Finite-dimensional subgroups of diffeomorphism groups - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T20:52:01Z http://mathoverflow.net/feeds/question/69895 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/69895/finite-dimensional-subgroups-of-diffeomorphism-groups Finite-dimensional subgroups of diffeomorphism groups Squark 2011-07-09T18:03:42Z 2011-08-02T20:38:27Z <p>This question is a generalization of <a href="http://mathoverflow.net/questions/69810" rel="nofollow">my previous question</a> about the circle to arbitrary manifolds.</p> <p>Is there a smooth manifold M with the following property.</p> <p>There exists a sequence of connected finite-dimensional subgroups Gi of M's diffeomorphism group G such that</p> <p>(a) Gi is contained in Gj for i &lt; j</p> <p>(b) The union of Gi is dense in G</p> <p>To remove doubt, "finite dimensional subgroup of M's diffeomorphism group" means a Lie group H with smooth faithful action on M.</p> <p>The answer to my previous question established that S1 is not such a manifold. I suspect that the answer to the general question is still "no". However, the proof would have to be more sophisticated since in the case of S1 we had essentially a closed list of possible "Hs".</p> <p>There is another closely related question. Fix a smooth manifold M. Consider connected Lie groups H with faithful and transitive smooth action on M. Is there an upper bound of H's dimension? For S1 the answer was "yes, 3".</p> http://mathoverflow.net/questions/69895/finite-dimensional-subgroups-of-diffeomorphism-groups/69898#69898 Answer by Alex Eskin for Finite-dimensional subgroups of diffeomorphism groups Alex Eskin 2011-07-09T18:56:39Z 2011-07-09T19:55:09Z <p>EDIT: Correctly state Zimmer's conjecture. </p> <p>This does not really answer the question, but the question of which (higher rank) Lie groups act by diffeomorphism on which smooth manifolds is called the "Zimmer Program". In particular, Zimmer conjectured that if $n \geq 3$ and $\Gamma \subset SL(n,\mathbb{R})$ is a lattice, and $\Gamma$ acts by diffeomorphisms on a compact manifold $M$ then the dimension of $M$ is at least $n-1$.</p> <p>There is a close connection between actions of Lie groups, and actions of their lattices. In particular if $\Gamma$ acts on $X$, then you can let $Y = G \times X$ mod the diagonal action of $\Gamma$. The space $Y$ looks like a fiber bundle with base $G/\Gamma$ and fiber $X$. Then we have an natural action of $G$ on $Y$ (by acting on the first factor); this is called the induced action. </p> <p>There is an enormous amount of literature on various aspects of the Zimmer program, but in particular, as far as I know, the above conjecture is still open. For a fairly recent survey of what is known, see this paper by David Fisher:</p> <p><a href="http://arxiv.org/abs/0809.4849v2" rel="nofollow">http://arxiv.org/abs/0809.4849v2</a></p> <p>(For details on the induction construction, see section 2.3 of the survey)</p> <p>There is no hope to understand the actions of $SO(n,1)$, because you can find lattices in $SO(n,1)$ which surject onto free groups. Then you can choose an arbitrary action of the free group on $X$ and do the induction construction, see Theorem 2.7 in the survey. The best question is for actions of Lie groups of real rank at least 2. </p> http://mathoverflow.net/questions/69895/finite-dimensional-subgroups-of-diffeomorphism-groups/71925#71925 Answer by Robert Bryant for Finite-dimensional subgroups of diffeomorphism groups Robert Bryant 2011-08-02T20:14:13Z 2011-08-02T20:38:27Z <p>I guess that the answer to your first question is <em>no</em>, based on the following: If the union of the $G_i$ were dense in $G=\mathrm{Diff}(M)$, then, presumably, for $i$ sufficiently large, the action of $G_i$ would be primitive (i.e., it would not preserve any nontrivial foliation) and locally transitive. The list of the effective, primitive, transitive Lie group actions is known, and, by examining this list, one sees that the dimension of such a group acting on an $n$-manifold is at most $n^2{+}2n$.</p> <p>The answer to your second question depends on the manifold. First, I suppose you have to restrict to the case in which $M$ actually <em>has</em> a transitive smooth action of a finite dimensional Lie group. (For example, any compact orientable surface $M^2$ of genus $2$ or more is not a homogeneous space.) </p> <p>It is not hard to come up with cases for which there is no upper bound. For example, let $M = \mathbb{R}^2$ and consider the group $G_d$ that consists of transformations of the form $\phi(x,y) = \bigl(x{+}a,\ y{+}p(x)\bigr)$ where $a$ is any constant and $p$ is any polynomial of degree $d$ or less. Then $G_d$ acts transitively on $M$ for all $d\ge0$ while $\dim G_d = d+2$. Thus, for $M=\mathbb{R}^2$ the dimension of such $H$ can be arbitrarily high. A similar argument with trig polynomials will provide such an example on the torus, which is compact. (N.B.: The action of $G_d$ is not primitive since it preserves the foliation by the lines $x=c$; thus, this example does not contradict my first paragraph.)</p> <p>On the other hand, for $M=S^2$, there is an upper bound for the dimension of a connected Lie group that acts faithfully and transitively on $M$. That upper bound is 8 ($=2^2+2\cdot2$) and is achieved by $SL(3,\mathbb{R})$ acting on $S^2$ regarded as the space of oriented lines in $\mathbb{R}^3$. In fact, you can say more: Any connected, transitive finite dimensional Lie subgroup of the diffeomorphism group of $S^2$ is conjugate to one of $SO(3)$, $PSL(2,C)$, or $SL(3,\mathbb{R})$. The latter two are maximal and contain the first one as maximal compact. (The easiest proof that I know of these statements uses the classification of primitive actions, at least in dimension $2$.)</p>