# Exists lagrangian subspace $L$ such that $L \cap W_j = \{0\}$.

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.

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 $$j$$.