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What type of compact manifolds, can be acted freely by symmetric group $S_{m}$ for some $m>2$?

Is there a compact manifold which can be act freely by all symmetric groups $S_{m}$?

This question have been asked already here and is indirectly related to this.

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The answer to the second question is more or less no by Theorem 2 in Popov's Finite subgroups of diffeomorphism groups: for every compact connected smooth manifold $M$ there is a constant $b_M$ such that if the alternating group $A_n$ acts (by diffeomorphisms, but not necessarily freely) on $M$, then $n \le b_M$.

I don't really understand what kind of answer you're expecting to the first question.

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  • $\begingroup$ Thank you for your answer. To start, I search for an example of a free action. then a possible classification. regarding your answer, what about continuous action? $\endgroup$
    – Ali Taghavi
    Commented Oct 23, 2014 at 22:35
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    $\begingroup$ Writing down examples is straightforward but I don't understand what kind of classification you're expecting. Somewhat more generally, let $M$ be a compact manifold, $G$ be a finite group, and $f : \pi_1(M) \to G$ be a surjective homomorphism; for example, we can always take $M$ to be a surface of sufficiently large genus. Then $f$ is the classifying map of a principal $G$-bundle on $M$ which is a compact manifold on which $G$ acts freely with quotient $M$. It is connected iff $M$ is. Every connected example arises in this way. $\endgroup$
    – Qiaochu Yuan
    Commented Oct 23, 2014 at 23:25
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    $\begingroup$ $M$ the frame bundle of a high-dimensional sphere would work, as well. You could ensure the action is free in this setting. $\endgroup$
    – Ryan Budney
    Commented Oct 24, 2014 at 4:53
  • $\begingroup$ @QiaochuYuan Could you please give a reference for your statements.(whare i can find such materials?) by classification I mean some thing as follows: for the moment lets forget "Free". We are interested to know about all actions of $S_{m}$ on $S^{m-1}$. Are there only a finite number of such actions up to topological equivalent? $\endgroup$
    – Ali Taghavi
    Commented Oct 24, 2014 at 12:39
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    $\begingroup$ If you think of the symmetric group as the group of motions of the $n$-simplex, the boundary of the $n$-simplex is canonically a smoothly-triangulated sphere. So the symmetry group is a group of diffeomorphisms of an $(n-1)$-sphere. It does not act freely on that sphere, but it does act freely on the unit tangent bundle, since these symmetries are isometries in the appropriate spherical riemann metric on the sphere. $\endgroup$
    – Ryan Budney
    Commented Oct 24, 2014 at 16:41

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