How do I complete this prove of $\Phi$ is linear and symplectic iff its graph is a Lagrangian subspace of $(V \times V,(−\omega) \oplus \omega)$?

Question

Let $(V, \omega)$ be a symplectic vector space. Prove that a map $\Phi : V \mapsto V $is linear and symplectic if and only if its graph $\{(v, \Phi v)| v \in V\}$ is a Lagrangian subspace of $V \times V$ , equipped with the symplectic form $(−\omega) \oplus \omega$.

$(\Leftarrow)$

If G is Lagrangian then, $(−\omega) \oplus \omega=0 $ on G $\iff 0 = ((−\omega) \oplus \omega)((v,\Phi(v)),(v',\Phi(v')))=-\omega(v,v')+w(\Phi(v),\Phi(v')), \forall v,v' \in V $

$\iff \omega(v,v')=w(\Phi(v),\Phi(v'))$ which is the definition of symplectic map

For the linearity: $\alpha,\beta \in \mathbb{R}; v,v'\in V \implies $

$\omega(\Phi(\alpha v+\beta v'),\Phi w )=\omega(\alpha v+\beta v',w)=\alpha\omega(v,w) +\beta \omega(v',w)=\omega(\alpha \Phi v +\beta \Phi v',\Phi w)$

$\omega(\Phi(\alpha v+\beta v') - (\alpha \Phi v +\beta \Phi v'),\Phi w )=0, \forall v,w\in V$

I would like to conclude that the left argument is 0 due to the non degeneracy of $\omega$, but I am unsure if I can do that, since I have $\Phi w$ as second argument instead of just $w$. I mean, if I had $\omega(\Phi(\alpha v+\beta v') - (\alpha \Phi v +\beta \Phi v'), w )=0, \forall v,w\in V$, I'd be done with linearity How do I conclude from here ? and how do I prove the other implication $(\Rightarrow)$ ?

Possible solution: I use that a subspace G is Lagragian if $G=G^\omega$

Let $\Phi$ be linear and symplectic Let $(v,\Phi v),(v',\Phi v')\in G \implies (v,\Phi v)+(v',\Phi v')=(v+v',\Phi(v+v'))\in G$, so $G$ is a linear subspace of $V \times V$

$G^\omega =\{(v,v')\in V\times V|((-\omega)\oplus\omega)((v,v'),(w,w'))=0 \forall (w,w') \in G\} =\{(v,v')\in V\times V| \omega(v',w')-\omega(v,w)=0 \forall (w,w') \in G\}=\{(v,v')\in V\times V| \omega(v',\Phi(w))-\omega(v,w)=0 \forall w \in V\}=\{(v,v')\in V\times V| \omega(v',\Phi(w))=\omega(\Phi(v),\Phi(w))=0 \forall w \in V\}$

In particular the equality inside the set also holds $\forall w=v$, so

$=\{(v,v')\in V\times V| \omega(v',\Phi(v))=\omega(\Phi(v),\Phi(v))=0 \forall v \in V\} =\{(v,\Phi(v))\in V\times V| v \in V\} = G$

However I am a bit unsure specially about the last part, when I set $w=v$

Is this correct?

Answer 1

Let $v,v',w \in V$ and $\alpha \in \mathbb{R}$. On the one hand, \begin{align} \alpha\omega(v,w)+\omega(v',w) & = \alpha \omega(\Phi(v),\Phi(w)) + \omega(\Phi(v'),\Phi(w)) \\ & = \omega(\alpha \Phi(v) +\Phi(v'),\Phi(w)) \end{align} using that $\Phi$ is symplectic and the bilinearity of $\omega$. On the other hand, \begin{equation} \alpha\omega(v,w) + \omega(v',w)=\omega(\alpha v+ v',w) = \omega(\Phi(\alpha v+v'),\Phi(w)). \end{equation} It follows that \begin{equation} \omega(\Phi(\alpha v+v'),\Phi(w)) = \omega(\alpha \Phi(v) +\Phi(v'),\Phi(w)) \quad \forall v,v',w \in V, \end{equation} and hence \begin{equation} \Phi(\alpha v+v') = \alpha \Phi(v) +\Phi(v'). \end{equation}

For the converse, note that the graph of $\Phi$ is an isotropic subspace of $V\times V$ since \begin{align} (-\omega \oplus \omega)((v,\Phi(v)),(w,\Phi(w))) & = -\omega(v,w)+\omega(\Phi(v),\Phi(w)) \\ & = -\omega(v,w)+\omega(v,w) \\ & = 0 \end{align} for all $(v,\Phi(v)),(w,\Phi(w))$ in the graph of $\Phi$. Then consider a map $V\to V\times V$ given by $v\mapsto (v,\Phi(v))$. It is clearly injective and by rank-nullity the dimension of $V$ equals the dimension of the image of this map, which is precisely the graph of $\Phi$. Obviously $\dim V = \frac{1}{2} \dim (V\times V)$ so the result follows.