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Math 250a: Higher Algebra Problem Set #6 (5 November 2004): Group and Galois cohomology
- Solve Exercises 2.1 and 2.2 in the Tate notes (identification of H^2 (G, A) with extensions of G by A).
Here’s an equivalent description of group cohomology that more closely resembles the simplicial (co)homology you may have seen in the context of algebraic topology:
- [Homogeneous cochains] Let G be a group, and A a G-module. To any r-cochain f : Gr^ → A associate the function fhomog : Gr+1^ → A defined by
fhomog(a 0 , a 1 ,... , ar ) := a 0 f (a− 0 1 a 1 , a− 1 1 a 2 ,... , a− r−^11 ar ).
Prove that fhomog is “homogeneous” in the sense that
g fhomog(a 0 , a 1 ,... , ar ) := fhomog(ga 0 , ga 1 ,... , gar )
for all g, ai ∈ G, and show that every homogeneous F : Gr+1^ → A is fhomog for a unique r-cochain f : Gr^ → A, namely the one defined by
f (x 1 ,... , xr ) = F (1, x 1 , x 1 x 2 ,... , x 1 x 2 x 3... xr ).
For any F : Gr+1^ → A, define a function dF : Gr+2^ → A by
(dF )(a 0 , a 1 ,... , ar+1) := F (a 1 , a 2 ,... , ar+1) − F (a 0 , a 2 , a 3 ,... , ar+1) +F (a 0 , a 1 , a 3 ,... , ar+1) − + · · · +(−1)r+1F (a 0 , a 1 , a 2 ,... , ar ).
Show that d^2 = 0, that is, that d(dF ) = 0 for any F. Check that if F : Gr+1^ → A is homogeneous then so is dF. Finally, show that d(fhomog) = (δf )homog for every r-cochain f : Gr^ → A, and conclude that δ^2 = 0.
Noncommutative group cohomology. Let G be a group acting by automorphisms on another group M. (As usual, if G has a topological structure, we’ll ask that this map G × M → M be continuous, and will mainly be concerned with the case that G is a Galois group and M has discrete topology.) We may then think of M G, the subgroup of M fixed by G, as “H^0 (G, M )”; but, unless M is commutative, we cannot build the full Hr^ (G, M ) edifice. Still we can construct a useful H^1 (G, M ). While this is not a group, it has a distinguished point (analogous to the origin in the commutative case) and a “long exact sequence” of sorts.
The set H^1 (G, M ) is still defined as “cocycles” modulo an equivalence relation. A “cocycle” is a (continuous) map f : G → M such that f (στ ) = f (σ) σf (τ ) holds for all σ, τ ∈ G. Two cocycles f, g are said to be equivalent (“cohomologous”) if there exists m ∈ M such that g(σ) = m−^1 f (σ)σ(m) for all σ ∈ G. The distinguished point in H^1 (G, M ) is the class of the cocycle sending every element of G to the identity of M.
- i) Verify that “is cohomologous to” is an equivalence relation, and that if σ 7 → f (σ) is a cocycle then so is σ 7 → m−^1 f (σ)σ(m) for every m ∈ M. ii) Suppose 1 → M ′^ → M → M ′′^ → 1 is a short exact sequence of groups with a (continuous) G-action. Verify that the induced sequence 1 → M ′G^ → M G^ → M ′′G^ is exact. Construct a map δ : M ′′G^ → H^1 (G, M ′) — which should specialize to the connecting homomorphism δ : H^0 (G, M ′′) → H^1 (G, M ′) when M is abelian — and show that an element of M ′′G^ is in the image of M G^ if and only if δ maps it to the distinguished element of H^1 (G, M ′). iii) Show further that the image of δ consists of those elements of H^1 (G, M ′) that map to the distinguished element of H^1 (G, M ), and that the image of the map H^1 (G, M ′) → H^1 (G, M ) consists of those elements of H^1 (G, M ) that map to the distinguished element of H^1 (G, M ′′).
In particular, if M is commutative then we have checked the exactness of the sequence
0 → H^0 (G, M ′) → H^0 (G, M ) → H^0 (G, M ′′) → H^1 (G, M ′) → H^1 (G, M ) → H^1 (G, M ′′).
We next generalize the theorem H^1 (K/k, K∗) = 0 from K∗^ = GL 1 (K) to GLn(K). That is, we prove that every continuous cocycle f : Gal(K/k) → GLn(K) is of the form f (σ) = Aσ(A−^1 ) for some A ∈ GLn(K).
- i) Explain why it is enough to prove this when the normal extension K/k is of finite dimension. ii) Prove that f (σ) = Aσ(A−^1 ) if and only if each of the column vectors v of A satisfies the condition v = f (σ)σ(v) for every σ ∈ G. iii) Assume [K : k] < ∞. Generalizing the proof of Theorem 4.29, construct for each w ∈ Kn^ a vector v(w) ∈ Kn^ such that v(w) = f (σ)σ(v(w)) for every σ ∈ G.
- i) Prove that the K-span of {v(w) : w ∈ Kn} is all of Kn. (If not, then some nonzero K-linear functional u sends each v(w) to zero. Apply this condition to aw for each a ∈ K to reach a contradiction.) This together with Problem 4 completes the proof of H^1 (K/k, GLn(K)) = 0. ii) What are H^1 (K/k, M ) where M is the ax + b group over K, or the group SLn(K)?
Finally, we complete the explanation of why the theorem H^1 (K/k, K∗) = 0 is called “Hilbert Satz 90” even though Hilbert’s theorem is equivalent only to the special case of cyclic Gal(K/k).
- Solve the first part of Exercise 6.4(i) in the Tate notes [exactness of the beginning
0 → H^1 (G/H, AH^ ) → H^1 (G, A) → H^1 (H, A)G/H
of the inflation-restriction sequence associated to a normal (closed) subgroup H of G].
- Use this to deduce the vanishing of H^1 (K/k, K∗) when Gal(K/k) is a (finite) p-group from the special case of a cyclic group and the fact that every finite p-group has a normal subgroup of index p.
By Exercise 4.2 it follows that H^1 (K/k, K∗) is trivial for any finite normal extension K/k, and we have seen already that by compactness of Gal(K/k) this is enough to prove H^1 (K/k, K∗) = 0 in general. This shows how, with enough cohomological tools, we can recover the general result “automatically” from Hilbert’s special case of a cyclic cover.
Problem set is due in class Friday, Novermber the 12th.
