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Let $(M, g)$ be a compact Riemannian manifold. Suppose $M$ admits a smooth free circle action (Denote the circle group by $G$. The action $G$ on $M$ is not necessarily isometric) and the orbit space $B$ is another closed manifold, i.e., we have a smooth principal circle bundle $G\hookrightarrow M\rightarrow B$.

My question is:

(1) With the Riemannian metric $g$ being fixed for $M$, can we find another free $G$-action on $M$ so that the new action is isometric?

(2) If so, can we even find a new free circle action on $M$ so that the map $M\rightarrow M/G$ is harmonic?

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I don't think so. I suspect that this action does not exists for generic metric $g$ on manifolds which admit circle actions.

Here is the thing that I have in mind.

Consider the two torus. This clearly has a free $G=S^1$ action. Now equip the torus with a metric $g$, and consider the scalar curvature $f$ of the metric. Suppose that there is an isometric action on the torus for the metric. Then the level sets of $f$ must be mapped to themselves by this action. So construct a metric which has a unique maximal curvature somewhere. This must be a fixed point of the action.

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  • $\begingroup$ As an extension of the answer: By Ebin's "The manifold of Riemannian metrics", the set of Riemannian metrics with a finite discrete group of isometries on a fixed compact smooth manifold is open and dense in the set of all Riemannian metrics that the manifold admits. That is, generic Riemannian metrics have few isometries. $\endgroup$
    – D. Corro
    Commented Jun 21 at 10:34

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