Metrical theorems on systems of affine forms

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General Section

Metrical theorems on systems of affine forms Mumtaz Hussain a,∗ , Simon Kristensen b , David Simmons c a

Department of Mathematics and Statistics, La Trobe University, P.O. Box 199, Bendigo, Victoria 3552, Australia b Department of Mathematics, Aarhus University, Ny Munkegade 118, DK-8000 Aarhus C, Denmark c University of York, Department of Mathematics, Heslington, York YO10 5DD, UK

a r t i c l e

i n f o

Article history: Received 3 October 2018 Received in revised form 3 November 2019 Accepted 7 November 2019 Available online xxxx Communicated by F. Pellarin Keywords: Khintchine–Groshev theorem Jarnik theorem Hausdorff measure Hausdorff dimension Schmidt game Hyperplane winning

a b s t r a c t In this paper we discuss metric theory associated with the affine (inhomogeneous) linear forms in the so called doubly metric settings within the classical and the mixed setups. We consider the system of affine forms given by q → qX + α, where q ∈ Zm (viewed as a row vector), X is an m × n real matrix and α ∈ Rn . The classical setting refers to the dist(qX + α, Zm ) to measure the closeness of the integer values of the system (X, α) to integers. The absolute value setting is obtained by replacing dist(qX + α, Zm ) with dist(qX + α, 0); and the more general mixed settings are obtained by replacing dist(qX +α, Zm ) with dist(qX +α, Λ), where Λ is a subgroup of Zm . We prove the Khintchine–Groshev and Jarník type theorems for the mixed affine forms and Jarník type theorem for the classical affine forms. We further prove that the sets of badly approximable affine forms, in both the classical and mixed settings, are hyperplane winning. The latter result, for

* Corresponding author. E-mail addresses: [email protected] (M. Hussain), [email protected] (S. Kristensen), [email protected] (D. Simmons). URL: https://sites.google.com/site/davidsimmonsmath/ (D. Simmons). https://doi.org/10.1016/j.jnt.2019.11.014 0022-314X/© 2019 Elsevier Inc. All rights reserved.

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the classical setting, answers a question raised by Kleinbock (1999). © 2019 Elsevier Inc. All rights reserved.

1. Background and statements of results 1.1. Classical affine forms Let ψ : N → R+ be a function tending to 0 at infinity, referred to as an approximation function. Let W (m, n; ψ) denote the set of pairs (X, α) ∈ Rmn ×Rn for which the system of inequalities |q1 x1i + q2 x2i + · · · + qm xmi − αi − pi | ≤ ψ(|q|)

for 1 ≤ i ≤ n

(1.1)

is satisfied for infinitely many q ∈ Zm \ {0} and p ∈ Zn . On the right-hand side, |q| denotes the max norm of q. In what follows, the system {q1 x1i + · · · + qm xmi : i = 1, . . . , n} of n linear forms in the m variables q1 , . . . , qm will be written more concisely as qX, where the matrix X is regarded as a point in Rmn . If the vector α is fixed then the study of the measure and dimension of the fiber def

Wα = {X ∈ Rmn : (X, α) ∈ W (m, n; ψ)} is referred as the singly metric theory, while the study of the entire set W (m, n; ψ) is called the doubly metric theory. Both the singly metric and the doubly metric theories are considered part of inhomogeneous Diophantine approximation. Inhomogeneous Diophantine approximation is in some respects different from homogeneous Diophantine approximation, which corresponds to the case α = 0. In particular, in the doubly metric case, the results are sometimes sharper and easier to prove than singly metric case due to the extra variable involved which offers an extra degree of freedom. The most fundamental result in metric Diophantine approximation is the Khintchine– Groshev theorem which gives an elegant answer to the question of the ‘size’ of the set in terms of Lebesgue or Hausdorff measure. For the modernised version of the Khintchine– Groshev theorem for W0 (m, n; ψ) we refer to [3] and for Wα (m, n; ψ) we refer to [1]. The doubly metric version of the Khintchine–Groshev theorem not only requires weaker assumptions than the homogeneous analogue but is also considerably easier to prove, see [40, Theorem 15], or [10, Theorem VII.II] for the special case of simultaneous approximation.

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Another set of significant interest is in a sense complementary to W (m, n; ψ), namely the set of inhomogeneous badly approximable affine forms. First we define the classical set of badly approximable affine forms as Bad(m, n) = {(X, α) ∈ Rmn × Rn :

inf

q∈Zm \{0}

|q|m/n qX − α > 0}

and for each α ∈ Rn we let Badα (m, n) = {X ∈ Rmn : (X, α) ∈ Bad(m, n)}. In his seminal paper [33], Kleinbock used ideas and techniques from the theory of dynamical systems to prove a doubly metric result, namely that the set Bad(m, n) is of full Hausdorff dimension. Essentially, his method is based on a deep connection between badly approximable systems of linear forms and orbits of certain lattices in Euclidean space under an appropriate action. One possible strengthening of such dimension results is to show that the sets in question are winning for what is called Schmidt’s game. Schmidt’s game was introduced by Schmidt in [38] and later used by himself in [39] to prove that the set Bad0 (m, n) of homogeneous badly approximable matrices is of full dimension. Actually this possible strengthening in the setting of this paper was conjectured by Kleinbock (see [33, pp. 101])1 as follows: “It seems natural to conjecture that Bad(m, n) is a wining subset of Rmn × Rn , and, moreover, that Badα (m, n) is a winning subset of Rmn for every α ∈ Rn . This seems to be an interesting and challenging problem in metric number theory.” Einsiedler and Tseng in [23] proved the latter half of the Kleinbock’s conjecture by proving that each fibre Badα (m, n) is winning for Schmidt’s game, which in-return is a considerable strengthening of just proving that the set in question has full Hausdorff dimension. We prove the former half of the conjecture in a stronger form. There have been a lot of developments recently not only in the usage of Schmidt’s game but also progress has been made in introducing variants of this game. The hyperplane game is a variant of a game defined by McMullen [36] which in turn is a variant of Schmidt’s game [38]. It was originally defined by Broderick, Fishman, Kleinbock, Reich, and Weiss [7]. Sets which are winning for the hyperplane game are called hyperplane winning. In [8], Broderick, Fishman, and the third-named author proved that each fibre Badα (m, n) is hyperplane winning, which is a strengthening of the result of Einsiedler and Tseng mentioned above. We prove the following result and thus fully settle Kleinbock’s conjecture. Theorem 1.1. The set Bad(m, n) is hyperplane winning. 1 In the quoted text, we have adopted notation of our paper to avoid ambiguity. We would like to thank Lovy Singhal for bringing this conjecture to our attention.

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Note that the hyperplane winning property is valid even if the games are played on certain fractals. Hence this theorem also implies that the set Bad(m, n) intersection with a large class of fractal sets has full Hausdorff dimension, that is, the dimension of the fractal sets. We refer the reader to [7] for details. 1.2. Mixed affine forms An extension of the classical problems which has drawn some attention recently is the following: what happens if one requires the pair (p, q) appearing in the definition of the set W (m, n; ψ) to belong to some fixed set of integer vectors? An example of such a result is a theorem of Dani, Laurent, and Nogueira [12], which is a doubly metric analogue of the Khintchine–Groshev theorem for the set W (m, n; ψ) under the additional restriction that, with respect to the direct sum decomposition of Rn corresponding to a fixed and sufficiently coarse partition π of the set of coordinates {1, . . . , m + n}, the components of the integer vector (p, q) are all primitive. In this paper we consider another natural restriction, namely that the vector p must lie in some sublattice of Zn . More precisely, fix an integer 0 ≤ u ≤ n, and consider sublattice Zu × {0}n−u ⊆ Zn . Let W (m, n, u; ψ) be the set of pairs (X, α) ∈ Rmn × Rn such that max{|q · x(1) − α1 − p1 |, . . . , |q · x(u) − αu − pu |, |q · x(u+1) − αu+1 |, . . . , |q · x(n) − αn |} ≤ ψ(|q|) (1.2) is satisfied for infinitely many q ∈ Zm \ {0} and p ∈ Zu × {0}n−u . Here, x(1) , . . . , x(n) are the column vectors of X. These linear forms are referred to as the absolute affine linear forms. The metrical theory for these sets and these particular sublattices contains the results for general sublattices, as the analogous set for a general sublattice is easily seen to be bi-Lipschitz equivalent to a set which can be bounded from inside and outside  for appropriate functions ψ (see e.g. [28]). Note that by sets of the form W (m, n, u; ψ) W (m, n; ψ) = W (m, n, n; ψ). Our new results on the metric properties of the sets W (m, n, u; ψ) and the fibres Wα (m, n, u; ψ) = {X ∈ Rmn : (X, α) ∈ W (m, n, u; ψ)} are proved using the methods developed in [29], including the appeal to the important tools of Beresnevich and Velani, the mass transference principle [4] and its generalization to linear forms [1,5]. The adaptation of the methods to the present setup requires some work, so rather than stating just the differences with the manuscript [29], we have chosen to present the present work in a more self-contained manner.

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In the case u = 0, α = 0, the set  W0 (m, n, 0; ψ) :=

X∈R

mn

:

max{|q · x(1) |, . . . , |q · x(n) |} ≤ ψ(|q|)



for infinitely many q ∈ Zm \ {0}

is well studied by various authors in [17,13,25,29,30]. The set arose in the literature in part due to its connections with the Kolmogorov–Arnold–Moser theory [2,20], linearization of germs of complex analytic diffeomorphisms of C m near a fixed point [2,21], operator theory [15,16] and recently discovered applications in signal processing [35,37]. Note that W (1, n, 0; ψ) = {(0, 0)}. Indeed, any (x1 , . . . , xn , α1 , . . . , αn ) ∈ W (1, n, 0; ψ) must satisfy the inequality    αi  ψ(q)  max xi −  < 1≤i≤n  q q for infinitely many q ∈ Z \ {0}. Since αi /q → 0 and ψ(q)/q → 0 as q → ∞, taking the limit as q → ∞ yields x1 = . . . = xn = 0, and since ψ(q) → 0 as q → ∞, multiplying by q and then taking the limit yields α1 = . . . = αn = 0. So (x, α) ∈ W (1, n, 0; ψ) only when (x, α) = (0, 0). In a similar way, it can be readily verified that W (1, n, u; ψ) = W (1, u; ψ) × {(0, 0)} for u < n and thus dim W (1, n, u; ψ) = dim W (1, u; ψ). Consequently, throughout this paper we assume that m > 1. Notation. Before proceeding with the description of our main results, we introduce some notation. The Vinogradov symbols  and will be used to indicate an inequality with an unspecified positive multiplicative constant depending only on m and n. If a  b and a b we write a b, and say that the quantities a and b are comparable. A dimension function is an increasing continuous function f : R+ → R+ such that f (r) → 0 as r → 0. Throughout the paper, Hf denotes the f -dimensional Hausdorff measure which will be fully defined in section 2.1. Finally, for convenience, if ψ is a given approximation function, then we write Ψ(q) :=

ψ(q) . q

The Hausdorff dimension of a set E will be denoted by dim E. The measure-theoretic results below for absolute value approximation crucially depend upon whether m + u > n or m + u ≤ n. The reason for a dichotomy should be clear from Lemma 1.6 below.

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Theorem 1.2. Let m+u > n and let ψ be an approximating function. Let f be a dimension function such that the maps r → r−(m+1)n f (r) and r → r−(m+u+1−n)n f (r) are both monotonic. Then for every nonempty open set U ⊆ Rmn × Rn we have

 f



H W (m, n, u; ψ) ∩ U =

⎧ ⎪ ⎪ ⎨

0

⎪ ⎪ ⎩ Hf (U )

if if

∞ q=1 ∞

f (Ψ(q))Ψ(q)−mn q m+u−1 < ∞, f (Ψ(q))Ψ(q)−mn q m+u−1 = ∞.

q=1

Note that we have not assumed the approximating function ψ to be monotonic for any m and n. This is possible due to the extra degree of freedom offered by the variable α. This is a doubly metric inhomogeneous version of the set considered in [29,30], where an analogous result was obtained for the homogeneous fibre obtained by fixing α = 0. As in most of the statements the convergence part is reasonably straightforward to establish and is free from any assumptions on m, u, n, the approximation function, and the dimension function f . It is the divergence statement which constitutes the main substance and this is where conditions come into play. The requirement that r → r−(m+1)n f (r) and other functions must be monotonic is a natural and not particularly restrictive condition. Essentially, the condition ensures that the Hausdorff measures cannot be too degenerate compared with the Lebesgue measure of the ambient Euclidean space, which is of dimension (m +1)n. In the case where f (r) := r(m+1)n the Hausdorff measure Hf is proportional to the standard (m + 1)n-dimensional Lebesgue measure of Rmn × Rn and the resulting special case of Theorem 1.2 is the natural analogue of the Khintchine–Groshev theorem for W (m, n, u; ψ): Corollary 1.3. Let m + u > n and let ψ be an approximating function. Then for every nonempty open set U ⊆ Rmn × Rn ,

  W (m, n, u; ψ) ∩ U 

(m+1)n

⎧ ⎪ ⎪ ⎨ =

0

⎪ ⎪ ⎩ |U |(m+1)n

if if

∞ q=1 ∞

Ψ(q)n q m+u−1 < ∞, Ψ(q)n q m+u−1 = ∞.

q=1

From this corollary, it can easily be seen that for τ

>

m+u n

− 1, the set

def

W (m, n, u; τ ) = W (m, n, u; q → q −τ ) is a null set. The following corollary of Theorem 1.2 gives the Hausdorff dimension of the set W (m, n, u; ψ) as well as its measure with respect to the dimension functions f (r) = rs , s > 0. Corollary 1.4. Let m + u > n, and let ψ be an approximating function. Then for all s ≥ 0 and for every nonempty open set U ⊆ Rmn × Rn ,

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 s



H W (m, n, u; ψ) ∩ U =

⎧ ⎪ ⎪ ⎨

0

if

⎪ ⎪ ⎩ Hs (U )

if

∞ q=1 ∞

7

Ψ(q)s−mn q m+u−1 < ∞, Ψ(q)s−mn q m+u−1 = ∞.

q=1

Consequently, if U is nonempty then   ∞

  s−mn m+u−1 dim W (m, n, u; ψ) ∩ U = sup mn ≤ s ≤ (m + 1)n : Ψ(q) q =∞ . q=1

Finally, for completeness, the dimension result for W (m, n, u; τ ) is given for m +u > n. This follows directly from the previous corollary. Corollary 1.5. For m + u > n and every nonempty open set U ⊆ Rmn × Rn ,   dim W (m, n, u; τ ) ∩ U =



mn +

m+u τ +1

(m + 1)n

if τ ≥ if τ ≤

m+u n m+u n

− 1, − 1.

In fact, Corollary 1.4 gives more than simply the Hausdorff dimension. It also give the Hausdorff measure at the critical exponent which is infinity in this case, except when W (m, n, u, τ ) is of full dimension. To state an analogue of Theorem 1.2 for the case m +u ≤ n, we will need to modify the conditions on the dimension function f . This change is due to the fact that if m + u ≤ n, then the set W (m, n, u; ψ) is contained in an algebraic variety of dimension strictly lower than (m + 1)n. To see this, first consider the case m = n, u = 0 and fix (X, α) ∈ W (m, m, 0; ψ) such that det X = 0. Multiplying the defining inequalities (1.2) by X −1 shows that we must have |q − αX −1 | ≤ C(X)ψ(|q|) for infinitely many q. As |q| tends to infinity, the left-hand side tends to infinity while the right-hand side tends to zero, and we get a contradiction. Hence, any element (X, α) ∈ W (m, m, 0; ψ) must satisfy det X = 0. In other words, W (m, m, 0; ψ) is contained in the hypersurface defined by the equation det X = 0. This logic can be generalized to cover the more general case m + u ≤ n. For this we first introduce more notation. For each m × n matrix X ∈ Rmn with column vectors  be the m × (n − u) matrix with column vectors x(u+1) , . . . , x(n) , x(1) , . . . , x(n) , let X  = (αu+1 , . . . , αn ). Then let Γ ⊆ Rmn × Rn be the set of (X, α) ∈ Rmn × Rn and let α  ⊕α  ⊕α  is 0. Here X  is the such that the determinant of every m × m minor of X   (m + 1) × (n − u) matrix whose first m rows are the rows of X and whose last row is α.

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Lemma 1.6. For m + u ≤ n, the set W (m, n, u; ψ) is contained in Γ. Moreover, def

dim Γ = γ = (m + 1)n − 2(n − m − u + 1). Proof. For the first statement, fix (X, α) ∈ W (m, n, u; ψ) and let Y be an m × m minor  Then if β is the vector whose coordinates are the coordinates of α corresponding of X. to the indices of the columns of Y , then it is easy to see that (Y, β) ∈ W (m, m, 0; ψ). We previously showed that this implies det Y = 0. Since Y was arbitrary, we have (X, α) ∈ Γ. Since, Γ ⊆ Rmn × Rn is defined to be the set of (X, α) ∈ Rmn × Rn such that the  α)  is 0. The second statement readily follows determinant of every m × m minor of (X, on subtracting the number of dependent coordinates coming from the condition that the  α)  is 0, from the full dimension i.e. (m +1)n. 2 determinant of every m ×m minor of (X, Remark 1.7. This Lemma is an analogue of [17, Lemma 1] but we should point out that  such that the calculation in [17] has a small error: if we consider the set of matrices X  are equal to certain prescribed values, then it the m − 1 certain prescribed columns of X is not true that the dimension of this set is always equal to the number claimed. It may be higher, e.g. if we prescribe that all the columns must equal zero. However, the claim is true in the generic case and this is why the computation still gives the correct answer. Theorem 1.8. Let m+u ≤ n and let ψ be an approximating function. Let f be a dimension function such that the maps r → r−m(n−m−u+1) f (r) and r−mn−m−u+1 f (r) are both monotonic. Let U be a nonempty open set. Then   Hf W (m, n, u; ψ) ∩ U ⎧ ∞ ⎪ f (Ψ(q))Ψ(q)−(m−1)n−m−u+1 q m+u−1 < ∞ if ⎪ ⎨0 q=1 = ∞ ⎪ ⎪ ⎩Hf (Γ ∩ U ) if f (Ψ(q))Ψ(q)−(m−1)n−m−u+1 q m+u−1 = ∞. q=1

Analogues of Corollaries 1.3 and 1.4 may be stated in a similar way. For the sake of brevity we only state the analogue of Corollary 1.5. Corollary 1.9. For m+u ≤ n and τ > 0. Then for every nonempty open set U ⊆ Rmn ×Rn we have 



dim W (m, n, u; τ ) ∩ U =

⎧  ⎨ (m + 1)n − 2 n −

m+u τ +1



⎩ (m + 1)n − 2(n − m − u + 1)

if

τ≥

1 m+u−1 ,

if

τ≤

1 m+u−1 .

Similar to the classical setting, next we consider the variant of the set Bad(m, n) complementary to W (m, n, u; ψ). To be precise, for each 0 ≤ u ≤ n let Bad(m, n, u)

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denote the set of all (X, α) ∈ Rmn × Rn for which there exists a constant C(X, α) such that max{|q · x(1) − α1 − p1 |, . . . , |q · x(u) − αu − pu |, |q · x(u+1) − αu+1 |, . . . , |q · x(n) − αn |} ≥ C(X, α) · |q|−

m+u n +1

for all integer vectors (p, q) ∈ Zu × Zm . It is an easy consequence of Theorem 1.2 that for m + u > n, Bad(m, n, u) is a null-set, i.e. |Bad(m, n, u)|(m+1)n = 0. On the other hand, by Lemma 1.6, for m + u ≤ n we have R(m+1)n \ Γ ⊆ Bad(m, n, u) and thus |Bad(m, n, u)|(m+1)n = 1. This raises the natural question of the Hausdorff dimension of Bad(m, n, u) whenever m + u > n, and of Bad(m, n, u) ∩ Γ whenever m + u ≤ n, as well as of the associated fibres Badα (m, n, u) = {X ∈ Rmn : (X, α) ∈ Bad(m, n, u)}. The set Bad0 (m, 1, 0) was first studied in [27], wherein it was proved to have maximal dimension. Later in [28], this result was extended to general linear forms Bad0 (m, n, u) allowing u = 0. We strengthen this result as follows: Theorem 1.10. The set Bad(m, n, u) is hyperplane winning, and in particular has full Hausdorff dimension. If m + u ≤ n, then the set Bad(m, n, u) ∩ Γ is hyperplane winning relative to Γ. It is worth noting that the method of proof of the present paper is immediately applicable to the situation when u = 0, and in that sense we are significantly strengthening the result of [28]. As stated before, the hyperplane winning property passes automatically to games played on certain fractals, therefore, this theorem implies that, for any fractal set K, the set Bad(m, n, u) ∩ K is hyperplane winning. This further implies that Bad(m, n, u) has full Hausdorff dimension when intersected with the fractal set K. 2. Preliminaries and auxiliary results To kick off this section we first define the basic concepts of Hausdorff measure and dimension. 2.1. Hausdorff measure and dimension Below is a brief introduction to Hausdorff f -measure and dimension. For further details see [6,24]. Let F ⊆ Rn . For any ρ > 0, a countable collection {Bi } of balls  in Rn with diameters diam(Bi ) ≤ ρ such that F ⊆ i Bi is called a ρ-cover of F . Define for a right continuous, monotonically increasing function f : R≥0 → R≥0 with f (t) > 0 for t > 0,

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Hρf (F ) = inf

f (diam(Bi )),

i

where the infimum is taken over all possible ρ-covers of F . The function f is called a dimension function, and the Hausdorff f -measure of F is Hf (F ) = lim Hρf (F ). ρ→0

In the particular case where f (r) = rs with s > 0, we write Hs for Hf and the measure is referred to as s-dimensional Hausdorff measure. The Hausdorff dimension of F is denoted by dim F and is defined as dim F := inf{s ∈ R+ : Hs (F ) = 0}. 2.2. Slicing We now state a result due to Beresnevich and Velani [5], referred to as the slicing lemma, which is a key ingredient in the proof of Theorem 1.2. We include the result mainly for completeness, as its application is identical to the one in [29]. Before we state the result it is necessary to introduce a little notation. Suppose that V is a linear subspace of Rk , V ⊥ will be used to denote the linear subspace of Rk orthogonal to V . Further V + a := {v + a : v ∈ V } for a ∈ V ⊥ . Lemma 2.1 (Slicing Lemma, [5, Lemma 4]). Let l, k ∈ N be such that l ≤ k and let f be def

a dimension function such that g(r) = r−l f (r) is also a dimension function. Let B ⊆ Rk be a Borel set and let V be a (k − l)-dimensional linear subspace of Rk . If for a subset S of V ⊥ of positive Hl measure Hg (B ∩ (V + b)) = ∞

∀ b ∈ S,

then Hf (B) = ∞. 2.3. Mass transference principle We describe the mass transference principle for linear forms tailored for our use. The actual framework is broad ranging and deals with the limsup sets defined by a sequence of neighbourhoods of ‘approximating’ planes. The mass transference principle for linear forms naturally enables us to generalize the Lebesgue measure statements of linear forms to the Hausdorff measure statements. In its original form, it was derived by Beresnevich and Velani from the mass transference principle for simultaneous approximation [4] using their ‘slicing’ technique introduced in [5] and described above. Let x(j) denote the jth column vector of X. For q ∈ Zm \ {0}, and p ∈ Zn , the resonant set Rp,q is defined by

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Rp,q = {(X, α) ∈ Rmn × Rn : qX − α = p} = Rp1 ,q × · · · × Rpn ,q and   Rpj ,q = (x(j) , αj ) : q · x(j) − αj = pj . It is then clear that the resonant sets are the affine subspaces of dimension mn, codimension n and are contained in W (m, n; ψ) for all functions ψ. Let R = {Rp,q : q ∈ Zm \ {0}, p ∈ Zn }. Given an approximating function ψ and a resonant set Rp,q , define the Ψ-neighbourhood of Rp,q as  Δ (Rp,q , Ψ(|q|)) =

(X, α) ∈ Rmn × Rn : dist ((X, α), Rp,q ) ≤

ψ(|q|) |q|

 ,

where dist(A, B) := inf{|a − b| : a ∈ A, b ∈ B}. Notice that if m = 1 then the resonant sets are points and the sets Δ (Rp,q , Ψ(|q|)) are balls centred at these points. Let Λ(m, n; Ψ) = {(X, α) ∈ Rmn × Rn : (X, α) ∈ Δ (Rp,q , Ψ(|q|)) for i.m. (p, q) ∈ Zn × Zm \ {0}} and define Δq (Ψ) :=



Δ(Rp,q , Ψ(|q|)).

|p|≤|q|

Then, Λ(m, n; Ψ) can be written as a limsup set so that Λ(m, n; Ψ) =

∞ 

∞ 

Δq (Ψ).

N =1 |q|=N

Theorem 2.2 (Mass transference principle, [1]). Let R and Ψ as above be given. Let f be a dimension function such that g(r) := r−mn f (r) is also a dimension function and such that the map r → r−(m+1)n f (r) is monotonic. Suppose for any ball B in R(m+1)n   1 H(m+1)n B ∩ Λ(m, n; g(Ψ) n ) = H(m+1)n (B). Then for any ball B ∈ R(m+1)n Hf (B ∩ Λ(m, n; Ψ)) = Hf (B). This theorem was originally proved with an additional hypothesis on the system of subspaces R, see [5, Theorem 3]. However, we need the stronger version in [1] in order to prove Theorems 1.2 and 2.3.

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2.4. An interlude on classical Diophantine approximation Below is the statement of a generalized form of a Khintchine–Groshev-type theorem for W (m, n; ψ) which to our knowledge is a new result and interesting in its own right. The reader will notice that it is a special case of Theorem 1.2, namely the case u = n, which correspond to the nearest integers lying in a full lattice and so the classical Diophantine approximation. We include it here, as it is a key ingredient in the proof of Theorem 1.2. Theorem 2.3. Let ψ be an approximating function. Let f and r → r−mn f (r) be dimension functions such that r → r−(m+1)n f (r) is monotonic. Then

Hf (W (m, n; ψ)) =

⎧ ⎪ ⎪ ⎨

0

if

⎪ ⎪ ⎩ Hf (Rmn × Rn )

if

∞ q=1 ∞

f (Ψ(q))Ψ(q)−mn q m+n−1 < ∞, f (Ψ(q))Ψ(q)−mn q m+n−1 = ∞.

q=1

Note that even in the one dimensional setting there is no condition of monotonicity imposed on the approximating function, in contrast to the case of estimating the dimension of a fibre corresponding to a fixed value of α (either zero or nonzero), where the monotonicity condition on the approximating function cannot be removed due to the Duffin–Schaeffer counterexample, see [3,22]. Note that if the dimension function f is such that r−(m+1)n f (r) → ∞ as r → 0 then f H (Rmn ×Rn ) = ∞ and Theorem 2.3 is the analogue of the classical result of Jarník (see [18,32]). In the case where f (r) := r(m+1)n the Hausdorff measure Hf is proportional to the standard (m + 1)n-dimensional Lebesgue measure supported on R(m+1)n and the result is the natural analogue of the Khintchine–Groshev theorem for W (m, n; ψ). A singly metric analogue of Theorem 2.3 can be found in Bugeaud’s paper [9]. We state the following special case of Theorem 2.3, which is a consequence of [40, Theorem 15]. Theorem 2.4 ([40, Theorem 15]). Let ψ be an approximating function. Then

|W (m, n; ψ)|(m+1)n =

⎧ ⎪ ⎪ ⎨0

if

⎪ ⎪ ⎩1

if

∞ q=1 ∞

ψ(q)n q m−1 < ∞, ψ(q)n q m−1 = ∞.

q=1

For the doubly metric case, the Hausdorff dimension for the set W (m, n; ψ) was established by Dodson in [19] and, for the singly metric case, by Levesley in [34]. A slightly more general form of Dodson’s result was established by Dickinson in [14].

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2.5. Proof of Theorem 2.3 2.5.1. The convergence case Notice that the set W (m, n; ψ) can be written in the following limsup form: W (m, n; ψ) =

∞  ∞  

Δ (Rp,q , Ψ(|q|)) .

N =1 h=N |q|=h

For each resonant set Rp,q the set Δ(Rp,q , Ψ(|q|)) can be covered by a collection of (mn +n)-dimensional closed hypercubes with disjoint interior and sidelength comparable with Ψ(|q|). It can be readily verified that the number C of such hypercubes satisfies C  Ψ(|q|)−mn |q|n . Thus, W (m, n; ψ) can be written as the limsup of a sequence of hypercubes whose total “f -dimensional cost” is ∞

Cf (Ψ(|q|)) 

h=1 |q|=h

∞

Ψ(|q|)−mn |q|n f (Ψ(|q|))

h=1 |q|=h



∞

hm+n−1 f (Ψ(h))Ψ(h)−mn < ∞.

h=1

Thus by the Hausdorff–Cantelli lemma [6, Lemma 3.10], we have Hf (W (m, n; ψ)) = 0, as required. 2.5.2. The divergence case The divergence case is an easy consequence of the mass transference principle discussed in the section §2.3. In view of this we shall use the divergence part of Theorem 2.4 (Cassels’ theorem) and the mass transference principle to prove the divergence part of Theorem 2.3. With reference to the framework of §2.3 we note that Λ(m, n;

1 Ψ) ⊆ W (m, n; ψ) ⊆ Λ(m, n; Ψ) m+1

Hence the divergence case follows. 3. Proof of Theorem 1.2 3.1. The convergence case The convergence part, although similar to the convergence part of Theorem 2.3, is included for completeness.

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For q ∈ Zm \ {0}, and pu = (p1 , . . . , pu ) ∈ Zu , the resonant set Rpu ,q is defined by Rpu ,q = {(X, α) ∈ Rmn × Rn : qX − α = pu } = Rp1 ,q × · · · × Rpu ,q × · · · × Rpn ,q , where   Rpj ,q = (x(j) , αj ) : q · x(j) − αj = pj

(1 ≤ j ≤ u)

and   Rpi ,q = (x(i) , αj ) : q · x(i) − αi = 0

(u + 1 ≤ i ≤ n).

It is then clear that the resonant sets are the affine subspaces of dimension mn, codimension n and are contained in W (m, n, u; ψ) for all functions ψ. Further, the set W (m, n, u; ψ) can be written in the following limsup form:

W (m, n, u; ψ) =

∞ ∞   

Δ (Rpu ,q , Ψ(|q|)) ,

N =1 h=N |q|=h

where  Δ (Rpu ,q , Ψ(|q|)) =

(X, α) ∈ Rmn × Rn : dist ((X, α), Rpu ,q ) ≤

ψ(|q|) |q|

 ,

is the Ψ-neighbourhood of Rpu ,q for any resonant set Rpu ,q . For each resonant set Rpu ,q the set Δ(Rpu ,q , Ψ(|q|)) can be covered by a collection of (mn + n)-dimensional closed hypercubes with disjoint interior and sidelength comparable with Ψ(|q|). It can be readily verified that the number C of such hypercubes satisfies C  Ψ(|q|)−mn |q|u . Thus, W (m, n, u; ψ) can be written as the limsup of a sequence of hypercubes whose total “f -dimensional cost” is ∞

Cf (Ψ(|q|)) 

h=1 |q|=h

∞

Ψ(|q|)−mn |q|u f (Ψ(|q|))

h=1 |q|=h



∞

hm+u−1 f (Ψ(h))Ψ(h)−mn < ∞.

h=1

Thus by the Hausdorff–Cantelli lemma, we have Hf (W (m, n, u; ψ)) = 0, as required.

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3.2. The divergence case As discussed earlier the statement of the Theorem essentially reduces to two cases, the finite measure case where r−(m+1)n f (r) → C > 0 as r → 0 and the infinite measure case in which r−(m+1)n f (r) → ∞ as r → 0. Therefore, we split the proof of the Theorem 1.2 into two parts, the finite measure case and the infinite measure case. Before proceeding, we will need the following key lemma, which will make our proofs work. Lemma 3.1. Let S ⊆ Mat(m+u−n)×n (R) × Rn of full Lebesgue measure. Let A ⊆ GLn×n (R) be a subset of a subspace of dimension (n − u)n such that A has positive (n − u)n-dimensional Lebesgue measure. Then, the set  Λ=

X YX





 , αX

∈ Mat(m+u)×n (R) × R : X ∈ A, (Y, α) ∈ S n

has full Lebesgue measure inside A × S, with the product structure being the one implicit in the affine system. Proof. By Fubini’s theorem, it suffices to show that for almost all X ∈ A, the fibre ΛX = {(Y X, αX) : (Y, α) ∈ S} has full measure. In fact, this is true for all X ∈ GLn×n (R); this follows from the invariance of Lebesgue measure under right multiplication by the invertible matrix X. 2 Remark 3.2. For a detailed proof for the analogous homogeneous situation we refer to Lemma 4.1 of [29]. 3.3. Finite measure case As in [29], in order to proceed, we will make some restrictions. Let ε > 0 and N > 0 be fixed but arbitrary. Define  Aε,N =

 < ε−1 , (X, α) ∈ Matm×n (R) × Rn : ε < det(X)

 max |xij | ≤ N

1≤i,j≤n

 denotes the n × n-matrix formed by the first n rows of the matrix X. The set where X is of positive measure for ε small enough and N large enough, and as ε decreases and N increases, the set fills up Mat(m+1)×n (R) with the exception of the null-set of matrices  is singular. X such that X

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We will prove that the divergence assumption implies that the set W (m, n, u; ψ) is full in Aε,N . We will translate the statements about W (m, n, u; ψ) into ones about usual Diophantine approximation, specifically regarding sets of the form Wα (m + u − n, n; cψ), c > 0. ˆ α) ˆ ∈ Consider the set of n affine forms in m + u − n variables defined by the pair (X, (m+u−n)n n × R . Suppose furthermore that these linear forms satisfy the inequalities R   ψ(|r|)  ˆ  ˆ ≤ , rX − α nN i

1 ≤ i ≤ n,

(3.1)

for infinitely many r ∈ Zm+u−n \ {0}, where xi denotes the distance from the ith coordinate of x to the nearest integer. A special case of Theorem 2.4 states that the ˆ α) ˆ is full in divergence condition of our theorem implies that the set of such pairs (X, n Mat(m+u−n)×n (R) × R , and hence in particular also in the image of Aε,N under the map 

 X ˆ  XX



 ˆX ,α

 ˆ α). ˆ → (X,

ˆ α) ˆ is in the set defined by (3.1), Now, suppose that (X, α) ∈ Aε,N is such that (X, ˆ α) ˆ is the uniquely determined pair such that where (X,  (X, α) =

 X ˆ  XX



  ˆX . ,α

We claim that (X, α) is in W (m, n, u; ψ). Indeed, let (rk )k be an infinite sequence of integer vector such that the inequalities (3.1) are satisfied for each k, and let pk be the ˆ − α. Now define qk = (pk , rk ). The inequalities defining nearest integer vector to rk X W (m, n, u; ψ) will be satisfied for these values of qk , since    qk Iu  Xu

0 X



        In    − α qk ˆ X − α X               , ˆ 1  , . . . , ± rk · x(n) + α ˆ n X

ε  ± rk · x(1) + α 1

n

  ˆ splits into n column vectors x(1) , . . . , x(n) , X = (Xu X  ). Where the implied where X constant may depend upon m, n, ε but not on qk . The ith coordinate of the first vector is at most ψ(|q|)/nN , so carrying out the matrix multiplication, using the triangle inequality and the fact that |xij | ≤ N for 1 ≤ i, j ≤ n shows that    qk Iu  Xu

0 X



  − α < ψ(|qk |). i

Applying Lemma 3.1, the divergence part of Theorem 1.2 follows in the case of Lebesgue measure.

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3.4. Infinite measure case The infinite measure case of the Theorem 1.2 can be easily deduced from the following lemma. Lemma 3.3. Let ψ be an approximating function and let f be a dimension function such def that g(r) = r−n(n−u) f (r) and r → r−(m+u−n)n g(r) are also dimension functions, r → r−(m+u+1−n)n g(r) is monotonic, and r−(m+1)n f (r) → ∞ as r → 0. If ∞

f (Ψ(q))Ψ(q)−mn q m+u−1 = ∞,

q=1

then Hf (W (m, n, u; ψ)) = ∞. Proof. The proof relies on a bi-Lipschitz injective map which preserves measure, just like the proof of Lemma 4.2 of [29]. As in the finite measure case, we fix ε > 0, N ≥ 1 and let  Aε,N =



 < ε−1 , (X, α) ∈ Matm×n (R) × R : ε < det(X)

max |xij | ≤ N

n

1≤i,j≤n

,

 denotes the n × n-matrix formed by the first n rows of X. We also define the where X set ε,N = A



 ∈ GLn (R) : ε < det(X)  < ε−1 , X

 max |xij | ≤ N

1≤i,j≤n

.

Consider a subspace   Iu  A0 = X ∈ Aε,N : X = Xu

0 X

 ,

ε,N , which is of positive (n − u)n-dimensional Hausdorff measure. For an appropriof A ately chosen constant c > 0 depending only on m, n, ε and N , we find that the map η : W (m + u − n, n, cψ) × A0 → W (m, n, u; ψ),    X ((Y, α), X) → , αX YX is a Lipschitz embedding. Indeed, it is evidently injective as X is invertible for all the domain. We have

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Hf (W (m, n, u; ψ)) ≥ Hf (η (W (m + u − n, n, cψ) × A0 ))

Hf (W (m + u − n, n, cψ) × A0 ) . Now after taking into account the conditions on the dimension functions as stated in the lemma along with the divergence part of Theorem 2.3 for W (m + u − n, n; cψ) and f = g, we apply the slicing Lemma 2.1 across A0 to get the infinite measure under the appropriate assumption of divergence. 2 4. Proof of Theorem 1.8 As in the preceding results, the convergence case is a rather dull affair, and boils down to constructing an appropriate fine cover and applying the Hausdorff–Cantelli lemma [6, Lemma 3.10]. We omit the details and proceed with the divergence case. Let p = m + u − 1. For clarity we only discuss the case m + u = n, or equivalently n − p = 1. Define the map η : Rmp × Rp × Rm−1 → Γ via the formula   η (x(1) , . . . , x(p) ), (α1 , . . . , αp ), (b1 , . . . , bm−1 ) ⎛ ⎞ m−1 m−1    

= ⎝ x(1) , . . . , x(p) , bj x(u+j) , α1 , . . . , αp , bj αu+j ⎠ . j=1

j=1

For the more general case m + u ≤ n, we need a slightly more complicated expression for η with more linearly dependent columns in the matrix. However, the ideas should be clear from the above simpler case. Fix C ≥ 1,   (4.1) (x(1) , . . . , x(p) ), (α1 , . . . , αp ) ∈ W (m, m + u − 1, u; C −1 ψ), m−1 and (b1 , . . . , bm−1 ) ∈ Rm−1 such that j=1 |bj | ≤ C. Note that by the triangle inequality,    m−1  m−1 m−1

    (u+j) q · bj x − bj αu+j  ≤ |bj | · q · x(u+j) − αu+j     j=1 j=1 j=1 ⎛ ⎞ m−1

≤⎝ |bj |⎠ C −1 ψ(|q|) ≤ ψ(|q|), j=1

for the infinitely many integer vectors q we obtain from condition (4.1). It follows that

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  m−1  C C η W (m, m + u − 1, u; C −1 ψ) × − m−1 , m−1 ⊆ W (m, n, u; ψ). In the more general case m + u ≤ n, we have   (m−1)(n−p)  C C η W (m, m + u − 1, u; C −1 ψ) × − m−1 , m−1 ⊆ W (m, n, u; ψ).

(4.2)

To proceed further, we will use the fact that η is a locally bi-Lipschitz embedding when restricted to the set Γ × R(m−1)(n−p) , where  ⊕ α)  = m − 1}. Γ = {(X, α) : rank(X Note that Γ is relatively open and dense in Γ. We will need the following lemma [31, Proposition 2.2]. Lemma 4.1. Let f be a dimension function, let L ⊆ Rl , and let η : L → Rk be a bi-Lipschitz embedding. Then Hf (L) Hf (η(L)). We are now in a position to prove the infinite measure case of Theorem 1.8. Lemma 4.2. Let m + u ≤ n. Let ψ be an approximating function and let f be a dimension function such that g(r) = r−(m−1)(n−p) f (r) is also a dimension function, and such that lim r−γ f (r) = ∞.

r→0

Let U ⊆ Rmn × Rn be a non empty open set. If ∞

f (Ψ(q))Ψ(q)−(m−1)n−p q p = ∞,

q=1

then Hf (W (m, n, u; ψ) ∩ U ) = ∞ = Hf (U ). Proof. Let V ⊆ Rmp × Rp be a nonempty open set. Since lim r−(m+1)p g(r) = ∞,

r→0

we have Hg (V ) = ∞. On the other hand, since ∞

g(Ψ(q))Ψ(q)−mp q p = ∞,

q=1

by Theorem 1.2, for every C ≥ 1 we have

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  Hg W (m, m + u − 1, u; C −1 ψ) ∩ V = Hg (V ) = ∞. Thus by [24, Corollary 7.12], for every nonempty open set V  ⊆ Rm−1 we have   Hf (W (m, m + u − 1, u; C −1 ψ) ∩ V ) × V  = ∞. Now choose V , V  , and C such that η(V × V  ) ⊆ U,

V ⊆



C C m−1 , m−1

(m−1)p

and such that η is bi-Lipschitz on V × V  . Then by (4.2) and Lemma 4.1,   Hf W (m, n, u; ψ) = ∞. Now suppose that lim inf r→0 r−γ f (r) < ∞. Then either Hf (Γ) = 0, or Hf is proportional to Hγ on Γ. Thus for the remainder of the proof, we assume that f (r) = rγ . As before, let g(r) = r−(m−1)(n−p) f (r), and suppose that ∞

f (Ψ(q))Ψ(q)−(m−1)n−p q p =

q=1

∞

g(Ψ(q))Ψ(q)−mp q p = ∞.

q=1

Again, by Theorem 1.2, for every C ≥ 1 and for every open set V ⊆ Rmp × Rp we have   Hg W (m, m + u − 1, u; C −1 ψ) ∩ V = Hg (V ) Since g(r) = r(m+1)p , this means that W (m, m + u − 1, u; C −1 ψ) is of full Lebesgue measure in Rmp × Rp . Thus, W (m, m + u − 1, u; C −1 ψ) × R(m−1)(n−p) is of full measure in Rmp × Rp × R(m−1)(n−p) . Since η is smooth, by (4.2) and Sard’s theorem [41, Theorem II.3.1], the set W (m, n, u; ψ) is of full measure in the image of η, which is in turn of full measure in Γ. So W (m, n, u; ψ) is of full measure in Γ. 2 5. Proof of Theorems 1.1 We begin with a description of the hyperplane game.

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5.1. Hyperplane winning Fix 0 < β < 1. The β-hyperplane game on Rk has two players, Alice and Bob, and is played as follows: (1) Bob chooses an initial closed ball B0 = B(x0 , ρ0 ). (2) After Bob’s nth move Bn , Alice ‘deletes a neighbourhood of a hyperplane An ’, i.e. she picks an affine hyperplane An ⊆ Rk , which must be avoided by Bob in his next move. (3) Bob now chooses a ball Bn+1 = B(xn+1 , ρn+1 ) satisfying Bn+1 ⊆ Bn \ An(βρn ) and ρn+1 ≥ βρn . (βρ )

Here, An n denotes a closed tubular neighbourhood of An of radius βρn . If this is not possible, Alice wins the game. A set S is said to be β-hyperplane winning if Alice has a strategy ensuring that ∞ 

Bn ∩ S = 

n=1

If S is β-hyperplane winning for all β > 0, it is said to be hyperplane winning. The game is a modification of the classical Schmidt game [38], which we will also need. In this game we have two parameters α > 0 and β > 0. In this game, instead of removing a neighbourhood of a hyperplane, Alice chooses a closed ball Wn ⊆ Bn of radius αρn . Also, at each step Bob must choose a closed ball Bn+1 ⊆ Wn of radius ρn+1 = βαρn . The conditions for winning the game are unchanged. A set S is (α, β)-winning if Alice has a winning strategy for this set of parameters. It is α-winning if it is (α, β)-winning for any β > 0. 5.2. Proof of Theorem 1.1 The idea is to modify the proof that Bad0 (m, n) is hyperplane winning found in [8]. Using the ideas of [23], this approach can be extended to prove that Badα (m, n) is hyperplane winning. This was stated as a result in a preprint of [8], though it did not make it to the final paper. In fact, the approach is sufficiently flexible to allow us to modify the proof to show that Bad(m, n) is hyperplane winning. Before starting the technical proof, let us explain the strategy in a little more detail. For the classical Schmidt game, Einsiedler-Tseng [23] modified Schmidt’s [39] original proof that Bad0 (m, n) is winning to show that Badα (m, n) is winning for any α. Proceeding in the same way, but with the hyperplane game in place of the classical game, one can extend the proof in [8] to show that Badα (m, n) is hyperplane winning. The

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strategy is devised so that from the kth step of the game, Bob is only able to chose from a set of affine forms which satisfy the inequality defining Badα (m, n) for all q with |q| bounded by an explicit parameter of the proof. We will provide a proof of this, based on the strategy devised in [8] with modifications according to the ideas of [23]. The key new insight is that in her kth move, Alice does not need to know the exact value of α, but only to know the value to within an accuracy of m

cρkm+n , which allows us to complete the proof of Theorem 1.1. We now proceed with an explicit description of Alice’s strategy. We will only describe the steps in the construction of the strategy and refer the reader to [8] for the full details. Initially, we describe how she plays in the case of Bad0 (m, n). This is the strategy shown to be hyperplane winning in [8]. Given a matrix X ∈ Rmn , we append a standard unit vector to each column, i.e. we consider the following matrix, = X



X In

 .

Note that X ∈ Rmn is an element of Bad0 (m, n) if ad only if there is a constant c > 0 such that for any (q, p) ∈ Zm × Zn ,  n > c. |x|n · |(q, p)X| 2

Let λ = n/(m + n), R > 1 and δ = Rn(n+m) . The strategy devised in [8] ensures that for any i ∈ N, the system of inequalities 0 < |q| < δRm(λ+i)  < δR−n(λ+i)−m , |(q, p)X|

(5.1)

has no solutions (q, p) ∈ Zm+n \ {0}. It is easily seen that this gives a winning strategy for Bad0 (m, n). Unfortunately, the strategy is not easily described, as one needs to consider simultaneously a dual setting. In this setup, one considers instead the matrices  XT

=

XT Im

 ,

and introduce a new parameter, δ T = R−m(m+n) . The condition which is ensured simultaneously to (5.1) is that for any j ∈ N, the system of inequalities 2

0 < |p| < δ T Rn(λ+j) |(p, q)X T | < δR−m(λ+i)−n ,

(5.2)

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has no solutions (p, q) ∈ Zn+m \ {0}. At the heart of constructing a winning strategy is the following lemma [8, Lemma 5.3]. Lemma 5.1. For R ∈ R sufficiently large, the following holds i) When the game is at stage ki , suppose that for any X ∈ B(Xki , ρki ), the system of inequalities (5.1) has no solutions (q, p) ∈ Zm+n \{0}; and the system of inequalities (5.2) has no solutions (p, q) ∈ Zn+m \ {0} with j = i − 1. Then, Alice has a strategy such that after finitely many stages at stage hi , say, the system of inequalities (5.2) has no solutions (p, q) ∈ Zn+m \ {0} with j = i for any X ∈ B(Xhi , ρhi ). ii) Dually, at stage hj , suppose that for any X ∈ B(Xhi , ρhi ), the system of inequalities (5.1) has no solutions (q, p) ∈ Zm+n \ {0}; and the system of inequalities (5.2) has no solutions (p, q) ∈ Zn+m \ {0} with j = i. Then, Alice has a strategy such that after finitely many stages at stage kj+1 , say, the system of inequalities (5.1) has no solutions (q, p) ∈ Zm+n \ {0} with i = j + 1 for any X ∈ B(Xkj+1 , ρkj+1 ) It clearly suffices to construct such a strategy, as one can then recursively construct a hyperplane winning strategy for Bad0 (m, n). This also explains why it is only necessary to know α to a certain precision in our later modification of the strategy, as we are successively avoiding more and more affine forms. We will be more precise later. The proofs of the two parts of the lemma are each others dual, as should be apparent from the definitions given. We will comment on the first part, so suppose that for all X ∈ B = B(Xki , ρki ), the non-existence of integer solutions to (5.2) with j = i − 1 and (5.1) is ensured as in the statement of the lemma. Define the set ! " S = (p, q) ∈ Zn+m : for some X ∈ B, (5.2) is satisfied with j = i . The R-span of S has dimension at most n. If the dimension is less than n, we extend this subspace in some arbitrary way to a space of dimension equal to n. Now, pick an orthonormal basis Y = {y1 , . . . , yn } for this subspace of Rn+m . Let bu denote the uth column of X T and define the matrix M (X, Y) = (bu · yv )1≤u,v≤n , and let D denote the determinant of this matrix. It is now shown using Cramer’s rule that under the given assumptions, √ |D| ≤ n nR−(m+n)(i+1) max{|D11 |, |D12 |, . . . , |Dnn |},

(5.3)

The idea is now to construct a strategy for Alice such that after a finite number of steps, (5.3) cannot hold. In this way, she will have avoided all solutions to (5.2) with j = i, and the game can continue with the dual construction.

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To prove that it is possible for Alice to ensure that (5.3) cannot hold after a finite number of steps, we will need additional notation. Let v ∈ {0, 1, . . . , n} and define the set Ωv = {(I, J) ∈ {0, 1, . . . , n} × {0, 1, . . . , n} : #I = #J = v}. For each ω = (I, J) ∈ Ωv , let Dω (X) = det(bik · yjl )1≤k,l≤v , where I = {i1 , . . . iv } and J = {j = 1, . . . , jv }, i.e. the minor of the determinant D corresponding to the choices I and J. Now, define the vector n2

M v (X) = (Dω (X))ω∈Ωv ∈ R

v

.

Finally, let Mv (X) = M v (X), where  ·  denotes the euclidean norm. For a ball B ∈ Rmn , we let Mv (B) = sup Mv (X). X∈B

Two cases require special treatment. For v = 0, Ω0 = {(, )}, and we let D(,) (X) = 1 for all X, so that M0 (B) = 1 identically. For v = −1, Ω−1 = , and M−1 (B) = 0 identically. The final ingredient in the proof is Lemma 6.1 from [8], which is now stated. Lemma 5.2. Let 0 < β < 1/3, σ ∈ R and v ∈ {0, 1, . . . , n}. There is a constant νv = νv (m, n, β, σ) > 0, such that for any μv ∈ (0, νv ] and any orthonormal system Y = {y1 , . . . , yn } ⊆ Rm+n , Alice can win the finite game with rules i) Bob chooses a closed ball B ⊆ Rmn of radius ρB < 1 and with maxX∈B X ≤ σ. ii) Bob and Alice play according to the rules of the hyperplane game until the radius of Bob’s ball Bv is less than μv ρB . iii) Alice wins if for any X ∈ Bv , Mv (X) > νv ρB Mv−1 (Bv ). The reader will note that the inequality of the lemma in the case v = N essentially is of the form (5.3), and indeed it is exactly what is required to obstruct existence of solutions. On choosing the parameter R large enough, the previous lemma ensures the validity of Lemma 5.1. The construction of the strategy is now based on some results of Schmidt [39], which are also proved in [8]. We condense them into a single lemma.

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Lemma 5.3. Fix X ∈ B and v ∈ {0, . . . , n}. Then, i) For any ω ∈ Ωv , ∇Dω (X)op ≤ vMv−1 (X). ii) For any ω ∈ Ωv , ∇∇Dω (X)op ≤ v 2 Mv−2 (X). iii) There exist constants ε1 , ε2 > 0 depending only on m, n and σ, such that if Mv (X) ≤ ε1 Mv−1 (X), then max ∇Dω (X)op > ε2 Mv−1 (X).

ω∈Ωv

Where  · op stands for the operator norm. We now proceed by induction in v. For v = 0, the claim is trivial, so suppose Lemma 5.2 is true for v −1. In other words, for some νv−1 > 0 and some μv−1 ∈ (0, νv−1 ], if Bv−1 is the first ball chosen by Bob with radius less than μv−1 ρB , then for all X ∈ Bv−1 , Mv−1 (X) > νv−1 Mv−2 (Bv−1 ). Applying inequality i) of Lemma 5.3 and using the mean value theorem, it is straightforward to show that Mv−1 (Bv−1 ) ≤ vMv−1 (X).

(5.4)

At this point, we are finally able to describe the strategy, Alice should follow to win the finite game. There are two cases. First, if Mv (X) > ε1 Mv−1 (X) for all X ∈ B, Alice will make arbitrary moves until ρ(Bv ) < μv ρB . This suffices, since (5.4) implies that Mv (X) ≥

ε1 ε1 Mv−1 (Bv−1 ) ≥ Mv−1 (Bv ), v v

since Bv ⊆ Bv−1 . Hence, choosing νv < εv1 suffices in this case. Hence, we need only worry about the case Mv (X) > ε1 Mv−1 (X) for some X ∈ B. In this case, Lemma 5.3 iii) holds, so that ∇Dω (X)op > ε2 Mv−1 (X) for some ω ∈ Ωv . We consider the linearization L of Dω around X, L(A ) = Dω (X) + ∇A −A Dω (A). Let L = L−1 (0), an affine subspace of Rmn . Alice removes the neighbourhood L(βρ(Bv−1 )) from Bv−1 and continues to make arbitrary moves until ρ(Bv ) < μv ρB . Because of Lemma 5.3 iii), if A ∈ Bv−1 \ L(βρ(Bv−1 )) , |L(A )| is fairly large. On the other hand, Lemma 5.3 iii) allows us to control how much L(A ) deviates from Dω (A ). Putting all

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1 2 estimates together, one finds that choosing νv = 2v β ε2 μv−1 will ensure the conclusion. See [8] for the calculations. This concludes our treatment of Bad0 (m, n). Now, to get from Bad0 (m, n) to Badα (m, n), we show how to modify the strategy using the ideas of [23]. In fact, their proof shows directly that Badα (m, n) is hyperplane winning. The added complication comes from the fact that they play their game on certain fractals. We describe the strategy in the following. We will use their notation, which again involve certain matrices. For X an m×n-matrix and α ∈ Rm , define

# LX (α) =

Im 0 0

X In 0

−α 0 1

$ .

The affine lattice associated to X and α is given by  LX (α)(Zm+n × {1}) =

# $  p LX (α) q : p ∈ Zm , q ∈ Zn ⊆ Rm+n × {1}. 1

Note that the last coordinate is always equal to 1, so by abuse of notation, we will also write LX (α)(Zm+n ) for the projection of the above set onto the first m + n coordinates, and we will refer to this set as the affine lattice associated to X and α. The space of all affine unimodular lattices Ωm+n,aff in Rm+n consists of all translates Λ + β, where Λ = gZm+n for some g ∈ SL(m + n, R) is a unimodular lattice and β ∈ Rm+n . Note that the affine lattices associated to some pair (X, α) are of this form. Finally, we will require a one-parameter action on the space Ωm+n,aff . The appropriate one is the left action by matrices of the form ⎛

et/m Im gt = ⎝ 0 0

0

e−t/n In 0

⎞ 0 0⎠ , 1

on Rm+n+1 and in particular on the invariant subspace Rm+n × {1} as above. Observe that the action defined on Rm+n × {1} induces an action on Ωm+n,aff . The key idea is to apply the following result of Dani [11, Theorem 2.20]. For an affine lattice Γ ∈ Ωm+n,aff , we let δ(Γ) denote the length of the shortest non-zero vector in Γ. Theorem 5.4. The pair (X, α) is in Badα (m, n) if and only if for some c > 0, inf δ(gt LX (α)Zm+n ) ≥ c.

t∈R

As we have seen, Alice has a β-winning strategy for the Bad0 (m, n)-game. By [7], this implies that Alice has an α0 -winning strategy for the usual Schmidt game for some α0 > 0. Suppose Bob has chosen the ball B = B(A , ρ ) at his last step of the hyperplane game. We will show how Alice should respond in her following two moves.

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27

Initially, let α = δα0 and for β > 0 let β0 = δ −1 β, where δ > 0 is chosen appropriately (δ = 2n /μ(B(0, 1)) will do, where μ is the n-dimensional Lebesgue measure, see [23]). Note that α0 β0 = αβ. The usual Schmidt game is (α0 , β0 )-winning. Hence, Alice may choose a ball B = B(A , α0 ρ ) according to this strategy. Next, t must be chosen so that for any D ∈ Rmn , gt LD (0) = L(α0 ρ )−1 D (0)gt for all D. This is done by setting  1   α0 ρ = exp − m + n1 t . We then define the affine lattice x by 1

x = gt LA (α)Zk = L0 (e n t α)gt LA (0)Zk . Now, let v be a vector in x of shortest norm. For a vector x ∈ Rm+n = Rm × Rn , let xp denote projection onto the first m components (the particle space in the language of [23]) and let xt denote projection onto the last n coordinates (the time space). There is a proper, affine subspace L of Rmn such that (LA (0)v)p = 0 if and only if A ∈ L. In fact, just as above in the α = 0 case, if A ∈ / L(ε) , then (LA (0)v)p  ≥ εvt .

(5.5)

Alice will remove this hyperplane for an appropriately chosen ε > 0. The value ε = 2δα0 ρ works. Note that in [23], the authors proceed by letting Alice choose yet another ball W inside B  but disjoint from L(ε) . This is due to the fact that they are concerned with the usual Schmidt game, but in the remainder of their argument, all that is really needed is (5.5). Hence, there is no problem in just removing the hyperplane neighbourhood and letting Bob pick the next ball. In her next move, Alice plays according to a winning strategy for the hyperplane game with target set Bad0 (m, n) with the same parameter, i.e. β = 2δα0 . This is to ensure that however the game ends, the resulting point will be an element in Bad0 (m, n). Let X be the final point when the above strategy is followed. Fix γ1 > 0 small enough that gt LX (0)Zk does not contain a non-zero element of norm ≤ γ1 by Theorem 5.4 with α = 0. Then, fix γ2 > 0 so that LA Zk does not contain any element with norm ≤ γ2 , which is possible since α ∈ / Zk . Finally, pick c > 0 so that Dvt  ≤ cvy  for all D ∈ B(0, 1). In [23], it is shown by induction that for any  ≥ 0, B(0, r) ∩ gt LX (α)Zk =  for r=

1 2

min{γ1 , γ2 }(1 + c)−2 (αβ)n/(m+n) .

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We will not repeat the argument here, but appealing to Theorem 5.4, this immediately implies that X ∈ Badα (m, n), so that this set is hyperplane winning. Finally, to complete the proof, note that in the above, when Alice is responding to Bob’s ball Bk = B(Xk , ρk ), she only needs know the value of α to within an accuracy of m

cρkm+n , where c > 0 is a small constant. Using the claim, we complete the proof as follows. Suppose that in the hyperplane game with target set Bad(m, n), Bob plays the ball Bk = B((Xk , αk ), ρk ). Then Alice can interpret this move as corresponding to the move Bk = B(Xk , ρk ) in the hyperplane game with target set Badα (m, n), where α ∈ B(αk , ρk ) is m

an unknown parameter. Since ρk < cρkm+n for all sufficiently large k, we see that α is m

known to within an accuracy of cρkm+n which is sufficient for Alice to use the strategy described above to generate a hyperplane-neighbourhood Ak = L(βρk ) with which to respond to Bob’s move Bk . In the hyperplane game with target set Bad(m, n), Alice plays the corresponding hyperplane-neighbourhood Ak = Ak × Rn = {(X, α) : X ∈ Ak , α ∈ Rn }. The game continues with an eventual result of (X, α) in the game with target set Bad(m, n) and an eventual result of X in the game with target set Badα (m, n). Since X ∈ Badα (m, n) if and only if (X, α) ∈ Bad(m, n), this shows that Bad(m, n) is hyperplane winning. 2 6. Proof of Theorem 1.10 We now proceed to deduce Theorem 1.10 from Theorem 1.1 by transporting a winning strategy from the classical setup to the new one. This is accomplished by appealing to the following result. Proposition 6.1. Let k, l ∈ N, let U ⊆ Rk be a nonempty open set whose complement is an algebraic variety, and let f : U → Rl be a differentially surjective map, locally C 1 conjugate to a projection π on U. Then the preimage under f of any hyperplane winning set in Rl is hyperplane winning. The proof of the Proposition depends on two older results, which we re-state here for easy reference. The first is [26, Lemma 3.9]. Lemma 6.2. Let k, D ∈ N and β ∈ (0, 1]. There exists a γ > 0, such that for any non-zero polynomial f : Rk → R of degree at most D, if Bob and Alice play the β-hyperplane game and if Bob’s first move is B0 = B(0, 1), then Alice has a strategy ensuring that the first ball chosen by Bob of radius less than γ is disjoint from the set

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!

x ∈ Rk : d(x, f −1 (0)) < γ

29

"

The other key ingredient is a special case of [7, Theorem 2.4], which we state as a lemma. Lemma 6.3. Let S ⊆ Rk be hyperplane winning, U ⊆ Rk be open and let f : U → Rk be C 1 non-singular map. Then, f −1 (S) ∪ U c is hyperplane winning. Proof. Alice begins by ensuring that we need not worry about the variety U c. This is accomplished by Lemma 6.2. Indeed, the variety U c is the zero set of some polynomial map, so by the Lemma, Alice has a strategy ensuring that after finitely many steps, Bob must choose balls at a positive distance γ > 0 from this zero set. Hence, there is no loss of generality in assuming that the first ball chosen by Bob is contained in U. By Lemma 6.3, applying a C 1 conjugation to a hyperplane winning set does not affect this property. Hence, in order to prove the proposition, it suffices to show that the preimage under a projection π of a hyperplane winning set is itself hyperplane winning. This is however easy. Any ball B(x, ρ) chosen by Bob in the preimage is sent to the ball B(π(x), ρ) in the image. Here, Alice plays according to the winning strategy for this move and picks a hyperplane A with neighbourhood A(βρ) , which is mapped to the hyperplane-neighbourhood π −1 (A)(βρ) in the preimage. Evidently, this is a hyperplane winning strategy. 2 With some additional care, one can extend Proposition 6.1 to the following more general form, which may have other applications. We will not pursue this in the present paper, as we will only need the simpler version. Proposition 6.4. Let M1 , M2 be manifolds, let U ⊆ M1 be a nonempty open set whose complement is an analytic variety and let f : U → M2 be a differentially surjective map, locally C 1 conjugate to a projection π on U. The preimage under f of any hyperplane winning set in M2 is hyperplane winning. In order to complete the proof of Theorem 1.10, we need only ensure that the conditions of Proposition 6.1 are satisfied. Before doing this we note that the defining inequalities take a particularly pleasing form. Namely, (X, α) ∈ Bad(m, n, u) if and only if    (p, q) Iu  Xu

0  X



  − m+u +1 − α ≥ C(X, α) |q| n

(6.1)

 for all (p, q) ∈ Zu × Zm \ {0}, where X is the matrix (Xu , X). We now split the proof into two parts depending upon the choices of m, u, and n.

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6.1. The case m + u ≤ n First, by way of motivating our proof when m + u ≤ n, we discuss the case where u = 0 and m = n. In this case, we consider the inequalities |qX − α| ≥ C(X) for all but finitely many q ∈ Zm . It is readily verified that unless there is linear dependence among the columns of X, this is trivially satisfied. Hence, in this simple subcase, 

 2 2 Rm \ {X ∈ Rm : x(1) , . . . , x(m) are linearly dependent} × Rm ⊆ Bad(m, m, 0).

In other words, the set Bad(m, m, 0) contains the set of matrices of full rank, which is hyperplane winning by [26, Lemma 3.9]. This proves our main theorem in the particular case m = n, u = 0. Note that in fact we get the stronger inequality |qX − α| > C(X)|q| ∀q ∈ Zm \ {0}

(6.2)

in the set considered. This follows as an invertible matrix can only distort the unit ball by a specified amount. This is much stronger than the defining inequality of the set Bad(m, m, 0) in this special case. This feature also applies to the more general setting when m + u ≤ n and further underlines the qualitative difference between the case m + u ≤ n and the converse m + u > n. We now give a full proof in the case m + u ≤ n. We will argue much in the spirit  in (6.1) has full rank. By [26, of the above. Fix X ∈ Rmn such that the matrix X Lemma 3.9], the set of such X is hyperplane winning. Performing Gaussian elimination on the columns of a matrix of the form of (6.1) implies the existence of an invertible (n × n)-matrix E(X) such that 

Iu Xu

0  X



 =

Iu ˆ X

0 Im

0 0

 E(X).

Applying this matrix from the right to a vector (p, q), we see that  (p, q)

Iu Xu

0  X



 = (p, q)

Iu ˆ X

0 Im

0 0



⎛

⎞T ˆ p + qX ⎠ E(X). E(X) = ⎝ q 0

Multiplication by the matrix E on the right hand side only serves to distort the unit ball in the absolute value to a different parallelepiped depending on X. This induces a different norm on the image, but by equivalence of norms on Euclidean spaces, this distortion can be absorbed in a positive constant. In other words,

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 #  Iu  (p, q)  Xu

0  X

$

31

⎛  ⎞T    ˆ  p + qX   ⎝   T ⎠ −α   − α ≥ C(X)  q    0  

(6.3)

 = αE(X)−1 . for all (p, q) ∈ Zu × Zm \ {0}, where α Finally, since the norm is the supremum norm, we get ⎛  ⎞T   ˆ  p + qX  ⎝  T ⎠ −α  |q| ,   ≥ C(α)  q   0   for sufficiently large q. By (6.3), almost every (X, α) is in Bad(m, n, u), and in fact with the stronger requirement from (6.2). This completes the proof of Theorem 1.10 in the case where m + u ≤ n. 6.2. The case m + u > n In this case, the measure of the set Bad(m, n, u) is zero, and the proof is not as elementary as above. Nevertheless, appealing to Proposition 6.1 and taking a cue from the elementary proof, it is easily accomplished as follows. For each (X, α) ∈ Rmn × Rn , we can write   = X



X = Xu



X X 

X  X 



where the columns are split into groups of size u and (n − u), and the rows are split into groups of size (n − u) and (m + u − n). Note that X  is a square matrix. Now let U be the def

set of pairs (X, α) such that det X  = 0. Define the map f : U → M2 = R(m+u−n)n ×Rn by letting # f (X, α) =

(X



X



 )

Iu X

0 X 

−1

 ,α

Iu X

0 X 

−1 $ .

Now it is easy to see that f is differentially surjective, and by Theorem 1.1, the set Bad(m + u − n, n) ⊆ M2 is hyperplane winning. Hence, to conclude by Proposition 6.1, it suffices to prove that the preimage of this set under f is contained in Bad(m, n, u). def

However, this is easy. Fix (X, α) ∈ Rmn × Rn , and suppose that (Y, β) = f (X, α) ∈ Bad(m + u − n, n). Then there is a constant C(Y, β) > 0 such that for any r ∈ Zn and s ∈ Zm+u−n \ {0},       m+u (r, s) In − β  ≥ C(Y, β) |s|− n +1 .   Y On the other hand, if we let

(6.4)

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 Z=

Iu X

0 X 

 ∈ GLn (R)

then 

Iu Xu

0  X



    In ,α = Z, βZ Y

and thus for p ∈ Zu and q ∈ Zm \ {0},    (p, q) Iu  Xu

0  X



         I − m+u +1 − α =  (p, q) n − β Z  ≥ C(X, α) |q| n , Y

with the last inequality following by multiplying (6.4) by the invertible matrix Z, and then using the fact that if (p, q) = (r, s), then |q| ≥ |s|. Thus (6.1) is satisfied and (X, α) ∈ Bad(m, n, u). Acknowledgments The first-named author is supported by La Trobe University’s start-up grant and the ARC DP (200100994), the second-named author is supported by the Danish Research Council for Independent Research and the third-named author was supported by the EPSRC Programme Grant EP/J018260/1, and is now supported by a fellowship from the Royal Society URF\R1\180649. The authors would like to thank Stephen Harrap for good discussions and excellent company. Finally, the authors would like to thank the anonymous referees for their detailed comments which has significantly improved the quality of the paper. References [1] Demi Allen, Victor Beresnevich, A mass transference principle for systems of linear forms and its applications, Compos. Math. 154 (5) (2018) 1014–1047, MR 3798593. [2] V.I. Arnol’d, Geometrical Methods in the Theory of Ordinary Differential Equations, Grundlehren der Mathematischen Wissenschaften (Fundamental Principles of Mathematical Science), vol. 250, Springer-Verlag, New York-Berlin, 1983, Translated from the Russian by Joseph Szücs, Translation edited by Mark Levi, MR 695786. [3] Victor Beresnevich, Detta Dickinson, Sanju Velani, Measure theoretic laws for lim sup sets, Mem. Am. Math. Soc. 179 (846) (2006), x+91 pp., MR 2184760. [4] Victor Beresnevich, Sanju Velani, A mass transference principle and the Duffin-Schaeffer conjecture for Hausdorff measures, Ann. Math. (2) 164 (3) (2006) 971–992, MR 2259250. [5] Victor Beresnevich, Sanju Velani, Schmidt’s theorem, Hausdorff measures, and slicing, Int. Math. Res. Not. (2006) 48794, MR 2264714. [6] Vasilii Bernik, Maurice Dodson, Metric Diophantine Approximation on Manifolds, Cambridge Tracts in Mathematics, vol. 137, Cambridge University Press, Cambridge, 1999, MR 1727177. [7] Ryan Broderick, Lior Fishman, Dmitry Kleinbock, Asaf Reich, Barak Weiss, The set of badly approximable vectors is strongly C 1 incompressible, Math. Proc. Camb. Philos. Soc. 153 (02) (2012) 319–339, MR 2981929. [8] Ryan Broderick, Lior Fishman, David Simmons, Badly approximable systems of affine forms and incompressibility on fractals, J. Number Theory 133 (7) (2013) 2186–2205, MR 3035957. [9] Yann Bugeaud, An inhomogeneous Jarník theorem, J. Anal. Math. 92 (2004) 327–349, MR 2072751.

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