Explicit $2$-cocycle for $2^{1+2n}_+$

Question

Let $p$ be a prime and $n\geq 1$, it is known that there are exactly two extraspecial groups of order $p^{1+2n}$, denoted by $p^{1+2n}_+$ and $p^{1+2n}_-$. They fit into central extensions $$0\to(\mathbb{Z}/p)\to E\to(\mathbb{Z}/p)^{2n}\to 0.$$ When $p$ is odd, $E=p^{1+2n}_+$ has exponent $p$ and an explicit $2$-cocycle $\beta:V\times V\to\mathbb{Z}/p$ for $E$ (where $V=(\mathbb{Z}/p)^{2n}$, seen as an $\mathbb{F}_p$-vector space) is given by $\beta(v,w)=\frac{1}{2}[v,w]$, with $$[v,w]:=\displaystyle\sum_{i=0}^n{v_iw_{n+i}-w_iv_{n+i}}.$$ Is there any explicit description for $\beta$ when $p=2$?

Answer 1

When $p=2$, the bilinear form given by commutators no longer specifies the extraspecial group. You need to use a quadratic form $q\colon V \to \mathbb{Z}/2$ giving the square map. A commutator in a group is always a product of three squares (via $x^{-1}y^{-1}xy = (x^{-1})^2(xy^{-1})^2y^2$ ) and so $q$ determines the commutator map via $[x,y]=q(x+y)+q(x)+q(y)$.

Given the quadratic form $q$ determining the extraspecial group, there are many cocycles representing the cohomology class. They satisfy $\beta(x,x)=q(x)$, and $\beta(x,y)+\beta(y,x)=q(x+y)+q(x)+q(y)$. To choose such a $\beta$, choose a basis of $V$, say $v_1,\dots,v_{2n}$. Then set $\beta(v_i,v_i)=q(v_i)$, choose $\beta(v_i,v_j)$ arbitrarily for $i<j$, and then set $\beta(v_j,v_i)=q(v_i+v_j)+q(v_i)+q(v_j)+\beta(v_i,v_j)$. Now extend bilinearly.

There are two types of nonsingular quadratic forms, distinguished by their Arf invariants, and the recipe above works fine for both cases.

Answer 2

The extraspecial group $P = 2_{+}^{1+2n}$ has generators $z$, $v_1$, $\ldots$, $v_n$, $w_1$, $\ldots$, $w_n$, with $Z(P) = \langle z \rangle$ cyclic of order $2$ and

\begin{align*} v_i^2 = w_i^2 = 1 & & \text{ for all } i\\ [v_i,v_j] = [w_i,w_j] = 1 && \text{ for all } i,j \\ [v_i,w_j] = 1 && \text{ for } i \neq j \\ [v_i,w_i] = z && \text{ for all } i \\ \end{align*}

So for $V = P/Z(P)$ you get the $2$-cocycle $f: V \times V \rightarrow Z(P)$ defined by $$f(v,v') = z^{\sum_{k = 1}^n i_k' j_k}$$

for $v = \prod_k v_k^{i_k} \prod_k w_k^{j_k}$ and $v' = \prod_k v_k^{i_k'} \prod_k w_k^{j_k'}$.

You can of course interpret this in terms of quadratic forms as in Dave Benson's answer. In this case $q: V \rightarrow \mathbb{F}_2$ with $q(v_i) = q(w_j) = 0$, with polarization $\beta$ defined by $\beta(v_i,v_j) = \beta(w_i,w_j) = 0$ and $\beta(v_i,w_j) = \delta_{i,j}$.

With some abuse of notation, the quadratic form on $V$ is defined by the formula $x^2 = z^{q(x)}$ for all $x \in P$.