25
$\begingroup$

Does there exist a (finite dimensional) smooth manifold $M$, such that every Riemannian metric on $M$ has no isometries except the identity?

Of course, such a manifold must not admit a diffeomorphism of finite order.

Since a surface $S$ admits a diffeomorphism of order $n$ iff its mapping class group (MCP) has an element of order $n$ (see here), it follows that if $S$ has the above property, then its MCP has only elements of infinite order.

Improve this question
$\endgroup$
1

2 Answers 2

Reset to default
26
$\begingroup$

The answer to the question in the first sentence is "yes". Let $M$ be a hyperbolic 3-manifold whose isometry group is trivial. Then by Theorem 1.1 of

Farb, Benson; Weinberger, Shmuel Hidden symmetries and arithmetic manifolds. Geometry, spectral theory, groups, and dynamics, 111–119, Contemp. Math., 387, Amer. Math. Soc., Providence, RI, 2005. (Reviewer: Bachir Bekka) 53C35 (53C23)

(which they attribute to Borel, though they do not give an original reference for it), the isometry group of every Riemannian metric on $M$ is isomorphic to a subgroup of the hyperbolic isometry group, and thus is trivial.

Along similar lines, you might be interested in Theorem H of

Dinkelbach, Jonathan and Leeb, Bernhard, Equivariant Ricci flow with surgery and applications to finite group actions on geometric 3-manifolds. Geom. Topol. 13 (2009), no. 2, 1129–1173.

It says that every finite-order diffeomorphism of a closed hyperbolic 3-manifold is smoothly conjugate to an isometry, so closed hyperbolic 3-manifolds with trivial isometry groups give examples of smooth manifolds with torsion-free diffeomorphism groups.

Improve this answer
$\endgroup$
4
  • $\begingroup$ I'm missing the other direction of the equivalence. It seems like you need an argument to show that an isometry is necessarily finite order... Or is it obvious? $\endgroup$
    – HJRW
    Commented Apr 18, 2016 at 17:49
  • $\begingroup$ @HJRW: Whoops, you're right! For instance, flat tori have infinite-order isometries. However, using a little more technology one can show that my examples still work. I'll edit the answer accordingly. $\endgroup$
    – Andy Putman
    Commented Apr 18, 2016 at 18:00
  • 7
    $\begingroup$ @HJRW: The isometry group of a compact Riemannian manifold is a compact Lie group. A compact Lie group is either finite or contains a copy of $U(1)$. The latter contains finite subgroups $Z/nZ$ for every $n$. Thus, the isometry group of a compact Riemannian manifold is either trivial or contains nontrivial finite order elements. $\endgroup$
    – Moishe Kohan
    Commented Apr 18, 2016 at 18:40
  • $\begingroup$ @studiosus -- that seems to do it! Thanks! $\endgroup$
    – HJRW
    Commented Apr 19, 2016 at 12:30
5
$\begingroup$

The survey paper "Do manifolds have little symmetry?" by Volker Puppe lists several known results about manifolds with no nontrivial finite group action. The ArXiv link is http://arxiv.org/pdf/math/0606714v1.pdf

In particular

  • aspherical manifolds with $Z(\pi_1M)=0$ and $Out(\pi_1M)$ torsionfree

have no finite group action (Borel)

Examples are

  • certain mapping tori of nilmanifolds (Conner-Raymond-Weinberger)
  • certain hypertoral manifolds, i.e., n-manifolds with a degree 1 map to the n-torus (Schultz)
  • certain 3-manifolds (Edwards)
  • a certain Bieberbach manifold (Waldmüller)
Improve this answer
$\endgroup$

You must log in to answer this question.

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