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Let $\Sigma$ be an oriented topological surface. For simplicity, assume that the genus of $\Sigma$ is at least $2$. There are a number of classical results on the homotopy types of various groups of self-maps of $\Sigma$:

1) Earle and Eells proved that the components of $\text{Diff}(\Sigma)$ are contractible.

2) Hamstrom proved that the components of $\text{Homeo}(\Sigma)$ are contractible.

3) Peter Scott proved that the components of $\text{Homeo}^{\text{PL}}(\Sigma)$ are contractible, where by $\text{Homeo}^{\text{PL}}(\Sigma)$ we mean the group of PL self-homeomorphisms of $\Sigma$.

Of course, in 1 and 3 we are fixing a $C^{\infty}$ or $\text{PL}$ structure on $\Sigma$, respectively.

Another standard fact is that every self homotopy-equivalence of $\Sigma$ is homotopic to a homeomorphism. This leads me to my question. Denote by $\text{HE}(\Sigma)$ the set of self homotopy-equivalences of $\Sigma$. Are the components of $\text{HE}(\Sigma)$ contractible?

Another related question is as follows. There is a beautiful alternate proof of the above theorem of Earle and Eells due to Earle and McMullen (see their paper "Quasiconformal Isotopies"; their proof uses complex analysis). The proofs of Hamstrom's theorem and Scott's theorem are very complicated -- are there any alternate approaches to them in the literature?

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A couple comments. For the result about diffeomorphism groups there is a very nice alternative proof due to A. Gramain in the Annales Scient. E.N.S. v.6 (1973), pp. 53-66, that uses no analysis, just basic differential topology. Another approach, which I'm not sure is written down anywhere in detail, is to take the proof of the corresponding result for Haken 3-manifolds (due independently to Ivanov and myself) and scale it back from 3-manifolds to surfaces. This too uses no analysis, just basic topology. I don't recall Scott's method for the PL case, but it might be similar to this.

For the result about homotopy equivalences it's best to look first at the space of homotopy equivalences that fix a basepoint. This has contractible components for any $K(\pi,1)$ space, by an elementary obstruction theory argument. (Just deform families of maps to the identity map, cell by cell, which is possible since the obstructions lie in the higher homotopy groups of the space.) By evaluating arbitrary homotopy equivalences at the basepoint one gets a fibration where the total space is the space of all homotopy equivalences, the base space is the $K(\pi,1)$ space, and the fiber is the earlier space of basepoint-preserving homotopy equivalences. Looking at the long exact sequence of homotopy groups for this fibration then gives the desired result. Triviality of the center of $\pi$ comes in at this point to show that the boundary map from $\pi_1$ of the base to $\pi_0$ of the fiber is injective.

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  • $\begingroup$ Thanks Allen! One beautiful property of the Earle-McMullen contraction of Diff_0 is that it "preserves symmetries" of the surface in the following sense. Let F_t:Diff_0-->Diff_0 be the E-M contraction, so F_0=id and F_1 is the constant map to the identity. Let f in Diff_0 commute with a finite subgroup G of Diff. Then F_t(f) will commute with G for all t. Do any of the other approaches to contacting Diff_0 have similar properties? $\endgroup$
    – Andy Putman
    Commented Mar 19, 2010 at 4:05
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    $\begingroup$ IMO "no analysis" is a tad misleading. You use Sard's theorem in a very significant way. $\endgroup$
    – Ryan Budney
    Commented Mar 19, 2010 at 8:09
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    $\begingroup$ Compared to the other proofs, there isn't that much analysis! The two proofs I mentioned both use the measurable Riemann mapping theorem (ie the solution to the Beltrami equation with measurable inputs). Also, the proof of Earle-Eells uses the contractibility of Teichmuller space, while the proof of Earle-McMullen uses the Douady-Earle extension theorem. Compared to all that, Sard's theorem is a triviality! $\endgroup$
    – Andy Putman
    Commented Mar 19, 2010 at 14:33
  • $\begingroup$ If I understand the Gramain construction Hatcher is referring to you also have to use Smale's theorem that Diff(S^2) has the homotopy-type of O(3). The main analytical ingredient is the Poincare-Bendixson theorem. $\endgroup$
    – Ryan Budney
    Commented Mar 19, 2010 at 17:16
  • $\begingroup$ Gramain's argument may use Smale's theorem, but there's a simple proof of Smale's theorem (without Poincaré-Bendixson) in the appendix to Cerf's "yellow book", SLN v.53, the $\Gamma_4=0$ theorem. Besides, isn't the Poincaré-Bendixson theorem a fairly simple topological argument: If a nonsingular flow in a square is vertical near the boundary, then every trajectory has to go from the bottom edge to the top edge, otherwise there would be an infinite trajectory spiraling close to itself somewhere, and this could be perturbed to a closed orbit which would have to enclose a singularity. $\endgroup$
    – Allen Hatcher
    Commented Mar 19, 2010 at 22:01
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The connected oriented surface $\Sigma_g$ of genus $g \ge 1$ is a $K(G_g, 1)$ where $G_g$ has a well-known presentation. For a general group $G$, the mapping space $Map(K(G, 1), K(G, 1))$ has homotopy type $$ \coprod_{f:G \to G} B(Z(f(G))) $$ where $f$ ranges over group endomorphisms of $G$ and $Z(f(G))$ denotes the centralizer of the image of $f$. In your case, you are interested in the case where $f$ is surjective, so the question reduces to whether $G_g$ has trivial center for $g \ge 2$, which I assume it does.

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  • $\begingroup$ Very nice! The center is indeed trivial for g at least 2. Do you know a nice reference for this result? $\endgroup$
    – Andy Putman
    Commented Mar 19, 2010 at 1:47
  • $\begingroup$ Which, the statement about the center? No, it just seems overwhelmingly likely due to the results 1-3, and I imagined it was a "simple matter of algebra", but didn't think about whether it's actually easy to prove. $\endgroup$
    – Reid Barton
    Commented Mar 19, 2010 at 1:52
  • $\begingroup$ No, I can do the group theory <grin>. I mean the statement about the mapping space that you quoted. $\endgroup$
    – Andy Putman
    Commented Mar 19, 2010 at 1:52
  • $\begingroup$ Oh. I don't have a reference offhand, though I'm sure someone could provide one. The way I would do the mapping space computation is to check that we can do it in groupoids, where it simply comes down to the definition of functor and natural transformation. To do this I think you just need the fact that groupoids = 1-truncated spaces is a reflexive sub-$(\infty,1)$-category of spaces. $\endgroup$
    – Reid Barton
    Commented Mar 19, 2010 at 1:59
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    $\begingroup$ I don't know if this helps, but you can interpret Reid's answer as a combination of (a refinement of) classifying spaces + covering space theory: Say $X$ is a pointed connected space. Then, $Map(X, K(G,1))$ turns out to be homotopy equivalent to the geometric realization of the category of $G$-bundles on $X$ (the $\pi_0$ statement is the usual classifying space story). And, covering space theory tells us this is equivalent to the category whose objects are homomorphisms $\pi_1 X \to G$, and whose morphisms are given by conjugation on $G$. Now plug in $X = K(G,1)$ to this computation. $\endgroup$
    – Anatoly Preygel
    Commented Mar 19, 2010 at 2:52
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Hi Andy,

Here is a proof for the case with marked points (see below for some ideas for the case of closed surfaces).

Proof: straight-line homotopy.

Less tersely: let $HE_0(\Sigma,\ast)$ be the identity component of the monoid of self-homotopy equivalences of $\Sigma$ fixing the basepoint; in particular, each $f\in HE_0(\Sigma,\ast)$ is homotopic rel $\ast$ to the identity. Fix a hyperbolic metric on $\Sigma$ and thus an identification of the universal cover $\widetilde{\Sigma}$ with the hyperbolic plane $\mathbb{H}^2$, and a basepoint $\ast$ in $\mathbb{H}^2$.

Each $f\colon \Sigma\to \Sigma$ has a unique lift to $\mathbb{H}^2$ fixing the basepoint. because $f$ acts trivially on $\pi_1(\Sigma)$, $f$ commutes with the deck transformations. Thus we may take the straight-line homotopy $f_t(x)=tx-(1-t)f(x)$, where by this convex combination I mean to move with unit speed along the geodesic from $f(x)$ to $x$. Since the deck transformations act by isometries on $\mathbb{H}^2$, this homotopy descends to $\Sigma$; each $f_t$ is still a homotopy equivalence. We can perform this straight-line homotopy for all $f$ simultaneously; since the lifts of $f$ are uniformly continuous, this homotopy is continuous on $HE_0(\Sigma,\ast)$ and gives a contraction to the identity.

There must be some work needed to get from this to the case for closed surfaces, because this proof works for a genus 1 surface with marked point, and of course $HE_0(T^2)$ is homotopy equivalent to $T^2$ itself. But it seems to me like a reduction should be possible; I think the important thing is that $\pi_1(\Sigma)$ is centerless.

Acknowledgement: I learned this idea from Rita Jimenez Rolland, based on conversations she had with Mladen Bestvina about the related case of $\text{Aut}(F_n)$.

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  • $\begingroup$ Thanks Tom! That's definitely the right proof for marked surfaces. $\endgroup$
    – Andy Putman
    Commented Mar 19, 2010 at 3:06

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