"Largest" finite-dimensional Lie subgroups of Diff(S^n), are they known? - MathOverflow

"Largest" finite-dimensional Lie subgroups of Diff(S^n), are they known?

2010-06-13T18:55:31Z

The group $Diff(S^n)$ ($C^\infty$-smooth diffeomorphisms of the $n$-sphere) has many interesting subgroups. But one question I've never seen explored is what are its "big" finite-dimensional subgroups?

For example, $Diff(S^n)$ contains a finite-dimensional Lie subgroup of dimension $n+2 \choose 2$, the subgroup of conformal automorphisms of $S^n$. Similarly it contains a compact Lie subgroup of dimension $n+1 \choose 2$, the isometry group of $S^n$.

Is it known that:

1) A finite-dimensional Lie subgroup of $Diff(S^n)$ having dimension at least $n+2 \choose 2$ is conjugate to a subgroup of the conformal automorphism group of $S^n$ ? (Answer, no, see Algori's answer below). Modified question: As Algori notes, $GL_{n+1}(\mathbb R) / \mathbb R_{>0}$ acts on $S^n$ and has dimension $n^2+2n$. So is a finite-dimensional Lie subgroup of $Diff(S^n)$ of dimension $n^2+2n$ (or larger) conjugate to a subgroup of this group?

2) A compact Lie subgroup of $Diff(S^n)$ having dimension at least $n+1 \choose 2$ is conjugate to a subgroup of the isometry group of $S^n$ ? (Answer: Yes, see Torsten Ekedahl's post below)

For example, arbitrary compact subgroups of $Diff(S^n)$ do not have to be conjugate to subgroups of the above two groups -- perhaps the earliest examples of these came from exotic projective and lens spaces. But I have little sense for how high-dimensional these "exotic" subgroups of $Diff(S^n)$ can be.

Answer by Torsten Ekedahl for "Largest" finite-dimensional Lie subgroups of Diff(S^n), are they known?

2010-06-13T19:11:46Z

For 2) I think the following is an answer: Suppose $K$ is a compact Lie subgroup of $\mathrm{Diff}(S^n)$ of dimension $\geq{n+1\choose 2}$. Being compact it is the group of isometries of some Riemannian metric of $S^n$ and we fix one such metric. The stabiliser of a point therefore has dimension at most $n\choose 2$ (the dimension of the orthogonal group of $\mathbb R^n$) and as an orbit has at most dimension $n$, $K$ has at most dimension $n+{n\choose 2}={n+1\choose 2}$ and hence we have equality. This means that the stabiliser contains $\mathrm{SO}_n$ and $K$ acts transitively. In particular the metric fixed by $K$ has constant curvature and then the curvature is positive and thus up to a constant conformal factor is conjugate to the standard metric. This gives a conjugation of $K$ into the isometry group of the standard sphere.

Addendum: Tom's example gives that the dimension of a non-compact group acting (faithfully) on $S^n$ is unbounded. One question seems to remain namely if the dimension of a semi-simple group action is bounded or not. By the above we get a bound on the dimension of a maximal compact subgroup. In many cases this seems to bound the dimension of the group itself. For instance, is the dimension bounded if the center is finite? In that case the situation is completely described by a Cartan decomposition of the Lie algebra and the question is whether the $-1$-part (usually denoted $\mathfrak p$) of it has dimension bounded by the dimension of the $+1$-part (the Lie algebra of the maximal compact). I myself don't know enough of the real Lie group theory to decide that.

An example beyond that is the universal cover $G$ of $\mathrm{SL}_2(\mathbb R)$. It is contractible and has no non-trivial compact connected subgroup. It also acts on the universal cover of $S^1$ (compatibly with the projective action of $\mathrm{SL}_2(\mathbb R)$ on $S^1$) and hence acts continuously on $[-1,1]$ (say) fixing the endpoints. However, the action at the endpoints is not flat so this action does not extend to a smooth action on $\mathbb R$ acting as the identity outside if $[-1,1]$. If it could be modified to do so one could use Tom's argument to get an action of any finite product of copies of $G$ (at least on $S^1$). One could try to find three vector fields with support in $[-1,1]$ fulfilling the defining relations of $\mathfrak{sl}_2(\mathbb R)$ but it looks tricky to me.

Addendum 1: To be more precise about the action on $[-1,1]$ we may use $\tanh$ as diffeomorphism from $\mathbb R$ to $(-1,1)$. The rotation group of $\mathrm{SL}_2(\mathbb R)$ lifts to the group of translations of $\mathbb R$ and they correspond under this diffeomorphism to $\varphi_\lambda(t)=(\lambda t+1)/(t+\lambda)$ of $(-1,1)$ for $\lambda>1$ or $\lambda<-1$ (we miss the identity map which corresponds to $\lambda=\pm\infty$). Now, one possibility of getting an action of $G$ on $\mathbb R$ which is the identity outside of $(-1,1)$ is by trying to conjugate the one we have by a diffeomorphism $\gamma$ of $(-1,1)$. This requires every conjugate diffeomorphism to be flat at $\pm1$ where a diffeomorphism $\psi$ of $(-1,1)$ is flat at $1$ if $\psi(x)=x+\mathcal{O}(|x-1|^n)$ for all $n$ and $x$ close to $1$ (and similarly for $-1$). Now suppose that we have a $\gamma$ such that conjugating the given action of $G$ by it gives an action all of whose elements are flat. In particular we have $\gamma(\varphi_\lambda(\gamma^{-1}(x)))=x+\mathcal{O}(|x-1|^n)$ for all $n$. In particular, putting $n=2$ we get that $\lim_{t\to1}\gamma'(\varphi_\lambda(\gamma^{-1}(t)))\varphi_\lambda'(\gamma^{-1}(t))\gamma'(\gamma^{-1}(t))^{-1}=1$ and putting $s=\gamma^{-1}(t)$, using that $\lim_{t\to1}\gamma^{-1}(t)=1$ and that $\lim_{s\to1}\varphi_\lambda'(s)=\varphi_\lambda'(1)=(\lambda-1)/(\lambda+1)$ this gives $\lim_{s\to1}\gamma'(\varphi_\lambda(s))/\gamma'(s)=(\lambda+1)/(\lambda-1)$. This puts a very stringent condition on $\gamma$ and then also the same condition must be fulfilled for $\varphi_\lambda$ replaced by any element of $G$ (as well as needing the flatness condition for all orders $n$).

Answer by algori for "Largest" finite-dimensional Lie subgroups of Diff(S^n), are they known?

2010-06-13T19:33:09Z

The quotient of $GL_{n+1}(\mathbf{R})$ by the positive scalars acts on $S^n$; it has dimension $n^2+2n$, so for $n>1$ the answer to the first question is no. For $n=3$ an alternative proof would be as follows: there are 3-manifolds that admit a projective structure, but no M\"obius structures, as explained in Agol's answer here: http://mathoverflow.net/questions/23567/mobius-and-projective-3-manifolds.

By the way, the answer seems to have been updated since I last looked there and it now contains an argument which can give a classification of maximal Lie groups acting smoothly and faithfully on manifolds of given (finite) dimension.

Answer by Tom Goodwillie for "Largest" finite-dimensional Lie subgroups of Diff(S^n), are they known?

2010-06-14T01:57:37Z

You can make big Lie groups act effectively on small manifolds by cheating: make the group a product of groups, with each factor acting by compactly supported diffeomorphisms on a different disjoint open subset. So the additive group $\mathbb R^N$ becomes a subgroup of $Diff(S^1)$ by flowing along $N$ commuting vector fields supported in $N$ disjoint arcs.

(added later) A similar cheat: let $a_1,\dots ,a_N$ be linearly independent functions of one variable. Then $a_1(x)\frac{\partial}{\partial y},\dots ,a_N(x)\frac{\partial}{\partial y}$ are independent commuting vector fields in the $x,y$ plane. Modify this example to make it compactly supported if you like.

(added still later) My proposed extension of the first cheat to a semisimple group (comment thread of Torsten's answer) is doomed: Choose a point on the circle and choose an element of SL_2(R) that fixes this point and acts on the tangent space there with eigenvalue c>1. Lifting the group element to the universal covering group in the right way, you get an element g of the latter group that fixes all the points above the given point in the universal covering space of the circle, in each case with eigenvalue c. But now if this line with this action could be embedded in a longer line with trivial action outside then there would be a sequence of fixed points of g with eigenvalue c converging to a fixed point of g with eigenvalue 1, contradiction.