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Mar 9, 2015 at 14:25 comment added Yunhyung Cho @Joseph O'Rourke : Hao Chen pointed out what I exactly want to say. Sorry for the confusion.
Mar 9, 2015 at 6:28 comment added Hao Chen I would rephrase the question as follows: can we always find a convex polytope $P'$ such that $P'$ is combinatorially equivalent to $P$ and $P'$ is index increasing w.r.t. a given vector (after all, we can fix the vector wlog)
Mar 8, 2015 at 23:38 comment added Joseph O'Rourke @TheMaskedAvenger: Wikipedia says, a "simple polytope is a $d$-dimensional polytope each of whose vertices are adjacent to exactly $d$ edges (also $d$ facets)."
Mar 8, 2015 at 23:16 comment added The Masked Avenger For the geometric incognoscenti among us, are there brief descriptions which would convey an appropriate meaning of "simple" for this context? My guess is no holes.
Mar 8, 2015 at 23:09 comment added Joseph O'Rourke @HaoChen: A deformation should be "preserving ... convexity," a stipulation I do not understand.
Mar 8, 2015 at 17:25 comment added Joseph O'Rourke @HaoChen: Ah, I was interpreting "simple" as in "simple polyhedron," rather than "simple polytope": all vertices degree $d$. Thanks for the clarifications.
Mar 8, 2015 at 16:52 comment added Hao Chen He said "simple convex polytope". I think "deformation" means any combinatorial equivalent realization.
Mar 8, 2015 at 15:44 history answered Joseph O'Rourke CC BY-SA 3.0