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Let $\Lambda_{n}$ be the set of all Lagrangian subspaces of $C^{n}$, and $P\in \Lambda_{n}$. Put $U_{P} = \{Q\in \Lambda_{n} : Q\cap (iP)=0\}$. There is an assertion that the set $U_{P}$ is homeomorphic to the real vector space of all symmetric endomorphisms of $P$. And then in the proof of it there is a fact that the subspaces $Q$ that intersect $iP$ only at $0$ are the graphs of the linear maps $\phi : P\to iP$. This is what I don't understand, any explanation or reference where I can find it would be helpful.

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up vote 3 down vote accepted

Remember these are Lagrangians and thus half-dimensional. It's easiest to see what is going on if you take $P = \mathbb{R}^n$. This simplifies notation and also is somewhat easier to understand, imo.

We are given a Lagrangian subspace $Q$, transverse to $i \mathbb{R}^n$. Then, consider the linear map $Q \to \mathbb{R}^n$ by taking a $z \in Q$ and mapping to the real part (i.e. projecting to $P$ along $iP$). This map is injective (by the transversality assumption) and is thus an isomorphism. The inverse of this map takes a point $x \in \mathbb{R}^n$ and constructs a $y(x) \in \mathbb{R}^n$ so $x + i y(x) \in Q$. The map $y \colon \mathbb{R}^n \to i \mathbb{R}^n$ is what you are looking for.

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