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For a convex polytope $P$ in $\mathbb R^4$, denote by $N_0,N_1,N_2,N_3$ respectively the number of vertices, edges, faces, cells. By Euler's formula, we know $N_0+N_2=N_1+N_3$, which means there is a sort of equilibrum among the $N_i$. But I wonder if there exist upper and/or lower bounds for $f(P):=\frac{N_1+N_2}{N_0+N_3}$ where $P\subset\mathbb R^4$ is any convex polytope.

E.g. for the regular one called 24-cell, we have $f(P)=4$.

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up vote 6 down vote accepted

As of 2003, I don't think it was known whether $f(P)$ is bounded. See Fat 4-polytopes and fatter 3-spheres by Eppstein, Kuperberg, and Ziegler (http://front.math.ucdavis.edu/0204.5007), which contains a beautiful construction that achieves $f(P)>5.048$, and shows that it is not bounded for strongly regular CW decompositions of the 3-sphere (instead of polytopes). I don't know whether there has been more recent progress.

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See also Ziegler's "Face Numbers of 4-Polytopes and 3-Spheres" (arxiv.org/abs/math/0208073), which describes the EKZ paper at a high level. –  Joseph O'Rourke Jan 20 '12 at 19:47
    
Thank you. Now knowing the terminology, I have also found Paffenholz' "New Polytopes from Products" (polymake.org/polytopes/paffenholz/data/preprints/…), giving a construction of polytopes whose fatness (that's the term!) comes arbitrarily close to $6$. –  Wolfgang Jan 20 '12 at 23:07
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Paffenholz in turn links to a paper by Ziegler that gets fatness close to 9: ams.org/journals/era/2004-10-14/S1079-6762-04-00137-4 –  David Eppstein Jan 21 '12 at 1:11

Each cell is a convex polytope in 3-space and so has at least 4 faces, whereas each face should have only 2 cells, or $N_3\geq 2N_4$. The same argument for the dual polytope gives $N_1\geq 2N_0$, which gives a lower bound of $f\geq 2$, attained by the $5$-simplex.

This is analogous to the polyhedral bound of $2E \geq 3F,3V$.

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