Smoothness of frame bundle of (global) orbifolds [reference request]

Question

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.

Answer 1

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.

Answer 2

I recommend looking at Moerdijk and MrCun's "Introduction to Foliations and Lie Groupoids". On page 42, they define the frame bundle and orthonormal frame-bundle of an orbifold and show it's a smooth manifold. To do this, they first establish this on orbifold chart $(U,G,\varphi)$ by essentially your proof. Then you need to take the filtered colimit of the smooth manifolds resulting from each chart. This becomes the frame bundle of the orbifold, and is of course, a manifold.

(So, applying this to canonical orbifold charts on $M/G$, this gives you the proof).

Answer 3

Jason has already explained the proof. The point is that, given a connected Riemannian manifold, any isometry is uniquely determined by how it acts on a single point and the tangent space at that point. The core reason, as explained by Jason, for this is the existence and uniqueness theorem for second order ODE's applied to geodesics radiating from the point.

As for references, I would be surprised if it is not in Kobayashi-Nomizu or Kobayashi's book, Transformation Groups in Differential Geometry. I'm sure it's in other places, as well. Look in any proof that the isometry group is finite-dimensional.