Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have recently heard about the hyperplane conjecture and I would like to understand better the problematic behind this conjecture.

The hyperplane conjecture: There exists a universal constant $C>0$ such that $$C\leq C_d:=\inf\Big( \max \big(V_{d-1}(H\cap K) : H\in\mathcal{H}_{d}\big) : \ K \text{ convex in }\mathbb{R}^d \text{ and } V_d(K)=1\Big )\quad \forall d\geq2$$ where $\mathcal{H_{d}}$ is the set of hyperplane in $\mathbb{R}^d$ and $V_d$ is the $d$-dimensional volume.

Basically it says that the maximal hyperplane section of any convex body cannot be really small compare to its volume. For me, it looks similar to an isoperimetric inequality.

I have several questions and instead of opening many different questions I post them all here hoping to get answer for at least some of them. Partial answers and comments related to the conjecture but not to one of the following questions are really welcome.

  1. Is there a simple justification why we should have $C_d>0$ for any $d$?
  2. Which convex body (of volume $1$) has the smallest maximal section in dimension $2$? (I guess the disk, but maybe I have wrong intuition).
  3. Which convex body (of volume $1$) has the smallest maximal section in dimension $3$?
  4. What is the maximal hyperplane section of a unit cube?
  5. Can you give other examples of maximal sections of some specific convex body in low dimension.

    5 bis:Can you give an example of convex body (of volume $1$) with a maximal hyperplane section smaller then the one of the ball (of volume $1$).

  6. Where does this conjecture comes from?
  7. Apart from giving a lower bound, do we know anything about the behaviour of the sequence $(C_d)_{d\geq2}$? Is it decreasing? Why?
  8. Is there any survey about this conjecture?

Answers:

  1. ..
  2. Emanuele Paolini explained why it is the isodiametric problem. The disc is the solution of this problem. It has been proved by Bieberbach in 1915 (in german), I found this reference in the introduction of the article Isodiametric Problems for Polygons by by Michael J. Mossinghoff.
  3. ..
  4. Theorem 6 of "Volumes of sections of cubes and related problems" of Keith Ball states the maximal section of a unit cube in $\mathbb{R}^d$ is $\sqrt{2}$. This sections is the generalization of the diagonal of a square. More precisely, it is spanned by a $(d-2)$-dimensional face of the cube and the diagonal of a $2$-dimensional orthogonal face.
  5. Einar Rødland gives the maximal section for the clear example of the sphere of dimension $d$ and volume $1$: $\frac{\Gamma(d/2+1)^{\frac{d-1}{d}}}{\Gamma(d/2+1/2)}$. Other less clear examples are really welcomed to complete the list.
share|improve this question
    
some interesting comments at the end of the following post: blogs.princeton.edu/imabandit/2013/06/22/… –  Gilles Bonnet Mar 30 at 21:31

2 Answers 2

About 2. In the plane the $1$-volume of the maximal hyperplane section, is nothing else than the diameter of the set. In fact since the set is convex, every section is a segment, and every diameter is a section. So the problem becomes the isodiametric problem, whose solution is known to be the disk.

share|improve this answer
    
1. Yes, but this ball can be arbitrarily small. Think of a rectangle with one tiny side and the other long. So your argument does not imply that $C_d$ is lower bounded. –  Gilles Bonnet Mar 26 at 15:50
    
you are right... I remove that part. –  Emanuele Paolini Mar 26 at 15:57
1  
Meta question: Should we delete the two comments above? –  Gilles Bonnet Mar 27 at 9:29

My gut feeling is that the shape that would give the smallest hyperplane section would be the $d$-ball. If that is the case, the $d$-volume of a $d$-ball of radius $r$ is $\nu_dr^d$ where $\nu_d=\pi^{d/2}/\Gamma(d/2+1)$. If $r_d=\nu_d^{-1/d}$ is the radius that gives volume 1 and we plug this into the $d-1$-volume of the equatorial hyperplane section $\nu_{d-1}r_d^{d-1}$, we get the ratio $$ C_d\stackrel{?}{=}\nu_{d-1}r_d^{d-1} =\frac{\Gamma(d/2+1)^{\frac{d-1}{d}}}{\Gamma(d/2+1/2)} \rightarrow e^{1/2} \text{ as }d\rightarrow\infty $$ unless I somehow messed up the calculations here.

The logic behind this gut feeling is that any perturbation of the $d$-ball which preserves the $d$-volume would have to increase the $d-1$-volume of at least some of the sections.

If this is true, it would make $C_d>0$ and $C>0$ with good margin.

Of course, I may be wrong and the hypothesis still be true...

share|improve this answer
1  
I actually had the same feeling as you about the sphere. But in the introduction of the following paper: arxiv.org/abs/math/0612517 they say that the conjecture remains open for more then two decades now. So I guess the problem is not that simple. And that why I posted this question here. –  Gilles Bonnet Mar 30 at 20:45
    
If instead of considering hyperplane section we consider line section then this is the isodiametric problem. This is the case in dimension 2 for example as noticed by @Emanuele Paolini. In dimension 2 I mentioned a reference in my question proving the disk minimise the isodiameter. I don't have reference for higher dimension but I guess the isodiameter problem have already been consider and that the sphere is the solution in any dimension. Now it seems to me much less clear that works as well for the hyperplane section... I don't see it. –  Gilles Bonnet Mar 30 at 20:52
1  
Of course, my gut feeling may be wrong. But it may also be that it is right, just that it's very hard to make a formally correct proof of it...just as with the sphere packing where the answer was "known" for centuries before anyone was able to prove it. –  Einar Rødland Mar 30 at 20:53

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.