Hamiltonian vector field proof? Given a manifold $M$ and a symplectic form $\omega \in \Omega^2(M)$, let $j:Y\to M$ be a symplectic submanifold (i.e, $j^*\omega \in \Omega^2(Y)$). Now, let $H \in C^{\infty}(M)$ be a smooth function such that $$\xi_{H}\lvert_x \in T_xY$$ for all $x \in Y$. 
How does one prove (or start to prove) that $\xi_H \lvert_Y$ is the Hamiltonian vector field of the function $H\lvert_Y \in C_{\infty}(Y)$ with respect to $j^*\omega$? Thanks for any help folks.
 A: I presume $j$ is an embedding. 
Let $\xi_Y \in TY$ be the unique vector field with $j_*\xi=\xi_H$.  You need to show (up to sign conventions) that for all $\eta$ tangent to $Y$ one has $[j^*\omega](\xi_Y, \eta)=[-d(j^*H)](\eta)=[-j^*(dH)](\eta)$ for all $\eta \in TY$; this is  the same as $\omega(\xi_H, j_* \eta)=-d(H)(j_*\eta)$. But since $\xi_H$ is Hamiltonian vector field of $H$, this is true for all $\nu$ (tangent to $M$ at points of $j(Y)$), hence in particular for $\nu=j_*\eta$.
In the more traditional order of mathematical argument:
1) Since $\xi_H$ is Hamiltonian vector field of $H$, we have (up to sign conventions) for all $\nu$ tangent to $M$ at points of $j(Y)$
$$\omega(\xi_H, \nu)=-d(H)(\nu)$$
2) In particular the above holds for $\nu=j_*\eta$, where  $\eta$ is any vector field tangent to $Y$. Plugging in we get
$$\omega(\xi_H, j_*\eta)=-d(H)(j_*\eta)$$
3) If we now define the vector field $\xi_Y$ tangent to $Y$ via $\xi_H=j_*\xi_Y$ (which we can, since  $\xi_H$ is tangent to $j(Y)$ and any vector field tangent to $j(Y)$ is a pushforward of unique tangent vector field of $Y$), then the above becomes
$$\omega(j_*\xi_Y, j_*\eta)=-d(H)(j_*\eta)$$
or 
$$(j^{*}\omega)(\xi_Y, \eta)=-d(j^* H)(\eta)$$
Since $\eta$ was arbitrary, we can write this as
$$(j^{*}\omega)(\xi_Y, \cdot)=-d(j^* H)(\cdot)$$
which precisely says that $\xi_Y$ is the Hamiltonian vector field of $j^* H$ with respect to the symplectic form  $j^{*}\omega$, as we wanted.
