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This is exercise 9, p.8 in Lectures on Symplectic Geometry by Ana Cannas da Silva.

Let $(V,\omega)$ be a symplectic vector space and $Y$ a lagrangian subspace of $V$, i.e. $Y = Y^\omega$. On $Y \oplus Y^*$ define $$\omega_0(u \oplus\alpha,v\oplus \beta) := \beta(u)-\alpha(v)$$ Then $(Y \oplus Y^*,\omega_0)$ is a symplectic vector space.

Prove that $(V,\omega)$ and $(Y \oplus Y^*,\omega_0)$ are symplectomorphic.

So we explicitely have to construct a symplectomorphism, say $\varphi: V \to Y \oplus Y^*$. I tried to deduce something from $$(\varphi^*\omega_0)(u,v) = \omega_0(\varphi(u),\varphi(v)) \overset{!}{=} \omega(u,v)$$ which leads to $$\varphi(v) := v \oplus-\frac{1}{2}\omega(v,\cdot)$$ which is obviously not well-defined since $v$ might not be in $Y$. I would appreciate a slight hint on how to define an appropriate $\varphi$.

TheGeekGreek
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    An easy way to prove they are symplectomorphic (without explicitly building a symplectomorphism) is to say that they have the same dimension. (Here I assume $V$ is finite dimensional, otherwise more data is needed on $V$ and what dual $Y^*$ we are talking about). – Noé AC Sep 08 '17 at 12:23

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Hint: since $\omega$ its symplectic, it induces an isomorphism $i:V\rightarrow V^*$ defined by $i(v)=\omega(v,.)$.

Define $f:Y\oplus Y^*\rightarrow V$ by $f(u+\alpha)=u+i^{-1}(\alpha)$. Here you identify $Y^*$ with a subspace of $V^*$ by completing a basis of $Y$ with the basis of a Lagrangian complement subspace.

Tsemo Aristide
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