# Minkowski's successive minima: A quantity not much larger than det(L)^(1/n) and not much smaller than λ_n(L)?

Let $\mathbf{v}_1, \mathbf{v}_2, ..., \mathbf{v}_n$ be $n$ linearly independent vectors in an $n$-dimensional lattice $\Lambda$ in $\mathbf{R}^n$ and let $\mathbf{v}^*_1 ,\mathbf{v}^*_2, ..., \mathbf{v}^*_n$ denote the Gram-Schmidt orthogonalization of $\mathbf{v}_1, \mathbf{v}_2, ..., \mathbf{v}_n$.

I'm interested in the following quantity:

$$f_n(\Lambda) =\min_{{\mathbf{v}_1, \mathbf{v}_2, ..., \mathbf{v}_n \in \Lambda}} \{\max_i \Vert\mathbf{v}^*_i\Vert \}$$

Notice that — in a definition similar to Minkowski's successive minima — $f_n(\Lambda)$ is the smallest radius of a ball in $\mathbf{R}^n$ that contains $n$ linearly independent Gram–Schmidt vectors generated by $\Lambda$.

In particular we have that $$(1)\;\;f_n(\Lambda)\leq \lambda_n(\Lambda),$$ where $\lambda_n(\Lambda)$ is the $n$-th successive minimum of $\Lambda$, and $$\;\;\;\;\; \ (2)\;\;\det(\Lambda)^{{1}/{n}} \leq f_n(\Lambda).$$

So I come to my question: Is it possible to bound (1) or (2) in the opposite direction to make a statement that qualitatively reads: $f_n(\Lambda)$ is not much larger than $\det(\Lambda)^{{1}/{n}}$ and/or not much smaller than $\lambda_n(\Lambda)$?

For example, say something like:

$$f_n(\Lambda)\leq c_1 \det(\Lambda)^{{1}/{n}} \;\;\;\mbox{ and/or }\;\;\; f_n(\Lambda)\geq \frac{1}{c_2}\lambda_n(\Lambda)$$

for some quantities $c_1,c_2$ that are "not too large" ?

• Since there are lattices in ${\bf R}^n$ with $\lambda_n(\Lambda)/\det(\Lambda)^{\frac 1 n}$ arbitrarily large (for example the ones gererated by the $(0,\ldots,1,\ldots,0)$ and $(0,\ldots,0,N)$ for $N\ge 1$ I doubt that the two inequalities could both be valid. Apr 5, 2013 at 11:52
• @Jean Thanks for the comment. Now that you mention it I see that there are lattices with $f_n(\Lambda)/det(\Lambda)$ arbitrarily large. E.g. in $\mathbb{R}^2$ the lattice generated by $(1/a , 0)$, $(0 , a)$ for $a>1$ has $f_n(\Lambda)/det(\Lambda) = a$. Apr 5, 2013 at 12:50

As already pointed out by Jean and Alexander, bound (2) does not hold in the opposite direction. For bound (1), the opposite inequality holds with $c_2=n$.

Theorem For any $n$-dimensional lattice $\Lambda$, $f_n(\Lambda)\geq \lambda_n(\Lambda)/n$.

Proof: Let ${\bf v}_1,\ldots,{\bf v}_n$ be a set of linearly independent vectors in a lattice minimizing $\max_i \|{\bf v}_i^*\|$. You can always turn ${\bf v}_1,\ldots,{\bf v}_n$ into a basis without increasing $\|{\bf v}_i^*\|$. So, assume it is a basis. Let ${\bf w}_1,\ldots,{\bf w}_n$ be a basis of the dual lattice $\Lambda^*$ such that ${\bf v_i}\cdot {\bf w_j}=\delta_{i,j}$.

Orthogonalizing ${\bf w}_1,\ldots,{\bf w}_n$ in reverse order gives $\|{\bf w}_i^*\| = \|{\bf v}_i^*\|^{-1}$. So, $\max_i \|{\bf v}_i^*\| = (\min_i \|{\bf w}_i^*\|)^{-1}$. Using the bound $\min_i \|{\bf w}_i^*\| \leq \lambda_1(\Lambda^*)$ and Banaszczyk's transference theorem $\lambda_n(\Lambda)\lambda_1(\Lambda^*)\leq n$ gives: $\max_i \|{\bf v}_i^*\| = (\min_i \|{\bf w}_i^*\|)^{-1} \geq \lambda_1(\Lambda^*)^{-1}\geq \lambda_n(\Lambda)/n$. So, $f_n(\Lambda)\geq \lambda_n(\Lambda)/n$. $\Box$