# Heegaard splitting, equivalent homeomorphisms, mapping class group of genus n-torus

Given a Heegaard splitting of genus $n$, and two distinct orientation preserving homeomorphisms, elements of the mapping class group of the genus $n$ torus, is there a method which shows whether or not these homeomorphisms, when used to identify the boundaries of the pair of handlebodies, will produce the same $3$-manifold?

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Yes: 1. You obtain the same Heegaard splitting iff your homeomorphisms are in the same double coset H\M/H where H is the subgroup of mapping classes that extend to the handlebody. 2. Reidemeister-Singer theorem says that two Heegaard splittings of the same manifold are the same up to stabilisation. – Maxime Bourrigan Mar 3 '11 at 11:16
Thanks a lot for that. – user6204 Mar 3 '11 at 12:05
Hi Maxime, can you please give me a reference for the first part of your answer ("You obtain the same Heegaard splitting iff your homeomorphisms are in the same double coset H\M/H where H is the subgroup of mapping classes that extend to the handlebody.") Moreover, I can't catch which is the relation between saying that two splitting give the same manifold (up to homeomorphism) and saying that two splitting are equivalent splittings for the same manifold. Someone can help? thanks in advance – Lor Apr 20 '11 at 7:48
I found a reference, look at page 134 of "Problems on mapping class groups and related topics", by Benson Farb, available on Google Books. It turns out that there may exist Heegaard splittings that are not in the double coset. It also explains what is meant by equivalent splittings. – user6204 Apr 21 '11 at 13:18
I added the tags "3-manifolds" and "geometric topology" because I had missed this question. – Daniel Moskovich Apr 24 '11 at 12:58

The original question was "Is there an algorithm that, given two genus g Heegaard splittings, decides if the resulting two manifolds are homeomorphic?" The answer to this question is "yes" but doesn't have anything to do with Heegaard splittings.

In theory the solution to the Geometrization Conjecture gives an algorithm to decide if two manifolds are homeomorphic.

In practice one converts the splitting to a triangulations and feeds the result to SnapPy. There are several other programs for three-manifold recognition, such as snap, orb, Regina, and the 3-Manifold Recognizer. However all use SnapPea as the engine for dealing with hyperbolic manifolds. See http://www.math.uiuc.edu/~nmd/computop/ for links. The Recognizer, for Windows, can be found at http://www.matlas.math.csu.ru/.

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Thanks a lot for that, I only set my comment as an answer because of the word-limit on comments.. That makes a lot of sense, thanks again Sam. – user6204 Apr 24 '11 at 11:10

Yes I did mean the article by Birman :) It seems that are a few different types of equivalence, namely strong equivalence, equivalence, and weak equivalence.

Let $X_g$ be the handlebody of genus g that is oriented, and $X'_g$ its homemorphic image, i.e. $X'_g = \tau(X_g)$. Let $\delta : \partial X_g \rightarrow \partial X_g$ be an arbitrary fixed orientation-reversing homeomorphism that extends to an orientation-reversing homeomorphism of the entire handle-body, $X_g \rightarrow X_g$, with the boundaries of the pair of handle-bodies identified by $\tau \delta \phi(p)=p$, for all elements, p of the handle-body $X_g$.

If two Heegaard splittings $X_g \cup_{\phi} X'_{g}$, $X_g \cup_{\psi} X'_{g}$ are homeomorphic, then this is denoted by $\phi \equiv \psi$, with $\phi$ and $\psi$ being in the same isotopy class.

Two Heegaard splittings $X_g \cup_{\phi} X'_{g}$, $X_g \cup_{\psi} X'_{g}$ are called strongly equivalent (denoted by $\phi \approxeq \psi$) if there is an orientation-preserving homeomorphism $h:X_g \cup_{\phi} X'_{g} \rightarrow X_g \cup_{\psi} X'_{g}$ such that $h(X_g)=X_g$ and $h(X'_g)= X'_g$.

Heegaard splittings are called equivalent ($\phi \approx \psi$) if there is an orientation-preserving homeomorphism $h:X_g \cup_{\phi} X'_{g} \rightarrow X_g \cup_{\psi} X'_{g}$ such that either $h(X_g)=X_g , h(X'_g)= X'_g$ ; or $h(X_g)=X'_g , h(X'_g)= X_g$.

Heegaard splittings are called weakly equivalent ($\phi$ $\sim$ $\psi$) if there is a homeomorphism $h:X_g \cup_{\phi} X'_{g} \rightarrow X_g \cup_{\psi} X'_{g}$ such that either $h(X_g)=X_g , h(X'_g)= X'_g$ ; or $h(X_g)=X'_g , h(X'_g)= X_g$.

And so $\phi \approxeq \psi \Rightarrow \phi \approx \psi \Rightarrow \phi \sim \psi \Rightarrow \phi \equiv \psi$

However, it is not the case that if there exists a homeomorphism between two Heegaard splittings that they are weakly equivalent, as there are examples of manifolds that have more than one weak equivalence class (which implies that homeomorphic Heegaard splittings need not be equivalent).

This is important as $\phi \approxeq \psi$ iff the isotopy classes of $\phi$ and $\psi$ respectively are in the same double coset of the group of isotopy classes of orientation-preserving self-homeomorphisms of $\partial X_g$ modulo the subgroup whose isotopy classes contain mappings that extend to homeomorphisms of $X_g$;

two Heegaard splittings are equivalent iff either the isotopy classes of $\phi$ and $\psi$ are in the same double-coset, or the isotopy class of $\psi$ is in the same double-coset as $\Delta \Phi \Delta ^{-1}$, where $\Delta$ is the isotopy class of $\delta$, the arbitrary fixed orientation-reversing homeomorphism, and $\Phi$ is the isotopy class of $\phi$;

and two Heegaard splittings are weakly equivalent iff the isotopy class of $\psi$ is in the same double-coset as the isotopy class of $\phi$ ($\Phi$), or $\Delta \Phi ^{-1} \Delta ^{-1}$ or $\Phi^{-1}$ or $\Delta \Phi \Delta ^{-1}$ (For a proof of this, see "On the equivalence of Heegaard splittings of closed, orientable 3-manifolds." in "Knots, groups, and 3-manifolds: papers dedicated to the memory of R. H. Fox" also by Joan Birman, from which this is a summary).

And so if there is a homeomorphism between two Heegaard splittings, that are not at least weakly equivalent, then the associated orientation-preserving homeomorphisms, $\psi$ and $\phi$ will not be in the same double-coset, as described above.

It seems difficult in general to determine whether two gluing maps will produce the same manifold, however there appear to be certain cases in which homological invariants can be used to show that particular manifolds have Heegaard splittings that are not homeomorphic (also in "On the equivalence of Heegaard splittings of closed, orientable 3-manifolds.").

There appear to be techniques to determine whether two Heegaard splittings are equivalent, which involve using the fundamental group of the handle-bodies, as there is a one-to-one correspondence between the equivalence classes of splitting homomorphisms and equivalence classes of Heegaard splittings; that is if the splitting homomorphisms induced by a pair of Heegaard splittings are equivalent, then the pair of Heegaard splittings are equivalent (See Jaco, William. Heegaard splittings and splitting homomorphisms. Trans. Amer. Math. Soc. 144 1969 365–379).

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wow, this is what I was looking for! Thank you sooooo much!! – Lor Apr 25 '11 at 13:06