# Bound on the shortest non-zero vector in any full rank n-dimensional lattice $\Lambda \subseteq \mathbb{R}^n$ with respect to the $1$-norm.

How can i prove $$\lambda_1 \; \leq \; (n! \; det(\Lambda))^{\frac{1}{n}} \approx \frac{n \; det{(\Lambda)}^\frac{1}{n}}{e}.$$

Here $$\Lambda_1$$ is shortest non-zero vector. My initial thought was using Minkowski theorem (choosing $$S =$$ n-Ball of radius $$\sqrt n \frac{\lambda_1}{n}$$) and proof by contradiction (assuming $$\lambda_1 \; > \; (n! \; det(\Lambda))^{\frac{1}{n}}$$ and contracting with minimality of $$\lambda_1$$).

[Minkowski’s convex body theorem] : Let $$\Lambda$$ be a full dimensional lattice. If $$S \subset \mathbb{R}^n$$ is a symmetric convex body of volume $$vol(S) > 2^n det(\Lambda)$$ , then $$S$$ contains a non-zero lattice point.

$$1-$$norm for a vector $$x$$ : $$\sum{}{}{|x_i|}$$.

• What is the volume of $\{ x, \|x\|_1 \le r\}$ ? Commented Nov 13, 2020 at 11:21
• anything unclear ? Commented Nov 14, 2020 at 10:19
• @ reuns No, thanks. I will accept your answer as correct one.Thanks
– mike
Commented Nov 14, 2020 at 14:47

$$C_n=Vol(\{ x, \|x\|_1\le 1\})= \int_{-1}^1 Vol(\{ y, \|y\|_1\le 1-x_1\})dx_1$$ $$=\int_{-1}^1 (1-|x_1|)^{n-1}C_{n-1}dx_1=\frac{2}n C_{n-1}=\frac{2^n}{n!}$$ For all $$r<\lambda_1/2$$ then $$\{x, \|x\|_1\le r\}$$ doesn't intersect its $$\Lambda$$-translates. Thus $$\det(\Lambda)\ge Vol(\{ x, \|x\|_1\le r\})=r^n\frac{2^n}{n!}$$ and hence $$\lambda_1^n\le n! \det(\Lambda)$$
I would like to add my answer too, somebody may find it useful. We show that every (full rank, n-dimensional) lattice $$\Lambda$$ always contains a non-zero vector $$x$$ such that $$||x||_1 \leq \; (n! \; det(\Lambda))^{\frac{1}{n}}$$. Let $$\lambda_1 = min\{||x||_1 : x \in \Lambda \setminus \{0\}\}$$ and assume for contradiction $$l > (n! \; det(\Lambda))^{\frac{1}{n}}$$. As @reuns mentioned take $$S = \{ x : ||x||_1 < l \}$$. Notice that $$S$$ is convex, symmetric and has volume $$vol(S) = 2^n \frac{l^n}{n!} > 2^n det(\Lambda)$$. So, by Minkowski's theorem, $$S$$ contains a non-zero lattice vector $$x$$. By definition of $$S$$ ,we have$$||x||_1 < l$$, a contradiction to the minimality of $$l$$.