For an action of a compact Lie group in a manifold, does the complement of the non-expansive maps set have positive Haar measure? Let $(M,g)$ be a riemannian manifold, it is smoothly acted by a compact Lie group $G$. Consider $\mu$ the Haar measure on $G$. Set $A=\{h\in G\, | \, |h_{*}v|_{g}\geq |v|_{g}, \, \forall\, v\in TM\}$. 
Is true that $\mu(A)>0$?
I know that the set $\{h\in G\, | \, |h_{*}v|_{g}>|v|_{g}, \, \forall\, v\in TM\}$ is open, because the expansive maps set forms a open set in $Diff^{\infty}(M)$. But, it can be empty, for example while $G$ is a finite group. Otherwise for finite groups $\mu(A)>0$.
I have tried to proof it looking to what happens in a neighborhood of the identity, concretely I tried to proof that interior of $A$ is not empty.   
 A: I'm a little rusty with Lie groups, so someone let me know if I've messed something up here. I'm assuming $G$ is connected for simplicity.
Take $g\in A$ and consider the subgroup $H$ generated by powers of $g$. Since $G$ is compact and $H$ is abelian, the closure $\bar H$ must be finite cyclic or a torus. In either case you should be able to show that the positive powers of $g$ alone are dense in $\bar H$, and thus we can find a sequence $\{g^{i_k}:i_k > 0\}$ converging to $g^{-1}$. Since $g$ is weakly expansive, the same is true of each $g^{i_k}$, and thus (by smoothness of the group action and closedness of the weak expansiveness condition) of $g^{-1}$. Thus $A$ is in fact closed under inversion, so it is a subgroup. 
It seems like this is a common argument: see e.g. this paper - we could have just cited the theorem "a closed subsemigroup of a compact group is a subgroup".
Knowing $A$ is a closed Lie subgroup, we have $\mu(A) > 0$ if and only if $A = G$. Since $g \in A$ and $g^{-1} \in A$ implies $g$ is an isometry, this means your claim is true if and only if the action of $G$ is isometric.
