Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

%This question is motivated by the little discussion here at the bottom.

The following thing are known about hyperbolic right-angled polytopes:

  1. Compact hyperbolic right-angled polytopes do not exist in dimensions greater than 4 (Vinberg, Potyagailo). Examples are known up to dimension 4 (folklore?).
  2. Hyperbolic right-angled polytopes of finite volume do not exist in dimensions greater than 12 (Dufour). Examples are known up to dimension 8 (Vinberg, Potyagailo).
  3. Ideal (i.e. all vertices are ideal) right-angled polytopes do not exist in dimensions greater than 6 (Sasha). Examples are known up to dimension 4 (folklore?).

All the results 1-2-3 use the Nikulin-Khovanskii inequality, bounding the average number $f^k_l$ of $k$-dimensional faces in $l$-dimensional ones for an $n$-dimensional convex polytope (not necessarily hyperbolic). Here $k,l \leq [n/2]$ is a condition.

The bound has the form $f^k_l < g(k,l,n)$, where $g$ is a decreasing function in $n$, if $k$ and $l$ are fixed. Thus, by getting a lower bound for $f^k_l$, we can bound $n$ from above.

The result by Vinberg Potyagailo uses the fact that $5 \leq f^1_2$, for a compact right-angled polytope, that it obtains the best possible bound on $n$ (w.r.t. the given $k,l$). The result by Dufour uses the fact that $27 \leq f^5_6$ for a finite-volume right-angled polytope. Thus, the best possible bound for $n$ with given $k,l$. Finally, I know that $24 \leq f^3_4$ for an ideal right-angled polytope and get the best possible bound with given $k,l$. Again.

It's clear that increasing $k,l$ will not improve the bound on $n$ in any of the above cases 1-2-3. However, decreasing $k,l$ (and finding the respective lower bounds for $f^k_l$ in each of the cases 1-2-3) does not improve it either. Thus, the only sharp bound coming from the use of the Nikulin-Khovanskii inequality happens in case 1.

My question is the following: Can one find a way of proving sharp bounds in cases 2-3 (let me have a liberty to conjecture that the respective bounds are 8 and 4) without making use of the Nikulin-Khovanskii inequality? Basically, I'm interested in any fact that helps settling the conjectural bounds, just at the moment I see that the Nikulin-Khovanskii inequality is not (directly) applicable any more.

share|improve this question
1  
By "Nikulin-Khovanski inequality", do you mean "the lower bound theorem", or something different? –  Igor Rivin Sep 4 '13 at 2:26
    
By "N.-K. inequality" I mean the inequality $f^k_l \leq g(k,l,n) = {n-k \choose n-l} \frac{{[n/2]\choose k}+{[(n+1)/2] \choose k}}{{[n/2]\choose l}+{[(n+1)/2] \choose l}}$, which is more an "upper bound theorem". The lower bound $f^k_l \geq a$ comes from some geometric considerations usually. As a rule one proves that the number of facets of $l$-dimensional (compact/finite-volume/ideal) polytope is not less than $a$. Thus $f^k_l \geq a$ with $k+1 = l > [n/2]$. Thus, for big $n$ and with fixed $k,l$, $a \leq f^k_l \leq g(k,l,n)$ will not be true, since $g(k,l,n)$ decreases with $n$. –  SashaKolpakov Sep 4 '13 at 4:15
    
Actually, $g(k,l,n) \sim f^k_l(H)$, where $H$ is a $n$-dimensional hypercube with $n$ big enough. Thus, the quantity $a$ above should be somewhat big in order to get a sensible upper bound for $n$. –  SashaKolpakov Sep 4 '13 at 4:17

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.