A Hamiltonian vector field on $\mathbb{R}^{4}$ which has closed orbit but does not have critical point

Question

Is there a polynomial function $H:\mathbb{R}^{4} \to \mathbb{R}$ without critical points but the corresponding hamiltonian vector field possess at least one closed orbit?

Answer 1

Assuming you are talking about the standard symplectic structure $dx_1 \wedge dy_1+dx_2 \wedge dy_2$, the Hamiltonian $H=x_1^2+y_1^2 + (x_1^2+y_1^2-1)^2 x_2$ works.

The Hamiltonian flow has closed orbits: Let $(x_1, y_1, x_2, y_2) = (\cos \theta, \sin \theta, x_2 , y_2)$. Then $\nabla H = 2 (x_1, y_1, 0,0)=2(\cos \theta, \sin \theta,0,0)$ so the Hamiltonian flow is in direction $(-2 \sin \theta, 2 \cos \theta, 0,0)$. We see that every circle of the form $x_1^2+y_1^2=1$, $x_2$ and $y_2$ constant, is a closed orbit.

There are no critical points: If $\partial H/\partial x_2=0$, then $x_1^2+y_1^2=1$. But then $\partial H/\partial x_1 = 2 x_1$ and $\partial H/\partial y_1 = 2 y_1$ can't both be zero.