Galois Cohomology (Study Group)
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Group Cohomology (by Chris Birkbeck)
1.1 Denition of group cohomology
Let G be a group
Denition 1.1. A
G-module M , is an abelian group with an action of G. I.e., there is an homomorphism
θ : G → Aut(M ). Usual, we take g ∈ G, m ∈ M and dene φ(g)m = g · m with the properties
• g(m1 + m2 ) = gm1 + gm2
• g(g 0 m) = (gg 0 )m
We can extend this action to an action of Z[G] on M .
Denition 1.2. For a G-module M , let M G = {m ∈ M |gm = m∀g ∈ G}
Fact. M G = HomZ[G] (Z, M ). We have a functor (−)G dened by HomZ[G] (Z, −).
Let A, B, C be G-modules with the short exact sequence 0 → A → B → C → 0. Apply (−)G to get an exact
sequence 0 → AG → B G → C G .
Denition 1.3. Let G be as above, M a G-module. For i ≥ 0, dene C i (G, A) to be Maps(Gi , M ) with G0 = {1}.
Note that C 0 (G, M ) ∼
= M . We give C i (G, M ) a group structure pointwise and (gφ)(x1 , . . . , xn ) = g(φ(x1 , . . . , xn )).
Denition 1.4. We dene co-boundary homomorphism
C (G, M ), gi ∈ G, then, d−1 = 0 and for n ≥ 0
n
dn (f )(g1 , . . . , gn+1 )
=
dn : C n (G, M ) → C n+1 (G, M ) as follows. Let f ∈
g1 f (g2 , . . . , gn+1 )
n
X
(−1)i f (g1 , . . . , gi−1 , gi gi+1 , gi+2 . . . , gn+1 )
+
i=1
+(−1)n+1 f (g1 , . . . , gn )
n=0
d0 f (g) = gf − f
n=1
d1 (f )(g1 , g2 ) = g1 f (g2 ) − f (g1 g2 ) + f (g1 )
Exercise: Show that 0
d−1
d
d
→ C 0 (G, M ) →0 C 1 (G, M ) →1 . . . is a cochain complex.
n
Let Z (G, M ) = ker dn . These are called n-cocycles
Let B n (G, M ) = im dn−1 . These are called n-coboundaries.
Dene the nth Cohomology group to be
H n (G, M ) =
Z n (G, M )
for n ≥ 0
B n (G, M )
n=0
We have B 0 (G, M ) = {0}, Z 0 (G, M ) = {f ∈ C 0 (G, M )|gf = f ∀g ∈ G}, hence H 0 (G, M ) = M G
n=1
We have B 1 (G, M ) = {f |f (g) = ga−a for some a ∈ M }, and Z 1 (G, M ) = {f |f (g1 , g2 ) = g1 f (g2 )+f (g1 )},
and H 1 (G, M ) = Z 1 /B 1 . If G acts trivially then H 1 = Hom(G, M ).
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1.2 Topological groups
Let G be a topological (pronite group) and T a topological G-module (i.e., T is an topological abelian group with
continuous G-action)
i
Denition 1.5. We dene Ccts
(G, T ) = CtsMaps(Gi , T ).
Take T, T 0 , T 00 to be topological G-modules and let 0 → T 0 → T → T 00 → 0 be a short exact sequence.
i
i
Furthermore let S : T 00 → T be a continuous section (as spaces). We get 0 → Ccts
(G, T 0 ) → Ccts
(G, T ) →
n+1
i
00
n
Ccts (G, T ) → 0 is exact for all i ≥ 0. We dene dn : Ccts (G, T ) → Ccts (G, T ) as before and then we get a long
exact sequence
n+1
n
n
n
· · · → Hcts
(G, T 0 ) → Hcts
(G, T ) → Hcts
(G, T 00 ) → Hcts
(G, T 0 ) → . . .
1.3 Maps between cohomology groups
Denition 1.6. Let G, G0 be groups and M (respectively M 0 ) be a G-module (respectively G0 -module). We say
φ : G0 → G and ψ : M → M 0 are compatible if for g 0 ∈ G0 , a ∈ M , ψ(φ(g 0 )a) = g 0 ψ(a).
These induces a map C i (G, M ) → C i (G0 , M 0 ) which in turn give a map H i (G, M ) → H i (G0 , M 0 )
Example.
• Let H ≤ G, φ : H ,→ G and ψ = id : M → M . We get (restriction ) Res : H n (G, M ) → H n (H, M ) dened by
[f ] 7→ [f |H ].
• Let H ≤ G, φ : G → G/H , ψ : M H ,→ M . Then we get (ination ) Inf : H n (G/H, M H ) → H n (G, M ) dened
by inf(f )(g) = f (g) for f ∈ Z n (G/H, M H ).
• We have an exact sequence
inf
res
trans
inf
0 → H 1 (G/H, M H ) → H 1 (G, M ) → H 1 (H, M )G/H → H 2 (G/H, M H ) → H 2 (G, M ) →
1.4 H 1 extension
∪
Let A, B be G-modules, give A⊗Z B a G-action. There exists a family of maps C n (G, A)⊗C n (G, B) → C n+m (G, A⊗
B). For n = 0, AG ⊗ B G → (A ⊗ B)G . This will extend to a map on H n .
We also get d(f ∪ g) = df ∪ g + (−1)n (f ∪ dg).
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