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Assume that $M$ is a simply connected closed Riemannian manifold with no boundary and nonnegative sectional curvaure Assume that ${\bf Z}_n=(g),\ n\geq 3$ acts on $M$ isometrically. Then if $gx=x$, i.e., it is a fixed point, clearly $g$ acts on ${\rm cut}\ x$ Here I have a question : $g\cdot x \in {\rm cut}\ x$ can happen for some $x$ ?

(Background : In the paper CRITICAL POINTS OF THE DISPLACEMENT FUNCTION OF AN ISOMETRY - VILNIS OZOLS, he consider an isometry whose displacement is small enough so that it takes each point into the complement of its cut locus. )

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    $\begingroup$ I am pretty sure that there are examples of a surface with a cyclic isometry moving some point into its cut locus. Take an equilaterlal triangle in the plane. Then double it, and glue the two triangles together along their edges. Each vertex is at maximal distance from each other vertex, so they lie in one another's "cut loci", although the surface is not smooth, so not quite an example. Blow a little air into it, to make a smooth balloon shaped very close to an equilateral triangle. Let $g$ rotate the vertices by $2 \pi/3$. I imagine the vertices stay at maximal distance. Proof may be hard. $\endgroup$
    – Ben McKay
    Commented Apr 17, 2016 at 10:39

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It can happen that $g.x\in cut(x)$ for some $x$. This is what happens for $S_{n+1}$ acting by permutation of homogeneous coordinates on $\mathbb{CP}^n$.

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    $\begingroup$ It seems this should work as well on real and quaternionic projective spaces. $\endgroup$
    – Sebastian Goette
    Commented Apr 17, 2016 at 12:43
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    $\begingroup$ He was looking for a simply connected example and real projective space is not simply connected. @SebastianGoette $\endgroup$
    – Mikhail Katz
    Commented Apr 18, 2016 at 9:08

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