A Geometric Face of Diophantine Analysis

Abstract

Geometry of numbers is a powerful tool in studying Diophantine inequalities. In geometry of numbers a basic question is to find a non-zero lattice vector from a convex subset in an n-dimensional space, say in $\mathbb {R}^n$. Hermann Minkowski answered this challenge with his convex body theorems. In these lectures we shall discuss how to apply Minkowski’s theorems to prove classical Diophantine inequalities and some variations of Siegel’s lemma. Further, we shall shortly discuss corresponding inequalities over imaginary quadratic fields. From the nature of the above results follows that a lower bound for the absolute value of an arbitrary non-zero linear form in m linearly independent numbers is not bigger than a certain positive function depending on the coefficients and the number of variables of the linear form. For a concrete set of numbers it is a big challenge to find such lower bounds. We will give a recent example on such lower bounds, namely a new generalised transcendence measure for e.

Appendix

1.1 Cauchy–Schwarz Inequality

Now we suppose that R is an integral domain with absolute value and the norm of $\overline{a}\in R^N$ is defined by

$$\begin{aligned} \Vert \overline{a}\Vert =\sqrt{\overline{a}\cdot \overline{a}} \end{aligned}$$

via the inner product $\cdot$ in $R^N$.

Lemma 14.6

(Cauchy–Schwarz inequality)

$$\begin{aligned} |\overline{a}\cdot \overline{b}|\le \Vert \overline{a}\Vert \Vert \overline{b}\Vert \quad \text{ for } \text{ all }\quad \overline{a}, \overline{b}\in R^{N}. \end{aligned}$$

(14.9)

The equality

$$\begin{aligned} |\overline{a}\cdot \overline{b}|=\Vert \overline{a}\Vert \Vert \overline{b}\Vert \end{aligned}$$

implies

$$\begin{aligned} \Vert \overline{b}\Vert \overline{a}=\omega \Vert \overline{a}\Vert \overline{b},\quad |\omega |=1,\ \omega \in R. \end{aligned}$$

(14.10)

In particular, when $R=\mathbb {Z}$, then $\omega =\pm 1$.

Proof

Write

$$\begin{aligned} \overline{w}:=\Vert \overline{b}\Vert ^2\overline{a}-(\overline{a}\cdot \overline{b})\overline{b}. \end{aligned}$$

Then

$$\begin{aligned} \overline{w}\cdot \overline{b}=0 \end{aligned}$$

and consequently

$$\begin{aligned} \Vert \overline{b}\Vert ^4\Vert \overline{a}\Vert ^2=\Vert \overline{w}\Vert ^2+|\overline{a}\cdot \overline{b}|^2\Vert \overline{b}\Vert ^2 \ge |\overline{a}\cdot \overline{b}|^2\Vert \overline{b}\Vert ^2 \end{aligned}$$

(14.11)

which implies (14.9).

Suppose now

$$\begin{aligned} |\overline{a}\cdot \overline{b}|=\Vert \overline{a}\Vert \Vert \overline{b}\Vert \quad \Leftrightarrow \quad \overline{a}\cdot \overline{b}=\omega \Vert \overline{a}\Vert \Vert \overline{b}\Vert ,\quad |\omega |=1,\ \omega \in R. \end{aligned}$$

Then (14.11) reads

$$\begin{aligned} \Vert \overline{b}\Vert ^4\Vert \overline{a}\Vert ^2=\Vert \overline{w}\Vert ^2+\Vert \overline{a}\Vert ^2\Vert \overline{b}\Vert ^4. \end{aligned}$$

Hence,

$$\begin{aligned} \Vert \overline{w}\Vert ^2=0\quad \Rightarrow \quad \overline{w}=\Vert \overline{b}\Vert ^2\overline{a}-(\overline{a}\cdot \overline{b})\overline{b}=\overline{0}\\ \quad \Rightarrow \quad \Vert \overline{b}\Vert ^2\overline{a}=(\overline{a}\cdot \overline{b})\overline{b}= \omega \Vert \overline{a}\Vert \Vert \overline{b}\Vert \overline{b}, \end{aligned}$$

which proves (14.10).   $\square$

1.2 Gram Determinant

The Gram determinant $\det [\overline{b}_m\cdot \overline{b}_n]_{1\le m,n\le M}$ satisfies the following estimates

$$\begin{aligned} \sqrt{\det (\mathcal {A}\mathcal {A}^t)}= & {} \sqrt{\det [\overline{b}_m\cdot \overline{b}_n]_{1\le m,n\le M}}\nonumber \\\le & {} \sqrt{\prod \limits _{m=1}^{M}\overline{b}_m\cdot \overline{b}_m}= \prod \limits _{m=1}^{M}\Vert \overline{b}_m\Vert _{2}\nonumber \\\le & {} \prod \limits _{m=1}^{M}\Vert \overline{b}_m\Vert _{1}\le \prod \limits _{m=1}^{M}N\underset{1\le n\le N}{\max }|a_{m,n}|\nonumber \\\le & {} \left( N\underset{1\le m,n\le N}{\max }|a_{m,n}|\right) ^M. \end{aligned}$$

(14.12)