**Background**


Let $(M,g)$ be a riemannian manifold and let $G$ be a finite group acting effectively and isometrically on $M$. Recall that this means that for all $x \in G$, the diffeomorphism $\gamma\_x$ is such that $\gamma\_x^\* g = g$ and that if $\gamma\_x(p) = p$ for all $p \in M$, then $x$ is the identity element.


If $G$ acts freely, then $M/G$ is a smooth manifold, otherwise let's call it a (global) orbifold.


Years ago I was told that even when the action of $G$ on $M$ is not free, its lift to an action on the orthonormal frame bundle $F(M)$ is free. This means that the quotient $F(M)/G$ is smooth and is the orthonormal frame bundle of $M/G$.


I can see this when $M = \mathbb{R}^n$ with the standard euclidean inner product, so that the action of $G$ is via orthogonal (hence in particular linear) transformations. Indeed, suppose that $x\in G$ fixes a point in the frame bundle. Such a point consists of a pair $(p,f)$ where $p$ is a point in $M$ and $f$ is a frame for the tangent space $T\_pM$ to $M$ at $p$. The action of $x$ on the pair $(p,f)$ is given by


$$(p,f)\mapsto (\gamma\_x(p),(D\gamma\_x)\_p f),$$


where $(D\gamma\_x)\_p$ is the derivative (i.e., the push-forward) of $\gamma\_x$ at $p$.


Now if $x$ fixes $(p,f)$, then $\gamma\_x(p)=p$ and $(D\gamma\_x)\_p$ is the identity endomorphism of $T\_pM$. But since $\gamma\_x$ is linear, it agrees with its derivative, which means that $\gamma\_x$ itself is the identity. Finally, since $G$ acts effectively, we conclude that $x$ is the identity.


**Question**



> 
> Is this still true for $M/G$, where $M$ is a riemannian manifold? And if so, can someone point me in the direction of a reference where this is proved?
> 
> 
> 


Many thanks in advance.



Smoothness of frame bundle of (global) orbifolds [reference request]
First, one can clearly assume $M$ is connected by simply applying the argument to each componenet of $M$.


The key fact is a generalization of your argument for $M=\mathbb{R}^n$: that if $f:M\rightarrow M$ is an isometry with $M$ connected and if there is a point $p\in M$ with $f(p) = p$ and $d\_pf = Id$, then $f$ itself is the identity map.


Assuming this fact for the moment, then if $\gamma\_x(p,f) = (p,f)$ one concludes $\gamma\_x = Id$, and then since the action is effective, x itself must have been the identity element. Thus, the only element of $G$ which fixes any element of the frame bundle is the identity, so the action on the frame bundle is free.


Now, why is the fact true?


Set $X = \{p\in M| f(p) = p$ and $D\_p f = Id\}$. $X$ is nonempty by assumption and clearly closed (by, say, a continuity argument). If we can show it's open, then we'll have $X=M$ by connectedness of $M$. Thus, in particular, $f(p) = p$ for all $p\in M$.


To see $X$ is open, let $q\in X$. Let $U\_q$ be a normal neighborhood around $q$. Normal means that every pair of points in $U\_q$ has a unique minimal geodesic between them. Normal neighborhoods always exist (any sufficiently small neighborhood is totally normal), though I don't immediately remember how to prove it, only that it uses the Gauss lemma. Do Carmo's Riemannian Geometry book proves it, if you need a reference for this.


I claim that $U\_q\subseteq X$. For, if $r\in U\_q$, there is a unique minimal geodesic $c$ with $c(0) = q$ and $c(1) = r$. Since $d\_q f = Id$, we must have $f(c(t)) = c(t)$, so in particular, $f(r) = f(c(1)) = c(1) = r$. In short, $f$ fixes all points in $U\_q$


So, we must simply establish that $d\_r f = Id$. But, if there is a $v\in T\_r M$ with $d\_r f v\neq v$, then the we have $f(exp(tv))\neq f(exp(t d\_r f v))$ for all sufficiently small $t$. But for all sufficiently small $t$, both curves lie in $U\_q$ and we know $f$ fixes all points in $U\_q$. Thus, $d\_f v = v$ and the result follows.


