Center of SO(V,q)

Let $V$ be finite dimensional vector spaces and $q$ is quadratic form. I'm looking for $Z(SO(V,q))$. where $SO(V,q)$ is special orthogonal group.

If $\operatorname{dim} V$ is odd then $Z(SO(V,q))={I}$ because $-\sigma_u\in SO(V,q)$ where $\sigma_u$ is reflection $$\sigma_u(x)=x-2\frac{b(x,u)}{q(u)}$$ But I couldn't calculate center when $\operatorname{dim} V$ is even.

I have not any idea how to deal with it.

Thanks.

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Do you know the signature of $q$? I'm confused about $\sigma_u$ in the case of $q(u) = 0$. – Jason DeVito Jun 4 '12 at 14:34
For example $q$ could be identically zero, in which case ${\rm SO}(V,q) = {\rm GL}(V)$ with centre all scalars. It also depends heavily on the field over which $V$. – Derek Holt Jun 4 '12 at 15:31
@JasonDeVito the definition reflection $\sigma_u$ is for $q(u)\neq0$ – Babak Miraftab Jun 4 '12 at 17:47
@DerekHolt You are right.Let $\mathbb F$ be finite field and $char(\mathbb F)\neq 2$. – Babak Miraftab Jun 4 '12 at 17:49

For finite fields of odd characteristic, there are two types of quadratic form when $n = {\rm dim} V$ is even and one type when $n$ is odd. When $n>2$, the groups ${\rm SO}(V,q)$ act absolutely irreducibly on $V$, and so only scalar matrices could be in the centre. But for the scalar $\lambda I_n$ to preserve $q$, we need $\lambda^2=1$, so $\lambda = \pm 1$. Hence $Z({\rm SO}(V,q))$ has order 1 when $n$ is odd and 2 when $n$ is even.
When $n=2$, ${\rm SO}(V,q)$ is a cyclic (and hence abelian) group of order $|F|-1$ when the form $q$ has plus-type and $|F|+1$ when $q$ has minus-type.
It is a consequence of Schur's Lemma in Representation Theory that the centralizer (in the matrix algebra) of the image of an irreducible representation is a division algebra over $F$ and, if the representation is absolutely irreducible, then it consists of scalar matrices only. – Derek Holt Jun 5 '12 at 8:32