Are critical points fixed?

Question

Let $M$ be a smooth manifold (compact, connected, without boundary and oriented if you wish) with a smooth action of $S^1$. Let $f:M\rightarrow\mathbb{R}$ be an invariant function $f$. I know how to prove that a fixed point of the action is a critical point of $f$. What I don't know is if the converse is true: I can prove that if a point is critical, then all the points in its orbit are critical as well, but that's all. So the questions are:

1) Are the critical points of $f$ necessarily fixed?

2) If the answer to 1) is "yes": any hint on the proof?

3) If the answer to 1) is "no": any counterexample? Is there any extra condition that makes the answer of 1) to be "yes"?

score 6 · Accepted Answer · answered Nov 07 '13 at 15:04

6

No; take the unit sphere, with $S^1$ acting by rotating the sphere about the $z$ axis, and let $f(x,y,z) = x^2+y^2.$ The two fixed points (poles) are critical points of $f$, but so is every point along the equator.

answered Nov 07 '13 at 15:04

user7530

49,280

Are you sure that this is the function you meant? I see your point if we take $f(x,y,z)=z^2$, though. – A. Bellmunt Nov 07 '13 at 15:11
1

Both should work: for my function, the equator is a maximum, and for yours, a minimum. – user7530 Nov 07 '13 at 15:13

score 0 · Answer 2 · answered Jan 31 '20 at 18:32

Two extra conditions for 1) to be true:

(i) If $(M,\omega)$ is symplectic.

(ii) The $S^1$-action is Hamiltonian.

Then it follows from nondegeneracy of the symplectic $2$-form $\omega$ that critical points are fixed points, since they are zeros of the fundamental vector field $\xi_M$.

Are critical points fixed?

2 Answers2