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Notes on the Definitions of Group Cohomology and Homology

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Notes on the Definitions of Group Cohomology and Homology Notes on the definitions of group cohomology and homology. Kevin Buzzard February 9, 2012 VERY sloppy notes on homology and cohomology. Needs work in several places. Last updated 3/12/07. 1 Derived functors. My goal is to define Ext and Tor but there's no harm talking about general nonsense for a bit. Because I want to derive both Hom and ⊗ with respect to both variables, we're going to have to talk about functors which are either left exact or right exact, and functors which are either covariant or contravariant. But there's a trick. If A is an abelian category then Aop is too, and so we can work out the theory just with right derived functors of a covariant left exact functor F : A ! B, and then apply the theory with functors A ! Bop, Aop ! B and Aop ! Bop! So first let's say A and B are abelian categories, and F is a covariant, left exact functor (that is, if 0 ! L ! M ! N ! 0 is exact then so is 0 ! F (L) ! F (M) ! F (N)). If A has enough injectives then left exact (covariant) functors have right derived functors, and they can be computed in the usual way: if X is an object of A then write down an injective resolution 0 ! X ! I0 ! I1 !··· , apply the functor and get a complex 0 ! F (I0) ! F (I1) ! ··· , and define the right derived functor Rn(F ) to be the cohomology of this complex at the point F (In). Note for example that R0(F ) = F by left exactness. We recall the basic property of the derived functor in this case: if 0 ! X ! Y ! Z ! 0 is exact, then so is 0 ! F (X) ! F (Y ) ! F (Z) ! R1F (X) ! R1F (Y ) !··· : Next, if F is a contravariant functor A ! B then think of F as a covariant functor F~ : Aop ! B; it seems that the standard terminology is that F is left exact if, by definition, F~ is left exact. If F is a contravariant functor which is left exact in this sense, then we will be able to pull off the same trick if Aop has enough injectives, which of course is the same as A having enough projectives. The functor F~ of course has right derived functors RnF~ : Aop ! B which can be thought of as contravariant functors RnF : A ! B|these are again called the right derived functors of the contravariant left exact functor F . Explicitly, if X 2 A and ···! P1 ! P0 ! X ! 0 is a projective resolution, and if we apply F , then we get 0 ! F (P0) ! F (P1) !··· and we take n the cohomology of this at the F (Pn) term to get R F (X). And the canonical exact sequence is that if 0 ! X ! Y ! Z ! 0 is exact then so is 0 ! F (Z) ! F (Y ) ! F (X) ! R1F (Z) ! R1F (Y ) !··· . As a sanity check: we computed using projective resolutions so if Z is projective then R1F (Z) = 0, but this is unsurprising because in this case the short exact sequence splits. A right exact covariant functor A ! B can be thought of as a left exact covariant functor Aop ! Bop, so the left derived functors of a right exact functor can be computed using projective resolutions in A, if A has enough projectives. Similarly a right exact contravariant functor A ! B, that is, a contravariant functor for which the associated covariant functor Aop ! B is right exact, will have left derived functors iff the associated map A ! Bop has right derived functors, so if A has enough injectives then the derived functors exist and you can compute them using injective resolutions in A. 1 2 Example of a projective resolution. This will come in handy later. Let G be a group, let A be the ring Z[G], and consider the category of left A-modules (that is, A acts on the left). Here is a \standard" projective resolution of Z, considered as a left G-module with G acting trivially. It's called the \unnormalised bar resolution". Let Bn (for n ≥ 0) be the free A-module on the set of all symbols (g1; g2; : : : ; gn). Thought of another way, Bn is the free Z-module on the set of all symbols [g0; g1; : : : ; gn] := g0(g1; g2; : : : ; gn). For n ≥ 1 define an A-linear map d : Bn ! Bn−1 by demanding that on A-generators it does this: d(g1; : : : ; gn) = g1(g2; : : : ; gn) − (g1g2; g3; : : : ; gn) + (g1; g2g3; g4; : : : ; gn) − · · · n−1 n ··· + (−1) (g1; g2; : : : ; gn−1gn) + (−1) (g1; g2; : : : ; gn−1): So, explicitly, d : B1 ! B0 sends the A-generator (g) to g(∗) − () with () the generator of B0 (which is free of rank 1) and d : B2 ! B1 sends (g; h) to g(h) − (gh) + (g). Finally define : B0 ! Z by sending g(∗), for g 2 G, to 1. The claim, which I won't prove, is that ···! B2 ! B1 ! B0 ! Z ! 0 is exact. I suspect that one relatively painless proof of this involves using some other basis where d has a much simpler definition! Another proof is hinted at in Exercise 6.5.1(1) of Weibel. But let's press on, I'll prove enough to be able to compute H1 and H1 group homology and cohomology. Checking that the sequence is exact at the Z and B0 terms can be done by hand: clearly B0 surjects onto Z, and if one thinks of B0 as the ring A with (∗) = 1, then the map B1 ! B0 sends (g) to g − 1, and such things generate (as a left ideal) the kernel of B0 ! Z, as is easily checked. Let me check exactness at the B1 term, by mimicing the general proof. To check exactness here 2 I firstly have to check that d = 0, and it suffices to check this on generators. If (g; h) 2 B2 then d(g; h) = g(h) − (gh) + (g) and d of this is gh(∗) − g(∗) − gh(∗) + (∗) + g(∗) − (∗) = 0, so 2 d = 0, and to check it exactness I just need to check that anything in the kernel of d : B1 ! B0 is in the image of d : B2 ! B1. Here's how I'll do this: define s : B1 ! B2, Z-linear but not A-linear, by s(g(h)) = (g; h), and define s : B0 ! B1, Z-linear but not A-linear, by s(g(∗)) = (g). I claim that ds + sd is the identity map B1 ! B1, and this will prove exactness at B1. It suffices to check on Z-generators, so it suffices to check that (ds + sd)(g(h)) = g(h). But this is easy: (ds)(g(h)) = d(g; h) = g(h) − (gh) + (g), and (sd)(g(h)) = s(gd(h)) = s(gh(∗) − g(∗)) = (gh) − (g) so ds + sd is the identity. But this is enough to prove that the kernel of d : B1 ! B0 coincides with the image of d : B2 ! B1! We can also make the Bn free right A-modules by switching the G-action in the usual way: b ∗ g := g−1b. Using this trick we also get a projective resolution of Z in the category of right A-modules. 3 Ext. Let A be a ring, not necessarily commutative. Write A -Mod for the category of left A-modules (that is, A acts on the left) and Mod- A for the category of right A-modules. Both of these categories have enough injectives and enough projectives. If N is a left A-module then HomA(N; ∗) is covariant and left exact, so it has right derived n n functors which are called ExtA(N; ∗), and ExtA(N; X) can be computed using an injective reso- lution of X. We deduce immediately that if 0 ! X ! Y ! Z ! 0 is a short exact sequence of left A-modules then we have a long exact sequence 1 1 0 ! HomA(N; X) ! HomA(N; Y ) ! HomA(N; Z) ! ExtA(N; X) ! ExtA(N; Y ) !··· : 1 As a sanity check: if X is injective then 0 ! X ! Y ! Z ! 0 splits so it looks like ExtA(N; X) should be zero, and it is because X is an injective resolution of X and this complex has no term in degree 1. 2 Now consider a left A-module X and the functor HomA(∗;X). This functor is contravariant and if M ! N is surjective then HomA(N; X) ! HomA(M; X) is injective, so HomA(∗;X) is left exact and contravariant, so it has right derived functors, and these functors can be computed using a projective resolution of N (because it's injective in the opposite category to A -Mod). n The funny thing is that these derived functors are just ExtA(∗;X) again! The reason for ∗ this is that if N and X are left R-modules and I is an injective resolution of X and P∗ is a ∗ projective resolution of N, then there is a double complex called Hom(P∗;I ) and the associated ∗ total complex has natural maps to Hom( P∗;X) and to HomA(N; I ) and both of these natural maps are quasi-isomorphisms. This is explained much more carefully in section 2.7 of Weibel's homological algebra book. We deduce that if 0 ! L ! M ! N ! 0 is a short exact sequence of left A-modules then we get a long exact sequence 1 1 0 ! HomA(N; X) ! HomA(M; X) ! HomA(L; X) ! ExtA(N; X) ! ExtA(M; X) !··· : 1 The sanity check: if N is projective then the sequence splits, so ExtA(N; X) should be zero, and it is because we computed it using a projective resolution of N.
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