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?

2$\begingroup$ 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. ReidemeisterSinger theorem says that two Heegaard splittings of the same manifold are the same up to stabilisation. $\endgroup$ – Maxime Bourrigan Mar 3 '11 at 11:16

$\begingroup$ 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 $\endgroup$ – Lor Apr 20 '11 at 7:48

$\begingroup$ 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. $\endgroup$ – user6204 Apr 21 '11 at 13:18

$\begingroup$ I didn't find anything at that page...anyway I already knew that book, so I think you are referring to the article of Birman (maybe we have different page numberings). The point is I'm not satisfied with this :) I mean: by Birman, we know that if a manifold admits two equivalent splittings, than they are obtain from gluing maps which are in the same double coset. But what about the converse? Suppose we are given two genus g handlebodies and we want to glue them along their boundaries to obtain a 3manifold, and we use two different maps, say $f$ and $g$. $\endgroup$ – Lor Apr 23 '11 at 18:12

1$\begingroup$ I added the tags "3manifolds" and "geometric topology" because I had missed this question. $\endgroup$ – 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 threemanifold recognition, such as snap, orb, Regina, and the 3Manifold 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/.

$\begingroup$ Thanks a lot for that, I only set my comment as an answer because of the wordlimit on comments.. That makes a lot of sense, thanks again Sam. $\endgroup$ – 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 orientationreversing homeomorphism that extends to an orientationreversing homeomorphism of the entire handlebody, $X_g \rightarrow X_g$, with the boundaries of the pair of handlebodies identified by $\tau \delta \phi(p)=p$, for all elements, p of the handlebody $X_g$.
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 orientationpreserving 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 orientationpreserving 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 orientationpreserving selfhomeomorphisms 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 doublecoset, or the isotopy class of $\psi$ is in the same doublecoset as $\Delta \Phi \Delta ^{1}$, where $\Delta$ is the isotopy class of $\delta$, the arbitrary fixed orientationreversing 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 doublecoset 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 3manifolds." in "Knots, groups, and 3manifolds: 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 orientationpreserving homeomorphisms, $\psi$ and $\phi$ will not be in the same doublecoset, 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 3manifolds.").
There appear to be techniques to determine whether two Heegaard splittings are equivalent, which involve using the fundamental group of the handlebodies, as there is a onetoone 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).

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