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A triangulation of a convex polytope $P\subset\Bbb R^n$ is a partition of $P$ into $n$-simplices $\{\Delta_1,...,\Delta_m\}$ each of which has all its vertices among the vertices of $P$. A polytope may have many different triangulations.

Question I: do all combinatorially equivalent polytopes have the same triangulations?

More precisely, let $P,Q\subset\Bbb R^n$ be combinatorially equivalent, $\phi:\mathcal F(P)\to\mathcal F(Q)$ be a face-lattice isomorphism, and $\{\Delta_1,...,\Delta_m\}$ a triangulation of $P$. If $\Delta_i$ has vertices $p_1,...,p_{n+1}\in\mathcal F_0(P)$, then let $\phi(\Delta_i)\subset Q$ be the simplex with vertices $\phi(p_1),...,\phi(p_{n+1})$. Do the simplices $\phi(\Delta_1),...,\phi(\Delta_m)$ form a triangulation of $Q$? That is, do they have disjoint interiors and cover all of $Q$?

Question II: if not, is there always a universal triangulation? That is, a special triangulation for $P$ that carries over to every combinatorially equivalent polytope in the way described above?

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    $\begingroup$ This seems to have been asked and aswered already at math.SE. $\endgroup$ May 28, 2022 at 15:56
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    $\begingroup$ @TobiasFritz Great find. The linked question is eerily similar and answer my question fully. $\endgroup$
    – M. Winter
    May 28, 2022 at 18:43

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Just to mark this question as answered, the comment by Tobias Fritz is spot on. In this answer to a Math Stack Exchange question, Francisco Santos completely resolves your questions. On the one hand, the answer to Question I is no: combinatorially equivalent polytopes can have different triangulations. On the other hand, the answer to Question II is: yes, there is at least one triangulation they all share (the so-called "pulling triangulations").

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