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A symplectic automorphism of a Hilbert space has the form $T=U(\cosh S+J\sinh S)$ for a unitary $U$, an antilinear involution $J$ and a positive operator $S$. In fact a version of this goes through in the unbounded case (see Squeezing Bogoliubov transformations on the infinite mode CCR-algebra, Honegger and Rieckers, Com. Math. Phys. 37 no. 9 (1996) )

Question: Does a similar result hold for a (not necessarily surjective) symplectic map $T:H\to K$ between Hilbert spaces?

(The reference I have relies pretty heavily on $T$ having dense range)

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(every separable h.sp. is necessarily isomorphic to $\ell^2(\mathbb{C})$, right? So, perhaps in this case the forumla you're looking for is just the one you posted, "conjugated" by the relevant isomorphisms..) – Qfwfq Aug 10 '11 at 21:21
(...I mean, if both $H$ and $\mathrm{range}(T)$ are both $\ell^2(\mathbb{C})$) – Qfwfq Aug 10 '11 at 21:23
The range of the symplectic map is only a real-linear subspace, so the corresponding isomorphisms may not be complex-linear, this is a problem for me. Also I'm interested in the finite dimensional case. – Ollie Margetts Aug 11 '11 at 17:19
(as well as the infinite dimensional..) – Ollie Margetts Aug 11 '11 at 17:20
In the case that $T$ is continuous, there holds for all $x,y \in H$ $$\langle Jx, y\rangle = \langle JTx,Ty\rangle = \langle T^\dagger J T x, y \rangle \Leftrightarrow T^\dagger J T = J.$$ (or do you have another definition of symplectic?) In that case you get $${T^\dagger}^{-1} = - J T J$$ and you see, that $T$ must be an automorphism. – kostja Aug 29 '11 at 12:53

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