TITLE: Bounding the product of lengths of basis vectors of a unimodular lattice
QUESTION [5 upvotes]: 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?

REPLY [2 votes]: 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 $A \le \prod_{i=1}^n \|b_i\|_2 \le \frac{\gamma_n^{n/2}}{2^n} \prod_{i=1}^n \sqrt{i + 3}$.