Is the ball reducible in some high dimension?

2012-06-27T17:50:45Z

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 :).

Answer by Greg Kuperberg for Is the ball reducible in some high dimension?

2012-06-30T19:10:28Z

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 $n(n+1)$.

Is the ball reducible in some high dimension? - MathOverflow

Is the ball reducible in some high dimension?

Answer by Greg Kuperberg for Is the ball reducible in some high dimension?