Symplectic group

Let $v=(x_1,\cdots ,x_a,y_1,\cdots ,y_a{)}^T$, and let $w=(x_1^{\prime },\cdots ,x_a^{\prime },y_1^{\prime },\cdots ,y_a^{\prime }{)}^T$. Define $B(v,w)=x_1{y}_1^{\prime }+\cdots +x_a{y}_a^{\prime }-x_1^{\prime }{y}_1-\cdots -x_a^{\prime }{y}_a$, i.e., $B(v,w)=v^{T}Jw$, where $J=\left(\begin{array}{cc}0& I_a\\ -I_a& 0\end{array}\right)$. In such a basis, define

$$\mathrm{Sp}(2a,q)=\{g\in \mathrm{GL}(2a,q):B(gv,gw)=B(v,w)\}=\{g\in \mathrm{GL}(2a,q):g^{T}Jg=J\}.$$

Simplicity

The only scalars in $\mathrm{Sp}(2a,q)$ are $\{\lambda I_{2a}:{\lambda }^2=1\}$. Hence

$$\mathrm{PSp}(V):=\mathrm{Sp}(V)/\{\lambda I_{2a}:{\lambda }^2=1\}=\mathrm{Sp}(V)/\mathrm{Z}(\mathrm{Sp}(V)),$$

where $Z(\mathrm{Sp}(V))=\{\lambda I_{2a}:{\lambda }^2=1\}$ and $|Z(\mathrm{Sp}(V))|=(2,q-1)$.

Theorem

If $a⩾2$, then the group $\mathrm{PSp}(2a,q)$ is simple, except for ${\mathrm{PSp}}_4(2)\cong S_6$.

$\mathrm{Aut}(\mathrm{Sp}(2a,p^f))$

Theorem

Let $G=\mathrm{Sp}(2a,p^f)$. Then

$$\mathrm{Out}G=\{\begin{array}{ll}⟨\delta ⟩\times ⟨\varphi ⟩={\mathbb{Z}}_{(2,q-1)}\times {\mathbb{Z}}_f& \text{if }G\ne \mathrm{Sp}(4,2^f),\\ ⟨\gamma ⟩={\mathbb{Z}}_{2f}& \text{if }G=\mathrm{Sp}(4,2^f).\end{array}$$

\begin{proof} See

\end{proof}

Normalizer

Definition

Let $\beta$ be a $\sigma$-sesquilinear form on $V$. Then $g$ is an isometry of $\beta$ if $\beta (ug,vg)=\beta (u,v)$ for all $u,v\in V$. The element $g$ is a similarity of $\beta$ if there is a $\lambda \in F\setminus \{0\}$ such that $\beta (ug,vg)=\lambda \beta (u,v)$ for all $u,v\in V$.

The isometry group of $\beta$ is the group of all isometries of $\beta$, and the similarity group of $\beta$ is the group of all similarities of $\beta$. The conformal group is the similarity group of $\beta$.

We refer to the conformal symplectic group $\mathrm{CSp}(2a,q)$.

The following lemma shows that the conformal group is equal to the normalizer in the corresponding linear group of the quasisimple group.

Lemma

Let $G⩽{\mathrm{GL}}_n(F)$ be an absolutely irreducible group consisting of isometries of a bilinear form $\beta$. Then ${\mathrm{N}}_{{\mathrm{GL}}_n(F)}(G)$ consists of similarities of $\beta$.

Corollary

$N_{\mathrm{GL}(2a,q)}(\mathrm{Sp}(2a,q))=\mathrm{CSp}(2a,q)$.

\begin{proof} By ^91tcc7, we have $N_{\mathrm{GL}(2a,q)}(\mathrm{Sp}(2a,q))⩽\mathrm{CSp}(2a,q)$.

Conversely, for any $x\in \mathrm{CSp}(2a,q)$, there exists $\lambda \in {\mathbb{F}}_q^{\ast }$ such that $\beta (ux,vx)=\lambda \beta (u,v)$. Notice that for any $u,v\in V$ and $s\in \mathrm{Sp}(2a,q)$, there is

$$\begin{array}{rl}\beta \left(u{x}^{-1}sx,v{x}^{-1}sx\right)& =\lambda \beta \left(u{x}^{-1}s,v{x}^{-1}s\right)\\ & =\lambda \beta \left(u{x}^{-1},v{x}^{-1}\right)\\ & =\beta (u,v)\end{array}$$

and so $x^{-1}sx\in \mathrm{Sp}(2a,q)$. \end{proof}

Definition

The conformal semilinear group of $V$ is the group of all semi-similarities of $V$. For $\mathrm{Sp}(2a,q)$, it is denoted by $\mathrm{C\Gamma Sp}(2a,q)$.

Theorem

$N_{\mathrm{GL}(n,p)}(\mathrm{Sp}(2a,q))=\mathrm{C\Gamma Sp}(2a,q)$.

\begin{proof} For any $x\in N_{\mathrm{GL}(n,p)}(\mathrm{Sp}(2a,q))$, we have $x^{-1}\mathrm{Sp}(2a,q)x=\mathrm{Sp}(2a,q)$. Then any $h\in \mathrm{Sp}(2a,q)$ can be written as $h=x^{-1}gx$ for some $g\in \mathrm{Sp}(2a,q)$. Take $c\in C_{\mathrm{GL}(n,p)}(\mathrm{Sp}(2a,q))$, then

$$\left(x^{-1}cx\right)h=x^{-1}cx{x}^{-1}gx=x^{-1}cgx=x^{-1}gcx=x^{-1}gx{x}^{-1}cx=h\left(x^{-1}cx\right).$$

Hence $x^{-1}cx\in C_{\mathrm{GL}(n,p)}(\mathrm{Sp}(2a,q))$ and so $x$ normalizes $C_{\mathrm{GL}(n,p)}(\mathrm{Sp}(2a,q))$.

Notice that $C_{\mathrm{GL}(n,p)}(\mathrm{Sp}(2a,q))={\mathbb{F}}_q^{\ast }$ (see here), then $x^{-1}{\mathbb{F}}_{q}x={\mathbb{F}}_q$. That is, for each $\lambda \in {\mathbb{F}}_q$, there exists ${\lambda }^{\sigma }\in {\mathbb{F}}_q$ such that $x^{-1}{m}_{\lambda }x=m_{{\lambda }^{\sigma }}$, where $m_{\lambda }:v\mapsto \lambda v$ and $\lambda ,{\lambda }^{\sigma }\in {\mathbb{F}}_q$. The map $\sigma :{\mathbb{F}}_q\to {\mathbb{F}}_q,\lambda \mapsto {\lambda }^{\sigma }$ is a field automorphism. It follows that $\lambda x(v)=x({\lambda }^{\sigma }v)$ and $x$ is ${\sigma }^{-1}$-semilinear.

Define ${\beta }_x(u,v)=\beta (ux,vx{)}^{\sigma }$, then by $\lambda x(v)=x({\lambda }^{\sigma }v)$, ${\beta }_x$ is a bilinear alternating form. Since $x$ normalizes $\mathrm{Sp}(2a,q)$, ${\beta }_x$ is a $\mathrm{Sp}(2a,q)$-invariant alternating form. Hence there exists ${\lambda }_0\in {\mathbb{F}}_q^{\ast }$ such that ${\beta }_x={\lambda }_0\beta$, and so $\beta (ux,vx)={\lambda }_0^{{\sigma }^{-1}}\beta (u,v{)}^{{\sigma }^{-1}}$. Therefore, $x\in \mathrm{C\Gamma Sp}(2a,q)$ and so $N_{\mathrm{GL}(n,p)}(\mathrm{Sp}(2a,q))⩽\mathrm{C\Gamma Sp}(2a,q)$.

Conversely, for any $x\in \mathrm{C\Gamma Sp}(2a,q)$, there exists $\sigma \in \mathrm{Gal}({\mathbb{F}}_q/{\mathbb{F}}_p)$ and $\lambda \in {\mathbb{F}}_q^{\ast }$ such that $\beta (ux,vx)=\lambda \beta (u,v{)}^{\sigma }$. It suffices to show that for any $s\in \mathrm{Sp}(2a,q)$, there is $x^{-1}sx\in \mathrm{Sp}(2a,q)$. Notice that

$$\begin{array}{rl}\beta \left(v{x}^{-1}sx,w{x}^{-1}sx\right)& =\lambda \beta {\left(v{x}^{-1}s,w{x}^{-1}s\right)}^{\sigma }\\ & =\lambda \beta {\left(v{x}^{-1},w{x}^{-1}\right)}^{\sigma }\\ & =\beta (v,w),\end{array}$$

then we have $x^{-1}sx\in \mathrm{Sp}(2a,q)$. Since every ${\mathbb{F}}_q/{\mathbb{F}}_p$-semilinear transformation of $V$ is ${\mathbb{F}}_p$-linear, we know $x\in \mathrm{GL}(n,p)$ and so $x\in N_{\mathrm{GL}(n,p)}(\mathrm{Sp}(2a,q))$. Now we finish the proof. \end{proof}