## Finite-dimensional subgroups of circle diffeomorphism group

Is there a sequence of connected finite-dimensional subgroups Gi of the circle diffeomorphism group G with the following properities:

(a) Gi is contained in Gj for i < j

(b) The union of Gi is dense in G

More rigorously "finite dimensional subgroup of circle diffeomorphism group" means a Lie group H with smooth faithful action on the circle.

In order to make sense of property (b) I have to specify a topology on G. I suspect that all reasonable topologies will yield the same answer, but for the sake of definiteness let's use the "sup-norm" topology. That is, given two diffeomorphism g1 and g2, I define the distance d(g1, g2) as

supremum over x in S1 of d(g1(x), g2(x))

Here the latter d is the usual distance on the circle. This is a metric and it induces a topology.

I suspect that the answer to my question is "no". Moreover, I suspect that there is no H as above with dimension > 3. But I might be wrong...

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Uou have Lie subgroups of arbitrary dimension, because you have action of $\mathbb R^k$. Choose $k$ disjoint subsets of the circle, in each of them consider a smooth vector field. Each of those vector fields generates an action of $\mathbb R$ and these actions commute with each other – Łukasz Grabowski Jul 8 2011 at 17:48
Good point, but it doesn't solve the main question – Squark Jul 8 2011 at 18:18
Also, I suspect that my guess regarding dimension 3 is correct if we restrict to analytic diffeomorphisms – Squark Jul 8 2011 at 18:25
This was first meant as a comment to an answer by Agol, but he deleted it before I was done typing: 1) analytic case: Given a Lie algebra $g\subset Vect(S^1)$ such that $g\cong sl_2\mathbb R$, the integrating Lie group will be an $n$-fold cover of $SL_2\mathbb R$ for some finite $n$. Moreover, any $n\ge 1$ can occur. 2) Smooth case: same story. Again, the universal cover of $SL_2\mathbb R$ does not embed in $Diff(S^1)$. 3) Differentiable case (but not $C^1$!): The universal cover of $SL_2\mathbb R$ can be made to act on $S^1$, in a way that fixes the complement of an interval. – André Henriques Jul 8 2011 at 22:15
@Andre: the argument I gave in the smooth case had a gap, so I deleted it. I may repost an answer when I have time to give a more complete argument. – Agol Jul 8 2011 at 22:40
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If $G$ is a connected Lie group acting transitively and faithfully on a connected smooth $1$-manifold, then $G$ is at most $3$-dimensional; in fact its Lie algebra embeds in that of $SL_2(\mathbb R)$. (Edit: Alex Eskin's answer says this in detail, with a reference.)

Each orbit of an action of a connected topological group on a $1$-manifold is either open or one point. Thus if $G_i$ has only one non-fixed orbit then $G_i$ is at most $3$-dimensional.

Let $F_i\subset S^1$ be the set of points fixed by the action of $G_i$. The intersection of all the $F_i$ is the set of points fixed by the union of the $G_i$, so surely it is empty if the union of the $G_i$ is dense. By compactness it follows that for all big enough $i$ $F_i$ is empty, making $G_i$ at most $3$-dimensional.

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The answer is indeed no, as described e.g. in the lecture notes by Ghys

http://www.math.ethz.ch/~bgabi/ghys%20groups%20acting%20on%20the%20circle.pdf

Section 4.1 has a list of all connected groups acting faithfully and transitively on the circle or the line. They are

1) $\mathbb{R}$ acting on itself,

2) the circle acting on itself,

3) the affine group of the line acting on the line, and

4) the k-fold cover of $PSL(2,\mathbb{R})$ acting on the circle.

Any faithful action of a connected Lie Group on the circle is made out of these: if $F$ is the set of fixed points, then on each connected component of the complement of $F$ the action must be conjugate to one of the above.

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