Symplectic form and hamilton's vector field Can somebody explain me this equation:
$$ \omega(X_H, Y) = \mathrm{d}H(Y), $$
where $\omega$ is symplectic form and $Y$ is any vector field on manifold $\mathcal{M}$. 
So, I know that $\mathrm{d}H$ is one-form, but why is vector field $Y$ for an argument? And what is this any vector field? Can somebody explain how should I understand this equation? 
 A: $\omega$ is non degenerated, this is equivalent to saying that for every $x\in M$, the application $h:T_xM\rightarrow T^*_xM$ defined by $h(X)(Y)=\omega_x(X,Y)$ is an isomorphism, we deduce that for
 every $\alpha$ in the dual of $T_xM$, there exists an element $X_{\alpha}\in T_xM$ such that for every $Y\in T_xM$, $\omega_x(X_{\alpha},Y)=\alpha(Y)$, take $\alpha=dH_x$.
A: Suppose for the moment that $M$ is $\mathbb{R}^{n}$, $n$ even, with coordinates and coordinates bases
$$(z^i,\mathbf{e}_i,\mathbf{e}^i)\in(\Gamma(M,\mathbb{R}),\Gamma(M,\mathrm{T}M),\Gamma(M,\mathrm{\Omega})).$$
Write
$$\omega=\tfrac{1}{2!}\omega_{ij}\mathbf{e}^{i}\wedge\mathbf{e}^j\text{,}$$
for the symplectic form,
$$X=v^i\mathbf{e}_i$$
for the Hamiltonian vector field of $H$, and
$$Y=y^i\mathbf{e}_i$$
for the universally quantified vector field. Then the equation given is
$$\tfrac{1}{2!}(\omega_{ij}-\omega_{ji})v^i y^j=\frac{\partial H}{\partial z^j}y^j$$
and the universal quantification over $Y$ means that we can equate coefficients on each side to get $n$ equations
$$\begin{align}\tfrac{1}{2!}(\omega_{ij}-\omega_{ji})v^i&=\frac{\partial H}{\partial z^j}& &(i=0,1,\ldots n-1)\text{.}\end{align}$$
Then what the author is really trying to say is that we have an equality of $1$-forms
$$\begin{align}\tfrac{1}{2!}(\omega_{ij}-\omega_{ji})v^i\mathbf{e}^j&=\frac{\partial H}{\partial z^j}\mathbf{e}^j \end{align}$$
and that to find the Hamiltonian vector field for $H$, we solve the $n$ scalar equations implied by matching coefficients of $\mathbf{e}^i$ ($i=0,1,\ldots,n-1$) for the $n$ unknowns $v^i$, $i=0,1,\ldots,n-1$, and combine them into to get the vector field $X=v^i\mathbf{e}_i$. In the case most familiar to physicists, the coordinates are split into position $x^{\alpha}$ and momentum $p_{\alpha}$ ($\alpha=0,1,\ldots, d-1$), the symplectic form and Hamiltonian field are written as
$$\begin{align}
\omega&=\mathrm{d}x^{\alpha}\wedge\mathrm{d}p_{\alpha} \\
\frac{\mathrm{d}}{\mathrm{d}t}&=\dot{x}^{\alpha}\frac{\partial}{\partial x^{\alpha}}+\dot{p}_{\alpha}\frac{\partial}{\partial p_{\alpha}}\text{,}
\end{align}$$
and the equality defining the Hamiltonian field is
$$\dot{x}^{\alpha}\mathrm{d}p_{\alpha}-\dot{p}_{\alpha}\mathrm{d}x^{\alpha}=\frac{\partial H}{\partial x^{\alpha}}\mathrm{d}x^{\alpha} +\frac{\partial H}{\partial p_{\alpha}}\mathrm{d}p_{\alpha}\text{,}$$
i.e.,
$$\begin{align}
\dot{x}^{\alpha}&=\frac{\partial H}{\partial p_{\alpha}} \\
\dot{p}_{\alpha}&=-\frac{\partial H}{\partial x^{\alpha}}\text{.}
\end{align}$$
Finally, it should be noted that this equation is usually notated in coordinate-free form using the interior product $$\iota_X\omega =\mathrm{d}H\text{.}$$
