Assume a finite group $G$ acts smoothly on a smooth manifold $M$. Let $p\in M^G$. Then there is the action of $G$ on $T_pM$ given by differentials of diffeomorphisms being the actions of group elements. Is it true that $$ T_p(M^G)=(T_pM)^G? $$


Yes, these sets are equal. I will use the notation $g\cdot$ to indicate the group action on $M$ and $g_\ast$ to indicate the action on $T_p M$.

One inclusion is easy: If $v\in T_p(M^G)$, then there is a curve $\gamma$ in $M^G$ with $\gamma(0) = p$ and $\gamma'(0) = v$. Now, if $g\in G$ is any group element, then $g\cdot \gamma(t) = \gamma(t)$ because $\gamma$ is in $M^G$. Thus, $g_\ast v = v$.

For the other direction, I don't know how to prove it without using some Riemannian geometry. By average an Riemannian metric on $M$ over the $G$ action, we may assume the $G$ is action is by isometries. Pick $v\in (T_p M)^G$ and consider the curve $\gamma(t) = \exp_p(tv)$.

By definition, this is a geodesic through $p$ with tangent vector $v$. We claim that for any $g\in G$, that $g \cdot \exp(tv) = \exp(tv)$. Believing this, it follows that $\exp(tv)\in M^G$ for all $t$ for which its defined (which includes a small neighborhood of $0\in \mathbb{R}$), so, in particular, $v\in T_p(M^G)$.

So, why is $g\cdot \exp(tv) = \exp(tv)$? Well, it's a standard fact from ODE theory, applied to Riemannian geometry that a geodesic is characterized by knowledge of a single point it goes through as well as the tangent vector at that point. So, it's enough to show that $g\exp_p(0) = \exp_p(0)$ and that $g_\ast \exp_p'(0) = \exp_p'(0)$.

But $g\cdot p = p$ and $g_\ast v = v$ follow since $v\in (T_p M)^G$, so we are done.

  • 1
    $\begingroup$ The same argument works any time you have an isometric action regardless of properties of $G$. Also, just as in the finite $G$ case, if $G$ is a compact Lie group we can always find a Riemannian metric making the group action isometric, so this result is true for all compact Lie groups as well (not just finite groups). $\endgroup$ Oct 28 '17 at 19:05

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