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...

  • 3
    $\begingroup$ 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 $\endgroup$ Jul 8 '11 at 17:48
  • $\begingroup$ Good point, but it doesn't solve the main question $\endgroup$
    – Vanessa
    Jul 8 '11 at 18:18
  • $\begingroup$ Also, I suspect that my guess regarding dimension 3 is correct if we restrict to analytic diffeomorphisms $\endgroup$
    – Vanessa
    Jul 8 '11 at 18:25
  • 1
    $\begingroup$ 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. $\endgroup$ Jul 8 '11 at 22:15
  • $\begingroup$ @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. $\endgroup$
    – Ian Agol
    Jul 8 '11 at 22:40

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.


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


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.

  • $\begingroup$ The infinite cover of $SL(2,R)$ acting on the line (infinite cover of $P^1$) is missing to this list. $\endgroup$
    – YCor
    Jan 24 '19 at 4:13
  • $\begingroup$ The link is broken. $\endgroup$
    – YCor
    Jan 24 '19 at 4:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.