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$\DeclareMathOperator\Conf{Conf}$Let $M$ be a manifold, and $\Conf_n M$ the ordered configuration space of n points on $M$. The symmetric group $S_n$ acts by permuting the points.

Is there a simple description of the homotopy fixed points of this action ? In particular I wonder if this is not empty. For $M = \mathbb R^2$ it seems to me that the existence of a coherent $S_n$-equivariant point on $\Conf_n M$ would imply that there is a section $S_n \to B_n$, the braid group on $n$ strands, because there would be a choice of a (homotopy class of) path from $(x_1,...,x_n)$ to $(x_\sigma(1),...,x_\sigma(n))$ that commutes with composition of permutations. And I know that it doesn't exist.

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  • $\begingroup$ A concrete model of the homotopy fixed points $X^{h\Sigma_n}$ is $\operatorname{Map}(\operatorname{Conf}_n(\mathbb{R}^\infty),X )^{\Sigma_i}$, and so a more general question are there obstructions to producing equivariant maps $\operatorname{Conf}_n(W)\rightarrow\operatorname{Conf}_n(M) $ when $\operatorname{dim}(W)> \operatorname{dim}(M)$? $\endgroup$
    – Connor Malin
    Commented Nov 22 at 15:57

2 Answers 2

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In general, A nice enough finite dimensional space with a free $\Sigma_i$-action does not admit any homotopy fixed points. This is because a homotopy fixed point provides a splitting of the equivariant map $X\rightarrow *$, and this is impossible by considering the homology of the homotopy orbits, since the group homology of the symmetric group is unbounded. In particular, the usual notion of a manifold (Hausdorff, second countable, etc) satisfies the niceness requirements, and the action on the configuration space is free.

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    $\begingroup$ that is a nice argument! $\endgroup$
    – Chris Schommer-Pries
    Commented Nov 22 at 19:58
  • $\begingroup$ Could you maybe include a little more detail about what the actual obstruction is? It's not clear to me why unboundedness prevents this map from having a section. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Nov 22 at 22:27
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    $\begingroup$ @R.vanDobbendeBruyn Functors preserve retractions, and so if a complex $X$ retracts onto a point in the homotopy category of $G$-spaces, it implies that $H_\ast (X_{hG})$ retracts onto $H_\ast(\ast_{hG})=H_\ast(BG)$. That implies that $H_\ast (X_{hG})$ is nonzero in all the degrees $H_\ast(BG)$ is. If the action of $G$ on $X$ is free and nice enough at the pointset level, then $X_{hG}\simeq X_G$. If $X$ is finite dimensional, this implies $H_\ast(X_{hG})$ is bounded, so if $H_\ast(BG)$ is unbounded such a map cannot exist. $\endgroup$
    – Connor Malin
    Commented Nov 23 at 8:44
  • $\begingroup$ Great answer, thank you! $\endgroup$
    – Nicolas Guès
    Commented Nov 23 at 14:13
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    $\begingroup$ This is a nice answer, but I found it a bit confusing since the point $*$ is not $G$-homotopy equivalent to $EG$. Would it not be better to say "provides an equivariant splitting of the classifying map $X\to EG$"? $\endgroup$
    – Mark Grant
    Commented Nov 23 at 15:08
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[UPDATE: Connor Malin's answer is definitely better than mine, but I will leave mine here in case the approach turns out to be useful for some other purpose.]

Put $X=\operatorname{Conf}_n(M)$ and $Y=\operatorname{Conf}_2(M)$. We want to show that $X^{h\Sigma_n}$ is empty. There is an evident projection $X\to Y$ which is equivariant for the evident copy of $\Sigma_2$ in $\Sigma_n$ so we get a projection $X^{h\Sigma_n}\to Y^{h\Sigma_2}$, so it will suffice to show that $Y^{h\Sigma_2}=\emptyset$. I think that this holds provided that $Y$ has the $\Sigma_2$-equivariant homotopy type of a finite $\Sigma_2$-CW complex, which I think should be true under mild conditions on $M$ (despite the fact that $Y$ is certainly not compact).

To see this, let $Y_2$ be the $2$-adic completion of $Y$ in the sense of Bousfield and Kan. This is obtained by applying a functorial simplicial construction to the singular complex $SY$, so we have maps $Y\xleftarrow{p}|SY|\xrightarrow{q}Y_2$ in which $p$ is a weak equivalence, and everything is $\Sigma_2$-equivariant. If we can show that $Y_2^{h\Sigma_2}=\emptyset$ then it will follow that $|SY|^{h\Sigma_2}=\emptyset$, and the homotopy fixed point functor preserves weak equivalences, so $Y^{h\Sigma_2}=\emptyset$.

We can now apply the Sullivan Conjecture, in the form proved by Lannes: see Theorem VIII.1.2 of the Alaska book, or Theorem 9.1.1 of the book Unstable Modules over the Steenrod Algebra and Sullivan's Fixed Point Conjecture. Assuming the stated finiteness condition, that says that the natural map $(Y^{\Sigma_2})_2\to Y_2^{h\Sigma_2}$ is a weak equivalence, but here $Y^{\Sigma_2}=\emptyset$, so $Y_2^{h\Sigma_2}=\emptyset$ as required. (Some versions of the Sullivan Conjecture require that $Y$ should be a nilpotent space, but the version proved by Lannes does not.)

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