Let $M$ be a (closed, smooth) manifold of dimension $d$. For $n$ a positive integer, fix $n$ points $x_1, \dots, x_n \in M$. The group of diffeomorphisms of $M$ that permutes the points $x_i$ surjects to the symmetric group $\Sigma_n$, at least for $d \geq 2$:

$$\Phi\colon \text{Diff}(M,\{x_1, \dots, x_n\}) \to \Sigma_n$$

Is it always true that $\Phi$ does not have a section for $n$ big enough?

How big do we have to choose $n$ for $M = S^d$? (The naive guess would be $d+3$.)

  • $\begingroup$ Why is $d + 3$ the naive guess? $\endgroup$ – Qiaochu Yuan Oct 22 '15 at 4:14
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    $\begingroup$ @QiaochuYuan, $d+2$ occurs with the simplex in $S^d$. $\endgroup$ – Mariano Suárez-Álvarez Oct 22 '15 at 4:17

I'm going to reverse the roles of $n$ and $d$ (since otherwise I will screw things up in this answer). If $M^n$ is a connected $n$-dimensional smooth manifold, then there is no section of the map $Diff(M^n,\{x_1,\ldots,x_d\}) \rightarrow \Sigma_d$ if $d \geq n+3$ (nb: you forgot in your question to assume that $M^n$ is connected). Indeed, assume that $\Sigma_d$ acts by diffeomorphisms on $M^n$ such that the action restricts to the permutation action on $\{x_1,\ldots,x_d\}$. The group $\Sigma_{d-1} \subset \Sigma_d$ then fixes $x_d$, so we get a linear action of $\Sigma_{d-1}$ on the tangent space $T_{x_d} M^n = \mathbb{R}^n$. Every element of $\Sigma_{d-1}$ must act nontrivially on this tangent space; indeed, by averaging we can find a Riemannian metric on $M^n$ that is preserved by the action of $\Sigma_d$, and an isometry of a connected Riemannian manifold that fixes a point and all tangent vectors at that point must be the identity. But the smallest dimensional vector space on which $S_{d-1}$ acts faithfully is $(d-2)$-dimensional. We thus deduce that $n \geq d-2$, so $d \leq n+2$, as desired.

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    $\begingroup$ Great answer! What happens when replacing diffeomorphisms by homeomorphisms? $\endgroup$ – Jens Reinhold Oct 22 '15 at 5:01
  • $\begingroup$ @JensReinhold: I expect that the same thing holds, but I don't know how to prove it. $\endgroup$ – Andy Putman Oct 26 '15 at 17:55

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