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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)?

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.

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    $\begingroup$ 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
    Commented Sep 22, 2019 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
    Commented Sep 22, 2019 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
    Commented Sep 23, 2019 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
    Commented Oct 1, 2019 at 11:04

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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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  • $\begingroup$ Sorry, but i couldn't found that example? (and I don't know how I accepted this answer (maybe I trusted to up-voters)) $\endgroup$
    – C.F.G
    Commented Aug 12, 2020 at 7:01

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