Bounding the product of lengths of basis vectors of a unimodular lattice - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T19:14:45Z http://mathoverflow.net/feeds/question/36329 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/36329/bounding-the-product-of-lengths-of-basis-vectors-of-a-unimodular-lattice Bounding the product of lengths of basis vectors of a unimodular lattice Dmitry Vaintrob 2010-08-21T21:18:30Z 2010-08-21T23:53:39Z <p>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?</p> http://mathoverflow.net/questions/36329/bounding-the-product-of-lengths-of-basis-vectors-of-a-unimodular-lattice/36332#36332 Answer by Richard Borcherds for Bounding the product of lengths of basis vectors of a unimodular lattice Richard Borcherds 2010-08-21T22:11:19Z 2010-08-21T22:45:41Z <p>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). </p> <p>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. </p> http://mathoverflow.net/questions/36329/bounding-the-product-of-lengths-of-basis-vectors-of-a-unimodular-lattice/36340#36340 Answer by felix for Bounding the product of lengths of basis vectors of a unimodular lattice felix 2010-08-21T23:53:39Z 2010-08-21T23:53:39Z <p>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 <a href="http://www.springerlink.com/content/yh7k451558438101/" rel="nofollow">this paper</a>, $\|b_i\|_2^2 \le \frac{i + 3}{4} \lambda_i(\Lambda)^2$, 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 <a href="http://en.wikipedia.org/wiki/Hermite_constant" rel="nofollow">$n$-th Hermite constant</a>, whence you get $A \le \prod_{i=1}^n \|b_i\|_2 \le \frac{\gamma_n^{n/2}}{2^n} \prod_{i=1}^n \sqrt{i + 3}$.</p>