Hamiltonian vector field

Remarks

The vector field exists and is unique because $\omega$ is nondegenerate, so $v \mapsto \iota_v \omega$ is an isomorphism from each tangent space onto the cotangent space.

Example: on $\R^{2n}$ with coordinates $(q_1, \ldots, q_n, p_1, \ldots, p_n)$ and $\omega = \sum_i dq_i \wedge dp_i$, set $$X_H = \sum_i \left( \frac{\partial H}{\partial p_i} \frac{\partial}{\partial q_i} - \frac{\partial H}{\partial q_i} \frac{\partial}{\partial p_i} \right).$$ Then $\iota_{X_H} \omega = \sum_i \left( dq_i(X_H)\, dp_i - dp_i(X_H)\, dq_i \right) = \sum_i \left( \frac{\partial H}{\partial p_i} dp_i + \frac{\partial H}{\partial q_i} dq_i \right) = dH$. So the integral curves of $X_H$ solve Hamilton's equations $q_i' = \partial H / \partial p_i$, $p_i' = -\partial H / \partial q_i$.

Non-example: on $\R^4$ the form $\omega = dq_1 \wedge dp_1$ is closed but degenerate. For $H = q_2$ there is no vector field $X$ with $\iota_X \omega = dq_2$, because $\iota_X \omega$ is always a combination of $dq_1$ and $dp_1$.

Some authors use the opposite sign, $\iota_{X_H} \omega = -dH$, together with $\omega = \sum_i dp_i \wedge dq_i$; this gives the same vector field on $\R^{2n}$. The function $H$ is constant along the flow of $X_H$, since $dH(X_H) = \omega(X_H, X_H) = 0$.