A Pachner complex for triangulated manifolds - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T21:15:36Z http://mathoverflow.net/feeds/question/35061 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/35061/a-pachner-complex-for-triangulated-manifolds A Pachner complex for triangulated manifolds Ryan Budney 2010-08-09T23:32:37Z 2010-08-10T16:17:42Z <p>A theorem of Pachner's states that if two triangulated PL-manifolds are PL-homeomorphic, the two triangulations are related via a finite sequence of moves, nowadays called "Pachner moves". </p> <p>A Pachner move has a very simple picture. If $N$ is a triangulated $n$-manifold and $C \subset N$ is a co-dimension zero subcomplex of $N$ which is equivalent to a subcomplex $D'$ of $\partial \Delta_{n+1}$ where $\Delta_{n+1}$ is an $(n+1)$-simplex equipped with its standard triangulation, then you can replace $N'$ in $N$ by $(\partial \Delta_{n+1}) \setminus D'$ from $\partial \Delta_{n+1}$, the gluing maps being the only ones available to you. </p> <p>One way to say Pachner's theorem is that there is a ''graph'' of triangulations of $N$, and it is connected. Specifically, the vertices of this graph consist of triangulations of $N$ taken up to the equivalence that two triangulations are equivalent if there is an automorphism of $N$ sending one triangulation to the other. The edges of this graph are the Pachner moves. </p> <p>This formulation hints at an idea. Is there a useful notion of "Pachner complex"? </p> <p>Of course this would lead to many further questions such as what geometric / topological properties does such a complex have, is it contractible for instance, are there "short" paths connecting any two points in the complex, and so on. </p> <p>I'm curious if people have a sense for what such a Pachner complex should be. For example, some Pachner moves commute in the sense that the subcomplexes $C_1, C_2 \subset N$ are disjoint, so you can apply the Pachner moves independantly of each other. Presumably this should give rise to a "square" in any reasonable Pachner complex. </p> <p>This feels related to the kind of complexes that Waldhausen and Hatcher used to use in the 70's but it's also a little different. </p> <ul> <li>Udo Pachner, P.L. homeomorphic manifolds are equivalent by elementary shellings, European J. Combin. 12 (1991), 129–145.</li> </ul> http://mathoverflow.net/questions/35061/a-pachner-complex-for-triangulated-manifolds/35117#35117 Answer by Jim Fowler for A Pachner complex for triangulated manifolds Jim Fowler 2010-08-10T14:31:16Z 2010-08-10T14:31:16Z <p>I think this is a really neat idea. If you haven't already, you might want to look at</p> <blockquote> <p>Nabutovsky, A. Geometry of the space of triangulations of a compact manifold. Comm. Math. Phys. 181 (1996), no. 2, 303--330. <a href="http://www.ams.org./mathscinet-getitem?mr=1414835" rel="nofollow">MR1414835</a></p> </blockquote> <p>This paper discusses the metric properties of the "graph of triangulations" you mention.</p> http://mathoverflow.net/questions/35061/a-pachner-complex-for-triangulated-manifolds/35136#35136 Answer by Kevin Walker for A Pachner complex for triangulated manifolds Kevin Walker 2010-08-10T16:17:42Z 2010-08-10T16:17:42Z <p>My vague recollection is that the way one proves Pachner's theorem is to prove the existence of a shellable triangulation of $M\times I$ extending the given triangulations on $M\times \{0,1\}$. Here "shellable" means that we can add $(n+1)$-simplices to $M\times\{0\}$ one at a time (eventually adding all of them to reach $M\times \{1\}$) so that the interface between what we've added and what we haven't is always a triangulated manifold homeomorphic to $M$.</p> <p>This suggests the following definition of a "Pachner complex". The 0- and 1-cells are of course triangulations of $M$ and Pachner moves, as you describe. There are two classes of 2-cells. The first class corresponds to changing the order in which $(n+1)$-simplices are added in the previous paragraph. It suffices (I think) to consider swapping the order of two $(n+1)$-simplices which are adjacent in "time". There are still many possibilities, since these two simplices could be adjacent (or disjoint) in $M$ in various ways, and also intersect the already-added simplices in various ways.</p> <p>The second class of 2-cell corresponds to changing the triangulation of $M\times I$ from the first paragraph above. In some sense, we do a higher order Pachner move on $M\times I$. More specifically, divide the boundary of an $(n+2)$-simplex into two $(n+1)$-balls $B_1$ and $B_2$ which intersect in an $n$-sphere $E$, then similarly divide $E$ into a pair of $n$-balls $C_1$ and $C_2$. After choosing orderings of $(n+1)$-simplices, $B_1$ and $B_2$ correspond to two different sequences of Pachner moves, $s_1$ and $s_2$, connecting the triangulated $n$-balls $C_1$ and $C_2$. Corresponding to all the above data we have a 2-cell in the Pachner complex whose boundary is $s_1 s_2^{-1}$.</p> <p>In general, the $k$-cells of the Pachner complex would correspond to higher order syzygies of the symmetric group, or ways of dividing the boundaries of higher dimensional simplices, or combinations of these.</p> <p>There are many details missing from the above sketch of a definition of a "Pachner complex", and even with those details nailed down one would have to show that the complex is contractible. But it seems to me this is the correct general idea for such a complex.</p>