2
$\begingroup$

Let $V$ be a symplectic vectorspace of dimension $2n$, and $r\mid n$. Is this statement true?"There is an isotropic spread of $r$ dimensional subspaces in $V$". By an isotropis subspace I mean a subspace $U$, that $U\subset U^{\perp}$.

any comment is appreciated.

$\endgroup$
8
  • 2
    $\begingroup$ What is an "isotropic spread"? $\endgroup$ Jul 3, 2015 at 12:24
  • $\begingroup$ a set of isotropic subspaces with trivial intersection, that covers $V$ $\endgroup$
    – user33209
    Jul 3, 2015 at 12:25
  • 1
    $\begingroup$ When you say "trivial intersection", do you mean that every pairwise intersection is trivial, or do you mean the common intersection of all specified subspaces is trivial? $\endgroup$ Jul 3, 2015 at 12:26
  • 2
    $\begingroup$ by the way in the literature, the subspaces $U$ with $U\subset U^\perp$ are called totally isotropic. $\endgroup$
    – Name
    Jul 3, 2015 at 13:25
  • 2
    $\begingroup$ You only need to prove the statement for $r=n$, as the rest will follow directly... And the statement is true when $2n=4$ and the field is ${\mathbb F}_3$ - see p.33 of this: math.lsu.edu/~hoffman/papers/spreads4.pdf (Whether that provides any evidence for the statement in general, I couldn't say. The group ${\rm Sp}_4(3)$ is a bit special.) $\endgroup$
    – Nick Gill
    Jul 27, 2015 at 10:00

2 Answers 2

1
$\begingroup$

Can you construct it by restriction of scalars? Namely, as Nick Gill says, it is enough to consider the case $r=n$ (i.e., Lagrangian subspaces). Secondly, let us fix a non-zero functional $\phi:\mathbb F_{q^n}\to\mathbb F_q$ and a symplectic space $(V,\omega)$ of dimension $2$ over $\mathbb F_{q^n}$. Then $(V,\phi\circ\omega)$ is a symplectic space of dimension $2n$ over $\mathbb F_q$. Any $\mathbb F_{q^n}$-line in $V$ is going to be Lagrangian (over $\mathbb F_{q^n}$, and therefore also over $\mathbb F_q$), and such lines form a spread.

P.S. I assume here that you are looking at $V$ over the finite field $\mathbb F_q$: this is mentioned in the title, but not in the body of the question.

$\endgroup$
0
$\begingroup$

What you can show is that, if $W$ is a maximal isotropic subspace of $V$ (and so the dimension of $W$ is $n$), then there exists another maximal isotropic subspace $U$ of $V$ such that $U\oplus W=V$. Now, the number $r$ dividing $n$, you can construct your spread taking $r$-dimensional subspaces of $U$ and $W$.

$\endgroup$
2
  • 2
    $\begingroup$ Aren't all elements of $V$ supposed to be contained in some element of the spread? Your suggestion would only cover $U\cup W$. $\endgroup$
    – j.p.
    Jul 15, 2015 at 16:09
  • $\begingroup$ You are right, I misread the definition of spread. $\endgroup$
    – user76083
    Jul 16, 2015 at 9:50

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.