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Suppose that $M$ is closed, connected PL $n$-manifold. We say that a triangulation of $M$ has local complexity at most $L$ if every zero-cell of $T$ meets at most $L$ $n$-simplices. (An alternative definition bounds the number of combinatorial types of "vertex links" that appear in $T$.)

Let $P(M)$ be the Pachner graph of $M$: the graph where vertices are (isotopy classes of) triangulations of $M$ (in the correct PL class) and edges are Pachner moves (aka bi-steller flips). Thus $P(M)$ is connected. Let $P_L(M)$ the the subgraph of triangulations with local complexity at most $L$.

Is there an $L'$ so that $P_L(M)$ is connected inside of $P_{L'}(M)$?

It would be particularly interesting to know this for the standard PL structure on $S^n$. Indeed, as a special case, we want to know how difficult it is to connect a triangulation of $S^n$, with bounded local complexity, to the boundary of the $n + 1$ simplex via triangulations with bounded local complexity.

My question was inspired by trying to understand Fedya's answer to this question.

(That is: I wanted to find a bounded geometry fillings of an $n$-sphere by finding a nice path in the space of triangulations. I don't see how Agol's technique does this. We are given a "bad" sequence of triangulations connecting two "good" triangulations. We can convert this into a riemannian metric on $M \times [0, 1]$, we can scale it up to make it almost flat, we can scatter points, and we can take the induced Delaunay triangulation. I can now imagine tricks to fix the beginning and end. But it seems very very difficult to recover the desired product structure...)

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    $\begingroup$ 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. $\endgroup$ Commented Dec 26, 2020 at 21:08
  • $\begingroup$ The $(n+1, 1)$ and $(1, n+1)$ moves remove and add a single vertex, respectively. $\endgroup$
    – Sam Nead
    Commented Dec 27, 2020 at 13:21
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    $\begingroup$ @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. $\endgroup$
    – Fedya
    Commented Dec 27, 2020 at 18:53
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    $\begingroup$ 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 $L'$? $\endgroup$
    – Andi Bauer
    Commented Jun 27, 2022 at 17:08
  • $\begingroup$ 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. $\endgroup$
    – Sam Nead
    Commented Jun 28, 2022 at 14:23

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