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Suppose $\Lambda\subset\mathbb{R}^n$ is a unimodular (i.e. volume $1$) lattice in Euclidean space. Let $v_1,\dots, v_n\in\Lambda$ be a basis of $\Lambda$ such that the product of lengths $A=|v_1|\cdots|v_n|$ is minimal. I'd like to bound this minimal product $A$ from above as $\Lambda$ varies (and $n$ is fixed). I can prove that such an upper bound exists - for instance for $n=2$, it's attained by the A2 ("hexagonal") lattice since any lattice has a basis such that the angle between the two basis vectors is between 60 and 120 degrees. I don't know what it is for general $n$. It seems like this should be known, but I can't find it. Does anyone know of a good bound?

flag	asked Aug 21 2010 at 21:18 Dmitry Vaintrob 1,002●3●10

Richard Borcherds · Answer 1 · 2010-08-21 22:45:41Z UTC

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There is such a bound due to Minkowski. More precisely he shows for any symmetric convex region C, one can find a basis in xC for sufficiently large x. I cant remember the exact bound offhand, but it is probably somewhere in "An Introduction to the Geometry of Numbers" by Cassels (or any other book on the geometry of numbers).

Addendum: on checking I realized that I misremembered Minkowski's result. He does indeed gives a bound for the product of the lengths of a basis in the lattice (in terms of successive minima), but it seems to be a basis for the real vector space rather than a basis for the lattice as you asked for.

edited Aug 21 2010 at 22:45

answered Aug 21 2010 at 22:11

Richard Borcherds
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	The keyword here is "successive minima". – Felipe Voloch Aug 21 2010 at 22:20
	That's what I thought, but on checking things seems more complicated than I remembered. Successive minima are what you need if you are looking for a basis of the real vector space, but if you want a basis for the lattice you might need something more. – Richard Borcherds Aug 21 2010 at 22:37
	Thank you for the reference! Interestingly, this bound also came up in number theory. What do Cassels and Minkowski use the bound for? I need it to bound from above the largest archimedean length of a fundamental unit under the Minkowski embedding (for a suitably chosen basis of fundamental units.) I wonder if such a bound is known - I'll make this another Math Overflow question. – Dmitry Vaintrob Aug 21 2010 at 22:52

felix · Answer 2 · 2010-08-21 23:53:39Z UTC

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I don't know how good the bound is you can obtain from this, but what about taking a Korkine-Zolotarev reduced basis of $\Lambda$, say $(b_1, \dots, b_n)$: then, by this paper, $\|b_i\|_2^2 \le \frac{i + 3}{4} \lambda_i(\Lambda)^2$, where $\lambda_i(\Lambda)$ is the $i$-th successive minimum of $\Lambda$. By Minkowski, $\prod_{i=1}^n \lambda_i(\Lambda) \le \gamma_n^{n/2} \det \Lambda = \gamma_n^{n/2}$ (in your case), $\gamma_n$ being the $n$-th Hermite constant, whence you get $A \le \prod_{i=1}^n \|b_i\|_2 \le \frac{\gamma_n^{n/2}}{2^n} \prod_{i=1}^n \sqrt{i + 3}$.

answered Aug 21 2010 at 23:53

felix
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	Note that if $\Lambda$ is a critical lattice, then $\prod_{i=1}^n \lambda_i(\Lambda) = \gamma_n^{n/2} \det \Lambda$. Moreover, if $(b_1, \dots, b_n)$ is any basis of $\Lambda$ with $\\|b_1\\|_2 \le \dots \le \\|b_n\\|_2$, then $\\|b_i\\|_2 \le \lambda_i(\Lambda)$ since $\\|b_i\\|_2 = \max\{ \\|b_1\\|_2, \dots, \\|b_i\\|_2 \} \ge \lambda_i(\Lambda)$. Hence, $\prod_{i=1}^n \\|b_i\\|_2 \ge \prod_{i=1}^n \lambda_i(\Lambda)$. – felix Aug 22 2010 at 19:29
	Since for any dimension $n$, there exists a critical lattice $\Lambda \subset \mathbb{R}^n$, we see that a bound for $A$ must be at least $\gamma_n^{n/2}$. So the question is: how good/bad is the factor $2^{-n} \prod_{i=1}^n \sqrt{i + 3}$? – felix Aug 22 2010 at 19:30

Bounding the product of lengths of basis vectors of a unimodular lattice

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