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

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?

• Just to check, you are asking for the standard symplectic structure on $\mathbb{R}^4$? – David E Speyer Jul 17 '15 at 13:16

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.
• I don't know. There is no topological obstacle to having two linearly independent vector fields in $\mathbb{R}^4$ which are tangent to a $2$-torus. I don't know how to make them commute. But that might just be me not knowing a lot of symplectic tricks. – David E Speyer Jul 18 '15 at 14:05
• This $H$ commutes with $x_2$, so it is completely integrable. But the flow corresponding to $x_2$ makes the bounded circles into unbounded cylinders. – David E Speyer Jul 19 '15 at 2:35