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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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  • $\begingroup$ There is where such asssertion? $\endgroup$ Commented Apr 23, 2013 at 7:09

2 Answers 2

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It is very elementary; the graph representation refers to the cartesian product $P\times iP$ coming from the real vector space direct sum decomposition $\mathbb{C}^n = P\oplus iP\sim P\times iP$.

Here the direct sum decomposition is possible because $\dim _ \mathbb{R}(P)=n $ and $P\cap iP=0$, from the definition of Lagrangian subspace (thus any $z$ writes uniquely as $z=x+iy$ with $x$ and $y$ in $P$).

The second projection $\pi_2: P\times iP \to iP$, restricted to a subspace $Q$, has kernel $Q\cap iP$, so it is injective exactly for $Q\in U_P$, in which case $Q=\operatorname{graph}\\ \pi_1({\pi_2}{|Q})^{-1}$.

For other details check e.g. the introductory pages of Hofer-Zehnder's book, Symplectic Invariants and Halmiltonian Dynamics.

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Take the projection $\mathbb C^n\to P$ with kernel $iP$ and restrict it to $Q$. Its inverse, composed with the projection $\mathbb C^n\to iP$ with kernel $P$ is the linear map $\Phi$ you look for. See 31.7 of here for more information.

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