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Consider an $n$-dimensional PL manifold $P$ together with a closed subpolyhedron $P_0$ such that there exists a finite combinatorial triangulation of $P$ that restricts to a triangulation of $P_0$.

Is it true that any two combinatorially equivalent combinatorial triangulations of $P$ that coincide on $P_0$ are related by a finite sequence of bistellar flips (Pachner moves) which are the identity on $P_0$?

The case of $P$ a $3$-manifold with boundary $\partial P= P_0$ is Theorem 2.1.2 of Quantum Invariants of Knots and Three Manifolds. My primary motivation for this question is to try to massage such a theorem, if it exists, to try to answer a previous MO question which I asked. But I also care about the answer for its own sake. Note that $P_0$ is not a submanifold, but only a subpolyhedron. Even if the statement isn't true in full generality, is it at least true if $P_0$ is somehow not "too pathological"?

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It's been a long time since I thought about these kinds of things, but perhaps you can use an argument similar to what we have in Theorem 2.16 of my paper with Banagl "Triangulations of 3–dimensional pseudomanifolds with an application to state-sum invariants". Here's the arxiv link: http://arxiv.org/abs/math/0408156

The basic idea is to connect the two triangulations by a series of Alexander star moves and then to show that each star move can be accomplished by a sequence of bistellar moves by working locally on a star neighborhood of the simplex being used for the star move and invoking a theorem of Maria Rita Casali about bistellar equivalence in manifolds with boundary.

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In his paper: http://arxiv.org/pdf/math/9911256.pdf Lickorish outlines a proof of theorem (Theorem 5.10) he credits to Pachner and Newman which says that two combinatorial $n$-manifolds with non-empty boundary are PL-homeomorphic if and only if they are related by a sequence of elementary shellings, inverse shellings and simplicial isomorphism.

I realize this isn't quite what you asked for, but it's the closest I've seen to what you've asked for.

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  • $\begingroup$ Thanks! So I can assume this statement doesn't actually appear anywhere? That's a shame... Anyway, I hope it's true at least. $\endgroup$ – Daniel Moskovich Dec 1 '13 at 10:27

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