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A convex polytope $P\subset\Bbb R^d$ is centrally symmetric if $-P=P$. It is self-dual (or better, self-polar?) if its polar dual $P^\circ$ is congruent to $P$, that is, there is a map $X\in\mathrm O(\smash{\Bbb R^d})$ with $\smash{P^\circ}=XP$.

Question: Are there centrally symmetric self-dual polytopes in dimension $d>4$?

Such exist in dimension $d=2$ and $d=4$:

  • for $d=2$ we have the regular 2n-gons,
  • for $d=4$ we have the regular 24-cell.
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  • $\begingroup$ Had you check this text? It does not seem to be about centrally symmetrical polytopes, but about self-dual. arxiv.org/pdf/1902.00784 $\endgroup$ – Fedor Petrov Dec 31 '20 at 12:25
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    $\begingroup$ @FedorPetrov Yes. The paper mainly deals with the case $P^\circ=-P$ (then $P$ cannot be centrally symmetric). The two mentioned constructions, pyramids and joins, do not yield centrally symmetric polytopes either. The add-and-cut construction works only if we already have a self-polar polytope in dimension $d>4$. $\endgroup$ – M. Winter Dec 31 '20 at 12:29
  • $\begingroup$ what is $X$ for a 24-cell? $\endgroup$ – Fedor Petrov Dec 31 '20 at 13:26
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    $\begingroup$ @FedorPetrov Let $P$ be the convex hull of the $24$ units in the ring of half integer quaternions. Let $Q$ be the convex hull of the $24$ integer quaternions with norm $2$. Then $P$ and $\tfrac{1}{\sqrt{2}} Q$ are dual polytopes, each of which is isomorphic to the $24$-cell. Multiplication (either left or right) by $\tfrac{1+i}{\sqrt{2}}$ is an isomorphism from $P$ to $\tfrac{1}{\sqrt{2}} Q$. $\endgroup$ – David E Speyer Dec 31 '20 at 14:08
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    $\begingroup$ I think that, with the standard inner product, the actual statement should be that $\sqrt[4]{2} P$ and $\tfrac{1}{\sqrt[4]{2}} Q$ are dual; the isomorphism is still given by multiplication by $\tfrac{1+i}{\sqrt{2}}$. $\endgroup$ – David E Speyer Dec 31 '20 at 16:25
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There are centrally symmetric self-dual polytopes in every dimension. This follows from Proposition 3.9 in Reisner, S., Certain Banach spaces associated with graphs and CL-spaces with 1- unconditional bases, J. Lond. Math. Soc., II. Ser. 43, No. 1, 137-148 (1991). ZBL0757.46030.

Moreover, in dimension $\geqslant 3$ the matrix $X$ can be chosen to be a permutation matrix.

Here is an example in dimension $3^d$ for every $d$. Start with Sztencel-Zaramba polytope $P$. This is the unit ball for the norm on $\mathbf{R}^3$ $$ \|(x,y,z)\| = \max \left( |y|+|z|, |x|+\frac 12 |z| \right)$$ whose dual norm satisfies $$ \|(x,y,z)\|_* = \|(z,y,x)\|. $$ We may now define inductively a sequence $\|\cdot\|_d$, which is norm on $\mathbf{R}^{3^d}$ (identified with $\mathbf{R}^{3^{d-1}}\times\mathbf{R}^{3^{d-1}}\times\mathbf{R}^{3^{d-1}}$). Chose $\|\cdot\|_1$ to be above norm, and use the recursive formula $$ \|(x,y,z)\|_{d+1} = \|( \|x\|_d ,\|y\|_d , \|z\|_d )\|_1 .$$ One checks by induction that there is a permutation matrix which maps the unit ball onto its polar.

To visualize the polytope $P$ you may use the Sage code

p1 = Polyhedron(vertices=[[0,1,1],[0,1,-1],[0,-1,1],[0,-1,-1],[1,0,1/2],[1,0,-1/2],[-1,0,1/2],[-1,0,-1/2]])
p1.projection().plot()
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  • $\begingroup$ This is amazing! Thank your for this answer. I had already mentally excluded the possibility that I could have overlooked an example in dimension three. $\endgroup$ – M. Winter Jan 7 at 13:43

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