Why is the isometry group of a closed negatively curved manifold finite?

Question

Why is the isometry group of a closed negatively curved manifold finite?

If the isometry group is infinite, then it must have a subgroup that is isomorphic to $S^1$, but how do I take advantage of this fact?

Answer 1

If the isotropy group had a subgroup isomorphic to $S^1$, then we would have a continuous family of isometries, which we can differentiate to get a (non-zero) Killing vector field $X$ on the manifold. This is very advantageous - we have now converted our airy-fairy information about the symmetry group into a concrete geometric object. In particular, we can now use the Bochner formula for Killing fields:

If $X$ is a Killing field, then the function $f = \frac12|X|^2$ satisfies $$\Delta f = |\nabla X|^2 - {\rm Ric}(X,X).$$

[Petersen, Riemannian Geometry 2ed., Prop 29]

Since we are assuming negative curvature and $X$ is not identically zero, this implies that $\Delta f$ is non-negative everywhere and positive somewhere, which contradicts the fact that any smooth function on a compact manifold satisfies $\int \Delta f = 0$.