Given a homomorphism of a group $G$ to the mapping class group of a manifold $M$, is there any condition that guarantees that it is defined by an action of $G$ on $M$? Thank you very much.
$\begingroup$
$\endgroup$
4
-
2$\begingroup$ Sounds a bit general. Even the case when $G$ is the mapping class group and the homomorphism is the identity is unclear. $\endgroup$– YCorCommented Feb 13, 2017 at 0:56
-
1$\begingroup$ As mentioned several times by other users, the definition of MCG is unclear in higher dimension (self-homeomorphisms or self-diffeomorphisms? what kind of isotopies?...) $\endgroup$– YCorCommented Feb 13, 2017 at 2:37
-
$\begingroup$ I'm mainly interested in the differentiable setting: self-diffeomorphisms and diffeotopies, but the continuous setting would be also interesting. The question is very general because I don't know any concrete situation with some hope of an answer. $\endgroup$– Jesús ÁlvarezCommented Feb 13, 2017 at 11:55
-
2$\begingroup$ What you are asking is a standard symmetry question. If you think of the mapping class group in terms of diffeomorphisms, you are asking the lifting problem for subgroups of $\pi_0 Diff(M)$, under the projection map $Diff(M) \to \pi_0 Diff(M)$. Generally you need to know something about the manifold to answer questions about this problem, there's little general-nonsense to rely on. Take a look at Sakuma's work on knot exteriors in $S^3$, it's quite beautiful. There is of course plenty to work with in low dimensions. $\endgroup$– Ryan BudneyCommented Feb 26, 2017 at 23:02
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
I'm not sure if this is exactly what you were asking about, but one way of parsing your question is to ask when a homomorphism from $G$ to $\operatorname{MCG}(M)$ is induced from a homomorphism $G \to \operatorname{Diff}(M)$ (or replace Diff with your favorite alternative). This is a very general (and very hard!) question, even in the case of $M$ a surface, where it goes by the name Nielsen realization problem. There's a lot of literature on this, so I'll give you some of the highlights.
The term "Nielsen realization problem" comes from the work of Nielsen in the 20's, who showed that if $G$ is a finite cyclic group, every homomorphism of $G$ into $\operatorname{Mod}(\Sigma_g)$ can be promoted into an action by homeomorphisms (indeed, all the way up to a group of isometries with respect to a hyperbolic metric; equivalently a group of automorphisms of some Riemann surface). In the '80s, Kerckhoff showed that the same result holds when $G$ is any finite group.
In the opposite direction, Morita showed that the full mapping class group of $\Sigma_g$ cannot lift to $\operatorname{Diff}(\Sigma_g)$. There's now zillions of proofs of this, but one nice thing about Morita's argument is that he produces cohomological obstructions for a subgroup of $\operatorname{Mod}(\Sigma_g)$ to lift to $\operatorname{Diff}(\Sigma_g)$. Precisely, he shows that the MMM classes $e_i$ for $i \ge 3$ obstruct such a lift, in the sense that if $e_i \ne 0$, no such lift exists.
This is a beautiful thread of ideas, but it doesn't come close to being a complete answer; in particular this doesn't say anything in the case that $G$ has cohomological dimension $5$ or lower. There has been some work dedicated to approaching this problem from a more dynamical point of view; see especially the work of Franks and Handel giving a proof of Morita's theorem using dynamics.
In general, this problem is really hard! We don't even know if every homomorphism from a surface group (i.e. $G \cong \pi_1(\Sigma_h)$) into $\operatorname{Mod}(\Sigma_g)$ admits a lift to $\operatorname{Diff}(\Sigma_g)$...
-
1$\begingroup$ "$Mod(\Sigma_g)$ cannot lift to $Diff(\Sigma_g)$" is not the same as "$Mod(\Sigma_g)$ cannot act by diffeomorphisms on that surface. What do you mean by the latter? I'm ready to believe that $Mod(\Sigma_g)$ has no injective homomorphism into $Homeo(S_{g'})$ for any $g'$, but I'm not sure you're claiming this. $\endgroup$– YCorCommented Feb 27, 2017 at 5:07
-
$\begingroup$ Yes, that's a good point. I'll edit my post to make this clearer. $\endgroup$ Commented Feb 27, 2017 at 15:33