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Are all points in the image of a Hamiltonian function regular value, Why? If not, is 0 a regular value?

Grey
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1 Answers1

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No, not in general although there are some examples where it can happen.

If the symplectic manifold is compact then it has a maximum and attains it at some point, just by the fact that it is a continuous function on a compact space.

An example where it does happen $\mathbb{R}^2 \setminus \{0 \}$, $\omega = dx \wedge dy$ and $H(x,y) = x^2+y^2$.

The question is $0$ a regular value doesn't really make sense since by changing coordinates it can be any point of the symplectic manifold. However if you assume it is a fixed point of a Hamiltonian circle action (as I think you are suggesting) then no it is a critical point, this follows from the Hamiltonian equation.

Nick L
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  • Thank you so much, It seems I made a stupid mistake. – Grey Apr 03 '22 at 12:33
  • In Salamon$McDuff's book they mentioned that every level set of a Hamiltonian function is a invariant submanifold, I am not sure they meant a regular submanifold or an immersed one? Could you please comment about this? Or, may you have a look at my another question:https://math.stackexchange.com/questions/4419323/question-about-tangent-space-of-a-level-set-of-function – Grey Apr 03 '22 at 12:56