Sorry if this question is belongs to MSE. I have no idea about it.

Question: Is there any Riemannian manifold of zero dimensional isometry group which its Ricci curvature is positive (or maybe zero) somewhere?

I know that due to Joachim Lohkamp every smooth manifold of dimension at least $3$ admits a complete metric whose Ricci curvature is bounded between two negative constants.

  • $\begingroup$ If you take any Ricci-positive metric and perturb the metric just a little in $C^2$, then the resulting metric is still Ricci-positive and generically admits no more isometries. You could change the question, replacing "of zero-dimensional isometry group" by "which does not admit a smooth effective action by a compact Lie group of dimension $\ge 1$". If a manifold $M$ admits a smooth action by a compact Lie group, then there exists a Riemannian metric on $M$ for which the action is isometric. $\endgroup$ – Sebastian Goette Sep 22 '19 at 16:20
  • $\begingroup$ @SebastianGoette: A perturbed Ricci-positive metric may admits no more isometries? really? do you have any concrete example in your mind? $\endgroup$ – C.F.G Sep 22 '19 at 18:30
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    $\begingroup$ You can take the following surface in $\mathbb R^3$: $(x^2+y^2+y^2)+10^{-10}(x^3+2y^3+3z^3)=1$. It's group of isometries is trivial. $\endgroup$ – Dmitri Panov Sep 23 '19 at 18:02
  • $\begingroup$ Sorry, I meant of course $(x^2+y^2+z^2)+...$ - a tiny perturbation of the unite sphere $\endgroup$ – Dmitri Panov Oct 1 '19 at 11:04

Such a manifold exists (with zero Ricci curvature): the automorphism group of the 3-dimensional flat manifold of Hantzsche and Wendt is finite.

See the book of Charlap: Bieberbach groups and flat manifold.


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