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I'd like to find a map $F$:Homeo($\mathbb{S}^{1}$)$\to O(2)$ which is both a group homomorphism and a homotopy equivalence. I know $O(2)$ is a deformation retract of Homeo($\mathbb{S}^{1}$), but I can't seem to come up with a map which also preserves the group structure.

One thing I thought to try was to map each orientation-preserving homeomorphism to its rotation number (or the unique rotation corresponding to it), but I don't think one generally has Rot($f\circ g$) = Rot($f$) + Rot($g$).

If anyone has a proof that such a map must exist (or that no such map exists), I would welcome that as well!

Missingno
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    You do not need a retraction to be a group homomorphism for this. Check your reference. – Moishe Kohan Jan 10 '25 at 02:19
  • The rotation number (or some other notion of the average rotation amount) seems like a good candidate. Is it really not a homomorphism? – Karl Jan 10 '25 at 08:58
  • I found this paper which seems to suggest not! https://people.mpim-bonn.mpg.de/zagier/files/doi/10.1007/BF01456075/fulltext.pdf?utm_source=chatgpt.com – Missingno Jan 10 '25 at 15:35
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    Here is an MSE link regarding why rotation number is not a homomrophism. – Lee Mosher Jan 10 '25 at 19:59

1 Answers1

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No, there is no map $\newcommand{\Homeo}{\operatorname{Homeo}}$ $F: \Homeo(S^1) \to O(2)$ which is both a group morphism and a homotopy equivalence. To see this, consider the subgroup $\Homeo_+(S^1)$ of orientation-preserving homeomorphisms. This is the path-component of the identity in $\Homeo(S^1)$. A map $F: \Homeo(S^1) \to O(2)$ which is a group morphism and a homotopy equivalence would necessarily restrict to a map with the same properties $\overline{F}: \Homeo_+(S^1) \to SO(2)$ between the identity components. Hence, it suffices to show that there is no group morphism $\Homeo_+(S^1) \to SO(2)$ which is a homotopy equivalence.

Recall that a group $G$ is called perfect if $G = [G,G]$, i.e. every element is a product of commutators. It follows easily from the definition that if $G$ is a perfect group and $A$ is an abelian group, then the only morphism $G \to A$ is the trivial morphism.

Proposition: $\Homeo_+(S^1)$ is a perfect group.

For a proof, see Proposition 5.11 of the following paper:

Ghys, É. Groups acting on the circle. Enseign. Math., II. Sér. 47, No. 3-4, 329-407 (2001).

Since $SO(2)$ is abelian, it follows from the proposition that the only morphism $\Homeo_+(S^1) \to SO(2)$ is the trivial morphism. On the other hand, the trivial morphism is not a homotopy equivalence since $SO(2)$ is not contractible. Therefore, there is no morphism $\Homeo_+(S^1) \to SO(2)$ which is a homotopy equivalence.

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