Finding the spinor norm of an element using a proposition in 'The Maximal Subgroups of the Low-Dimensional Finite Classical Groups'.

Question

Let $F=\mathbb{F}_{q}$, where $q$ is an odd prime power. Let $e,f,d$ be a standard basis for the $3$-dimensional orthogonal space $V$, i.e. $(e,e)=(f,f)=(e,d)=(f,d)$ and $(e,f)=(d,d)=1$. I have an element $g\in SO_{3}(q)$ defined by: $g: e\mapsto -e$, $f\mapsto \frac{1}{2}e -f +d$, $d\mapsto e+d$. I would like to determine the spinor norm of this element using Proposition 1.6.11 in the book 'The Maximal Subgroups of the Low-Dimensional Finite Classical Groups' by Bray, Holt and Roney-Dougal.

The proposition is quite long to state so it would be handy if someone who can help already has a copy of the book to refer to. If not, then please let me know and I can post what the proposition says.

I have followed the proposition and have that the matrices $$A:=I_{3}-g=\left( \begin{array}{ccc} 2 & -\frac{1}{2} & -1 \\ 0 & 2 & 0 \\ 0 & -1 & 0 \end{array} \right)$$ $$F= \textrm{matrix of invariant symmetric bilinear form =}\left( \begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right).$$

If $k$ is the rank of $A$, the proposition says to let $B$ be a $k\times 3$ matrix whose rows form a basis of a complement of the nullspace of $A$. I have that $ker\, A=<(1,0,2)^{T}>$. Now by the way the proposition is stated it seems as if it does not make a difference as to what complement is taken. However, I have tried 3 different complements and I get contradictory answers each time.

Try 1) Orthogonal complement of $Ker\, A$, where $B=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & -2 & 1 \end{array} \right)$. This gives me $det(BAFB^{T})=-25$.

Try 2) $B=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \end{array} \right)$. This gives me $det(BAFB^{T})=-4$.

Try 3) $B=\left( \begin{array}{ccc} 0 & 0 & 1 \\ 0 & 1 & 0 \end{array} \right)$. This gives me $det(BAFB^{T})=0$.

The problem is that whatever complement I take the determinants should all be non-zero squares at the same time, but this is not the case.

I am not sure if I have misunderstood how to use the proposition. Any help will be appreciated. Thanks.

Answer 1

The convention in the book is that matrices act on row vectors from the right. So the kernel of $A$ is $\langle (0,1,2) \rangle$ and Try 3) is not a complement. (You should never get $0$ for $\det(BAFB^T)$.)

Try 1) and Try 2) are either both squares or both nonsquares depending on whether $-1$ is a square.

Also, following the convention of the book, the matrix of $g$ would be the transpose of what you have, but that is not the cause of the problem!

Answer 2

In characteristic 5, Try (1) does not actually yield a complement. The bottom row, $(0,-2,1)=-2(0,1,2)$, is in the kernel in this case. So again, no contradiction.