Skew of 2-cocycle for trivial group action of abelian group is alternating bihomomorphism

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Statement

In terms of 2-cocycles

Let f be a 2-cocycle for trivial group action (?) of an abelian group G on an abelian group A. In other words, f:G \times G \to A is a function such that:

\! f(g_2,g_3) + f(g_1,g_2 + g_3) = f(g_1,g_2) + f(g_1 + g_2,g_3)

Then, the function:

\! h(g_1,g_2) := f(g_1,g_2) - f(g_2,g_1)

is an alternating bihomomorphism from G \times G to A.

Interpretation in terms of second cohomology group

Since any 2-coboundary for the trivial group action is a symmetric 2-cocycle for trivial group action, its skew is zero. Thus, the skew of a 2-cocycle depends only on its cohomology class. The above statement can thus be described by saying that \operatorname{Skew} gives a homomorphism from the second cohomology group to the group of alternating bihomomorphisms:

\! H^2(G,A) \to \bigwedge^2(G,A)

Related facts

Converse

Applications

  • Class two implies commutator map is endomorphism: In a group of nilpotency class two, the commutator map is an endomorphism in either variable. This is a corollary of the fact stated here, if we interpret A as the center, G as the inner automorphism group, and f is a 2-cocycle representing the extension. The skew of f is the commutator map.

Proof

Given: An abelian group G and an abelian group A. A 2-cocycle f:G \times G \to A.

To prove: The function \! h(g_1,g_2) := f(g_1,g_2) - f(g_2,g_1) is an alternating bihomomorphism.

Proof: Clearly, h(g,g) = 0 for all g, so it suffices to show that:

\! h(g_1, g_2 + g_3) = h(g_1,g_2) + h(g_1,g_3)

and

\! h(g_1 + g_2,g_3) = h(g_1,g_3) + h(g_2,g_3)

Since f is a 2-cocycle from G to A, we have:

\! f(g_1,g_2 + g_3) + f(g_2,g_3) = f(g_1 + g_2,g_3) + f(g_1,g_2) \qquad (1)

Since this is true for all g_1,g_2,g_3 \in G, the corresponding statement is true with g_1,g_2,g_3 cyclically permuted:

\! f(g_2,g_3 + g_1) + f(g_3,g_1) = f(g_2 + g_3,g_1) + f(g_2,g_3) \qquad (2)

Interchanging g_1 and g_2 in the original expression, we get:

\! f(g_2,g_1 + g_3) + f(g_1,g_3) = f(g_2 + g_1,g_3) + f(g_2,g_1) \qquad (3)

We now do (1) + (2) - (3) to obtain:

\! h(g_1, g_2 + g_3) = h(g_1,g_2) + h(g_1,g_3)

Since the variables are universally quantified, this proves the right linearity. Because of the left right symmetry, it also proves right linearity.