COHOMOLOGY OF GROUPS
J. WARNER
Abstract. Notes on Kenneth Brown’s book Cohomology of Groups.
1. Some Homological Algebra 1.1. Review of Chain Complexes. Let R be a ring, and let (C, d) and (C0, d0) be two chain complexes 0 of left R-modules. Define a complex of abelian groups HR(C,C ) as follows. Let
0 Y 0 HR(C,C )n = HomR(Cq,Cq+n) q∈Z 0 n and define the boundary map Dn by Dn(f) = d f − (−1) fd.
1.5. The Standard Resolution. For any group G, we can always form the following free resolution of Z over ZG. Let Fn be the free ZG module with basis given by the (n+1)-tuples of elements of G whose first component is 1: (1, g1, g2, . . . , gn). The G-action on Fn is defined on basis elements component-wise. We introduce the shorthand bar notation:
[g1|g2| ... |gn] = (1, g1, g1g2, . . . , g1g2 . . . gn) If n = 0, there is only one basis element which we denote by [ ]. Define the boundary morphisms by n X i ∂ = (−1) di i=0 where g [g | ... |g ] i = 0 1 2 n di([g1|g2| ... |gn]) = [g1|g2| ... |gigi + 1| ... |gn] 0 < i < n [g1|g2| ... |gn−1] i = n
For F0 = ZG, the boundary morphism is the standard augmentation. We will refer to this resolution as the standard resolution or bar resolution
2. The Homology of a Group 2.1. Generalities. In homological algebra, we can construct invariants of algebraic objects using the fol- lowing procedure. Let R be a ring, and let T be a covariant additive functor from the category of R-modules to the category of abelian groups. Recall, this means that the map induced by T on Hom sets
HomR(M,N) → HomZ(TM,TN) is a homomorphism of abelian groups. Now, let ε : F → M be a free (or projective) resolution of M. Applying T to F we obtain a chain complex of abelian groups. Since T is additive, it preserves homotopy equivalences, so that the homology groups Hn(TF ) depend only on T and M. Notice that if T is exact, then Hn(TF ) = 0 for n > 0 and H0(TF ) = TM. So, in some sense, the higher homology groups measure the failure of T to be exact.
Date: Fall 2013. 1 2.2. Co-invariants. Here we introduce the co-invariants functor, which we will use in the context of the previous section to define the homology groups of a group G.
Definition 2.1. The group of co-invariants of M, denoted MG, is the quotient of M by the additive subgroup of elements of the form gm − m.
MG is the largest quotient of M on which G acts trivially.
Exercise 2.2. If f : M → N is a map of ZG-modules, show that f : MG → NG which maps m to f(m) is a well-defined map of abelian groups.
We thus obtain a covariant, additive (check!) functor (·)G from ZG-modules to abelian groups, called the co-invariants functor. The following proposition will reveal some information (·)G. ∼ Proposition 2.3. MG = Z ⊗ZG M as abelian groups, where Z is a trivial right ZG-module.
Proof. The map M → Z ⊗ZG M defined by m 7→ 1 ⊗ m vanishes on elements of the form gm − m, so it factors through MG to a map of the form m 7→ 1 ⊗ m. The map Z × M → MG defined by (a, m) 7→ am is ZG biadditive, so by the universal property of tensor products, we obtain a well-defined map Z ⊗ZG M → M given by a ⊗ m 7→ am. These two maps are inverses of each other.
By properties of the tensor product, we see that (·)G is right exact, and that it take a free ZG-module with basis (ei) to a free Z-module with basis (ei). Also, here the map f : MG → NG sends a⊗m to a⊗f(m).
2.3. The Definition of H∗G. We now use the co-variants functor to define the homology of a group G. The following definition is independent of choice of projective resolution, as described in §2.1.
Definition 2.4. Let ε : F → Z be a projective resolution of Z by ZG-modules, and define the homology groups of G by HiG = Hi(FG) 2 n−1 Example 2.5. Let G = Zn = hti, let N = 1 + t + t + ... + t and consider the projective resolution: t−1 N t−1 ε ··· −−→ ZG −→ ZG −−→ ZG −→ Z → 0
Applying (·)G we obtain the complex 0 n 0 ··· −→ Z −→ Z −→ Z → 0 from which it follows that i = 0 Z HiG = Zn i odd 0 i even, i > 0.
Let Fn be the standard resolution of §1.5, and let Cn(G) = (Fn)G. By our observation of the functor (·)G, Cn(G) is a free Z-module with basis [g1|g2| ... |gn]. Notice here by abuse of notation we have
[g1|g2| ... |gn] = (1, g1, g1g2, . . . , g1g2 . . . gn)
Since G acts trivially on (Fn)G, the boundary morphisms for C∗(G) only differ for i = 0. They are given by:
n X i ∂ = (−1) di i=0 where [g | ... |g ] i = 0 2 n di([g1|g2| ... |gn]) = [g1|g2| ... |gigi + 1| ... |gn] 0 < i < n [g1|g2| ... |gn−1] i = n
The beginning of the resolution C∗(G) looks like ∂ 0 C2(G) −→ C1(G) −→ Z → 0 where ∂[g|h] = [h] − [gh] + [g]. It follows that H0G = Z for any group G. 2 ∼ 0 Proposition 2.6. H1G is isomorphic to the abelianization of G: H1G = Gab = G/G .
Proof. Let g denote the homology class of the cycle [g] ∈ C1(G), and define a map ϕ : H1G → Gab by g 7→ gG0 and extend additively to make a homomorphism of abelian groups (that is g + h 7→ ghG0). The map is well-defined because if g = h, then [g]−[h] = [a]+[b]−[ba], so that g is mapped to haba−1b−1G0 = hG0. The map is surjective, and injectivity follows from the identity gh = g + h. 0 0 2.6. Functoriality. Suppose α : G → G is a map of groups, and define Hn(α): Hn(G) → Hn(G ) via
[g1| ... |gn] 7−→ [α(g1)| ... |α(gn)]
3. Homology and Cohomology with Coefficients
3.0. Preliminaries on HomG. Let G be a group, and let M and N be ZG-modules. Exercise 3.1. Show that the formula (gu)(m) = g · u(g−1m)
defines a left action of G on HomZ(M,N), called the diagonal action. Notice that gu = u if and only if the action of G commutes with u, so it follows that G HomZG(M,N) = HomZ(M,N) ⊂ HomZ(M,N) Exercise 3.2. If F is a projective ZG-module, and M is a ZG-module which is free as a Z-module, then F ⊗ M with diagonal action g · f ⊗ m = (g · f) ⊗ (g · m) is a projective ZG-module. 3.1. Definition of H∗(G, M). Let G be a group, and let F → Z be a projective resolution of Z over ZG. For a G-module M, consider M as a chain complex concentrated in degree 0. Then HZG(F,M)n = HomZG(F−n,M), so we can reindex to consider the cochain complex defined by n HZG(F,M) = HZG(F,M)−n = HomZG(Fn,M)
Since all boundary maps in the complex defined by M are 0, the boundary map of HZG(F,M) is n+1 (δnu)(x) = (−1) u(∂x)
for u ∈ HomZG(Fn,M) and x ∈ Fn+1. Remark 3.3. The sign convention here is used to follow the book. Changing the signs of coboundary maps does not affect cohomology. Definition 3.4. The cohomology groups of G with coefficients in M are the cohomology groups of the
cochain complex HZG(F,M): n n H (G, M) = H (HZG(F,M)) Exercise 3.5. ∼ G G (1) Show that HomZG(Z,M) = M as abelian groups, where M is the subgroup of invariants, ie, of fixed points of M under the action of G. 0 G (2) Use the left exactness of HomZG(·,M) to show that H (G, M) = M . Example 3.6. Let G be an infinite cyclic group with generator t, and consider the projective resolution t−1 0 → ZG −−→ ZG → Z → 0
The cochain complex HZG(F,M) has the form: 0 → M −−→1−t M → 0 It follows that H0(G, M) consists of all m such that tm = m, ie, H0(G, M) = M G, and H1(G, M) is the quotient of M by elements of the form gm − m. This quotient, denoted MG, is called the group of co-invariants. 3 Example 3.7. Let G be a finite cyclic group of order n with generator t, let N = 1 + t + t2 + ... + tn−1, and consider the projective resolution N t−1 N t−1 ··· −→ ZG −−→ ZG −→ ZG −−→ ZG → Z → 0 For a ZG-module M, the associated chain complex is then 0 → M −−→1−t M −→N M −−→1−t M −→·N · · G Since Ng = N = gN for any g ∈ G, we have a well-defined map, called the Norm map, N : MG → M sending [m] to Nm. Exercise 3.8. Check that Hn(G, M) = ker N for odd n and Hn(G, M) = coker N for even n ≥ 2. (We already know that H0(G, M) = M G)
3.2. Ext. We obtain an immediate generalization of H∗(G, ·) by taking projective resolutions of ZG-modules other than Z Definition 3.9. Let M and N be ZG-modules, and choose a projective resolution F → M over ZG. Define Extn (M,N) = Hn( (F,N)) ZG HZG If M is free as a Z-module, then in some sense Ext contains no new information. Proposition 3.10. If M is Z-free, then Extn (M,N) =∼ Hn(G, Hom (M,N)) ZG Z
where G acts diagonally on HomZ(M,N). Proof. Let F → Z be a projective resolution, so that F ⊗ M → M is a resolution of M. By Exercise 3.2, the resolution is projective, so we have Extn (M,N) = Hn( (F ⊗ M,N)) ZG HZG n G = H (HZ(F ⊗ M,N) ) n G = H (HZ(F, HomZ(M,N) ) n = H (HZG(F, HomG(M,N)) n = H (G, HomZ(M,N)) 3.3. Extension and Co-extension of Scalars. Given a ring map α : R → S, we obtain a functor from S- modules to R-modules by restricting along α. Here we define the left and right adjoint functors of restriction. First, let M be a left R-module, and consider S as a right R-module via s · r = sα(r). Then S ⊗R M is a left S-module via s · (s0 ⊗ m) = (ss0 ⊗ m). This S-module is obtained from M by extension of scalars. Notice we have a canonical map of R-modules sending m to 1 ⊗ m, where S ⊗R M is an R-module by restricting scalars. We also obtain the following universal property of extension of scalars: If N is any S-modules, and f : M → N is any R-module map, then there is a map g : S ⊗R M → N of S-modules such that gi = f. The universal property shows that
HomR(M, Res(N)) = HomS(S ⊗R M,N) so that extension of scalars is left adjoint to restriction. If we apply the universal property with M = N and f = idN we obtain a surjective map of S-modules s ⊗ n 7→ s · n, which splits as a map of R-modules. 0 Exercise 3.11. Define co-extension of scalars by the S-module HomR(S, M) under the action (sf)(s ) = 0 f(s s). Show there is a natural R-module map π : HomR(S, M) → M, and that given any S-module N which maps to M as an R-module via f, there is an S-module map g : N → HomR(S, M) such that πg = f. Show then that co-extension of scalars is right adjoint to restriction, and that taking M = N yields a canonical injective S-module map N → HomR(S, N), which is split as an R-module map, and which maps n to the map s 7→ sn. 4 Notice that if α : ZG → Z is the augmentation map, then extension of scalars is the functor M → MG and co-extension of scalars is the functor M → M G.
3.4. Injective Modules.
Definition 3.12. An R-module I is injective if HomR(·,I) is exact. Theorem 3.13. Any R-module can be embedded in an injective module. 3.5. Induced and Co-induced Modules. Here we apply the methods of §3.3 to the injective map of rings α : ZH,→ ZG for some subgroup H ≤ G. G Definition 3.14. If M is a ZH-module, then define the induced module IndH M := ZG⊗ZHM and the G co-induced module CoindH M := HomZH (ZG, M).
Let gi be a set of representatives of the left cosets of H in G, and notice that M ZG = giZH gi so that ZG is a free right ZH module. From above, it follows that G M IndH M = ZG ⊗ZH M = gi ⊗ M gi as an abelian group, where g ⊗ M = {g ⊗ m | m ∈ M}. Notice that g ⊗ M =∼ M as abelian groups. If we G choose the representative of H to be 1, then the map of H-modules M → IndH sending m to 1 ⊗ m maps M isomorphically onto the H-submodule 1 ⊗ M.
G Proposition 3.15. The G-module IndH M contains M as an H-submodule, and is a direct sum of the transforms gM as g ranges over a set of representatives of the left cosets of H in G.
G M IndH M = gM g∈G/H The previous proposition actually completely characterizes induced G-modules, as we now see. Proposition 3.16. Let N be a G-module such that M N = Mi i∈I as an abelian group. Suppose the G action on N transitively permutes the summands, that is, there is a transitive action of G on I such that gMi = Mgi. Choose M to be one of the summands, and let H ⊂ G be ∼ G the subgroup fixing M. Then M is a ZH-module and N = IndH M. Proof. Since H fixes M, M is an ZH-submodule. By the universal property of extension of scalars, the G ZH-module map M,→ N extends to a ZG-module map f : IndH M → N. Since f is a map of ZG-modules, the universal property also shows that f maps gMi to Mgi isomorphically as abelian groups, so that f is an isomorphism.
Corollary 3.17. If G doesn’t act transitively on I, Gi is the subgroup fixing Mi, and E is a set of repre- sentatives of I under the G-action, then Mi is a ZGi-module and M N ∼ IndG M = Gi i i∈E
Proof. Group the summands Mi by their orbit, and apply the proposition to each group.
Example 3.18. Consider the permutation module Z[G/H], which is free as a Z-module with basis given 0 0 ∼ G by the left cosets of H in G, and has G-action given by g · g H = gg H. Then Z[G/H] = IndH Z where H acts trivially on Z. Here the distinguished copy of Z inside of Z[G/H] is the mathbbZ span of the coset H. 5 G Remark 3.19. Notice that if M is the distinguished H-module in IndH M, then the transform gM is a gHg−1-module, and the bijection f : gM → M given by multiplication by g−1 is an isomorphism of gHg−1 modules if we consider M a gHg−1-module via restriction along the map α(k) = g−1kg where k ∈ gHg−1. This follows from the formula f(kn) = α(k)f(n) G ∼ G Proposition 3.20. If (G : H) < ∞ then IndH M = CoindH M 3.6. H∗ as a Functor of the Coefficient Module. Let ε : F → Z be a projective resolution of Z over G. ∗ A map of G-modules f : M → N gives a map of cochain complexes of abelian groups f : HomG(F,M) → ∗ ∗ ∗ HomG(F,N), which gives a map on cohomology H (G, M) → H (G, N). It follows that H (G, ·) is a covariant functor from G-modules to abelian groups. Proposition 3.21. (i) There is a natural isomorphism H0(G, M) =∼ M G. j (ii) For any exact sequence 0 → M 0 −→i M −→ M 00 → 0 of G-modules and any integer n there is a natural map δ : Hn(G, M 00) → Hn+1(G, M 0) such that the sequence
0 → H0(G, M 0) → H0(G, M) → H0(G, M 00) −→δ H1(G, M 0) → H1(G, M) → · · · is exact. (iii) If Q is an injective G-module, then Hn(G, Q) = 0 for n > 0. Remark 3.22. The naturality assertion above means for any commutative diagram
0 / M 0 / M / M 00 / 0
0 / N 0 / N / N 00 / 0 with exact rows, the following square commutes:
Hn(G, M 00) δ / Hn+1(G, M 0)
Hn(G, N 00) δ / Hn+1(G, N 0)
Proof. (i) follows from exercise 3.5. (ii) follows from the Snake lemma and the exact sequence of cochain complexes 0 → H (F,M 0) → H (F,M) → H (F,M 00) → 0 where F is a projective resolution of Z over G. To see the naturality assertion, imagine the exact sequence of cochain complexes described above, and the corresponding sequence of cochain complexes with N 0, N, and N 00 ’hovering’ above. This gives a 3-dimensional commutative diagram. Considering how each se- quence of complexes is used to define the map δ, the commutativity of this 3-dimensional diagram gives the commutativity of the square. Finally, the exactness of HomG(·,Q) yields (iii). Property (iii) is expressed by saying that injective modules are H∗-acyclic. Definition 3.23. A functor T from R-modules to abelian groups is co-effaceable if every module M can be embedded in a module M such that T (M) = 0. By theorem 3.13, it follows that Hn(G, ·) is co-effaceable for n > 0. Proposition 3.24 (Shapiro’s Lemma). If H ≤ G and M is an H-module, then ∗ ∼ ∗ G H (H,M) = H (G, CoindH M) 6 Proof. Any projective resolution F → Z of Z over ZG can be regarded as a projective resolution over ZH, and the universal property of co-induction gives an isomorphism of chain complexes: ∼ G HH (F,M) = HG(F, CoindH M)
Corollary 3.25. If A is an abelian group, then the G module HomZ(ZG, A) with G-action (gf)(h) = f(hg) is H∗-acyclic.
G n Proof. The G-module HomZ(ZG, A) is CoindH A with H = {1}, and H ({1},A) = 0 for all n > 0 by id considering the projective resolution of Z over Z given by 0 → Z −→ Z → 0.
Since any G-module M embeds into the co-induced module HomZ(ZG, M) via Exercise 3.11, the corollary gives another proof that Hn(G, ·) is co-effaceable for n > 0. By Proposition 3.20 and Example 3.18, if (G : H) < ∞ we have ∗ ∗ H (H, Z) =∼ H (G, Z[G/H]) ∼ Also, since ZG ⊗ZH ZH = ZG, in the case of finite index it follows that ∗ ∗ H (H, ZH) =∼ H (G, ZG) 3.7. Dimension Shifting.
6. Cohomology Theory of Finite Groups 6.7. A Duality Theory. Recall: Complete Resolution (of finite type), Tate Cohomology Existence of complete resolutions of finite type =⇒ Hˆ i(G, Z) is finitely generated. |G| annihilates Hˆ i(G, Z) =⇒ Hˆ i(G, Z) is finite. Definition 6.1. Let A be an abelian group, and define its dual by A0 := Hom(A, Q/Z). Exercise 6.2. 0 ∼ (i) If nA = 0, then A = Hom(A, Zn) ∼ 0 ∼ (ii) If A = Zn then A = Zn (iii) (A × B)0 =∼ A0 × B0 (iv) If A is finite, A =∼ A0 (non-canonically), and A =∼ A00 canonically.
Given a map ρ : A ⊗ B → Q/Z, we have the induced mapρ ¯ : A → B0 which maps a to {b 7→ ρ(a ⊗ b). Definition 6.3. ρ is a duality pairing ifρ ¯ is an isomorphism.
Example 6.4. Evaluation A0 ⊗ A → Q/Z. −1 ∼ Note: If nA = nB = 0, then imρ ⊂ (n Z)/Z = Zn so we have duality pairings A ⊗ B → Zn Theorem 6.5. The cup product Hˆ i(G, Z) ⊗ Hˆ −i(G, Z) → Hˆ 0(G, Z) = Z/|G| · Z is a duality pairing. 6.8. Cohomologically Trivial Modules.
Definition 6.6. A G-module is cohomologically trivial if Hˆ i(H, M) = 0 for all i ∈ Z and all subgroups H ⊂ G. Exercise 6.7. (i) Induced modules are c.t. (ii) Free modules are c.t. (iii) Projective modules are c.t. Proposition 6.8. If G is a p-group and M is a G-module in which every element has order a power of p, then M G 6= 0
Proof. 7 Corollary 6.9. If k is a field of characteristic p, G is a p-group, and M is a simple kG-module, then M =∼ k, the trivial module.
Proof. Another way of viewing the corollary: J(kG) = I. Corollary 6.10. I is a nilpotent ideal
Proof. In general, the Jacobson radical of an Artinian Ring is nilpotent. Example 6.11. G finite cyclic of order pa.
Corollary 6.12. If M is a kG-module with MG = 0, then M = 0 Proof. This is also just Nakayama’s Lemma (kG is local, with unique maximal ideal I). Theorem 6.13. Let G be a p-group, k a field of characteristic p, and M a kG-module. TFAE: (i) M is free. (ii) M is projective. (iii) M is cohomologically trivial. (iv) Hˆ i(G, M) = 0 for some i ∈ Z. Proof. Proof of (iv) ⇒ (i) uses dimension shifting and the previous corollary. Theorem 6.14. Let G be a p-group, and M a ZG-module. Then M is cohomologically trivial if and only if Hˆ i(G, M) = 0 for two consecutive integers i.
Proof. Uses the long exact sequence associated to the short exact sequence 0 → M → M → M/pM → 0. Proposition 6.15. A ZG-module M is cohomologically trivial if and only if its restriction to G(p) is cohomologically trivial for each p.
Proof. Theorem 6.16. A G-module M is cohomologically trivial if and only if for each prime p there are two consecutive integers i such that Hˆ i(G(p),M) = 0.
Theorem 6.17. If M is a Z-free, cohomologically trivial G-module, then M is projective. Theorem 6.18. Let M be a G-module. TFAE: (i) M is cohomologically trivial. (ii) proj dim M ≤ 1. (iii) proj dim M < ∞.
Proof.
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