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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.

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    $\begingroup$ What is an "isotropic spread"? $\endgroup$
    – Jason Starr
    Commented Jul 3, 2015 at 12:24
  • $\begingroup$ a set of isotropic subspaces with trivial intersection, that covers $V$ $\endgroup$
    – user33209
    Commented Jul 3, 2015 at 12:25
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    $\begingroup$ 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? $\endgroup$
    – Jason Starr
    Commented Jul 3, 2015 at 12:26
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    $\begingroup$ by the way in the literature, the subspaces $U$ with $U\subset U^\perp$ are called totally isotropic. $\endgroup$
    – Name
    Commented Jul 3, 2015 at 13:25
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    $\begingroup$ 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
    Commented Jul 27, 2015 at 10:00

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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.

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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 $W$.

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    $\begingroup$ 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.
    Commented Jul 15, 2015 at 16:09
  • $\begingroup$ You are right, I misread the definition of spread. $\endgroup$
    – user76083
    Commented Jul 16, 2015 at 9:50

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