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Consider $S^2\times S^2$ with its standard symplectic form. I have seen the following statement made that $\Psi(x_1,y_1,z_1,x_2,y_2,z_2)=(-x_1,y_1,-z_1,x_2,-y_2,-z_2)$ is an Hamiltonian symplectomorphism.

Well after thinking about this for a while I conviced myself that this is true by using Banyaga's theorem and the fact that the first homology group vanishes.

However I was not able to construct an Hamiltonian function $H_t$ such that it's Hamiltonian flow $\Psi_t$ would give me that $\Psi_1$=$\Psi$?

Therefore I was wondering if anyone has any suggestions in order to find this $H_t$? It could even be time-independent.

Thanks in advance.

Someone
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1 Answers1

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For a sphere with the standard symplectic form, the height function $H(x,y,z) = y$ generates a periodic Hamiltonian flow that rotates the sphere.

The symplectomorphism that you wrote is just a $180$ degrees rotation of the first sphere around its $y$ axis, and a $180$ degrees rotation of the second sphere around its $x$ axis.

Therefore, the Hamiltonian $H(x_1, y_1, z_1, x_2, y_2, z_2) = y_1 + x_2$ generates a periodic Hamitonian flow that rotates the spheres around the desired axes at the same speed. After normalizing $H$, you would get the desired symplectomorphism as the time-$1$ map of the Hamiltonian flow.

Rei Henigman
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  • Thanks for the answer! I have been trying to compute in cylindrical coordinates the expression for the Hamiltonian flow generated by $H(x,y,z)=y$ in the $2-$sphere, however I obtain a vector field of the likes $X_{H}=-\frac{r}{\sqrt{1-r^2}}\sin(\theta)\frac{\partial}{\partial \theta}+\sqrt{1-r^2}\cos(\theta)\frac{\partial}{\partial r}$, and I'm not being able to solve the ODE to find the flow. Do you know an expression for the Hamiltonian flow of the function? – Someone Dec 08 '22 at 07:36
  • It's straightforward when using the symplectic form $\omega = d\theta \wedge dh$ and $H=h$. Then $(\iota_{X_H}\omega)(v)= (d\theta \wedge dh)(X_H, v) = (X_H)\theta v_h - (X_H)_h v\theta$, but also $dH(v) = v_h$. Hence $X_H = \frac{\partial}{\partial \theta}$. The flow $\Psi_t$ is then rotation by degree $t$. – Rei Henigman Dec 08 '22 at 12:38
  • See Example 3.1.4 in Mcduff's book for this form of the symplectic form on the sphere, and see Example 3.1.7 for this periodic Hamiltonian flow. – Rei Henigman Dec 08 '22 at 12:40
  • Not sure that the notation of my first comment is clear, so just to clarify - $(X_H)_\theta$ and $(X_H)_h$ are the coefficients of $X_H$ when written in coordinates $(\theta, h)$. More explicitly:

    $X_H = (X_H)\theta \frac{\partial}{\partial \theta} + (X_H)_h \frac{\partial}{\partial h}$. Similarly $v = v\theta \frac{\partial}{\partial \theta} + v_h \frac{\partial}{\partial h}$.

     – Rei Henigman Dec 08 '22 at 12:44
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    Oh yeah those are way better coordinates to work with, thanks ! – Someone Dec 08 '22 at 14:51