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Since I did not quite get the responses I would have liked when I asked this question four months ago, let me reformulate it slightly:

What are some examples of $\text{Isom}(M)$ and $\text{Conf}(M)$?

For example, Aaron mentions in his (very helpful) answer that $\text{Isom}(\mathbb{S}^n) \cong O(n+1)$. What about hyperbolic n-space? Or the n-torus? The more examples the better.

For precision, I am hoping that we can express $\text{Isom}(M)$ or $\text{Conf}(M)$ as some sort of "recognizable" Lie group, by which I mean a product, quotient, and/or connected sum of linear groups, euclidean spaces, or spheres.

Original Question: (What are some examples of automorphism groups of manifolds which turn out to be Lie groups?)

I recently read that the group of diffeomorphisms of a smooth manifold which preserve some sort of geometric structure (e.g. Riemannian structure, conformal structure, etc.) frequently turn out to be a Lie group. What are some examples of this?

I've read about the euclidean group $E(n)$ which are the isometries of $\mathbb{R}^n$, and also about the conformal automorphisms of the complex plane, upper half plane, and unit disc. What others are there? Do the groups Diff(M), Iso(M), etc. frequently turn out to be something recognizable? What are the isometry groups or conformal groups of n-spheres, say, or some common 2-manifolds?

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At least when we're talking about diffeomorphism groups, these often turn out to be infinite dimensional... and then you have to ask questions like "What do we mean by an infinite dimensional Lie group?" –  Dylan Wilson Dec 25 '10 at 10:42
The conformal example generalizes: Automorphisms which preserve a complex sructure $J$ on $M$ (i.e. biholomorphisms of the complex manifold $X = (M,J)$) form a Lie group, which is complex as soon as $M$ is compact. –  Gunnar Magnusson Dec 26 '10 at 11:10
Generally speaking it's a very rare event for the diffeomorphism group of a manifold to have the homotopy-type of a Lie group (or any finite-dimensional CW-complex). The reference is: Antonelli, P.L.; Burghelea, D.; Kahn, P.J. The nonfinite type of some $Diff\sb 0M\sp n$. (English) [J] Bull. Am. Math. Soc. 76, 1246-1250 (1970). –  Ryan Budney Apr 10 '11 at 21:48

2 Answers 2

up vote 4 down vote accepted

Dylan's right that diffeomorphism groups are usually infinite dimensional. There's a theory of infinite dimensional Lie groups, I've heard, but I think it's a lot harder to work out things like convergence. Worth noting, though, is that in this analogy, if $Diff(M)$ is a Lie group then its Lie algebra should be $Vect(M)$, the (infinite-dimensional) vector space of smooth vector fields on $M$. The Lie algebra bracket is just the usual vector field bracket (possibly up to a sign? but if so, then it's only because of stupid clashing conventions). Heuristically, to look for finite dimensional Lie groups we're going to want to put some more structure on $M$ and require that our diffeomorphisms preserve this. If you're looking at things that aren't so homogeneous you get into the territory of $G$-bundles, which I think you should definitely look into if you don't already know about -- you don't need to know too much Lie theory to get the basics, and it's really cool (I think you'll like it too). I don't have a reference offhand (please, anyone, post as a comment if you do) but I took an excellent class on them from Robert Bryant last year and a friend of mine is working on TeXing up the notes, so hopefully those will be available (at least in part) before too long.

By the way, $Aut(M)/$isotopy is called the mapping class group. This can be written $MCG(M)=Aut(M)/Aut_0(M)$, where $Aut_0(M)$ is the connected component of the identity map, which if you think about it is all isotopies are anyways. This has been extensively studied. I don't remember much from a class I took with Jeff Brock a few years ago, but checking wikipedia reminds me that e.g. $MCG(\mathbb{T}^n)\simeq GL(n,\mathbb{Z})$. It also says that the MCGs of surfaces have been extensively studied. In any case, you can see that this is just the group $\pi_0(Aut(M))$, so assuming you understand $MCG(M)$ (!) then you only need to figure out $Aut_0(M)$ (!) to have the whole picture. Which brings us back to the original question.

I don't know much, but like I said it's definitely easier (and finite-dimensionaler) if we restrict ourselves to manifolds with structure. We can start by saying something simple: $Isom(S^n)=O(n+1)$. One potentially fruitful way to think about this is to just start with a (matrix) group and find an invariant subspace of the (affine) space on which it acts, hopefully freely -- very often this should be a manifold, I think. Otherwise things will probably be significantly more difficult. If you put the flat metric on a torus and assume its fundamental domain is a square, then you should be able to look at isometries by what they do to your fundamental square sitting on a latticed plane. To represent an isometry it'll need to be an isometry of the square, and to be a continuous map you'll need the four corners to be mapped to preimages of the same point under $\mathbb{R}^2\rightarrow \mathbb{T}^2$. I'm pretty sure this ends up meaning $Isom_0(\mathbb{T}^2)\cong \mathbb{T}^2$. But maybe that's not so surprising, since it's a group itself. I guess all it's really telling you is that there are no isometries other than the "multiply-by-$g$" map for $g\in \mathbb{T}^2$. The group $Conf(\mathbb{T}^2)$ should be pretty different, but I think you can't continuously dilate because that would eventually change the degree of your map (which is a discrete invariant), so a similar analysis should go through but you won't get any interesting Lie-type behavior. More interesting might be to try this all with the fundamental domain for an $n$-holed torus ($n\geq 2$) sitting on a latticed hyperbolic plane. But it's late and I'm tired, so I'm not going to. You should though! If you do, let me know what you come up with.

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Oops, what I said about $Conf(\mathbb{T}^2)$ is probably wrong -- I was imagining that everything got dilated by the same scale factor, which is obviously not necessarily true. In any case, I don't really know how to get a handle on this. –  Aaron Mazel-Gee Dec 26 '10 at 11:38
By any chance did your friend ever finishing TeXing up Bryant's notes? I've seen Bryant give several talks, and he always does a masterful job. (Also, he's now become active on Mathoverflow and has posted many beautifully written answers.) –  Jason DeVito Feb 6 '13 at 19:58
Also, in your last paragraph, it's not generally true that (identity component of) the isometry group of a homogeneous space $G/H$ is $G$. For example, if you pick the correct metric on $SU(n)$, then $SU(n)/SU(n-1) \cong S^{2n-1}$ has isometry group $O(2n)$, strictly bigger than $SU(n)$. (Also, for the case of a Lie group, the identity component of the isometry group (with a choice of bi-invariant metric) is usually $(G\times G)/ \Delta Z(G)$, so is usually much bigger than $G$. The case of $G = \mathbb{T}^2$ is works out as you said because $Z(\mathbb{T}^2) = \mathbb{T}^2)$. –  Jason DeVito Feb 6 '13 at 20:08
Yeah, Bryant is great. I don't think the texing job ever got even close to finished, but in any case I have some pretty carefully handwritten notes in pdf format -- email me if you're interested. Also, thanks for the correction. I haven't thought about this stuff in a long time (probably about 25 months or so). –  Aaron Mazel-Gee Feb 9 '13 at 7:24

$\text{Isom}(\mathbb{S}^n) \cong O(n+1)$

$\text{Isom}(\mathbb{R}^n) \cong E(n)$

$\text{Isom}(\mathbb{H}^n) \cong O(n,1)$

$\text{Conf}(\mathbb{S}^2) \cong \text{PGL}(2,\mathbb{C}) \cong \text{PSL}(2,\mathbb{C})$

$\text{Conf}(U) \cong \text{PSL}(2,\mathbb{C})$

Sources: "Riemannian Manifolds: An Introduction to Curvature" (Lee) (Chapter 3), and my own understanding of Wikipedia (and my complex analysis lecture notes).

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Did you forget a quotient in the fourth one? –  Alexei Averchenko Apr 14 '11 at 9:08
@Alexei: Did I? Doesn't the "P" in "PGL" or "PSL" already imply a quotient of "GL" (or "SL" respectively)? Or am I missing something silly? –  Jesse Madnick Apr 14 '11 at 17:28

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