Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I'm trying to work through calculating the order of orthogonal groups in characteristic $\neq 2$. However there is one proof by induction used that i can't quite follow. Could someone help me understand where the formula for $z_{m+1}$ comes from and how we know $U$ must contain $2q-1$ vectors with norm $0$ and $q-1$ vectors of every non-zero norm in the following extract:

Let $z_m$ denote the number of non-zero isotropic vectors in an orthogonal space with dimension $2m$ or $2m+1$. We claim that:

$z_m = q^{2m}-1$ for dimension $2m+1$

$z_m = (q^m-1)(q^{m-1}+1)$ for plus type with dimension $2m$

$z_m = (q^m-1)(q^{m-1}-1)$ for minus type with dimension $2m$

For our inductive step we look at a $n+2$ dimensional space $V$ to ensure all spaces remain of the same type. Split V into the direct sum of $U$ and $W$ where $U$ is a $2$-dimensional space of plus type and $W$ is an $n$-dimensional space with the same type as $V$. Any isotopic vector in $V$ can be written as $u+w$ for isotropic vectors $u\in U$ and $w\in W$. Either $u$ and $w$ both have norm $0$ (with one being non-zero) or $u$ has norm $\lambda \neq 0$ and $w$ has norm $-\lambda$. Since $U$ contains $2q-1$ vectors of norm $0$ (including the zero vector) and $q-1$ vectors of every non-zero norm we count:

$z_{m+1}=(2q-1)(1+z_m)+(q-1)(q^n-1-z_m)-1$

The other three cases are similar.

Thanks

share|improve this question

1 Answer 1

up vote 1 down vote accepted

It is the case that each isotropic vector in $V$ has the form $u+w$ where $u\in U$ and $w\in W$ but $u$ and $w$ need not be isotropic.

To see where $2q-1$ and $q-1$ come from, the quadratic form on $U$ has norm given by $(x,y)\mapsto xy$. Now count how many pairs of elements of $\mathbb{F}_q$ give zero, and any given nonzero element.

share|improve this answer
    
I was under the impression you could ensure $u$ and $w$ are isotropic. Also how would such a counting argument go in generality? I'm hoping if i can understand this part it'll help me derive the formula for $z_{m+1} –  Leigh Bell May 7 '10 at 17:26
    
Isotropy of $u$ and $w$ means that $q(u)=q(w)=0$ where $q$ is the quadratic form in question. In your posting you actually discuss the case where both are nonzero. For the counting argument, think about how many pairs $(x,y)$ of elements in $\mathbb{F}_q$ satsify $xy=0$? $xy=1$? etc. –  Robin Chapman May 7 '10 at 18:48

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.