Differential Geometry – Is the Isometry Group of a Complete Finite Volume Negatively Curved Manifold Finite?

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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.

Best Answer

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.)

  1. 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.

  1. 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.

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