action of the extra-special group

Question

I'm reading a paper which has this line:

A direct computation shows that $P\Omega_8$($\mathbb K$) has an elementary abelian subgroup $X = 2^2$ such that $C_{P\Omega_8(\mathbb K)}(X) = T_4.2^{1+4}_+$. Now the action of $2^{1+4}_+$ on $T_4$ is also explicitly determined.

$\mathbb K$ is algebraically closed with char not equal 2 and $P\Omega_8$($\mathbb K$) is the adjoint algebraic group of type D.

What is the action exactly here? I suppose the four 2 invert the four $T_1$? What about the center? Not sure...

Answer 1

Take $T_4$ as the diagonal torus, which is composed by 4-tuples of 2x2 rotation matrices $a,b,c,d$: $$\begin{bmatrix}a & & & \\ & b & &\\ & & c &\\ & & & d\end{bmatrix}$$.

Let $T$ denote the matrix $\begin{bmatrix} 1 & 0 \\ 0 & -1\end{bmatrix}$ and $I$ the $2\times2$ identity matrix.

Then the extraspecial group $E=2^{1+4}_+$ acts on $T_4$ by conjugation of the following matrices:

$\begin{bmatrix}X & & & \\ & X & &\\ & & X &\\ & & & X\end{bmatrix}$, $\begin{bmatrix}& X & & \\ X & & &\\ & & & X\\ & & X & \end{bmatrix}$, $\begin{bmatrix}& & X & \\ & & & X\\ X & & &\\ & X & & \end{bmatrix}$ and$\begin{bmatrix}& & & X \\ & & X &\\ & X & &\\ X & & &\end{bmatrix}$.

where $X$ represents an element that is either $T$ or $I$, and the number of $T$s is always even.

This group has $32$ elements, because every matrix above represent $8$ elements. There is a matrix representation of $Q_8 \circ Q_8$ on GroupNames. By checking the matrices on the GroupNames page and exchanging $T$ for $-1$ and $I$ for $1$, it is evident that $Q_8 \circ Q_8$ is a subgroup of $E$. Since $E$ and $Q_8 \circ Q_8$ both have $32$ elements, we have $E=Q_8 \circ Q_8$.

Bonus: The group $X$ is generated by the following elements of the maximal torus.

$\begin{bmatrix}-I & & & \\ & -I & &\\ & & I &\\ & & & I\end{bmatrix}$ and $\begin{bmatrix}-I & & & \\ & I & &\\ & & -I &\\ & & & I\end{bmatrix}$

Another way of seeing this group is via the projections from the universal cover to the orthogonal group and then the projective orthogonal group, i.e. $Spin_8^+(K) \rightarrow \Omega_8^+(K) \rightarrow P\Omega_8^+(K)$.

Let $\hat{T}_4$ be the Cartan subgroup of $Spin_8^+(K)$ and $T'_4$ its image in $\Omega_8^+(K)$. Then the image of the exponent-2 subgroup of $\hat{T}_4$ in $\Omega_8^+(K)$ are composed of the elements in $T'_4$ that have order $2$ and reversing an even number of indicies, i.e. the elements $(a, b, c, d) \rightarrow (s_1 a,s_2 b,s_3 c, s_4 d)$ where the $s_n$ are either $+1$ or $-1$ and $s_1s_2s_3s_4=1$. There are $8$ such elements.

In the map $\Omega_8^+(K) \rightarrow P\Omega_8^+(K)$, the element $(a, b, c, d) \rightarrow (-a,-b,-c,-d)$ is identified with the identity element, so the image of the subgroup in $T'_4$ in $T_4$ has only 4 elements.