# Mapping class groups in high dimension

$$\DeclareMathOperator\MCG{MCG}\DeclareMathOperator\Diff{Diff}\DeclareMathOperator\Homeo{Homeo}$$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.

Edit: regarding A. Kupers' remark on commensurability, here is a nice note https://comptes-rendus.academie-sciences.fr/mathematique/item/CRMATH_2020__358_4_469_0/

• 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?
• 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. – David C Oct 1 '15 at 12:44
• It's worth pointing out that Sullivan did not prove that $MCG(M)$ is commensurable with an arithmetic group, if we use the modern definitions of these words: comptes-rendus.academie-sciences.fr/mathematique/item/… – skupers Feb 1 at 0:19
• I disagree that this is the "modern definition of these words". As far as I know, commensurability has always meant the larger equivalence relation -- what the authors of the note call "commensurability up to finite kernels". However, in many settings -- specifically groups that are "good in the sense of Serre" -- the smaller equivalence relation and the larger one coincide, so the distinction is often not important. Of course it is important for arithmetic groups, since Deligne's examples are both residually finite and not "good in the sense of Serre". – HJRW Feb 1 at 22:13
• A google search indicates that most authors define two groups to be (abstractly) commensurable if they admit isomorphic finite index subgroups, see for instance Morris' introductory book on arithmetic groups. In particular, goodness is preserved by commensurability (see e.g. Lemma 3.2 of Grunewald et al's 'Cohomological Goodness and the profinite completion of Bianchi groups'). – archipelago Feb 2 at 12:39
• More recent works seem to follow this as well, such as sciencedirect.com/science/article/pii/S0021782417300624 – archipelago Feb 2 at 13:02

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 by Kervaire--Milnor.
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: $$K_{*}(\mathbb{Z})\otimes \mathbb{Q}\cong \begin{cases}\mathbb{Q}&\text{if }*\equiv 1 (4)\\0&\text{else }\end{cases}$$ 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.

Later edit: The identification of $$\pi_i(Top(d)/O(d))\otimes\mathbb{Q}$$ in the range $$i\lesssim \frac{4}{3}d$$ mentioned above holds in fact for $$i\lesssim 2d$$ by Corollary 4.2 of Randal-Williams and Corollary C of Krannich. Beyond this range, progress in even dimensions has been made by Kupers--Randal-Williams.