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Jun 28, 2022 at 14:23 comment added Sam Nead I don't see how a triangulation of $M \times [0, 1]$ gives a sequence of Pachner moves. I tried to say this in the last (parenthetical) paragraph of my post.
Jun 27, 2022 at 17:08 comment added Andi Bauer I asked a very similar question. I was wondering: If it was true that there is $L'$ such that any PL $n+1$-manifold can be triangulated with local complexity at most $L'$, would that imply the answer to your question? Since any triangulation of $M\times [0,1]$ triangulated by $T_1$ on $M\times 0$ and $T_2$ on $M\times 1$ yields a sequence of Pachner moves connecting $T_1$ and $T_2$ where all intermediate triangulations have complexity at most $L'$?
Dec 27, 2020 at 18:53 comment added Fedya @MoisheKohan, I think this is orthogonal to the work of Nabutovsky and Weinberger. The smooth version of this question is: let's say we want to stay in the realm of Riemannian manifolds with sectional curvatures <1 and injectivity radius >1. From a given metric on S^n we can (by rescaling) walk to the standard one through such metrics, but (according to N-W) the intermediate metrics may have vastly larger volume. I suspect the same thing works in PL: we might have to vastly increase the number of vertices along the way, but we can retain bounded geometry. But it's no longer trivial.
Dec 27, 2020 at 13:21 comment added Sam Nead The $(n+1, 1)$ and $(1, n+1)$ moves remove and add a single vertex, respectively.
Dec 26, 2020 at 21:08 comment added Moishe Kohan You should also allow barycentric subdivision since Pachner moves proserve the number of vertices. I do not really understand the question, but it is likely related to the work of Nabutosvky and Weinberger on the “landscapes” of spaces of Riemannian metrics.
Dec 26, 2020 at 16:36 history asked Sam Nead CC BY-SA 4.0