18
$\begingroup$

Let $K$ be a bounded symmetric ($-K=K$) open convex body in $\mathbb R^n$. The critical determinant $d(K)$ of $K$ is the least possible volume $|\operatorname{det}(a_1\dots a_n)|$ of the fundamental parallelepiped of a lattice $\Lambda=\{\sum_{j=1}^n m_j a_j: m_j\in\mathbb Z\}$ such that $K\cap \Lambda=\{0\}$. Clearly, if $K'\subset K$ is another symmetric convex shape, then $d(K')\le d(K)$. The convex symmetric shape $K$ is called irreducible if the inequality is strict for every proper subset $K'$ of $K$.

It is not hard to see that the unit ball $B$ is irreducible in the dimensions 1,2,3. Moreover, if $B'\subset B$ and the radius of the largest ball contained in $B'$ is $1-\delta$, then $d(B')\le d(B)-c\delta$. However, this breaks in dimension $4$. When we cut off two opposite caps of depth $\delta$ from the unit ball in $\mathbb R^4$, we still get $d(B')<d(B)$ but the difference is now merely of order $\delta^2$. This suggests that when the dimension goes up and the number of touching points increases, we may end up in the situation when the ball is no longer irreducible. My question is whether this is really the case or whether the ball stays irreducible all the way up and just gets "less ans less" so in some sense.

The reason I'm asking is that we've just proved with Yoav Kallus that the ball in $\mathbb R^3$ is a local minimizer of the optimal lattice packing density. The corresponding statement is known to be false in $\mathbb R^2$ and I wonder if there is a trivial reason (namely, reducibility) for it to be false in some or, better, all high dimensions.

Any (relevant) ideas and/or references are welcome :).

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ at the risk of appearing foolish, I suggest dimensions 8 and 24, where we get surprisingly small determinants, scaled versions of $\mathbb E_8$ and the Leech lattice. In the latter case, there are no roots, the minimum norm is 4, so one scales by a rather large factor. In order to get no roots, considerable symmetry is forced, there is a very nearly regular simplex of lattice points of circumradius $\sqrt 2.$ Meaning it may be possible to shave small caps from the unit ball without allowing a lattice of smaller determinant. $\endgroup$
    – Will Jagy
    Jun 27, 2012 at 22:16
  • $\begingroup$ hmmm...maybe the simplex in question is regular. Also, I do not know what the extreme symmetry implies for your question. Something worth knowing, maybe. $\endgroup$
    – Will Jagy
    Jun 27, 2012 at 22:33
  • 2
    $\begingroup$ Looking at E8 is a good idea. I'll certainly try that. Thanks! :) $\endgroup$
    – fedja
    Jun 28, 2012 at 20:58
  • 1
    $\begingroup$ OK, so E8 has touching points of the kind $e_i\pm e_j$ ($i\ne j$) where $e_k$ is the standard orthonormal basis in $R^8$ plus plenty of others. Now it looks like it suffices to check that if $A$ is a quadratic form close to $I$ and $(Ax,x)\ge 2$ for all $x=e_i\pm e_j$, then $det\,A\ge 1$. If $a_{ii}=1+b_i$, we should have $2|a_{ij}|\le b_i+b_j$ whence, $b_i+b_j\ge 0$ for all $i\ne j$ and $a_{ij}^2\le\frac 12(b_i^2+b_j^2)$. That gives the linear term of at least $B=\max_j b_j=\max_j |b_j|$ and the quadratic terms of order $B^2$, giving the desired result. Am I missing anything? $\endgroup$
    – fedja
    Jun 29, 2012 at 1:42
  • $\begingroup$ fedja, I trust you have it correct. I should not pretend to know anything useful about real valued quadratic forms. i just thought that the two examples I suggested might be some of the few where the smallest possible determinant would actually occur at a (scaled) integral lattice. But i will read your comment some more and see if anything else comes to mind. $\endgroup$
    – Will Jagy
    Jun 29, 2012 at 3:36

1 Answer 1

Reset to default
7
$\begingroup$

According to a talk that I found on the web, it is a theorem of Voronoi that every indecomposable root lattice is extreme. Also the $E_8$ lattice is the union of two copies of the $D_8$ lattice with the same sphere radius. And, of course, the rotational symmetry group of the $E_8$ is transitive on roots. So I think that that gives it to you: If you put two small, flat dents in a round ball in $\mathbb{R}^8$, then you cannot deform the $E_8$ lattice packing, because there is a $D_8$ lattice inside that is far away from the dents.

The same argument works for the Leech lattice, which contains a $D_{24}$ sublattice of index 8192. It also works for $E_7$, because it contains $A_7$. (Also $E_8$ contains $A_8$, but not in the same way, since $[E_8:A_8] = 3$ while $[E_7:A_7] = 2$.)

However, I do not think that it is known, nor even a conjecture with strong evidence, that the kissing number of the best lattice sphere packing in dimension $n \to \infty$ is more than the bare minimum $n(n+1)$.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Yeah, this is pretty much the same as what I wrote in my comment (only stated in a higher level language). Alas, I also have no idea why the kissing number of the best lattice should be high in high dimensions. So, it looks like I have no choice but to accept the current state of affairs by accepting this answer. :) $\endgroup$
    – fedja
    Jul 1, 2012 at 11:50
  • $\begingroup$ For what it's worth, this type of argument seems to show that the ball in $R^4$ is not a local minimizer either despite it is irreducible (you need to cut a noticeable spherical slice of depth $\delta$ now instead of just a cap). $\endgroup$
    – fedja
    Jul 1, 2012 at 11:57
  • $\begingroup$ Yeah, I concede that my answer is not entirely independent from the one that you already know; but it may still have some conceptual value. Your approach is to make a direct stability calculation for a sphere with two polar dents, simplified with the remaining available symmetry. It seems to show that in dimensions 5-8 and 24, a sphere with two dents packs only with the same lattice determinant as a perfect sphere. And in dimensions such as 12, modulo a conjecture that a highly symmetric lattice is the best one. $\endgroup$
    – Greg Kuperberg
    Jul 2, 2012 at 4:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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