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

I'm sorry for the title, it was little difficult to phrase..

Let us consider a matrix lattice $L\subset M_n(\mathbb{C})$. By this I mean a discrete additive group in $M_n(\mathbb{C})$. Let us also suppose that this lattice $L$ has a peculiar property that $|det(X)|\geq 1$, for all non-zero elements in $L$. I'm interested about the asymptotic behavior (as a function of R) of the sums $$ \sum_{ X \neq 0, ||X||_F\leq R, X\in L}\frac{1}{|\det(X)|^s}, $$ where $s$ is a positive number and $||..||_F$ is the Frobenius norm.

A natural example of such lattices appear if we consider a totally real field extension $L/\mathbb{Q}$. We can now define the following map (using the classical Minkowski embedding), $\psi: L \mapsto M_n(\mathbb{R})$, where $\psi(a)=\mathrm{diag}(\sigma_1(a),\dots, \sigma_n(a))$ and where $\sigma_i$ are the $\mathbb{Q}$-embeddings. Now $\psi(O_L)$, where $O_L$ is the ring of algebraic integers, is an $n$-dimensional lattice in $M_n(\mathbb{C})$ (well actually in $M_n(\mathbb{R})$ ). The corresponding sum is then

$$ \sum_{x \neq 0, ||\psi(x)||_F \leq R } \frac{1}{| det(\psi(x)) |^s} =\sum_{x \neq 0, ||\psi(x)||_F \leq R }\frac{1}{|N_{L/\mathbb{Q}}(x)|^s}, $$

where $N_{L/\mathbb{Q}}$ is the algebraic norm.

These sums naturally appear when analyzing performance of space-time block codes in multiple antenna channels. In the case of number fields and in some other cases I can give a good answer. These are based on approximation of truncated Dedekind zeta-function and on density results of units. In the general case I can give some loose bounds. As a general problem this is simply a question about summing values of a function over lattice points. However, these determinantial sums are very sensitive on the chosen lattices. I would be very interested on any references where this kind of sums have been considered before.

share|improve this question
add comment

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer


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

Browse other questions tagged or ask your own question.