Group cohomology

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 01 February 2012

Group cohomology

Let $G$ be a group and let $M$ be an abelian group with an action of $G$ by automorphisms, i.e. let $M$ be a $G -$ module. Define abelian groups $C^{0} (G, M) = M and C^{n} (G, M) = {f : G \times \dots \times G \to M}, for n \in ℤ_{> 0},$ where the operation in $C^{n} (G, M)$ is given by $(f_{1} + f_{2}) (g_{1} ... g_{n}) = f_{1} (g_{1} ... g_{n}) + f_{2} (g_{1} ... g_{n}) .$ Define $0 \to C^{0} (G, M) \overset{\partial_{1}}{\to} C^{1} (M) \overset{\partial_{2}}{\to} C^{2} (M) \overset{\partial_{3}}{\to} \dots such that \partial_{n + 1} \partial_{n} = 0,$ by defining $(\partial_{n} f) (g_{1} ... g_{n}) = g_{1} f (g_{2} ... g_{n}) + \sum_{i = 1}^{n} {(- 1)}^{i} f (g_{1} ... g_{i - 1} g_{i} g_{i + 1} g_{i + 2} ... g_{n}) + {- 1}^{n} f (g_{1} ... g_{n - 1}),$ for all $f \in C^{n - 1} (G, M)$ and $g_{1} ... g_{n} \in G .$

The $n -$ cochains are the elements of $C^{n} (G, M) .$
The $n -$ coboundary map is $\partial_{n} : C^{n} (G, M) \to C^{n + 1} (G, M) .$
The $n -$ cocycles are the elements of $\ker \partial_{n + 1} .$
The $n -$ coboundaries are the elements of $im \partial_{n} .$
The $n^{th}$ cohomology group of $G$ with coefficients in $M$ is the abelian group $H^{n} (G, M) = \frac{\ker \partial_{n + 1}}{im \partial_{n}} .$

Let $𝔼$ be a field and let $G$ be a finite subgroup of $Aut (𝔼)$ .

$H^{1} (G, 𝔼)$ is trivial.
$H^{1} (G, 𝔼^{*})$ is trivial.

		Proof.
	b. Let $f : G \to E^{*}$ be a cocycle so that $f (g_{1} g_{2}) = f (g_{1}) (g_{1} f (g_{2})), for all g_{1}, g_{2} \in G .$ By Dedekind's lemma we may choose $x \in 𝔼$ such that $b = \sum_{g \in G} f (g) g (x)$ is nonzero. If $h \in G$ then $h b = \sum_{g \in G} (h f (g)) \cdot h g (x) = \sum_{g \in G} f {(h)}^{- 1} f (h g) \cdot h g (x) = f {(h)}^{- 1} b .$ So $f (h) = b / h b$ and $f$ is a coboundary. $□$

(Dedekind's Lemma.) Let $𝔽$ be a subfield of $𝔼$ .

Distinct embeddings of $𝔽$ into $𝔼$ are linearly independent.
Distinct characters $χ : G \to 𝔼^{*}$ are linearly independent.
Distinct elements of $Aut (𝔼)$ are linearly independent in ${End}_{𝔽} (𝔼)$ .

		Proof.
	Let $𝔼$ be a finite extension of $𝔽$ . Let $g_{i} : A \to 𝔼, 1 \leq i \leq n$ be algebra homomorphisms. Let $c_{1} ... c_{n} \in 𝔼$ be such that $\sum_{i = 1}^{n} c_{i} g_{i} = 0 .$ For any $x, y \in A,$ $0 = \sum_{i = 1}^{n} c_{i} g_{i} (x y) - g_{n} (x) \sum_{i = 1}^{n} c_{i} g_{i} (y) = \sum_{i = 1}^{n - 1} c_{i} (g_{i} (x) - g_{n} (x)) g_{i} (y) .$ So, by induction, $c_{i} (g_{i} (x) - g_{n} (x)) = 0, for all 1 \leq i \leq n - 1, and all x \in A .$ For each $1 \leq i \leq n - 1,$ we may choose $x$ such that $g_{i} (x) \neq g_{n} (x),$ and conclude that $l_{i} = 0 for all 1 \leq i \leq n - 1 .$ So $\sum_{i = 1}^{n} l_{i} g_{i} = l_{n} g_{n} = 0 .$ Thus $0 = l_{n} g_{n} (1) = l_{n} .$ $□$

Notes and References

Where are these from?

References

References?

page history