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Let us say that a (closed, connected) manifold has a symmetry if it admits a non-trivial action by a finite group. Note that I am not asking the action to be free. So for example rotating the 2-sphere by $\frac{2 \pi}{n}$ generates a non-trivial action by the cylic group of order $n$.

My question now is whether the following is true. (I remember discussing this some time ago in Bonn, but my memory is fuzzy. My internet searching has failed me, and so I ask it here).

Statement: Most manifolds do not admit any symmetries.

Question: Is the above statement true?

I think most naturally occurring manifolds, the ones we are familiar with and typically work with, do admit symmetries and so this would give another answer to this recent MO question: Situations where “naturally occurring” mathematical objects behave very differently from “typical” ones

Of course to answer this one needs to make precise what "most" means. I am quite flexible on how to interpret this. Also we better restrict to connected manifolds, and maybe also to simply connected manifolds. Is anything known about this question? I am likewise curious about anything known in this sort of direction.

This question can be asked in the smooth, PL, and Top categories.

In each case it can be rephrased in terms of torsion in the automorphism group of the manifold. For example in the smooth case the statement can be rephrased:

Alternative Statement: For most manifolds, their diffeomorphism group is torsion free.

In any manifold we can always find a disk. A boundary preserving automorphism of a disk always can be extended by the identity to the whole manifold. Thus if these admit torsion elements, then the automorphism group of any manifold of this dimension will admit torsion elements. I don't think this is a problem in the PL and Top cases. But in the Smooth case we have for $d \geq 5$:

$$ \pi_0(\textrm{Diff}_\partial(D^n)) \cong \Theta_{d+1}$$

where $\Theta_{d+1}$ is the group of exotic spheres.

Question 2: Can the finite group $\Theta_{d+1}$ be realized as a subgroup of $\textrm{Diff}_\partial(D^n)$? Is the latter group torsion free?

If $\textrm{Diff}_\partial(D^n)$ has torsion, then the main question is really most interesting in the PL and Top cases, I think.

I am aware that for Reimannian manifolds there are positive results in this direction. I am less surprised by these as I imagine a generic Reimanniant manifold to be quite "lumpy" and not even have local symmetries.

A closely related counter point which is also related to this question:

Counter-question: Are there any examples of manifolds without any symmetry? Do all manifolds admit a symmetry?

archipelago
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Chris Schommer-Pries
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    Counter-question. Yes, I believe the first example is due to Borel ("On periodic maps of certain K(pi,1)"), which proves that any aspherical manifold with centerless fundamental group and torsion-free outer automorphism group supports no finite order diffeomorphisms. For hyperbolic manifolds I believe every finite group action is conjugate to an action by hyperbolic isometries, so you're just asking (by Mostow) for hyperbolic manifolds with Out(pi_1) tors-free, just like in Borel's statement. – mme Jul 27 '21 at 19:57
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    Any finite order diffeomorphism of a closed manifold that fixes a point and the tangent space at that point is the identity (choose an invariant Riemannian metric and shoot geodesics away from that point). In particular $Diff_\partial(D^d)$ is torsion-free (extend the diffeomorphism to the sphere and use the above). – archipelago Jul 27 '21 at 20:06
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    "...Also we better restrict to connected manifolds, and maybe also to simply connected manifolds." I missed this sentence.

    I think there is likely to be substantial difference between the s.c. case and the general case: one reason to believe your Statement is that one might believe that most manifolds satisfy Borel's hypotheses.

     – mme Jul 27 '21 at 20:07
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    There's a paper by Volker Puppe with a title very reminiscent of the question, and there he proves that "most" 3-manifolds have no finite order diffeomorphisms. – Denis T Jul 27 '21 at 20:29
  • Did you discuss it with D.Fuchs? – markvs Jul 27 '21 at 20:43
  • The answer to your question is going to be sensitive to dimension. There aren't any compact simply-connected 1-manifolds. In dimension 2 there's only the one manifold $S^2$, so your answer will be negative. In dimension 3, again you run into the problem of not having interesting simply-connected manifolds. I think your question is somewhat bogged-down in the issue of what "most" should mean. The question still makes sense for manifolds with fundamental groups. In that case, you start seeing symmetryless manifolds in dimension 3. – Ryan Budney Jul 28 '21 at 04:44
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    For a specific reference for the simply connected case of "counter-question" is Theorem 4 of https://arxiv.org/pdf/math/0606714.pdf. Puppe constructed a simply connected $6$-manifold on which no finite group can act effectively by orientation preserving transformations. – Nick L Jul 28 '21 at 10:45
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    Furthermore, the question does a generic simply connected, smooth $6$-manifold admit no finite group actions makes perfect sense, because such manifolds are uniquely determined by their cohomolgy ring and characteristic classes (Wall-Jupp-Jubr) so the set of diffeomorphism classes sits in a space suitable for such quantitative questions. Puppe proved several striking statements in the case when the manifolds are spin (a substantial subset), for example that the set of manifolds having a cohomologically non-trivial finite group action has density 0. See Theorem 3 of the paper mentioned above. – Nick L Jul 28 '21 at 10:59

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