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My question is easy to formulate:

What is known about the homotopy groups of Homeo($\mathbb{R}^n$)? Especially, what is its fundamental group? (A guess would be $\mathbb{Z}$ for $n=2$ and $\mathbb{Z}/2$ for large $n$.)

  1. It is well-known that it has two components, determined by the orientation behaviour.
  2. Stably a lot is known: One can use either Sullivans results about the homotopy type of $G/TOP$ and that the homotopy groups of $G$ are the stable homotopy groups of spheres in degrees $i\geq 1$. Also, $TOP/PL$ is famously 2-connected, being a $K(\mathbb{Z}/2,3)$. If I understand it correctly, in [Brumfiel, On the homotopy groups of $BPL$ and $PL/O$] it is also mentioned in the introduction that $PL/O$ is at least $2$-connected(*). This would show $\pi_1(TOP) = \mathbb{Z}/2 $.
  3. Maybe there are results on the connectivity of the map Homeo($\mathbb{R}^n$)$\rightarrow TOP$?

(*) In the paper it is mentioned that $\pi_k(PL/O)$ is isomorphic to the group of concordance classes of smoothings of the $k$-sphere. I guess low dimensional smoothing theory implies that any two smooth structures on $S^k$ ($k\leq 2$) are concordant, since a structure on $S^k\times \{0,1\}$ extends to a structure on $S^k \times I$.

Christian
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    The map $\operatorname{Homeo}(\mathbb{R}^n ) =TOP(n)\rightarrow TOP$ should be (n-1)-connected because by Kirby-Siebenmann the map $Top(n)/O(n) \rightarrow Top/O$ is n-connected and the map $O(n) \rightarrow O$ is (n-1)-connected. So, for example, since $Top(3)$ is equivalent to $O(3)$ (I think this follows from Smale conjecture in dimension 3), the fundamental group $\pi_1(Top(n))=\mathbb{Z}/2$ for $n \geq 3$. – Connor Malin Dec 04 '21 at 23:30
  • Ah, I did not know the Kirby-Siebenmann result. I suppose I can find it in the "Foundational Essays"-book? I don't really see how what you mentioned follows from the Smale conjecture, but I also only read the Wikipedia entry. Is there a way to compare $TOP(3)$ to $Diff(S^3)$? – Christian Dec 05 '21 at 10:17
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    $Top(3)$ can be identified with the homeomorphisms of $S^3$ fixing a point. There is a fiber sequence then $Top(3) \rightarrow Homeo(S^n) \rightarrow S^3$ with the last map given by evaluation at the point at infinity. This splits since $S^3$ is a Lie group. In dimension 3, $Homeo$ is the same as $Diff$ so we can apply the Smale conjecture. Hence we get $O(4) \simeq S^3 \times Top(3)$ which probably implies that $Top(3)\simeq O(3)$. – Connor Malin Dec 05 '21 at 16:03

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