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Let $X$ be a topological space, and equip its homeomorphism group $\text{Homeo}(X)$ with the compact-open topology. I probably need to restrict $X$ to be locally compact for it to play off with $\text{Homeo}(X)$ nicely.

The usual definition of $X$ being homogeneous is that given $x,y \in X$ there exists $f \in \text{Homeo(X)}$ with $f(x) = y$. My question: When can this choice of $f$ be made continuous? More precisely, is there a natural class of homogeneous spaces for which there is always a continuous map $$\theta: \{(x,y): x \neq y\} \rightarrow \text{Homeo}(X)$$ such that $\theta(x,y)$ carries $x$ to $y$. Is it true for connected manifolds?

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  • $\begingroup$ I guess your question is sufficiently hard, so if you’ll not obtain a satisfying answer here then I recommend you to ask it at MathOverflow. $\endgroup$ – Alex Ravsky Nov 22 '17 at 8:17
  • $\begingroup$ I am not sure what "natural" would mean here, since this fails already for closed connected oriented surfaces of genus $\ne 1$. $\endgroup$ – Moishe Kohan Nov 24 '17 at 14:28
  • $\begingroup$ @MoisheCohen Could you explain or give a reference? Also did you mean genus >1? I think it's true for spheres in any dimension. $\endgroup$ – Cihan Nov 24 '17 at 18:12
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One of such natural classes are topological groups $G=X$ with $\theta(x,y)(t)=yx^{-1}(t)$ for each $t\in G$. Indeed, clearly that $\theta(x,y)(x)=y$ for each $x,y\in G$. To prove the continuity of the map $\theta: G\times G\to \operatorname{Homeo}(X)$ consider a subbase $\mathcal B$ of the space $\operatorname{Homeo}(X)$ consisting of sets $[K,O]=\{f\in\operatorname{Homeo}(X): f(K)\subset O,$ $K$ is a compact subset of $G$ and $O$ is an open subset of $G \}$. It suffices to show that $\theta^{-1}(B)$ is open for each $[K,U]\in\mathcal B$. Let $\theta(x,y)\in [K,O]$, that is $yx^{-1}K\subset O$. Continuity of operations of topological group imply that for each $t\in K$ there exists open neighborhoods $U_t$, $V_t$, and $W_t$ of points $y$, $x$, and $t$ respectively such that $U_t^{-1}V_tW_t\subset O$. A family $\{W_t:t\in K\}$ is an open cover of the set $K$. Since the set $K$ is compact, there exists a finite subset $F$ of $K$ such that $K\subset \{W_t:t\in F\}$. Put $U=\bigcap \{U_t:t\in F\}$ and $V=\bigcap \{V_t:t\in F\}$. Then $U^{-1}VK\subset O$, so $\theta(U,V)\in [K,O]$.

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You are asking the following: Pick $x\in X$ and for $G=Homeo(X)$ consider the orbit map $o_x: G\to X$, $o_x(g)=g(x)$. Assuming that $o_x$ is surjective, when does it have a (continuous) section $s: X\to G$?

A general observation is that if $X$ is not contractible, then no section $s: X\to G$ is homotopic to a constant map.

Here is what happens in the case when $X$ is a closed connected oriented 2-dimensional manifold.

  1. If $\chi(X)=0$, then $X\cong S^1\times S^1$ and, hence, is homeomorphic to a Lie group and hence, $o_x$ has a section.

  2. If $\chi(X)=2$, then $G=Homeo(X)$ is homotopy-equivalent to $O(3)$, hence, $\pi_2(G)=0$. However, if there is a section $s: X\to G$, then $\pi_2(G)\ne 0$, proving that a section does not exist.

  3. If $\chi(X)<0$ then $G$ is homotopy equivalent to ${\mathbb Z}$ (the identity component of $G$ is contractible). Hence, you cannot have a section $s: X\to G$.

I suspect that when $X$ is the $n$-dimensional sphere then a section $s: X\to G$ exists if and only if $n=1, 3, 7$, i.e. when $S^n$ has structure of a Lie group.

See

M.-E. Hamstorm, Homotopy groups of the space of homeomorphisms on a 2-manifold, Illinois J. Math. Volume 10, Issue 4 (1966), 563-573

for the details concerning homotopy types of the homeomorphism groups of compact surfaces. (The same conclusion applies when one works with the groups of diffeomorphisms, resp. PL homeomorphisms: Earle and Eells, 1966; resp. Scott, 1970.)

Edit: I just realized that for some reason, you are interested in the existence of a section which is continuous away from one point (I am not sure why though). This changes the answer. Namely, for $X=S^2$ (or sphere of any dimension for this matter), the map $\Theta$ exists, it is a rotation in the plane passing through $x, y$. On the other hand, if $\chi(X)<0$ the map $\Theta$ still does not exist since $X-\{x\}$ is not contractible and the argument in the case 3 still applies.

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  • $\begingroup$ I have something in mind about configuration spaces but I'm not sure I have the right formulation yet (it will probably be a more specialized question, for MO). I figured the full continuity would be too strong here and wondered about this case. $\endgroup$ – Cihan Nov 25 '17 at 3:53

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