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Let $(V, \omega)$ be a finite-dimensional real symplectic vector space, i.e. $\omega : V \times V \to \mathbb{R}$ is a non-degenerate skew-symmetric bilinear map. A linear subspace $L \subset V$ is called Lagrangian if $L = L^\perp$, where $L^\perp = \{v \in V : \omega(v, L) = \{0\}\}$.

Let $U \subset V$ be a linear subspace such that $$\dim U = \frac{1}{2} \dim V.$$ Question. Why is there always a Lagrangian subspace $L \subset V$ such that $V = U \oplus L$?

This is easy if $U$ is also Lagrangian, and there are plenty of sources explaining this (we can set $L = I(U)$ where $I$ is a compatible complex structure). But I didn't find any reference explaining this more general fact, although I have seen it used in some research papers. It is used, for instance, for the existence of local Lagrangian bisections in symplectic groupoids. Any hints on the proof would be appreciated.

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    $\begingroup$ This follows from the general fact that, given an $m$-dimensional subspace $U\subset V$, there is a symplectic basis $e_i,f^i$ for $1\le i\le n =\frac12\dim V$ and an integer $k\le \frac12 m$ such that $U$ has basis $\{\,e_i\,|\,1\le i\le m-k\,\}\cup\{\,f^j\,|\,1\le j\le k\,\}$. When $m=n$, let a Lagrangian $L\subset V$ have basis $g^j = f^j + s^{ij}e_i$ where $s^{ij}=s^{ji}$. When the upper right $k$-by-$(n{-}k)$ submatrix of $(s^{ji})$ has rank $k$ (true generically), $L$ is transverse to $U$. (Symplectic basis means $\omega(e_i,e_j)=\omega(f^i,f^j)=0$ and $\omega(e_i,f^j)=\delta_i^j$.) $\endgroup$ – Robert Bryant May 9 at 14:15
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    $\begingroup$ Sorry, that should be "upper right $k$-by-$k$ submatrix", not "upper right $k$-by-$(n{-}k)$ submatrix", in the above comment. $\endgroup$ – Robert Bryant May 9 at 19:00
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If $U$ is non-degenerate, then choose a symplectic isomorphism $f : {-U} \to U^\perp$, where $-U$ is $U$ equipped with the opposite $-\langle\cdot, \cdot\rangle$ of the original symplectic pairing, and put $L = \{f(u) - u \mathrel: u \in U\}$. (I think that this is closely related to your idea of choosing a compatible complex structure, but I'm trying to avoid assuming anything about the ground field.)

If $U$ is not non-degenerate (!), then choose a non-$0$ vector $x \in U \cap U^\perp$, and then a vector $y \in V \setminus x^\perp$. By induction, there is a totally isotropic complement $\tilde M/x^{\perp\perp}$ to $U/x^{\perp\perp}$ in $x^\perp/x^{\perp\perp}$. Put $L = (\tilde M \cap y^\perp) + y^{\perp\perp}$.

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    $\begingroup$ I am more used to "isotropic" rather than "totally degenerate". $\endgroup$ – F Zaldivar May 8 at 23:52
  • $\begingroup$ @FZaldivar, that works for me! $\endgroup$ – LSpice May 9 at 1:40
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    $\begingroup$ By the way, nice answer! $\endgroup$ – F Zaldivar May 9 at 1:51
  • $\begingroup$ Thanks, @FZaldivar! I didn't know this result, and thought it sounded too strong but saw from ‍@‍abx's answer that it was true, so thought I'd see if I could come up with an elementary proof. $\endgroup$ – LSpice May 9 at 10:31
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    $\begingroup$ I'm not sure if I understand why $\dim (U \cap x^\perp / x^{\perp\perp}) = \frac{1}{2} \dim(x^\perp / x^{\perp\perp})$. In fact, I think that this is equivalent to $\dim(U \cap x^\perp) = \frac{1}{2} \dim V$ which implies that $U \cap x^\perp = U$ and hence $x \in U^\perp$, contrary to assumption. $\endgroup$ – Spenser May 9 at 10:45
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This might be an overkill, but you can argue as follows. The variety $\mathbb{G}$ of Lagrangian subspaces has dimension $N:=\frac{1}{2}n(n+1) $, where $\dim V=2n$. The Lagrangian subspaces which contain a line $\ell\subset V$ correspond bijectively to the Lagrangian subspaces of $\ell^{\perp}/\ell$, of dimension $2n-2$, hence they form a subvariety $\mathbb{G}_{\ell}$ of $\mathbb{G}$ of codimension $n$. When $\ell$ varies in $U$, the various $\mathbb{G}_{\ell}$ form a Grassmann bundle $\mathscr{G}\rightarrow \Bbb{P}(U)$, of dimension $N-1$. Now the locus of Lagrangian subspaces $L$ such that $L\cap U\neq \varnothing$ is the union of the $\mathbb{G}_{\ell}$, hence the image of $\mathscr{G}$, so it is a strict subvariety (in fact a hypersurface) in $\mathbb{G}$.

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    $\begingroup$ If OP disprefers the algebraic geometry language this can be phrased as a transversality argument. $\endgroup$ – mme May 8 at 23:31

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