Finiteness of fixed points of a Lie group action

Question

Let $\psi: G\rightarrow \mathrm{Diff}(M)$ be a smooth non-trivial action of a compact connected Lie group $G$ on a compact connected smooth manifold $M$.

Under which assumptions there will be a finite number of fixed points on $M$ by the action?

Answer 1

I don't have a complete answer, but I think the problem can be reduced to understanding Lefschetz fixed points of isometries. Here's why: The set of fixed points, $M^G$, is closed. A closed subset of a compact manifold is finite if and only if it's discrete, so we need to make $M^G$ discrete --- i.e. we want to make sure every $p \in M^G$ has a neighborhood disjoint from the rest of $M^G$. Fortunately, we can find a nice Riemannian metric on $M$ that is invariant under left multiplication $L_g: M \to M$, so each $L_g$ is an isometry. We can then use the Riemannian exponential map to find a neighborhood $U \subset M$ of fixed point $p \in M^G$ such that $q \in U$ lies in $M^G$ if and only if the velocity vector of the geodesic adjoining $p$ to $q$ is an eigenvector of the map $(dL_g)_p : T_p M \to T_p M$ with eigenvalue 1. In other words, we must make sure that if $p \in M^G$, then there exists $g\in G$ such that $p$ is a Lefschetz fixed point of the isometry $L_g : M \to M$. Some elaboration is below.

$\boldsymbol{M^G}$ is closed: Let $M^G:= \{ p \in M \mid L_g(p)=p \text{ for all } g \in G\}$, the set of points fixed by $G$. If we let $L_g : M \to M$ be $L_g(p)=g\cdot p$ and $\ell_g: M \to M \times M$ be $\ell_g(p)=(p,g\cdot p)$, we have $$M^{\{g\}}:=\{p \in M \mid L_g(p)=p\}=\{ p \in M \mid \ell_g(p) \in \Delta(M)\}=\ell_g^{-1}(\Delta(M)),$$ where $\Delta(M)$ is the diagonal in $M \times M$. Since the diagonal is closed (since $M$ is Hausdorff) and $\ell_g$ is continuous, $M^{\{g\}}$ is closed. Therefore $M^G=\cap_{g \in G} M^{\{g\}}$ is closed.

How to get a nice neighborhood of a point in $\boldsymbol{M^G}$: We'll show that $M^G$ is a smooth submanifold. By compactness, $G$ possesses a left-invariant measure. Then we can take any Riemannian metric on $M$ and average it over $G$ to obtain a $G$-invariant Riemannian metric on $M$. Thus $L_g: M \to M$ is an isometry for all $g \in G$. Denote the corresponding exponential map on $M$ by $\exp_p: T_p M \to M$. Since $L_g$ is an isometry, it takes a geodesic at $p$ with initial velocity vector $v \in T_p M$ (with $v$ sufficiently small) to a geodesic at $L_g(p)$ with velocity $(dL_g)_p (v)$. This implies that $L_g \circ \exp_p(v)= \exp_{L_g(p)}((dL_g)_p v)$. If $p \in M^G$, then it follows that $\exp_p (v)$ lies in $M^G$ if and only if $(dL_g)_p(v)=v$ for all $g \in G$. Thus the preimage $\exp_p^{-1}(M^G)$ is the intersection of a linear subspace with a neighborhood of $0 \in T_p M$, since $\exp_p (\alpha v +w ) \in M^G$ for any $\exp_p(v),\exp_p(w)\in M^G$ and $\alpha \in \mathbb{R}$. Since $\exp_p$ is a diffeomorphism near $0 \in T_p M$, it forms a local chart for $M^G$.

Answer 2

If G is discrete group. Because, this action is already proper (Lee Introduction to smooth manifolds, 2nd edition, Corollary 21.6). Then the stabilizer is a compact subgroup of G (Lee again, Corollary 21.8). Hence G is finite. I guess.