A Langrangian in a symplectic vector space determines the symplectic form completely. This is a linear analogue of Weinstein's neighborhood theorem.

Recall that, given a vector space $L$ with dual $L^*$, the vector space $L\oplus L^*$ can be endowed with a symplectic form via the formula

$$ \omega_L((v,\xi),(v',\xi'))=\xi(v')-\xi'(v). $$ Note that $L$ and $L^*$ are Lagrangian subspaces of $L\oplus L^*$. Now let $L$ be a Lagrangian inside a symplectic vector space $(V,\omega)$. Then $L$ admits a vector space complement $L'$ that is also Lagrangian. This defines a bijection $\varphi:L'\rightarrow L^*$ via the formula $\varphi(v)(w)=\omega(v,w)$. Then the map $\Phi:V=L\oplus L'\rightarrow L\oplus L^*$ with $\Phi(v,v')=v+\varphi(v')$ is a symplectomorphism between $(V,\omega)$ and $(L\oplus L^*,\omega_L)$. Hence the Lagrangian $L$ completely determines the symplectic structure on $V$.

If $L$ is a Langrangian in a symplectic vector space $V$, then $V$ is isomorphic to $L\oplus L^*$ with the standard symplectic form. This is a linear analogue of Weinstein's neighborhood theorem.

Recall that, given a vector space $L$ with dual $L^*$, the vector space $L\oplus L^*$ can be endowed with a symplectic form via the formula

$$ \omega_L((v,\xi),(v',\xi'))=\xi(v')-\xi'(v). $$ Note that $L$ and $L^*$ are Lagrangian subspaces of $L\oplus L^*$. Now let $L$ be a Lagrangian inside a symplectic vector space $(V,\omega)$. Then $L$ admits a vector space complement $L'$ that is also Lagrangian. This defines a bijection $\varphi:L'\rightarrow L^*$ via the formula $\varphi(v)(w)=\omega(v,w)$. Then the map $\Phi:V=L\oplus L'\rightarrow L\oplus L^*$ with $\Phi(v,v')=v+\varphi(v')$ is a symplectomorphism between $(V,\omega)$ and $(L\oplus L^*,\omega_L)$.

A Langrangian in a symplectic vector space determines the symplectic form completely. This is a linear analogue of Weinstein's neighborhood theorem.

Recall that, given a vector space $L$ with dual $L^*$, the vector space $L\oplus L^*$ can be endowed with a symplectic form via the formula

$$ \omega_L((v,\xi),(v',\xi'))=\xi(v')-\xi'(v). $$ Note that $L$ and $L^*$ are Lagrangian subspaces of $L\oplus L^*$. Now let $L$ be a Lagrangian inside a symplectic vector space $(V,\omega)$. Then $L$ admits a vector space complement $L'$ that is also Lagrangian. This defines a map $\varphi:L'\rightarrow L^*$ via the formula $\varphi(v)(w)=\omega(v,w)$. Then the map $\Phi:V=L\oplus L'\rightarrow L\oplus L^*$ with $\Phi(v,v')=v+\varphi(v')$ is a symplectomorphism between $(V,\omega)$ and $(L\oplus L^*,\omega_L)$. Hence the Lagrangian $L$ completely determines the symplectic structure on $V$.

A Langrangian in a symplectic vector space determines the symplectic form completely. This is a linear analogue of Weinstein's neighborhood theorem.

Recall that, given a vector space $L$ with dual $L^*$, the vector space $L\oplus L^*$ can be endowed with a symplectic form via the formula

$$ \omega_L((v,\xi),(v',\xi'))=\xi(v')-\xi'(v). $$ Note that $L$ and $L^*$ are Lagrangian subspaces of $L\oplus L^*$. Now let $L$ be a Lagrangian inside a symplectic vector space $(V,\omega)$. Then $L$ admits a vector space complement $L'$ that is also Lagrangian. This defines a bijection $\varphi:L'\rightarrow L^*$ via the formula $\varphi(v)(w)=\omega(v,w)$. Then the map $\Phi:V=L\oplus L'\rightarrow L\oplus L^*$ with $\Phi(v,v')=v+\varphi(v')$ is a symplectomorphism between $(V,\omega)$ and $(L\oplus L^*,\omega_L)$. Hence the Lagrangian $L$ completely determines the symplectic structure on $V$.