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

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

• Alternating bihomomorphism of finitely generated abelian groups arises as skew of 2-cocycle

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.