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

While trying to use Minkowski's theorem to calculate the (left) class number of a noncommutative ring, I ran into the following problem:

What is the volume of the largest symmetric convex subset $S$ of $\mbox{Mat}_3(\mathbb{R})$ such that every matrix in this subset has determinant at most $1$? In particular, is there such a set with volume greater than 885?

The best subset that I have been able to find so far is

$S_1 = \{M\in\mbox{Mat}_3(\mathbb{R})\mid \forall O\in O(3,\mathbb{R}), \mbox{ Tr}(MO) \le 3\}$.

If we apply Gram-Schmidt to the rows of $M$ and use the resulting orthonormal basis for the columns of an orthogonal matrix $O$, then by the AM-GM inequality we can conclude that $\det M \le 1$ from $\mbox{Tr}(MO)\le 3$. Furthermore, if $S$ is any symmetric convex set of matrices of determinant at most $1$ which contains an orthogonal matrix $O$, then every $M\in S$ must satisfy $\mbox{Tr}(MO^{-1}) \le 3$ by considering the tangent hyperplane to the collection of matrices with determinant $1$. Thus $S_1$ is the largest such set which contains all orthogonal matrices.

Unfortunately, it's incredibly difficult to calculate the volume of $S_1$. All I've been able to do so far is to note that it contains the set $S_r$ of matrices such that the sum of the norms of the rows is at most $3$, and that the volume of $S_r$ is $\frac{972\pi^3}{35}\approx 861.089$. If we let $S_c$ be the similar set obtained by considering the columns, then Monte-Carlo integration indicates that the volume of $S_r\cup S_c$ is around $1050$, but I haven't been able to prove this.

What is the volume of $S_1$? If this is too hard, what is the volume of $S_r\cup S_c$?

Motivation

The ring I am dealing with is $\mathbb{Z}\langle a,x\rangle/(a^3 = a-1, x^3 = 3x-1, xa = a(x^2-2)+2-x)$. It is an order in the central simple algebra over $\mathbb{Q}$ with invariants $1/3$ at $3$, $2/3$ at $2$, and $0$ at every other prime. The discriminant of this ring is $-6^6$ (as one can verify by working out the trace form by hand).

Since the associated central simple algebra has nontrivial invariants at $2$ and $3$, and since $\mbox{Nm}(x+1) = -3$ and $(a+xa-1)^3 = 2$, all left ideals of norm at most $4$ are principal. If we tensor this ring up to $\mathbb{R}$ it becomes $\mbox{Mat}_3(\mathbb{R})$, so we can think of it as a lattice in $\mbox{Mat}_3(\mathbb{R})$ of covolume $\sqrt{6^6}$. By Minkowski's bound, every left ideal class contains a representative of norm at most $\left(\frac{2^9\sqrt{6^6}}{\mbox{Vol}(S)}\right)^{1/3}$, so if we can find a set $S$ with volume at least $885$ then we can conclude that this ring has left class number $1$.

share|improve this question
    
Why is $\left| \det M \right| \leq 1$ for all $M \in S_1$? It's not immediately obvious to me (though I have the feeling it should be easy). –  Noam D. Elkies Jul 6 '13 at 21:07
add comment

1 Answer

up vote 8 down vote accepted

The volume of $S_1$ is $\frac{16767 \pi^4}{560} \approx 2916.53$.

The idea is to realize that the set $S_1$ can be equivalently described as the nuclear norm (sum of the singular values $\sigma_i$) of $M$ being less than or equal to 3, i.e., $S_1 = \{M : \sum_i \sigma_i(M) \leq 3 \}$.

This is an orthogonally invariant set, so from the singular value decomposition $M = U \Sigma V^T$ we can do the integral pretty explicitly, using the fact that the Jacobian is given by $(\sigma_1^2-\sigma_2^2) (\sigma_1^2-\sigma_3^2)(\sigma_2^2-\sigma_3^2)$, and the integral is over the polytope $\{(\sigma_1,\sigma_2,\sigma_3) : \sigma_1 \geq \sigma_2 \geq \sigma_3 \geq 0, \, \sigma_1 + \sigma_2 + \sigma_3 \leq 3\}$ (this integral is $\frac{16767}{17920}$). The other factor corresponds to the orthogonal matrices $U,V$, and is $\frac{1}{2^n} Vol(O^n)^2= \frac{1}{2^n} (\prod_{i=1}^n Vol(S^{i-1}))^2 = \frac{1}{2^n} (\prod_{i=1}^n \frac{2 \pi^{i/2}}{\Gamma(i/2)})^2$, which for $n=3$ gives $32 \pi^4$.

share|improve this answer
    
@Noam, an easy way to see this is $|\det M | = \prod_i \sigma_i \leq (\frac{1}{n}\sum_i \sigma_i)^n$. –  coma Jul 6 '13 at 23:08
add comment

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.