Alternating forms on abelian groups

Question

Let $G$ and $H$ be abelian groups. By an alternating form, I mean a bilinear function $A\colon G\times G\to H$ such that $A(x,x)=0$ for all $x\in G$.

Question. If $A\colon G\times G\to H$ is an alternating form, must there exist a bilinear function $B\colon G\times G\to H$ such that $A(x,y)=B(x,y)-B(y,x)$?

I think that the answer is NO, but I can't find a counterexample.

Of course, the answer would be YES if these were vector spaces: represent $A$ as a matrix, and use the upper triangular part to get $B$.

Here are some things I know.

• If 2 is invertible in $H$, the answer is YES ($B(x,y)=\tfrac12 A(x,y)$).
• If $G=\bigoplus G_i$, then to find $B$ it suffices to find a $B_{ii}$ for each restriction $A_{ii}\colon G_i\times G_i\to H$ of $A$ (same idea as the linear algebra proof).
• In particular, the answer is always YES if $G$ is a direct sum of cyclic groups.

You can reformulate the question in various ways. E.g., the answer is yes if

• the image of $\mathrm{id}-\sigma\colon G\otimes G\to G\otimes G$ is a summand of its codomain, or
• $H^2(C_2, \mathrm{Hom}(G\otimes G,H)) = \mathrm{Hom}(G/2, H)$.

Added: Here's some more context on this question. It is known that there exists a 2-cocycle $C\colon G\times G\to H$ such that $A(x,y)=C(x,y)-C(y,x)$. This is a theorem of NJS Hughes ("The Use of Bilinear Maps in the classification of groups of class 2", 1951), also proved as Thm 26.2 in Eilenberg-MacLane, "On the groups H(Pi,n), II", 1953. The idea is that, given a central extension $H\to E\to G$ with $G,H$ abelian, the commutator defines an alternating form $A:G\times G\to H$. Hughes showed that alternating forms always arise from such a central extension.

So you can formulate my question as: can you always do this using a central extension defined by a bilinear form?

Another way to say it: given an extension $0\to H\to F\to G\to 0$ of abelian groups, and a bilinear form $B\colon G\times G\to H$, you can define a new (possibly non-abelian) extension $E$ with group law $x*y:= x+y+B(x,y)$ (this $B$ is the restriction of the form from $G$ to $F$). Does every central extension of abelian groups arise this way?

Answer 1

The answer is no. I give an example with a rank $3$ torsion-free abelian group $L$. We have the alternating map$$L \times L \to \wedge^2 L, \quad x \otimes y \mapsto x \wedge y.$$We have the alternating map $a: \wedge^2 L \to \otimes^2 L$ which is injective, with torsion-free cokernel. The question is whether there is a left inverse $b: \otimes^2 L \to \wedge^2 L$. We have$$ab' = x \otimes y \mapsto x \otimes y - y \otimes x$$such that $b'a = 2$. It suffices to arrange that all endomorphisms of the abelian group $\otimes^2 L$ are linear combinations of the identity and the switch $\sigma: x \otimes y \mapsto y \otimes x$.

Lemma. Let $V$ be a vector space of dimension $n > m \ge 2$ over a field $k$. Then there are finitely many surjections $V \underset{p_i}{\to} W_i$, $\dim W_i = m$ such that if $T \in \text{End}\otimes^2 V$ preserves $\text{Ker}(\otimes^2 V \to \otimes^2 W_i)$ for all $i$, then $T$ is a linear combination of $\text{id}$ and $\sigma$.

Proof. Let $V$ have basis $\{e_i\}$. Using projections defined by subsets of the coordinates, verify that for $i \neq j$, $T(e_i \otimes e_j)$ is a linear combination of $e_i \otimes e_j$ and $e_j \otimes e_i$. For $k \neq i, j$, use another basis containing $(e_i, e_j \neq e_k)$ to compare the coefficients in $T(e_i \otimes e_j)$ and $T(e_i \otimes e_k)$. Use a basis containing $(e_1, e_1 + e_2, e_3)$ to control $T(e_1 \otimes e_1)$.

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The proposed construction runs as follows. Let $L_0 = \mathbf{Z}^3$, $V = L_0 {}_\mathbf{Q} = \mathbf{Q}^3$, applying the lemma with $n = 3$, $m = 2$, let $V \twoheadrightarrow W_i$ be the surjections, $\text{Ker}(p_i) \cap L_0 = \mathbf{Z} u_i$.

Choose distinct auxiliary odd primes $p_i$ and let $L = L_0 + \sum \mathbf{Z}[1/p_i] u_i \subset V$.

Then $\text{Ker}(\otimes^2 V \to \otimes^2 W_i)$ is spanned by the infinitely $p_i$-divisible vectors in $\otimes^2 L$. Thus for $T \in \text{End}\otimes^2L$, the condition of the lemma is verified for $T_\mathbf{Q} \in \text{End}\otimes^2 V$. Thus $T_\mathbf{Q}$ is a rational combination of $\text{id}$ and $\sigma$. To check that the coefficients are integral, it's enough to work over each localization $\mathbf{Z}_{(p)}$, we have that $L \otimes \mathbf{Z}_{(p)}$ is isomorphic to $\mathbf{Z}_{(p)}^3$ or $\mathbf{Z}_{(p)}^2 \otimes \mathbf{Q}$ and the verification of integrality is straightforward.

Remark. One may wonder whether it is enough to work only locally at $2$. This becomes more technical. Let $\mathbf{Z}_2$ be the $2$-adic integers. $\mathbf{Q}_2$ is of infinite transcendence degree over $\mathbf{Q}$. In particular, we have a vector $(1, u, v) \in \mathbf{Z}_2^3$ with $u$, $v$ algebraically independent over $\mathbf{Q}$. One can do the above construction replacing $L_0$ by $(\mathbf{Z}_2^3 + \mathbf{Q}_2(1, u, v)) \cap \mathbf{Z}[1/2]^3$. The lemma holds (up to a suitable increase in the number $N$ of $W_i$'s) when the $V \twoheadrightarrow W_i$ define a point in some nonempty Zariski open subset of the $N$th power of the relevant Grassmanian.