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Throughout, by finite triangulation I mean a triangulation consisting of a finite number of triangles.

Suppose $T$ and $T'$ are finite triangulations of a 3-manifold $M$. We will say that $T'$ is simpler than $T$ iff $T'$ consists of the same number or fewer triangles than $T$ and that $T'$ is a simplest triangulation of $M$ iff $\forall$ triangulation $T$ of $M$, $T'$ is simpler than $T$.

Note: If a 3-manifold $M$ has a finite triangulation, then clearly it has a simplest triangulation.

By a theorem of Pachner (Theorem A.1.1. in 'The geometry of dynamical triangulations') any two triangulations of a manifold can be transformed from one to another by a finite number of stellar subdivisions. As we are only dealing with 3-manifolds, there are only 4 stellar subdivisions; known as the $1 \to 4$, $2 \to 3$, $3 \to 2$ and $4 \to 1$ moves as described in http://at.yorku.ca/t/a/i/c/45.pdf and hereafter called the Pachner moves. So clearly, there exists a finite sequence of Pachner moves from any finite triangulaiton $T$ of $M$ to $T'$, a simplest triangulation of $M$.

If $T$ is a finite triangulation of $M$, does the greedy algorithm of just applying as many $4 \to 1$ and $3 \to 2$ Pachner moves to $T$ as possible always result in a simplest triangulation of $M$?

Or alternatively,

Is there a finite triangulation $T$ of a 3-manifold $M$ such that repeatedly applying only the $4 \to 1$ and $3 \to 2$ Pachner moves does not eventually result in a simplest triangulation of $M$?

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    $\begingroup$ A tangential remark: If the moves are restricted to $2 \rightarrow 3$ and $3 \rightarrow 2$, it is open whether or not the space of triangulations is connected. cs.smith.edu/~orourke/TOPP/P28.html#Problem.28 $\endgroup$
    – Joseph O'Rourke
    Commented Aug 31, 2010 at 14:26
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    $\begingroup$ Moves $2 \to 3$ and $3 \to 2$ cannot be enough since they do not change the number of vertices in the triangulation. However, I think they are enough to connect every couple of triangulations having the same number of vertices with at least 2 tetrahedra thanks to a theorem of Matveev-Piergallini. (This should be checked, however.) $\endgroup$
    – Bruno Martelli
    Commented Aug 31, 2010 at 15:10
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    $\begingroup$ I think that the following is an open problem, however: given two triangulations $T_1, T_2$ with the same number of tetrahedra, is there a triangulation $T$ which is obtained from both only by moves $2 \to 3$ (and not its inverse)? $\endgroup$
    – Bruno Martelli
    Commented Aug 31, 2010 at 15:25
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    $\begingroup$ Forgive me for responding to an eight year old comment, but @JosephO'Rourke Rubinstein, Segerman, and Tillmann arxiv.org/pdf/1812.02806.pdf have a relatively recent arxiv preprint which directly addresses your comment and tries to address the history brought up by Bruno Martelli. It's my understanding that as part of their lit review they came across a paper of Gennaro Amendola, "A calculus for ideal triangulations of three-manifolds with embedded arcs" which addresses your concerns head on. Additionally, their paper addresses the question. $\endgroup$
    – Neil Hoffman
    Commented Jan 28, 2019 at 19:05

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There are many such examples, depending upon what you mean by "triangulation". If a triangulation is just a glueing of tetrahedra along faces, then the simplest one is probably the following: the 3-sphere has a triangulation with 1 tetrahedron, and a triangulation with 2 tetrahedra (it is a nice excercise to find them). However, you cannot apply any $4 \to 1$ or $3 \to 2$ move to a triangulation having only 2 tetrahedra.

If by "triangulation" you mean a "simplicial triangulation", then it suffices to construct a triangulation of the 3-sphere such that every edge meets at least 4 distinct tetrahedra. In this case, no $4 \to 1$ and no $3 \to 2$ moves can apply.

For instance, you can take a triangulation of the 2-sphere such that every vertex is incident to at least 4 triangles (as an example, the octahedron). By doing a suspension of this triangulation you get a triangulation of the 3-sphere such that every edge is incident to at least 4 tetrahedra.

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    $\begingroup$ Martelli would know better than me but my impression is that there are "many" of these local minima where one has to do long chains of simplex-increasing Pachner moves before one can descend to a global minimum (number of top-dimensional simplices). In dimension $n \geq 4$ Nabutovsky has shown the number of these "deep" local minima grows exponentially the further you get from a minimial triangulation (in the Pachner graph of a manifold). $\endgroup$
    – Ryan Budney
    Commented Aug 31, 2010 at 18:32
  • $\begingroup$ By "Triangulation" I did mean the former, i.e. a glueing of tetrahedra along faces. $\endgroup$
    – Mark Bell
    Commented Aug 31, 2010 at 19:28
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If I were to bet on the question, I would expect that expanding and contracting moves would be needed. My guess to construct an example would be to take an unknot that required a type II move that increased the number of crossings, and follow that simplification. To get a 3-mfd from such an example, try a Dehn filling of the knot (or better just work with the manifold with boundary). I think that it is not too hard to write down a triangulation from the diagram (several CW complexes and handle decompositions are apparent). Examine what the Reidemeister moves do to the triangulation, and decompose each in terms of Pachner moves.

For example the number of 2-handles differs on the two sides of a type II move.

I certainly haven't worked out the details, but it should now be a good exercise in handle/cell/triangulation moves.

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Here is a concrete example of an ideal triangulation of a trefoil knot with isomorphism signature `dLQabccbrwn'. The gluing pattern is: Gluing pattern for dLQabccbrwn

One can check that it has 1-ideal vertex and no degree 3 edges so it is locally minimal. To recognize the manifold, one can truncate the ideal vertices in Regina and see that it is the trefoil complement (or rather SFS [D: (2,1) (3,-2)]).

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