Download Group cohomology - of Alexey Beshenov

Group cohomology
These are my notes from class. I put together the exercise sessions and lectures and also slightly
rearranged the exposition. All typos and mistakes are mine, so please write me about them:
[email protected].
Version 2013-12-02. What is missing is the last exercise session with Dajano.
The latest edition is available at http://176.58.104.245/ALGANT/EREZ/erez.pdf
Contents
1 Motivating examples (lower cohomology groups)
1.1 G-invariants and G-coinvariants . . . . . . . . . . . . . .
1.2 Group H 1 (G, A) and split extensions . . . . . . . . . . .
1.3 Group H 2 (G, A) and extensions . . . . . . . . . . . . . .
1.4 H 1 (Gal(L/K), L× ) and Hilbert’s Theorem 90 . . . . . . .
1.5 Group H 2 (Gal(L/K), L× ) and crossed product algebras
1.6 Bar-resolution . . . . . . . . . . . . . . . . . . . . . . . . .
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3
. 4
. 6
. 8
. 9
. 9
. 11
2 Preliminaries from algebraic topology
3 Preliminaries from homological algebra
3.1 Projective modules . . . . . . . . . . . . . .
3.2 Injective modules . . . . . . . . . . . . . . .
3.3 Projective resolutions . . . . . . . . . . . .
3.4 Injective resolutions . . . . . . . . . . . . .
3.5 Right derived functor Ext•R and H • (G, A)
3.6 Left derived functor Tor•R and H• (G, A) .
3.7 Long exact sequences . . . . . . . . . . . .
13
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14
14
16
17
20
21
22
22
4 Group (co)homology via resolutions of Z
24
4.1 Cyclic groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
4.2 Free resolutions of Z coming from topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
4.3 Bar-resolution revisited . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
5 Cup-product on cohomology
31
6 Cohomology of cyclic groups H • (C, Z/`)
33
6.1 Groups H n (C, Z/`) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
6.2 Multiplication on H • (C, Z/`) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
1
7 Restriction, inflation, corestriction
7.1 H • (−, −) as functors of pairs . . . . .
7.2 Inflation-restriction exact sequence .
7.3 Restriction and corestriction . . . . .
7.4 Induction, coinduction, and Shapiro’s
. . . . .
. . . . .
. . . . .
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37
37
39
41
42
8 Cohomology H • (GLn (Fq ), Z/`Z). Connections with K-theory
43
8.1 K0 of a ring (Dajano) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
9 Class field theory and group cohomology
9.1 Introduction . . . . . . . . . . . . . . . . .
9.2 Galois cohomology . . . . . . . . . . . . .
9.3 Tate cohomology Ĥ n (G, A) . . . . . . . .
9.4 Kummer theory . . . . . . . . . . . . . . .
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2
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44
44
44
45
48
1
Motivating examples (lower cohomology groups)
Let G be a group and let A be a (left) G-module. This means that A is an abelian group equipped with
an action by G on the left.
G×A
→
A,
(g, a)
7Ï
g · a.
We ask that 1 ∈ G acts trivially, and that
(g h) · a = g · (h · a),
g · (a + a0 ) = g · a + g · a0 .
A G-module A can be regarded as a Z[G]-module (in the usual sense of modules over a ring), where
Z[G] is the group ring of G. This is because a Z[G]-module structure is a morphism Z[G] → EndZ A to
the ring of automorphisms of A, but there is an adjunction
HomRing (Z[G], R) ' HomGrp (G, R× ).
So a morphism Z[G] → EndZ A corresponds to a morphism G → AutZ A.
Example 1.1. Recall that the group ring K[G] is a free K-module over G:
M
K[G] :=
K · g,
g∈G
and multiplication in K[G] extends linearly the multiplication in G. If G is not commutative, then K[G]
is a noncommutative ring.
For example, for G = Z one has the group ring K[G] ' K[X, X −1 ].
For G = Z/n one has the group ring K[G] ' K[X]/(X n − 1).
We will be dealing with K = Z. For brevity we will write “ZG” instead of “Z[G]”.
The operation in the G-module A is denoted additively as “+” and the action of G is denoted multiplicatively as “g · a”. One should be careful though: the inner operation in A can be multiplication as
well. Here is a very important example: if L/K is field extension, then the Galois group G = Gal(L/K)
acts on L× , making L× into a G-module.
Our goal is to define functors
ZG-mod
→
Families of abelian groups
(G, A)
7Ï
(H i (G, A))i≥0
(cohomology)
(G, A)
7Ï
(Hi (G, A))i≥0
(homology)
H i (G, A) are to be thought as invariants of the group action. They can be defined by “explicit”
formulas, but also as derived functors of the functors A 7Ï H 0 (G, A) and A 7Ï H0 (G, A).
3
1.1
G-invariants and G-coinvariants
Definition 1.1. Let A be a G-module. We define the fixed points (G-invariants) in A by
AG := H 0 (G, A) := {a ∈ A | g · a = a for all g ∈ G}.
Similarly the G-coinvariants of a G-module A are by definition
AG := H0 (G, A) := A/IG A.
Here IG is the augmentation ideal in ZG, defined to be the kernel of the augmentation map
X
ε
ZG
Ï
−
ag g
7Ï
g∈G
Z,
X
ag .
g∈G
One has IG A = ha (g −
P1) | a ∈ A, gP∈ Gi. Indeed,
P if x ∈ IG , then x =
so that we can write x = g∈G ag g − g∈G ag = g∈G ag (g − 1).
P
g∈G
ag g with
P
g∈G
ag = 0,
Proposition 1.1. Let IG2 be the subgroup of IG given by all products x y where x, y ∈ IG . The factorgroup IG /IG2 is the abelianization of G, i.e. IG /IG2 ' G/[G, G].
Proof. Define a map
φ : G/[G, G] →
g
7Ï
IG /IG2 ,
g − 1.
It is well-defined since IG /IG2 is an abelian group (any morphism from G to an abelian group must
factor through G/[G, G]). It is a group homomorphism since modulo IG2
φ(g h) = (g h − 1) = (g − 1) (h − 1) +(g − 1) + (h − 1) ≡ (g − 1) + (h − 1) = φ(g) + φ(h).
{z
}
|
∈IG2
The inverse is given by φ−1 : g − 1 7Ï g.
The meaning of invariants and coinvariants is the following:
• AG is the largest G-submodule of A on which the action of G is trivial.
• AG is the largest G-quotient of A on which the action of G is trivial.
The functor A 7Ï H 0 (G, A) is a left exact functor in A, basically because the fixed points H 0 (G, A) can
be identified with HomZG (Z, A) where Z is a ZG-module with the trivial action of G. (Indeed, picking a
map Z → A is just picking an element a ∈ A; to be G-equivariant for this map means that a is a fixed
point.)
However, A 7Ï H 0 (G, A) is not a right exact functor. If we have a short exact sequence
0→A→B→C→0
then we get an exact sequence
0 → H 0 (G, A) → H 0 (G, B) → H 0 (G, C)
and in general we can’t continue with “→ 0”. This is in fact the main reason to have this course :-) To
continue such a sequence on the right, one has to introduce right derived functors.
4
Similarly, the coinvariants A 7Ï H0 (G, A) is a right exact functor, sending a short exact sequence
0→A→B→C→0
to an exact sequence
H0 (G, A) → H0 (G, B) → H0 (G, C) → 0
But in general it is not left exact, and to continue it on the left one needs left derived functors.
Actually H0 (G, −) is right exact because it is the tensor product functor.
Proposition 1.2. H0 (G, A) ' Z ⊗ZG A, where Z is considered to be a right ZG-module with the trivial
action.
Proof. The isomorphism is given by
AG
a
→
Z ⊗ZG A,
7Ï 1 ⊗ a
where a is the class of a ∈ A in the quotient A/IG A. It is well-defined, since for a − g · a ∈ IG A
a − g · a 7Ï 1 ⊗ a − 1 ⊗ g · a = 1 ⊗ a − 1 · g ⊗ a = 0.
The inverse to this map is given by
Z ⊗ZG A
→
AG ,
n⊗a
7Ï
n a.
Example 1.2. Consider the action of S3 on the vector space R3 by permuting coordinates. That is,
σ · (x1 , x2 , x3 ) := (xσ(1) , xσ(2) , xσ(3) ). Let L be the line generated by (1, 1, 1). We have a short exact
sequence of groups
0 → L → R3 → R3 /L → 0.
There is an induced action of S3 on R3 /L, and it is actually a short exact sequence of ZS3 -modules.
The fixed points are the following:
H 0 (G, L) = L,
H 0 (G, R3 ) = L,
H 0 (G, R3 /L) = 0.
Later on we will see that any short exact sequence of ZG-modules
0→A→B→C→0
as above gives a long exact sequence in cohomology
δ
0 → H 0 (G, A) → H 0 (G, B) → H 0 (G, C) Ï
− H 1 (G, A) → H 1 (G, B) → H 1 (G, C) → · · ·
But first let’s see at least what are the cohomology groups “H 1 (G, A)”.
5
1.2
Group H 1 (G, A) and split extensions
Definition 1.2. Let A be a group. We say that E is an extension of A by G if we have a short exact
sequence
i
π
1→AÏ
− E−
Ï G → 1.
A can be identified with a normal subgroup i(A) / G, and so G acts on a ∈ A by conjugation:
i(g · a) = g · i(a) · g −1 ,
where π(g) = g
Conversely, given a prescribed action of G on A, one may ask what are the extensions giving rise
to this action, up to some reasonable equivalence.
Definition 1.3. We say that two extensions are equivalent if there is an morphism φ : E → E 0 making
the diagram commute:
>E
i
/A
1
π
>G
' φ
i0
E0
/1
π0
Note that φ is automatically an isomorphism (by five lemma).
There is a special class of extensions: the split ones.
Definition 1.4. We say that a short exact sequence
1→A→E→G→1
is split if there is a section s : G → E, a group homomorphism such that π ◦ s = 1G :
/A
1
/E
\
π
/G
/1
s
If an extension is split, then each element of E can be uniquely written as i(a) s(g) for some a ∈ A
and g ∈ G. So as a set we can identify E with A × G. Further, we can derive the group operation, using
the identity
i(g a) = s(g) i(a) s(g)−1 .
(i(a) s(g)) (i(a0 ) s(g 0 )) = i(a) s(g) i(a0 ) s(g)−1 s(g) s(g 0 ) = i(a) i(g a0 ) s(g) s(g 0 ) = i(a + g a0 ) s(g g 0 ).
|
{z
}
(The operation in A is written additively!)
So the group operation is given by
(a, g) · (a0 , g 0 ) = (a + g · a0 , g g 0 ).
This is called the semidirect product A o G.
So if an extension of A by G (with a prescribed action of G on A) splits, then this extension is
isomorphic to A o G. But we can ask what are the possible splittings s : G → E of a given split
extension, up to some conjugacy.
6
Definition 1.5. Let s, s0 : G → E be two group homomorphisms giving sections. We say that they are
conjugate if there exists a ∈ A such that s(g) = i(a) s0 (g) i(a)−1 .
Claim. Splittings correspond to crossed homomorphisms d : G → A such that
d(g g 0 ) = d(g) + g · d(g 0 ).
(1)
Proof. Consider a split extension
1
/ AoG
`
/A
π
/G
/1
s
The splitting s must have form g 7Ï (d(g), g) for some d : G → A. We should have s(g g 0 ) = s(g) s(g 0 ),
which means
(d(g g 0 ), g g 0 ) = (d(g), g) · (d(g 0 ), g 0 ) = (d(g) + g · d(g)0 , g g 0 ).
Thus the identity d(g g 0 ) = d(g) + g · d(g)0 .
Claim. Two crossed homomorphisms corresponding to conjugate sections s, s0 differ by a function
G
→ A,
g
7Ï g a − a,
for some fixed a ∈ A (it is indeed a crossed homomorphism). That is,
d(g) = d0 (g) + (g a − a).
Proof. Let s(g) = (d(g), g) and s0 (g) = (d0 (g), g) be two splittings. The conjugacy means that for some
a∈A
s(g) i(a) = i(a) s0 (g).
So we have
(d(g), g) · (a, 1) = (a, 1) · (d0 (g), g),
(d(g) + g a, g) = (a + d0 (g), g),
which is equivalent to d(g) − d0 (g) = a − g a.
A homomorphism d : g 7Ï a − g a is also a crossed homomorphism:
d(g) + g · d(g 0 ) = a − g a + g (a − g 0 a) = a − g g 0 a = d(g g 0 ).
Now the crossed homomorphisms d : G → A form an abelian group (with the obvious “pointwise”
addition), and the homomorphisms of the form g 7Ï g a − a form its subgroup.
Proposition 1.3. The splittings of a given split extension
1→A→AoG →G →1
up to conjugacy classes are given by the abelian group
H 1 (G, A) :=
“crossed homomorphisms” d : G → A such that d(g g 0 ) = d(g) + g · d(g 0 )
.
maps of the form g 7Ï (g · a − a)
We say that the crossed homomorphisms are 1-cocycles, while the homomorphisms g 7Ï g · a − a
are 1-coboundaries. So cohomology is given by cocycles modulo coboundaries.
7
1.3
Group H 2 (G, A) and extensions
Again, we look at the extensions
1→A→E→G→1
giving rise to a fixed action of G on A. Now we do not ask such extension to be split.
We consider a set-theoretical section s : G → E (we still ask π ◦ s = 1G ), which is not necessarily a
homomorphism. One can measure to which extent it fails to be a homomorphism:
s(g) s(h) = i(f(g, h)) · s(g h),
where f : G × G → A is some set-theoretical map.
We assume “normalization” s(1) = 1, which implies
f(g, 1) = f(1, g) = 0A .
(2)
As a set E = A × G, since all elements are of the form i(a) s(g), and the group law on A × G is given
by
(a, g) · (b, h) = (a + g b + f(g, h), g h).
So a function f : G × G → A together with an action of G on A determine an extension.
One can’t start with an arbitrary map f : G ×G → A since the multiplication may fail to be associative.
So we need to impose some conditions on f:
((a, g) · (b, h)) · (c, k) = (a + gb + f(g, h), gh) · (c, k) = (a + gb + f(g, h) + ghc + f(gh, k), ghk),
(a, g) · ((b, h) · (c, k)) = (a, g) · (b + h c + f(h, k), hk) = (a + gb + ghc + g f(h, k) + f(g, hk), ghk),
a + gb + f(g, h) + ghc + f(gh, k) = a + gb + ghc + g f(h, k) + f(g, hk),
g f(h, k) − f(gh, k) + f(g, h k) − f(g, h) = 0.
(3)
Now one can do a routine check that starting with f : G × G → A satisfying (3), one makes A × G into
a group. So extensions 1 → A → E → G → 1 with normalized section are classified by such “2-cocycles”
f : G × G → A.
For an extension 1 → A → E → G → 1 We had before a section s : G → E, and now consider another
section s0 : G → E of the same extension. They must differ by some function u : G → A, namely:
s0 (g) = i(u(g)) s(g).
The normalization imposes u(1) = 0A . This section s0 gives rise to another 2-cocycle f 0 . A routine
computation gives f 0 in terms of f:
s0 (g) s0 (h)
=
(i(u(g)) s(g)) (i(u(h)) s(h))
= i(u(g)) s(g) i(u(h)) s(g)−1 s(g) s(h)
{z
}
|
= i(u(g)) i(g u(h)) s(g) s(h)
= i(u(g) + g u(h)) i(f(g, h)) s(g h)
= i(u(g) + g u(h) + f(g, h)) i(u(g h))−1 i(u(g h)) s(g h)
|
{z
}
= i(u(g) + g u(h) + f(g, h) − u(g h)) s0 (g h).
8
Thus the difference between the two cocycles is
f 0 (g, h) − f(g, h) = u(g) + g u(h) − u(g h).
(4)
So the extensions are classified by the following group:
H 2 (G, A) :=
“2-cocycles” f : G × G → A satisfying (3) and normalization (2)
.
“2-coboundaries” (g, h) 7Ï u(g) + g u(h) − u(g h) for some normalized u : G → A
(We can check that 2-coboundaries satisfy (3)!)
1.4
H 1 (Gal(L/K), L× ) and Hilbert’s Theorem 90
Probably the first result in group cohomology is due to Hilbert (1897).
Theorem 1.1 (Hilberts Satz 90). Let L/K be a finite Galois extension. Then H 1 (Gal(L/K), L× ) = 0.
Proof. We need to show that any 1-cocycle d : G → L× is automatically a 1-coboundary.
In multiplicative notation being a 1-cocycle means that
d(σ σ 0 ) = σ(d(σ 0 )) · d(σ).
Being a 1-coboundary in multiplicative notation means that d has form
d(σ) =
a
σ(a)
for some a ∈ L× ,
d(σ) 6= 0, and the automorphisms σ ∈ G are linearly independent over L, so there is an element
b ∈ L× such that
X
a :=
d(σ) · σ(b) 6= 0.
σ∈G
Now for σ ∈ G one has
σ(a)
X
=
σ(d(σ 0 )) · σ σ 0 (b)
σ 0 ∈G
X d(σ σ 0 )
· σ σ 0 (b)
d(σ)
σ 0 ∈G
1 X
=
d(σ σ 0 ) · σ σ 0 (b)
d(σ) 0
σ ∈G
1 X
a
=
d(σ 00 ) · σ 00 (b) =
.
d(σ) 00
d(σ)
=
σ ∈G
1.5
Group H 2 (Gal(L/K), L× ) and crossed product algebras
Let L/K be a field extension. Let G be the automorphism group of L as a K-algebra. We have K ≤ LG .
The extension is called Galois if K = LG .
Example 1.3. Not every extension is Galois. Consider K = Q and f(X) = X 3 − 2. Take an extension
L = K[X]/(f). Since the result contains one root of X 3 − 2 but not the other two, we have K 6= LG .
9
Example 1.4. Consider an extension Q(ζn )/Q where ζn is the n-th root of unity. It is a Galois extension
with Galois group G ' (Z/nZ)× .
If G is a Galois group of L/K, then G acts on L. It also descends to actions on certain subsets. For
instance, in case of a number field L/Q we have an action on the ring of integers OL .
Recall that we have the quaternion R-algebra H.
H = R1 ⊕ Ri ⊕ Rj ⊕ Rk.
It is a skew-field, with multiplication given by
1
i
j
k
1
1
i
j
k
i
i
−1
−k
+j
j
k
j
k
+k −j
−1 +i
−i −1
C is the maximal commutative subfield of H. Also dimR H = (dimR C)2 .
Moreover, H ⊗R C ' M2 (C) with dim M2 (C) = 22 .
H
n
C
n2
n
R
In general, if we start from a finite Galois extension L/K of degree [L : K] = |G|, then we can define
an algebra
M
V :=
L · eσ
σ∈G
with a product
eσ · eτ := f(σ, τ) eστ ,
eσ · x := σ(x) eσ .
Here f : G × G → L× , and σ(x) denotes the Galois action of σ on x ∈ L.
A tedious verification shows that the associativity of the product above imposes the same associativity
condition (3) on f as we have seen before.
This construction leads to crossed product algebras (L/K, f). Two such algebras (L/K, f) and
(L/K, f 0 ) are isomorphic iff the 2-cocycles f and f 0 differ by a 2-coboundary element as in (4).
Example 1.5. H is a crossed product algebra (C/R, f).
If one takes f(g, h) = 1 for all g, h ∈ G, then (L/K, f) ' Mn (K).
The cross product algebras up to isomorphism correspond to H 2 (Gal(L/K), L× ).
Later on we will see how to calculate such cohomology groups. In the example with quaternions
actually H 2 (Gal(C/R), C× ) ' Z/2, so there are two algebras: the trivial cocycle gives M2 (R) and the
nontrivial cocycle gives H.
Note that this example shows that H 2 (Gal(L/K), L× ) is nontrivial, while H 1 (Gal(L/K), L× ) is trivial by
Hilbert’s theorem 90.
10
1.6
Bar-resolution
We have seen explicit definitions of cohomology groups H 0 (G, A), H 1 (G, A), H 2 (G, A), coming from some
algebraic problems. Now we are going to look at certain formulas giving all H n (G, A) for n = 0, 1, 2, . . .
Consider Z as a ZG-module with the trivial action of G. Now we will build an exact sequence of
ZG-modules
d
d
ε
2
1
· · · → B2 −
Ï
B1 −
Ï
B0 Ï
− Z→0
(5)
Here B0 := ZG, and the “augmentation” ε is given by
→ G,
ZG
g
7Ï 1
For n > 0 by Bn we denote the free ZG-module generated by symbols [[g1 | · · · | gn ]] for gi ∈ G (the
name “bar-resolution” comes from this notation). The differential dn : Bn → Bn−1 is defined as follows:
dn [[g1 | · · · | gn ]]
:= g1 [[g2 | · · · | gn ]] − [[g1 g2 | g3 | · · · | gn ]]
+[[g1 | g2 g3 | · · · | gn ]] − · · · + (−1)n [[g1 | · · · | gn−1 ]]
X
g1 [[g2 | · · · | gn ]] +
(−1)i [[g1 | · · · | gi gi+1 | · · · | gn ]] + (−1)n [[g1 | · · · | gn−1 ]].
=
1≤i≤n−1
In particular, ZG is generated by symbol [[ ]], and the differential d1 : B1 → ZG is given by
d1 [[g1 ]] := g1 [[ ]] − [[ ]].
One can check that (5) is indeed an exact sequence, but we postpone it for §4 when it will be clear
why do we need that and what is going on.
Now let A be a ZG-module. We apply HomZG (−, A) to the sequence (5) and obtain a chain complex
(but not necessarily exact anymore, which is the key point)
d0
d1
0 → HomZG (B0 , A) −Ï HomZG (B1 , A) −Ï HomZG (B2 , A) → · · ·
Since Bn is just the free module generated by all possible tuples (g1 , . . . , gn ) ∈ G
· · × G}, we see
| × ·{z
n
that the elements of C i (G, A) := HomZG (Bi , A) can be identified with G-equivariant maps
f : G n := G
· · × G} → A.
| × ·{z
n
Further, we can write down explicitly the maps dn : f 7Ï f ◦ dn+1 :
dn+1
G
/A
=
f
GO n
dn f
n+1
We see that the coboundary maps
d0
d1
0 → C 0 (G, A) −Ï C 1 (G, A) −Ï C 2 (G, A) → · · ·
are given by formulas
11
(dn f)(g1 , . . . , gn+1 ) := g1 f(g2 , . . . , gn+1 ) +
X
(−1)i f(g1 , g2 , . . . , gi gi+1 , . . . , gn+1 ) + (−1)n+1 f(g1 , . . . , gn )
1≤i≤n
(remember that f are G-equivariant).
(d0 a)(g1 )
(d f)(g1 , g2 )
1
(g f)(g1 , g2 , g3 )
2
= g1 · a − a,
= g1 f(g2 ) − f(g1 g2 ) + f(g1 ),
= g1 f(g2 , g3 ) − f(g1 g2 , g3 ) + f(g1 , g2 g3 ) − f(g1 , g2 ),
..
.
Now define
n-cocycles Z n (G, A)
n-coboundaries Bn (G, A)
:=
:=
ker dn ,
im dn−1 .
We see that Z 1 (G, A), B1 (G, A), Z 2 (G, A), B2 (G, A) are given by the same formulas that appeared in
the examples of §1.
One can check that “d2 ” = dn ◦ dn−1 = 0 (because of the exactness of (5)), which means Bn (G, A) ≤
n
Z (G, A), and we can consider factor-modules
H n (G, A) :=
Z n (G, A)
.
Bn (G, A)
In particular,
H 0 (G, A) = ker d0 = {a ∈ A | g a = a for all g ∈ G}.
We are going to understand why this definition is so and see how one computes H n (G, A) in some
cases.
12
2
Preliminaries from algebraic topology
The main reference is J. P. May, A Concise Course in Algebraic Topology.
Were “defined” π1 , covering spaces, singular homology, cellular homology. The only point where
we need it is §4.2.
13
3
Preliminaries from homological algebra
Here we collect some homological algebra that is needed for group cohomology. For full generalities
see Weibel, An introduction to homological algebra and Grothendieck’s “Tohoku paper”. We will work
exclusively with the category of R-modules. More specifically, we will have in mind R = ZG.
3.1
Projective modules
Definition 3.1. An R-module P is called projective if one of the following equivalent conditions hold:
(P1) HomR (P, −) is an exact functor. That is, for any short exact sequence
p
i
0→AÏ
− BÏ
− C→0
one gets a short exact sequence
p∗
i
∗
0 → HomR (P, A) −
Ï
HomR (P, B) −
Ï HomR (P, C) → 0
(P2) There exists an R-module Q such that P ⊕ Q is a free R-module.
(P3) For each morphism of R-modules P → C and each surjection B C there is a lifting P → B:
P
∃
B

//C
/0
(P4) Any short exact sequence
0→A→B→P→0
splits.
Let us understand these definitions.
(P2) Ñ (P3) It is clear that the free modules and their direct summands enjoy the lifting property.
(P3) Ñ (P4) Consider a short exact sequence 0 → A → B → P → 0. Apply the lifting property to get a
section:
P
s
0
/A
/B

id
/P
//0
(P4) Ñ (P2) Consider an exact sequence 0 → A → F → P → 0 where F is a free R-module surjecting
to P. It splits, and so F ' A ⊕ P.
The covariant functor HomR-mod (M, −) is automatically left exact for any R-module M. More precisely, we have the following.
14
Proposition 3.1. Let
p
i
0→AÏ
− BÏ
− C
be a sequence of R-modules. Then it is exact iff for any R-module M the induced sequence
p∗
i
∗
0 → HomR-mod (M, A) −
Ï
HomR-mod (M, B) −
Ï HomR-mod (M, C)
exact.
Remark. Recall that we have an adjunction − ⊗R M a HomR-mod (M, −):
HomR-mod (N ⊗R M, L) ' HomR-mod (N, HomR-mod (M, L)).
Whenever two additive functors are adjoint, the left adjoint is right exact and the right adjoint is left exact.
So the fact that HomR-mod (M, −) is left exact follows from this adjunction. We also see that − ⊗R M is right exact.
However, HomR-mod (M, −) is not always right exact! Consider for instance the short exact sequence
of abelian groups
n
0→Z−
Ï Z → Z/n → 0
Apply to this HomZ-mod (Z/n, −):
0 → Hom(Z/n, Z) → Hom(Z/n, Z) → Hom(Z/n, Z/n) → 0
The problem is that Z/n is a torsion group, while Z is torsion-free. So Hom(Z/n, Z) = 0, and the last
sequence is far from being exact.
Example 3.1. Modules over a field k are all free, so they are projective as well.
Example 3.2. If R is a PID, then any submodule of a free R-module is free. So by (P2) we conclude
that over a PID every projective R-module is a free module.
Example 3.3. It is trivial to see that not every projective module is free. Consider two rings R and S
and their product R × S. Now R × 0 is a summand of R × S, so it is projective as an R × S-module.
However, it is not free. Indeed, the R × S-action is not faithful:
(0, s) · (r, 0) = 0 for any r ∈ R, s ∈ S.
Example 3.4. We will show that Q is not a projective Z-module using the splitting criterion (P4).
Splitting means that we have a section
0
/A
/B
c
p
/P
/0
s
such that p ◦ s = idP . So to show that Q is not a projective Z-module, we can find a surjective map
A Q having no section.
Let F be the free Z-module generated by positive integers 1, 2, 3, . . ., and consider the map
F
→
[n] 7Ï
Q,
1
.
n
It is a surjection. However, there is no section s : Q → F. Suppose for the sake of contradiction a
section exists. Then
15
s(1) =
X
ai · [i].
i
Take n > |ai | and let
X
1
bi · [i].
s( ) =
n
i
Then one has
s(1) =
X
ai · [i] = s(n ·
X
1
)=n
bi · [i].
n
So ai = n bi , contradicting n > |ai |.
3.2
Injective modules
Definition 3.2. An R-module I is called injective if one of the following equivalent conditions hold:
(I1) HomR (−, I) is an exact functor. That is, for any short exact sequence
p
i
0→AÏ
− BÏ
− C→0
one gets a short exact sequence
p∗
i
∗
0 → HomR (C, I) −
Ï HomR (B, I) −
Ï
HomR (A, I) → 0
(I2) If I is a submodule of some other R-module M, then there exists another submodule J of M such
that I ⊕ J = M (that is, I + J = M and I ∩ J = {0}).
(I3) For each morphism of R-modules C → I and each injection C ,Ï C there is a morphism B → I
making the diagram commute:
IO _
∃
/ C 
0
/B
(I4) Any short exact sequence
0→I→K→L→0
splits.
We do not check the equivalence of (I1)–(I4), since they are dual to (P1)–(P4).
Example 3.5. For k-vector spaces, say, (I2) is obvious. Whenever we have a subspace V ⊆ U, there is
its complement V ⊥ such that V ⊕ V ⊥ = U (any basis for V can be completed to a basis of U).
Proposition 3.2 (Baer criterion). An R-module I is injective iff every map a → I extends to a map
R → I, where a ⊆ R is a left ideal of R.
IO _
0
/ a

16
/R
Proof. Of course if I is injective, then this property must be satisfied.
In the other direction . . . . . (TODO: fill in)
Corollary 3.1. If I is an injective module, then it is divisible: for any r ∈ R \ {0} the multiplication
by r is a surjection I → I.
Over a PID the other direction is true as well: if I is a divisible module, then it is injective. Hence
over a PID, injective modules coincide with divisible modules.
Example 3.6. Q and Q/Z are divisible Z-modules, hence injective.
Z is not divisible, hence it is not an injective Z-module.
The contravariant functor HomR-mod (−, M) is automatically left exact for any R-module M. More
precisely, we have the following.
Proposition 3.3. Let
p
i
AÏ
− BÏ
− C→0
be a sequence of R-modules. Then it is exact iff for any R-module M the induced sequence
p∗
i
∗
0 → HomR-mod (C, M) −
Ï HomR-mod (B, M) −
Ï
HomR-mod (A, M)
exact.
This is dual to proposition 3.1.
However, HomR-mod (−, M) is not always right exact! Consider our favorite short exact sequence of
abelian groups
n
0→Z−
Ï Z → Z/n → 0
Apply to this HomZ-mod (−, Z/n):
0
/ Hom(Z/n, Z/n)
/ Hom(Z, Z/n)
f
/ Hom(Z, Z/n)
/ f ◦n
/0
Having in mind that the last map is induced by multiplication n : Z → Z, we see that it is 0. Indeed,
any homomorphism f : Z → Z/n should take nZ to 0.
3.3
Projective resolutions
Let M be an R-module. Then a free resolution of M is simply an exact sequence of R-modules
ε
· · · → P2 → P1 → P0 Ï
− M
Where Pi ’s are free R-modules. One can always obtain such a free resolution: take P0 to be the
ε
free module generated by the elements of M and the obvious surjection P0 Ï
− M. Then take P1 to be
the free module surjecting to ker ε, and so on.
More generally, a projective resolution is a resolution where P0 , P1 , P2 , . . . are projective R-modules.
Such a resolution always exists because the category of R-modules has enough projectives, i.e. for
any module M there is a projective module P with a surjection P M.
Forming a resolution step-by step, using the fact that there are enough projectives, can be depicted
by the following commutative diagram:
17
0
···
"
<0
= M3 p
d
0
!
/ P3
"
==
< M2
M1 p
/ P2
=
d
!!
<0
.
!
/ P1
d
d
!!
"
0
0
< M0
.
/ P0
=
ε
//M
/0
"
0
0
The diagonal sequences are exact. The differentials are defined by compositions, and the resulting
sequence is exact.
We claim that different projective resolutions of M are homotopy equivalent. Recall some definitions
from homological algebra.
Suppose (C• , d• ) and (C•0 , d•0 ) are two complexes. A morphism between complexes f• : C• → C•0 is a
morphism commuting with the differentials
/ Cn+1
···
fn+1
/ ···
fn
/ C0
n+1
···
/ Cn
d
/ Cn0
d0
/ ···
Such a map induces a map on homology
H• (f) : H• (C• ) → H• (C•0 ).
Now we say that two morphisms f, g : C• → C•0 are homotopic (f ∼ g) if there is a map h =
0
(hn : Cn → Cn+1
)
···
/ Cn+1
/ Cn
h
···
|
/ C0
n+1
f
d0
g
d
/ Cn−1
/ ···
/ C0
n−1
/ ···
h
|
/ Cn0
such that
d0 h + hd = f − g.
If f ∼ g then H• (f) = H• (g).
Theorem 3.1. Assume we have two projective resolutions
···
/ P2
/ P1
/ P0
···
/ P0
2
/ P0
1
/ P0
0
ε
ε0
/M
/0
/M
/0
Then there exist a unique up to homotopy map of chain complexes f• : P• → P•0 .
18
···
···
/ P2
/ P1
/ P0
f2
f1
f0
/ P0
2
/ P0
1
ε
/ P0
0
ε0
/M
/0
/M
/0
To obtain these fn one essentially uses the lifting property that projective modules posses (by their
definition).
Proposition 3.4 (Lifting property). Let P be a projective module. Consider a diagram with j ◦ φ = 0
where the row is exact at M:
P
0
φ
M0
/M
i
j
!
/ M 00
Then there exists a morphism f : P → M 0 making the diagram commute:
P
f
M0
0
φ
}
/M
i
j
!
/ M 00
One sees that from the lifting property follow the following two:
(A) Let P be a projective module. Assume in the following diagram d2 f d = 0 and the row M 0 →
M → M 00 is exact.
d
P
f0
/Q
f
M0
d1
/M
d2
/ M 00
Then there exists a map f 0 making the square commute.
(Apply the lifting property to φ = f d.)
(B) Let P be a projective module. Assume in the following diagram d2 h d = 0 and the row M 0 →
M → M 00 is exact.
P
h
M
0
/0
d
0
f
}
d1
}
/M
h
d2
Then there exists a map h0 such that d1 h0 + h d = f.
(Apply the lifting property to φ = f − h d.)
19
/ M 00
One can obtain the required maps fn between projective resolutions by induction using (A).
/ Pn+1
···
?
/ Pn0
d0
/ ···
/ Pn−1
d
fn
/ P0
n+1
···
/ Pn
d
fn−1
/ P0
n−1
d0
/ ···
By induction each fn commutes with the differentials, and so one has d0 fn d = 0. Now by (A), there
exists fn+1 .
Given two such extensions fn+1 , gn+1 , put τn := fn − gn . Consider
/ Pn+1
···
hn+1 ?
···
/ P0
n+2
{
τn+1
}
/ P0
n+1
/ Pn
d
hn
d0
τn
}
/ Pn0
d
/ Pn−1
/ ···
hn−1
d0
/ ···
(τn − hn−1 d) d = τn d = d0 τn+1 .
Now by (B) there exists hn+1 , so by induction we have a homotopy between f• and g• .
ε
ε0
Assume we have two projective resolutions P• Ï
− M and P•0 −
Ï M. Then we have unique up to
homotopy maps f• : P• → P•0 and f•−1 : P•0 → P• . By uniqueness up to homotopy, f• ◦ f•−1 ∼ 1P•0 and
f•−1 ◦ f• ∼ 1P• . Thus a projective resolution is unique up to homotopy.
3.4
Injective resolutions
We say that an abelian category A has enough injectives if for any object X ∈ Ob(A ) there is a
monomorphism 0 → X → I to an injective object I ∈ Ob(A ). What we need is the fact that the category
of R-modules has enough injectives. This is standard, but not so immediate as enough projectives; see
[Lang, §XX.4].
So for an R-module M we can also construct its injective resolution
0 → M → I0 → I1 → I2 → · · ·
where I 0 , I 1 , I 2 , . . . are injective R-modules.
Such a resolution is unique up to chain homotopy. We omit the details since we already saw this
for projective resolutions.
Remark. “Derived Functor Rap” by Paul Bressler (October 14, 1988):
This is a new rap on the oldest of stories
Functors on abelian categories.
If the functor is left exact
You can derive it and that’s a fact.
But first you must have enough injective
Objects in the category to stay active.
If that’s the case no time to loose;
Resolve injectively any way you choose.
Apply the functor and don’t be sore
The sequence ain’t exact no more.
Here comes the part that is the most fun, Sir,
Take homology to get the answer.
On resolution it don’t depend:
All are chain homotopy equivalent.
Hey, Mama, when your algebra shows a gap
Go over this Derived Functor Rap.
20
3.5
Right derived functor Ext•R and H • (G, A)
Let M and A be R-modules. For A consider an injective resolution
0 → A → I0 → I1 → I2 → · · ·
Apply to this the functor HomR (M, −)
0 → HomR (M, A) → HomR (M, I 0 ) → HomR (M, I 1 ) → HomR (M, I 2 ) → · · ·
Since HomR (M, −) is left exact, we get a cochain complex (not necessarily exact anymore), and we
can take its cohomology:
ExtnR (M, A) := H n (HomR (M, I • )).
This means (by definition) that
ExtnR (M, A) := (Rn HomR (M, −))(A)
is the right derived functor of a left exact functor HomR (M, −).
Now we can define the group cohomology as follows:
H n (G, A) := ExtnZG (Z, A),
where Z is a ZG-module with the trivial action of G.
So to compute H n (G, A), we should take an injective resolution of A by ZG-modules:
0 → A → I0 → I1 → I2 → · · ·
Then we should consider the complex
0 → HomZG (Z, A) → HomZG (Z, I 0 ) → HomZG (Z, I 1 ) → HomZG (Z, I 2 ) → · · ·
and calculate its cohomology
H n (G, A) := H n (HomZG (Z, I • )).
To calculate ExtnR (M, A) we can also start with a projective resolution of M
· · · → P2 → P1 → P0 → M → 0
and then apply the contravariant functor HomR (−, A):
0 → HomR (M, A) → HomR (M, P0 ) → HomR (M, P1 ) → HomR (M, P2 ) → · · ·
Now this is a (co)complex, and its cohomology is isomorphic to Ext•R as it was defined before:
ExtnR (M, A) ' H n HomR (−, A))(M).
According to this,
H n (G, A) ' H n (HomZG (P• , A)).
21
3.6
Left derived functor Tor•R and H• (G, A)
Let M and A be R-modules. For M consider a projective resolution
· · · → P2 → P1 → P0 → M → 0
Apply to this the functor − ⊗R A:
· · · → P2 ⊗R A → P1 ⊗R A → P0 ⊗R A → M ⊗R A → 0
Since − ⊗R A is right exact, we get a chain complex (not necessarily exact anymore), and we can
take its homology:
TornR (M, A) := Hn (P• ⊗R A).
This means that
TornR (M, A) := (Ln (− ⊗R A))(M)
is the left derived functor of a right exact functor − ⊗R A.
Now we can define the group homology as follows:
Hn (G, A) := TornZG (Z, A),
where Z is a ZG-module with the trivial action of G.
So to compute Hn (G, A), we should take an projective resolution of Z by ZG-modules:
· · · → P2 → P1 → P0 → Z → 0
Then we should consider the complex
· · · → P2 ⊗ZG A → P1 ⊗ZG A → P0 ⊗ZG A → Z ⊗ZG A → 0
and calculate its homology
Hn (G, A) := Hn (P• ⊗ZG A).
3.7
Long exact sequences
For any right exact functor F the left derived functors L• F form together a homological δ-functor.
This means that for any short exact sequence
0→A→B→C→0
δ
n
there are natural connecting morphisms Ln F(C) −Ï
Ln−1 F(A) such that the following long sequence is
exact:
δ
δ
2
1
··· −
Ï
L1 F(A) → L1 F(B) → L0 F(C) −
Ï
L0 F(A) → L0 F(B) → L0 F(C) → 0
This is actually the whole point of the derived machinery! Since we start with F right exact, the
sequence “F(A) → F(B) → F(C) → 0” is automatically exact, but to continue this to the left, one has to
introduce left derived functors of F!
Similarly, for any left exact functor F the right derived functors R• F form together a cohomological
δ-functor. That is, for any short exact sequence
0→A→B→C→0
22
δn
there are natural connecting morphisms Rn F(C) −Ï Rn+1 F(A) such that the following long sequence
is exact:
δ0
0 → R0 F(A) → R0 F(B) → R0 F(C) −
Ï R1 F(A) → R1 F(B) → R1 F(C) → · · ·
For details see [Weibel, Chapter 2] and [Tohoku]. We are going to use the following consequence
of this general fact.
Proposition 3.5. Consider a short exact sequence of ZG-modules
0→A→B→C→0
Then there is a long exact sequence in cohomology
δ0
δ1
δ
δ
0 → H 0 (G, A) → H 0 (G, B) → H 0 (G, C) −
Ï H 1 (G, A) → H 1 (G, B) → H 1 (G, C) −
Ï ···
and a long exact sequence in homology
1
2
0 Î H0 (G, C) Î H0 (G, B) Î H0 (G, A) Î
−
H1 (G, C) Î H1 (G, B) Î H1 (G, A) Î
−
···
The connecting morphisms δ• and δ • are natural, etc.
The naturality means that δ behaves well for morphisms of short exact sequences. Assume we have
a commutative diagram with exact rows
0
/A
/B
/C
/0
0
/ A0
/ B0
/ C0
/0
Then the connecting homomorphisms must commute (e.g. for group cohomology; the same for
any left/right derived functor):
/ H n+1 (G, A)
δn
H n (G, C)
H n (G, C 0 )
/ H n+1 (G, A0 )
δ 0n
Remark. Sometimes this fancy stuff is not enough and one wants to know the connecting homomorphism. Here
is the answer for cohomology using the bar-resolution.
p
Suppose 0 → A → B Ï
− C → 0 is an exact sequence. Consider an element [φ] ∈ H n (G, C) represented by a cocycle
n
e : G n → B:
φ : G → C. Consider a lift of this to φ
>B
e
φ
Gn
φ
p
/C
e takes values in A ⊆ B, and we can take
Since dn φ = 0, the cocycle dn φ
δ n : H n (G, C)
→
[φ]
7Ï
23
H n+1 (G, A),
e
[dn φ].
4
Group (co)homology via resolutions of Z
Let us summarize again what we have seen in §3.5 and 3.6. Let A be a ZG-module. One can compute
H • (G, A) and H• (G, A) as follows. Consider Z as ZG-module with the trivial action of G and take its
projective resolution via ZG-modules:
· · · → P2 → P1 → P0 → Z → 0
Then the group (co)homology is given by
H n (G, A)
:=
H n (HomZG (P• , A)),
Hn (G, A)
:=
Hn (P• ⊗ZG A).
Example 4.1. In particular,
H 0 (G, A) = HomZG (Z, A),
H0 (G, A) = Z ⊗ZG A.
Example 4.2. If P is a projective ZG-module, then Hn (G, P) = 0 for all n > 0.
If I is an injective ZG-module, then H n (G, I) = 0 for n > 0.
This follows immediately from the definitions and the fact that − ⊗ZG P and HomZG (−, I) are exact
functors.
Remark. Of course the property of being injective depends on the category. For instance, Q/Z is an injective
Z-module, but it is not an injective ZG-module. To see this, take G = Z/2 and calculate (taking the trivial action on
Q/Z)
H 1 (Z/2, Q/Z) = HomZ (Z/2, Q/Z) ' {a ∈ Q/Z | 2 a = 0}.
The latter is not trivial, it contains 21 . All calculations of cyclic group (co)homology will be explained below.
Example 4.3. Consider Z as a ZG-module with the trivial action of G. Then H1 (G, Z) ' G/[G, G].
Proof. Consider a short exact sequence (which one has by definition of IG )
ε
0 → IG → ZG Ï
− Z→0
The homology long exact sequence gives
δ
0 Î H0 (G, Z) Î H0 (G, ZG) Î H0 (G, IG ) Î
− H1 (G, Z) Î H1 (G, ZG) Î · · ·
Now ZG is projective, so H1 (G, ZG) = 0, and thus δ is an injection.
The group H0 (G, ZG) is ZG/IG ZG ' im ε = Z. We have that δ is a surjection as well.
We conclude H1 (G, Z) ' H0 (G, IG ) = IG /IG2 ' G/[G, G] (cf. proposition 1.1).
Here is an application of the abstract nonsense:
Proposition 4.1 (Dimension shifting). Let A be a ZG-module. Then for any i > 1 there exists a
ZG-module A0 such that H i (G, A) ' H i−1 (G, A0 ).
24
Proof. A is a submodule of some injective ZG-module I (we have enough injectives!), and we take
A0 := I/A:
0 → A → I → A0 → 0
Now the cohomology long exact sequence gives
δ
· · · → H i−1 (G, A) → H i−1 (G, I) → H i−1 (A0 ) Ï
− H i (G, A) → H i (G, I) → · · ·
Since I is injective, H i−1 (G, I) = 0 and H i (G, I) = 0, thus H i−1 (A0 ) ' H i (G, A).
Observe that this has nothing to do with group cohomology, it’s a fact about right derived functors.
(Exercise: how dimension can be shifted in the left derived case?)
4.1
Cyclic groups
Let us see on the easiest examples how one can do computations.
Example 4.4. Let G = hti ' Z be an infinite cyclic group. Then there is a free resolution
t−1
ε
· · · → 0 → ZG −−Ï ZG Ï
− Z→0
After taking − ⊗ZG A, we get a complex
t−1
· · · → 0 → A −−Ï A → 0
After taking HomZG (−, A), we get a co-complex
t−1
0 → A −−Ï A → 0 → · · ·
This allows to compute
 G
 A , i = 0,
AG , i = 1,
H i (G, A) =

0,
i > 1.

 AG , i = 0,
AG , i = 1,
Hi (G, A) =

0,
i > 1.
Example 4.5. Consider a finite cyclic group G = ht | t n = 1i ' Z/n. Take the resolution
N
t−1
N
ε
F• : · · · → ZG −
Ï ZG −−Ï ZG −
Ï ZG Ï
− Z→0
Here
N :=
X
g = 1 + t + t 2 + · · · + t n−1 .
g∈G
Taking − ⊗ZG A gives a complex
N
t−1
N
t−1
··· −
Ï A −−Ï A −
Ï A −−Ï A → 0
Taking HomZG (−, A) gives a co-complex
t−1
N
t−1
N
0 → A −−Ï A −
Ï A −−Ï A −
Ï ···
Observe that N(g · a) = N(a) for all g ∈ G and a ∈ A and N · A ⊆ AG , so N descends to a norm map
N : AG → AG .
From this we compute
25
 G
 A ,
ker N,
H i (G, A) =

coker N,

i = 0,
 AG ,
AG /N · A = coker N, i ≥ 1 odd,
Hi (G, A) =

ker N,
i ≥ 2 even.
i = 0,
i ≥ 1 odd,
i ≥ 2 even.
Note that ker N = ker N/DA, where D : a 7Ï t · a − a.
In particular, if the action is trivial,
AG = AG = A,
ker N ' {x ∈ A | n x = 0},
•
coker N ' A/nA.
×
Example 4.6. For instance, one can calculate H (Gal(C/R), C ) and verify what we have seen in §1.4
and §1.5.
The Galois group is Z/2Z. Now N : C× → C× is given by N(z) = z · z = |z|2 (do not be confused with
multiplicative notation!)
Next AG = {z ∈ C× | z = z} = R× and NA = {|z|2 | z ∈ C× } = R≥0 , so H 2 (G, A) ' AG /NA ' Z/2.
Further ker N = {z ∈ C× | |z|2 = 1} = S 1 and DA = { zz | z ∈ C× } = S 1 , so that ker N = 0 (as it must
be according to Hilbert’s 90).
i
Hi (Gal(C, R), C× )
0
R×
1
0
2
Z/2
3
0
4
Z/2
5
0
···
···
For the rest of this section G will denote a finite cyclic group. Now we see that (co)homology of G
is periodic: H n (G, A) ' H n+2 (G, A) for n ≥ 1. This suggests the following.
Definition 4.1. Let G be a finite cyclic group and let A be a G-module such that all positive cohomology
groups are finite. The Herbrand quotient of A is
h(A) :=
|H 2 (G, A)|
h0 (A)
.
=
h1 (A)
|H 1 (G, A)|
P Thisi number h(A) is an analogue of the Euler–Poincaré characteristic of a chain complex χ(C• ) :=
i (−1) rk Ci . The Euler–Poincaré characteristic is additive in short exact sequences: if we have
0 → A• → B• → C• → 0,
then
χ(B• ) = χ(A• ) + χ(C• ).
The similar property of the Herbrand quotient is the following.
Proposition 4.2. Let 0 → A → B → C → 0 be an exact sequence of G-modules. Suppose the
Herbrand quotient is defined for two of the three modules. Then it is defined for the third, and
h(B) = h(A) · h(C).
Proof. We look at the long exact sequence in cohomology. It becomes cyclic starting from H 1 , so we
draw an “exact hexagon”:
fA
H 1 (G, B)
8
fB
/ H 1 (G, C)
fC
&
H 2 (G, A)
H 1 (G, A)
f
gC
H (G, C) o
2
gB
26
x
H (G, B)
2
gA
By exactness, if for two of A, B, C the cohomology groups are finite, then it is so for the third. Now
we compute the relation between h(A), h(B), h(C). Looking e.g. at exactness in the term H 1 (G, A), we
have
| ker fA | = | im gC | = | coker gB | =
|H 2 (C)|
.
| im gB |
Similar identities for all six terms give
|H 1 (A)| =
| im fA | · | im gC |,
|H (C)| =
| im fC | · | im fB |,
|H (B)| =
| im gB | · | im gA |.
1
2
|H 2 (A)| = | im gA | · | im fC |,
|H 2 (C)| = | im gC | · | im gB |,
|H 1 (B)| = | im fB | · | im fA |.
And from this we see
|H 1 (A)| · |H 1 (C)| · |H 2 (B)| =
|H 2 (B)|
|H 1 (B)|
|H 2 (A)| · |H 2 (C)| · |H 1 (B)|,
|H 2 (A)| |H 2 (C)|
·
.
|H 1 (A)| |H 1 (C)|
=
Proposition 4.3. Let A be a finite G-module. Then h(A) = 1.
Proof. We have an exact sequence
0
/ AG
/A
/A
D
x
/ A/DA
/ t·x−x
/0
Now |AG | = |A/DA|, since
|A|
= |DA|,
|AG |
|A|
= |A/DA|.
|DA|
(For this we need to assume that A is finite!)
Now H 2 (G, A) = AG /NA and H 1 (G, A) = ker N/DA:
N
Ï AG → H 2 (G, A) → 0
1 → H 1 (G, A) → A/DA −
From this we have
|H 2 (G, A)| =
|AG |
,
| im N|
|A/DA|
= | im N|.
|H 1 (G, A)|
Since |AG | = |A/DA|, this implies |H 2 (G, A)| = |H 1 (G, A)|.
Proposition 4.4. Let f : A → B be a homomorphism of G-modules with finite kernel and cokernel.
Then if one of h(A), h(B) is defined, the other is also defined, and h(A) = h(B).
27
Proof. We look at exact sequences.
0 → ker f → A → im f → 0
0 → im f → B → B/ im f → 0
Suppose h(B) is defined. Then
h(B) = h(im f) · h(B/ im f).
So h(im f) is defined. Since ker f is finite, also h(ker f) = 1 is defined. Thus
h(A) = h(ker f) · h(im f) = h(B).
Similarly, if we assume that h(A) is defined, we do the same with the other short exact sequence.
What we have proved thus far has an interesting application.
Proposition 4.5. Let K = Fq be a finite field. Consider a finite extension L/K. Then H 2 (Gal(L/K), L× ) =
0.
Proof. The Galois group Gal(L/K) is a finite cyclic group (generated by the Frobenius), so we can apply
our results above about Herbrand quotients. We should have h(L× ) = 1, so that h1 (L× ) = h2 (L× ). But
h1 (L× ) = 0 by Hilbert’s 90.
Remark. H 2 (L/K, L× ) is actually isomorphic to the relative Brauer group Br(L/K) (cf. §1.5). The absolute Brauer
group Br(K) = limL/K Br(L/K) classifies central simple algebras over K.
Î−
In Br(K) algebras are taken modulo an equivalence. Namely, any simple algebra is a matrix algebra Mn (D) over
some skew-field D (by “Artin–Wedderburn theorem”). For a skew-field D we identify matrix algebras Mn (D) ∼
Mm (D) for all m, n ≥ 1. Multiplication in Br(K) is the tensor product of algebras: [A1 ] ⊗K [A2 ] := [A1 ⊗K A2 ]. Thus
the identity element is [K] ∼ [Mn (K)]. The inverse is obtained by taking the opposite algebra (with multiplication
x ? y := y · x). Indeed, one checks A ⊗K Aop ' Mn (K). Cf. [Rowen, Ring theory, vol. II, Chapter 7].
What we just computed above is that Br(Fp ) = 0 is trivial. This is essentially equivalent to the Wedderburn’s
theorem: any finite skew-field is automatically commutative.
As we have seen, Br(R) = H 2 (Gal(C/R), C× ) ' Z/2. The representatives of the central simple algebras over R in
Br(R) are the quaternions H and R itself (this result is essentially due to Frobenius).
4.2
Free resolutions of Z coming from topology
e be a CW-complex with an action of G permuting the cells (so that each g ∈ G gives a homeLet X
e →X
e mapping a cell to a cell). This gives rise to action of G on the cellular chain
omorphism g : X
e permuting the Z-basis of each Cn (X).
e
complex C• (X),
ε
e → C1 (X)
e → C0 (X)
e Ï
· · · → C2 (X)
− Z→0
e
If the action of G is free (meaning that g · σ 6= σ unless g = 1), then G acts freely on each Cn (X),
e is a free ZG-module spanned by the orbits of the action on the n-cells. If X
e is also
and each Cn (X)
contractible, then we have an exact sequence of free ZG-modules.
28
e be a contractible CW-complex with a free action of G on its cells. Then the
Proposition 4.6. Let X
augmented cellular chain complex
ε
e → C1 (X)
e → C0 (X)
e Ï
· · · → C2 (X)
− Z→0
gives a free resolution of Z by ZG-modules.
Definition 4.2. A CW-complex X is called the Eilenberg–MacLane space K(G, 1) for a group G if
1. X is connected.
2. π1 (X) ' G
e of X is contractible
3. The universal cover X
e
(equivalently, Hi (X) = 0 for i ≥ 0; or πi (X) = 0 for i ≥ 2).
e and G acts freely on X.
e So one has the
The cell structure on X induces a cell structure on X,
following.
e of its
Proposition 4.7. Let X be a K(G, 1)-space. The augmented cellular chain complex C• (X)
universal cover gives a free resolution of Z by ZG-modules.
Now we have a free resolution
ε
e → C1 (X)
e → C0 (X)
e Ï
· · · → C2 (X)
− Z→0
If we take the co-invariants, we get a complex
e G → C1 (X)
e G → C0 (X)
e G →0
· · · → C2 (X)
e
But this is isomorphic to the complex of orbits X/G,
so the cellular complex of the K(G, 1)-space X:
· · · → C2 (X) → C1 (X) → C0 (X) → 0
Thus the homology Hn (G, Z) is the usual cellular homology of the corresponding K(G, 1)-space.
Hn (G, Z) ' HnCW (K(G, 1), Z).
Sometimes one takes this as the definition for group (co)homology.
1
Example 4.7. For G = Zn the K(G, 1)-space is the n-torus T n := S
· · × S}1 . Its universal cover is Rn ,
| × ·{z
n
it is contractible.
We can compute the homology groups of the tori T n := S 1 ×· · ·×S 1 using the Künneth isomorphism
M
Hk (T n ) '
Hi (T k−1 ) ⊗ Hk−i (S 1 ).
i
Since S has only homology H0 (S ) ' H1 (S 1 ) ' Z, we get
1
1
Hk (T n ) ' Hk (T n−1 ) ⊗ H0 (S 1 ) ⊕ Hk−1 (T n−1 ) ⊗ H1 (S 1 ) ' Hk (T n−1 ) ⊕ Hk−1 (T n−1 ),
n−1
and we see that rk Hk (T n ) =rk Hk (Tn−1 ) + rk
) is the same recursive relation as we have for
Hk−1 (T
n
n−1
n−1
the binomial coefficients: k = k + k−1 (le triangle de Pascal / il triangolo di Tartaglia), and the
initial values rk H0 (T 1 ) = rk H1 (T 1 ) = 1 are the same, so rk Hk (T n ) = nk .
29
H0
Z
Z
Z
Z
..
.
T1
T2
T3
T4
..
.
H1
Z
Z2
Z3
Z4
..
.
H2
0
Z
Z3
Z6
..
.
H3
0
0
Z
Z4
..
.
H4
0
0
0
Z
..
.
···
···
···
···
···
..
.
Thus
n
Hi (Zn , Z) = Hi (T n ) = Z( i ) .
1
Example 4.8. For a free group Fn on n generators one has K(G, 1) = S
· · ∨ S}1 , the bouquet of n
| ∨ ·{z
n
circles. Thus we see
H0 (Fn , Z) ' Z,
4.3
H1 (Fn , Z) ' Zn ,
Hi (Fn , Z) = 0 for i > 1.
Bar-resolution revisited
The K(G, 1)-space for a given G is unique up to homotopy equivalence. Let us see the easiest way to
produce a K(G, 1)-space for a given group.
Consider a “simplicial complex”, having as its n-simpleces the (n + 1)-tuples [g0 , g1 , . . . , gn ] ∈ G n+1 .
This n-simplex is being attached to (n − 1)-simpleces [g0 , . . . , gbi , . . . , gn ] as to its vertices. The resulting
e is contractible (it’s a usual simplex if G is finite; otherwise it’s a “very big simplex”, but it is
complex X
still contractible). G acts freely on the simpleces by multiplication on the left:
g · [g0 , . . . , gn ] := [g g0 , . . . , g gn ].
(6)
e
e is its universal cover. This construction
Now the space of orbits X := X/G
is a K(G, 1)-space, and X
is monstrous, but we know that our K(G, 1) is homotopy equivalent to any other.
e generated by symbols [g0 , . . . , gn ] with an
So we can take a resolution by free Z-modules C• (X)
action of G by (6). The differentials in this resolution are the usual differentials in simplicial homology:
X
dn [g0 , . . . , gn ] =
(−1)i [g0 , . . . , gbi , . . . , gn ].
0≤i≤n
A resolution with such differentials is called homogeneous.
e can be written as
Now note that every n-simplex in X
[g0 , g0 g1 , g0 g1 g2 , . . . , g0 g1 · · · gn ] = g0 · [1, g1 , g1 g2 , . . . , g1 · · · gn ] =: g0 · [[g1 | g2 | · · · | gn ]].
So simpleces in X look like [[g1 | g2 | · · · | gn ]], and the boundary of such a simplex is
dn [[g1 | g2 | · · · | gn ]] = [[g2 | · · · | gn ]] +
X
(−1)i [[g1 | · · · | gi gi+1 | · · · | gn ]] + (−1)n [[g1 | · · · | gn−1 ]].
1≤i≤n−1
e as a free ZG-module. Now look at the
These symbols [[g1 | g2 | · · · | gn ]] form a ZG-basis of C• (X)
bar-resolution (§1.6) and observe it uses exactly the same differentials! One could make a routine
computation of exactness of the bar-resolution, but now this is essentially because the universal covering
e for K(G, 1) is contractible.
space X
30
5
Cup-product on cohomology
Now assume A is a commutative ring with an action of G. On cochains a ∈ C n (G, A), b ∈ C m (G, A) of
the standard homogeneous (!) resolution we define the cup-product as follows:
(a ^ b) (g0 , . . . , gn+m ) := a(g0 , . . . , gm ) · b(gm , . . . , gm+1 ),
here · is the usual product in the ring A.
L
This actually gives a product on the graded commutative algebra H • (G, A) := n≥0 H n (G, A). It is
essentially the same as the cup-product on cellular cohomology (of the corresponding K(G, 1) space),
so what follows can be found in standard books on algebraic topology (cf. [Hatcher, §3.2]).
Proposition 5.1. For a ∈ C n (G, A) and b ∈ C m (G, A) one has
∂(a ^ b) = (∂a) ^ b + (−1)m a ^ (∂b).
Proof. We compute directly the summands
(∂a ^ b)(g0 , . . . , gm+n+1 )
X
=
a(g0 , . . . , gbi , . . . , gn+1 ) · b(gn+1 , . . . , gn+m+1 ),
0≤i≤n+1
X
n
(−1) (a ^ ∂b)(g0 , . . . , gm+n+1 )
(−1)i a(g0 , . . . , gn ) · b(gn , . . . , gbi , . . . , gn+m+1 ).
=
n≤i≤n+m+1
If we add two expressions, everything cancels and remains
∂(a ^ b)(g0 , . . . , gn+m+1 ) = (a ^ b)(d(g0 , . . . , gn+m+1 )),
since
d(g0 , . . . , gn+m+1 ) =
X
(−1)i (g0 , . . . , gbi , . . . , gn+m+1 )
0≤i≤n+m+1
Thanks to the identity ∂(a ^ b) = ∂a ^ b ± a ^ b, we know
• cup product of two cocycles is a cocycle,
• cup product of a cocycle and a coboundary is a coboundary, since a ^ ∂b = ±∂(a ^ b) if ∂a = 0
and ∂a ^ b = ∂(a ^ b) if ∂b = 0.
So the cup product is defined on the level of cohomology:
^ : H n (G, A) × H m (G, A) → H n+m (G, A).
This product is graded commutative:
Proposition 5.2. if α ∈ H m (G, A) and β ∈ H n (G, A), then for α ^ β ∈ H m+n (G, A)
α ^ β = (−1)m n β ^ α.
31
Proof. We introduce an involution σ which induces the identity on the level of cohomology. For a
cochain a : G n+1 → A we define
n+1
(σa)(g0 , . . . , gn ) = (−1)( 2 ) a(gn , gn−1 , . . . , g1 , g0 ).
We claim that σ : C • (G, A) → C • (G, A) defines an involution on the complex, i.e. ∂(σa) = σ(∂a). This
is a direct computation:
∂(σa)
=
n+1
(−1)( 2 )
X
(−1)i a(gn , gn−1 , . . . , gd
n−i . . . , g1 , g0 ),
i
!
σ(∂a)
X
= σ
i
(−1) a(g0 , g1 , . . . , gbi . . . , gn−1 , gn )
0≤n+1
n
=
(−1)( 2 )
X
(−1)n−i a(gn , gn−1 , . . . , gd
n−i , . . . , g1 , g0 ).
i
Now observe that
n+1
n
(−1)( 2 ) = (−1)n · (−1)( 2 ) .
One can show that this σ is homotopical to the identity map 1 : C • (G, A) → C • (G, A). The sign of the
involution is chosen so that
(−1)(
n+m+1
2
m+1
) = (−1)m n · (−1)(n+1
2 ) · (−1)( 2 ) .
TODO: see [Hatcher, Theorem 3.14].
32
Cohomology of cyclic groups H • (C, Z/`)
6
TODO: we already know the cohomology of cyclic groups, but the computations of Prof. Erez explicitly
use two resolutions simultaneously to compute the products (this is probably an overkill). I will go
through the stuff below later to correct it. Please skip this section.
. . . . . . . . .
In this section let C be a finite cyclic group hs | sg = 1i of order g. We will show that for all n ≥ 0
and ` a prime divisor of g one has
H n (C, Z/`) ' Z/`.
6.1
Groups H n (C, Z/`)
We have two resolutions of Z by free ZG-modules: the standard and the special resolution for cyclic
groups:
0o
Zo
L0 o
0o
Zo
ZC o
L1 o
D
ZC o
···
N
···
where D = 1 − s and N = 1 + s + · · · + sg−1 .
We compare these two: define two chain maps ρ• : L• → (ZC)• and τ : (ZC)• → C• that are mutually
inverse up to homotopy.
0o
Zo
LO 0 o
ρ0
0o
Zo
ρ1
σ0
ZC o
LO 1 o
d
D
d
···
σ1
ZC o
N
···
The definition of ρ is the following (extending ZG-linearly):
ρ2n (1)
X
:=
(sα1 +1 , sα1 , sα2 +1 , sα2 , . . . , sαn , 1),
α1 ,...,αn
ρ2n+1 (1)
X
:=
(sα1 +1 , sα1 , sα2 +1 , sα2 , . . . , sαn , s, 1).
α1 ,...,αn
We verify that these are chain maps, that is d ρ2n = ρ2n−1 N and d ρ2n+1 = ρ2n D.
d ρ2n (1)
X X
=
α1 ,...,αn
i
−
α1 +1 , . . . , sαn , 1)
(sα1 +1 , sα1 , . . . , s[
X X
i
X X
+
i
α1 , . . . , sαn , 1)
(sα1 +1 , sα1 , . . . , sc
α1 ,...,αn
(sα1 +1 , sα1 , . . . , sα1+1 , . . . , sαn )
α1 ,...,αn
X X
=
(sαn +1 , sαn , . . . , sαn−1 +1 , sαn −1 , s, 1) sα
α α1 ,...,αn
X
=
ρ2n−1 (sα ) = ρ2n−1 N(1).
α
33
We define σ• step-by-step.
c0 : ZC
s
i
→ CC,
7Ï
−(1 + · · · + si−1 )
(1 ≤ i ≤ g − 1),
1 7Ï 0.
i
(c0 D + N c1 ) s =
c1 : ZC
→
CC,
s
i
7Ï
sg−1
0
7Ï
1.
(0 ≤ i ≤ g − 2),
c0 (si − si+1 ),
c0 (sg−1 − 1) + (1 + s + · · · + sg−1 )
0≤i ≤g −2
i =g −1
= si .
c0 D + Nc1 = id : ZC → ZC. Also c1 N + Dc0 = id. We let σ0 := id and
σ1 (x0 , 1) := c0 σ0 d(x0 , 1) = c0 (1 − x0 ) = −c0 (x0 ).
More generally
σ1 (x0 , x1 ) := −c0 (x0 x1−1 ) x1 .
It is a chain map: D σ1 = σ0 d.
Next we define
σ2 (x0 , x1 , 1) := −c1 [c0 (x0 x1−1 ) x1 ],
and so on.
σ induce isomorphisms on cohomology, so we can compute H • (C, Z/`) by the complex with D and
N.
Now
HomZC (ZC, Z/`)
f
'
Z/`,
7Ï f(1).
Also D∗ = 0 and N ∗ = 0 since ` | g. Hence for all n ≥ 0
H n (C, Z/`) ' Z/`.
In HomZC ((D, N)• , Z/`) the homology has all kernels of differentials = Z/` and all images = 0.
6.2
Multiplication on H • (C, Z/`)
As before, let C be a cyclic group of order g and let ` be a prime number dividing g.
As we saw, there is a graded Z/`-algebra structure on
M
H • (G, Z/`) =
H n (G, Z/`).
n≥0
And we can show that as a Z/`-algebra,
34
H • (C, Z/`) ' (Z/`)[u, v]/(v 2 = 0),
where u ∈ H 1 (C, Z/`) and v ∈ H 1 (C, Z/`).
n 0
H n (C, Z/`) Z/`
generator 1
1
Z/`
v
2
Z/`
u
3
Z/`
uv
···
···
···
4
Z/`
u2
We know already that for all n ≥ 0 the n-th cohomology group H n (C, Z/`) is isomorphic to Z/`. We
define elements v ∈ H 1 (C, Z/`), u ∈ H 2 (C, Z/`) on cochains of the (D, N)-resolution:
v : ZC
→
1 7Ï
u : ZC
Z/`,
1.
→ Z/`,
1 7Ï 1.
By anti-commutativity of the product, we should have (assuming ` is odd)
v ^ v = (−1) v ^ v = 0
We show that multiplication by u in H • (C, Z/`) is an isomorphism:
H n (C, Z/`)
−^u
−−−Ï
x
7Ï
H n+2 (C, Z/`),
x ^ u.
We show that the following diagram commutes:
HomZC (L2n , Z/`)
φ
/ HomZC (L2 , Z/`)
∗
ρ2n+2
∗
ρ2n
Z/`
Z/`
Here φ := σ2∗ u; it is a standard cochain while u is a (D, N)-cochain.
A similar diagram holds for 2n + 1 and 2n − 1 (the odd case)
HomZC (L2n , Z/`)
φ
/ HomZC (L2 , Z/`)
∗
ρ2n+1
∗
ρ2n−1
Z/`
Z/`
∗
∗
Let us check that ρ2n+2
(f ^ φ) = ρ2n
(f). Since everything is invariant under the action of C, we
check on 1.
∗
ρ2n+2
(f ^ φ)
(f ^ φ) ρ2n+2 (1)
X
=
f(sα1 +1 , sα1 , . . . , sαn , sαn+1 +1 ) · φ(sαn+1 +1 , sαn+1 , 1 )
=
α1 ,...,αn+1
X
=
f(αα1 +1 , sα1 , . . . , sαn , 1) ·
X
α1 ,...,αn
φ(sα+1 , sα , 1) = f ρ2n (1).
α
|
{z
1
35
}
P
∗
∗
To check ρ2n+1
(f ^ φ) = ρ2n−1
(f), we need a similar equality α φ(sα , s, 1).
X
X
X
X
φ(sα+1 , sα , 1) =
φ(si+1 , si , 1) = −
u(c1 (c0 (s)) si ) =
u(c1 (si )) = 1.
α
i
0≤i≤g−1
i
Here we use that
φ(sα , sβ , 1) = (σ2∗ u)(sα , sβ , 1) = −u(c1 (c0 (sα−β ) sβ )),
c0 (s) = −1,
u(1) = 1.
Hence (?)
H • (C, Z/`) ' (Z/`)[u, v]/(v 2 = 0).
36
7
Restriction, inflation, corestriction
Note: this was done both by Dajano and Prof. Erez. I put here the exposition of Dajano.
7.1
H • (−, −) as functors of pairs
We introduced H • (G, −) as the right derived functors of a left exact functor A 7Ï AG =: H 0 (G, A).
Rather than working with H • (G, −) as functors on the category of ZG-modules, it is convenient to
consider H • (−, −) as functors on a certain category of pairs. Such a pair (G, A) consists of a group G
and a ZG-module A. A morphism of pairs (G, A) → (G 0 , A0 ) is given by the following data:
1. A morphism of groups φ : G 0 → G (going in the opposite direction).
2. A morphism of ZG 0 -modules f : φ∗ A → A0 , where φ∗ A is the ZG 0 -module made from A with the
action of G 0 induced by φ:
g 0 · a := φ(g 0 ) · a.
The condition that f is a morphism of ZG 0 -modules means that for all g 0 ∈ G and a ∈ A
g 0 · f(a) = f(φ(g 0 ) · a).
f
AO
φ(g 0 )
(7)
/ A0
O
g0
A
f
/ A0
0
0
Now φ induces a map H n (G, A) → H n (G 0 , φ∗ A). From (φ∗ A)G = Aφ(G ) we have an inclusion AG ,Ï
0
∗
(φ A)G , and it is in fact natural in A:
A
H 0 (G, A) B

H 0 (G, B) 
/ H 0 (G 0 , φ∗ A)
/ H 0 (G 0 , φ∗ B)
Thus there are morphisms of all derived functors H n (G, A) → H n (G 0 , φ∗ A) for n > 0.
37
Remark. Recall a general result from homological algebra. For any two left exact functors F, G : A → B any
natural transformation F Ñ G
X
F(X)
f
F(f)
Y
ηX
F(Y )
/ G(X)
ηY
G(f)
/ G(Y )
uniquely extends to a natural transformation of right derived functors Rn F Ñ Rn G for all n > 0, compatible with
the connecting morphisms δ (see [Weibel, Chapter 2] and [Tohoku])
0→A→B→C→0
/ Rn G(C)
Rn F(C)
δ
δ
/ Rn+1 G(A)
Rn+1 F(A)
Further the map f : φ∗ A → A induces maps H n (G 0 , φ∗ A) → H n (G 0 , A) by functoriality of H n (G 0 , −).
One can check that this is also natural in A.
A
H n (G 0 , φ∗ A)
/ H n (G 0 , A)
B
H n (G 0 , φ∗ B)
/ H n (G 0 , B)
(B should also come with a map φ∗ B → B making all commutative, so this is a tautology?)
We compose the two induced morphisms to get
+
/ H n (G 0 , A0 )
/ H n (G 0 , φ∗ A)
H n (G, A)
So this is by definition the map in cohomology corresponding to a morphism of pairs (G, A) →
(G 0 , A0 ).
All this can be made explicit for the usual cocycles. Given φ : G 0 → G and f : φ∗ A → A0 , the map
H (G 0 , A) → H n (G 0 , A0 ) is induced by the map transforming a cocycle h : G × · · · × G → A into φn ◦ h ◦ f n
(as one would expect):
n
G × · ·O · × G
φn
h
/A
fn
G0 × · · · × G0
/ A0
It is a routine verification that starting from such a map on cocycles, one really has a map H n (G 0 , A) →
H (G 0 , A0 ), which is functorial, commutes with connecting homomorphisms δ, etc.
n
In what follows we are going to consider three specific morphisms.
• Inflation. If H E G is a normal subgroup in G and A is a ZG-module, then we can consider
the canonical projection φ : G → G/H and the inclusion f : AH ,Ï A. This gives a morphism
n
H
n
inf G
H : H (G/H, A ) → H (G, A).
38
• Restriction. If we have a subgroup inclusion H ,Ï G, then a ZG-module A can be viewed as a
n
n
ZH-module by restricting the action. So there is an obvious morphism resG
H : H (G, A) → H (H, A)
(which is a special case of the functors-on-pairs stuff).
• Corestriction (also called transfer). In the same situation H ≤ G, assuming the subgroup index
n
n
is finite, one can obtain a morphism in the opposite direction corG
H : H (H, A) → H (G, A). Unfor0
tunately, this is not induced by a morphism of pairs! On the level of H this is obtained by the
“norm map”
NG/H : AH
→ AG ,
X
g · a.
7Ï
a
g∈G/H
Such norms play an important role in finite group cohomology.
7.2
Inflation-restriction exact sequence
Proposition 7.1. For any subgroup G, normal subgroup H E G, and a ZG-module A there is an
exact sequence
inf G
resG
H
H
0 → H 1 (G/H, AH ) −−Ï
H 1 (G, A) −−Ï
H 1 (H, A)
There is a way to derive things from the Hochschild–Serre spectral sequence, but we can do an
explicit verification. For a standard cocycle h : G/H → AH the inflation and restriction are given by
G/H
OO
G
GO
?
H
/ AH
_
h
inf G
H (h)
/A
/A
h
resG
H (h)
/A
G
1
H
1
Now fill in yourself the details showing that inf G
H : H (G/H, A ) → H (G, A) is injective and ker resH =
G
im inf H .
Claim. If I is an injective ZG-module, then I H is an injective ZG/H-module.
Proof. Let M and N be ZG/H-modules. Let π : G → G/H be the canonical projection. Then π ∗ M and
π ∗ N are ZG-modules. Consider the following diagram:
0
/ M 
~
I H _
I
39
/N
?
f
We want to prove that for n ∈ N one has f(n) ∈ I H . Indeed,
h · f(n) = f(π(h) · n) = f(n).
Proposition 7.2. If H q (H, A) = 0 for all 1 ≤ q ≤ n − 1, then one has an exact sequence
0 → H n (G/H, AH ) −Ï H n (G, A) −−Ï H n (H, A)
inf
res
Proof. For n = 1 this is the usual inflation-restriction sequence.
The induction step is by dimension shifting. For some injective module I take a short exact sequence
0 → A → I → I/A → 0
By the assumption H 1 (H, A) = 0 we get a short exact sequence
0 → AH → I H → (I/A)H → 0
(This is not true in general, of course!)
Now we have a commutative diagram (where A0 := I/A)
0
/ H n (G/H, AH )
/ H n−1 (G, A)
inf
δ '
0
/ H n+1 (G/H, (A0 )H )
res
δ '
/ H n+1 (G, A0 )
inf
/ H n−1 (H, A)
δ '
res
/ H n+1 (H, A0 )
The first row is exact by the induction hypothesis, and so is the second.
Corollary 7.1. If G has a normal subgroup H / G such that AH is trivial, then H 1 (G, A) is a subgroup
of H 1 (H, A).
Proof. In this case we have a short exact sequence
0 → H 1 (G/H, AH ) −Ï H 1 (G, A) −−Ï H 1 (H, A)
inf
res
and H 1 (G/H, AH ) = 0.
Corollary 7.2. If there exists a normal subgroup H / G such that H 1 (H, A) = 0, then H 1 (G/H, AH ) '
H 1 (G, A).
Proof. That’s because if H q (H, A) = 0 for all 1 ≤ q ≤ n − 1, then H q (G/H, AH ) ' H q (G, A) for all
1 ≤ q ≤ n − 1.
Now let us see how to apply this. Let L/K be a finite Galois extension with a Galois group G,
containing a subextension L0 /K with a Galois group G/H. Then there is an exact sequence
0 → H 2 (G/H, (L0 )× ) → H 2 (G, L× ) → H 2 (H, L× )
This is the inflation–restriction sequence for n = 2. It exists because H 1 (G, L× ) = 0 by Hilbert 90.
40
7.3
Restriction and corestriction
Proposition 7.3. Let H ≤ G be a finite index subgroup. Let A be a ZG-module. The composition of
restriction and corestriction is the multiplication by [G : H]:
H i (G, A)
resG
H
/ H i (H, A)
corG
H
[G:H]
%
H i (G, A)
Observe that we did not write explicit formulas for the corestriction H i (H, A) → H i (G, A): we know
what it is for H 0 (H, A) → H 0 (G, A), and the rest is given automatically. So the proof, as one expects,
boils down to i = 0. [NOTE: Erez really gives corestriction by an obscure formula.]
H
Proof. Let i = 0. Then for an element a ∈ AG its restriction resG
is essentially the same
H (a) ∈ A
G
H
G
element, because A ,Ï A . Now, keeping in mind that a comes from A ,
X
X
G
corG
g ·a =
a = [G : H] · a.
H ◦ resH (a) =
g∈G/H
g∈G/H
Assume the statement is true for i. For i + 1 one has a commutative diagram by dimension shifting
(cf. proposition 4.1)
H i+1 (G, A)
'
H i (G, A0 )
cor ◦ res/
H i+1 (G, A)
'
/ H i (G, A0 )
cor ◦ res
i
i
Corollary 7.3. If ` is coprime with [G : H], then we have an injection resG
H : H (G, Z/`) ,Ï H (H, Z/`).
Corollary 7.4. If H i (H, A) = 0, then H i (G, A) is annihilated by n = [G : H].
Proof. Indeed, in this case multiplication by n factors through 0.
H i (G, A)
res
/ H i (H, A)
cor
/ H i (G, A)
4
n
In particular, if G is a group of order n, then H i (G, A) is killed by n (to see this consider restriction
and corestriction with H being the trivial group). This leads to an important fact:
Corollary 7.5. Let A be a finitely generated abelian group and G be a finite group. Then H i (G, A)
are finite groups for all i > 0.
Proof. In this case H i (G, A) are finitely generated abelian groups, but they are killed by n, so they are
also torsion groups.
Proposition 7.4. Let G be a finite group and let i ≥ 0. For any prime p take a Sylow p-subgroup
Gp ≤ G.
1. The kernel of res : H i (G, A) → H i (Gp , A) does not contain nontrivial elements of p-torsion.
41
2. If res : H i (G, A) → H i (Gp , A) is trivial for any p then H i (G, A) = 0.
Proof.
1. Let n := [G : Gp ]. Then (n, p) = 1 since Gp is a Sylow subgroup.
Suppose x ∈ ker(res : H i (G, A) → H i (Gp , A)). Then cor ◦ res(x) = n x = 0. Suppose that x is an
element of p-torsion, so that p` x = 0 for some `. Since p and n are relatively prime, we have the
Bézout’s identity 1 = p` a + n b for some a, b ∈ Z. Thus x = p` a · x + n b · x = 0.
2. Suppose res : H i (G, A) → H i (Gp , A) is trivial for any p. Assume x ∈ H i (G, A) is a nonzero element.
Then by the previous point p` ·x 6= 0 for any p and any `. On the other hand, cor ◦ res(x) = n x = 0
where n = p1`1 · · · pk`k for some pi and `i . So p1`1 · · · pk`k · x = 0, contradiction.
7.4
Induction, coinduction, and Shapiro’s lemma
NOTE: it is not used later, but Prof. Erez explained “H • (H, A) ' H • (G, coindG
H A)” and used in proving
G
corG
◦
res
=
[G
:
H].
H
H
. . . . . . . .
Let H ≤ G be a subgroup, so that ZH ,Ï ZG. For a ZH-module A we can take its G-induction and
G-coinduction
indG
HA
:=
ZG ⊗ZH A,
A
:=
HomZH (ZG, A).
coindG
H
Proposition 7.5 (“Shapiro’s lemma”).
H • (H, A)
H• (H, A)
' H • (G, coindG
H A),
' H• (G, indG
H A).
Proof. For a ZH-module A the cohomology H n (G, A) is obtained by taking a projective resolution
ε
P• Ï
− Z by ZH-modules and then calculating H n (HomZH (P• , A)).
Observe that ZG is a free ZH-module, so we can take resolutions by ZG-modules as well.
There is an adjunction
HomZH (P• , A) ' HomZG (P• , HomZH (ZG, A)).
So
H n (HomZH (P• , A)) ' H n (HomZG (P• , HomZH (ZG, A))) ' H n (G, coindG
H A).
As for homology,
P• ⊗ZH A ' P• ⊗ZG (ZG ⊗ZH A).
Hn (P• ⊗ZH A) ' Hn (P• ⊗ZG (ZG ⊗ZH A)) ' Hn (H, indG
H A).
42
Cohomology H • (GLn (Fq ), Z/`Z). Connections with K-theory
8
1. It was stated that H • (GLn (Fq ), Z/`Z) injects in H • (C m , Z/`) for a cyclic group C, but there was no a
complete calculation of H • (GLn (Fq ), Z/`Z). See Quillen’s papers or [Benson, Representations and
Cohomology, vol. II, §2.9].
2. The only lecture about K-theory wasn’t very detailed.
8.1
K0 of a ring (Dajano)
Let M be a commutative monoid. We are looking for a universal abelian group M + together with a
morphism i : M → M + such that each monoid morphism φ : M → A to an abelian group A factors
uniquely through M + :
/A
=
φ
M
i
!
∃!
M+
This M + is called the Grothendieck group corresponding to M.
This means that we need a functor AbMon → Ab which is left adjoint to the forgetful functor
Ab → AbMon .
HomAb (M + , A) ' HomAbMon (M, A).
The construction of M is clear: we take the free abelian group generated by elements [m] for each
m ∈ M and factor out the subgroup generated by
[m] + [n] − [m + n].
If M is not a monoid with cancellation (a + c = b + c Ñ a = c), then the map i : M → M + is not
necessarily an inclusion!
Example 8.1. For the monoid (R≥1 , ·, 1) the Grothendieck group is R>0 .
Consider the monoid (N, +, 0). Then the corresponding Grothendieck group is Z.
Now consider a monoid (N, ·, 1), with respect to multiplication. Multiplication by 0 is a problem: it is
not a monoid with cancellation. One sees that the Grothendieck group is {0}, trivial.
Definition 8.1. For a ring R consider the abelian monoid P(R) consisting of the isomorphism classes
of finitely generated projective R-modules, with respect to the direct sum ⊕. Then the 0-th K-group of
R is
K0 (R) := P(R)+ .
Example 8.2. For a field k the finitely generated projective k-modules are finite dimensional k-vector
spaces. So the monoid P(k) is just (N, 0, +). The corresponding Grothendieck group is K0 (k) ' Z.
Similarly K0 (Z) ' Z and the same holds for any PID: K0 (R) ' Z.
This is functorial. If we have a ring morphism f : R → S, then for each projective R-module P there
is a projective S-module P ⊗R S. This induces a map K0 (R) → K0 (S).
K0 is a functor Ring → Ab .
TODO: one can define explicitly K1 and K2 , and then there is a general Quillen’s definition of Kn .
43
9
Class field theory and group cohomology
9.1
Introduction
The class field theory describes abelian extensions of number fields. That is,
L
G
K
Q
Where K/L is a Galois extension with abelian Galois group G (so that K = LG ).
The class field theory was completed in the 1920’s by Takagi, Artin, Hasse, etc. In the 1950’s a new
approach emerged using the cohomology theory.
The central result is the Artin’s reciprocity isomorphism, relating on the one hand, the (abelianized)
Galois group of L/K, and on the other hand, certain generalized ideal class group of K related to the
arithmetic of L/K.
Example 9.1. The usual class group is defined as follows:
Cl(K) :=
fractional ideals in OK
.
principal ideals in OK
The Artin’s reciprocity says in this case that Cl(K) ' Gal(H/K), where H is the maximal abelian
extension of K.
Example 9.2. For K = Q there are no nontrivial unramified extensions.
Example 9.3. For the cyclotomic extensions K = Q and L = Q(ζm ) the Galois group is (Z/mZ)× .
Namely, to an element a ∈ (Z/mZ)× corresponds an automorphism ζ 7Ï ζ a .
A prime p ramifies in L/Q iff p | m.
E.g. for a Galois extension of global fields L/K the Artin reciprocity law is the isomorphism
Gal(L/K)ab '
CK
,
NL/K (CL )
where CK is the idèle class group (introduced by Chevalley and Weil) of K and NL/K is the norm map.
For a Galois extension of local fields Lv /Kv the Artin reciprocity is the isomorphism
Gal(Lv /Kv )ab '
Kv×
.
NLv /Kv (Lv× )
The local case is easier and motivates some results.
9.2
Galois cohomology
Let G = Gal(Ksep /K) be the absolute Galois group of K. In this case subextensions L/K correspond to
subgroups of G, but this correspondence is not 1-1 as in the finite Galois theory. For this reason one
puts on G a topology and instead of looking at all its subgroups, one considers its closed subgroups.
The topology arises naturally (it is called the profinite topology): on every finite quotient G/H we
put the discrete topology, and then we pass to projective limit:
44
G = lim G/H.
Î−
H/G
For a G-module A with G being a topological group, the right definition of group cohomology
H • (G, A) should consider continuous actions G × A → A, where A has the discrete topology. So the
cocycles in the standard complex for H • (G, A) should be continuous, etc.
Equivalently, one can consider the limit of the usual cohomology groups
H • (G, A) = lim H • (G/H, AH ).
Î−
The class field theory is about Galois cohomology of certain special G-modules: like L× for a local
extension L/K or the idèle class group CL .
9.3
Tate cohomology Ĥ n (G, A)
References: [Brown, Chapter VI] and the chapter by Atiyah and Wall in [Cassels and Fröhlich].
If G is a finite group, then there are similarities between the homology groups and cohomology
groups. When G is finite, then a G-module M is called induced if M ' ZG ⊗ A for an abelian group A.
Further, M is called coinduced if M ' Hom(ZG, A). For a finite group G being induced and coinduced
is the same. Indeed, we have an isomorphism
Hom(ZG, A)
f
→ ZG ⊗Z A,
X
7Ï
g ⊗ f(g).
g∈G
N : H0 (G, A) → H 0 (G, A)
If G is a finite group and A
Pis a G-module, then there is a natural norm map
n
taking a representative a to g∈G g(a). The Tate cohomology groups Ĥ (G, A) are given by
• Ĥ n (G, A) := H n (G, A) for n ≥ 1.
• Ĥ 0 (G, A) := ker N.
• Ĥ −1 (G, A) := coker N.
• Ĥ n (G, A) := H−(n+1) (G, A) for n ≤ −2.
The point of this is to put together homology and cohomology. A short exact sequence of G-modules
0→A→B→C→0
gives a “very long exact sequence” in Tate cohomology:
· · · → Ĥ −1 (G, A) → Ĥ −1 (G, B) → Ĥ −1 (G, C) → Ĥ 0 (G, A) → Ĥ 0 (G, B) → Ĥ 0 (G, C) → Ĥ 1 (G, A) → · · ·
Recall that the usual (co)homology groups are given by taking a projective resolution of Z by ZGmodules
· · · → P2 → P1 → P0 → Z → 0
and applying the functor HomZG (−, A) to it.
Now the Tate cohomology arises from some “very long projective resolution”
· · · → P 2 → P 1 → P 0 → P −1 → P −2 → · · ·
45
See [Brown, Ch. VI] for the construction.
One can define ^-products on the Tate cohomology Ĥ • one can prove the existence, uniqueness,
and some properties, but it is not really given by explicit formulas. Here is one of the results about
^-products in Tate cohomology:
Theorem 9.1 (Tate). Let G be a finite group, let A be a G-module and a ∈ H 2 (G, A). For each prime
p let Gp be the p-Sylow subgroup of G and assume
1. H 1 (Gp , A) = 0.
2. H 2 (Gp , A) is generated by resG
Gp (a) and has order |Gp |.
2
Then for all subgroups H ≤ G and for all n the cup product with resG
H (a) ∈ H (H, A) induces an
isomorphism
−^resG (a)
Ĥ n (H, Z) −−−−−H−Ï Ĥ n+2 (H, A).
In class field theory one is interested in n = −2 and a specific G-module A. The reciprocity isomorphism has form
−^cL/K
Ĥ −2 (G, Z) −−−−Ï Ĥ 0 (G, A),
where cL/K is an element of Ĥ 2 (G, A) is some “canonical class”.
For extensions of local fields H 2 (G, L× ) is a part of the absolute Brauer group Br(K), and cL/K can
be obtained as a class corresponding to a certain central simple algebra.
In the case of a local field extension L/K this can be worked out as follows: for K we have its maximal
unramified extension Kur /K, and Br(Kur ) ' H 2 (Gal(Kur /K), (Kur )× ). The Galois group Gal(Kur /K) is
given by the profinite completion of Z:
sep
b
Gal(Kur /K) ' Gal(FK /FK ) ' lim Z/n ' Z.
Î−
We have also a valuation v : (Kur )× → Z, and it induces an isomorphism
H 2 (Gal(Kur /K), (Kur )× )
'
b Z)
/ H 2 (Z,
invK
'
( Q/Z
invK : Br(K) ' Q/Z.
For an extension L/K of degree n the group H 2 (Gal(L/K), L× ) is cyclic of degree n, generated by an
element uL/K ∈ Br(K) with invK (uL/K ) = n1 in Q/Z. This is the canonical class in the local case.
Check the Neukirch’s book on class field theory to see how one can define a valuation and a relevant
b
Z-extension
in the global case as well.
Cohomologically trivial modules (Dajano)
Definition 9.1. A ZG-module A is said to be cohomologically trivial if Ĥ i (H, A) = 0 for any i ∈ Z and
any subgroup H ≤ G.
TODO: see [Brown, §VI.8] and [Local Fields, Ch. IX] for a better exposition.
Lemma 9.1. Let G be a p-group (i.e. a group with each element of order pk ). Let A be a ZG-module
such that p A = 0. Then AG 6= 0, unless A = 0.
46
Proof. We may assume that A is finitely generated as a ZG-module and as an abelian group. This
means A is finite of order pk . Every G-orbit in A has order p` for some ` ≥ 0. So |AG | ≡ |A| ≡ 0
(mod p), which implies AG 6= 0.
Proposition 9.1. Let G be a p-group. Let A be a ZG-module such that p A = 0. Then A = 0 iff
H0 (G, A) = 0.
Proof. If p A = 0, then A is a Fp -vector space. One has
H0 (G, A)∨ ' H 0 (G, A∨ ).
Indeed, H0 (G, A) ' Z ⊗ZG A and H 0 (G, A) ' HomZG (Z, A). We have an adjunction
HomFp (Z ⊗ZG A, Fp ) ' HomZG (Z, HomFp (A, Fp )).
So if H0 (G, A) = 0, then H 0 (G, A∨ ) = 0, which means A∨ = 0 (by the lemma above). Finally,
A ' (A∨ )∨ = 0.
Proposition 9.2. Let G be a p-group. Let A be a ZG-module such that p A = 0. The following are
equivalent:
(a) A is free as an Fp G-module.
(b) A is cohomologically trivial.
(c) H1 (G, A) = 0.
Proof. The implication (b) Ñ (c) is trivial.
(c) Ñ (a). Since p A = 0, we can view A as an Fp -vector space. Consider a basis for an Fp -vector space
A/IG A ' H0 (G, A). Let it be {αi }i∈I . Now lift this basis to A via the map A A/IG A. We claim that the
lifted vectors {e
αi }i∈I span the whole A. Indeed, this is because
e =
H0 (G, A/A)
e
A/A
= 0,
e
IG (A/A)
e = 0.
which by the lemma above gives A/A
So there is a surjection L A where L is a free Fp G-module. It induces an isomorphism L/IG L '
A/IG A. Let L0 be the kernel of this map. Consider a short exact sequence
0 → L0 → L → A → 0
Write down the corresponding cohomology long exact sequence:
· · · → H1 (G, A) → H0 (G, L0 ) → H0 (G, L) → H0 (G, A) → 0
By our hypothesis H1 (G, A) = 0; we also have H0 (G, L) ' H0 (G, A), so H0 (G, L0 ) = 0, which by the
lemma implies L0 = 0 and so A ' L is a free Fp G-module.
Proposition 9.3. Let G be a p-group and let A be a G-module without p-torsion. Then T.F.A.E.:
(1) A is cohomologically trivial.
(2) A/pA is a free Fp G-module.
47
Proof. Since A has no p-torsion, multiplication by p is an injective map A → A, and we have a short
exact sequence
p
0→AÏ
− A → A/pA → 0
Look at the corresponding cohomology long exact sequence.
· · · → Ĥ q (G, A) → Ĥ q (G, A) → Ĥ q (G, A/pA) → Ĥ q+1 (G, A) → · · ·
If A is cohomologically trivial, then from this sequence Ĥ q (G, A/pA) = 0, which means A/pA is a
free Fp G-module (the previous proposition).
Now assume A/pA is a free Fp G-module. Then for each H ≤ G we have
· · · → Ĥ q−1 (H, A/pA) → Ĥ q (H, A) → Ĥ q (H, A) → Ĥ q (H, A/pA) → · · ·
p∗
Ĥ q (H, A/pA) = 0 since A/pA is free, and so the induced map Ĥ q (H, A) −Ï Ĥ q (H, A) is an isomorphism. But H is killed by pk , so multiplication by pk induces the zero map Ĥ q (H, A) → Ĥ q (H, A). This
means p∗ is zero, so that Ĥ q (H, A) = 0.
Proposition 9.4. Let G be a finite group acting over Q. Then for any i > 0 one has Ĥ i (G, Q) = 0.
'
Proof. Let n := |G|. The multiplication by n is an isomorphism Q −
Ï Q. On the other hand, G is killed
by n, so Ĥ i (G, Q) = 0.
n
A more general statement: if |G| = nk and multiplication by n gives an isomorphism A −
Ï A, then
Ĥ (G, A) = 0.
i
9.4
Kummer theory
The Kummer theory is about extensions of the form K(a1/n )/K for some a ∈ K× . Further we assume
that n-th roots of unity are in the field: µn ⊆ K× (so that char K - n). Such an extension is called a
Kummer extension.
Consider the separable closure Ksep and the Galois group G := Gal(Ksep /K). It is a projective of
finite Galois groups Gal(L/K).
G acts on (Ksep )× naturally, and it fixes µn by assumption. Consider a short exact sequence
1
/ µn
/ (Ksep )×
x
/ (Ksep )×
/ xn
/1
It induces a long exact sequence in cohomology
1 → H 0 (G, µn ) → H 0 (G, (Ksep )× ) → H 0 (G, (Ksep )× ) → H 1 (G, µn ) → H 1 (G, (Ksep )× ) → · · ·
But now H 0 (G, (Ksep )× ) ' K× and H 1 (G, (Ksep )× ) ' 1 by Hilbert 90, so we conclude that
Hom(G, µn ) ' H 1 (G, µn ) ' K× /(K× )n .
The isomorphism comes from the connecting homomorphism. It can be made explicit:
'
K× /(K× )n
−
Ï
a
7Ï
Hom(G, µn ),
σ(a1/n )
σ 7Ï
.
a1/n
48