1
$\begingroup$

Let $F$ be a field, $n$ be a positive integer. Denote by $h_{F}(n)$ the maximal dimension of a subspace $X\subset F^n$ such that $(x,y)=0$ for any two (not necessary distinct) vectors $x,y\in F^n$, where $(x,y)=x_1y_1+\dots+x_ny_n$ for $x=(x_1,\dots,x_n)$, $y=(y_1,\dots,y_n)$. For example, $h_{\mathbb{R}}(n)=0$ for any $n$.

It is clear that $h_F(n)$ is non-decreasing by $n$, that $h_F(n)\leq n/2$ (such $X$ is contained in $X^{\perp}$, hence $\dim(X)\leq \dim(X^{\perp})=n-\dim(X)$) and that always $h_{F}(n+k)\geq h_F(n)+h_F(k)$. It allows to get $h_{F}(n)=[n/2]$ for $F=\mathbb{C}$ or $F=\mathbb{F}_p$ with prime $p=4k+1$ or $p=2$.

Now take $p=4k+3$ and $F=\mathbb{F}_p$.

It is easy to get $h_{F}(2)=0$, $h_{F}(3)=1$, $h_{F}(4)=2$ (take span of $(a,b,c,0)$ and $(0,-c,b,a)$ with $a^2+b^2+c^2=0$). Hence $h_{F}(4n)=2n$, $h_{F}(4n+1)=2n$, $h_{F}(4n+3)=2n+1$. But what about $h_F(4n+2)$? If it equals $2n+1$ for some $n$, then also for all greater $n$. But does there always exist such $n$ and if it does exist, how to find it as a function of $p$? I managed only to observe by hands that $h_{F}(6)=2$ for $p=3$.

$\endgroup$

1 Answer 1

2
$\begingroup$

This is the question of finding maximal isotropic subspaces of an inner-product space. The results for finite fields of odd characteristic are well-known and can be found in Serre's Course in Arithmetic.

Let's consider the quadratic form $Q=x_1^2+\cdots+x_n^2$. When $p\equiv 3$ (mod $4$) the dimension of the maximum isotropic subspace is $2k$ for $n=4k$, $4k+1$, $4k+2$ and $2k+1$ for $n=4k+3$.

To see that it isn't $2k+1$ for $n=4k+2$ we argue as follows. If a nonsingular quadratic form has a dimension $r$ isotropic subspace then $V$ it has a subspace $W$ or dimension $2r$ on which the form restricts to $y_1 y_2+y_3y_4+\cdots+\cdots y_{2r-1}y_{2r}$. If $r=n/2$ the discriminant of this form is $(-1)^r$. But this is not the same as the discriminant of $Q$ (namely $1$) modulo squares if $r$ is odd (since $p\equiv 3$ (mod $4$).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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