**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.