Note: These are notes live-tex’d from a graduate course in Cohomology in Representation Theory taught by Dan Nakano at the University of Georgia in Spring 2022. As such, any errors or inaccuracies are almost certainly my own.
Last updated: 2022-05-29
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:
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:
Simple objects in \({\mathsf{k}{\hbox{-}}\mathsf{Mod}}\) are isomorphic to \(k\), and indecomposables are also isomorphic to \(k\) if we restrict to finite dimensional modules.
Simple objects in \({\mathsf{{\mathbb{Z}}}{\hbox{-}}\mathsf{Mod}}\) are cyclic groups of prime order, \(C_p\). Indecomposables are \({\mathbb{Z}}, C_{p^k}\), using the classification theorem to rule out composites.
For \(A\in {\mathsf{Alg}}^{\mathrm{fd}}_{/ {k}}\), the simple objects in \({\mathsf{A}{\hbox{-}}\mathsf{Mod}}\) are hard to determine in general. The same goes for indecomposables, and is undecidable in many cases (equivalent to the word problem in finite groups).
See finite, tame, and wild representation types.
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:
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:
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.
Last time:
Today: projective modules and their resolutions.
See Krull-Schmidt theorem.
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:
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
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.
If \(R=k\in \mathsf{Field}\), then every \(M\in {\mathsf{k}{\hbox{-}}\mathsf{Mod}}\) is free and thus projective since \(M \cong \bigoplus_{i\in I} k\) with \(k\) free in \({\mathsf{k}{\hbox{-}}\mathsf{Mod}}\).
If \(R={\mathbb{Z}}\), let \(P\in {\mathsf{{\mathbb{Z}}}{\hbox{-}}\mathsf{Mod}}\) be projective and \(F\) free and consider \(0\to K\to F\to P\to 0\). Since \(F\cong P \oplus K\), \(P\) is a submodule of \(F\), making \(P\) free since \({\mathbb{Z}}\) is a PID. So projective implies free.
Not every \(M\in {\mathsf{{\mathbb{Z}}}{\hbox{-}}\mathsf{Mod}}\) is projective: take \(C_6\in {\mathsf{{\mathbb{Z}}}{\hbox{-}}\mathsf{Mod}}\), then \(C_6 \cong C_2 \oplus C_3\) so \(C_2, C_3\) are projective in \({\mathsf{C_6}{\hbox{-}}\mathsf{Mod}}\) but not free here.
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:
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 \(\psi: M\otimes_R N\to P\):
Uniqueness follows from the standard argument on universal properties:
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:
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 \(z=x\otimes y\)!
For \(R=k\in \mathsf{Field}\), \(M\otimes_k N \in ({k}, {k}){\hbox{-}}\mathsf{biMod}\). If \(M = \left\langle{m_i}\right\rangle\) and \(N = \left\langle{n_j}\right\rangle\), then \(M\otimes_k n = \left\langle{m_i\otimes n_j}\right\rangle\) and \(\dim_k M\otimes_k N = \dim_k M \cdot \dim_k N\).
For \(A\in {\mathsf{Ab}}{\mathsf{Grp}}\), \(A\otimes_{\mathbb{Z}}{\mathbb{Z}}\cong A\) since \(x\otimes y = xy\otimes 1\).
\(M\mathrel{\vcenter{:}}= C_p\otimes_{\mathbb{Z}}{\mathbb{Q}}= 0\). It suffices to check on simple tensors: \begin{align*} x\otimes y &= x\otimes{p\over p} y \\ &= x\otimes p\qty{1\over p} y \\ &= px\otimes\qty{1\over p} y \\ &= 0\otimes{1\over p}y \\ &= 0 .\end{align*}
More generally, if \(A\in {\mathsf{Ab}}{\mathsf{Grp}}\) is torsion then \(A\otimes_{\mathbb{Z}}{\mathbb{Q}}= 0\).
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
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 \(F(\operatorname{id}_A) = \operatorname{id}_{F(A)}\).
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:
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:
\(\mathsf{C} = {\mathsf{Ab}}{\mathsf{Grp}}\) and \(\mathsf{D} = {\mathsf{{\mathbb{Z}}}{\hbox{-}}\mathsf{Mod}}\) by taking \(G:\mathsf{D}\to \mathsf{C}\) the forgetful functor, and for \(F\), using the same underlying set and defining the \({\mathbb{Z}}{\hbox{-}}\)module structure by \(n\cdot m \mathrel{\vcenter{:}}= m + m + \cdots + m\).
\(\mathsf{C}={\mathsf{R}{\hbox{-}}\mathsf{Mod}}\) and \(\mathsf{D} = {\mathsf{\operatorname{Mat}_{n\times n}(R)}{\hbox{-}}\mathsf{Mod}}\). For \({\mathsf{k}{\hbox{-}}\mathsf{Mod}}\), the simple objects are \(k\), but for \({\mathsf{\operatorname{Mat}_{n\times n}(R)}{\hbox{-}}\mathsf{Mod}}\), the simple objects are \(k^n\), so these categories are not isomorphic. However, it turns out that they are equivalent.
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 \({\mathsf{R}{\hbox{-}}\mathsf{Mod}} \simeq{\mathsf{S}{\hbox{-}}\mathsf{Mod}}\).
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
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.
Faithfulness: if \(F(f_1) = F(f_2)\), then \(f_1(x_j) = f_2(x_j)\) for all \(j\), making \(f_1=f_2\).
Fullness: let \(g\in \mathop{\mathrm{Hom}}_S(M^n, N^n)\) for \(M, N \in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}\) and \(e_{ij}\) be the elementary matrix with a 1 only in the \(i, j\) position. Check that \(e_{11} M^n = \left\{{{\left[ {x,0,\cdots} \right]} {~\mathrel{\Big\vert}~}x\in M}\right\}\), \(e_{11} N^n = \left\{{{\left[ {y,0,\cdots} \right]}{~\mathrel{\Big\vert}~}y\in N}\right\}\), and \(\operatorname{diag}(x)\) be a matrix with only copies of \(x\) on the diagonal. Then \(g(e_{11} M^n) \subseteq e_{11} g(M^n) \subseteq e_{11}N^n\) and \(g{\left[ {x, 0, \cdots} \right]} = {\left[ {y, 0, \cdots} \right]}\). Define \(f:M\to N\) by \(f(x) = y\), then on one hand, \begin{align*} g(\operatorname{diag}(a) {\left[ {x, 0,\cdots} \right]}) = g{\left[ {ax, 0, \cdots} \right]} = {\left[ {f(ax), 0, \cdots} \right]} ,\end{align*} but since \(g\) is a morphism of \(S{\hbox{-}}\)modules, this also equals \(\operatorname{diag}(a)\cdot g{\left[ {x,0,\cdots} \right]} = {\left[ {ay,0,\cdots} \right]}\). Then \(f(ax) = ay = af(x)\), so \(f\) is a morphism of \(R{\hbox{-}}\)modules.
Note that \(e_{j1} \mathbf{x} = {\left[ {0, \cdots, x,\cdots 0} \right]}\) with \(x\) in the \(j\)th position. Check that \(g(e_{j1}\mathbf{x}) = g{\left[ {0, \cdots, x, \cdots, 0} \right]}\). The LHS is \begin{align*} e_{j1} g(\mathbf{x}) = e_{j1}{\left[ {f(x), 0, \cdots} \right]} = {\left[ { 0,\cdots, f(x), \cdots, 0} \right]} \end{align*} with \(f(x)\) in the \(j\)th position. Hence \(g(\mathbf{x}) = {\left[ {f(x_1), \cdots, f(x_n)} \right]}\), making \(F\) full.
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 \(M\otimes A \xrightarrow{\operatorname{id}\otimes f} M\otimes B \xrightarrow{\operatorname{id}\otimes g} C\).
Show \(\operatorname{id}\otimes g\) is surjective: write \(m\in M\otimes C\) as \(m=\sum x_i\otimes y_j\), pull back the \(y_j\) via \(g\) to get \(z_j\) with \(g(z_j) = y_j\). Then \begin{align*} (\operatorname{id}\otimes g)(\sum x_i \otimes z_J) = \sum x_i\otimes g(z_j) = \sum x_i \otimes y_j .\end{align*}
Exactness, \(\operatorname{im}(\operatorname{id}\otimes f) = \ker (\operatorname{id}\otimes g)\): Use that \(gf=0\) by exactness of the original sequence, and \((\operatorname{id}\otimes g)\circ (\operatorname{id}\otimes f) = \operatorname{id}\otimes(g\circ f) = 0\), so \(\operatorname{im}(\operatorname{id}\otimes f) \subseteq \ker(\operatorname{id}\otimes g)\).
Why is this well-defined? Check \(g(z_1) = y = g(z_2)\) implies \(z_1 -z_2\in \ker g = \operatorname{im}f\), so write \(f(y) = z_1-z_2\) for some \(y\). Then \(x\otimes z_1 + \operatorname{im}f = x\otimes z_2 + \operatorname{im}f\).
Why does this factor through the tensor product? Check that \(\Psi\) is a balanced product, this yields \(\mkern 1.5mu\overline{\mkern-1.5mu\Psi\mkern-1.5mu}\mkern 1.5mu: M\otimes C\to {M\otimes B\over \operatorname{im}(\operatorname{id}\otimes f)}\). Now check that \(\mkern 1.5mu\overline{\mkern-1.5mu\Psi\mkern-1.5mu}\mkern 1.5mu, \Gamma\) are mutually inverse: \begin{align*} \Gamma\Psi(x\otimes y) &= \Gamma(x\otimes z + \operatorname{im}(\operatorname{id}\otimes f)) = x\otimes g(z) = x\otimes y \\ \Psi\Gamma(x\otimes z + \operatorname{im}(\operatorname{id}\otimes f)) &= (x\otimes g(z) ) = x\otimes z + \operatorname{im}f .\end{align*}
When is \(M\otimes_R ({-})\) exact?
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 \(0_{AB} \mathrel{\vcenter{:}}= 0_{B} \circ 0_A \in \mathop{\mathrm{Hom}}_{\mathsf{C}}(A, B)\).
A category \(\mathsf{C}\) is additive iff
A morphism: \(k:K\to A\) is monic iff whenever \(g_1, g_2: L\to K\), \(kg_1 = kg_2 \implies g_1 = g_2\):
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 \(f\in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}(A, B)\),
Some notes:
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:
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:
Now inductively, we want to construct the following lift:
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:
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:
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:
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 \(0 \to M \xhookrightarrow{\eta} { {D}_{\scriptscriptstyle \bullet}}\)
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 \((L_n F)(M) \mathrel{\vcenter{:}}= H_n(F({ {C}_{\scriptscriptstyle \bullet}} ))\).
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.
Show that any two lifts \(\alpha, \alpha'\) induce the same map on homology.
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:
Show that such resolutions exist. This involves constructing \({\varepsilon}: C_0 \to M\):
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
If \(g,h\) are mono (resp. epi, resp. iso) then \(f\) is mono (resp. epi, resp. iso).
Let \(x\in M\)
Set \(y=\sigma(x)\)
Find \(z\in C_0\) such that \({\varepsilon}'' p_0 (z) = y\).
Consider \({\varepsilon}(z) - x\) and apply \(\sigma\): \begin{align*} \sigma({\varepsilon}(z) - x) &= \sigma {\varepsilon}(x) - \sigma(x) \\ &= {\varepsilon}'' p_0(x) - \sigma(x) \\ &= y-y \\ &= 0 .\end{align*}
So \({\varepsilon}(z) - x\in \ker \sigma = \operatorname{im}\gamma\)
Pull back to \(w\in C_0'\) such that \(\gamma {\varepsilon}'(w) = {\varepsilon}(z) - x\)
Check \({\varepsilon}i_0 (w) = \gamma {\varepsilon}'(w) = {\varepsilon}(z) - x\), so \({\varepsilon}(i_0(w) - z) = -x\).
The setup:
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
Missed! Please send me notes. :)
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!
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:
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
\(\impliedby\): \(I_S \subseteq R_S\) 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
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
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.
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:
Any irreducible subset of \(X\) is entirely contained in a single irreducible component.
Any space is a union of its irreducible components.
Coming up:
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
There exists \(K > 0\) such that \(a_n \leq K n^{\gamma-1}\) for \(n\geq 0\)
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.
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:
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:
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.
For \(A = C_p\) with \(\operatorname{ch}k = p > 0\), then \begin{align*} R \mathrel{\vcenter{:}}= H^0(C_p; k) \cong \begin{cases} k[x,y]/\left\langle{y^2}\right\rangle, {\left\lvert {x} \right\rvert} = 2, {\left\lvert {y} \right\rvert} = 1 & p\geq 3 \\ k[x], {\left\lvert {x} \right\rvert} = 1 & p = 2. \end{cases}, \qquad \operatorname{mSpec}R \cong {\mathbb{A}}^1_{/ {k}} .\end{align*}
Dan’s favorite: \(A = u({\mathfrak{g}})\) for \({\mathfrak{g}}= {\mathfrak{sl}}_2\) with \(\operatorname{ch}k = p \geq 3\) for \(u\) the small enveloping algebra. Friedlander-Parshall show \(\operatorname{mSpec}R = k[{\mathcal{N}}]\) for \({\mathcal{N}}\mathrel{\vcenter{:}}=\left\{{M { \begin{bmatrix} {a} & {b} \\ {c} & {-a} \end{bmatrix} } {~\mathrel{\Big\vert}~}M\text{ is nilpotent}}\right\}\). This can be presented as \begin{align*} k[{\mathcal{N}}] \cong k[x,y,z] / \left\langle{z^2 + xy}\right\rangle, {\left\lvert {x} \right\rvert}, {\left\lvert {y} \right\rvert}, {\left\lvert {z} \right\rvert} = 2 ,\end{align*} and we’ll see how finite generation is used in this setting.
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
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
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 \(V_G(M) \cong V_G(M {}^{ \vee })\).
\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.
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 \(\dim u({\mathfrak{g}}) = p^{\dim {\mathfrak{g}}}\).
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.
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 \(u_q( {\mathfrak{g}})\).
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 \(N\in P_x\).
A space \(X\in {\mathsf{Top}}\) is a Zariski space iff
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:
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
We’ll need two more properties for the Balmer classification:
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 \(\mathsf{Loc}(C)\).
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:
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.
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
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:
Let \(\mathsf{K}\) be a compactly generated tensor-triangulated category and let \(X\) be a Zariski space. Suppose that
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.
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:
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.
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.
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
There is a bijective correspondence \begin{align*} \adjunction{\Gamma}{\Theta}{{\mathsf{kG}{\hbox{-}}\mathsf{stMod}}}{{\mathcal{X}}_{\mathrm{sp}}} .\end{align*}
\(\operatorname{Spc}({\mathsf{kG}{\hbox{-}}\mathsf{stMod}}) \underset{{\mathsf{Top}}}{\cong} X\).
Some remarks:
This theorem is an indication of why cohomology is central in understanding the tensor structure of representation categories. If \(G\in {\mathsf{Fin}}{\mathsf{Grp}}{\mathsf{Sch}}_{/ {k}}\) then the coordinate ring \(k[G]\) is a commutative Hopf algebra, so \(A = kG = k[G] {}^{ \vee }\) is a finite dimensional cocommutative Hopf algebra. So there is an equivalence of categories between \({\mathsf{Rep}}G\) and \({\mathsf{Rep}}A\) for \(A\) such a Hopf algebra. By a result of Friedlander-Suslin, \(R\) is finitely generated.
The realization of \(\mathbf{V}\) and \({\mathcal{V}}\) for a general group scheme involve so-called \(\pi{\hbox{-}}\)points developed be Friedlander-Pevtsovz and the construction of explicit rank varieties.
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,
There is a correspondence \begin{align*} \adjunction{}{}{{\mathsf{u({\mathfrak{g}})}{\hbox{-}}\mathsf{stMod}}}{{\mathcal{X}}_{\mathrm{sp}}} .\end{align*}
There is a homeomorphism \begin{align*} \operatorname{Spc}\qty{ {\mathsf{u({\mathfrak{g}})}{\hbox{-}}\mathsf{stMod}} } \underset{{\mathsf{Top}}}{\cong} \mathop{\mathrm{Proj}}(k[{\mathcal{N}}]) .\end{align*}
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.
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.
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
Next time:
