Skew-symmetric non-degenerate bicharacters over abelian groups

Question

I'm struggling to solve a problem in group cohomology theory which could seem immediate to some more expert mathematicians here. Suppose to have a non-degenerate, skew-symmetric bicharacter $b\colon G\times G\to\mathbb{T}$ on an Abelian group $G$, namely $$b(gg',h)=b(g,h)b(g',h),\quad g,g',h\in G$$ $$b(g,hh')=b(g,h)b(g,h'),\quad g,h,h'\in G$$ $$b(g,h)=\overline{b(h,g)},\quad g,h\in G$$ $$\text{rad}(b):=\{g\in G\,\colon\,b(g,h)=1,\quad h\in G\}=(e)$$ Then, $b(g,g)\in\{\pm1\}\cong\mathbb{Z_2}$ for every $g\in G$, the map $\varphi(g):=b(g,g)\in\mathbb{Z}_2$ ($g\in G$) is a group homomorphism and the isotropy set $\Delta_+:=\{g\in G\colon b(g,g)=1\}=\ker(\varphi)$ is a subgroup of $G$ of index either 1 (i.e. $\Delta_+=G$ and $b$ is alternating) or 2 (in which case $G/\Delta_+\cong\mathbb{Z}_2$). I am interested in the second scenario, where the restriction $b|_{\Delta_+\times\Delta_+}$ is indeed an alternating bicharacter on $\Delta_+$. It seems that $b|_{\Delta_+\times\Delta_+}$ may well be constantly $1$, but I currently have only two situations when this happens:

1. $G=\mathbb{Z}_2=\{0,1\}$, $b(x,y)=(-1)^{xy}$. Here, $\Delta_+=\{(0,0)\}$ and $b|_{\{(0,0)\}\times\{(0,0)\}}\equiv1$;
2. $G=\mathbb{Z}_2\times\mathbb{Z}_2$, $b((x_1,x_2),(y_1,y_2))=(-1)^{x_1y_1+x_2y_2}$. Here, $\Delta_+=\{(0,0),(1,1)\}$ and $b|_{\Delta_+\times\Delta_+}\equiv1$. One can actually build two other bicharacters on $\mathbb{Z}_2\times\mathbb{Z}_2$ doing the same trick, but they are equivalent/congruent to the one written here.

Do you think is it possible to build other examples, at least in the finite Abelian group setting, or they are the only cases occurring, due to some hidden theory I'm not currently aware of?

Answer 1

Let us assume $G$ is finite to simplify things.

A nondegenerated bicharacter such as in your question (skew-symmetric or not) is the same as an isomorphism $\varphi: G\to \hat{G}$ where $\hat{G}=\operatorname{Hom}(G,\mathbb{T})$ is the Pontryagin dual.

Let $H$ be any index $2$ subgroup of $G$ (it is supposed to represent $\Delta_+$ in your question). Then by duality there is a canonical surjection $\hat{G}\to \hat{H}$ induced by the inclusion map $H\to G$, and the kernel of $\hat{G}\to \hat{H}$ has order $2$ (it is dual to $G/H$).

In you setup, you ask that the composition $H\to G\to \hat{G}\to \hat{H}$ is the trivial morphism (this is what corresponds to the restriction of $b$ being trivial on $\Delta_+$). But the composition $H\to G\to \hat{G}$ is injective, and it has to land into the kernel of $\hat{G}\to \hat{H}$, so $|H|\leqslant 2$.

Since $[G:H]=2$, this means that $|G|\leqslant 4$, so your examples are indeed the only ones.