# Minimum of the product of linear forms over a lattice

In Chapter [IX.1] of Siegel's Lectures on the Geometry of Numbers it is shown that if we have $n$ linear forms $y_{j}=\sum_{k=1}^{n}{a_{jk}x_{k}},\quad j=1,\ldots,n$, with the coefficient matrix $(a_{jk})$ being non-singular and having determinant $D$, then

$$\min_{(x_{i})_{i}^{n} \in \mathbb{Z}^{n}-\{0\}}|y_{1}\ldots y_{n}| \leq\frac{n!|D|}{n^{n}}.$$

This bound is not, however, best possible. Siegel goes on to prove that for $n=2$ the best possible upper bound on $|y_{1}y_{2}|$ is $\sqrt{\frac{1}{5}}|D|$.

My question is: let $|y_{1}\ldots y_{n}| \leq c_{n}|D|$ be the best possible upper bound, what is known about $c_{n}$?

Siegel's result is $c_{2}=\frac{1}{\sqrt{5}}$ and he also credits Davenport with showing that $c_{3} \leq \frac{1}{7}$.

What else has been found since?

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I am probably misinterpreting the notation. What if $A$ is the identity matrix. Then $D=1$, and the product of the $y$s can be made arbitrarily large. Some control on the $x_i$ needs to be specified? –  Suvrit Nov 18 '13 at 19:28
@Suvrit Yes, sorry, I garbled it a bit. Please see the edit. –  Felix Goldberg Nov 18 '13 at 19:33
Thanks Felix. Just one more clarification: If $A=I_n$, then $D=1$, which $y_j=x_j$. The minimum of the product on the lhs will be when $|x_j|=1$, which however is $\not\le$ the specified rhs...what am I missing? –  Suvrit Nov 18 '13 at 22:02
@Suvrit You can take $x_{1}=0$ and $|x_{j}|=1, j \geq 2$ and get $0$. –  Felix Goldberg Nov 19 '13 at 0:08
But you said that $(x_i)_i^n \in Z^n-0$; I assumed that meant all the $x_i$ are nonzero....confusing notation. –  Suvrit Nov 19 '13 at 2:18