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First I want to review some concept from quadratic form. Let $V$ be quadratic space over finite field $F$ and $char(F)\neq 2$ with quadratic form $q$. For exmaple

$q:V\rightarrow V$ and $|F|=q$ and $\dim V=n$.

We may assume the vectors of $V$ are represented as column vectors $v=(a_1,b_1,…)\in V$, and that $q$ takes one of the three forms

1.$q(v)={a_1}^2-{b_1}^2+…+{a_k}^2-{b_k}^2$

2.$q(v)={a_1}^2-{b_1}^2+…+{a_k}^2-{db_k}^2$

3.$q(v)={a_1}^2-{b_1}^2+…+{a_k}^2-{b_k}^2-{a_{k+1}}^2$

or if we want to have matrix representation we have

1.$q=diag(1,-1,1,-1,…,1,-1)$

2.$q=diag(1,-1,1,-1,…,1,-1,1)$

3.$q=diag(1,-1,1,-1,…,1,-1,1,-d)$

Finally for subspace $W$ we have $W^{\perp}=\{v\in V\mid b(v,w)=0 ,\forall w \in W \}$ where $b(x,y)=(\frac{1}{2})[q(x+y)-q(x)-q(y)]$.

The problem I am interested in :

Compute the number of $k$-dimensional subspaces $W$ such that $W\subseteq W^{\perp}$; this subspace is called totally isotropic subspace (T,I). For $k=1$, it is prominent theorem. That comes back to find solution of these equations

1.$q(v)={x_1}^2-{y_1}^2+…+{x_k}^2-{y_k}^2=0$

2.$q(v)={x_1}^2-{y_1}^2+…+{x_k}^2-{dy_k}^2=0$

3.$q(v)={x_1}^2-{y_1}^2+…+{x_k}^2-{y_k}^2-{x_{k+1}}^2=0$ So the number of 1-dimensional (T,I) is

1.$\frac{q^{2k-1}+q^{k}-q^{k-1}-1}{q-1}$

2.$\frac{q^{2k-1}-q^{k}+q^{k-1}-1}{q-1}$

3.$\frac{q^{2k}-1}{q-1}$

Now what can I do for $k\geq 2$.

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evidently from page79 classic group and geometric algebra by Larry Grove, books.google.com/… –  Will Jagy Jul 19 '12 at 5:21
2  
anyway, copied from math.stackexchange.com/questions/172026/… –  Will Jagy Jul 19 '12 at 5:27
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1 Answer

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Let $q$ denote the order of the finite field, which assume has odd characteristic. If $Q$ is a non-degenerate quadratic form in $n$ variables, for $n$ odd (this is your case 3), we have that the number of totally isotropic subspaces of dimension $k$ (for $k \leq \frac{n-1}{2}$, otherwise the number is $0$) is $$ \frac{\prod_{i=0}^{k-1} (q^{2n-2i-1 } -1)}{\prod_{i=1}^{k} (q^i -1)}.$$

If the number of variables, $n$, is even then the answer depends on the signature of the quadratic form. Let $\epsilon = \{+1,-1\}$ denote this signature. The number of totally isotropic subspaces of dimension $k$ is then

$$ \frac{(q^{n-k} - \epsilon q^{n/2-k} + \epsilon q^{n/2} -1) \prod_{i=1}^{k-1} (q^{n-2i} -1 ) }{\prod_{i=1}^{k} (q^i -1) }.$$

Again, we assume that $k\leq n/2-1$ if $\epsilon =-1$ and $k \leq n/2$ if $\epsilon = +1$ (otherwise the number is $0$). Note that $\epsilon=+1$ corresponds to your case $1$ and $\epsilon=-1$ corresponds to your case $2$.

A short proof of these identities can be found in:

V. Pless, The number of isotropic subspaces in a finite geometry. Atti Accad. Naz. Lincei Rend. CI. Sci. Fic. Mat. Natur. (8) 39 1965 418-421.

although I believe these results were known previously.

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