Let $K$ be a compact oriented 3-dimensional handlebody of genus $g$. The group $H_g$ of isotopy classes of diffeomorphisms of $K$ is called the handlebody group. (It embeds as a subgroup of the mapping class group of the genus $g$ surface $\partial K$.) The fundamental group of $K$ is a free group of rank $g$, so there is a homomorphism

$H_g \to Out(F_g).$

I've been thinking about this homomorphism and its kernel, and I've come to suspect that the kernel is generated by Dehn twists around curves in $\partial K$ that bound discs in $K$. These elements are all clearly contained in the kernel, but do they generate the entire kernel?

Does anyone know of a reference, proof, or counter example?

  • $\begingroup$ May one ask why you are interested in this homomorphism? $\endgroup$
    – Sam Nead
    Apr 29, 2010 at 19:23
  • $\begingroup$ Well, I've been thinking about this map quite a lot recently in the context of modular operads. These sorts of groups make some very interesting operads. (For example, algebras over the operad of surface MCGs are infinite loop spaces by a theorem of Tillmann). These operads are related to topological conformal field theory and string topology and various other things. $\endgroup$ Apr 30, 2010 at 6:45

2 Answers 2


This result is due to Luft; see "Actions of the homeotopy group of an orientable 3-dimensional handlebody".

McCullough, in "Twist groups of compact 3-manifolds", proves that the twist group is not finitely generated and gives further references.

EDIT. Ninja'ed (well, at least the first sentence).


The answer is yes. I'm not sure who this is due to, but one nice reference is section 3.9 of Fomenko-Matveev's book "Algorithmic and Computer Methods for 3-Manifolds". It's not stated in precisely the form you stated it, but it can be easily extracted from there.

  • $\begingroup$ Thanks! After looking in Fomenko-Matveev, I managed to find Luft, Actions of the homeotopy group of an orientable 3-dimensional handlebody, Math. Ann. 234 (1978) 279-292 which seems to prove it. $\endgroup$ Apr 29, 2010 at 16:49

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.