Is the isometry group of a complete finite volume negatively curved manifold finite?

Question

Is the isometry group of a complete finite volume negatively curved manifold finite?

This question is based off of Why is the isometry group of a closed negatively curved manifold finite?

and the comment from Is the isometry group of a finite volume negatively curved manifold finite? that, although a finite volume negatively curved manifold may have infinite isometry group if it is not complete, if the manifold is complete then the isometry group must be finite.

Answer 1

Here is what I know:

1. Suppose that $M$ is a complete Riemannian manifold of finite volume and negative sectional curvature, $K<0$. Then the isometry group of $M$ is discrete.

To prove this claim note that $Isom(M)$ is a Lie group and it is nondiscrete if and only if its Lie algebra is nonzero, equivalently, if and only if $M$ admits a nonzero Killing field. But a combination of Propositions 2.2 and 5.5 in

Bishop, R. L.; O’Neill, B., Manifolds of negative curvature, Trans. Am. Math. Soc. 145, 1-49 (1969). ZBL0191.52002.

implies that in our situation every Killing field on $M$ is (identically) zero. (This is noted on page 20 of the paper.)

2. Suppose now that, additionally, the sectional curvature of $M$ is also bounded from below, i.e. there exists a negative constant $b$ such that $b\le K<0$ on $M$. Then $M$ satisfies the Margulis Lemma and, accordingly, admits a thick-thin decomposition. You can find an extended discussion of this in Chapter (Lecture) II of

Ballmann, Werner; Gromov, Mikhael; Schroeder, Viktor, Manifolds of nonpositive curvature, Progress in Mathematics, 61. Boston-Basel-Stuttgart: Birkhäuser. iv, 263 pp. (1985). ZBL0591.53001.

Let $\mu$ denote the Margulis constant of $M$ (it depends only on dimension and lower curvature bound.) Since $M$ has finite volume, the thick part $M_{[\mu/2,\infty)}$ of $M$ (the subset where the injectivity radius is $\ge \mu/2$) is compact and nonempty, see Theorem on page 102 of the book.

Given this, take some $x\in M_{[\mu/2,\infty)}$. Then for every isometry $g\in Isom(M)$, $g(x)\in M_{[\mu/2,\infty)}$. By compactness of $M_{[\mu/2,\infty)}$, it follows that $Isom(M)$ is a compact group. Since $Isom(M)$ is discrete, it follows that $Isom(M)$ is finite.

3. Edit. Here is a finiteness argument which works in general. Suppose that $M$ is a complete connected Riemannian manifold of finite volume with a discrete subgroup $\Gamma < Isom(M)$. I claim that $\Gamma$ is finite. Suppose not. Then, by the Arzela-Ascoli theorem, there exists a sequence of isometries $\gamma_i\in \Gamma$ such that for some (equivalently, every) $x\in M$, $\lim_{i\to \infty}d(x, \gamma_i(x))=\infty$, where $d$ is the Riemannian distance function on $M$. Let $\epsilon$ denote the injectivity radius of $M$ at $x$ and let $B=B(x,\epsilon)$ denote the $\epsilon$-ball in $M$ centered at $x$. After passing to a further subsequence, we can assume that all the balls $B(\gamma_i x, \epsilon), i\in {\mathbb N}$, are pairwise disjoint. But then $M$ has infinite volume (since all the balls $B(\gamma_i x, \epsilon)$ have the same positive volume). A contradiction.

To conclude: If $M$ is a complete finite volume Riemannian manifold of negative curvature, then $M$ has finite isometry group.