0
$\begingroup$

It is known since Bochner that for M compact with negative Ricci curvature, the group of isometries is discreet and hence finite. Are there any generalizations to compact non-positive sectional curvature manifolds? Of course, $T^n$ is a counterexample for strictly 0 curvature case. But what about sectional curvature <0 on a non-empty open set, and vanishing elsewhere?

As I understand Bochner's argument does break down in this case.

Known counterexamples?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

Bochner's theorem extends to nonpositive Ricci to give:

If $(M,g)$ is compact and has $\textrm{Ric}\leq 0$ then any Killing vector $X$ is parallel and $\textrm{Ric}(X,X) = 0$.

See Petersen (3rd ed) Theorem 8.2.2.

Thus if $(M,g)$ is compact and has $\textrm{Ric}\leq 0$ with $<0$ then $X = 0$.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Thanks, I will need to think about that for a bit. What about if M has (totally geodesic) boundary? Sorry if these are elementary questions, not really my subject. $\endgroup$
    – Yasha
    Commented Jun 18, 2023 at 16:05
  • $\begingroup$ Ok I read Petersen. Looks good. It seems like in case the manifold has boundary the argument would have problems, since it is using the maximum principle for the Laplacian. $\endgroup$
    – Yasha
    Commented Jun 18, 2023 at 16:53
  • $\begingroup$ Do you want to assume anything about the killing vector at the boundary? I think if the KVF is tangent to the boundary then you should be able to make it work. But with no boundary conditions there must be a counterexample. $\endgroup$
    – Otis Chodosh
    Commented Jun 18, 2023 at 17:08
  • $\begingroup$ No additional assumptions on KVF. $\endgroup$
    – Yasha
    Commented Jun 19, 2023 at 14:33

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .