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Let $M,N$ be two closed differential manifolds and let $G$ be a compact Lie group. Assume that $G$ acts on both manifolds $M,N$ nicely (i.e. free and proper so that $M/G$ and $N/G$ have the structure of manifolds, but I don't think that this is important for my question). Furthermore, assume that there exists open neighborhoods $U_1\subset M$ and $V_1 \subset N$ and a map $$s_1:U_1\rightarrow V_1.$$

Question: Is it possible to construct $G$-invariant open subset $U\subset M$ and $V\subset N$ and a map $s:U\rightarrow V$ such that $\forall g\in G$ and $\forall p \in U$ we have $$s(g\bullet_{M} p)=g\bullet_{N}s(p),$$ where $\bullet_{M}$ and $\bullet_{N}$ denote the $G$-action in both $M$ and $N$, respectively and such that $U_{1} \subset U$, $V_{1} \subset V$ and there exists some subset $A\subset U_{1}$ such that $s|_{A} \equiv s_{1}$. If this is in general not possible, then under what conditions on $s_1$ can this be realized?

For an answer I would be thankful!

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  • $\begingroup$ I forgot to mention that all open subsets here are supposed to be connected. But I don't know if this is important for the question. $\endgroup$
    – Stan
    Commented Jun 14, 2019 at 9:13
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    $\begingroup$ How does $s_1$ relate to $s$? Are you aware of the slice theorem? This seems to give what you want $\endgroup$
    – Thomas Rot
    Commented Jun 14, 2019 at 9:16
  • $\begingroup$ I am not sure how they relate. I am thinking that one can "push" $s_1$ in direction of the group action in order to define $s$. But I do not know how to make this precise. $\endgroup$
    – Stan
    Commented Jun 14, 2019 at 9:19
  • $\begingroup$ But as you formulate it $s_1$ does not seem to be doing anything (You can always define a map from $M$ to $N$, e.g. a constant map). $\endgroup$
    – Thomas Rot
    Commented Jun 14, 2019 at 9:20
  • $\begingroup$ Yes you are right. I wish that $U_1 \subset U$ and $V_1 \subset V$ and that $s|_{U_{1}} \equiv s_1$. I will change this in the question. Thanks I forgot to mention. $\endgroup$
    – Stan
    Commented Jun 14, 2019 at 9:24

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If $A$ is just one point $p$ then you can define $s$ by the required equivarianace property on orbit of $p$ and then you can extend it to whole $M$ since you said that $M/G$ is a manifold.

On the other hand, if $g \in G$ is sufficiently close to identity, then $g\cdot p$ will be close to $p$ and your $s_1$ must itself satisfy equivariance with respect to some small neighborhood of identity element of $G$. So if you want bigger $A$ you need some assumptions on $s_1$.

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