A note on badly approximable systems of linear forms over a field of formal series

Abstract

We prove that the set of badly approximable systems of M linear forms in N variables over the field of formal power series is hyperplane absolute winning. A consequence of this winning property is that the set has full Hausdorff dimension.

1 Introduction

Metric Diophantine approximation is concerned with the study of approximating irrational numbers by rational numbers from a measure theoretic point of view. Given any irrational number x it is possible to find rational numbers p/q that approximate x to any prescribed accuracy $\psi (q) > 0$, but as $\psi (q)\rightarrow 0$, the rational approximations necessarily have larger and larger denominators. As a consequence of Dirichlet’s theorem (1842), for any irrational number x, there exists infinitely many rational numbers p/q such that $|x-p/q|<1/q^2$. Improving this rate of approximation leads to investigate two sets: the set of $\psi$-approximable numbers and the set of badly approximable numbers. Khintchine (1924) and Jarník (1931) theorems gives a full measure theoretic picture for the size (Lebesgue measure and Hausdorff measure respectively) of $\psi$-approximable set

$$\begin{aligned} W(\psi ):=\{x\in \mathbb {R}: |x-p/q|<\psi (q) \ \text {for infinitely many } (p, q)\in \mathbb {Z}\times {\mathbb {N}}\}. \end{aligned}$$

We refer the reader to [7] for the most recent result in this direction. The numbers x for which there exists a constant $c(x)>0$ such that $|x - p/q| \ge c(x)|q|^{-2}$ for all $p/q\in {\mathbb {Q}}$ are called the badly approximable numbers. From Khintchine’s theorem, it follows that the Lebesgue measure of the set of badly approximable numbers is zero. Jarník (1932) proved that the Hausdorff dimension of this set is 1, that is, there are too many badly approximable numbers. Some examples of badly approximable numbers include the quadratic irrationals.

The one dimensional notions extends naturally to higher dimensions, therein, one can consider either simultaneous (approximation by rational points) or dual (approximation by rational hyperplanes) setting. Combining both approaches leads to the theory of approximating systems of linear forms. Historically, the main focus has been on establishing the metrical theory for the set of well and badly approximable systems of linear forms. Let

$$\begin{aligned}\textrm{Bad}(M,N)=\left\{ A \in \mathbb {R}^{MN}: \inf _{\textbf{q}\in \mathbb {Z}^N\setminus \{\textbf{0}\}} |\textbf{q}|^{N/M}|\textbf{q}A -\textbf{p}|>0\right\} \end{aligned}$$

denote the set of badly approximable systems of linear forms. Here and throughout, the matrix A is regarded as a point in $\mathbb {R}^{MN}$, the norm $|\textbf{q}|=\max \{|q_1|,\ldots ,|q_N|\}$ is the supremum norm, and the system

$$\begin{aligned} \{q_1x_{1i}+ \dots + q_{N}x_{Ni} \;: \; i = 1,\ldots , M\} \end{aligned}$$

of M linear forms in N variables $q_1,\ldots ,q_N$ is written more concisely as $\textbf{q}A$.

Wolfgang Schmidt [12] introduced Schmidt’s game for showing that the set of badly approximable linear forms is a winning set, thus proving a stronger property than just full Hausdorff dimension. The basic scheme appearing in the definition of a winning set has seen many variants. Some of these include the notions of absolute winning and strong winning sets [11], modified winning sets [6], hyperplane absolute winning, [2], potential winning [4], and Cantor winning [1].

The aim of this paper is to prove that the set of badly approximable linear forms over the field of formal power series is Hyperplane Absolute Winning (HAW). The definition of a Hyperplane game is given below and from it one can easily see that HAW property is in fact stronger than $\alpha$-winning in a Schmidt sense [2, Proposition 2.3]. The hyperplane game was defined in [2] and it was proved in [3] that systems of badly approximable linear forms over real numbers are hyperplane winning. See also [5, Theorem 1.1] for a proof of $\textrm{Bad}(M, N)$ to be HAW.

Let $\mathbb {F}$ denote the finite field of $q=p^r$ elements, where p is a prime and r is a positive integer. We define the field of Laurent series with coefficients from $\mathbb {F}$ or the field of formal power series with coefficients from $\mathbb {F}$ to be

$$\begin{aligned} \mathcal {L} = \left\{ \sum _{i=-n}^\infty a_{-i}X^{-i}\,: \, n\in \mathbb {Z},\ a_i\in \mathbb {F},\ a_n\ne 0\right\} \cup \{0\}. \end{aligned}$$

An absolute value $\Vert \cdot \Vert$ on $\mathcal {L}$ can be defined as

$$\begin{aligned} \left\| \sum _{i=-n}^\infty a_{-i}X^{-i} \right\| =q^n,\quad \,\left\| 0\right\| =0. \end{aligned}$$

Under the induced metric $d(x,y)=\Vert x-y\Vert$, the space $(\mathcal {L},d)$ is a complete metric space. Furthermore the absolute value satisfies for any $x,y\in \mathcal {L},$

$$\begin{aligned}{} & {} \Vert x\Vert \ge 0 \ \ \forall x, \text { and } \Vert x\Vert =0 \text { if and only if } x=0, \\{} & {} \Vert xy\Vert =\Vert x\Vert \, \Vert y\Vert , \end{aligned}$$

$$\begin{aligned} \Vert x+y\Vert \le \max (\Vert x\Vert ,\Vert y\Vert ). \end{aligned}$$

(1)

The last property is known as the non-Archimedean property. In fact, equality holds whenever $\Vert x\Vert \ne \Vert y\Vert$.

As we will be working in the finite dimensional vector spaces over $\mathcal {L}$, we need an appropriate extension of the one-dimensional absolute value.

Definition 1.1

Let $h\in \mathbb {N}$. For any $\textbf{x}=(x_1,\ldots , x_h)\in \mathcal {L}^h$, we define the height of x to be

$$\begin{aligned} \Vert \textbf{x}\Vert _{\infty } = \max \{\Vert x_1\Vert ,\ldots ,\Vert x_h\Vert \}. \end{aligned}$$

One can easily see that the property (1) is also true for $\Vert \textbf{x}\Vert _\infty$.

In $\mathcal {L}$, the polynomial ring $\mathbb {F}[X]$ plays a role analogous to the one played by the integers in the field of real numbers. We define the polynomial part of a non-zero element by

$$\begin{aligned} \left[ \sum _{i=-n}^\infty a_{-i}X^{-i} \right] = \sum _{i=-n}^0 a_{-i}X^{-i} \end{aligned}$$

whenever $n\ge 0$. When $n<0$, the polynomial part is equal to zero. Likewise, the polynomial part of the zero element is itself equal to zero.

Using the above definitions, let us define the distance to $\mathbb {F}[X]^h$ for a point $\textbf{x}\in \mathcal {L}^h$ as

$$\begin{aligned} |\langle \textbf{x}\rangle | = \min _{\textbf{p}\in \mathbb {F}[X]^h} \Vert \textbf{x}-\textbf{p}\Vert _{\infty }. \end{aligned}$$

Also, for $\textbf{x}=(x_1,\ldots ,x_h)$ and $\textbf{y}=(y_1,\ldots ,y_h) \in \mathcal {L}^h$, we define

$$\begin{aligned} \textbf{x}\cdot \textbf{y}= \sum \limits _{i=1}^{h} x_i y_i. \end{aligned}$$

We will make use of the following inequality

$$\begin{aligned} \Vert \textbf{x}\cdot \textbf{y}\Vert \le \Vert \textbf{x}\Vert _\infty \, \Vert \textbf{y}\Vert _\infty . \end{aligned}$$

Proof

$$\begin{aligned} \Vert \textbf{x}\cdot \textbf{y}\Vert = \left\| \sum \limits _{i=1}^{h} x_i y_i \right\| \le \max _{i} \Vert x_i y_i \Vert = \max _i \Vert x_i\Vert \, \Vert y_i \Vert \le \max _i \Vert x_i \Vert \max _j \Vert y_j \Vert =\Vert \textbf{x}\Vert _\infty \, \Vert \textbf{y}\Vert _\infty . \end{aligned}$$

$\square$

We identify $M\times N$-matrices with coefficients from $\mathcal {L}$ with $\mathcal {L}^{MN}$ in the usual way. Matrix products are defined as in the real case. A system of M linear forms in N variables given by $A\in \mathcal {L}^{MN}$ is called badly approximable if there exists a constant $K=K(A)>0$ such that for every $\textbf{q}\in \mathbb {F}[X]^N\backslash \{\textbf{0}\}$,

$$\begin{aligned} |\langle A\textbf{q}\rangle |^M > \frac{K}{\Vert \textbf{q}\Vert _\infty ^N}. \end{aligned}$$

Let $\textrm{Bad}(M, N)$ denotes the set of all badly approximable linear forms $A\in \mathcal {L}^{MN}$. In [9], Kristensen proved that the set $\textrm{Bad}(M, N)$ is of full Hausdorff dimension by proving that this set is winning in terms of the Schmidt $(\alpha ,\beta )$-game. For further details about the $(\alpha ,\beta )$-game, we refer to Schmidt’s paper [13]. For completeness, we state that, Kristensen [8] proved results on Haar measure and Hausdorff dimension of the set of well-approximable matrices over a field of formal series which implies that the Haar measure of $\textrm{Bad}(M, N)$ is zero.

As stated before, HAW is a stronger variant of Schmidt’s $(\alpha ,\beta )$-game, hence a natural question is to investigate the set $\textrm{Bad}(M, N)$ for HAW property. We prove the following theorem.

Theorem 1.2

For every $M,N\in \mathbb {N}, \textrm{Bad}(M, N)$ is hyperplane absolute winning.

1.1 Hyperplane Absolute Winning

Let $\** =\mathcal {L}^{MN} \times \mathbb {R}_{\ge 0}$. We call $\**$ the space of formal balls in $\mathcal {L}^{MN}$, where $\xi =(C,\rho )\in \**$ is said to have centre C and radius $\rho$. Define the map $\psi$ from $\**$ to the subsets of $\mathcal {L}^{MN}$, assigning a real $\Vert \cdot \Vert _\infty$-ball to the abstract one defined above. That is, for $\xi =(C,\rho )\in \**$,

$$\begin{aligned} \psi (\xi ) = \{ J \in \mathcal {L}^{MN}: \, \Vert J-C\Vert _{\infty }\le \rho \}. \end{aligned}$$

For two formal balls $B_1, B_2\in \**$ we say that $B_1=(C_1,\rho _1) \subseteq B_2=(C_2,\rho _2)$ if

$$\begin{aligned} \rho _1+\Vert C_1-C_2 \Vert _\infty \le \rho _2. \end{aligned}$$

Note that if $B_1 \subseteq B_2$ in $\**$, then $\psi (B_1) \subseteq \psi (B_2)$¹ as subsets of $\mathcal {L}^{MN}$.

Fix $0<\beta <\frac{1}{3}$ and a target set $S\subseteq \mathcal {L}^{MN}$. Hyperplane game is played as follows:

• Bob chooses initial ball $B_0 \subseteq \**$.

• Alice chooses a $(MN-1)$-dimensional affine subspace $L_1$ in $\mathcal {L}^{MN}$ and a number $\epsilon _1$, satisfying $0< \epsilon _1 \le ~\beta \rho (B_0)$, where $\rho (B_0)$ is the radius of $B_0$.

• Bob must choose a ball $B_1 \subseteq B_0$ with $\rho (B_1) \ge \beta \rho (B_0)$ and such that

  $$\begin{aligned} \psi (B_1) \cap L_1^{(\epsilon _1)} = \emptyset , \end{aligned}$$

  where $L_1^{(\epsilon _1)}$ is the $\epsilon _1$-thickening of the subspace $L_1$.

• And so on ad infinitum.

We obtain a nested sequence of balls $B_0 \supseteq B_1 \supseteq \ldots$ and declare Alice the winner if and only if

$$\begin{aligned} S \cap \bigcap _i \psi (B_i) \ne \emptyset . \end{aligned}$$

If Alice has a strategy to win regardless of Bob’s play, we say that S is Hyperplane Absolute Winning (HAW).

2 Proof of Theorem 1.2

To prove this theorem, the plan is to apply strategy developed in [3] to the formal series setting. Let

$$\begin{aligned} H=M N \text { and } L=M+N. \end{aligned}$$

We shall be playing a hyperplane game on $\mathcal {L}^H$. For $k\in \mathbb {N}$ we denote the $k^{th}$ ball chosen by Bob by B(k). Let $\rho =\rho (B(0))$, and set $\sigma =\max \{\Vert J\Vert _\infty :\,J\in B(0)\}$. We use boldface lower case letter ($\textbf{x}, \textbf{y},$ etc.) to denote points in $\mathcal {L}^N$ and $\mathcal {L}^M$. We use boldface upper case letters (${\textbf{P}}, \textbf{B},$ etc.) to denote points in $\mathcal {L}^L$. Finally, upper case letters (A, Y, etc.) denote points in $\mathcal {L}^H$.

For any matrix

where every $\gamma _{ij}$ is from $\mathcal {L}$, let

$$\begin{aligned} \textbf{A}_1 = ( \gamma _{11} \dots \gamma _{1N},1,0,\ldots , 0), \qquad \textbf{B}_1 = ( \gamma _{11} \dots \gamma _{M1},1,0,\ldots , 0), \\ \textbf{A}_2 = ( \gamma _{21} \dots \gamma _{2N},0,1,\ldots , 0), \qquad \textbf{B}_2 = ( \gamma _{12} \dots \gamma _{M2},0,1,\ldots , 0), \\ \ldots \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\ \textbf{A}_M = ( \gamma _{M1} \dots \gamma _{MN},0,0,\ldots , 1),\qquad \textbf{B}_N = ( \gamma _{1N} \dots \gamma _{MN},0,0,\ldots , 1). \end{aligned}$$

For ${\textbf{Q}}\in \mathcal {L}^L$, let $\textbf{q}\in \mathcal {L}^N$ be the projection of ${\textbf{Q}}$ onto first N coordinates. Similarly, for ${\textbf{P}}\in \mathcal {L}^L$, let $\textbf{p}\in \mathcal {L}^M$ be the projection of ${\textbf{P}}$ onto the first M coordinates.

Set

$$\begin{aligned} \mathcal {A}({\textbf{Q}})=(\textbf{A}_1\cdot {\textbf{Q}},\ldots , \textbf{A}_M\cdot {\textbf{Q}}) \end{aligned}$$

and

$$\begin{aligned} \mathcal {B}({\textbf{P}})=(\textbf{B}_1\cdot {\textbf{P}},\ldots , \textbf{B}_M\cdot {\textbf{P}}). \end{aligned}$$

We notice that a matrix A lies in $\textrm{Bad}(M, N)$ if and only if there exists a constant c such that for all ${\textbf{Q}}\in \mathbb {F}[X]^L$ with ${\textbf{Q}}\ne 0$,

$$\begin{aligned} \Vert {\textbf{Q}}\Vert _\infty ^N\cdot \Vert \mathcal {A}({\textbf{Q}})\Vert _\infty ^M > c. \end{aligned}$$

Let $\lambda = N/L$. Given $R\in \mathbb {R}_{>1}$, let $\delta =R^{-NL^2}$ and $\delta ^T=R^{-ML^2}$. Consider the two systems of inequalities

$$\begin{aligned} {}&0<\Vert \textbf{q}\Vert _\infty <\delta R^{M(\lambda +i)}, \end{aligned}$$

(2)

$$\begin{aligned} {}&\Vert \mathcal {A}({\textbf{Q}})\Vert _{\infty }<\delta R^{-N(\lambda +i)-M} \ \end{aligned}$$

(3)

and

$$\begin{aligned}&{}0<\Vert \textbf{p}\Vert _\infty <\delta ^T R^{N(1+j)}, \end{aligned}$$

(4)

$$\begin{aligned}&{}\Vert \mathcal {B}({\textbf{P}})\Vert _{\infty }<\delta ^T R^{-M(1+j)-N}. \end{aligned}$$

(5)

It is easy to see that if we fix $A\in \mathcal {L}^H$ and if for each $i\in \mathbb {N}$, the system of inequalities (2), (3) has no solution ${\textbf{Q}}\in \mathbb {F}[X]^L$, then $A\in \textrm{Bad}(M, N)$.

Next two lemmas highly rely on the theory of the geometry of numbers. For vector spaces over the field of formal power series the theory was comprehensively developed by Mahler in [10]. A brief summary was given in [9], as well as the equivalent form of these lemmas, we take it from there without a proof. This is also an adaptation of Schmidt’s lemmas [13] to the formal series case.

Lemma 2.1

There exists a constant $R_1 = R_1(M,N,\sigma )$ such that for every $i\in \mathbb {N}$ and for every $R\ge R_1$, if a ball B satisfies

$$\begin{aligned} \rho (B)<R^{-L(\lambda +i)} \end{aligned}$$

(6)

and if for all $A\in B$ the system (2), (3) has no solution ${\textbf{Q}}\in \mathbb {F}[X]^L$, then the set of all vectors ${\textbf{P}}\in \mathbb {F}[X]^L$ satisfying (4) with $j=i$ such that there exists $A\in B$ satisfying (5) with $j=i$ spans a subspace of $\mathcal {L}^L$ whose dimension is at most N.

Lemma 2.2

There exists a constant $R_2 = R_2(M,N,\sigma )$ such that for every $j\in \mathbb {N}$ and for every $R\ge R_2$, if a ball B satisfies

$$\begin{aligned} \rho (B)<R^{-L(1+j)} \end{aligned}$$

(7)

and if for all $A\in B$ the system (4), (5) has no solution $P\in \mathbb {F}[X]^L$, then the set of all vectors ${\textbf{Q}}\in \mathbb {F}[X]^L$ satisfying (2) with $i=j+1$ such that there exists $A\in B$ satisfying (3) with $i=j+1$ spans a subspace of $\mathcal {L}^L$ whose dimension is at most M.

Suppose that a game has already been played; for each $i\in \mathbb {N}$, let $k_i\in \mathbb {N}$ be the minimal number such that (6) holds with $B=B(k_i)$, and for each $j\in \mathbb {N}$, let $h_j\in \mathbb {N}$ be the minimal number such that (7) holds with $B=B(h_j)$. Alice will try to play the game in such a way so that for all $A\in B(k_i)$, the system of inequalities (2) and (3) has no solution ${\textbf{Q}}\in \mathbb {F}[X]^L$ and so that for all $A\in B(h_j)$, the system of inequalities (4), (5) has no solution ${\textbf{P}}\in \mathbb {F}[X]^L$. The following lemma guarantees that she can always continue to play in such a way.

Lemma 2.3

If $R\in \mathbb {R}$ is sufficiently large, then the following hold:

• (i) When the game is at a stage $k_i$, if, for all $A\in B(k_i)$, the system of inequalities (2) and (3) has no solution ${\textbf{Q}}\in \mathbb {F}[X]^L$, and the system of inequalities (4), (5) with $j=i-1$ has no solution ${\textbf{P}}\in \mathbb {F}[X]^L$, then Alice has a strategy ensuring that when the game reaches stage $h_i$, then for all $A\in B(h_i)$ the system of inequalities (4), (5) with $j=i$ will have no solution ${\textbf{P}}\in \mathbb {F}[X]^L$.

• (ii) Dually, when the game is at a stage $h_j$, if, for all $A\in B(h_j)$, the system of inequalities (4), (5) has no solution ${\textbf{P}}\in \mathbb {F}[X]^L$, and the system of inequalities (2), (3) with $i=j$ has no solution ${\textbf{Q}}\in \mathbb {F}[X]^L$, then Alice has a strategy ensuring that when the game reaches stage $k_{j+1}$, then for all $A\in B(k_{j+1})$ the system of inequalities (2), (3) with $i=j+1$ will have no solution ${\textbf{Q}}\in \mathbb {F}[X]^L$.

The proofs of the two parts of the lemma are dual to each other, so we only prove the first claim.

Proof of Lemma 2.3 (Part 1)

For convenience of notation let $B=B(k_i)$. Suppose that $R\ge R_1$. Then by Lemma 2.1, there are at most N linearly independent integer points ${\textbf{P}}'_1,\ldots ,{\textbf{P}}'_u,\ u\le N$ in the set

$$\begin{aligned} S:= \{ {\textbf{P}}\in \mathbb {F}[X]^L: \text { there exists }A\in B\text { satisfying } (4), (5) \text { with }j=i\}. \end{aligned}$$

(8)

If $u<N$, we will add $N-u$ more integer vectors to get a set $\mathcal {P}'= \{ {\textbf{P}}'_1,\ldots ,{\textbf{P}}'_N \}$, which spans a subspace of $\mathcal {L}^L$ of dimension N.

Using $\mathcal {P}'$ one can easily construct a basis $\mathcal {P}=\{ {\textbf{P}}_1,\ldots , {\textbf{P}}_N \}$ of the same subspace, such that if

$$\begin{aligned} {\textbf{P}}_j = (P_{j,1},\ldots , P_{j,N+M}),\ P_{j,i}\in \mathcal {L}, \end{aligned}$$

then the matrix

(9)

has vectors $(1,0\ldots ,0)^T,(0,1,\ldots ,0)^T,\ldots ,(0,\ldots ,0,1)^T$ in some order as its N columns (not necessarily the first N columns), while every other element of the matrix has norm $\Vert P_{i,j} \Vert \le 1$ for $i=1,\ldots ,N, j=1,\ldots ,N+M$. This can be done by applying elementary operations (dividing row on the element with the maximum norm will give us ‘1’s and subtracting rows with a suitable coefficient from another ones will give us ‘0’s) on our integer vectors $\mathcal {P}'= \{ {\textbf{P}}'_1,\ldots ,{\textbf{P}}'_N \}$, guaranteed by Lemma 2.1. We will call a set of vectors with these properties a normalized set of vectors.

Now suppose that ${\textbf{P}}\in S$, and let $A\in B$ be the corresponding matrix. By the hypothesis, the pair $({\textbf{P}},A)$ do not satisfy the system of inequalities (4), (5) with $j=i-1$. But $({\textbf{P}},A)$ do satisfy (5) with $j=i$, which implies (5) with $j=i-1$. Thus they must not satisfy (4) with $j=i-1$, i.e.

$$\begin{aligned} \Vert \textbf{p}\Vert _\infty \ge \delta ^T R^{N(1+(i-1))}. \end{aligned}$$

Since ${\textbf{P}}\in S$, we can write ${\textbf{P}}= t_1 {\textbf{P}}_1 +\dots +t_n {\textbf{P}}_n,\, t_1,\ldots , t_N \in \mathcal {L}$. We have that

$$\begin{aligned} \delta ^T R^{Ni} \le \Vert \textbf{p}\Vert _\infty \le \Vert {\textbf{P}}\Vert _\infty = \max (\Vert t_1\Vert ,\ldots ,\Vert t_N\Vert ). \end{aligned}$$

So,

$$\begin{aligned} \max (\Vert t_1\Vert ,\ldots ,\Vert t_N\Vert ) \ge \delta ^T R^{Ni}. \end{aligned}$$

(10)

On the other hand, we may write (5) with $j=i$ as

$$\begin{aligned} \Vert t_1 (\textbf{B}_1 \cdot {\textbf{P}}_1) + \ldots + t_N (\textbf{B}_1 \cdot {\textbf{P}}_N) \Vert< \delta ^T R^{-M(1+j)-N} \\ \vdots \\ \Vert t_1 (\textbf{B}_N \cdot {\textbf{P}}_1) + \ldots + t_N (\textbf{B}_N \cdot {\textbf{P}}_N) \Vert < \delta ^T R^{-M(1+j)-N}. \end{aligned}$$

Let $D=D(A,\mathcal {P})$ be the determinant of the matrix

$$\begin{aligned} M(A,\mathcal {P}) {\mathop {=}\limits ^{\text {def}}}(\textbf{B}_u \cdot {\textbf{P}}_v)_{1\le u,v\le N}, \end{aligned}$$

(11)

and let $D_{uv}=D_{uv}(A,\mathcal {P})$ be the cofactor of the entry $\textbf{B}_u \cdot {\textbf{P}}_v$ in this matrix. We can solve this matrix inequality using Cramer’s Rule for every $1\le v \le N$

$$\begin{aligned} \Vert t_v \Vert \, \Vert D \Vert \le \delta ^T R^{-M(1+j)-N} \max \left( \Vert D_{1v}\Vert ,\ldots , \Vert D_{Nv}\Vert \right) . \end{aligned}$$

Together with (10) we get

$$\begin{aligned} \Vert D\Vert \le R^{-L(1+i)} \max \left( \Vert D_{11}\Vert ,\Vert D_{12}\Vert ,\ldots , \Vert D_{NN}\Vert \right) . \end{aligned}$$

(12)

We want now to construct a strategy for Alice such that inequality (12) cannot hold after finite number of steps. This way Alice will avoid all solutions to (4), (5) with $j=i$, and we can continue with the dual construction.

At this point, we introduce some notation. For each $v\in \{0,\ldots , N\}$ and for each $A\in \mathcal {L}^H$, consider the set of $v\times v$ minors of the matrix $M(A,\mathcal {P})$ defined by (11). Each minor can be described by a pair of sets $I,J\in \{1,\ldots , N\}$ satisfying $\#(I)=\#(J)=v$. For each such pair, we define the map

$$\begin{aligned} D_{(I,J)}:\mathcal {L}^H\rightarrow \mathcal {L}, \\ D_{(I,J)}{\mathop {=}\limits ^{\text {def}}}\det (\textbf{B}_{i_k}\cdot {\textbf{P}}_{j_l})_{k,l=1,\ldots ,v}, \end{aligned}$$

where $I=\{i_k:\, k=1,\ldots ,v\}$ and $J=\{j_l:\, l=1,\ldots , v\}$. In other words, $D_{(I,J)}$ is the determinant of the $v\times v$ minor of $M(A,\mathcal {P})$ described by the pair (I, J). For short, let $\omega =(I,J)$ and let $\Omega _v=\{(I,J):v=\#(I)=\#(J)\}$. The special cases $v=0$ and $v=-1$ are dealt with as follows: $\Omega _0=\{\theta \}$ and $D_\theta (A)=1$ for all A, where $\theta =(\emptyset ,\emptyset )$; $\Omega _{-1}=\emptyset$.

$$\begin{aligned} \vec {M}_v(A) = \vec {M}_{v,\mathcal {P}}(A){\mathop {=}\limits ^{\text {def}}}\left( D_\omega (A)\right) _{\omega \in \Omega _v}\in \mathcal {L}^{\left( {\begin{array}{c}N\\ v\end{array}}\right) ^2}, \end{aligned}$$

and for short let $M_v(A)=\Vert \vec {M}_v(A)\Vert _\infty$. For a set $B\in \mathcal {L}^H$ let

$$\begin{aligned} M_v(B){\mathop {=}\limits ^{\text {def}}}\sup _{A\in B} M_v(A). \end{aligned}$$

$\square$

To prove the second part of Lemma 2.3, we would need the following important lemma about the so-called finite game:

Lemma 2.4

For all $0<\beta <\frac{1}{3}$ and for all $0 \le v \le N$ there exists

$$\begin{aligned} \nu _v = \nu _v(M,N,\beta , \sigma ) >0 \end{aligned}$$

such that for any $0<\mu _v \le \nu _v$ and for any normalized set of linearly independent vectors $\mathcal {P}=\{{\textbf{P}}_1,\ldots ,{\textbf{P}}_N\}\subseteq \mathcal {L}^L$, Alice can win the following finite game:

• Bob plays a closed ball $B\subseteq \mathcal {L}^H$ satisfying $\rho _B{\mathop {=}\limits ^{\text {def}}}\rho (B)<1$ and $\max _{A\in B} \Vert A\Vert \le \sigma .$

• Alice and Bob play the hyperplane game until the radius of Bob’s ball $B_v$ is less than $\mu _v\rho _B$.

• Alice wins if for all $A\in B_v$, we have

  $$\begin{aligned} M_v(A)>\nu _v\rho _B M_{v-1}(B_v). \end{aligned}$$

  (13)

We will prove this lemma later, but for now let us assume Lemma 2.4 and use it to complete the proof of Lemma 2.3(ii).

Proof of Lemma 2.3 (Part 2)

Let $\nu _N>0$ be the number guaranteed by Lemma 2.4 for $v=N$. Suppose $R\ge R_1$ is large enough so that

$$\begin{aligned} R^{-M} \le \beta \nu _N. \end{aligned}$$

Now suppose that the game has progressed to the stage $k_i$, and let $\rho _{k_i}=\rho (B(k_i))$. If $k_i>0$, then since $k_i$ is the minimal integer such that (6) is satisfied with $B=B(k_i)$, we have

$$\begin{aligned} \rho _{k_i} \ge \beta R^{-L(\lambda +i)}. \end{aligned}$$

We can ensure that $k_i>0$ for all $i\ge 0$ by requiring that

$$\begin{aligned} R^{-M}\le \rho _0, \end{aligned}$$

where $\rho _0$ is the radius of Bob’s first ball. In particular, we have

$$\begin{aligned} \mu _N{\mathop {=}\limits ^{\text {def}}}\frac{R^{-L(1+i)}}{\rho _{k_i}}\le \frac{1}{\beta R^M}\le \nu _N. \end{aligned}$$

This means that this is a valid choice of $\mu _N$.

Let $\mathcal {P}=\{{\textbf{P}}_1,\ldots ,{\textbf{P}}_N\}$ be a set of normalized linearly independent vectors, which spans a subspace of $\mathcal {L}^L$ containing the set S defined by (8). This sets the stage for the finite game described in Lemma 2.4, which is the lemma says Alice can win. Now, the finite game is played until Bob’s ball B satisfies

$$\begin{aligned} \rho (B)<\mu _N\rho _{k_i}=R^{-L(1+i)}, \end{aligned}$$

in other words, the last ball that Bob plays in the finite game is exactly $B(h_i)$. Since Alice wins the finite game, we have that for every $A\in B(h_i)$, (13) holds with $v=N$. In particular, we have $\omega =(\{1,\ldots ,N\},\{1,\ldots ,N\})\in \Omega _N$ and so for all $A\in B(h_i)$

$$\begin{aligned} \Vert D(A)\Vert > \nu _N \rho _{k_i} M_{N-1}(B(h_i)) \ge R^{-L(1+i)} \max ( \Vert D_{11}(A)\Vert ,\Vert D_{12}(A)\Vert ,\ldots ,\Vert D_{NN}(A)\Vert ). \end{aligned}$$

That is, (12) is not satisfied for any point $A\in B(h_i)$. On the other hand, fixing $A\in B(h_i)$, we see by Lemma 2.4 that if there exists a point ${\textbf{P}}\in \mathbb {F}[X]^L$ satisfying (4) and (5) with $j=i$, then (12) would also be satisfied, a contradiction. Thus the system of inequalities (4), (5) with $j=i$ has no solution ${\textbf{P}}\in \mathbb {F}[X]^L$.

With the proof of Lemma 2.3, we are now in a position to complete the proof of Theorem 1.2.

2.1 Completing the proof of Theorem 1.2

Let $R\in \mathbb {R}$ be chosen large enough so that Lemma 2.3 holds. Note that when $i=0$, inequality (2) has no solution ${\textbf{Q}}\in \mathbb {F}[X]^L$, and when $j=0$, inequality (4) has no solution ${\textbf{P}}\in \mathbb {F}[X]^L$. Thus Alice can make dummy moves until stage $h_0$, at which point the hypotheses of Lemma 2.3(ii) hold. Then Alice has a strategy to ensure that for all $A\in B(k_1)$, the system of inequalities (2), (3) with $i=1$ has no solution ${\textbf{Q}}\in \mathbb {F}[X]^L$. By continuing in this way, Alice ensures that if A is the intersection point of the balls $(B(k))_{k\in \mathbb {N}}$, then for all $i\in \mathbb {N}$, the system of inequalities (2), (3) has no solution ${\textbf{Q}}\in \mathbb {F}[X]^L$. This implies that A is badly approximable by what we stated before. Thus the set of badly approximable systems of linear forms over formal series is HAW. $\square$

2.2 Proof of Lemma 2.4

Finally, the only thing remaining is to prove Lemma 2.4. For this we will need two lemmas, originally developed by Schmidt. To properly formulate them we need some notation. Let $\omega \in \Omega _v$, $v\in \{0,\ldots ,N\}$. We define the discrete gradient of $D_\omega (A)$ to be the vector

where $E_{ij}\in \mathcal {L}^H$ is a matrix with value ‘1’ as the ijth entry and zeroes everywhere else.

We also define discrete directional derivative of $D_\omega$ in direction $A'$ at point A as

$$\begin{aligned} \nabla _{A'} D_{(I,J)}(A) = D_\omega ( A+A') - D_\omega (A). \end{aligned}$$

Also let

$$\begin{aligned} \Vert \nabla D_\omega (A) \Vert _{op} = \inf \{ c\ge 0: \Vert \nabla _{A'} D_{(I,J)}(A) \Vert \le c \Vert A'\Vert _\infty \text { for all } A' \in \mathcal {L}^H \}. \end{aligned}$$

Then we have the following

Lemma 2.5

Fix $A\in B$, $v\in \{0,\ldots , N\}$, and $\omega \in \Omega _v$. Then

$$\begin{aligned} \Vert \nabla D_\omega (A) \Vert _{op} \le M_{v-1}(A), \end{aligned}$$

(14)

$$\begin{aligned} \Vert \nabla \nabla D_\omega (A) \Vert _{op} \le M_{v-2}(A). \end{aligned}$$

(15)

Proof

Write $\omega =(I,J)$. For any matrix $A' \in \mathcal {L}^H$, we compute

$$\begin{aligned} \nabla _{A'} D_{(I,J)}(A) = \sum _{k,l=1}^v \pm [ A' \textbf{e}_{i_k}\cdot {\textbf{P}}_{j_l}] D_{(I\backslash \{i_k\}, J\backslash \{j_l\})} (A). \end{aligned}$$

(16)

Here $\textbf{e}_1,\ldots ,\textbf{e}_N$ are the standard basis vectors for $\mathcal {L}^L$. It follows that

$$\begin{aligned} \Vert \nabla _{A'} D_{(I,J)}(A) \Vert&\le M_{v-1}(A) \max _{1\le k,l \le v} \Vert A' \textbf{e}_{i_k}\cdot {\textbf{P}}_{j_l} \Vert \\ {}&\le M_{v-1}(A) \max _{1\le k,l \le v} \Vert A' \textbf{e}_{i_k}\Vert _\infty \, \Vert {\textbf{P}}_{j_l} \Vert _\infty \\&=M_{v-1} (A) \Vert A'\Vert _\infty , \end{aligned}$$

that implies equation (14). Now for $A'' \in \mathcal {L}^H$, we get the formula

$$\begin{aligned} \nabla _{A''} \nabla _{A'} D_{(I,J)}(A) = \sum _{k,l=1}^v \sum _{k',l'=1}^v \pm [ A' \textbf{e}_{i_k}\cdot {\textbf{P}}_{j_l}] [ A'' \textbf{e}_{i_k'}\cdot {\textbf{P}}_{j_l'}] D_{(I\backslash \{i_k,i_{k'}\}, J\backslash \{j_l,j_{l'}\})} (A) \end{aligned}$$

and by a similar computation we get (15). $\square$

For the next lemma we will also need a notion of a polar (or reciprocal) set of vectors to $\mathcal {P}$. In the formal power series setting this was used by Mahler in [10]. For a set of vectors $\mathcal {P}=\{{\textbf{P}}_1,\ldots ,{\textbf{P}}_N\}, {\textbf{P}}_i \in \mathcal {L}^L$, linearly independent over $\mathcal {L}$, we call a set $\mathcal {T} =\{\textbf{T}_1,\ldots ,\textbf{T}_N\},\textbf{T}_i \in \mathcal {L}^L$ of N vectors polar to $\mathcal {P}$ if

$$\begin{aligned} {\textbf{P}}_i \cdot \textbf{T}_j = \delta _{ij}, \end{aligned}$$

(17)

where $\delta _{ij}$ is a Kronecker symbol. A polar set can be explicitly found by solving the system of linear equations (17) with $i=1,\ldots ,N$ for each $j=1,\ldots ,N$. Recall that the matrix (9) of $\mathcal {P}$ has standard unit vectors $\textbf{e}_{j_i}\in \mathcal {L}^N,i=1,\ldots ,N$ as some N columns of it. Using matrix (9) of $\mathcal {P}$, one can easily get a polar set $\mathcal {T}$. For this let us fix elements in those N columns and set every other element of this matrix to be equal to 0. Define $\textbf{T}_j$ as rows of this new matrix. Basically, $\textbf{T}_j,j=1,\ldots ,N$ are some of the standard unit vectors in $\mathcal {L}^{L}$. With this choice we have property (17) and $\Vert \textbf{T}_j \Vert =1$ for each $j=1,\ldots ,N$.

Lemma 2.6

Fix $A \in B$ and $v\in \{0,\ldots , N\}$. There exist constants $\epsilon _1, \epsilon _2>0$ depending only on M, N and $\sigma$ such that if

$$\begin{aligned} M_v(A)\le \epsilon _1 M_{v-1} (A) \end{aligned}$$

then

$$\begin{aligned} \max _{\omega \in \Omega _v} \Vert \nabla D_\omega (A) \Vert _{op} > \epsilon _2 M_{v-1} (A). \end{aligned}$$

Proof

Let $\omega '=(I',J')\in \omega _{v-1}$ be the value which maximizes the expression $\Vert D_{w'} (A) \Vert$. Choose any $i_0\in \{1,\ldots ,N\}\backslash I', j_0\in \{1,\ldots , N\}\backslash J'$, and let $\omega =(I,J)=(I'\cup \{i_0\},J'\cup \{j_0\})\in \Omega _v$. We will let $A'$ be such that $A'\textbf{e}_i=\textbf{0}$ for all i except $i=i_0$ and so (16) reduces to

$$\begin{aligned} \nabla _{A'} D_{(I,J)}(A) = \sum _{l=0}^{v-1} \pm [ A' \textbf{e}_{i_0}\cdot {\textbf{P}}_{j_l}] D_{(I\backslash \{i_0\}, J\backslash \{j_l\})} (A). \end{aligned}$$

We may choose any value for $A'\textbf{e}_{i_0}$ which lies in $\mathcal {L}^M$. In particular, we may let

$$\begin{aligned} A'\textbf{e}_{i_0} = \textbf{T}_{j_0} - \sum _{i=1}^N [\textbf{T}_{j_0}\cdot \textbf{e}_{M+i}]\textbf{B}_i \end{aligned}$$

since computation verifies $A'\textbf{e}_{i_0}\cdot \textbf{e}_{M+i}=0$ for all $i=1,\ldots , N$. Now

$$\begin{aligned} \nabla _{A'} D_{(I,J)}(A)&= \sum _{l=0}^{v-1} \pm [ \textbf{T}_{j_0}\cdot {\textbf{P}}_{j_l}] D_{(I\backslash \{i_0\}, J\backslash \{j_l\})} (A) \\&\quad - \sum _{i=1}^N [\textbf{T}_{j_0}\cdot \textbf{e}_{M+i}] \sum _{l=0}^{v-1} \pm [ \textbf{B}_i\cdot {\textbf{P}}_{j_l}] D_{(I\backslash \{i_0\}, J\backslash \{j_l\})} (A) \\&= \pm D_{(I\backslash \{i_0\}, J\backslash \{j_0\})} (A) - \sum _{i=1}^N [ \textbf{T}_{j_0}\cdot \textbf{e}_{M+i}] D_{(I\cup \{i\}\backslash \{i_0\}, J)} (A). \end{aligned}$$

Thus

$$\begin{aligned} \Vert \nabla _{A'} D_\omega (A)\Vert&\ge \Vert D_{\omega '}(A)\Vert - M_v(A) \left\| \sum _{i=1}^N [\textbf{T}_{j_0} \cdot \textbf{e}_{M+i}]\right\| \\ {}&\ge M_{v-1} (A) - M_v (A). \end{aligned}$$

Whereas,

$$\begin{aligned} \Vert A'\Vert _\infty \le \sigma \end{aligned}$$

and the lemma follows. $\square$

We can now prove Lemma 2.4 by induction on v. When $v=0$ the lemma is trivial (by convention we say that $\max (\emptyset )=0$). Suppose that the lemma has been proven for $v-1$, and we want to prove it for v. Let $\nu _{v-1}>0$ be given by the induction hypothesis. Fix $0<\mu _{v-1}\le \nu _{v-1}$ and $\nu _{v}>0$ to be determined. Suppose that we are given $0<\mu _v\le \nu _v$ and a sequence of normalized vectors ${\textbf{P}}_1,\ldots , {\textbf{P}}_N$, and let B be the first ball played by Bob in the finite game. By the induction hypothesis, Alice can play in a way such that if $B_{v-1}$ is first chosen by Bob satisfying $\rho (B_{v-1})<\mu _{v-1}\rho _B$, then for all $A\in B_{v-1}$ we have

$$\begin{aligned} M_{v-1}(A)>\nu _{v-1}\rho _B M_{v-2}(B_{v-1}). \end{aligned}$$

We must describe how Alice will continue her strategy so as to satisfy (13). We begin with the following observation:

Lemma 2.7

For all $A\in B_{v-1}$ we have

$$\begin{aligned} M_{v-1} (B_{v-1}) \le 2 M_{v-1}(A). \end{aligned}$$

Proof

Fix $A'= A+C \in B_{v-1}$ and $\omega ' \in \Omega _{v-1}$. We use (14) to bound the following difference:

$$\begin{aligned} \Vert D_{\omega '}(A') - D_{\omega '} (A) \Vert \le \Vert C\Vert _\infty M_{v-2}(B_{v-1}) \le \mu _{v-1} \rho _B M_{v-2}(B_{v-1}) \le M_{v-1} (A). \end{aligned}$$

Thus $\Vert D_{\omega '}(A')\Vert \le 2 M_{v-1}(A)$. Taking the supremum over all $\omega '\in \Omega _{v-1}$ and all $A' \in B_{v-1}$ completes the proof. $\square$

To complete the proof of Lemma 2.4, we divide it into two cases:

Case 1: $M_v(A)> \epsilon _1 M_{v-1}(A)$ for all $A \in B$. In this case, Alice will make dummy moves until $\rho (B_v)<\mu _v\rho _B$. By Lemma 2.7 we have $M_v(A)>(\epsilon /2)M_{v-1}(B_{v-1}) \ge (\epsilon /2)M_{v-1}(B_{v})$, so (13) holds as long as $\nu _v\le \epsilon _1/(2\sigma )$.

Case 2: $M_v(A)\le \epsilon _1 M_{v-1}(A)$ for some $A \in B$. In this case, by Lemma 2.6 we have

$$\begin{aligned} \Vert \nabla D_\omega (A) \Vert _{op} > \epsilon _2 M_{v-1} (A) \end{aligned}$$

(18)

for some $\omega \in \Omega _v$. Let

$$\begin{aligned} F(A') = \nabla _{A'-A} D_\omega (A) \end{aligned}$$

and let $F_{k} = F^{-1}(0)$. Alice’s strategy will be to delete the neighbourhood $F_{k}^{(\beta \rho (B_{v-1}))}$ of $F_{k}$, and then make dummy moves until $\rho (B_v)< \mu _v \rho _B$. The gradient condition (18) implies that

$$\begin{aligned} \Vert F(\hat{A})\Vert&\ge \beta \rho (B_{v-1}) \epsilon _2 M_{v-1}(A) \nonumber \\&\ge \beta ^2 \mu _{v-1} \rho _B \epsilon _2 M_{v-1}(A) \end{aligned}$$

(19)

for all $\hat{A} \in B_{v-1}\backslash F_{k}^{(\beta \rho (B_{v-1}))}$. On the other hand, letting $C=\hat{A}-A$, we get

$$\begin{aligned} \Vert \nabla _C D_\omega (A) \Vert= & {} \Vert \nabla _C D_\omega (\hat{A}) + \nabla _C D_\omega (A) - \nabla _C D_\omega (\hat{A})\Vert \\\le & {} \Vert \nabla _C D_\omega (\hat{A}) \Vert + \Vert \nabla _C D_\omega (A) - \nabla _C D_\omega (\hat{A}))\Vert \end{aligned}$$

or

$$\begin{aligned} \Vert \nabla _C D_\omega (\hat{A}) \Vert \ge \Vert \nabla _C D_\omega (A) \Vert - \Vert \nabla _C D_\omega (A) - \nabla _C D_\omega (\hat{A})\Vert . \end{aligned}$$

(20)

Next, we have

$$\begin{aligned} \Vert \nabla _C D_\omega (\hat{A}) \Vert \le \Vert C \Vert _{\infty } \Vert \nabla D_\omega (\hat{A}) \Vert _{\infty }< \rho (B_{v-1}) \{\Vert \nabla D_\omega (\hat{A}) \Vert _{\infty }, \Vert D_\omega (\hat{A}) \Vert \} \le \epsilon _3 \Vert D_\omega (\hat{A}) \Vert ,\nonumber \\ \end{aligned}$$

(21)

where $\epsilon _3$ depends only on $\sigma , M$ and N. Also by (15) we have

$$\begin{aligned} \Vert \nabla _C D_\omega (A) - \nabla _C D_\omega (\hat{A})\Vert\le & {} \Vert C \Vert ^2 M_{v-2}(A)\le (\mu _{v-1} \rho _B)^2 M_{v-2} (B_{v-1})\nonumber \\ {}\le & {} \frac{\mu _{v-1}^2}{\nu _{v-1}} \rho _B M_{v-1}(A). \end{aligned}$$

(22)

Combining (19), (20), (21) and (22) together gives us

$$\begin{aligned} \epsilon _3 \Vert D_\omega (\hat{A}) \Vert \ge \Vert \nabla _C D_\omega (\hat{A}) \Vert&\ge \beta ^2 \mu _{v-1} \rho _B \epsilon _2 M_{v-1}(A) - \frac{\mu _{v-1}^2}{\nu _{v-1}} \rho _B M_{v-1}(A) \\&\ge \left( \beta ^2\epsilon _2 - \frac{\mu _{v-1}}{\nu _{v-1}} \right) \mu _{v-1} \rho _B M_{v-1}(A) \\&\ge \frac{1}{2} \left( \beta ^2\epsilon _2 - \frac{\mu _{v-1}}{\nu _{v-1}} \right) \mu _{v-1} \rho _B M_{v-1}(B_v). \end{aligned}$$

Applying the trivial bound $\Vert D_\omega (\hat{A}) \Vert \le M_v (\hat{A})$ and letting

$$\begin{aligned} \mu _{v-1} = \frac{\beta ^2 \epsilon _2 \nu _{v-1}}{2} \,\,\, \text { and }\,\,\, \nu _v = \frac{1}{4\epsilon _3} \beta ^2 \epsilon _2 \mu _{v-1}, \end{aligned}$$

we see that (13) is satisfied for $A=\hat{A}$ for every element of $B_{v-1}\backslash F_{k}^{(\beta \rho (B_{v-1}))}$. By the definition of the game, Bob will have to choose his next ball inside $B_{v-1}\backslash F_{k}^{(\beta \rho (B_{v-1}))}$, so that $B_v \subseteq B_{v-1}\backslash F_{k}^{(\beta \rho (B_{v-1}))}$. Thus (13) holds for every element of $B_v$. $\square$