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?
On
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 $$\iota_X\omega =\mathrm{d}H\text{.}$$
$\omega$ 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 $\alpha=dH_x$.