[[Central extension of an abelian group]]
# 2 central extension of an elementary abelian 2-group
Let $E = 2^n$ an an [[Elementary abelian group|elementary abelian]] [[p-group|2-group]] ($\mathbb{Z}_{2}$-[[vector space]]) of rank $n$ and consider a [[Central extension of an abelian group|central extension]]
$$
\begin{align*}
1 \to \mathbb{Z}_{2}^+ \stackrel{\exp}{\hookrightarrow} \hat{E} \stackrel{\pi}{\twoheadrightarrow} E \to 1
\end{align*}
$$
where $\pi x = \overline{x}$
Given a $\Set$-[[Split epimorphism|section]] $s_{(-)} : E \hookrightarrow \hat{E}$ of $\pi$, the **associated squaring map** is then
$$
\begin{align*}
q : E &\to \mathbb{Z}_{2} \\
a &\mapsto \ln (s_{a}^2)
\end{align*}
$$
which is a [[quadratic form]] independent of $s_{(-)}$.
The [[Correspondence between quadratic forms and symmetric bilinear forms away from 2|polar form]] of $q$ is the [[Central extension of an abelian group|associated commutator map]] $c_{0}$,
and we have a [[Surjectivity, injectivity, and bijectivity|bijection]] between (arbitrary) central extensions of the above form and quadratic forms on $E$.
Furthermore, a group $\hat{E}$ is an [[Extraspecial p-group|extraspecial 2-group]] iff it is a central extension of the above form for which $q$ is [[Quadratic space#^nondegenerate]].[^1988] #m/thm/group
> [!check]- Proof
> Clearly equivalent extensions determine the same squaring map.
> Noting that $\pi (s_{a}^2) = 2a = 0$, it follows that $q$ is well defined,
> and since for any $p \in \mathbb{Z}_{2}$,
> $$
> \begin{align*}
> \ln(s_{a}\mathrm{e}^{p} s_{a} \mathrm{e}^{p}) =
> \ln(s_{a}^2 \mathrm{ e}^{2p}) = \ln(s_{a}^2) + 2p = \ln(s_{a}^2)
> \end{align*}
> $$
> so $q$ is independent of the section chosen.
> Now let $s_{a}s_{b}\mathrm{e}^p = s_{a+b}$.
> Then
> $$
> \begin{align*}
> b_{q}(a,b) &= q(a + b) - q(a) - q(b) \\
> &= \ln(s_{a+b}^2) - \ln(s_{a}^2) - \ln(s_{b}^2) \\
> &= \ln(s_{a}s_{b}\mathrm{e}^ps_{a}s_{b}\mathrm{e}^ps_{a}^{-2}s_{b}^{-2}) \\
> &= \ln(s_{a}s_{b}s_{a}^{-2}s_{a}s_{b}s_{b}^{-2}) \\
> &= \ln[s_{a},s_{b}] = c_{0}
> \end{align*}
> $$
> as claimed.
>
> Let $q : E \to \mathbb{Z}_{2}$ be a quadratic form, $\{ x_{i} \}_{i=1}^n$ be a $\mathbb{Z}_{2}$-[[Vector basis|basis]] of $E$, and define a unique bilinear map so that
> $$
> \begin{align*}
> \varepsilon_{0} : E \times E &\to \mathbb{Z}_{2} \\
> (x,x) &\mapsto q(x) \\
> (x_{i}, x_{j}) &\mapsto 0 & i<j
> \end{align*}
> $$
> Then by the [[Central group extension#Correspondence between 2-cocycles and central extensions]] there is a central extension
> $$
> \begin{align*}
> 1 \to \mathbb{Z}_{2}^+ \hookrightarrow \hat{E} \twoheadrightarrow E \to 1
> \end{align*}
> $$
> with the 2-cocycle $\varepsilon_{0}$ and thus the squaring map $q$.
>
> Now for uniqueness, suppose
> $$
> \begin{align*}
> 1 \to \mathbb{Z}_{2}^+ \hookrightarrow B \stackrel{\varphi}\twoheadrightarrow E \to 1
> \end{align*}
> $$
> is a central extension with squaring map $q$.
> Then the associated bilinear map is the polar form $b_{q}$.
> Defining a $\Set$-section $s_{(-)}$ of $\varphi$ so that
> $$
> \begin{align*}
> s_{(-)} : \sum_{k=1}^n m_{k}x_{k} = \prod_{k=1}^n s_{x_{k}}^{m_{k}}
> \end{align*}
> $$
> it is easily shown that $\varepsilon_{0}$ is the corresponding 2-cocyle and so $(B,\varphi)$ is equivalent. <span class="QED"/>
[^1988]: 1988\. [[Sources/@frenkelVertexOperatorAlgebras1988|Vertex operator algebras and the Monster]], §5.3, pp. 108–110
## Properties
### Automorphisms
Letting
$$
\begin{align*}
\Aut(\hat{E}; \mathrm{e}) &= \{ \varphi \in \Aut \hat{E} : \varphi \exp = \exp \} \\
\Aut(E; q) &= \{ \psi \in \Aut E : q\psi = q \} \\
\Aut(E; c_{0}) &= \{ \psi \in \Aut E : c_{0}(\psi,\psi) = c_{0} \}
\end{align*}
$$
it follows $\Aut(E;q) \leq \Aut(E;c_{0})$,
and we have the [[group extension]]
$$
\begin{align*}
1 \to \Ab(E, \mathbb{Z}_{2}) \stackrel{*}{\hookrightarrow} \Aut(\hat{E}; \mathrm{e}) \stackrel{\pi}{\twoheadrightarrow} \Aut(E; c_{0}) \to 1
\end{align*}
$$
where for $\lambda \in \Ab(A, \mathbb{Z}_{2}) \cong \mathbb{Z}_{2}^n \cong E$, #m/thm/group
$$
\begin{align*}
\lambda^* : \hat{A} &\to \hat{A} \\
x &\mapsto x \mathrm{e}^{\lambda \bar{x}}
\end{align*}
$$
cf. [[Cyclic central extension of a free abelian group|the analogous result for free abelian groups]].[^5.4.5]
Furthermore, if $\hat{E}$ is extraspecial then $\Aut \hat{E} = \Aut(\hat{E}; \mathrm{e})$,
and the [[Inner group automorphism|inner automorphisms]] are given by
$$
\begin{align*}
\Inn \hat{E} = \ker \pi = \Ab(E, \mathbb{Z}_{2})^* \cong E
\end{align*}
$$
where the isomorphism is natural, giving the short exact sequences
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[^5.4.5]: 1988\. [[Sources/@frenkelVertexOperatorAlgebras1988|Vertex operator algebras and the Monster]], ¶5.4.5, p. 114
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