A famous result of Sullivan (closely related to work of Wilkerson) says that the group of isotopy classes of diffeomorphisms of a simply-connected closed smooth manifold of dimension $\geq 5$ is commensurable with an arithmetic group. Sullivan gave an outline of its proof, and several later papers filled in more details treating some additional cases. These in particular give more of a hint about the required results from the theory of algebraic groups.

One such paper, Triantafillou's The Arithmeticity of Groups of Automorphisms of Spaces contains the following statement regarding a useful general result. It concerns maps from a short exact sequence of groups $$1 \to A \to A' \to A'' \to 1$$ to a short exact sequence of groups obtained by taking $\mathbb{Q}$-points of an extension of algebraic groups $$1 \to G \to G' \to G'' \to 1.$$

To complete the proof one needs an argument to the effect that the middle vertical map of a diagram of short exact sequences involving the $\mathbb{Q}$-points of algebraic groups is arithmetic if the outer maps are arithmetic and the kernel of the algebraic extension is unipotent. Here a map from a group into an algebraic group over $\mathbb{Q}$ is called arithmetic if its kernel is finite and its image is arithmetic. It appears that no such result exists in print nor is it folklore in the subject. Upon our inquiry, A. Borel furnished a proof of this algebraic fact which involves basic structure theorems of algebraic groups and the behaviour of lattices under the maps $\exp$ and $\log$ between the Lie algebra and the unipotent group.

However, she does not provide Borel's proof and instead uses a weaker result sufficient for her purposes. Is there a proof of this statement available in the literature somewhere?

  • $\begingroup$ In the 1st sentence, I guess "isotopy classes" means "isotopy classes of self-diffeomorphisms". $\endgroup$
    – YCor
    Commented Jul 4, 2019 at 15:58
  • 1
    $\begingroup$ When $G$ is a linear algebraic $\mathbf{Q}$-group, one way to define an arithmetic subgroup of $G_\mathbf{Q}$ is as a subgroup commensurate with $G_\mathbf{Z}$ (the latter being defined according to some choice of faithful representation, but the commensuration class doesn't). Here two subgroups of a groups are said to be commensurate if the intersection has finite index in both. Is it the intended definition? $\endgroup$
    – YCor
    Commented Jul 4, 2019 at 16:01
  • $\begingroup$ Thanks, and yes, that is the intended definition. $\endgroup$
    – skupers
    Commented Jul 4, 2019 at 16:23
  • $\begingroup$ Another technical point: if $1\to G\to G'\to G''\to 1$ is an exact sequence of $\mathbf{Q}$-groups, then $1\to G_\mathbf{Q}\to G'_\mathbf{Q}\to G''_\mathbf{Q}$ is exact, but the right-hand map can fail to be surjective. One might assume that it is surjective, but this drastically reduces the generality of the statement. Otherwise one might just ignore this, and just assume that exactness at the level of algebraic groups (i.e., surjectivity of the map between $\mathbf{C}$-points), which seems a natural assumption. $\endgroup$
    – YCor
    Commented Jul 4, 2019 at 17:07
  • $\begingroup$ I guess I would like the most general statement which is true; the paper is not very precise about what "algebraic fact" was proven. $\endgroup$
    – skupers
    Commented Jul 4, 2019 at 17:35

1 Answer 1


A proof appears in Section 2.3 of my paper Mapping class groups of manifolds with boundary are of finite type. As pointed out in the comments, it is crucial that $G$ is unipotent.


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