Let $M$ be a $1$-connected, closed, smooth manifold with $dim(M)>4$ and let us set $MCG(M)=\pi_0(Diff(M))$. Dennis Sullivan proved that $MCG(M)$ is commensurable to an arithmetic group.

  • I was wondering if the same type of result holds for topological manifolds and $\pi_0(Homeo)$.
  • Let us consider the canonical morphism $i:Diff(M)\rightarrow Homeo(M)$, what do we know about the morphism induced on $\pi_0$?
  • And more generally about the homotopy groups of the homotopy fiber?
  • For which classical manifolds this homotopy fiber has been determined?
  • For spheres, this should be well known and related to exotic spheres, and what do we know about complex projective spaces?
  • $\begingroup$ In the proof of the Sullivan-Wilkerson theorem it is shown that the group of homotopy self-equivalences of such a manifold is commensurable to an arithmetic group. I learned about this from the survey article of Arkowitz on The Group of Self-Homotopy Equivalences. The group of self-homeomorphisms or diffeomorphisms embeds in this group as a subgroup. I am unfamiliar with the commensurability relation -- is this enough for you to draw conclusions about your first point? $\endgroup$ – James Schwass Oct 1 '15 at 12:20
  • $\begingroup$ James I do not see why $MCG(M)$ embeds into $\pi_0(hAut(M))$ (the connected components of the space of self homotopy equivalences), This is not the case of the $n$-torus (with $n>5$), do you have a reference for the simply-connected case? Your remark about the Sullivan-Wilkerson theorem is very good, because it is the first step to prove the arithmicity of $MCG(M)$, the second step is to use surgery theory. $\endgroup$ – David C Oct 1 '15 at 12:44
  • $\begingroup$ Ah, I was carelessly ignoring the distinction between the spaces of various self equivalences and the associated groups consisting of path components. $\endgroup$ – James Schwass Oct 1 '15 at 14:04
  • $\begingroup$ Have you found the answers to any of these questions since this was posted? $\endgroup$ – Mike Miller Jun 23 '16 at 20:55
  • $\begingroup$ Mike Miller, no I have not found any answer. $\endgroup$ – David C Jun 23 '16 at 22:05

Let me assume that M is at least 5-dimensional.

Sullivan's proof only uses surgery theory and properties of O(n) that also hold for Top(n), so the answer to your first question is yes.

Regarding your other questions: the homotopy fiber Homeo(M)/Diff(M) of the map BDiff(M) -> BTop(M) is subject of smoothing theory. There is a map from Homeo(M)/Diff(M) to the space of sections of the bundle over the d-manifold M with fiber Top(d)/O(d) which is associated to the frame bundle of M using the action of O(d) on Top(d)/O(d). This map is injective on path components and an isomorphism on all homotopy groups. This can be found for instance in the book of Kirby and Siebenmann.

Up to a question about components, this reduces the study of the homotopy fiber you asked about to understanding the homotopy type of Top(d)/O(d) (and the twist of the bundle), which is hard. The homotopy groups of Top(d)/O(d) are object of study in geometric topology for a long time and can be divided into three ranges:

  1. For $i\lesssim d$, we have $\pi_i(Top(d)/O(d))\cong \Theta_i$, where $\Theta_i$ is the group of homotopy spheres which is understood in terms of the stable homotopy groups of spheres.
  2. For $d \lesssim i\lesssim \frac{4}{3}d$, the groups $\pi_i(Top(d)/O(d))$ can in principle be understood in terms of Waldhausen's algebraic $K$-theory of spaces. Rationally, this can be used to compute that in this range $\pi_*(Top(d)/O(d))\otimes \mathbb{Q}$ vanishes for $d$ even and is isomorphism to $K_{*-d+1}(\mathbb{Z})\otimes \mathbb{Q}$ if $d$ is odd (that's a calculation of Farrell--Hsiang). The rational $K$-groups of the integers were computed by Borel: $$ K_{*}(\mathbb{Z})\otimes \mathbb{Q}=\begin{cases}\mathbb{Q}&\text{if }*\equiv 1 (4)\\0&\text{else }\end{cases}$$ in positive degrees.
  3. In the range $i \gtrsim \frac{4}{3}d$ very little is known, but some progress has been made in recent years. For instance, work of Watanabe (building on ideas of Kontsevich) and Weiss shows that there are plenty of nontrivial rational classes.

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