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I've been trying to understand the proof of the following lemma:

Lemma. Given any finite collection of Lagrangian subspaces $M_1, \dots, M_r$, one can find a Lagrangian subspace $L$ with $L \cap M_j=\{0\}$ for all $j$.

Proof. Let $L$ be a isotropic subspace with $L\cap M_j=\{0\}$ and not properly contained in a larger isotropic subspace with this property. We claim that $L$ is Lagrangian. If not, $L^\omega$ is a coisotropic properly containing $L$. Let $\pi:L^\omega \rightarrow L^\omega / L$ be the quotient map. Chose any $1$-dimensional subspace $F \subset L^\omega/L$, such that both $F$ is transversal to all $\pi(M_j \cap L^\omega)$. This is possible, since $\pi(M_j \cap L^\omega)$ is isotropic and therefore has positive codimension. Then $L^\prime = \pi^{-1}(F)$ is an isotropic subspace with $L \subset L^\prime$ and $L^\prime \cap M_j = \{0\}$. This contradiction shows $L=L^\omega$.

Which is in the Mainrenken notes, more precisely on pages $7-8$.

And I ran into some problems:

  • First with the existence of $L$. But I believe that since $\{0 \}$ is isotropic, in the worst case it will be equal to $L$.

  • Secondly, so far I have not understood why it is possible to choose a $1$-dimensional space $F$ transverse in $L^\omega/L$. In particular, why does $\pi(M_j \cap L ^\omega)$ having a positive codimension imply the existence of $F$? Since there are $j$'s $\pi(M_j\cap L^\omega)$, what guarantees that $\pi$ will not take $L \cap W_j$ in space $F$?

Looking for explanations, I found this post. But it didn't help me much. So if anyone can help, I'll be grateful.

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The problem is solved similarly in both cases. If you have a collection $M_j$ of isotropic subspaces, then their dimension is at most half the dimension of $V$, in particular, there exists $v \in V \setminus M_1 \cup ...\cup M_r$ since a union of such subspaces cannot cover the entire space. Now remember that any one dimensional subspace is isotropic, therefore taking $L$ as the space generated by $v$, we have that $L \cap M_j = \emptyset$ for each $j$.

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