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Let $k$ be a field of characteristic $0$, let $V$ be a finite-dimensional vector space over $V$, and let $\omega(-,-)$ be a symplectic bilinear form on $V$. In other words, $\omega(-,-)$ is an alternating form that is nondegenerate in the sense that it identifies $V$ with its dual $V^{\ast}$.

Now let $I,J \subset V$ be isotropic subspaces, that is, subspaces on which $\omega$ vanishes identically. The form $\omega(-,-)$ then induces symplectic forms on $I^{\perp}/I$ and $J^{\perp}/J$. Define a map $f\colon I^{\perp} \cap J^{\perp} \rightarrow I^{\perp}/I \oplus J^{\perp}/J$ via the formula $f(x) = (x,-x)$. The image of $f$ is then an isotropic subspace of $I^{\perp}/I \oplus J^{\perp}/J$.

A paper I am reading claims that the image of $f$ is a Lagrangian, that is, that $\text{Image}(f)^{\perp} = \text{Image}(f)$. I can't figure out how to prove this. Can anyone help?

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  • $\begingroup$ One probably-easy way to see it is to count dimensions. $\endgroup$
    – LSpice
    Commented Jul 12 at 19:36
  • $\begingroup$ The dimension here that I don't know how to count is that of $I^{\perp} \cap J^{\perp}$. It's not even clear to me that it only depends on the dimensions of $I$ and $J$ and $I \cap J$. $\endgroup$
    – Linda
    Commented Jul 12 at 19:46

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$\DeclareMathOperator\im{im}$If you know a subspace of a finite-dimensional vector space is isotropic, to prove it is Lagrangian, it suffices to show it has half the total dimension. The rest is an easy linear algebra exercise: $$\dim \im(f)=\dim(I^\perp\cap J^\perp)-\dim \ker(f)=\dim((I\oplus J)^\perp)-\dim(I\cap J)=\dim V-\dim(I\oplus J)-\dim(I\cap J)=\dim V-\dim I-\dim J=(\dim I^\perp/I+\dim J^\perp/J)/2.$$

By the way, you need to use the symplectic form induced by $-\omega$ on the $J^\perp/J$ factor for $f$ to have isotropic image. Also, I think this question is more suitable for MathStackExchange.

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    $\begingroup$ Note that I used the antidiagonal embedding taking $x$ to $(x,-x)$. That's why I don't need $-\omega$ on that factor. $\endgroup$
    – Linda
    Commented Jul 12 at 19:49
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    $\begingroup$ The negation signs cancel out: $\omega(-x,-y)=\omega(x,y)$. $\endgroup$
    – Qiuyu Ren
    Commented Jul 12 at 19:58
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    $\begingroup$ Ah, that's a good point. I was being overly clever in attempting to avoid introducing more notation. $\endgroup$
    – Linda
    Commented Jul 12 at 20:04
  • $\begingroup$ If you find a question more appropriate for MSE, then it is better to vote to migrate (or I think you can flag if you don't yet have privileges to cast such a vote) and answer there, rather than answering in what you feel to be the wrong place. $\endgroup$
    – LSpice
    Commented Jul 12 at 21:45
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    $\begingroup$ @LSpice I'm new to this site. Thanks for letting me know! $\endgroup$
    – Qiuyu Ren
    Commented Jul 12 at 22:06

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