!

12.2 Localization

Recall the definition of \({\mathbb{Q}}\) as \({\mathbb{Z}}{ \left[ { \scriptstyle \frac{1}{S} } \right] }\) where \(S = {\mathbb{Z}}\setminus\left\{{0}\right\}\) using the arithmetic of fractions. More generally, for \(D\) an integral domain, there is a field of fractions \(F\) with \(D \hookrightarrow F\) satisfying a universal property and thus uniqueness. Recall the definition of localization and the universal property: if \(\eta: R\to R'\) with \(\eta(S) \subseteq (R')^{\times}\) then \(\exists \tilde\eta: R \left[ { \scriptstyle { {S}^{-1}} } \right] \to R'\).

Next time:

13 Tuesday, March 01

Recall the definition of the localization of an \(R\in \mathsf{CRing}^{\operatorname{unital}}\) at a submonoid \(S \leq (M, \cdot)\), written \(R \left[ { \scriptstyle { {S}^{-1}} } \right]\). Similarly for \(M\in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}\), one can form \(M \left[ { \scriptstyle { {S}^{-1}} } \right]\), and \(({-}) \left[ { \scriptstyle { {S}^{-1}} } \right]\) is a functor where the induced map on \(M \xrightarrow{f} N\) is \(f_S(m/s) \mathrel{\vcenter{:}}= f(m)/s\).

For \(I\in \operatorname{Id}(R)\), let \(j(I) \mathrel{\vcenter{:}}=\left\{{a\in R{~\mathrel{\Big\vert}~}a/s\in I \text{ for some } s\in S}\right\}\) which is again an ideal in \(R\). Then

  1. \(j(I)_S = I\),
  2. \(I_S = R_S \iff I\) contains an element of \(S\).

\(\impliedby\): \(I_S \subseteq R_S\) is clear. Let \(x/t\in R_S\) and \(s\in I \cap S\), then \({sx\over st} = {x\over t}\in I_S\).

\(\implies\): Write \(1=i/s\) to produce \(t\in s\) with \(t(s-i) = 0\). Then \(z=ts \in S\) and \(z=it\in I\) so \(z \in I \cap S\).

Let \(P\in \operatorname{Spec}R\) with \(S \cap p = \emptyset\), then \(j(P_S) = P\).

\(\supseteq\): Clear.

\(\subseteq\): Let \(a\in j(P_S)\), so \(a/s=p/t\) for \(s,t\in S, p\in P\) and \(\exists u\in S\) such that \(u(at-sp)=0\in P\), so \(uat - usp\in P\) where \(usp\in P\). Thus \(uat\in P \implies a(ut)\in P\implies a\in P\), since \(ut\in S\) and \(ut\not\in P\).

There is an order-preserving correspondence \begin{align*} \left\{{p\in \operatorname{Spec}R {~\mathrel{\Big\vert}~}p \cap S = \emptyset}\right\} &\rightleftharpoons\operatorname{Spec}R \left[ { \scriptstyle { {S}^{-1}} } \right] \\ P &\mapsto P \left[ { \scriptstyle { {S}^{-1}} } \right] \\ j(P') &\mapsfrom P' .\end{align*}

We need to show

  1. \(P \left[ { \scriptstyle { {S}^{-1}} } \right] \in \operatorname{Spec}R \left[ { \scriptstyle { {S}^{-1}} } \right]\) is actually prime.
  2. If \(P'\in \operatorname{Spec}R \left[ { \scriptstyle { {S}^{-1}} } \right]\) then \(j(P')\in \operatorname{Spec}R\) with \(j(P') \cap S = \emptyset\).

For one: \begin{align*} {x\over t}, {y\over t} \in P_S &\implies {xy\over st} \in P_S \\ &\implies xy \in j(P_S) = P \\ &\implies x\in P \text{ or } y\in P \\ &\implies x/s\in P \text{ or } y/s\in P .\end{align*}

For two: \begin{align*} xy\in j(P') &\implies {xy\over s}\in P' \\ &\implies {x\over 1}{y\over s}\in P' \\ &\implies {x\over 1}\in P' \text{ or } {y\over s}\in P' \\ &\implies {x}\in P' \text{ or } {y}\in P' \\ .\end{align*}

If \(x\in j(P') \cap S\) then \({x\over t}\in P'\) so \({t\over x}{x\over t}\in P'\). \(\contradiction\)

One can then check that these two maps compose to the identity.

Show that if \(p\in \operatorname{Spec}R\) then \(R_p \in \mathsf{Loc}\mathsf{Ring}\) is local. Use that the image of \(p\) in \(R_p\) is \(P_p = R_p\setminus R_p^{\times}\), making it maximal and unique.

Show that

  1. \(M=0 \iff M_S = 0\) for all \(S\),
  2. \(M=0 \iff M_p = 0\, \forall p\in \operatorname{mSpec}R\),
  3. \(M=0 \iff M_p = 0\, \forall p\in \operatorname{Spec}R\), noting that this is a stronger condition than maximal.

For (2), use that \(\operatorname{Ann}_R(x)\) is a proper ideal and thus contained in a maximal, and show by contradiction that \(x/1\neq 0\in M_p\).

Show that if \(f\in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}(M, N)\) then

Recall that for \(A \subseteq R\), \(V(A) \mathrel{\vcenter{:}}=\left\{{p\in \operatorname{Spec}R{~\mathrel{\Big\vert}~}p\supseteq A}\right\}\). Letting \(I(A)\) be the ideal generated by \(A\), then check that \(V(I(A)) = V(A)\) and \(V(I) = V(\sqrt I)\).

Check that defining closed sets as \(\left\{{V(A) {~\mathrel{\Big\vert}~}A \subseteq R}\right\}\) forms the basis for a topology on \(\operatorname{Spec}R\), and \(V(p) \cap V(q) = V(pq)\).

Next time: generic points, idempotents, irreducible sets.

14 Tuesday, March 15

See https://www.math.ucla.edu/~balmer/Pubfile/TTG.pdf

Recall that \(V(B) \mathrel{\vcenter{:}}=\left\{{p\in \operatorname{Spec}R {~\mathrel{\Big\vert}~}p\supseteq B}\right\}\) are the closed sets for the Zariski topology, and \(V(B) = V(\left\langle{B}\right\rangle)\). Write \(I(A) = \displaystyle\bigcap_{p\in A} p\) for the vanishing ideal of \(A\), and note \(V(I(A)) = { \operatorname{cl}} _{\operatorname{Spec}R} A\). Recall \(\sqrt{J} = \displaystyle\bigcap_{p\supseteq J} = \left\{{x\in R {~\mathrel{\Big\vert}~}\exists n\, \text{ such that } x^n \in J}\right\}\), so \(\sqrt{0}\) is the nilradical, i.e. all nilpotent elements. An ideal \(J\) is radical iff \(\sqrt J = J\).

For \(X=\operatorname{Spec}R\), \(I(V(J)) = \sqrt{J}\), and there is a bijection between closed subsets of \(X\) and radical ideals in \(R\).

\begin{align*} I(V(J)) = \displaystyle\bigcap_{p\in V(J)} p = \displaystyle\bigcap_{p\supseteq J} p = \sqrt{J} ,\end{align*} and \begin{align*} J \xrightarrow{V} V(J) \xrightarrow{I} I(V(J)) = \sqrt{J} = J .\end{align*}

Recall that \(X\) is reducible iff \(X= X_1 \cup X_2\) with \(X_i\) nonempty proper and closed.

For \(R\in \mathsf{CRing}\), a closed subset \(A \subseteq X\) is irreducible iff \(I(A)\) is a prime ideal.

\(\implies\): Suppose \(A\) is irreducible, let \(fg\in I(A) = \displaystyle\bigcap_{p\in A} p\). Then \(fg\in p\implies f\in [\) without loss of generality for all \(p\in A\), and \(A = (A \cap V(f)) \cup(A \cap V(g))\) so \(A \subseteq V(f)\) or \(A \subseteq V(g)\). Thus \(f\in \sqrt{\left\langle{f}\right\rangle} = I(V(f)) \subseteq I(A)\) (similarly for \(g\)).

\(\impliedby\): Suppose \(I(A)\) is a prime ideal and \(A = A_1 \cup A_2\) with \(A_j\) closed, so \(I(A) \subseteq I(A_j)\). Then \begin{align*} I(A) = I(A_1 \cup A_2) = I(A_1) \cap I(A_2) .\end{align*} If \(I(A_j) \subsetneq I(A)\) are proper containments, then one reaches a contradiction: if \(x\in I(A_1)\) and \(y\in I(A_2)\), use that \(xy\in I(A)\) to conclude \(x\in I(A)\) or \(y\in I(A)\).

Let \(X\in {\mathsf{Top}}\); TFAE:

  1. \(X\) is irreducible.
  2. Any two open nonempty sets intersect.
  3. Any nonempty open is dense in \(X\).
  1. Any irreducible subset of \(X\) is entirely contained in a single irreducible component.

  2. Any space is a union of its irreducible components.

Coming up:

15 Tuesday, March 22

15.1 Hilbert-Serre

Setup: \(V\in {\mathsf{gr}\,}_{\mathbb{Z}}{\mathsf{k}{\hbox{-}}\mathsf{Mod}}\) a graded vector space, so \(V = \bigoplus _{r\geq 0} V_r\) with \(\dim_k V_r < \infty\). Define the Poincare series \begin{align*} p(V, t) = \sum_{r\geq 0} \dim V_r t^r .\end{align*}

Let \(R\in {\mathsf{gr}\,}_{\mathbb{Z}}\mathsf{CRing}\) be of finite type over \(A_0\) for \(A\in {{k}{\hbox{-}}\mathsf{Alg}}\) and suppose \(R\) is finitely generated over \(A_0\) by homogeneous elements of degrees \(k_1,\cdots, k_s\). Supposing \(V\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\), \begin{align*} p(V, t) = {f(t) \over \prod_{1\leq j\leq s} 1-t^{k_j} }, \qquad f(t) \in {\mathbb{Z}}[t] .\end{align*}

Suppose that \begin{align*} p(V, t) = {f(t) \over \prod_{1\leq j\leq s} 1-t^{k_j} } = \sum_{r\geq 0} a_r t^r, \qquad f(t) \in {\mathbb{Z}}[t], a_r\in {\mathbb{Z}}_{\geq 0} .\end{align*} Let \(\gamma\) be the order of the pole of \(p(t)\) at \(t=1\). Then

  1. There exists \(K > 0\) such that \(a_n \leq K n^{\gamma-1}\) for \(n\geq 0\)

  2. There does not exist \(k > 0\) such that \(a_n \leq k n^{\gamma - 2}\).

Let \(V\) be a graded vector space of finite type over \(k\). The rate of growth \(\gamma(V)\) of \(V\) is the smallest \(\gamma\) such that \(\dim V_n \leq C n^{\gamma-1}\) for all \(n\geq 0\) for some constant \(C\).

Compare this to the complexity \(C_G(M) = \gamma(P_0)\) where \(P^0 \rightrightarrows M\) is a minimal projective resolution.

15.2 Finite Generation of Cohomology

Fix \(G \in {\mathsf{Fin}}{\mathsf{Grp}}\). Recall that \({ {H}^{\scriptscriptstyle \bullet}} (G; k) { {\operatorname{Ext} }^{\scriptscriptstyle \bullet}} _{G}(k, k)\) has an algebra structure given by concatenation of LESs:

Link to Diagram

Recall that \(\operatorname{Ext} ^n_{G}(k, k) = \mathop{\mathrm{Hom}}_{kG}(P_n, k)\), providing the additive structure. Moreover, \(\operatorname{Ext} _{kG}(M, M)\) is a ring, and if \(N\in {\mathsf{kG}{\hbox{-}}\mathsf{Mod}}\), then \(\operatorname{Ext} _{kG}{N, M} \in {\mathsf{\operatorname{Ext} _{kG}(M, M)}{\hbox{-}}\mathsf{Mod}}\). Similarly \(\operatorname{Ext} ^0_{kG}(N, M) \in {\mathsf{ { {\operatorname{Ext} }^{\scriptscriptstyle \bullet}} (k, k)}{\hbox{-}}\mathsf{Mod}}\) by tensoring LESs.

There is a coproduct \begin{align*} kG &\xrightarrow{\Delta} kG \otimes_k kG \\ g &\mapsto g\otimes g .\end{align*} There is a cup product:

Link to Diagram

It is a theorem that this coincides with the Yoneda product.

Quillen described \(\operatorname{mSpec} { {H}^{\scriptscriptstyle \bullet}} (G, k)^{ \text{red} }\) in the 70s. Idea: look at \(E \hookrightarrow G\) the elementary abelian subgroups, so \(E \cong C_p{ {}^{ \scriptscriptstyle\times^{m} } }\) where \(p = \operatorname{ch}k\), and consider \(V_G(k) = \displaystyle\bigcup_{E\leq G} V_E(k)/\sim\) the union of all elementary abelian subgroups, where \(V_G(k) \mathrel{\vcenter{:}}=\operatorname{mSpec} { {H}^{\scriptscriptstyle \bullet}} ^{}(G; k)^{ \text{red} }\). Note that in characteristic zero, this is semisimple and only \(H^0=k\) survives.

16 Tuesday, March 29

Setup: for \(G \in {\mathsf{Fin}}{\mathsf{Grp}}, k\in \mathsf{Field}\) with \(\operatorname{ch}k = p \divides {\sharp}G\). For \(M\in {\mathsf{kG}{\hbox{-}}\mathsf{Mod}}\), we associate \(V_G(M) \subseteq \operatorname{mSpec}(R)\) for \(R\mathrel{\vcenter{:}}= H^0(G; k)\). There is a ring morphism \(\Phi_M: R\to \operatorname{Ext} ^0_{kG}(M, M)\), we set \(I_G(M) = \left\{{x\in R {~\mathrel{\Big\vert}~}\Phi_M(x) = 0}\right\}\) and define the support variety as \(V_G(M) = \operatorname{mSpec}(R/I_G(M))\).

Let \(G = C_p{ {}^{ \scriptscriptstyle\times^{n} } }\), then

16.1 Rank Varieties

For \(kG = k[z_1,\cdots, z_n]/\left\langle{z_1^p,\cdots, z_n^p}\right\rangle\), let \(x_{\mathbf{a}} \mathrel{\vcenter{:}}=\sum a_i z_i\) for \(a_i\in k\). Define the rank variety \begin{align*} V_E^r(M) = \left\{{\mathbf{a} {~\mathrel{\Big\vert}~}\mathop{\mathrm{Res}}^{kG}_{ \left\langle{x_{\mathbf{a}}}\right\rangle } \text{ is not free} }\right\} \cup\left\{{0}\right\} .\end{align*}

\begin{align*} V_E(M) \cong V_E^r(M) .\end{align*}

Note that \(\operatorname{Ext} ^0(M, M)\curvearrowright\operatorname{Ext} ^0(M', M)\) by splicing, so we can define \(I_G(M', M) \mathrel{\vcenter{:}}=\operatorname{Ann}_R \operatorname{Ext} _{kG}^1(M', M)\) and the relative support variety \(V_G(M', M) = \operatorname{mSpec}(R/ I_G(M', M))\). This recovers the previous notion by \(V_G(M, M) = V_G(M)\).

Since \(I_G(M', M) \supseteq I_G(M) + I_G(M')\), \begin{align*} V_G(M', M) \subseteq V_G(M) \cap V_G(M') ,\end{align*} which relates relative support varieties to the usual support varieties.

If \(0\to A\to B\to C\to 0\) is a SES, there is a LES in \(\operatorname{Ext} _{kG}\) and by considering annihilators we have \begin{align*} I_G(A, M)\cdot I_G(B, M) \subseteq I_G(C, M) \implies V_G(C, M) \subseteq V_G(A, M)\cup V_G(C, M) .\end{align*}

Let \(M\in \mathsf{kG}{\hbox{-}}\mathsf{Mod}\), then \begin{align*} V_G(M) \subseteq \displaystyle\bigcup_{S\leq M \text{ simple}} V_G(S, M) .\end{align*}

Take the SES \(0\to S_1 \to M \to M/S_1\to 0\), then \(V_G(M) = V_G(M, M) \subseteq V_G(S_1, M) \cup V_G(M/S_1, M)\). Continuing this way yields a union of \(V(T, M)\) over all composition factors \(T\). Conversely, by the intersection formula above, this union is contained in \(V_G(M)\), so these are all equal.

Let \(M \in {\mathsf{kG}{\hbox{-}}\mathsf{Mod}}\), then

  1. \(c_G(M) = \dim V_G(M)\)
  2. \(V_G(M) = \left\{{0}\right\}\) (as a conical varieties) iff \(M\) is projective.

Note (2) follows from (1), since complexity zero modules are precisely projectives. Consider \(\Phi_M: R\to \operatorname{Ext} ^0_{kG}(M, M)\), which induces \(R/I_G(M) \hookrightarrow\operatorname{Ext} _{kG}^0(M, M)\) which is finitely generated over \(R/I_{G}(M)\). A computation shows \begin{align*} c_G(M) &= \gamma(\operatorname{Ext} _{kG}^0(M, M)) \\ &= \gamma( R/I_G(M) ) \\ &= \operatorname{krulldim}(R/I_G(M)) \\ &= \dim V_G(M) .\end{align*}

Consider a LES \(0\to M\to M_1\to \cdots \to M_n \to M\to 0 \in \operatorname{Ext} _{kG}^n(M, M)\). Apply \(\Omega^n({-})\), which arises from projective covers \({ {P}^{\scriptscriptstyle \bullet}} \rightrightarrows M\) and truncating to get \(0\to \Omega^n \to P^{n-1}\to \cdots \to P_0 \to M\to 0\). Similarly define \(\Omega^{-n}\) in terms of injective resolutions. There is an isomorphism \(\operatorname{Ext} _{kG}^n(M, M) \cong \operatorname{Ext} _{kG}^n(\Omega^s M, \Omega^s M)\) which is compatible with the \(R\) action. Thus \(V_G(M) \cong V_G (\Omega^s M)\) for any \(s\). Since \(kG\) is a Hopf algebra, dualizing yields \(\operatorname{Ext} _{kG}^n(M, M) \cong \operatorname{Ext} _{kG}^n(M {}^{ \vee }, M {}^{ \vee })\) and thus \(V_G(M) \cong V_G(M {}^{ \vee })\).

16.2 Properties of support varieties

\begin{align*} V_G(M_1 \bigoplus M_2) \cong V_G(M_1)\cup V_G(M_2) .\end{align*}

Distribute: \begin{align*} \operatorname{Ext} _{kG}^0(M_1 \oplus M_2, M_1 \oplus M_2) & \cong \operatorname{Ext} _{kG}^0(M_1, M_1) \oplus \operatorname{Ext} _{kG}^0(M_1, M_2) \oplus \operatorname{Ext} _{kG}^0(M_2, M_1) \oplus \operatorname{Ext} _{kG}^0(M_3, M_2) .\end{align*}

Now \(I_G(M_1 \bigoplus M_2) \subseteq I_G(M_1) \oplus I_G(M_2)\), so \(V_G(M_1) \cup V_G(M_2) \subseteq V_G(M_1 \oplus M_2)\). Applying the 2 out of 3 property, \(V_G(M_1 \oplus M_2) \subseteq V_G(M_1) \cup V_G(M_2)\) since there is a SES \(0\to M_1 \to M_1 \oplus M_2 \to M_2\to 0\).

Let \(M, N\in {\mathsf{kG}{\hbox{-}}\mathsf{Mod}}\), then \begin{align*} V_G(M\otimes_k N) = V_G(M) \cap V_G(N) .\end{align*}

Conjectured by Carlson, proved by Arvrunin-Scott (82). Prove for elementary abelians, piece together using the Quillen stratification.

Let \(X = \operatorname{mSpec}R\), which is a conical variety, and let \(W \subseteq X\) be a closed conical subvariety (e.g. a line through the origin). Then there exists an \(M\in {\mathsf{kG}{\hbox{-}}\mathsf{Mod}}\) such that \(V_G(M) = W\).

Take \(\zeta: \Omega^n k \to k\), so \(\zeta\in R/I_G(M)\), and define certain \(L_\zeta\) modules and set \(Z(\zeta) \mathrel{\vcenter{:}}= V_G(L_\zeta)\).

Let \(M \in {\mathsf{kG}{\hbox{-}}\mathsf{Mod}}\) be indecomposable. Then the projectivization \(\mathop{\mathrm{Proj}}V_G(M)\) is connected.

16.3 Supports using primes

As before, set \(R = H^{\text{even}}(G; k), X= \operatorname{Spec}R\), and now define \begin{align*} V_G(M) = \left\{{p\in X{~\mathrel{\Big\vert}~}\operatorname{Ext} _{kG}^0(M, M)_p \neq 0}\right\} .\end{align*} All of the theorems mentioned today go through with this new definition.

Let \(I_G(M) = \operatorname{Ann}_R \operatorname{Ext} _{kG}^0(M, M) {~\trianglelefteq~}R\), and show \begin{align*} V_G(M) = \left\{{p\in X{~\mathrel{\Big\vert}~}p\supseteq I_G(M) }\right\} = V(I_G(M)) \end{align*} is a closed set.

Let \({\mathfrak{g}}\in \mathsf{Lie}{\mathsf{Alg}}_{/ {k}}\) with \(\operatorname{ch}k = p > 0\), e.g. \({\mathfrak{g}}= {\mathfrak{gl}}_n(k)\). Then there is a \(p\)th power operation \(x^{{\left\lceil p \right\rceil}} = x\cdot x\cdots x\). The pair \(({\mathfrak{g}}, {\left\lceil p \right\rceil})\) forms a restricted Lie algebra. Consider the enveloping algebra \(U({\mathfrak{g}})\), and define \begin{align*} u({\mathfrak{g}}) \mathrel{\vcenter{:}}= U({\mathfrak{g}})/ \left\langle{x^p - x{ {}^{ \scriptstyle\otimes_{k}^{p} } } {~\mathrel{\Big\vert}~}x\in {\mathfrak{g}}}\right\rangle ,\end{align*} which is a finite-dimensional Hopf algebra:

The dimension is given by \(\dim u({\mathfrak{g}}) = p^{\dim {\mathfrak{g}}}\).

17 Tuesday, April 05

17.1 Lie Theory

Setup: \(k = { \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu }\), \(\operatorname{ch}k = p > 0\), \({\mathfrak{g}}\) a restricted Lie algebra (e.g. \({\mathfrak{g}}= \mathsf{Lie}(G)\) for \(G\in{\mathsf{Aff}}{\mathsf{Alg}}{\mathsf{Grp}}_{/ {k}}\)). Write \(A^{{\left\lceil p \right\rceil} } = AA\cdots A\) and set \(A = u({\mathfrak{g}}) = U({\mathfrak{g}})/ J\) where \(J = \left\langle{x{ {}^{ \scriptstyle\otimes_{k}^{p} } } - x^{{\left\lceil p \right\rceil}}}\right\rangle\) which is an ideal generated by central elements. Note that \(A\) is a finite-dimensional Hopf algebra.

Proved last time: \(H^0(A; k) \in {\mathsf{Alg}_{/k} }^{\mathrm{fg}}\), using a spectral sequence argument. From the spectral sequence, there is a finite morphism \begin{align*} \Phi: S({\mathfrak{g}}^+)^{(1)} \to H^0(A; k) ,\end{align*} making \(H^0(A; k)\) an integral extension of \(\operatorname{im}\Phi\). This induces a map \begin{align*} \Phi: \operatorname{mSpec}H^0(A; k) \hookrightarrow{\mathfrak{g}} .\end{align*}

\begin{align*} \operatorname{mSpec}H^0(A; k) \cong {\mathcal{N}}_p \mathrel{\vcenter{:}}=\left\{{x\in {\mathfrak{g}}{~\mathrel{\Big\vert}~}x^{{\left\lceil p \right\rceil}}}\right\} .\end{align*}

For \({\mathfrak{g}}= {\mathfrak{gl}}_n\), \({\mathcal{N}}_p \leq {\mathcal{N}}\) is a subvariety of the nilpotent cone. Moreover \({\mathcal{N}}_p\) is stable under \(G = \operatorname{GL}_n\), and there are only finitely many orbits. There is a decomposition into finitely many irreducible orbit closures \begin{align*} {\mathcal{N}}_p = \displaystyle\bigcup_i \mkern 1.5mu\overline{\mkern-1.5muGx_i\mkern-1.5mu}\mkern 1.5mu .\end{align*} This corresponds to Jordan decompositions with blocks of size at most \(p\).

Using spectral sequences one can show that if \(M, N \in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}\) then \(\operatorname{Ext} ^0_A(M, N)\) is finitely-generated as a module over \(R\mathrel{\vcenter{:}}= H^0(A; k)\). So one can define support varieties \(V_{{\mathfrak{g}}}(M) = \operatorname{mSpec}R/J_M\) where \(I_M = \operatorname{Ann}_R \operatorname{Ext} ^0_A(M, M)\). Some facts:

Given \(M\in {\mathsf{u({\mathfrak{g}})}{\hbox{-}}\mathsf{Mod}}\), \begin{align*} V_{{\mathfrak{g}}}(M) \cong \left\{{x\in {\mathfrak{g}}{~\mathrel{\Big\vert}~}x^{[p]} = 0, M \downarrow_{U(\left\langle{x}\right\rangle)} \text{ is not free over } u(\left\langle{x}\right\rangle) \leq u({\mathfrak{g}}) }\right\} \cup\left\{{0}\right\} ,\end{align*} which is similar to the rank variety for finite groups, concretely realize the support variety.

Here \(\left\langle{x}\right\rangle = kx\) is a 1-dimensional Lie algebra, and if \(x^{[p]} = 0\) then \(u(\left\langle{x}\right\rangle) = k[x] / \left\langle{x^p}\right\rangle\) is a PID. We know how to classify modules over a PID: there are only finitely many indecomposable such modules.

17.2 Reductive algebraic groups

For type \(A_n \sim \operatorname{GL}_{n+1}\), \(\alpha_0 = \tilde \alpha_n = \sum_{1\leq i \leq n} \alpha_i\) and \(h=n+1\). For \({\mathsf{G}}_2\), \(\tilde \alpha_n = 3\alpha_1 + 2\alpha_2\) and \(h=6\).

If \(p\geq h\) then \({\mathcal{N}}_p({\mathfrak{g}}) = {\mathcal{N}}\).

A prime is bad if it divides any coefficient of the highest weight. By type:

Type Bad primes
\(A_n\) None
\(B_n\) 2
\(C_n\) 2
\(D_n\) 2
\(E_6\) 2,3
\(E_7\) 2,3
\(E_8\) 2,3,5
\(F_4\) 2,3
\(G_2\) 2,3

\({\mathcal{N}}_p = \mkern 1.5mu\overline{\mkern-1.5mu{\mathcal{O}}\mkern-1.5mu}\mkern 1.5mu\) is an orbit closure, where \({\mathcal{O}}\) is a \(G{\hbox{-}}\)orbit in \({\mathcal{N}}\). Hence \({\mathcal{N}}_p({\mathfrak{g}})\) is an irreducible variety.

Let \(X = X(T)\) be the weight lattice and let \(\lambda \in X\), then \begin{align*} \Phi_\lambda \mathrel{\vcenter{:}}=\left\{{ \alpha\in \Phi {~\mathrel{\Big\vert}~}{\left\langle {\lambda + \rho},~{\alpha {}^{ \vee }} \right\rangle} \in p{\mathbb{Z}}}\right\} .\end{align*} Under the action of the affine Weyl group, this is empty when \(\lambda\) is on a wall (non-regular) and otherwise contains some roots for regular weights. When \(p\) is a good prime, there exists a \(w\in W\) with \(w(\Phi_\lambda) = \Phi_J\) for \(J \subseteq \Delta\) a subsystem of simple roots. In this case, there is a Levi decomposition \begin{align*} {\mathfrak{g}}= u_J \oplus \ell_J \oplus u_J^+ .\end{align*}

On Levis: consider type \(A_5 \sim \operatorname{GL}_6\) with simple roots \(\alpha_i\).

Consider induced/costandard modules \(H^0( \lambda) = \operatorname{Ind}_B^G \lambda = \nabla(\lambda)\), which are nonzero only when \(\lambda \in X_+\) is a dominant weight. Their characters are given by Weyl’s character formula, and their duals are essentially Weyl modules which admit Weyl filtrations. What are their support varieties?

Let \(\lambda\in X_+\) and let \(p\) be a good prime, and let \(w\in W\) such that \(w(\Phi_\lambda ) = \Phi_J\) for \(J \subseteq \Delta\). Then \begin{align*} V_{{\mathfrak{g}}} H^0( \lambda) = G\cdot u_J = \mkern 1.5mu\overline{\mkern-1.5mu{\mathcal{O}}\mkern-1.5mu}\mkern 1.5mu \end{align*} is the closure of a “Richardson orbit.”

Natural progression: what about tilting modules (good filtrations with costandard sections and good + Weyl filtrations)? We’re aiming for the Humphreys conjecture.

Let \(T( \lambda)\) be a tilting module for \(\lambda \in X_+\). A conjecture of Humphreys: \(V_{{\mathfrak{g}}} T( \lambda)\) arises from considering 2-sided cells of the affine Weyl group, which biject with nilpotent orbits.

In type \(A_2\):

There are three nilpotent orbits corresponding to Jordan blocks of type \(X\alpha_1: (1,0)\) and \(X_\mathrm{reg}: (1,1)\) in \({\mathfrak{gl}}_3\). Three cases:

The computation of \(V_G T( \lambda)\) is still open. Some recent work:

What about simple \(G{\hbox{-}}\)modules? Recall \(L(\lambda) = \mathop{\mathrm{Soc}}_G \nabla( \lambda) \subseteq \nabla( \lambda)\) – computing \(V_G L( \lambda)\) is open.

Let \(p > h\) and \(w( \Phi_ \lambda) = \Phi_J\), then \begin{align*} V_{u_q({\mathfrak{g}})} L( \lambda) = G u_J ,\end{align*} i.e. the support varieties in the quantum case are known. This uses that the Lusztig character formula is know for \(u_q( {\mathfrak{g}})\).

18 Tuesday, April 12

18.1 Tensor triangular geometry

Last time: tensor categories and triangulated categories. Idea due to Balmer: treat categories like rings.

A tensor triangulated category (TTC) is a triple \((K, \otimes, 1)\) where

We’ll have notions of ideals, thick ideals, and prime ideals in \(K\). Define \(\operatorname{Spc}K\) to be the set of prime ideals with the following topology: for a collection \(C \subseteq \operatorname{Spec}K\), define \(Z(C) = \left\{{p\in \operatorname{Spc}K {~\mathrel{\Big\vert}~}C \cap p = \emptyset}\right\}\). Note that there is a universal categorical construction of \(\operatorname{Spc}K\) which we won’t discuss here.

TTC philosophy: let \(K\) be a compactly generated TTC with a generating set \(K^c\). Note that \(K\) can include “infinitely generated” objects, while \(K^c\) should thought of as “finite-dimensional” objects. Problems:

Although not all objects can be classified, there is a classification of thick tensor ideals. Idea: use the algebraic topology philosophy of passing to infinitely generated objects to simplify classification.

We’ll need a candidate space \(X\cong_{\mathsf{Top}}\operatorname{Spc}(K^c)\), e.g. a Zariski space: Noetherian, and every irreducible contains a generic point. We’ll also need an assignment \(V: K^c\leadsto X_{{ \operatorname{cl}} }\) (the closed sets of \(X\)) satisfying certain properties, which is called a support datum. For \(I\) a thick tensor ideal, define \begin{align*} \Gamma(I) \mathrel{\vcenter{:}}=\displaystyle\bigcup_{M\in I} V(M) \in X_{\mathrm{sp}} ,\end{align*} a union of closed sets which is called specialization closed. Conversely, for \(W\) a specialized closed set, define a thick tensor ideal \begin{align*} \Theta(W) \mathrel{\vcenter{:}}=\left\{{M\in K^c {~\mathrel{\Big\vert}~}V(M) \subseteq W}\right\} .\end{align*}

One can check that a tensor product property holds: if \(M\in K^c\) and \(N\in \Theta(W)\), check \(V(M\otimes N) = V(M) \cap V(N) \subseteq W\). Under suitable conditions, a deep result is that \(\Gamma \circ \Theta = \operatorname{id}\) and \(\Theta \circ \Gamma = \operatorname{id}\). This yields a bijection \begin{align*} \left\{{\substack{ \text{Thick tensor ideals of } K^c }}\right\} &\rightleftharpoons \left\{{\substack{ \text{Specialization closed sets of } X }}\right\} \\ I &\mapsto \Gamma(I) \\ \Theta(W) &\mapsfrom W \end{align*}

Define \begin{align*} f: X\to \operatorname{Spc}K^c \\ x &\mapsto P_x \mathrel{\vcenter{:}}=\left\{{M \in K^c {~\mathrel{\Big\vert}~}x\not\int V(M)}\right\} .\end{align*} This is a prime ideal: if \(M\otimes N\in P_x\), then \(x\not \in V(M\otimes N) = V(M) \cap V(N)\), so \(M\in P_x\) or \(N\in P_x\).

18.2 Zariski spaces

A space \(X\in {\mathsf{Top}}\) is a Zariski space iff

  1. \(X\) is a Noetherian space, and
  2. Every irreducible closed set has a unique generic point.

Note that since \(X\) is Noetherian, it admits a decomposition into irreducible components \(X = \displaystyle\bigcup_{1\leq i \leq t} W_i\).

The basic examples:

Notation:

18.3 Support data

Recall

Note that \(V_G(P) = \emptyset\) for any projective and \(V_G(k) = \emptyset\). In general, we’ll similarly want \(V_G(0) = \emptyset\) and \(V_G(1) = X\).

A support datum is an assignment \(V: K \to {\mathcal{X}}\) such that

  1. \(V(0) = \emptyset\) and \(V(1) = X\).
  2. \(V\qty{\bigoplus _{i\in I} M_i = \displaystyle\bigcup_{i\in I} V(M_i) }\)
  3. \(V(\Sigma M) = V(M)\) (similar to \(\Omega\))
  4. For any distinguished triangle \(M\to N\to Q\to \Sigma M, V(N) \subseteq V(M) \cup V(Q)\).
  5. \(V(M\otimes N) = V(M) \cap V(N)\).

We’ll need two more properties for the Balmer classification:

  1. Faithfulness: \(V(M) = \emptyset \iff M \cong 0\).
  2. Realization: for any \(W\in {\mathcal{X}}_{{ \operatorname{cl}} }\) there exists a compact \(M\in K^c\) with \(V(M) = W\).

Note that (6) holds for group cohomology, and (7) is Carlson’s realization theorem.

Let \(K\) be a TTC which is closed under set-indexed coproducts and let \(V:K\to {\mathcal{X}}\) be a support datum. Let \(C\) be a collection of objects in \(K\) and suppose \(W \subseteq X\) with \(V(M) \subseteq W\) for all \(M\in C\). Then \(V(M) \subseteq W\) for all \(M\) in \(\mathsf{Loc}(C)\).

Note that \(\mathsf{Loc}(C)\) is closed under

These follow directly from the properties of support data and properties of \(\mathsf{Loc}(C)\).

18.4 Extension of support data

Let \(X\) be a Zariski space and let \(K\supseteq K^c\) be a compactly generated TTC. Let \(V: K^c\to {\mathcal{X}}_{{ \operatorname{cl}} }\) be a support data on compact objects, we then seek an extension: a support datum \({\mathcal{V}}\) on \(K\) forming a commutative diagram:

Link to Diagram

Let \(K\) be a compactly generated TTC and \(V: K^c\to {\mathcal{X}}_{{ \operatorname{cl}} }\) be a support datum. Then \({\mathcal{V}}: K\to {\mathcal{X}}\) extends \(V\) iff

We’ll need Hopkins’ theorem to analyze such extensions.

19 Tuesday, April 19

19.1 Hopkins’ Theorem

Let \(\mathsf{K}\) be a compactly generated tensor triangulated category with \(\mathsf{K}^c\) a subcategory of compact objects. Goal: classify \(\operatorname{Spc}\mathsf{K}^c\). A candidate for its homeomorphism type: we’ll build a Zariski space \(X\) and a homeomorphism \(\operatorname{Spc}\mathsf{K}^c \to X\). We’ll use support data \(\mathbf{V}: \mathsf{K}^c\to {\mathcal{X}}_{{ \operatorname{cl}} }\) which satisfies the faithfulness and realization properties. We’ll extend this to \(\mathcal{V}: \mathsf{K} \to {\mathcal{X}}\). So we need

19.2 Localization functors

Let \(\mathsf{C} \leq \mathsf{K}\) be a thick subcategory for \(\mathsf{K}\in {\mathsf{triang}}\mathsf{Cat}\). A mysterious sequence: \begin{align*} \Gamma_c(M) \to M \to L_c(M) .\end{align*}

Suppose \(W\in {\mathcal{X}}_{{\mathrm{irr}}}\) is nonempty and let \(Z = \left\{{x\in X{~\mathrel{\Big\vert}~}w\not\subseteq { \operatorname{cl}} _X\left\{{x}\right\}}\right\}\). Define a functor \(\nabla_W = \Gamma_{I_W} L_{I_Z}\) and \({\mathcal{V}}(M) \mathrel{\vcenter{:}}=\left\{{x\in X {~\mathrel{\Big\vert}~}\nabla_{\left\{{x}\right\}} (M) = 0}\right\}\).

Let \(\mathsf{K}\) be a compactly generated tensor triangulated category, \(X\) a Zariski space, and \({\mathcal{X}}_{{ \operatorname{cl}} }\) the closed sets. Given a compact object \(M\in \mathsf{K}^c\), let \(\left\langle{M}\right\rangle_{\mathsf{K} ^c}\) be the thick tensor ideal in \(\mathsf{K}^c\) generated by \(M\). Let \(\mathbf V: \mathsf{K}^c\to {\mathcal{X}}_{{ \operatorname{cl}} }\) be support data satisfying the faithfulness condition and suppose \({\mathcal{V}}: \mathsf{K}\to {\mathcal{X}}\) is an extension. Set \(W = \mathbf V(M)\) and \(I_W = \left\{{N\in \mathsf{K}^c {~\mathrel{\Big\vert}~}V(N) \subseteq W}\right\}\). Then \begin{align*} I_W = \left\langle{M}\right\rangle_{\mathsf{K}^c} ,\end{align*} i.e. this is generated by a single object.

Let \(I \mathrel{\vcenter{:}}= I_W\) and \(I' \mathrel{\vcenter{:}}=\left\langle{M}\right\rangle_{\mathsf{K}^c}\).

\(I' \subseteq I\): If \(N\in I'\), then \(N\) is obtained by taking direct sums, direct summands, distinguished triangles, shifts, etc. These all preserve support containment, so \(\mathbf V(N) \subseteq W\) and \(N\in I = I_W\).

\(I \subseteq I'\): Let \(N\in {\mathsf{K^c}}\). Apply the functorial triangle \(\Gamma_{I'} \to \operatorname{id}\to L_{I'}\) to \(\Gamma_I(N)\) to obtain \begin{align*} \Gamma_{I'} \Gamma_I N\to \Gamma_I(N) \to L_{I'} \Gamma_I N .\end{align*} From above, \(I' \subseteq I\) so the first term is in \(\mathsf{Loc}(I)\). Since the second term is as well, the 2-out-of-3 property guarantees that the third term satisfies \(L_{I'} \Gamma_I N \in\mathsf{Loc}(I)\). By the lemma, \(V(L_{I'} \Gamma_I N) \subseteq W\). There are no nonzero maps \(I' \to VL_{I'}\Gamma_I N\), therefore for \(S\in {\mathsf{K^c}}\), noting that \(S\otimes M \in I'\), \begin{align*} 0 = \mathop{\mathrm{Hom}}_{\mathsf{K}}(S\otimes M, L_{I'} \Gamma_I M) = \mathop{\mathrm{Hom}}_{\mathsf{K}}(S, M {}^{ \vee }\otimes L_{I'} \Gamma_I N) ,\end{align*} and since \(S\) is an arbitrary compact object, this forces \(M {}^{ \vee }\otimes L_{I'} \Gamma_I N = 0\). By faithfulness, and the tensor product property, \begin{align*} \emptyset &= {\mathcal{V}}(M {}^{ \vee }\otimes L_{I'} \Gamma_I N)\\ &= {\mathcal{V}}(M {}^{ \vee }) \cap{\mathcal{V}}(L_{I'}\Gamma_I N)\\ &= \mathbf{V}(M) \cap{\mathcal{V}}(L_{I'} \Gamma_I N)\\ &= W \cap{\mathcal{V}}(L_{I'} \Gamma_I N)\\ &= {\mathcal{V}}(L_{I'} \Gamma_I N) ,\end{align*} so by faithfulness (again) \(L_{I'} \Gamma_I N = 0\). Thus by the localization triangle, \(\Gamma_{I'} \Gamma_I N \cong \Gamma_I N\).

Now specialize to \(N\in I\); the localization triangle yields \begin{align*} \Gamma_I N \to N \xrightarrow{0} L_I(N) \implies \Gamma_I N \cong N .\end{align*} Now replacing \(I\) with \(I'\) yields \(\Gamma_{I'} N \cong N\) since \(L_{I'} N \cong L_{I'} \Gamma_I N \cong 0\) by the previous part. Thus \(N\in \mathsf{Loc}(I')\) by applying a result of Neeman, implying \(N\in I'\) and \(I \subseteq I'\).

Many different takes on classification of thick tensor ideals:

Let

Let \(\operatorname{Id}({\mathsf{K^c}})\) be the set of thick tensor ideals in \({\mathsf{K^c}}\), then there is a bijection

\begin{align*} \operatorname{Id}({\mathsf{K^c}}) &\rightleftharpoons {\mathcal{X}}_{\mathrm{sp} } \\ I &\mapsto \Gamma(I) \mathrel{\vcenter{:}}=\displaystyle\bigcup_{M\in I} \mathbf{V}(I) \\ \Theta(W) = I_W \mathrel{\vcenter{:}}=\left\{{N\in {\mathsf{K^c}}{~\mathrel{\Big\vert}~}\mathbf V(N) \subseteq W}\right\} &\mapsfrom W .\end{align*}

Show that \(I_W\in \operatorname{Id}({\mathsf{K^c}})\) is in fact a thick tensor ideal.

\(\Gamma \circ \Theta = \operatorname{id}\): Check that \begin{align*} \Gamma\Theta W = \Gamma(I_W) = \displaystyle\bigcup_{M\in I_W} \mathbf{V}(M) \subseteq W .\end{align*} For the reverse inclusion, let \(W = \displaystyle\bigcup_{j\in W} W_j\) where \(W_j\in {\mathcal{X}}_{{ \operatorname{cl}} }\). By the realization property, there exist \(N_j \in {\mathsf{K^c}}\) such that \(\mathbf{V}(N_j) = W_j\), so \(N_j\in I_W\). Now \(W \subseteq \displaystyle\bigcup_{M\in I_W} \mathbf{V}(M)\), so \(W = \displaystyle\bigcup_{M\in I_W} \mathbf{V}(M)\).


\(\Theta \circ \Gamma = \operatorname{id}\): For \(I\in \operatorname{Id}({\mathsf{K^c}})\), set \(W \mathrel{\vcenter{:}}=\Gamma(I) = \displaystyle\bigcup_{M\in I} \mathbf{V}(M)\), then \begin{align*} \Theta\Gamma I = \Theta(W) = I_W \supseteq I .\end{align*} For the reverse inclusion \(I_W \subseteq I\): let \(N\in I_W\). Since \(X\) is a Zariski space, \(X\) is Noetherian and there is an irreducible component decomposition \(V(N) = \displaystyle\bigcup_i W_i\) with each \(W_i\) irreducible with a unique generic point, so \(W_i = { \operatorname{cl}} _{W_i} \left\{{x_i}\right\}\). Since each \(W_i \subseteq W\), each \(x_i\in W = \displaystyle\bigcup\mathbf{V}(M)\), so there exist \(M_i\) with \(x_i \in \mathbf{V}(M_i)\). Since supports are closed, \(W_i = { \operatorname{cl}} _{W_i}\left\{{x_i}\right\} \subseteq \mathbf{V}(M_i)\). Setting \(M\mathrel{\vcenter{:}}=\bigoplus _i M_i\in I\) yields \(V(N) \subseteq \displaystyle\bigcup V(M_i) = V(M) \subseteq W\).

\begin{align*} N \in \left\langle{M}\right\rangle_{{\mathsf{K^c}}} .\end{align*}

Proving the claim will complete the proof, since \(I\) is a thick ideal containing \(M\), so \(\left\langle{M}\right\rangle_{{\mathsf{K^c}}} \subseteq I\) and \(N\in I\).

By Hopkins’ theorem, \(\left\langle{M}\right\rangle_{{\mathsf{K^c}}} = I_Z\) where \(Z = \mathbf{V}(M)\). Since \(V(N) \subseteq V(M) = Z\), we have \(N\in I_Z = \left\langle{M}\right\rangle_{{\mathsf{K^c}}}\).

Next time:

20 Thursday, April 21

20.1 Classification theorem

Let \(\mathsf{K}\) be a compactly generated tensor-triangulated category and let \(X\) be a Zariski space. Suppose that

  1. \(\mathbf{V}: \mathsf{K}^c\to {\mathcal{X}}_{{ \operatorname{cl}} }\) is a support datum,
  2. \(\mathbf{V}\) satisfies the faithfulness property,
  3. \({\mathcal{V}}: \mathsf{K}\to {\mathcal{X}}\) extends \(\mathbf{V}\).

Then there exists a bijective correspondence \begin{align*} \adjunction{\Gamma}{\Theta}{\operatorname{Id}(\mathsf{K}^c) }{{\mathcal{X}}_{\mathrm{sp}} } \end{align*} where \(\Gamma(I) \mathrel{\vcenter{:}}=\displaystyle\bigcup_{M\in I} \mathbf{V}(M)\) and \(\Theta(W) \mathrel{\vcenter{:}}=\left\{{N\in\mathsf{K}^c{~\mathrel{\Big\vert}~}\mathbf{V}(N) \subseteq W}\right\}\).

This relies on Hopkins’ theorem.

20.2 Balmer spectrum

Let \(\mathsf{K}\) and \(X\) be as in the previous theorem, satisfying the same assumptions. Then there exists a homeomorphism \(f: X\to \operatorname{Spc}\mathsf{K}^c\).

Since \(\mathbf{V}: \mathsf{K}^c\to {\mathcal{X}}_{{ \operatorname{cl}} }\) is a support datum, Balmer shows there exists a continuous map \begin{align*} f: X &\to \operatorname{Spc}\mathsf{K}^c \\ x &\mapsto P_x \mathrel{\vcenter{:}}=\left\{{M{~\mathrel{\Big\vert}~}x\not\in\mathbf{V}(M) }\right\} .\end{align*}

Note that \(P_x\) is a prime ideal:

\begin{align*} M\otimes N\in P_x &\implies x\not\in\mathbf{V}(M\otimes N) \\ &\implies x\not\in\mathbf{V}(M) \cap\mathbf{V}(N) \\ &\implies x\not\in \mathbf{V}(M) \text{ or } x\not\in \mathbf{V}(N) \\ &\implies M\in P_x \text{ or } N\in P_x .\end{align*}

Applying the classification theorem, this yields a bijection.

Examples of classification:

For \(G\in{\mathsf{Fin}}{\mathsf{Grp}}, \operatorname{ch}k = p\divides {\sharp}G\), take \(\mathsf{K} = {\mathsf{kG}{\hbox{-}}\mathsf{stMod}}\), \(R = H^{\mathrm{even}}(G; k)\), and \(X = \mathop{\mathrm{Proj}}R = \mathop{\mathrm{Proj}}(\operatorname{Spec}R)\). Checking that this satisfies the 4 properties in the theorem:

  1. For \(M\in \mathsf{K}^c\), we take \(\mathbf{V}(M) = \left\{{p\in X{~\mathrel{\Big\vert}~} { {\operatorname{Ext} }^{\scriptscriptstyle \bullet}} _{kG}(M, M) \left[ { \scriptstyle { {p}^{-1}} } \right] \neq 0 }\right\}\). This yields a support datum.

  2. The tensor product property holds because \(\mathbf{V}_E(M) = \mathbf{V}_E^r(M)\) (the rank variety), and we showed that \(\mathbf{V}\) satisfies faithfulness and (Carlson) realization properties.

  3. We can use localization functors to define \({\mathcal{V}}: \mathsf{K}\to {\mathcal{X}}\) which satisfies the same support data properties. For this to be an extension, one should check that

To prove these properties, Benson-Carlson-Rickard start with \(E\) elementary abelian, so \(E = \left\langle{x_1,\cdots, x_n}\right\rangle \cong C_p{ {}^{ \scriptscriptstyle\times^{n} } }\) with \(o(x_i) = p\) for all \(i\). Set \(y_i = x_i-1 \in kE\), so \(y_i^p=0\), and define cyclic subgroups \(\mathbf{\alpha }= {\left[ {\alpha_1,\cdots, \alpha_n} \right]} \in L^n\) where \(L/k\) is a field of large transcendence degree. Define \(y_{\mathbf{\alpha}} \mathrel{\vcenter{:}}=\sum_{1\leq i\leq n} \alpha_i y_i\) and define a rank variety \begin{align*} {\mathcal{V}}_E^r(M) = \left\{{ \mathbf{\alpha }\in L^n {~\mathrel{\Big\vert}~}L\otimes_k M \downarrow_{\left\langle{y_{\mathbf{\alpha}}}\right\rangle} \text{ is not free } }\right\}\cup\left\{{0}\right\} .\end{align*}

Let \(E\) be as above and suppose \(\operatorname{trdeg}(L/k) \geq n\). Then if \(M\in \mathsf{K}\), \({\mathcal{V}}_E(M) \cong {\mathcal{V}}_E^r(M)\), and the three properties for (3) above hold for \(E\).

Let \(A = kG\) for \(G\) a finite group scheme, and let \(R = H^{\mathrm{even}}(G; k)\) and \(X = \mathop{\mathrm{Proj}}(R)\). Then

Some remarks:

A special case: let \({\mathfrak{g}}= \mathsf{Lie}G\) for \(G\in{\mathsf{Alg}}{\mathsf{Grp}}_{/ {k}}\) reductive and \(k\) positive characteristic. Let \(A = u({\mathfrak{g}})\), which is a finite-dimensional cocommutative Hopf algebra. If \(p > h\) for \(h\) the Coxeter number, \begin{align*} {\mathcal{N}}_p = \left\{{x\in {\mathfrak{g}}{~\mathrel{\Big\vert}~}x^{[p]} = 0 }\right\} = {\mathcal{N}}, \text{ the nilpotent cone} ,\end{align*} \(R = H^{\mathrm{even}}(u({\mathfrak{g}}); k) = k[{\mathcal{N}}]\), and \(X = \mathop{\mathrm{Proj}}(k[{\mathcal{N}}])\), then applying the theorem,

Let \(\tilde {\mathcal{N}}\to {\mathcal{N}}\) be the Springer resolution. There is an equivalence of derived categories \begin{align*} {\mathbb{D}}^b {\mathsf{ u_\zeta({\mathfrak{g}})_0}{\hbox{-}}\mathsf{Mod}} { \, \xrightarrow{\sim}\, }{\mathbb{D}}^b {\mathsf{Coh}}^{G\times {\mathbb{C}}^{\times}} k[\tilde{\mathcal{N}}] { \, \xrightarrow{\sim}\, }{\mathbb{D}}^b \mathsf{Perv}({\Omega}{\operatorname{Gr}}) .\end{align*} where \(\mathsf{Perv}({-})\) is the category of perverse sheaves and \({\Omega}{\operatorname{Gr}}\) is the loop Grassmannian.

For \(M\) a \(u_\zeta({\mathfrak{g}}){\hbox{-}}\)module and \(R = H^{\mathrm{even}}(u_\zeta({\mathfrak{g}}); M) = {\mathbb{C}}[{\mathcal{N}}] \cong {\mathbb{C}}[\tilde {\mathcal{N}}]\). There is an action of \(R\) on \({ {H}^{\scriptscriptstyle \bullet}} (u_\zeta({\mathfrak{g}}); M)\). Next time: examples for Lie superalgebras and Thomason’s reconstruction theorem for rings.

21 Tuesday, April 26

See Boe-Kujawa-Nakano, Adv. Math 2017.

Setup: \(\mathsf{K}^c \leq \mathsf{K}\in {\mathsf{TTC}}\), \(X\) a Zariski space, \(V:\mathsf{K}^c\to {\mathcal{X}}_{{ \operatorname{cl}} }\) with an extension \({\mathcal{V}}:\mathsf{K}\to {\mathcal{X}}\). Let \({\mathfrak{g}}= {\mathfrak{g}}_{0} \oplus {\mathfrak{g}}_1\) be a Lie superalgebra with a \(C_2\) grading over \(k= {\mathbb{C}}\) where \({\mathfrak{g}}_0\curvearrowright{\mathfrak{g}}_1\), e.g. \({\mathfrak{gl}}_{m, n} = {\mathfrak{gl}}_m \times {\mathfrak{gl}}_n\) with matrices \({ \begin{bmatrix} {{\mathfrak{g}}_0} & {{\mathfrak{g}}_1} \\ {{\mathfrak{g}}_1} & {{\mathfrak{g}}_0} \end{bmatrix} }\) with the bracket action. Write \(\mathsf{Lie}G_0 = {\mathfrak{g}}_0\), and note that \(G_0\) is reductive. Let \({\mathcal{F}}({\mathfrak{g}}, {\mathfrak{g}}_0)\) be the category of finite-dimensional \({\mathfrak{g}}{\hbox{-}}\)supermodules which are completely reducible over \({\mathfrak{g}}_0\). Take \(\mathsf{K}^c = {\mathsf{{\mathcal{F}}({\mathfrak{g}},{\mathfrak{g}}_0)}{\hbox{-}}\mathsf{stMod}} \leq \mathsf{K} = {\mathsf{C({\mathfrak{g}}, {\mathfrak{g}}_0)}{\hbox{-}}\mathsf{stMod}}\), where for \(C\) we drop the finite-dimensional condition.

Set \(R = H^0({\mathfrak{g}}_1, {\mathfrak{g}}_0; {\mathbb{C}}) = \operatorname{Ext} ({\mathbb{C}},{\mathbb{C}}) \cong S({\mathfrak{g}}_0 {}^{ \vee })^{G_0}\). By a theorem of Hilbert, \(\operatorname{Ext} (M, M)\) is finitely generated over \(R\). Write \(V_{{\mathfrak{g}},{\mathfrak{g}}_0}(M) = \mathrm{sp}ec R/J_M\) – for Kac modules \(K(\lambda) = U({\mathfrak{g}}) \otimes_{U(P^0)} L_0(\lambda)\), \(V = 0\) but not every \(K(\lambda)\) is projective.

Idea: use detecting subalgebras. For \({\mathfrak{g}}= {\mathfrak{gl}}_{n,n}\), let \(f_1\) be the “torus”:

Then define \(f_0 = [f_1, f_1]\).

Let \(X = N\mathop{\mathrm{Proj}}(S^*(f_1 {}^{ \vee }))\) where \(S^*(f_1 {}^{ \vee }) \cong \operatorname{Ext} _{f_1, f_0}({\mathbb{C}}, {\mathbb{C}}) = R'\) and \(N = { N }_{G_0}(f_1)\), which is a reductive algebraic group. Define a support datum by \(\mathbf{V}(M) = \left\{{p\in X{~\mathrel{\Big\vert}~}\operatorname{Ext} _{f, f_0}(M,M)_p = 0}\right\}\). The goal is to construct \({\mathcal{V}}: K\to {\mathcal{X}}\) using localization functors – one needs to show the tensor product formula, and the faithfulness and realization properties, which follows from Dede’s lemma. It turns out that \(f_1\cong {\mathfrak{sl}}(1,1){ {}^{ \scriptscriptstyle\times^{m} } }\) and it suffices to define the rank variety on \(f_1\). Define \begin{align*} V_{f_1}^{\operatorname{rank}}(M) = \left\{{\mkern 1.5mu\overline{\mkern-1.5mux\mkern-1.5mu}\mkern 1.5mu = \tilde K\otimes_{\mathbb{Q}}f_1 {~\mathrel{\Big\vert}~}K\otimes_{\mathbb{C}}M\downarrow{\left\langle{\mkern 1.5mu\overline{\mkern-1.5mux\mkern-1.5mu}\mkern 1.5mu}\right\rangle} \text{ is not projective} }\right\} \end{align*} where \(\tilde K\supseteq{\mathbb{C}}\) is an extension with \(\operatorname{trdeg}_{\mathbb{C}}\tilde K \geq \dim f_1\). A theorem shows \({\mathcal{V}}(M) = V_{f_1}^{\operatorname{rank}}(M)\) for \(M\in K\). This yields a classification for \({\mathfrak{gl}}_{m, n}\) of thick tensor ideals in \(K^c\) in terms of \({\mathcal{X}}_{\mathrm{sp}}\).

What is the classification of other Lie superalgebras? This is an open problem.

21.1 Noncommutative theory

How does one extend this theory to noncommutative TTCs? See Nakano-Vashaw-Yakomov, to appear in Amer J. Math.

Let \(K\) be a compactly generated monoidal triangulated category, not necessarily symmetric. One approaches this via noncommutative ring theory, where e.g. even the definition of prime ideals differs. We’ll only consider 2-sided ideals.

A thick triangulated subcategory \(P\) is a completely prime ideal iff \(M\otimes N\in P\implies M\in P\) or \(N\in P\). The ideal \(P\) is prime iff \(I\otimes J\subseteq P \implies I \subseteq P\) or \(J \subseteq P\), where \(I,J\) are themselves ideals. Define \(\mathrm{sp}c K\) to be prime ideals and \(\mathrm{CP}\operatorname{Spc}K\) to be completely prime ideals.

Let \(A\in \mathsf{Hopf}{\mathsf{Alg}_{/k} }^{{\mathrm{fd}}}\) where the coproduct \(\Delta: A\to A{ {}^{ \scriptstyle\otimes_{k}^{2} } }\) is not necessarily commutative, e.g. in the setting of quantum groups. Some remarks:

Let \(K\) be a monoidal triangulated category, \(X\) a Zariski space, and \({\mathcal{X}}= 2^X\) the subsets of \(X\). A map \(\sigma: K\to{\mathcal{X}}\) is a weak support datum iff

Set \(\Phi_\sigma(I) \mathrel{\vcenter{:}}=\displaystyle\bigcup_{M\in I} \sigma(I)\); Then \(\sigma\) is a support datum if additionally

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22 Bibliography

1.
Jacobson, N.: Basic algebra II. Dover Publications, Inc. (2009)