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Are there any interesting results, if a Morse function $M \rightarrow \mathbb{R}$ happens to be the moment map, for some $S^1$ action on $M$ (equipped with $\omega$) as well? Thank you very much.

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    $\begingroup$ A moment map is always Morse-Bott, so being told it's actually Morse is just saying that the S^1-fixed points are isolated. $\endgroup$
    – Allen Knutson
    Commented Oct 21, 2010 at 2:01

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In this case, if you pick a $S^1$-invariant metric, the Morse flow will commute with the $S^1$ action, and so you can think of them together as an action of $\mathbb{C}^*$ where the unit circle acts by the original $S^1$, and the real line acts by Morse flow.

In particular, the critical points will be the fixed points of the action, and the Morse index of a fixed point will be determined by how many of the $S^1$-weights on the tangent space at the point are positive, and how many are negative.

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