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Let $f$ be a continuous function on $S^3$ and let $\xi^{\perp}=\{x\in S^3:\,x\cdot\xi=0\}$ be a two-dimensional equator of $S^3$ orthogonal to the direction $\xi\in S^3$ (here $x\cdot\xi$ stands for a usual inner product in ${\mathbb R^4}$). We call an action of $Z_2$ on $S^3$ natural, if it is a reflection with respect to the origin, i.e., $T(x)=-x$, $T\in Z_2$, $x\in S^3$.

Assume that for every equator $\xi^{\perp}$, the function $f$ is invariant under the non-natural $Z_2$-action on $\xi^{\perp}$, i.e., $f(T_{\xi}(x))=f(x)$ for all $x\in \xi^{\perp}$, $T_{\xi}\in Z_2$. Does it follow that $f$ is identically constant on $S^3$?

We are especially interested in the case when $T_{\xi}$ is just a rotation by the angle $\pi$ around some (unknown) axis $l_{\xi}\in \xi^{\perp}$.

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  • $\begingroup$ What do you mean by "the non-natural $\mathbf{Z}_2$-action"? There are many possible actions apart from the antipodal one. Does the action vary continuously with $\xi$? Is the action assumed to be by an isometry? In the case that $T_{\xi}$ is always trivial, I can certainly answer your question! Then $f$ could be any function. $\endgroup$
    – John Harvey
    Commented Aug 21, 2015 at 22:32
  • $\begingroup$ Please assume that the action is by a rotation by the angle $\pi$ around some unknown axis $l_{\xi}$ in $\xi^{\perp}$. You might also assume that the action varies continuously, we are interested in the general case though. $\endgroup$
    – Dmitry Ryabogin
    Commented Aug 22, 2015 at 2:53

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