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Fix very tiny error
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Andy Putman
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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.

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 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.

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.

Source Link
Andy Putman
  • 44.8k
  • 14
  • 186
  • 272

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 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.