The Smallest Solution of an Isotropic Quadratic Form

University of Kentucky UKnowledge

Theses and Dissertations--Mathematics Mathematics

2021

Deborah H. Blevins University of Kentucky, [email protected] Author ORCID Identifier: https://orcid.org/0000-0002-6815-5290 Digital Object Identifier: https://doi.org/10.13023/etd.2021.173

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Recommended Citation Blevins, Deborah H., "The Smallest Solution of an Isotropic Quadratic Form" (2021). Theses and Dissertations--Mathematics. 80. https://uknowledge.uky.edu/math_etds/80

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The document mentioned above has been reviewed and accepted by the student’s advisor, on behalf of the advisory committee, and by the Director of Graduate Studies (DGS), on behalf of the program; we verify that this is the final, approved version of the student’s thesis including all changes required by the advisory committee. The undersigned agree to abide by the statements above.

Deborah H. Blevins, Student

Dr. David Leep, Major Professor

Dr. Benjamin Braun, Director of Graduate Studies The Smallest Solution of an Isotropic Quadratic Form

DISSERTATION

A dissertation submitted in partial fulﬁllment of the requirements for the degree of Doctor of Philosophy in the College of Arts and Sciences at the University of Kentucky

By Deborah H. Blevins Lexington, Kentucky

Director: Dr. David Leep, Professor of Mathematics Lexington, Kentucky 2021

The Smallest Solution of an Isotropic Quadratic Form Pn Pn An isotropic quadratic form f(x1, . . . , xn) = i=1 j=1 fijxixj defined on a Z- lattice has a smallest solution, where the size of the solution is measured using the infinity norm (k k∞), the `1 norm (k k1), or the Euclidean norm (k k2). Much work has been done to find the least upper bound and greatest lower bound on the smallest solution, beginning with Cassels in the mid-1950’s. Defining F := (f11, . . . , f1n, f21, . . . , f2n, . . . , fn1, . . . , fnn), upper bound results have the form kxki ≤ θ CkFki , with i ∈ {1, 2, ∞} and C a constant depending only on n. Aside from Cassels and Davenport, authors have concentrated more on finding the smallest exponent θ and less on C. Since Cassels’ publication, others have generalized his result and an- swered related questions. In particular, Schlickewei and Schmidt considered cases in which a quadratic form vanishes on a space of dimension greater than 1 and whether or not there is a bound on the product of the norms of multiple linearly independent nontrivial solutions. Schulze-Pillot explored similar questions, in addition to considering bounds on the determinant of the matrix whose columns are linearly independent nontrivial solutions of a quadratic form. Another goal has been to determine the best possible bound in each of these cases. In particular, Kneser demonstrated Cassels’ bound is best possible and Schlickewei and Schmidt have shown their results, along with several of Schulze-Pillot’s, are best possible. This dissertation gives a detailed, comprehensive exposition of these results. In particular, many of the results of Schulze-Pillot and Schlickewei and Schmidt are re- examined, leading to greater insight into the details of the proofs. The main methods used to solve this problem come from quadratic form theory, lattice theory, and the geometry of numbers.

KEYWORDS: isotropic quadratic forms, lattice, geometry of numbers, successive minima

Deborah H. Blevins

May 12, 2021 The Smallest Solution of an Isotropic Quadratic Form

By Deborah H. Blevins

Dr. David Leep Director of Dissertation

Dr. Benjamin Braun Director of Graduate Studies

May 12, 2021 Date “Therefore I run thus: not with uncertainty. Thus I ﬁght: not as one who beats the air.” 1 Corinthians 9:26 To Logan, Daddy, Mama, and Daniel – Thank you for running the race with me and teaching me what matters most. ACKNOWLEDGMENTS

First and foremost I would like to thank Dr. David Leep, whose guidance, direction, and support made this possible. After our ﬁrst conversation, my impression was that you were kind and thoughtful. The years since have proven this to be true over and over. Thank you for the many hours you have spent helping me, telling me stories, making me laugh, and for your endless advice – both about mathematics and about life. I am a better mathematician, and a better human, for it. I am deeply grateful to the remaining members of my committee – Drs. Corso, Shao, Srinivasan, and Jaromczyk – for your support and commitment. Thank you to Dr. Ben Braun, our Director of Graduate Studies, for helping me decide which prelims to take when I was in despair after so many failed attempts and for always responding kindly to panicked emails. Thank you to Dr. erica Whitaker, our Director of Service Courses, for advocating for graduate students, helping me grow as a teacher, always having your door open, and having candy in your oﬃce. Thank you to Christine Levitt, Rejeana Cassady, and Sheri Rhine for holding our department together, and each of us along with it. Dr. Manning, Dr. Newquist, and Dr. Ramey – thank you for believing I could do this before it was a thought in my mind. You all showed me what it looks like to invest in your students both inside and outside the classroom and pushed me to be a leader when I didn’t want to be. I hope to be like you. I would also like to thank Dr. Robert Bridges for giving me the opportunity to participate in the NSF Mathematical Sciences Graduate Internship at Oak Ridge National Laboratory. What I thought would be a ten-week internship turned into a much longer collaboration and richer experience than I could have anticipated. You have gone above and beyond in mentorship and I am so grateful. Thank you also

iii to Miki Verma – for making a space in your office for me, for answering a thousand questions, and for understanding my need for a regular schedule. To my dear husband, Logan, thank you for being brave enough to marry a graduate student, for believing in me more than I believe in myself, for keeping me moti- vated when my tank is empty, and for being excited about the next adventure. Your steadiness has been my life raft the past three years. This is all better with you by my side. To my parents and brother, David, Donna, and Daniel Wilkerson, there are not words adequate enough to express my gratitude nor enough space to list all the ways you’ve encouraged and supported me. You are my original support system, my first phone call, and the reason I am in graduate school. Stephen and Kathy Blevins – thank you for caring for me as your own, for asking all the questions in order to understand what I do in graduate school, and for indulging my affinity for infinity. You all are such a precious addition to my life and to this support team. My time as a graduate student would have come to a quick end if it were not for the most spectacular collection of friends I made in the department. Kalila, Julie, Darleen, Kevin, and Rachel P. – thank you for cooking me so many meals, going for walks, laughing with me, crying with me, and cheering me on. Thank you for all of the little moments of humanity we shared in the midst of the graduate school machine – they carried me through. To Katie, Rachel S., Morgan, and Kay – you are the very best of friends. If we really are an average of the people we spend the most time around, I’m glad it’s you all. You sharpen me as iron sharpens iron. You are my fiercest supporters and greatest inspirations. Having a roommate in graduate school is no easy task but Krystal, Rebecca, Kristen, and Aimee – you all are champs. Thank you for putting up with countless

iv study sessions, for never complaining when school came before keeping the kitchen clean, and for letting me “decorate” the apartment for “prelim season.” Thank you to the Young Adult group at Immanuel Baptist Church. If you tired of me requesting prayer for prelims, you never showed it. Your prayers and support were unmatched, particularly during those difficult first three years. And finally, thank you to my sweet cats, Izzy and Ashe, who have brought me more joy than I thought possible and have been my constant dissertation-writing companions. If they could talk, they would tell you all about quadratic forms.

v TABLE OF CONTENTS

Acknowledgments ...... iii

List of Tables ...... viii

Chapter 1 Introduction ...... 1 1.1 One Nontrivial Solution ...... 1 1.2 More Than One Nontrivial Solution ...... 5 1.3 Generalizations ...... 6

Chapter 2 Background ...... 11 2.1 Quadratic Maps ...... 11 2.2 Quadratic Forms Over the Real Numbers ...... 16 2.3 Identities and Inequalities ...... 20 2.4 Norms ...... 22

Chapter 3 A Treatise on Lattices ...... 25

Chapter 4 Cassels 1955 ...... 42

Chapter 5 Davenport 1956 ...... 52

Chapter 6 Watson 1957 ...... 60

Chapter 7 Birch & Davenport 1958 ...... 69

Chapter 8 Davenport 1971 ...... 75

Chapter 9 Schulze-Pillot 1983 ...... 81

Chapter 10 Schlickewei 1985 ...... 113

Chapter 11 Schmidt 1985 ...... 128

Chapter 12 Schlickewei & Schmidt 1988 ...... 136

Appendices ...... 149 Appendix A: Basis of a Projection Lattice ...... 149 Appendix B: Hermite’s Constant ...... 151 Appendix C: Solutions of Nondegenerate Quadratic Forms ...... 160 Appendix D: Supplement to Chapter 9 ...... 162

Bibliography ...... 164

vi Vita ...... 166

vii LIST OF TABLES

1.1 A selection of known values of the exponent θ...... 4

viii Chapter 1 Introduction

Pn Pn Consider a quadratic form f(x) = i=1 j=1 fijxixj ∈ R[x1, . . . , xn] where fij = fji n2 and deﬁne F := (f11, f12, . . . , f1n, f21, f22, . . . , f2n, . . . , fn1, fn2, . . . , fnn) ∈ R . For convenience, we write “form” instead of quadratic form. The form f is called isotropic if there is some nonzero x ∈ Rn such that f(x) = 0. Such a solution is an isotropic vector. Often there are many nontrivial solutions, and if they lie in a lattice there is a smallest such solution. The question then arises if there are bounds on this smallest solution. Many have sought to answer this question, and their results will be presented here. Throughout work on this subject, there are three frequently used n norms. For x = (x1, . . . , xn) ∈ R , we use the following notation:

kxk∞ := max |xi| i X kxk1 := |xi| i q 2 2 kxk2 := x1 + ··· + xn. It is important to note that √ kxk∞ ≤ kxk1 ≤ nkxk2 ≤ nkxk∞, which shows that these norms all induce the same metric topology on Rn. Therefore in the results that follow, changing the norm affects only the constants found in the bounds and never the exponents. P 0 Some authors express the quadratic form f as f(x) = 1≤i≤j≤n fijxixj. In this 0 0 0 0 0 0 0 0 n(n+1) case, we define F := (f11, f12, . . . , f1n, f22, f23, . . . , f2n, . . . , fnn) ∈ R 2 . Noting 0 1 0 that when i = j, fii = fii and when i 6= j, fij = fji = 2 fij, it is straightforward to show 0 0 kFk∞ ≤ kF k∞ ≤ 2kFk∞ and kFk1 = kF k1. √ 0 For the third norm, we will show later that kFk2 ≤ kF k2 ≤ 2kFk2. Therefore results using one expression of f can be transformed to results using the other ex- 2 pression and affect only the constants. Since most authors use F ∈ Rn , this is what we will use unless otherwise stated.

1.1 One Nontrivial Solution

The ﬁrst result came in 1955 from Cassels in [2] (with a correction in [3]) who was able to prove the following.

Theorem 1.1. Let f ∈ Z[x1, . . . , xn] be an isotropic form with isotropic vector in Zn. Then there is some nonzero x ∈ Zn such that f(x) = 0 and n−1 1 2 kxk ≤ (3n2 + n − 10)(n − 1)2[(n − 2)!]2kF0k . ∞ 2 ∞

1 Later in [4, pg. 87], Cassels improved the bound to the following:

0 (n−1)/2 kxk∞ ≤ (3kF k1) . (1.1)

n−1 He also showed that the exponent of 2 cannot be improved by considering the following quadratic form, introduced by M. Kneser,

n 2 X 2 f(x) = x1 − (xj − bxj−1) , j=2 with b a suﬃciently large positive integer. If x is a nontrivial solution of f, it can be shown that 1 1 · kF0k(n−1)/2 < kxk < (3kF0k )(n−1)/2. 2 n(n−1)/2 1 ∞ 1 In 1957, Davenport published the following similar, albeit more natural, result in [5].

n Theorem 1.2. Let f ∈ Z[x1, . . . , xn] be an isotropic form over Z . Then there is a nonzero x ∈ Zn such that f(x) = 0 and

n−1 2 n−1 2 2 kxk2 ≤ γn−1 2kFk2 , (1.2) where γn is Hermite’s constant (deﬁned in Deﬁnition 5.3). Thus, after taking the √ n−1 n−1 2 square root, kxk2 ≤ (γn−1 2) 2 kFk2 . Instead of considering the constant in the bound, in [23] Watson explored the exponent. Proposition 2.19 gives that every quadratic form f ∈ Z[x1, . . . , xn] can be written 2 2 2 2 f = L1 + ··· + Lr − (Lr+1 + ··· + Lr+s) where r, s are uniquely determined by f, r + s ≤ n, and L1,...,Lr+s are linearly independent linear forms in R[x1, . . . , xn]. We write sgn(f) = (r, s). In particular, r + s = n when f is nondegenerate by Corollary 2.23. Without loss of generality, one can assume r ≥ s because multiplying f by −1 does not change the set of isotropic vectors. In [23, Thm. 1] Watson showed:

Theorem 1.3. Let f ∈ Z[x1, . . . , xn] be a nondegenerate isotropic form with sgn(f) = (r, s). If f has an isotropic integral solution, then there is a nontrivial solution x = n (x1, . . . , xn) ∈ Z with 0 θ kxk∞ ≤ knkF k∞ (1.3) r where θ = max(2, 2 ) and kn depends only on n. Note that Kneser’s example is the case (r, s) = (n − 1, 1), and it was shown in [4] that the exponent cannot be improved in this case. However, Watson’s result gives insight into many other cases. He also considered whether or not a lower bound on θ r r exists. Let h = b s c. Then h ≤ s < h + 1 and therefore h is the unique integer such that h · s ≤ r < (h + 1) · s.

2 Watson showed in [23, Thm. 2] that θ cannot be replaced by any exponent less than h 1 r 2 (see Theorem 6.2 for the full statement), so that θ ≥ 2 b s c. Note that s 6= 0, for otherwise f would be a deﬁnite form and have no nontrivial zero. However, when the diﬀerence between r and s is small and r + s = n is relatively large, there is a wide h gap between Watson’s lower bound exponent of 2 and Cassels’ upper bound exponent n−1 of 2 . For this reason, Watson improved the result in the cases (r, s) = (2, 2) and 3 (r, s) = (3, 2). For (2, 2), he showed that Cassels’ exponent of 2 is best possible in eq. (1.3), while for (3, 2) he showed the exponent in eq. (1.3) cannot be reduced below 3 2 . In 1985, Schlickewei supplied improvements in [14, Cor. 3] to the exponent θ used by Watson. Here we use the notation n to indicate the existence of a constant depending only on n in the bound. He demonstrated the following:

Theorem 1.4. Let f be a quadratic form with integer coeﬃcients with signature (r, s) and with n = r + s ≥ 5 and r ≥ s > 0. Then there exists a nonzero x ∈ Zn such that θ f(x) = 0 and kxk∞ n kFk2 where

 1 r  2 · s when r ≥ s + 3  1 s+2 θ = 2 · s−1 when r = s + 2 or r = s + 1  1 s+1 2 · s−2 when r = s. Also in 1985, Schmidt improved in [19, Thm. 1] the exponent in the lower bound that was originally given by Watson. For this result, one needs additional notation. n kxk∞ For nonzero x = (x1, . . . , xn) ∈ R , let hxi = minxi6=0{|xi|} and q(x) = hxi . Then the theorem states:

Theorem 1.5. Suppose n = r + s, r ≥ s > 0. Given C ≥ 1 there is a quadratic form f with integer coeﬃcients with the following properties:

(i) sgn(f) = (r, s)

(ii) kFk∞ ≤ C and

n r (iii) for every point x ∈ R \{0} with f(x) ≤ 0, q(x) n C 2s .

n If x ∈ Z , hxi ≥ 1 and therefore q(x) ≤ kxk1. Hence

r r 2s kxk∞ ≥ q(x) n C 2s ≥ kFk∞ .

1 r 1 r This implies θ ≥ 2 · s for Watson’s exponent θ (it is straightforward to show 2 · s ≤ θ for each value of θ in Theorem 1.4). In fact, combining this with Theorem 1.4 gives 1 r that an exponent of 2 · s is best possible when r ≥ s+3. The best possible value of θ in the cases r = s + 2, r = s + 1, and r = s are still open questions in general. However, the answer is known for small values of n. For example, Schmidt’s Theorem 2 in [19] implies that θ = 2 is best possible when (r, s) = (3, 2), which is an improvement on Watson’s work. With this and the results given above, the best possible exponent is n−1 known for n ≤ 5 and in all cases it is Cassels’ exponent 2 . For n = 6, the best

3 Table 1.1: A selection of known values of the exponent θ. n (r, s) pair Best exponent θ Result by: 1 2 (1, 1) 2 Kneser, [4, p. 87] 3 (2, 1) 1 Kneser, [4, p. 87] 3 4 (3, 1) 2 Kneser, [4, p. 87] 3 4 (2, 2) 2 Watson, [23, p. 543] 5 (4, 1) 2 Kneser, [4, p. 87] 5 (3, 2) 2 Schmidt, [19, Thm. 2] 5 6 (5, 1) 2 Kneser, [4, p. 87] 6 (4, 2) 1 ≤ θ ≤ 2 Schmidt, [19, Thm. 1] (lower); Schlickewei, [14, Cor. 3] (upper) 1 6 (3, 3) 2 ≤ θ ≤ 2 Schmidt, [19, Thm. 1] (lower); Schlickewei, [14, Cor. 3] (upper) 7 (6, 1) 3 Kneser, [4, p. 87] 5 7 (5, 2) 4 Schmidt, [19, Thm. 1] (lower); Schlickewei, [14, Cor. 3] (upper) 2 5 7 (4, 3) 3 ≤ θ ≤ 4 Schmidt, [19, Thm. 1] (lower); Schlickewei, [14, Cor. 3], (upper) possible exponent is not that of Cassels in all cases. The results are summarized in Table 1.1, where “best exponent θ” indicates the smallest θ such that the inequality n−1 holds. Recall that in all cases, Cassels’ exponent of 2 is an upper bound. In 1958, Birch and Davenport generalized in [1, Thm. A] the result of Theorem 1.2 for quadratic forms having integral values when evaluated on a general lattice Λ. In particular, they showed

Theorem 1.6. Let Λ ⊂ Rn be an n-dimensional lattice and ∆ = det(Λ). Assume f ∈ R[x1, . . . , xn] is a quadratic form with a nontrivial zero in Λ and f(Λ) ⊆ Z. Then there exists x ∈ Λ, x 6= 0, such that f(x) = 0 and

n−1 2 n−1 2 2 2 kxk2 ≤ γn−1 (2kFk2) ∆ . (1.4)

√ n−1 n−1 2 Thus, after taking the square root, kxk2 ≤ (γn−1 2) 2 kFk2 ∆. Birch and Davenport provided an equivalent statement in Theorem B of the same publication that takes a more algebraic approach. Theorem 1.7. Let f be an indefinite quadratic form with integer coefficients and g be a positive definite quadratic form with real coefficients. Let F = (fij) be the n × n symmetric matrix associated to f and G = (gij) be the n × n symmetric matrix associated to g. If f has a nontrival solution in Zn, then there is x ∈ Zn such that −1 2 n−1 0 < g(x) n (2Tr(FG ) ) 2 · det(G). (1.5)

4 1.2 More Than One Nontrivial Solution

The next steps were also taken by Davenport, who considered whether there is a bound for the product of the norms of two linearly independent nontrivial solutions of a quadratic form. In [6, Thm. 1] he was able to show the following:

Theorem 1.8. Let f ∈ R[x1, . . . , xn] be a quadratic form such that f(Λ) ⊆ Z for a lattice Λ ⊂ Rn. If f has an isotropic vector in Λ, then there exist linearly independent x, y ∈ Λ such that f(x) = f(y) = 0 and

n−1 2 kxk2 kyk2 n kFk∞ ∆ . (1.6) This prompted the question of whether or not a similar statement could be made for the product of more than two linearly independent nontrivial solutions of a quadratic form. In 1983 Schulze-Pillot showed in [20] that for n ≥ 3 and any function G dependent on n, ∆, and F, one can always ﬁnd a quadratic form f and lattice Λ such that for any x1,..., xn ∈ Λ linearly independent isotropic zeros of f, kx1k∞ · · · kxnk∞ > G(n, ∆, F). Hence, no such statement exists for the product of the norms of three or more nontrivial solutions. In Theorem 2 of the same publication, Schulze-Pillot was able to show that a bound does exist when it depends on an extra variable, namely the norm of one of the nontrivial solutions. Speciﬁcally, he showed the following:

Theorem 1.9. Let f be a nondegenerate quadratic form in R[x1, . . . , xn] and let Λ ⊂ Rn be a lattice with f(Λ) ⊆ Z. If f has a nontrivial solution, there exist linearly independent vectors x1,..., xn ∈ Λ such that f(x1) = ··· = f(xn) = 0 and

n−1 (n−1)2 2(n−1) kx1k∞ kx2k∞ · · · kxnk∞ n kFk∞ ∆ . (1.7)

He was also able to demonstrate that the exponents on kFk∞ and ∆ are best possible. A corollary to the result above shows that there is in fact a bound when Λ = Zn. In this case, n2 2 −1 kx1k∞ · · · kxnk∞ n kFk∞ . (1.8) At the time of publication of [20], it was not known whether the exponent in this result was best possible. Schulze-Pillot also explored a bound on the determinant of the matrix whose columns are nontrivial solutions x1,..., xn ∈ Λ of the nondegenerate form f. In [20, Thm. 1] he proved:

Theorem 1.10. Let f ∈ R[x1, . . . , xn] be a nondegenerate isotropic quadratic form and Λ ⊂ Rn a lattice such that f(Λ) ⊆ Z. Let ∆ = det(Λ) and let A be the n × n matrix with ij-th entry fij. Then

n−1 n | det(x1,..., xn)| n | det(A)| 2 ∆ . (1.9) It was also unknown at the time of publication of [20] if the exponents in this result were best possible.

5 1.3 Generalizations

After the foundations had been laid, Schlickewei and Schmidt were able to generalize many of the results from Section 1.1 and Section 1.2, as well as to show that some of Schulze-Pillot’s results are best possible. In [14, Thm. 1], Schlickewei gave the following result: Pn Pn Theorem 1.11. Let f(x) = i=1 j=1 fijxixj be a nonzero quadratic form with n fij = fji. Let Λ be a lattice in R with determinant ∆. Assume f(Λ) ⊆ Z. Let d ≥ 1 be maximal such that there exist d linearly independent lattice points x1,..., xd ∈ Λ such that f vanishes on the linear space generated by x1,..., xd. Then there exist d such points such that n−d 2 kx1k∞ · · · kxdk∞ n kFk2 ∆. (1.10) Consequently, as stated in [14, Cor. 2], there is one isotropic vector x ∈ Λ such that

n−d 1 2d d kxk∞ n kFk2 ∆ . (1.11)

He also presented the following result in [14, Thm. 2], which is equivalent to Theo- rem 1.11:

Theorem 1.12. Let f be a nonzero quadratic form with integer coefficients and associated n × n matrix F = (fij) and g a positive definite quadratic form with real coefficients and associated n×n matrix G = (gij). Consider d ≥ 1 maximal such that f vanishes on a rational linear subspace of dimension d. Then there exist d linearly n independent x1,..., xd ∈ Z such that f vanishes on the linear subspace generated by x1,..., xd and

−1 2 n−d 0 < g(x1) ··· g(xd) n (Tr (FG ) ) 2 · det(G). (1.12)

Schlickewei conjectured that the exponents in eqs. (1.10) and (1.11) were best possible. The following theorem, found in [16, Thm. 1], implies that in the case n Λ = Z the exponent on kFk2 is best possible and hence eqs. (1.10) and (1.11) are best possible. Indeed, the theorem states:

Theorem 1.13. Suppose n ≥ 2d > 0. Given C ≥ 1, there is a nondegenerate quadratic from f with integral coeﬃcients in n variables with the following properties:

(i) f vanishes on a d-dimensional rational subspace.

(ii) kFk∞ ≤ C.

(iii) Every point x ∈ Rn\{0} with f(x) ≤ 0 has

n−d q(x) n C 2d .

6 Hence for any nonzero integral solution x of f,

n−d n−d 2d kxk∞ ≥ q(x) n C 2d ≥ kFk∞ and the exponent on kFk2 in eqs. (1.10) and (1.11) are best possible. In addition, both [15] and [17] mention a generalization of Theorem 1.11, stating that there is a d-dimensional sublattice Γ ⊂ Λ such that f(Γ) = 0 and

n−d 2 det(Γ) n kFk2 ∆. (1.13)

Indeed, choosing Γ to be the lattice generated by x1,..., xd from Theorem 1.11, one has Γ ⊂ Λ and f(Γ) = 0. Since det(Γ) n kx1k∞ · · · kxdk∞, as seen in the proof of Theorem 3.26, eq. (1.13) follows directly from eq. (1.10). In [14] Schlickewei was also able to answer the question of what bound exists when 0 f vanishes on a subspace generated by d ≥ 1 linearly independent vectors x1,..., xd0 , where d0 is not necessarily maximal (in contrast to before, in which d was maximal). To do so, the hypotheses of Theorem 1.11 are needed, in addition to at least one of the following:

(a) f is nondegenerate,

(b) there exists c ∈ R such that c 6= 0 and Λ = cZn,

Then [14, Cor. 1] gives:

Corollary 1.14. Let Λ, ∆, and f be as in Theorem 1.11. Suppose one of (a), (b), (c) 0 0 holds. Let d ≥ 1 so that there exist d linearly independent points x1,..., xd0 such that f vanishes on the subspace generated by x1,..., xd0 . Then there exist x1,..., xd0 such that n−d0 2 kx1k∞ · · · kxd0 k∞ n kFk2 ∆. (1.14) If (a) holds and d0 = 1, then this is the result of Birch and Davenport in [1, Thm. A] (Theorem 1.6 above). In [15, Thm. 2], Schlickewei and Schmidt generalized Schulze-Pillot’s Theorem 1 (here Theorem 1.10) in the following way: Pn Pn Theorem 1.15. Let f(x) = i=1 j=1 fijxixj be a nondegenerate quadratic form where fij = fji. Let A be the n × n matrix A = (fij). Suppose f(Λ) ⊆ Z and that for d > 0 there is a d-dimensional sublattice Γ ⊂ Λ such that f(Γ) = 0. Then there are linearly independent vectors x1,..., xn ∈ Λ such that f(x1) = ··· = f(xn) = 0 and

n−d n | det(x1,..., xn)| n | det(A)| 2d ∆ d . (1.15)

7 Note that the case d = 1 is exactly Schulze-Pillot’s Theorem 1 as seen in Theo- rem 1.10. Yet another generalization of Schulze-Pillot’s work came in [17, Thm. 2]. Schlick- ewei and Schmidt proved: Pn Pn Theorem 1.16. Let f(x) = i=1 j=1 fijxixj be a quadratic form with fij = fji such that f(Λ) ⊆ Z. Let 0 < t ≤ d < n and rank(f) = n − d + t. Suppose f(Γ) = 0 for a d-dimensional sublattice Γ of Λ. Then there are linearly independent x1,..., xn ∈ Λ such that f(x1) = ··· = f(xn) = 0 and

2(n−d) n−d t (kx1k∞ · · · kxd−tk∞) (kxd−t+1k∞ · · · kxdk∞) (kxd+1k∞ · · · kxnk∞) (n−d)2 2(n−d) n kFk2 ∆ . (1.16) In particular, in [17, Cor. 1] Schlickewei and Schmidt proved that with the hypotheses of Theorem 1.16, the case Λ = Zn is

(n−d+t)2 2t −t kx1k∞ · · · kxnk∞ n kFk2 . (1.17) Note that Theorem 1.9 and its corollary, both proven by Schulze-Pillot, are the d = t = 1 case of these results. Also consider that, based upon Theorem 1.2, Theorem 1.8, and eq. (1.11), one might have suspected the exponent on kFk2 in the nondegenerate n(n−d) n2 n (i.e. d = t) case to be 2d = 2d − 2 . Since nondegeneracy implies 2d ≤ n by n2 Theorem 2.16, this exponent is less than the given exponent of 2d − d, except in the case n = 2d. However Theorem 2 of [16] by Schlickewei and Schmidt states: Theorem 1.17. Suppose n ≥ 2d > 0. Given C ≥ 1, there is a nondegenerate quadratic form f with integral coeﬃcients in n variables with the following properties:

(i) kFk∞ ≤ C. (ii) There is a d-dimensional rational subspace S on which f vanishes, such that any point x ∈ S\{0} has n−2d q(x) n C 2d .

(iii) Any point x ∈ Rn\S with f(x) ≤ 0 satisﬁes

n q(x) n C 2d .

Therefore, given n linearly independent solutions x1,..., xn of f, at most d of them will live in S. So

2 n−2d n n2 n d n−d −d 2d −d q(x1) ··· q(xn) n (C 2d ) · (C 2d ) = C 2d ≥ kFk∞ .

n If x1,..., xn ∈ Z ,

n2 2d −d kx1k∞ · · · kxnk∞ ≥ q(x1) ··· q(xn) n kFk∞ .

8 Here the exponent on kFk∞ is identical to the exponent on kFk2 in eq. (1.17) when f is nondegenerate (i.e d = t), hence implying this exponent is best possible in the nondegenerate case. It is not certain if the exponent is best possible when f is nondegenerate. In addition, in [17] they showed that the exponent on kFk2 in eq. (1.17) comes from the nondegenerate part of f, as follows. Assume the hypotheses of Theorem 1.16 with Λ = Zn. Consider the case t < d (i.e. f is degenerate). Rank(f) = n − d + t implies that dim(rad(f)) = d − t and f can be written

2 2 f(x) = g(x1, . . . , xn−d+t) + 0 · xn−d+t+1 + ··· + 0 · xn where g(x1, . . . , xn−d+t) is a nondegenerate quadratic form in x1, . . . , xn−d+t. Let d0 be maximal such that f vanishes on a d0-dimensional subspace. Then d0 ≥ d and Lemma 2.7 gives that g vanishes on a subspace of dimension d0 − (d − t) = d0 − d + t ≥ t. Hence g vanishes on a t-dimensional subspace. By Corollary C.3 there are x1,..., xn−d+t linearly independent solutions of g. Then applying eq. (1.17) to g, noting that kGk2 = kFk2, gives

[(n−d+t)−t+t]2 (n−d+t)2 2t −t 2t −t kx1k∞ · · · kxn−d+tk∞ n kFk2 = kFk2 .

Thus, when considering f, x1,..., xn−d+t are linearly independent solutions, in addition to xn−d+t+1 = en−d+t+1,..., xn = en where the ei’s with n − d + t + 1 ≤ i ≤ n are standard basis vectors of Rn. Hence eq. (1.17) is recovered. In [15, Thm. 1], Schlickewei and Schmidt proved the following, which summarizes the results of Theorems 1.1, 1.6, 1.8 and 1.9: Pn Pn Theorem 1.18. Let f be a quadratic form f(x) = i=1 j=1 fijxixj ∈ R[x1, . . . , xn] and Λ ⊂ Rn an n-dimensional lattice such that ∆ = det(Λ) and f(Λ) ⊂ Z. Let 0 < d < n and suppose rank(f) > n − d. Suppose moreover that f vanishes on a d-dimensional sublattice Γ ⊂ Λ. Then there exist d-dimensional sublattices Γ0, Γ1,..., Γn−d of Λ with the following properties:

(i) For 0 ≤ i ≤ n − d, f vanishes on Γi.

(ii) For each 1 ≤ j ≤ n − d, Γ0 ∩ Γj has dimension d − 1.

n (iii) The union of Γ0, Γ1,..., Γn−d spans R .

n−d 2 (iv) For each 1 ≤ j ≤ n − d, det Γ0 · det Γj n kFk2 ∆ . From this they showed, in [15, Cor.], that, assuming the same hypotheses,

n−d (n−d)2 2(n−d) det Γ0 · det Γ1 ··· det Γn−d n kFk2 ∆ . (1.18)

Using a quadratic form very similar to that of M. Kneser mentioned above, they were able to show that both Theorem 1.18 and its corollary are best possible. In [17, Thm. 1], Schlickewei and Schmidt proved the following result. Here we let dae denote the greatest integer z such that z ≥ a. For example, d−2.7e = −2.

9 Pn Pn Theorem 1.19. Let 0 < d < n and 0 < t ≤ d. Let f(x) = i=1 j=1 fijxixj ∈ R[x1, . . . , xn] with fij = fji and rank(f) = n − d + t. Let f be such that f(Λ) ⊂ Z. Assume there is a d-dimensional lattice Γ ⊂ Λ such that f vanishes on Γ. Then there n−d 0 exist d t e pairs Γi, Γi of d-dimensional sublattices with the following properties: n−d 0 (i) For 1 ≤ i ≤ d t e, f vanishes on Γi and Γi. n−d 0 (ii) For 1 ≤ i ≤ d t e, Γi ∩ Γi has dimension d − t. 0 0 n (iii) The union of Γ1, ··· , Γd n−d e, Γ1, ··· , Γ n−d spans R . t d t e

n−d 0 n−d 2 (iv) For 1 ≤ i ≤ d t e, det Γi · det Γi n kFk2 ∆ . In the special case when t = 1, this is similar, but not identical, to Theorem 1.18.

10 Chapter 2 Background

2.1 Quadratic Maps

Deﬁnition 2.1. Let V be a vector space and k be a ﬁeld with char(k) 6= 2. A function f : V → k is a quadratic map if

(i) f(av) = a2v for all a ∈ k and v ∈ V and

1 (ii) the map Bf : V × V → k deﬁned by Bf (v, w) = 2 f(v + w) − f(v) − f(w) is a symmetric bilinear form.

Note that, in particular, Deﬁnition 2.1 gives f(v) = Bf (v, v) since 1 B (v, v) = f(v + v) − f(v) − f(v) f 2 1 = f(2v) − 2f(v) 2 1 = 4f(v) − 2f(v) 2 = f(v).

Proposition 2.2. Let k be a ﬁeld with char(k) 6= 2 and let V be an n-dimensional vector space over k with basis v1,..., vn. Let f : V → k be a quadratic map. Then

n X 2 X f(x1v1 + ··· + xnvn) = f(vi)xi + 2Bf (vi, vj)xixj i=1 1≤i

Proof.

n ! n ! X X f(x1v1 + ··· + xnvn) = f(x1v1) + 2Bf x1v1, xivi + f xivi i=2 i=2 n n ! X X = f(x1v1) + 2Bf (x1v1, xivi) + f xivi i=2 i=2 n n ! 2 X X = x1f(v1) + x1xi2Bf (v1, vi) + f xivi i=2 i=2 Pn Iterating this process starting with f ( i=2 xivi) gives the ﬁrst equality. The second equality holds since f(vi) = Bf (vi, vi) and Bf (vi, vj) = Bf (vj, vi).

11 Definition 2.3. Let Qf (x1, . . . , xn) denote the quadratic form f(x1v1 + ··· + xnvn), and call Qf the quadratic form associated to the quadratic map f and the basis v1,..., vn. Pn Pn Definition 2.4. Let Q(x1, . . . , xn) = i=1 j=1 qijxixj be an arbitrary quadratic form in k[x1, . . . , xn] with qij = qji. Let A = (qij) be the n × n symmetric matrix whose (i, j)-entry is given by qij. We call A the symmetric matrix associated to the quadratic form Q. Note that Q(x) = xT Ax. Let k be a field with char(k) 6= 2 and V an n-dimensional vector space over k with basis v1,..., vn. Let f : V → k be a quadratic map. Then it follows from Proposition 2.2 and Definitions 2.3 and 2.4 that the n×n symmetric matrix A = (aij) associated to Qf is given by ( Bf (vi, vj) when i 6= j aij = f(vi) when i = j. Proposition 2.5. Let k be a field with char(k) 6= 2. Let V be an n-dimensional vector space over k with basis v1,..., vn. For any quadratic form Q ∈ k[x1, . . . , xn], Q = Qf for some uniquely determined quadratic map f : V → k with respect to the basis v1,..., vn. Pn Pn Proof. Let Q(x1, . . . , xn) = i=1 j=1 qijxixj with qij = qji and let A be the n × n symmetric matrix associated to Q. For v = c1v1 + ··· + cnvn ∈ V , where c1, . . . , cn ∈ k, define f(v) = Q(c1, . . . , cn). Then for a ∈ k 2 2 f(av) = Q(ac1, . . . , acn) = a Q(c1, . . . , cn) = a f(v).

T Let w = d1v1 + ··· + dnvn, with d1, . . . , dn ∈ k. Let c = (c1, . . . , cn) and d = T T (d1, . . . , dn) . Then since f(v) = Q(c1, . . . , cn) = c Ac,

f(v + w) = Q(c1 + d1, . . . , cn + dn) = (c + d)T A(c + d) = cT Ac + dT Ad + cT Ad + dT Ac = f(v) + f(w) + cT Ad + dT Ac

1 Consider the map Bf : V × V → k given by Bf (v, w) = 2 f(v + w) − f(v) − f(w) . Since A = AT and cT Ad is a 1 × 1 matrix, cT Ad = (cT Ad)T = dT AT c = dT Ac. T Then Bf (v, w) = c Ad. It is straightforward to show Bf is a symmetric bilinear form. Thus f is a quadratic map and Qf (x1, . . . , xn) = f(x1v1 + ··· xnvn) = Q(x1, . . . , xn) with respect to the basis v1,..., vn. The quadratic map is uniquely determined by Q and the basis v1,..., vn because for each v = c1v1 + ··· + cnvn ∈ V , we must have f(v) = Q(c1, . . . , cn). Definition 2.6. Let k be a field with char(k) 6= 2 and let V be a vector space over k. Let f : V → k be a quadratic map. The radical of f is defined by rad(f) = {v ∈ V |Bf (v,V ) = 0}.

12 Note that f(v) = Bf (v, v) = 0 for all v ∈ rad(f). It is straightforward to show that rad(f) is a subspace of V . We can write V = W ⊕rad(f) for some subspace W of V . Suppose that dim(V ) = n and dim(rad(f)) = n−m. Let v1,..., vm be a basis of W , and let vm+1,..., vn be a basis of rad(f). Let g = f|W : W → k be the quadratic map deﬁned by restricting the domain of f to W . Then f(x1v1 + ··· + xnvn) = f(x1v1 + ··· + xmvm) = g(x1v1 + ··· + xmvm). Then Qf (x1, . . . , xn) = Qg(x1, . . . , xm). Lemma 2.7. Let k be a ﬁeld with char(k) 6= 2 and let V be a vector space over k. Let f : V → k be a quadratic map. Let V = W ⊕ rad(f). Suppose that f vanishes on a d-dimensional subspace of V where d is maximal. Then f|W vanishes on a subspace of dimension d − dim(rad(f)).

Proof. Suppose that Y is a subspace of V with dim(Y ) = d such that f vanishes on Y . Then f vanishes on Y + rad(f). Since Y is maximal, it follows that Y + rad(f) = Y and thus rad(f) ⊆ Y . We have Y = (Y ∩ W ) ⊕ rad(f). Thus dim(Y ∩ W ) = dim(Y )−dim(rad(f)) = d−dim(rad(f)) and since Y ∩W ⊆ Y , f|W (Y ∩W ) = 0. Deﬁnition 2.8. Let k be a ﬁeld with char(k) 6= 2 and let V be a vector space over k. Let f : V → k be a quadratic map. Then rank(f) := dim(V ) − dim(rad(f)).

Definition 2.9. Let k be a field with char(k) 6= 2 and let V be a vector space over k. Let f : V → k be a quadratic map. The map f is nondegenerate if rad(f) = 0. Otherwise, it is degenerate. A quadratic form Q is nondegenerate if Q = Qf for some nondegenerate quadratic map f : V → k and basis v1,..., vn of V . Otherwise, Q is degenerate. Definition 2.10. Let k be a field of char(k) 6= 2 and let V be a vector space over k. Let Q(x1, . . . , xn) ∈ k[x1, . . . , xn] be a quadratic form. Let f : V → k be the quadratic map such that Q = Qf . Then rank(Qf ) := rank(f).

Proposition 2.11. Let Q(x1, . . . , xn) ∈ k[x1, . . . , xn] be a quadratic form with associated symmetric matrix A. Then rank(Q) = rank(A).

Proof. By Proposition 2.5, there is a quadratic map f : V → k with respect to basis v1,..., vn of V such that Q = Qf . Let v = c1v1 + ··· + cnvn ∈ V and w = d1v1 + ··· + dnvn ∈ V. Then

w ∈ rad(f) ⇐⇒ Bf (V, w) = 0 T n ⇐⇒ c Ad = 0 for all c ∈ R ⇐⇒ Ad = 0 ⇐⇒ d ∈ ker(A).

n There is an isomorphism V → K , depending on the basis v1,..., vn, deﬁned by v 7→ (c1, . . . , cn) where v = c1v1 + ··· + cnvn. This isomorphism takes rad(f)

13 to ker(A). Therefore, dim(rad(f)) = dim(ker(A)). Since rank(Q) = rank(f) by deﬁnition, we have

rank(Q) = rank(f) = dim(V ) − dim(rad(f)) = dim(V ) − dim(ker(A)) = dim(col(A)) = rank(A).

Corollary 2.12. Let f : V → k be a quadratic map with associated quadratic form Qf . Let A be the n×n symmetric matrix associated to Qf and suppose rank(Qf ) < n. Then Qf has a nontrivial solution.

Proof. By Proposition 2.11, rank(Qf ) = rank(A) and therefore rank(A) < n. Hence, n T there is some nonzero x ∈ R such that Ax = 0. Then Qf (x) = x Ax = 0 and Qf has a nontrivial solution. Pn Pn Proposition 2.13. Let f(x1, . . . , xn) = i=1 j=1 fijxixj ∈ k[x1, . . . , xn] be a quadratic form with not all fij = 0. Then there exist linearly independent linear forms L1,...,Lt ∈ 2 2 k[x1, . . . , xn], with t ≤ n, such that f = a1L1 + ··· + atLt , where a1, . . . , at ∈ k. n Proof. If f11 = 0, consider that since f 6= 0, there is some a = (a1, . . . , an) ∈ k such that f(a) 6= 0. Then there exists a linear change of variables such that, when applied to f, f(1, 0,..., 0) 6= 0 and hence f11 = f(1, 0,..., 0) 6= 0. 2 2 We proceed by induction on n. If n = 1, f = f11x1. Then f = f11L1 where L1 = x1. Now assume by induction that the theorem holds for quadratic forms in ` variables, with 1 ≤ ` ≤ n − 1. Let

n X 2 X f(x1, . . . , xn) = fiixi + fijxixj. i=1 1≤i

We can assume f11 6= 0. Then

n ! X fii X fij f = f x2 + x x 11 f i f i j i=1 11 i=1≤i

14 gives 2 2 2 f = f11L1 + a2L2 + ··· + atLt .

Note that L1,...,Lt is a linearly independent set as L1 has a nonzero x1 term that none of L2,...,Lt have. Definition 2.14. Let k be a field with char(k) 6= 2 and let V be a finite-dimensional vector space over k. Let f : V → k be a quadratic map and W ⊆ V a subspace. We define ⊥f W = {v ∈ V | Bf (v,W ) = 0}. Lemma 2.15. Let k be a field with char(k) 6= 2 and let V be a finite-dimensional vector space over k. Let f : V → k be a nondegenerate quadratic map. Then dim(W ⊥f ) = dim(V ) − dim(W ).

Proof. Let V ∗ = Hom(V, k) and W ∗ = Hom(W, k). Define the map ϕ : V → V ∗ by ∗ ϕ(v) = fv where fv ∈ V and fv(y) = Bf (v, y) for all y ∈ V . Then ϕ is a linear map and we will show it is an isomorphism. Since dim(V ) = dim(V ∗), it is sufficient to ∗ show ϕ is injective. Let v ∈ ker(ϕ). Then fv = 0 ∈ V , which means for all y ∈ V , we have 0 = fv(y) = Bf (v, y) and therefore v ∈ rad(f). Since f is nondegenerate, rad(f) = 0 and therefore v = 0. Hence ϕ is injective. ∗ ∗ ∗ Now, define the linear map π : V → W by π(g) = g |W for g ∈ V . We will show π is surjective. We can write V = W ⊕ Z for a subspace Z of V . Let h ∈ W ∗. For v ∈ V , v = w + z, where w ∈ W and z ∈ Z are uniquely determined. Define ∗ the map g : V → k by g(v) = g(w + z) = h(w). Then g ∈ V and g |W = h. So π is surjective Now, consider π ◦ ϕ : V → V ∗ → W ∗. This map is surjective since both ϕ and π are surjective. If v ∈ ker(π ◦ ϕ), then 0 = π ◦ ϕ(v) = π(fv)) = fv |W , that is ⊥f Bf (v,W ) = 0. Hence ker(π ◦ ϕ) = W . Because ϕ is an isomorphism, we have

dim(W ⊥f ) = dim(ker(π ◦ ϕ)) = dim(ker(π)) = dim(V ∗) − dim(im(π)) = dim(V ∗) − dim(W ∗) = dim(V ) − dim(W ).

Theorem 2.16. Let k be a ﬁeld with char(k) 6= 2 and let V be a ﬁnite-dimensional vector space over k. Let f : V → k be a nondegenerate quadratic map that vanishes on a subspace W ⊆ V . Then 2 dim(W ) ≤ dim(V ).

⊥f Proof. Since f vanishes on W , Bf (W, W ) = 0. This implies W ⊆ W . Therefore, dim(W ) ≤ dim(W ⊥f ). By Lemma 2.15, dim(W ⊥f ) = dim(V ) − dim(W ). Therefore dim(W ) ≤ dim(V ) − dim(W ) and hence 2 dim(W ) ≤ dim(V ).

15 Lemma 2.17. Let k be a ﬁeld with char(k) 6= 2 and let V be a ﬁnite-dimensional vector space over k. Let f : V → k be a quadratic map. Let W ⊆ V be a subspace ⊥f and suppose f|W is nondegenerate. Then V = W ⊕ W .

⊥f Proof. Suppose x ∈ W ∩ W . Then Bf (x, y) = 0 for all y ∈ W . Since f|W is nondegenerate, x = 0. Recall that W ∗ = Hom(W, k). Now consider the map

ϕ : W → W ∗

w 7→ gw where gw(u) = Bf (w, u) with u ∈ W . We wish to show that ϕ is injective. Assume gw = 0. Then gw(u) = Bf (w, u) = 0 for all u ∈ W . Since rad(f|W ) = 0, this implies w = 0. Therefore ϕ is injective. Now since dim(W ) = dim(W ∗), this also implies that ϕ is surjective and therefore an isomorphism. ∗ Choose v ∈ V . Deﬁne h : W → k by h(w) = Bf (v, w). Since h ∈ W and ϕ is ∗ an isomorphism, there exists gw0 ∈ W such that h = gw0 . Hence for all w ∈ W ,

h(w) = Bf (v, w) = Bf (w0, w) = gw0 (w).

Now, consider that v = w0 + (v − w0). For any w ∈ W ,

Bf (v − w0, w) = Bf (v, w) − Bf (w0, w) = 0.

⊥f ⊥f Hence v−w0 ∈ W and v can be written as the sum of w0 ∈ W and v−w0 ∈ W . ⊥ Hence V = W ⊕ W f . Corollary 2.18. Let k be a ﬁeld with char(k) 6= 2 and let V be a ﬁnite-dimensional vector space over k. Let f : V → k be a quadratic map. Let W ⊆ V be a subspace ⊥f and suppose f|W is anisotropic. Then V = W ⊕ W .

Proof. If f|W is anisotropic, then rad(f|W ) = 0, and the proof follows from Lemma 2.17.

2.2 Quadratic Forms Over the Real Numbers

Pn Pn Proposition 2.19. Let f(x1, . . . , xn) = i=1 j=1 fijxixj ∈ R[x1, . . . , xn] be a nonzero quadratic form. Then there exist uniquely determined r, s ∈ Z and linearly independent linear forms L1,...,Lr+s ∈ R[x1, . . . , xn], with r + s ≤ n, such that 2 2 2 2 f = L1 + ··· + Lr − (Lr+1 + ... + Lr+s).

2 2 Proof. By Proposition 2.13, we know f can be written f = a1L1 + ··· atLt where a1, . . . , at ∈ R and L1,..., Lt are linearly independent linear forms in R[x1, . . . , xn]. Re-index a1, . . . , at so that a1, . . . , ar > 0 and ar+1, . . . , ar+s < 0, where r + s = t. √ p Let Li = aiLi for 1 ≤ i ≤ r and Li = |ai|Li for r + 1 ≤ i ≤ r + s. Then

2 2 2 2 f = L1 + ··· + Lr − (Lr+1 + ··· + Lr+s).

16 Now we show that r, s are unique to f. Suppose that f can be written f = 2 2 2 2 M1 + ··· + Mt − (Mt+1 + ··· + Mt+u), where M1,...,Mt+u are linearly independent forms in x1, . . . , xn and t+u ≤ n. We will show r = t and s = u. By Proposition 2.22, we have rank(f) = r+s and rank(f) = t+u and thus r+s = t+u. Let V ⊂ Rn be the subspace defined by M1 = ··· = Mt = 0. Then dim(V ) = n − t, since M1,...,Mt are n linearly independent. Let Y ⊂ R be the subspace defined by Lr+1 = ··· = Lr+s = 0. Then, since Lr+1,...,Lr+s are linearly independent, dim(Y ) = n−s. Let W = rad(f). Then dim(W ) = n−(r+s) = n−(t+u) and W is defined both by L1 = ··· = Lr+s = 0 and M1 = ··· = Mt+u = 0. So W ⊆ V ∩ Y . We will show W = V ∩ Y . Let a = (a1, . . . , an) ∈ V ∩ Y . Then a ∈ Y implies that f(a) ≥ 0 and a ∈ W implies that f(a) ≤ 0. Hence f(a) = 0. Now, since a ∈ Y , this implies L1(a) = ··· = Lr(a) = 0 and therefore a ∈ W . Therefore V ∩ Y ⊆ W and thus V ∩ Y = W . Then n ≥ dim(V + Y ) = dim(V ) + dim(Y ) − dim(V ∩ Y ) = n − t + n − s − (n − (r + s)) = n − t + r. Hence, t ≥ r. A symmetric argument, defining V as the subspace defined by Mt+1 = ··· = Mt+u = 0 and Y as the subspace defined by L1 = ··· = Lr = 0, gives t ≤ r. Hence t = r and s = u.

Let f : V → R be a quadratic map with Bf : V × V → R the corresponding symmetric bilinear form. For a subspace W ⊆ V , let f|W denote the restriction of f to W . We write V = Y ⊥ Z if Y,Z are subspaces of V with V = Y ⊕Z and Bf (y, z) = 0 for all y ∈ Y and z ∈ Z. Deﬁnition 2.20. We say that the signature of f is (r, s, t), and write sgn(f) = (r, s, t), if there exist subspaces of V , possibly zero, such that V = V1 ⊥ V2 ⊥ V3 with dim(V1) = r, dim(V2) = s, dim(V3) = t, and  f(v) > 0 for all nonzero v ∈ V ,  1 f(v) < 0 for all nonzero v ∈ V2,  f(v) = 0 for all v ∈ V3. The triple (r, s, t) is well deﬁned by Proposition 2.19. If sgn(f) = (r, s, t), it is straightforward to verify that with the notation above we have rad(f) = V3 and dim(rad(f)) = t, and in Proposition 2.22 we show rank(f) = r + s. If V = Y ⊥ Z , then sgn(V ) = sgn(Y ) + sgn(Z). When f is nondegenerate and hence t = 0, we often write sgn(f) = (r, s).

Definition 2.21. Let f(x1, . . . , xn) ∈ R[x1, . . . , xn] be a quadratic form with signature (r, s, t). The form f is definite if (r, s, t) = (n, 0, 0) or (r, s, t) = (0, n, 0). It is positive definite if r = n and negative definite if s = n. It is semi-definite if r = 0 and s < n, or s = 0 and r < n. It is indefinite if 0 < r, s < n.

17 Proposition 2.22. Let Q(x1, . . . , xn) ∈ R[x1, . . . , xn] be a quadratic form where 2 2 2 2 Q(x1, . . . , xn) = L1 + ··· + Lr − (Lr+1 + ··· + Lr+s), where L1,...,Lr+s are linearly independent linear forms in x1, . . . , xn and r + s ≤ n. Then rank(Q) = r + s.

Proof. Let Q(x1, . . . , xn) ∈ k[x1, . . . , xn] be a quadratic form with associated n × n symmetric matrix A. By Proposition 2.19, there exist L1,...,Lr+s linearly independent linear forms in k[x1, . . . , xn], such that

2 2 2 2 Q(x1, . . . , xn) = L1 + ··· + Lr − Lr+1 − · · · − Lr+s.

Write Li = ai1x1 + ··· + ainxn for 1 ≤ i ≤ r + s and define ai = (ai1, . . . , ain). Since the Li’s are linearly independent, a1,..., ar+s are linearly independent and therefore can be extended to a basis a1,..., ar+s, ar+s+1,..., an. Define the matrix   − a1 −  .  D =  .  − an − and let C be a diagonal matrix where the first r diagonal entries are 1, the next s diagonal entries are −1 and the last n − r − s diagonal entries are 0. Then for x = (x1, . . . , xn),

T T 2 2 2 2 x D CDx = L1(x) + ··· + Lr(x) − Lr+1(x) − · · · − Lr+s(x) = Q(x) = xT Ax.

Since this holds for every x ∈ Rn, DT CD = A. Since D is invertible, multiplying C on the right by D is equivalent to performing elementary column operations on C. Similarly, DT is invertible and multiplying C on the left by DT is equivalent to performing elementary row operations on C. Therefore rank(A) = rank(DT CD) = rank(C) = r + s. Corollary 2.23. Suppose f : Rn → R is a quadratic map with respect to the basis n v1,..., vn of R and let Qf be the associated quadratic form of signature (r, s). Then the following are equivalent:

(i) Qf is nondegenerate

(ii) rank(Qf ) = n (iii) r + s = n.

n Proof. (i) ⇐⇒ (ii) Since rank(Qf ) = rank(f) = dim(R ) − dim(rad(f)), it follows n that rank(Qf ) = dim(R ) = n if and only if dim(rad(f)) = 0, i.e. Qf is nondegenerate. (ii) ⇐⇒ (iii) By Proposition 2.22, rank(Qf ) = r + s. Hence rank(Qf ) = n if and only if r + s = n.

18 Corollary 2.24. Let W ⊂ Rn. Deﬁne W ⊥ := {x ∈ Rn | x · y = 0 for every y ∈ W }, where x · y is the standard inner product. Then Rn = W ⊕ W ⊥.

2 2 Proof. This follows directly from Lemma 2.17 with f = x1 + ··· + xn.

Lemma 2.25. Let f : V → R be a quadratic map with Bf : V × V → R the corresponding symmetric bilinear form. Assume that dim(V ) = n and let W be a subspace of V with dim(W ) = n − 1. Assume that rank(f) = n. Then n − 2 ≤ rank(f|W ) ≤ n − 1.

Proof. Since dim(W ) = n−1, we have rank(f|W ) ≤ n−1. Suppose that rank(f|W ) ≤ n − 3. Then

dim(rad(f|W )) = dim(W ) − rank(f|W ) ≥ (n − 1) − (n − 3) = 2.

Let v ∈ V , v ∈/ W , and let U = v⊥f ∩ W . Then dim(U) ≥ n − 2. Since U has codimension at most 1 in W and dim(rad(f|W )) ≥ 2, there exists 0 6= y ∈ ⊥f U ∩ rad(f|W ). We have V = (R · v) ⊕ W . Since y ∈ U ⊆ v , we have Bf (y, v) = 0. Since y ∈ rad(f|W ), we have Bf (y,W ) = 0. Thus Bf (y,V ) = 0, and so y ∈ rad(f), which is a contradiction because rad(f) = 0. Therefore, n − 2 ≤ rank(f|W ).

Proposition 2.26. Let f : V → R be a quadratic map with Bf : V × V → R the corresponding symmetric bilinear form. Assume that dim(V ) = n and let W be a subspace of V with dim(W ) = n−1. Assume that rank(f) = n and sgn(f) = (r, s, 0).

1. If rank(f|W ) = n − 2, then sgn(f|W ) = (r − 1, s − 1, 1).

2. If rank(f|W ) = n − 1, then either sgn(f|W ) = (r, s − 1, 0) or sgn(f|W ) = (r − 1, s, 0).

Proof. (1) If rank(f|W ) = n − 2, then dim(rad(f|W )) = 1. There exists a subspace U ⊂ W such that W = (R · y) ⊕ U where y ∈ rad(f|W ) and dim(U) = n − 2. Then ∼ n since V = R , by Corollary 2.24 there exists a subspace Y ⊂ V such that V = Y ⊥ U where y ∈ Y and dim(Y ) = 2. There exists v ∈ Y such that {v, y} is a basis of Y . We have Bf (y,W ) = 0 because y ∈ rad(f|W ). Since v ∈/ W , we have V = (R · v) ⊕ W . Since y ∈/ rad(f), it follows that Bf (v, y) 6= 0. 2 Then f(x1v + x2y) = f(v)x1 + 2Bf (v, y)x1x2 because f(y) = 0. For convenience, let f(v) = a and Bf (v, y) = b. Then

2 ax1 + 2bx1x2 = x1(ax1 + 2bx2) (a + 1)x + 2bx 2 (a − 1)x + 2bx 2 = 1 2 − 1 2 . 2 2

(a+1)x1+2bx2 (a−1)x1+2bx2 Since b 6= 0, the linear forms 2 and 2 are linearly independent over R. Thus sgn(f|Y ) = (1, 1, 0). Since sgn(f) = (r, s, 0), we have sgn(f|U ) = (r − 1, s − 1, 0), and thus sgn(f|W ) = (r − 1, s − 1, 1).

19 (2) Since dim(W ) = n−1 and rank(f|W ) = n−1, it follows that f|W is nondegenerate. Again by Corollary 2.24, there exists v ∈ V such that V = (R · v) ⊥ W . Since rad(f) = (0), we have f(v) 6= 0. Thus either sgn(R · v) = (0, 1, 0) or sgn(R · v) = (1, 0, 0). Therefore either sgn(f|W ) = (r, s − 1, 0) or sgn(f|W ) = (r − 1, s, 0). Pn Pn Lemma 2.27. Let f(x1, . . . , xn) = i=1 j=1 fijxixj ∈ R[x1, . . . , xn] be a quadratic n form such that f(Z ) ⊆ Z. Then f(x1, . . . , xn) ∈ Z[x1, . . . , xn].

Proof. Let e1,..., en be the standard basis vectors. Then f(ei) = fii ∈ Z for 1 ≤ i ≤ n. Also, for i 6= j, f(ei + ej) = fii + fjj + fij + fji = fii + fjj + 2fij ∈ Z. Since 2 fii, fjj ∈ Z, it follows that 2fij ∈ Z. Since fii is the coefficient on xi and 2fij is the coefficient on xixj for 1 ≤ i, j ≤ n, we have shown all coefficients are integers.

2.3 Identities and Inequalities

Lemma 2.28.

2 2 2 2 2 X 2 (x1 + ··· + xn)(y1 + ··· yn) = (x1y1 + ··· + xnyn) + (xiyj − xjyi) . 1≤i

|x · y| ≤ kxk2kyk2 Proof. By Lemma 2.28,

2 2 2 X 2 |x · y| = (x1y1 + ··· + xnyn) ≤ (x1y1 + ··· + xnyn) + (xiyj − xjyi) 1≤i

Lemma 2.30. Let x1, . . . , xn ∈ R. Then 2 2 2 (x1 + ··· + xn) ≤ n(x1 + ··· + xn). Proof. We have

2 2 2 X (x1 + ··· + xn) = x1 + ··· + xn + 2xixj i

20 T T An n × n matrix Q is an orthogonal matrix if QQ = Q Q = In, where In is the n × n identity matrix. It follows easily that an orthogonal matrix Q satisﬁes det(Q) = ±1. For v, w ∈ Rn, let v · w denote the standard dot product in Rn. The next proposition states the Gram-Schmidt procedure in an eﬃcient way.

Proposition 2.31. Let A be an n × n matrix with entries in R. Assume that the columns of A are linearly independent. Then A = QR where Q is an n×n orthogonal matrix and R is an n × n upper triangular matrix.

Proof. Let v1,..., vn be the columns of A. Deﬁne y1,..., yn as follows.

y1 = v1, y1 · v2 y2 = v2 − y1, y1 · y1 y1 · v3 y2 · v3 y3 = v3 − y1 − y2 y1 · y1 y2 · y2 . . k−1 X yi · vk y = v − y k k y · y i i=1 i i . . n−1 X yi · vn y = v − y n n y · y i i=1 i i

It is straightforward to show that the vectors y1,..., yn are linearly independent. One proves by induction on k ≥ 2 that yj · yk = 0 for all j < k ≤ n. The main calculation is that if j < k, then induction gives

k−1 ! X yi · vk yj · vk y · y = y · v − y = y · v − (y · y ) = 0. j k j k y · y i j k y · y j j i=1 i i j j

yi Let wi = , 1 ≤ i ≤ n. Then w1,..., wn is an orthonormal set of vectors. kyik2 2 That is, wi · wj = 0 for all i 6= j and kwik2 = wi · wi = 1 for 1 ≤ i ≤ n. Therefore, w1,..., wn is an orthonormal set of basis vectors of the column space of A. Let Q be the n × n matrix whose columns are w1,..., wn. Then Q is an orthogonal matrix. For 1 ≤ k ≤ n, we have vk = c1w1 + ··· + cnwn with each ci ∈ R. Then wj · vk = cj(wj · wj) = cj, 1 ≤ j ≤ n. Thus

vk = (w1 · vk)w1 + (w2 · vk)w2 + ··· + (wn · vk)wn.

21 This gives the following set of equations.

v1 = (w1 · v1)w1

v2 = (w1 · v2)w1 + (w2 · v2)w2

v3 = (w1 · v3)w1 + (w2 · v3)w2 + (w3 · v3)w3 . .

vn = (w1 · vn)w1 + (w2 · vn)w2 + (w3 · vn)w3 + ··· + (wn · vn)wn Let   w1 · v1 w1 · v2 w1 · v3 ···  0 w2 · v2 w2 · v3 ··· R =   .  0 0 w3 · v3 ···  . . . .  . . . .. Then R is an n × n upper triangular matrix and the equations above show that A = QR. Proposition 2.32 (Hadamard’s Inequality). Let A be an invertible n×n matrix with Qn entries in R. Let v1,..., vn be the n columns of A. Then | det(A)| ≤ j=1 kvjk2. Proof. Use Proposition 2.31 to write A = QR where Q is an orthogonal matrix and R is an upper triangular matrix. Let R = (rij) where rij is the (i, j)-entry of R. Let w1,..., wn denote the columns of Q. Since Q is an orthogonal matrix, we have wi · wi = 1 for 1 ≤ i ≤ n and wi · wj = 0 if i 6= j. Since A = QR, we have Pj vj = i=1 rijwi, 1 ≤ j ≤ n. We have j ! j ! j j 2 X X X 2 X 2 2 kvjk2 = vj · vj = rijwi · rijwi = rij(wi · wi) = rij ≥ rjj. i=1 i=1 i=1 i=1

Thus |rjj| ≤ kvjk2 for 1 ≤ j ≤ n. Since R is an upper triangular matrix, we have det(R) = r11 ··· rnn. This gives n n Y Y | det(A)| = | det(Q)|| det(R)| = | det(R)| = |rjj| ≤ kvjk2. j=1 j=1

2.4 Norms

n n There are three norms on R that we will consider. For x = (x1, . . . , xn) ∈ R deﬁne the norms k k∞, k k1, and k k2 as follows:

kxk∞ = max |xi| 1≤i≤n n X kxk1 = |xi| i=1 q 2 2 kxk2 = x1 + ··· + xn

22 It is straightforward to show that all three of these norms obey the triangle inequality, that is, kx + yki ≤ kxki + kyki for i ∈ {1, 2, ∞}. Note that k k∞ is the standard inﬁnity-norm, k k1 is the standard 1-norm, and k k2 is the standard 2-norm (or Euclidean norm).

Lemma 2.33. The norms k k∞, k k1, and k k2 induce the same metric topology on n √ R , as kxk∞ ≤ kxk1 ≤ n · kxk2 ≤ n · kxk∞. In addition, kxk∞ ≤ kxk2.

Proof. It is clear that kxk∞ ≤ kxk1. By Lemma 2.30, we have

2 2 2 (|x1| + ··· + |xn|) ≤ n(|x1| + ··· + |xn| ) √ q 2 2 |x1| + ··· + |xn| ≤ n x1 + ··· + xn √ kxk1 ≤ nkx2k.

Also √ √ q 2 2 n · kxk2 = n · x1 + ··· + xn √ p 2 2 ≤ n · kxk∞ + ··· + kxk∞ √ p 2 = n · n · kxk∞

= n · kxk∞.

In addition

p 2 kxk∞ = kxk∞ q 2 2 ≤ x1 + ··· + xn

= kxk2. √ Therefore kxk∞ ≤ kxk1 ≤ n · kxk2 ≤ n · kxk∞ and kxk∞ ≤ kxk2. P 0 Some authors express the quadratic form f as f(x) = 1≤i≤j≤n fijxixj. In this 0 0 0 0 0 0 0 0 n(n+1) case, we deﬁne F := (f11, f12, . . . , f1n, f22, f23, . . . , f2n, . . . , fnn) ∈ R 2 . Noting 0 1 0 that when i = j, fii = fii and when i 6= j, fij = fji = 2 fij, it is straightforward to show 0 0 kFk∞ ≤ kF k∞ ≤ 2kFk∞ and kFk1 = kF k1.

23 √ 0 For the third norm, we show that kFk2 ≤ kF k2 ≤ 2kFk2.

2 X 2 kFk2 = fij 1≤i,j≤n X 2 X 2 = fii + 2fij 1≤i≤n 1≤i

Therefore results using one expression of f can be transformed to results using the other expression and aﬀect only the constants. Of the three norms, k k∞ is used most frequently in this paper. For convenience, we will let kxk := kxk∞ for the remainder of the paper.

24 Chapter 3 A Treatise on Lattices

n Deﬁnition 3.1. Let b1,..., bs be s linearly independent vectors in R . Then the lattice generated by b1,..., bs is the set

( s )

X xibi xi ∈ Z . i=1

Deﬁnition 3.2. A subset Λ in Rn is an s-dimensional lattice if there exist s linearly Ps independent vectors b1,..., bs such that Λ = i=1 Zbi.

Deﬁnition 3.3. Let Λ be the lattice in Rn generated by linearly independent vectors n×s b1,..., bs. Let B = [b1 ··· bs] ∈ R be the matrix whose columns are b1,..., bs with respect to the standard basis vectors. Then the determinant of Λ, denoted det(Λ), is given by det(Λ) = p| det(BT B)|. Note that when n = s, this gives det(Λ) = | det(B)|.

n Proposition 3.4. Let 1 ≤ s ≤ n, let b1,..., bs ∈ R be linearly independent vectors n n×s written with respect to the standard basis of R , and let B = [b1,..., bs] ∈ R . Then det(BT B) > 0.

Proof #1. Since BT B is a symmetric s × s matrix, there exists an invertible s × s T T matrix P such that P (B B)P = D where D = diag(c1, . . . , cs) is a diagonal matrix th s whose (i, i)-entry is ci. Let ei denote the i standard basis vector of R . Then T T T T T ci = ei Dei = ei P B BP ei = (BP ei) (BP ei) ≥ 0. We have BP ei 6= 0 because P is invertible and the columns of B are linearly independent. Thus ci > 0. It T T 2 T follows that det(P B BP ) = (det(P )) det(B B) = det(D) = c1 ··· cs > 0, and T thus det(B B) > 0. Proof #2. Let v ∈ Rs be an arbitrary nonzero vector. Then Bv 6= 0 because the columns of B are linearly independent. Thus vT (BT B)v = (Bv)T (Bv) > 0. It T T follows that B B is a positive deﬁnite s × s matrix, and thus det(B B) > 0. Lemma 3.5. Let Λ be an s-dimensional lattice in Rn and Γ be an s-dimensional lattice such that Γ ⊆ Λ. Then there is a positive integer c such that

det(Γ) = c · det(Λ).

Proof. Let x1,..., xs be a basis of Γ and y1,..., ys a basis of Λ. Then xi = ai1y1 + ··· + aisys where aij ∈ Z and 1 ≤ i, j ≤ s. Let M := (aij), N be the s × n matrix whose rows are y1,..., ys, and P be the s × n matrix whose rows are x1,..., xs.

25 Then MN = P and

det(Γ) = pdet(PP T ) = pdet (MN(MN)T ) = pdet(MNN T M T ) = pdet(MM T )pdet(NN T ) = pdet(M)2 · det(Λ) = det(M) · det(Λ).

Since the entries of M are integers, det(M) ∈ Z. n Theorem 3.6. Let v1,..., vs be linearly independent vectors in R , where s ≤ n. th Let Λ be the lattice generated by v1,..., vs and let C be the n × s matrix whose j n column is given by vj, written in terms of the standard basis of R . Let

R = {x1v1 + ··· + xsvs | 0 ≤ xj ≤ 1, 1 ≤ j ≤ s} and let vol(R) denote the s-dimensional volume of R. Then vol(R) = det(Λ) = pdet(CT C). In particular, if s = n, then vol(R) = | det(C)|.

n ⊥ Proof. Let W ⊆ R be the subspace spanned by v1,..., vs. Let W be the orthogonal complement of W with respect to the standard inner product on Rn. Then Rn = W ⊕ W ⊥ by Corollary 2.24. ⊥ Let w1,..., wn−s be an orthonormal basis of W (that is, wk · wl = 0 for k 6= l and wk · wk = 1 for each k). th Let D denote the n × (n − s) matrix whose j column is given by wj written with n T respect to the standard basis of R . It follows that D D = In−s. Let d denote the (n − s)-dimensional volume determined by the points {y1w1 + ··· + yn−swn−s | 0 ≤ yj ≤ 1, 1 ≤ j ≤ n − s}. We have d = 1 because {w1,..., wn−s} is an orthonormal set. Let E = [CD] denote the n × n matrix whose ﬁrst s columns are given by C and whose last n − s columns are given by D. Let e denote the n-dimensional volume determined by the points {x1v1 + ··· + xsvs + y1w1 + ··· + yn−swn−s | 0 ≤ xj, yk ≤ 1, 1 ≤ j ≤ s, 1 ≤ k ≤ n − s}. p T It is known that e = | det(E)| = det(E E). Since vj · wk = 0 for each j and k, and since {w1,..., wn−s} is an orthonormal set, it is straightforward to check that ET E is a matrix whose upper left s × s submatrix is CT C, whose lower right (n − s) × (n − s) submatrix is In−s, and with all other entries equal to 0. We have vol(R) · d = e because W and W ⊥ are orthogonal and since d = 1, it follows that

vol(R) = e = pdet(ET E) = pdet(CT C) det(DT D) = pdet(CT C). p By Deﬁnition 3.3, det(Λ) = det(CT C) so vol(R) = det(Λ).

26 n Let Bj(0, r) = {x ∈ R |kxkj ≤ r}, the closed ball of radius r with respect to the norm k kj, where j ∈ {1, 2, ∞}.

Deﬁnition 3.7. Let Λ be a lattice of dimension s in Rn. For 1 ≤ i ≤ s we deﬁne the ith successive minimum with respect to the norm k kj as

λi(Λ)j = inf{r| dim(span(Λ ∩ Bj(0, r))) ≥ i}.

Lemma 3.8. Let Λ ⊂ Rn be an s-dimensional lattice. Then for each j ∈ {1, 2, ∞},

λ1(Λ)j = inf{kvkj | v ∈ Λ, v 6= 0}. Proof. By deﬁnition,

λ1(Λ)j = inf{r| dim(span(Λ ∩ Bj(0, r))) ≥ 1}

= inf{r| Λ ∩ Bj(0, r) 6= ∅}

= inf{kvkj| v ∈ Λ, v 6= 0}.

n Theorem 3.9. Let Λ ⊂ R be an s-dimensional lattice with basis b1,..., bs. Let ˜ ˜ b1,..., bs be the Gram-Schmidt orthogonalization of the ordered set b1,..., bs. Then ˜ λ1(Λ)2 ≥ min kbik2 > 0. 1≤i≤s

s Proof. For every nonzero vector v ∈ Λ, there is some nonzero x = (x1, . . . , xs) ∈ Z Ps such that v = i=1 xibi. Let j ∈ {1, . . . , s} be the largest index such that xj 6= 0. Then * j +

X ˜ ˜ ˜ ˜ ˜ 2 xibi, bj = |xj|hbj, bji = |xj|hbj, bji = |xj| kbjk2 i=1 Now by Corollary 2.29, we also know

* j + j

X ˜ X ˜ xibi, bj ≤ xibi · kbjk2 i=1 i=1 2 Therefore * j + j

˜ 2 X ˜ X ˜ |xj| kbjk2 = xibi, bj ≤ xibi · kbjk2 i=1 i=1 2 so that j

˜ X |xj| kbjk2 ≤ xibi . i=1 2

Since xj is a nonzero integer,

j ˜ ˜ ˜ X min kbik2 ≤ kbjk2 ≤ |xj| kbjk2 ≤ xibi . 1≤i≤s i=1 2

27 ˜ Thus min1≤i≤s kbik2 is a lower bound on the norm of all lattice elements. Since λ1(Λ)2 is the greatest lower bound of the same set by Lemma 3.8, ˜ λ1(Λ)2 ≥ min kbik2 > 0. 1≤i≤s

n Corollary 3.10. Let Λ ⊂ R be an s-dimensional lattice with basis b1,..., bs. Let ˜ ˜ b1,..., bs be the Gram-Schmidt orthogonalization of the ordered set b1,..., bs. Then

1 ˜ λ1(Λ)1 ≥ min kbik1 > 0, n 1≤i≤s 1 ˜ λ1(Λ)∞ ≥ min kbik∞ > 0. n 1≤i≤s Proof. By Lemma 2.33 and Theorem 3.9, √ ˜ ˜ min kbik1 ≤ n min kbik2 1≤i≤s 1≤i≤s √ ≤ n λ1(Λ)2

≤ n λ1(Λ)∞

≤ n λ1(Λ)1, and √ ˜ ˜ min kbik∞ ≤ n min kbik2 1≤i≤s 1≤i≤s √ ≤ n λ1(Λ)2

≤ n λ1(Λ)∞.

Corollary 3.11. Let Λ ⊂ Rn be an s-dimensional lattice. Then for each j ∈ {1, 2, ∞}, there exists some j > 0 such that for x, y ∈ Λ with x 6= y, kx−ykj > j. In particular, if = min{1, 2, ∞}, then for j ∈ {1, 2, ∞}, kx − ykj ≥ λ1(Λ)j > > 0. Proof. Since x, y ∈ Λ, it follows that x − y ∈ Λ. Also x − y 6= 0 since x 6= y. By Lemma 3.8, kx − ykj ≥ λ1(Λ)j and by Theorem 3.9 and Corollary 3.10, λ1(Λ)j > 0. Thus, there is some j > 0 such that

kx − ykj ≥ λ1(Λ)j > j > 0.

Moreover, deﬁne := min{1, 2, ∞}. Then

kx − ykj ≥ λ1(Λ)j > > 0.

28 n Deﬁnition 3.12. A subgroup G of R is discrete with respect to k kj, where j ∈ {1, 2, ∞}, if there exists > 0 such that Bj(0, ) contains no nonzero element of G. That is, the subspace topology on G with respect to k kj is discrete.

If G is discrete with respect to one of k k1, k k2, or k k∞, it is discrete with respect to each of the others by Lemma 2.33.

Proposition 3.13. A subgroup of Rn is a lattice if and only if it is discrete. Proof. First, assume Λ ⊂ Rn is an s-dimensional lattice. Then by Corollary 3.11, there is some > 0 such that for every nonzero x ∈ Λ, kxk2 > . Hence B2 (0, ) contains no nonzero lattice element. So Λ is discrete. For the other direction, see [9, Prop. 7.3, p. 150]. Here we show a diﬀerent way of proving that a lattice Λ is discrete. Theorem 3.14. Suppose Λ is a lattice. Then Λ is discrete.

n Proof. Let b1,..., bn be a basis of Λ. Then it is also a basis of R and we can n n define the function ϕ : R → R so that ϕ(a1b1 + ··· + anbn) = (a1, . . . , an), where a1, . . . , an ∈ R. Then ϕ is a continuous function. Choose r ∈ R>0 such that Λ ∩ B2(0, r) contains a nonzero lattice point (for example, choose r such that kb1k2 < r). Now B2(0, r) is closed and bounded and therefore compact. Therefore ϕ B2(0, r) is also compact. Hence it is bounded, and it follows that there is some positive M ∈ R such that ϕ B2(0, r) ⊂ B2(0,M). Let v = a1b1 + ··· + anbn ∈ B2(0, r). Then ϕ(v) = (a1, . . . , an) ∈ B2(0,M). So 2 2 2 a1 + ··· + an ≤ M and therefore |ai| ≤ M for 1 ≤ i ≤ n. Now, there are only finitely many ai ∈ Z with |ai| ≤ M and therefore only n finitely many (a1, . . . , an) ∈ Z with |ai| ≤ M for 1 ≤ i ≤ n. Hence, there are only finitely many points in Λ ∩ B2(0, r). In particular, Λ ∩ B2(0, r) contains at least one nonzero lattice point. Therefore one of these nonzero lattice points has minimal 1 length d. Then B2 0, 2 d contains no nonzero element of Λ. By Definition 3.12, Λ is discrete. Proposition 3.15. Suppose Λ ⊂ Rn is an s-dimensional lattice. For each j ∈ {1, 2, ∞}, there are vectors ˜x1,..., ˜xs ∈ Λ, which depend on j, such that k˜xikj = λi(Λ)j, 1 ≤ i ≤ s.

Proof. Suppose for some i, with 1 ≤ i ≤ s, that Bj(0, λi(Λ)j) contains infinitely many points in Λ. Since any infinite subset in a bounded subset of Rn has a limit point, for any > 0, there are x, y ∈ Λ with x 6= y such that kx − ykj < . This contradicts Corollary 3.11. Hence there are only finitely many elements of Λ in B(0, λi(Λ)j). By definition of λi(Λ)j, one of them must have length λi(Λ)j. Lemma 3.16. Suppose Λ is an s-dimensional lattice in Rn where s ≤ n. Let ˜x1,..., ˜xs be the successive minimum vectors of Λ with respect to one of the three norms k k1, k k2, k k∞. Let y1,..., ys be linearly independent vectors in Λ. For j ∈ {1, 2, ∞} and 1 ≤ i ≤ s,

29 (1) after re-indexing so that ky1kj ≤ · · · ≤ kyskj, one has k˜xikj ≤ kyikj

(2) k˜x1kj · · · k˜xskj ≤ ky1kj · · · kyskj.

Proof. Proceed by induction. By deﬁnition of a successive minimum vector, k˜x1kj ≤ ky1kj. Assume, by way of induction, k˜xikj ≤ kyikj for 1 ≤ i ≤ m. Suppose kym+1kj < k˜xm+1kj. Since ky1kj ≤ · · · ≤ kym+1kj, it follows that y1,..., ym+1 ∈ Λ ∩ B(0, kym+1kj) ⊂ Λ ∩ B(0, k˜xm+1kj). Therefore, since y1,..., ym+1 are linearly independent, k˜xm+1kj 6= inf{r| dim(span(Λ ∩ B(0, r))) ≥ m + 1} and is hence is not the (m + 1)-th successive minimum vector, a contradiction. Therefore k˜xm+1kj ≤ kym+1kj and k˜x1kj · · · k˜xm+1kj ≤ ky1kj · · · kym+1kj. Induction on m gives the desired results.

Notation: n indicates a constant depending only on n.

Corollary 3.17. Suppose Λ is an s-dimensional lattice in Rn where s ≤ n. Let ˜x1,..., ˜xs be the successive minimum vectors of Λ with respect to k kj for some j ∈ {1, 2, ∞}. If y1,..., ys are successive minimum vectors of Λ with respect to k kk, for some k ∈ {1, 2, ∞} and j 6= k, then k˜xikj n kyikk for 1 ≤ i ≤ s.

Proof. By Lemma 3.16, k˜xikj ≤ kyikj and by Lemma 2.33 kyikj n kyikk. Theorem 3.18 (Blichfeld’s Theorem). For any n-dimensional lattice Λ ⊂ Rn and n measurable set S ⊂ R with vol(S) > det(Λ), there exist two distinct points z1, z2 ∈ S such that z1 − z2 ∈ Λ.

Proof (following [12]). Let b1,..., bn be a basis of lattice Λ. Let B be the n × n matrix whose i-th column is bi. Deﬁne the fundamental parallelepiped of the lattice as n P(B) := {Bx | x ∈ R and 0 ≤ xi < 1 for 1 ≤ i ≤ n}. We claim that as v ranges over all of Λ, the sets

v + P(B) := {v + w | w ∈ P(B)}

n n form a partition of R . To see this, ﬁrst consider some element r = (r1, . . . , rn) ∈ R . Since the columns of B are linearly independent, B is invertible, and hence there n exists s = (s1 . . . , sn) ∈ R such that Bs = r. Note that si can be written si = ai + bi where ai ∈ Z and 0 ≤ bi < 1. Then r = Bs = B(a + b) = Ba + Bb and by deﬁnition Ba ∈ Λ and Bb ∈ P(B). Hence every r ∈ Rn is in v + P(B) for some v ∈ Λ. Now, assume v + P(B) ∩ w + P(B) 6= ∅ for some v, w ∈ Λ. Write n v = Ba1 and w = Ba2 for a1, a2 ∈ Z . Then there exist b1, b2 with 0 ≤ b1i, b2i < 1 for 1 ≤ i ≤ n such that Ba1+Bb1 = Ba2+Bb2 and hence B(a1−a2) = Ba1−Ba2 = Bb2 − Bb1 = B(b2 − b1). Since B is invertible, this gives a1 − a2 = b2 − b1. Since

30 0 ≤ b1i, b2i < 1, it is true that −1 < b2i − b1i < 1 and hence −1 < a1i − a2i < 1 for 1 ≤ i ≤ n. But since a1i, a2i ∈ Z, necessarily a1i −a2i = 0. Hence a1 = a2 and v = w. Therefore, the sets v + P(B) are disjoint and form a partition of Rn. Now deﬁne

Sv := S ∩ (v + P(B)). P Since S = ∪v∈ΛSv, vol(S) = v∈Λ vol(Sv). Deﬁne ˆ Sv := Sv − v. ˆ ˆ Then Sv ⊂ P(B) and vol(Sv) = vol(Sv). So

X ˆ X vol(Sv) = vol(Sv) = vol(S) > det(Λ) = vol(P(B)). v∈Λ v∈Λ ˆ ˆ ˆ ˆ Therefore, there exist v, w ∈ Λ, v 6= w such that Sv ∩ Sw 6= ∅. Let z ∈ Sv ∩ Sw. Then z + v ∈ Sv ⊆ S and z + w ∈ Sw ⊆ S so that (z + v) − (z + w) = v − w ∈ Λ. Let z1 := z + v and z2 := z + w. Then, as desired, z1, z2 ∈ S and z1 − z2 ∈ Λ. n Deﬁnition 3.19. A set C ∈ R is convex if for any c1, c2 ∈ C we have

λ1c1 + λ2c2 ∈ C with λ1 ≥ 0, λ2 ≥ 0, λ1 + λ2 = 1.

Deﬁnition 3.20. A set C ∈ Rn is symmetric about 0 if −s ∈ C whenever s ∈ C . Theorem 3.21 (Minkowski’s Generalized Convex Body Theorem). Let Λ be an n- dimensional lattice in Rn. Then for any convex set S that is symmetric about 0, if vol(S) > 2n det(Λ), then S contains a non-zero lattice point.

Proof (following [12]). Deﬁne 1 Sˆ := S = {x | 2x ∈ S}. 2

Then vol(Sˆ) = 2−nvol(S). Therefore

vol(S) > 2n det(Λ) 2−nvol(S) > det(Λ) vol(Sˆ) > det(Λ).

ˆ By Theorem 3.18, there exist z1, z2 ∈ S, z1 6= z2 such that z1 − z2 ∈ Λ. Now by deﬁnition, 2z1, 2z2 ∈ S and since S is centrally symmetric, −2z2 ∈ S. And since S is 1 1 convex, it follows that 2 (2z1) − 2 (2z2) = z1 − z2 ∈ S. So z1 − z2 is a nonzero lattice point contained in S.

31 Theorem 3.22 (Minkowski’s First Theorem). For any n-dimensional lattice Λ in Rn, √ 1/n λ1(Λ)2 ≤ n(det(Λ)) .

Proof (following [12]). Write x = (x1, . . . , xn) and consider the set n λ1(Λ)2 S = x ∈ |xi| < √ . R n Then for x ∈ S, q 2 2 kxk2 = x1 + ··· + xn s λ (Λ) 2 λ (Λ) 2 < 1√ 2 + ··· + 1√ 2 n n s λ (Λ) 2 = n 1√ 2 n

= λ1(Λ)2.

2λ (Λ) √1 2 Hence S ⊂ B2(0, λ1(Λ)2). Now S is an n-dimensional cube with side-length n . n 2λ (Λ) √1 2 Hence n < vol(B2(0, λ1(Λ)2)). Since B2(0, λ1(Λ)2) is open, it does not con- n tain any lattice vectors. Therefore by Theorem 3.21, vol(B2(0, λ1(Λ)2)) ≤ 2 det(Λ). n 2λ (Λ) √1 2 n Hence n < vol(B2(0, λ1(Λ)2)) ≤ 2 det(Λ) implies λ (Λ) n 1√ 2 < det(Λ) n √ 1/n λ1(Λ)2 < n(det(Λ)) , as desired. Theorem 3.23 (Minkowski’s Second Theorem). For any n-dimensional lattice Λ in Rn, n !1/n Y √ 1/n λi(Λ)2 ≤ n · (det(Λ)) . i=1 Proof (following [12]). By Proposition 3.15, we can ﬁnd linearly independent vectors x1,..., xn ∈ Λ achieving the successive minima, i.e. kxik2 = λi(Λ)2. For ease of notation, we will write λi in place of λi(Λ)2. Let ˜x1,..., ˜xn be their Gram-Schmidt orthogonalization, so that ˜xi · ˜xj = 0 when i 6= j and the R−span(x1,..., xi) = R−span(˜x1,..., ˜xi). Consider the open ellipsoid T which has axes ˜x1,..., ˜xn with Pn ˜xi n lengths λ1, . . . , λn. Then y = ai ∈ is in the ellipsoid T if and only if i=1 k˜xik2 R 2 Pn ai i=1 2 < 1. Note that λi

n ! 2 X ˜xi ˜xj · ˜xj k˜xjk y · ˜x = a · ˜x = a = a 2 = a k˜x k j i k˜x k j j k˜x k j k˜x k j j 2 i=1 i 2 j 2 j 2

32 so that y · ˜xj aj = k˜xjk2 and n 2 n 2 X a X y · ˜xi 1 i = . λ2 k˜x k λ2 i=1 i i=1 i 2 i Hence we can write

( n 2 ) X hy, ˜xii T = y ∈ n < 1 . R k˜x k · λ i=1 i 2 i We wish to show that T does not contain any non-zero lattice points. Let y 6= 0 ∈ Λ and 1 ≤ k ≤ n be the maximum k such that kyk2 ≥ λk. If y ∈/ span(˜x1,..., ˜xk) = span(x1,..., xk), then x1,..., xk, y is a set of k + 1 linearly independent vectors of length less than λk+1, which is impossible by the deﬁnition of successive minima. Pk ˜xi Hence, y ∈ span(˜x1,..., ˜xk), giving y = ai and hy, ˜xii = 0 for k + 1 ≤ i ≤ i=1 k˜xik2 n. Therefore

n 2 k 2 X hy, ˜xii X hy, ˜xii = k˜x k · λ k˜x k · λ i=1 i 2 i i=1 i 2 i k 2 1 X hy, ˜xii ≥ λ2 k˜x k k i=1 i 2 k 1 X = a2 λ2 i k i=1 2 kyk2 = 2 λk ≥ 1,

2 Pk 2 where the last equality comes from kyk2 = y · y = i=1 ai . Therefore y ∈/ T . Using the contrapositive of Theorem 3.21, we have

vol(T ) ≤ 2n det(Λ).

Now, we also know

n ! Y vol(T ) = λi · volume of n-dimensional unit ball. i=1

√2 Since the n-dimensional unit ball contains a cube of side length n , we can say

n ! n Y 2 vol(T ) ≥ λ √ . i n i=1

33 Hence

n ! n Y 2 λ √ ≤ 2n det(Λ) i n i=1 n !1/n Y 2 λ √ ≤ 2 · (det(Λ))1/n i n i=1 n !1/n Y √ 1/n λi ≤ n · (det(Λ)) . i=1

Corollary 3.24. Let Λ be an s-dimensional lattice in Rn. Then

s !1/s Y √ 1/s λi(Λ)2 ≤ s · (det(Λ)) . i=1

Proof. Let v1,..., vs be a basis of Λ. Let M be the n × s matrix whose columns n are v1,..., vs written with respect to the standard basis of R . By Deﬁnition 3.3, we know det(Λ) = pdet(M T M). Theorem 3.6 implies that det(Λ) is equal to the s-dimensional volume determined by the set R = {x1v1 + ··· + xsvs|0 ≤ xi ≤ 1, 1 ≤ i ≤ s}. s Let W be the vector subspace generated by v1,... vs. We may assume W = R because W has an orthonormal basis. Therefore, since Λ ⊂ W , Λ is a full rank lattice in s. For λ (Λ) , . . . , λ (Λ) the successive minima of Λ, Theorem 3.23 gives R √1 2 s 2 √ Qs 1/s 1/s 1/s ( i=1 λi(Λ)2) ≤ s · (vol(R)) = s · (det(Λ)) .

Now, Theorem 3.23 and Corollary 3.24 hold for k k1 and k k∞ as well, with only a change in constants, which is best expressed with n. Recall that k k = k k∞.

Lemma 3.25. Suppose Λ ⊂ Rn is an s-dimensional lattice with s ≤ n. Suppose ˜x1,..., ˜xs ∈ Λ are successive minimum vectors of Λ with respect to the norm k k. Then there exists a basis x1,..., xs of Λ such that

1. the ˜xi’s can be written

˜x1 = a11x1

˜x2 = a21x1 + a22x2 . .

˜xs = as1x1 + ··· + assxs

with aij ∈ Z and |aij| < |aii| for j < i and

2. k˜xik n kxik n k˜xik for 1 ≤ i ≤ s.

34 Proof. Let z1,..., zs be an arbitrary basis of Λ. Then ˜x1 = b11z1 + ··· + b1szs 0 0 with b1i ∈ Z. Deﬁne a11 := gcd(b11, . . . , b1s) so that ˜x1 = a11(b11z1 + ··· + b1szs). 0 0 By the unimodular row lemma, there is a Z-basis x1, z2 ,..., zs of L where x1 = 0 0 b11z1 + ··· + b1szs. 0 0 By induction, assume there is a basis x1,..., xm, xm+1,..., xs of Λ such that ˜xi = ai1x1 +···+aiixi, with aij ∈ Z, 1 ≤ i ≤ m, and |aij| < |aii| for j < i. We have ˜xm+1 = 0 0 0 0 0 am+1,1x1 +···+am+1,mxm +bm+1,m+1xm+1 +···+bm+1,sxs with am+1,j, bm+1,j ∈ Z. Let 0 0 0 am+1,m+1 = gcd(bm+1,m+1, . . . , bm+1,s) so that ˜xm+1 = am+1,1x1 + ··· + am+1,mxm + 0 0 0 0 0 am+1,m+1(bm+1,m+1xm+1 +···+bm+1,sxs). By the unimodular row lemma, there exists 0 0 0 0 a Z-basis zm+1, zm+2,..., zs of Zxm+1 +···+Zxs with zm+1 = bm+1,m+1xm+1 +···+ 0 0 0 0 0 bm+1,sxs. Thus ˜xm+1 = am+1,1x1 + ··· + am+1,mxm + am+1,m+1zm+1. 0 By the division algorithm, there is a unique kj ∈ Z such that 0 ≤ |am+1,j − 0 0 kjam+1,m+1| < |am+1,m+1| for 1 ≤ j ≤ m. Replace zm+1 with xm+1 := zm+1 + k1x1 + ··· + kmxm so that

0 0 0 0 0 ˜xm+1 = (am+1,1 − k1am+1,m+1)x1 + ··· + (am+1,m − kmam+1,m+1)xm + am+1,m+1xm+1.

0 0 Let am+1,j = am+1,j − kjam+1,m+1. Then |am+1,j| < |am+1,m+1| for j < m + 1. By induction we obtain a basis x1,..., xs of L having the form stated. For property 2, consider that ˜x1 = a11x1. Since k˜x1k is minimal, |a11| = 1. Otherwise, kx1k < k˜x1k, which is a contradiction. Therefore, k˜x1k = kx1k. Also, since ˜x1 = a11x1, Span(˜x1) = Span(x1). By way of induction, assume kxik n k˜xik for 1 ≤ i ≤ m and Span(˜x1,..., ˜xm) = Span(x1,..., xm). Now since ˜xm+1 = am+1,1x1 + ··· + am+1,m+1xm+1 with |am+1,j| < |am+1,m+1| with 1 ≤ j ≤ m, we have

am+1,1 am+1,m 1 xm+1 = − x1 − · · · − xm + ˜xm+1. am+1,m+1 am+1,m+1 am+1,m+1

Combining this with k˜x1k ≤ · · · ≤ k˜xsk, since they are successive minimum vectors, gives

kxm+1k ≤ kx1k + ··· + kxmk + k˜xm+1k

n k˜x1k + ··· + k˜xmk + k˜xm+1k

n (m + 1) · k˜xm+1k

n k˜xm+1k. Hence, by induction, the right-hand side of property 2 holds. For the left-hand side of property 2, we know from property 1 that x1,..., xs is a basis. Thus they are linearly independent. In particular, xm+1 is not in the span of x1,..., xm and therefore not in the span of ˜x1,..., ˜xm. Hence k˜xm+1k ≤ kxm+1k and property 2 holds by induction. So x1,..., xs is the desired basis. Theorem 3.26. Suppose Λ ⊂ Rn is an s-dimensional lattice with s ≤ n. Suppose ˜x1,..., ˜xs ∈ Λ are successive minimum vectors of Λ with respect to the norm k k. Let 0 0 x1,..., xs be as in Lemma 3.25. Then there exists a basis x1,..., xs of Λ such that the following statements hold.

35 0 1. k˜xik n kxik n k˜xik, 1 ≤ i ≤ s. 0 0 0 2. kx1k ≤ kx2k ≤ · · · ≤ kxsk. 0 0 3. det(Λ) n kx1k · · · kxsk n det(Λ). 0 0 Proof. (1) and (2): We will construct by induction a basis x1,..., xs of Λ satisfying 0 (1) and (2). Let x1 = x1. Then (1) holds for i = 1 and (2) holds vacuously. Suppose 0 0 that x1,..., xm have been constructed such that the following three statements hold. 0 0 (a) x1,..., xm, xm+1,..., xs is a basis of Λ. 0 (b) k˜xik n kxik n k˜xik, 1 ≤ i ≤ m. 0 0 0 (c) kx1k ≤ kx2k ≤ · · · ≤ kxmk. 0 0 Suppose that kxmk ≤ kxm+1k. Then let xm+1 = xm+1. Then (a) - (c) hold for 0 0 x1,..., xm+1 and the inductive step is complete. 0 0 0 0 0 Now suppose that kxm+1k < kxmk. Let xm+1 = 2xm + xm+1. Then x1,..., xm, 0 xm+1, xm+2,..., xs is a basis of Λ, which means that (a) holds at level m + 1. 0 0 0 We have kxm+1k ≤ 2kxmk + kxm+1k < 3kxmk. This gives 0 0 kxm+1k n kxmk n k˜xmk ≤ k˜xm+1k. 0 In particular, we have kxm+1k n k˜xm+1k. This gives part of (b) for i = m + 1. 0 0 0 0 0 Since 2xm = xm+1 − xm+1, we have 2kxmk ≤ kxm+1k + kxm+1k. Thus kxm+1k ≥ 0 0 0 0 0 0 2kxmk−kxm+1k > 2kxmk−kxmk = kxmk. This gives kxmk < kxm+1k, which proves (c) at level m + 1. 0 0 0 Since kxm+1k < kxmk and kxmk < kxm+1k from above, we now have k˜xm+1k n 0 0 0 kxm+1k < kxmk < kxm+1k. Thus k˜xm+1k n kxm+1k, which gives the remaining part of (b) for i = m + 1. The proof of statements (1) and (2) is now finished by induction on m. 0 0 0 0 (3): Let M = [x1 ··· xs], an n×s matrix whose columns are x1,... xs with respect to the standard basis vectors. For the left-hand side, recall that by Definition 3.3, p T 0 0 T det(Λ) = det(M M) since x1,..., xs is a basis of Λ. Note that M M = (cij) 0T 0 where cij = xi xj. Hence by the definition of the determinant, s T X Y det(M M) = sgn(σ) ci,σ(i)

σ∈Ss i=1 s X Y ≤ |ci,σ(i)|

σ∈Ss i=1 s X Y 0 0 ≤ nkxik2 kxσ(i)k2 σ∈Ss i=1 s 0 2 0 2 = s! · n kx1k2 · · · kxsk2 n 0 2 0 2 ≤ n! · n kx1k2 · · · kxsk2 0 2 0 2 n kx1k2 · · · kxsk2.

36 p T 0 0 Therefore det(Λ) = det(M M) n kx1k2 · · · kxsk2. By Lemma 2.33, 0 0 0 0 det(Λ) n kx1k2 · · · kxsk2 n kx1k · · · kxsk.

For the right hand side, we introduce vectors y1,..., ys that are successive minimum vectors of Λ with respect to the norm k k2. Then

0 0 kx1k · · · kxsk n k˜x1k · · · k˜xsk (by (1) )

n ky1k2 · · · kysk2 (by Corollary 3.17) ≤ ss/2 det(Λ) (by Corollary 3.24) ≤ nn/2 det(Λ)

n det(Λ), as desired. Theorem 3.27. Let Λ ⊂ Rn be an n-dimensional lattice. Let 1 ≤ s ≤ n and N > 0. Then there exist only ﬁnitely many s-dimensional sublattices with determinant ≤ N. Proof. If there are no s-dimensional sublattices with determinant ≤ N, we are done. Suppose there is one and call it Λs. Then by Theorem 3.26 there is a basis b1,..., bs of Λs such that kb1k · · · kbsk n det(Λs). Let cn be the constant implied in n. Therefore kb1k · · · kbsk ≤ cn det(Λs) ≤ cnN.

Without loss of generality, assume kb1k ≤ · · · ≤ kbsk. Let ` = λ1(Λs)∞. Then s−1 ` ≤ kb1k · · · kbs−1k. Therefore

cnN cnN kbsk ≤ ≤ s−1 . kb1k · · · kbs−1k `

cnN Hence kbik ≤ `s−1 for 1 ≤ i ≤ s. By Theorem 3.9, there are only finitely many cnN elements of Λs in B∞ 0, `s−1 . Hence there are only finitely many s-element subsets cnN in B∞ 0, `s−1 . Since any s-dimensional lattice with determinant ≤ N has a basis cnN lying in B∞ 0, `s−1 , it follows that there are only finitely many s-dimensional lattices with determinant ≤ N. Lemma 3.28. Let Λ ⊂ Rn be an n-dimensional lattice and W ⊂ Rn an s-dimensional subspace containing linearly independent lattice points x1,..., xs. Then W ∩ Λ is an s-dimensional lattice.

Proof. By Lemma 3.4 on page 105 of [4], there is a basis b1,..., bn of Λ such that

x1 = a11b1

x2 = a21b1 + a22b2 . .

xs = as1b1 + ··· + assbs

37 where aij ∈ Z and aii 6= 0 for 1 ≤ i ≤ s, 1 ≤ j ≤ i. We will show that b1,..., bs is an integer basis of W ∩ Λ. To this end, consider w ∈ W ∩ Λ which, as a lattice point, can be written w = r1b1 + ··· + rnbn where r1, . . . , rn ∈ Z. Since x1,... xs ∈ W and aii 6= 0 for 1 ≤ i ≤ s, b1,..., bs ∈ W . Since b1,..., bs are linearly independent, they form a basis of W . Therefore, w = t1b1 + ··· + tsbs with t1, . . . , ts ∈ R. Then

w = r1b1 + ··· + rnbn = t1b1 + ··· + tsbs.

By linear independence of b1,..., bn, it follows that ti = ri for 1 ≤ i ≤ s and ti = 0 for s + 1 ≤ i ≤ n. Since r1, . . . , rs ∈ Z, the vectors b1,..., bs form an integer basis for W ∩ Λ. Lemma 3.29. Let Λ ⊂ Rn be an n-dimensional lattice of determinant ∆ and W ⊂ Rn be an s-dimensional subspace such that W ∩Λ is an s-dimensional sublattice. Let W ⊥ be the orthogonal complement of W with respect to the standard inner product on Rn. Write Rn = W ⊕ W ⊥ and let π be the projection map π : W ⊕ W ⊥ → W ⊥.

Then π(Λ) is a lattice of dimension n − s.

Proof. Consider v1,..., vs a basis of the sublattice W ∩Λ. Suppose a1v1+···+asvs ∈ Λ with a1, . . . , as ∈ R. Then a1v1 + ··· + asvs ∈ W ∩ Λ and hence there exist b1, . . . , bs ∈ Z such that b1v1 + ··· + bsvs = a1v1 + ··· + asvs. Since v1,..., vs are linearly independent, this implies ai = bi for 1 ≤ i ≤ s. Hence, by [4, p. 106, n Thm. 3.1], we can extend v1,..., vs to a basis v1,..., vn of Λ. Now, since R = ⊥ ⊥ W ⊕ W and v1,..., vs is a basis of W , we can choose a basis ws+1,..., wn of W n and conclude v1,..., vs, ws+1,..., wn is a basis of R . Write each basis vector v1,..., vn of Λ in terms of the basis v1,..., vs, ws+1,..., wn so that s n X X vj = cijvi + dijwi, i=1 i=s+1 where cij, dij ∈ R. Note that dij = 0 when 1 ≤ j ≤ s. Let

n X zj := dijwi. i=s+1 Then n X π(vj) = dijwi = zj i=s+1 and hence zj ∈ π(Λ). Consider that for every y ∈ π(Λ), there is some x ∈ Λ such Pn that π(x) = y. Then x can be written x = j=1 ajvj where aj ∈ Z. Then

n ! n n n X X X X y = π(x) = π ajvj = ajπ(vj) = ajπ(vj) = ajzj. j=1 j=1 j=s+1 j=s+1

38 Hence every y ∈ π(Λ) can be written as an integer linear combination of the vectors zs+1,..., zn ∈ π(Λ). Pn To see that these vectors are linearly independent, assume j=s+1 bjzj = 0. Ps Rewriting zj as vj − i=1 cijvi gives n s ! X X bj vj − cijvi = 0 j=s+1 i=1 n n s X X X bjvj − bjcijvi = 0. j=s+1 j=s+1 i=1

Since v1,..., vn are linearly independent, their coeﬃcients in the sums above must be zero. In particular, bj = 0 for s + 1 ≤ j ≤ n. Hence, zs+1,..., zn are linearly independent. Corollary 3.30. Let Λ ⊂ Rn be a d-dimensional lattice. Let Y be the space generated by Λ and let W ⊂ Rn be an s-dimensional subspace such that W ∩Λ is a t-dimensional sublattice. Let (W ∩ Y )⊥Y be the orthogonal complement of W ∩ Y in Y with respect to the standard inner product on Rn. Write Y = (W ∩ Y ) ⊕ (W ∩ Y )⊥Y and let π be the projection map

π :(W ∩ Y ) ⊕ (W ∩ Y )⊥Y → (W ∩ Y )⊥Y . Then π(Λ) is a (d − t)-dimensional sublattice. ∼ d Proof. Note that Y = R and Λ is a d-dimensional lattice in Y . Since Y is generated by Λ and W ∩ Λ has dimension t, dim(W ∩ Y ) = t. Then it follows from Lemma 3.29 that π(Λ) is a (d − t)-dimensional sublattice. Theorem 3.31. Let Λ ⊂ Rn be an n-dimensional lattice of determinant ∆ and W ⊂ Rn be an s-dimensional subspace such that W ∩Λ is an s-dimensional sublattice. Let W ⊥ be the orthogonal complement of W with respect to the standard inner product on Rn. Write Rn = W ⊕ W ⊥ and let π be the projection map π : W ⊕ W ⊥ → W ⊥. Then ∆ = det(W ∩ Λ) · det(π(Λ)).

Proof. Let v1,..., vs be a basis for W ∩ Λ. Let N := [v1 ··· vn] be the matrix whose columns are v1,..., vn with respect to the standard basis vectors. Then q p T det(W ∩ Λ) = det(N N) = det(vi · vj), 1 ≤ i, j ≤ s.

Now, recall from the proof of Lemma 3.29 that we can extend v1,..., vs to a basis v1,..., vn of Λ with s n s ! X X X vj = cijvi + dijwi = cijvi + zj, i=1 i=s+1 i=1

39 ⊥ such that cij, dij ∈ R and ws+1,..., wn a basis of W . It was shown that zs+1,..., zn form a basis of π(Λ). Hence, deﬁning P := [zs+1 ··· zn] as the matrix whose columns are zs+1,..., zn with respect to the standard basis vectors gives q p T det(π(Λ)) = det(P P ) = det(zi · zj) with s + 1 ≤ i, j ≤ n. Let M := [v1 ··· vn] be the matrix whose columns are v1,..., vn with respect to the standard basis vectors. and note that ∆ = pdet(M T M). Now let

0 Ps Ps M := v1 ··· vs (vs+1 − i=1 ci,s+1vi) ··· (vn − i=1 ci,nvi) = v1 ··· vs zs+1 ··· zn . Ps Note that each i=1 cijvi with s + 1 ≤ j ≤ n is a linear combination of the ﬁrst s columns of M so that det(M 0) = det(M). Then

∆ = pdet(M T M) = pdet(M)2 = pdet(M 0)2 = pdet((M 0)T M 0).

Consider that

v · v v · z (M 0)T M 0 = i j i j , vj · zi zi · zj where the upper left hand block has dimension s × s, the upper right hand block has dimension s×(n−s), the lower left hand block has dimension (n−s)×s, and the lower ⊥ right hand block has dimension (n − s) × (n − s). Since zj ∈ W for s + 1 ≤ j ≤ n, vi · zj = 0 when 1 ≤ i ≤ n. Then

v · v 0 (M 0)T M 0 = i j 0 zi · zj 0 T 0 2 2 so that det((M ) M ) = det(vi · vj) det(zi · zj) = det(W ∩ Λ) det(π(Λ)) . Therefore

∆ = pdet((M 0)T M 0) = pdet(W ∩ Λ)2 · det(π(L))2 = det(W ∩ Λ) · det(π(Λ)).

Corollary 3.32. Let Λ ⊂ Rn be a d-dimensional lattice of determinant ∆. Let Y be the space generated by Λ and let W ⊂ Rn be an s-dimensional subspace such that W ∩Λ is a t-dimensional sublattice. Let (W ∩Y )⊥Y be the orthogonal complement of W ∩Y in Y with respect to the standard inner product on Rn. Write Y = (W ∩Y )⊕(W ∩Y )⊥Y and let π be the projection map

π :(W ∩ Y ) ⊕ (W ∩ Y )⊥Y → (W ∩ Y )⊥Y .

Then ∆ = det(W ∩ Λ) · det(π(Λ)).

40 ∼ d Proof. Note that Y = R and Λ is a d-dimensional lattice in Y . Since Y is generated by Λ and W ∩ Λ has dimension t, dim(W ∩ Y ) = t. Then by Theorem 3.31, ∆ = det(W ∩ Λ) · det(π(Λ)).

Lemma 3.33. Suppose ˜x1,..., ˜xs have the property that k˜xik < max((i−1)k˜xi−1k, 1) for i ≥ 2. Then k˜xik < max((i − 1)!k˜x1k, (i − 1)!) for i ≥ 2.

Proof. First consider when i = 2. Then by assumption k˜x2k < max(k˜x1k, 1). Now, by way of induction, assume k˜xmk < max((m − 1)!k˜x1k, (m − 1)!) for 2 ≤ m < s. Then

k˜xm+1k < max(mk˜xmk, 1)

≤ max(m(m − 1)! k˜x1k, m(m − 1)!, 1)

= max(m! k˜x1k, m!).

Hence by induction, k˜xik < max((i − 1)!k˜x1k, (i − 1)!) for i ≥ 2. Lemma 3.34. Let Λ ⊂ Rn be an s-dimensional lattice. Let ϕ :Λ → Z be a non-trivial linear map. Then ker(ϕ) contains s − 1 linearly independent elements of Λ.

Proof. Since Λ is an s-dimensional lattice, it is a free Z-module of rank s and hence ∼ s ∼ s Λ = Z . Then ker(ϕ) ⊂ Λ = Z is a submodule. By [7, p. 460, Thm. 4], ker(ϕ) is free of rank m ≤ s and there is a basis y1,..., ys of Λ such that a1y1, . . . , amym is a basis of ker(ϕ), where a1, . . . , am ∈ Z. Now, m 6= s, for if m = s, ker(ϕ) = Λ, which is a contradiction since ϕ is non-trivial. Suppose m ≤ s − 2. Then ϕ(ys−1) 6= 0 and ϕ(ys) 6= 0. Let b := ϕ(ys−1) and c := ϕ(ys). Then

ϕ(cys−1 − bys) = cϕ(ys−1) − bϕ(ys) = 0.

This implies that cys−1 − bys ∈ ker(ϕ), which is a contradiction since ys−1, ys ∈/ ker(ϕ). Thus, m = s − 1.

41 Chapter 4 Cassels 1955

This section is based upon Cassels’ exposition in [4], which is an improvement on his original publication [2] (and a correction in [3]). Here we use X f(x) = fijxixj, 1≤i≤j≤n

n(n+1) so that F = (f11, . . . , f1n, f22, . . . , f2n, . . . , fnn) ∈ R 2 .

Theorem 4.1. Let Lj(x) ∈ R[x1, . . . , xn] be linear forms for 1 ≤ j ≤ n. Let N be the n × n matrix whose j-th row is the coeﬃcients of Lj(x). Let D = det(N) and suppose L1(x),...,Ln(x) are such that D 6= 0. Assume tj > 0, 1 ≤ j ≤ n, satisﬁes

t1 ··· tn ≥ |D|.

Then there is an a ∈ Zn, a 6= 0, such that

|L1(a)| ≤ t1

|Lj(a)| < tj (2 ≤ j ≤ n). Proof (following [4], page 71). Let 0 < < 1. Deﬁne

C := {x : |L1(x)| < t1(1 + ), |Lj(x)| < tj for (2 ≤ j ≤ n)}.

By deﬁnition of N, for x ∈ Rn,   L1(x)  .  Nx =  .  . Ln(x)

Now let c ∈ C. Deﬁne y := Nc. Then     y1 L1(c)  .   .  y =  .  =  .  yn Ln(c) and so |y1| < t1(1 + ) and |yj| < tj for (2 ≤ j ≤ n). Then N(C) is an n-dimensional open box with side lengths 2t1(1 + ), 2t2,..., 2tn. Thus

n vol(N(C)) = 2 t1 ··· tn(1 + ).

Now since vol(N(C)) = |D| · vol(C) (see the proof of [9, Prop. 7.5, pp. 151-152]), we have

n −1 vol(C) = 2 |D| t1 ··· tn(1 + ) ≥ 2n(1 + ) > 2n.

42 Now, since C is an n-dimensional parallelepiped, it is convex and symmetric. Then n n by Theorem 3.21 with Λ = Z , C contains an a ∈ Z such that a 6= 0. There is a continuous map ϕ : Rn → Rn defined by ϕ(x) = Nx. Since the rows of N are linearly independent, this is a bijection. ϕ−1 is also continuous, and hence ϕ is a −1 homeomorphism. N(C ) is a closed box and therefore compact. Then ϕ (N(C )) = C is compact and therefore bounded and hence contains a finite number of nonzero a ∈ Zn by Proposition 3.13. We have just shown that there is such an element in each C, and so there must be at least one, say a, in infinitely many C. Within this set of C containing a, there is a subsequence of the ’s that converges to 0. Therefore, T a ∈ C. Thus |L1(a)| < t1(1 + ) for every 0 < < 1 and thus |L1(a)| ≤ t1.

Corollary 4.2. Let n ≥ 1, M > 1 be integers and let θ1, . . . , θn be real numbers. Then there is an integer m with

0 < m < M and integers l1, . . . , ln such that

−1/n |mθj − lj| < M (1 ≤ j ≤ n − 1) and −1/n |mθn − ln| ≤ M . Proof (following [4], page 71). We will apply Theorem 4.1 to the following n+1 linear forms:

Lj(x) = θjxn+1 − xj (1 ≤ j ≤ n)

Ln+1(x) = xn+1.

The matrix associated to these linear forms is   −1 0 0 ··· 0 θ1  0 −1 0 ··· 0 θ2  . . .   .. . .   0 0 . .  N :=  . .   0 0 0 .. 0 .     0 0 0 0 −1 θn 0 ········· 0 1

−1/n and so D := det(N) = ±1. Let tj = M for 1 ≤ j ≤ n and tn+1 = M. Then

−1/n n t1 ··· tn+1 = M · (M ) = 1 = |D|.

n+1 1 Then, by Theorem 4.1, there is some nonzero a = (l1, . . . , ln, m) ∈ Z such that

−1/n |Lj(a)| = |θjm − lj| < M for 1 ≤ j ≤ n − 1 1This is stronger than necessary, but follows directly from Theorem 4.1.

43 −1/n |Ln(a)| = |θnm − ln| ≤ M and |Ln+1(a)| = |m| < M. −1/n If m = 0 then |θjm−lj| = |−lj| ≤ M < 1 since M > 1. Since lj ∈ Z, this implies lj = 0 for 1 ≤ j ≤ n and so a = 0, which is a contradiction. If m > 0, we are done. If m < 0, we use −a instead, since this gives −m > 0 and

−1/n |Lj(−a)| = | − Lj(a)| = |Lj(a)| ≤ M for 1 ≤ j ≤ n − 1

−1/n |Ln(−a)| = | − Ln(a)| = |Ln(a)| < M and |Ln+1(−a)| = | − Ln+1(a)| = |Ln+1(a)| < M. Since |m| < M, it follows that 0 < −m = | − m| < M.

Recall that k k = k k∞. Theorem 4.3. Let X f(x) = fijxixj ∈ Z[x1, . . . , xn] 1≤i≤j≤n be an isotropic form over Z in n variables. Then there is an a ∈ Zn, a 6= 0 with f(a) = 0 such that (n−1)/2 kak ≤ (3kFk1) .

Proof (following [4], page 87). Choose a ∈ Zn, a 6= 0 such that f(a) = 0 and kak is n minimal. Note that we can do this since kxk ∈ Z≥0 for all x ∈ Z . Since f(−a) = (−1)2f(a) = f(a) and kak = k−ak, we can assume, by permuting indices if necessary, that a1 = kak.

If a1 = 1, then (n−1)/2 1 = kak ≤ (3kFk1) (n−1)/2 since, by deﬁnition of kFk1, kFk1 ≥ 1 and therefore (3kFk1) ≥ 1. Now suppose a1 ≥ 2. Let θ2, . . . , θn be any real numbers. By Corollary 4.2, there is an integer b1 such that 0 < b1 < a1 and integers b2, . . . , bn such that

−1/(n−1) |b1θj − bj| < a1 (2 ≤ j ≤ n − 1) and −1/(n−1) |b1θn − bn| ≤ a1 .

44 Then for 2 ≤ j ≤ n

|bj| = |bj − b1θj + b1θj|

≤ |bj − b1θj| + |b1θj|

= |b1θj − bj| + |b1θj| −1/(n−1) ≤ a1 + |b1θj|

aj Let θj = . Since |aj| ≤ |a1|, |θj| ≤ 1. Then the above inequality gives a1

−1/(n−1) −1/(n−1) |bj| ≤ |b1θj| + a1 ≤ b1 + a1 < b1 + 1 for 2 ≤ j ≤ n. Therefore we have

kbk = b1 < a1 = kak.

Since kak is minimal, this gives that f(b) 6= 0. Now we wish to choose λ, µ ∈ Z such that for a∗ := λa + µb we have f(a∗) = 0. Consider that

f(a∗) = f(λa + µb)

= Bf (λa + µb, λa + µb)

= Bf (λa, λa) + 2Bf (λa, µb) + Bf (µb, µb) 2 2 = λ f(a) + 2λµBf (a, b) + µ f(b) 2 = 2λµBf (a, b) + µ f(b).

Consider that since a, b ∈ Zn, it follows that f(a), f(b), f(a + b) ∈ Z. Therefore, since f(a + b) = f(a) + 2Bf (a, b) + f(b), we have 2Bf (a, b) ∈ Z. Thus, choosing

λ = f(b) ∈ Z and µ = −2Bf (a, b) ∈ Z gives the desired result:

∗ 2 f(a ) = 2λµBf (a, b) + µ f(b) 2 2 = −4Bf (a, b) f(b) + 4Bf (a, b) f(b) = 0.

Now we note that a∗ 6= 0, for we see that a, b are linearly independent since f(a) = 0 while f(b) 6= 0. Hence a∗ = λa + µb = 0 if and only if λ = µ = 0. But λ = f(b) 6= 0, so a∗ 6= 0.

45 b Now deﬁne φ := 1 . Then a1

aj |b1θj − bj| = b1 − bj = |φaj − bj| = |bj − φaj|. a1

Let δj := bj − φaj for 1 ≤ j ≤ n. Then for 2 ≤ j ≤ n − 1

−1/(n−1) |δj| = |bj − φaj| = |b1θj − bj| < a1 and −1/(n−1) |δn| = |bn − φan| = |b1θn − bn| ≤ a1 . −1/(n−1) Also δ1 = b1 − φa1 = 0 and |δ1| ≤ a1 since a1 > 0. We can write bj = φaj + δj for 1 ≤ j ≤ n so that b = φa + δ. Then we have

Bf (a, b) = Bf (a, φa + δ) = Bf (a, φa) + Bf (a, δ) = φf(a) + Bf (a, δ) = Bf (a, δ) and

f(b) = Bf (φa + δ, φa + δ)

= Bf (φa, φa) + 2Bf (φa, δ) + Bf (δ, δ) 2 = φ f(a) + 2φBf (a, δ) + f(δ)

= 2φBf (a, δ) + f(δ).

We now have

a∗ = λa + µb

= f(b)a − 2Bf (a, b)b

= [2φBf (a, δ) + f(δ)]a − 2Bf (a, δ)[φa + δ]

= 2φaBf (a, δ) + af(δ) − 2φaBf (a, δ) − 2δBf (a, δ)

= af(δ) − 2δBf (a, δ).

Then for 1 ≤ k ≤ n

Therefore ∗ 2 ka k ≤ 3kFk1kakkδk

46 and since kak ≤ ka∗k, this gives

2 kak ≤ 3kFk1kakkδk

2 1 ≤ 3kFk1kδk . −1/(n−1) Since |δj| ≤ a1 for 1 ≤ j ≤ n, we have

−1/(n−1) −1/(n−1) kδk ≤ a1 = kak .

Therefore 2 −2/(n−1) 1 ≤ 3kFk1kδk ≤ 3kFk1kak . Then

2/(n−1) kak ≤ 3kFk1 (n−1)/2 kak ≤ (3kFk1) as desired. Note: The estimate in the proof above is rather crude, so it is believed that the constant 3 can be improved. However, the exponent (n − 1)/2 cannot. Here we show why. Let n ≥ 2 be given. Let Λ ⊂ Rn be an n-dimensional lattice of determinant ∆. Suppose there exist C, θ ∈ R>0 such that for every isotropic quadratic form f ∈ R[x1, . . . , xn] with f(Λ) ⊆ Z, there exists some nonzero a ∈ Λ such that f(a) = 0 and θ kaki < C · kFkj ∆, (∗) where i, j ∈ {1, 2, ∞}. We will say (∗) holds for (C, θ) if the inequality above holds for every such lattice Λ and isotropic quadratic form f as above. We will frequently try to show that a given value of θ is the smallest value such that there exists a constant C such that (∗) holds for (C, θ). In this case we say θ is best possible.

Theorem 4.4. Let j ∈ {1, 2, ∞}. Let G be an inﬁnite family of quadratic forms G in R[x1, . . . , xn] with the following two properties:

(i) kGkj is arbitrarily large for g ∈ G and

(ii) There is an n-dimensional lattice Λ ⊂ Rn of determinant ∆ such that g(Λ) ⊆ Z for each g in G.

Let i ∈ {1, 2, ∞}. Suppose there is a positive constant B such that every isotropic vector a of g satisﬁes θ B · kGkj ∆ < kaki for every g ∈ G. Then θ is best possible.

47 Proof. Suppose m < θ and (∗) holds for (D, m). Then for any g ∈ G, there exists an isotropic vector a such that

θ m B · kGkj ∆ < kaki < D · kGkj ∆. Therefore θ m B · kGkj < D · kGkj for every g ∈ G. This gives D kGkθ−m < j B for every g ∈ G, which is a contradiction since θ −m > 0 and kGkj is arbitrarily large for g ∈ G. Hence m ≥ θ and θ is the best possible exponent.

Theorem 4.5. There is an inﬁnite family of quadratic forms F in Z[x1, . . . , xn] with the following properties:

(i) kFk1 is arbitrarily large for f ∈ F,

(ii) f(Zn) ⊆ Z for each f ∈ F, and (iii) for every isotropic vector a ∈ Zn of f,

1 1 n−1 kFk 2 < kak. 2 n(n−1)/2 1

Proof (following [4], p. 87). Let b be a large positive integer (to be determined later) and let n 2 X 2 f(x) = x1 − (xj − bxj−1) j=2 n 2 X 2 2 2 = x1 − (xj − 2bxj−1xj + b xj−1). j=2

Then 2

2 2 kFk1 = |1 − b | + | − 1 − b |(n − 2) + | − 1| + |2b|(n − 1) = |b2 − 1| + |1 + b2|(n − 2) + 1 + 2b(n − 1) = b2 − 1 + (1 + b2)(n − 2) + 1 + 2b(n − 1) = n − 2 + 2b(n − 1) + b2(n − 1).

Suppose there is some nonzero a ∈ Z such that f(a) = 0. Then

n 2 X 2 f(a) = a1 − (aj − baj−1) = 0. j=2 2 Cassels incorrectly calculated kFk1 in [4], leaving off the −2. This, however, does not affect the final result.

48 Suppose a1 = 0. Then n X 2 f(a) = − (aj − baj−1) = 0. j=2 Then aj − baj−1 = 0 2 ≤ j ≤ n.

In particular, this gives a2 − ba1 = 0, which implies a2 = 0. Continuing this process for all i, we see that ai = 0 for all i and therefore a = 0, which is a contradiction. So a1 6= 0. Deﬁne λj = aj − baj−1. Then f(a) = 0 gives n n 2 X 2 X 2 a1 = (aj − baj−1) = λj . j=2 j=2 Note also that

n−2 n−1 λn + λn−1b + ... + λ2b + a1b n−2 n−1 = (an − ban−1) + (an−1 − ban−2)b + ... + (a2 − ba1)b + a1b = an

Now

n−1 n−1 |a1b | = |a1b − an + an| n−1 ≤ |a1b − an| + |an| n−2 = | − λn − λn−1b − · · · − λ2b | + |an| n−2 = |λn + λn−1b + ··· + λ2b | + |an|.

n−2 n−2 Since |λn + λn−1b + ··· + λ2b | ≤ |λn| + |λn−1|b + ··· + |λ2|b , it follows that

n−1 n−2 n−1 n−2 |a1|b −|λn|−|λn−1|b−· · ·−|λ2|b ≤ |a1|b −|λn +λn−1b+···+λ2b | ≤ |an|.

n−1 n−2 Ultimately we will show b − b ≤ |an|. To this end, we see that

n−1 n−2 n−1 n−2 b − b ≤ |a1|b − |λn| − |λn−1|b − · · · − |λ2|b ≤ |an| if and only if

n−1 n−2 n−3 0 ≤ (|a1| − 1)b + (1 − |λ2|)b − |λ3|b − · · · − |λn−1|b − |λn|. (4.1)

Since f(a) = f(−a), we can assume a1 > 0. So either a1 = 1 or a1 ≥ 2. Case 1: a1 = 1. 2 2 Then λ2 + ··· + λn = 1. If |λ2| = 1, then λi = 0 for 3 ≤ i ≤ n. Then eq. (4.1) becomes 0 ≥ 0, which is clearly true. If λ2 = 0, then one λi = 1 and all others are 0. So eq. (4.1) becomes bn−2 − bn−i ≥ 0,

49 which is clearly true, since 2 < i. Case 2: a1 ≥ 2. n−1 n−2 This gives that a1 − 1 ≥ 1. Let h(b) = (a1 − 1)b + cn−2b + ··· + c1b + c0 for any c0, c1, . . . , cn−2 ∈ R. Then limb→∞ h(b) = ∞. Hence for all b suﬃciently large

n−1 n−2 n−3 (|a1| − 1)b + (1 − |λ2|)b − |λ3|b − · · · − |λn−1|b − |λn| ≥ 0.

Then since eq. (4.1) holds in all cases,

n−1 n−2 b − b ≤ |an| ≤ ||a||. (4.2)

Now choose b such that the above inequality and the following three inequalities hold: n − 2 b > , (4.3) 2

b > 2n, (4.4) and b > 2. (4.5) Thus we choose b > 2n and the remaining two inequalities hold. Then by eqs. (4.3) and (4.4) n − 2 + 2b(n − 1) < 2b + 2b(n − 1) = 2bn < b2 So 2 2 2 2 kFk1 = n − 2 + 2b(n − 1) + b (n − 1) < b + b (n − 1) = b n and therefore

2 kFk1 < b n (n−1)/2 (n−1)/2 n−1 kFk1 < n b 1 (n−1)/2 kFk(n−1)/2 · < bn−1 1 n 1 1 1 · kFk(n−1)/2 < bn−1 (4.6) 2 n(n−1)/2 1 2 Now by eq. (4.5),

b > 2 1 b > 1 2 1 b − b > 1 2 1 b − 1 > b 2 1 bn−1 − bn−2 > bn−1. (4.7) 2

50 Combining eqs. (4.2), (4.6) and (4.7), and Theorem 4.3 we have 1 1 1 · kFk(n−1)/2 < bn−1 < bn−1 − bn−2 ≤ kak. 2 n(n−1)/2 1 2

n−1 Therefore, the exponent 2 in Theorem 4.3 is best possible.

51 Chapter 5 Davenport 1956

This section gives a detailed exposition of the results in [5]. Given a quadratic form f(x1, . . . , xm), let A be its associated m × m matrix. We deﬁne

∆(f) := det(A).

Definition 5.1. Let f(x1, . . . , xm) be a positive definite quadratic form with real coefficients. Let L(f) be the minimum value attained by f when evaluated on Zm − {0}. Let ∆(f) be the determinant of f. Define

L(f) A := : f is as above ∆1/m(f) L(f) B := : f is as above, ∆(f) = 1 ∆1/m(f) L(f) C := : f is as above,L(f) = 1 ∆1/m(f) (a) Deﬁne γm := sup A.

(b) Deﬁne 0 γm := sup B. (c) Deﬁne 00 γm := sup C.

0 00 Lemma 5.2. γm = γm = γm

0 Proof. First we show γm = γm. Note that

B ⊂ A. Choose f such that ∆(f) 6= 1. Then

1 1 ∆ · f = · ∆(f) = 1 ∆1/m(f) ∆(f) and 1 L(f) L · f = . ∆1/m(f) ∆1/m(f) Then L 1 · f ∆1/m(f) L(f) = ∈ B. 1/m 1/m 1 ∆ (f) ∆ ∆1/m(f) · f

52 0 00 So A ⊂ B and therefore A = B so γm = γm. Now we show γm = γm. Note ﬁrst that

C ⊂ A.

Choose f such that L(f) 6= 1. Then

1 L(f) L · f = = 1 L(f) L(f) and 1 ∆(f) ∆ · f = . L(f) Lm(f) Then 1 L L(f) · f 1 L(f) = = ∈ C. 1/m 1/m ∆1/m 1 · f ∆(f) ∆ (f) L(f) Lm(f) 00 So A ⊂ C and therefore A = C so γm = γm.

Definition 5.3. Hermite’s constant, γm, is the least upper bound of the minima of positive definite quadratic forms in m variables with real coefficients, evaluated on m 0 Z − {0}, of determinant 1. Note that this corresponds to γm from Definition 5.1.

Note: γ1 = 1. To see this, consider B from Deﬁnition 5.1. There is only one quadratic 2 form in one variable with determinant 1, and that is f = x11. Then L(f) = 1 also so B = {1} and so γ1 = 1. Lemma 5.4. m−1 m γm−1 ≤ γm for m ≥ 2 where γm is Hermite’s constant.

Proof. Let φ(x1, . . . , xm−1) ∈ R[x1, . . . , xm−1] be a positive deﬁnite quadratic form such that ∆(φ) = 1. Suppose L(φ) = a. Then by deﬁnition of γm−1

a ≤ γm−1 m−1 m−1 a ≤ γm−1 .

Now, since γm−1 is the supremum of the set of minima, we can ﬁnd a sequence {φi}, each with ∆(φi) = 1 and minimum ai, such that limi→∞ ai = γm−1 and therefore

m−1 m−1 lim ai = γm−1 . i→∞

2 Let f := φ + axm. Then ∆(f) = a and since φ is positive deﬁnite and L(φ) = a, we m have a > 0. Therefore φ(c1, . . . , cm−1) ≤ f(c1, . . . , cm) for every (c1, . . . , cm) ∈ Z and therefore L(φ) = a ≤ L(f). Since f(0,..., 0, 1) = a, it follows that L(f) = a. Then 1 1 ∆ · f = ∆(f) = 1 a1/m a

53 and 1 a L · f = . a1/m a1/m

Then by deﬁnition of γm, a ≤ γ a1/m m am ≤ γm a m m−1 m a ≤ γm .

2 Since we can apply this reasoning to a sequence {fi} where fi = φi + aixm, we have

m−1 m−1 m γm−1 = lim ai ≤ γm . i→∞

Deﬁnition 5.5. Let Λ be an n-dimensional lattice with basis b1,..., bn. Then, for c1, . . . , cn ∈ Z, v = c1b1 + ··· + cnvn ∈ Λ is primitive if gcd(c1, . . . , cn) = 1. Lemma 5.6. Let Λ ⊂ Rn be an n-dimensional lattice with det(Λ) = ∆. Let a ∈ Λ be primitive. Then there is some z ∈ Λ such that a, z are linearly independent and 1/(n−1) the perpendicular distance from z to a does not exceed γ1/2 ∆ . n−1 kak2

Proof. Let W = Ra. Then Rn = W ⊕W ⊥ by Corollary 2.24. Let π : W ⊕W ⊥ → W ⊥ be the projection deﬁned by the standard inner product. Then Λ0 := π(Λ) is an (n − 1)-dimensional lattice by Lemma 3.29 and W ∩ Λ is a 1-dimensional lattice by Lemma 3.28. Let det(Λ0) = ∆0. By Theorem 3.31, we know that ∆ ∆0 = . det(W ∩ Λ)

Since W ∩ Λ is 1-dimensional, it has a basis b and a = cb with c ∈ Z. Since a is primitive, |c| = 1. Without loss of generality, we can assume c = 1 and therefore a is p T a basis of W ∩ Λ. Then det(W ∩ Λ) = | det(Za)| = det(a a) = kak2 and hence ∆ ∆0 = . kak2

Let f : Rn → R be the quadratic map whose corresponding quadratic form 2 2 with respect to the standard basis is Qf (x1, . . . , xn) = x1 + ··· + xn. It follows easily that Bf (x, y) = x · y, the standard dot product. Let v1,..., vn−1 be a 0 basis of Λ . Then Deﬁnition 2.3 gives that, with respect to this basis, Qf |Λ0 = Pn−1 Pn−1 i=1 j=1 Bf (vi, vj)xixj. Then the associated matrix A = (Bf (vi, vj)) = (vi · vj). Let M be the n × (n − 1) matrix whose columns are v1,..., vn−1 with respect to the standard basis vectors. Then ∆0 = det(Λ0) = pdet(M T M). Therefore

T 0 2 det (Qf |Λ0 ) = det(A) = det(vi · vj) = det M M = (∆ ) .

54 Now by Theorem B.14, we know that there exists some nonzero u = (u1, . . . , un) ∈ Λ0 such that

2 1/(n−1) kuk2 = f(u) = Qf |Λ0 (u1, . . . , un) ≤ γn−1 (det(Qf |Λ0 )) . So

2 0 2 1/(n−1) kuk2 ≤ γn−1((∆ ) ) 2 0 2/(n−1) kuk2 ≤ γn−1(∆ ) 1/2 0 1/(n−1) kuk2 ≤ γn−1(∆ ) 1/(n−1) 1/2 ∆ kuk2 ≤ γn−1 . kak2 Now, there is some z ∈ Λ such that π(z) = u. Since Rn = W ⊕ W ⊥, we can write z = ra + u, where r ∈ R and u ∈ W ⊥. Hence, the perpendicular distance from a to z is kuk2 and we have the result. Lemma 5.7. Let Λ ⊂ Rn be an n-dimensional lattice of determinant ∆. Let a = 2 (a1, . . . , an) ∈ Λ with a 6= 0 such that kak2 is minimal. Then there exists z = (z1, . . . , zn) ∈ Λ such that a, z are linearly independent and

1 X 2 21− n−1 2/(n−1) 0 < (aizj − ajzi) ≤ γn−1 kak2 ∆ . i,j Proof. Here the dot product denotes the standard inner product. For any z ∈ Λ the vector projection of z onto a is z · a proj z = a. a a · a Let c = z·a . Then a · (z − ca) = 0. So the perpendicular distance from z to a is a·a p kz − cak2 = (z − ca) · (z − ca). Now (z − ca) · (z − ca) = z · z − 2ca · z + c2a · a z · a (z · a)2 = z · z − 2 (a · z) + (a · a) a · a (a · a)2 −2(z · a)(a · z) + (a · z)2 = z · z + a · a (a · z)2 = z · z − a · a (z · z)(a · a) − (a · z)2 = . a · a 2 Now, since kak2 is minimal, a must be primitive. Recalling that kz − cak2 is the perpendicular distance from z to a, by Lemma 5.6, we can ﬁnd a z such that kz − 1/2 ∆ 1/(n−1) cak2 ≤ γ ( ) . That is n−1 kak2

r 2 1/(n−1) (z · z)(a · a) − (a · z) 1/2 ∆ ≤ γn−1 . a · a kak2

55 So we see that, by Lemma 2.28, (z · z)(a · a) − (a · z)2 ∆ 2/(n−1) ≤ γn−1 a · a kak2 2 2 2 2 2 2/(n−1) (z1 + ··· + zn)(a1 + ··· + an) − (a1z1 + ··· + anzn) ∆ 2 ≤ γn−1 kak2 kak2 P 2 2/(n−1) 1≤i

akbl + albk = 2Bf (akz − zka, alz − zla).

Proof. Let u := f(z) and v := −2Bf (a, z) so that b = ua + vz. Consider the right hand side, recalling that Bf (a, a) = f(a) = 0 and Bf (z, z) = f(z) = u :

2Bf (akz − zka, alz − zla) = 2(akalBf (z, z) − zkalBf (a, z) − akzlBf (z, a)

+ zkzlBf (a, a)) −v = 2 a a u − (z a + a z ) + z z · 0 k l k l k l 2 k l

= 2akalu + (akzl + alzk)v. Now, consider the left hand side:

akbl + albk = ak(ual + vzl) + al(uak + vzk)

= 2akalu + (akzl + alzk)v.

Pn Pn n Lemma 5.9. Let f(x1, . . . , xn) = i=1 j=1 fijxixj be a quadratic form. Let Λ ⊂ R be an n-dimensional lattice. Suppose a, z ∈ Λ, both nonzero, such that f(a) = 0 and a, z linearly independent. Let b = f(z)a − 2Bf (a, z)z. Suppose f does not vanish at any x = (x1, . . . , xn) ∈ Λ with some xi = 0. Assume ak 6= 0 for all 1 ≤ k ≤ n. Then bk 6= 0 for all 1 ≤ k ≤ n.

56 Proof. Since ak 6= 0 for all 1 ≤ k ≤ n, linear independence of a, z gives

akz − zka 6= 0.

Note that the kth coordinate of akz − zka is akzk − zkak = 0. This implies that f(akz − zka) 6= 0 by our assumption. By setting k = l in Lemma 5.8, we have

2akbk = 2Bf (akz − zka, akz − zka) = 2f(akz − zka) 6= 0.

Since 2ak 6= 0, bk 6= 0 for 1 ≤ k ≤ n. Pn Pn Theorem 5.10. Let f(x1, . . . , xn) = i=1 j=1 fijxixj be a quadratic form with integer coeﬃcients. If f has a nontrivial integral zero, then there is some x ∈ Zn such that f(x) = 0 and

2 n−1 2(n−1)/2 0 < kxk2 ≤ γn−1 2kFk2 where γn−1 is Hermite’s constant, as defined in Definition 5.3. Proof (following [5]). First we show that it is sufficient to prove the result for quadratic forms f such that for all x such that f(x) = 0, xk 6= 0 for all 1 ≤ k ≤ n. Assume the result in this case. Now consider some f with a nontrivial solution x where at least one xk = 0. Consider a truncated form of this quadratic form, 2 2 g(x1, . . . , xr) := f(x1, . . . , xr, 0,..., 0). Then kGk2 ≤ kFk2 so that this inequality, along with our assumption and Lemma 5.4 give

r (r−1) (n−1) 2 X 2 r−1 2 2 n−1 2 2 0 < kxk2 = xi ≤ γr−1 2kGk2 ≤ γn−1 2kFk2 . i=1 Hence we need only consider cases in which nontrivial solutions have no zero entries.

n 2 Assume n > 1. Let a = (a1, . . . , an) ∈ Z , a 6= 0 be such that f(a) = 0 and kak2 is minimal. Let z ∈ Zn be an arbitrary nonzero vector such that z, a are linearly independent. Let tij = aizj − ajzi. Then

n n n X X X Bf (akz − zka, alz − zla) = fij(akzi − zkai)(alzj − zlaj) = fijtkitlj. i=1 j=1 i,j=1

Let u = f(z) and v = −2Bf (a, z). Set b = ua + vz. Then

f(b) = f(ua + vz) 2 2 = u f(a) + 2uvBf (a, z) + v f(z) −v = 0 + 2uv + v2u 2 = 0

57 Lemma 5.9 gives b 6= 0. By Lemma 5.8, we have

n X akbl + albk = 2Bf (akz − zka, alz − zla) = 2 fijtkitlj. i,j=1

Recall det(Zn) = 1. So by Lemma 5.7 there exists some nonzero z ∈ Zn such that when b = ua + vz,

n n n !2 X 2 X X (akbl + albk) = 4 fijtkitlj k,l=1 k,l=1 i,j=1 n n ! n ! X X 2 X 2 2 ≤ 4 fij tkitlj (by Corollary 2.29) k,l=1 i,j=1 i,j=1 n ! n n ! X 2 X X 2 2 = 4 fij tkitlj i,j=1 k,l=1 i,j=1 n ! n ! 2 X 2 X 2 = 4 kFk2 tki tlj k,i=1 l,j=1 n !2 2 X 2 = 4 kFk2 tki k,i=1 1 2 2 21− n−1 ≤ 4 kFk2 γn−1 kak2 (by Lemma 5.7)

2 2 2 22− n−1 = 4 kFk2 γn−1 kak2 2(n−2) 2 2 2 n−1 = 4 kFk2 γn−1 kak2 . (5.1)

Now we also have

n n X 2 X 2 2 2 2 (akbl + albk) = (akbl + 2akblalbk + al bk) k,l=1 k,l=1 n ! n ! n X 2 X 2 X = 2 ak bl + 2 akbkalbl k=1 l=1 k,l=1 n ! n ! 2 2 X X = 2 kak2 kbk2 + 2 akbk albl k=1 l=1 n !2 2 2 X = 2 kak2 kbk2 + 2 akbk (5.2) k=1

58 Combining eqs. (5.1) and (5.2), we have

n !2 2 2 2 2 X 2 kak2 kbk2 ≤ 2 kak2 kbk2 + 2 akbk k=1 n X 2 = (akbl + albk) k,l=1 2(n−2) 2 2 2 n−1 ≤ 4 kFk2 γn−1 kak2 .

Therefore we have 2(n−2) 2 2 2 2 n−1 −1 kbk2 ≤ 2 kFk2 γn−1 kak2 . Recall that b 6= 0 and f(b) = 0. In addition, we can show b ∈ Zn. Since a, z ∈ Zn, it follows f(a), u = f(z), f(a + z) ∈ Z. Hence we have

v = −2Bf (a, z) = f(a) + f(z) − f(a + z) ∈ Z

n 2 2(n−2) n−3 and therefore b = ua + vz ∈ Z . Since kak2 is minimal and n−1 − 1 = n−1 , it follows that n−3 2 2 2 2 2 n−1 kak2 ≤ kbk2 ≤ 2 kFk2 γn−1 kak2 .

n−3 2 2 n−1 Since kak2 6= 0, we divide both sides of the inequality by (kak2) to see that

2 2 n−1 2 2 kak2 ≤ 2 kFk2 γn−1

n−1 2 2 2 n−1 0 < kak2 ≤ 2kFk2 γn−1 as desired.

59 Chapter 6 Watson 1957

This section contains an exposition of reference [23]. Let X f(x) = fijxixj ∈ Z[x1, . . . , xn]. 1≤i≤j≤n

n(n+1) In this section, we let F = (f11, . . . , f1n, f22, . . . , f2n, . . . , fnn) ∈ R 2 . Recall that by Proposition 2.13, we can write

2 2 2 2 f = L1 + ··· + Lr − (Lr+1 + ··· + Lr+s) where L1,...,Lr+s are real, linearly independent linear forms in x1, . . . , xn. In The- orem 4.5, a quadratic form of signature (1, n − 1) was used to prove that Cassels’ n−1 exponent of 2 on kFk1 was best possible. Here, we consider Watson’s results from [23], where he improves the exponent on kFk1 (by way of kFk) for quadratic forms of signature (r, s) 6= (1, n − 1). P Theorem 6.1. Suppose f(x) = 1≤i≤j≤n fijxixj ∈ Z[x1, . . . , xn] is nondegenerate n 2 2 2 2 and isotropic over Z and f(x) = L1 +···+Lr −(Lr+1 +···+Lr+s) where L1,...,Lr+s n are linearly independent linear forms in R[x1, . . . , xn]. Then there exists a ∈ Z such that f(a) = 0 and θ 0 < kak n kFk where r s θ = max 2, , , 2 2 and kn, the constant implied in n, is at least 1.

n−1 Proof (following [23]). First consider n ≤ 5. Then 2 ≤ 2 ≤ θ and

(n−1)/2 kak ≤ (3kFk1) (by Theorem 4.3) 3n(n + 1) (n−1)/2 ≤ kFk (by Lemma 2.33) 2 3n(n + 1)(n−1)/2 = kFk(n−1)/2 2 θ ≤ knkFk

(n−1)/2 3n(n+1) and we take kn = 2 . Now assume n ≥ 6. By way of induction, assume the result for (n − 1)-ary forms. Consider g = f(x1, . . . , xn−1, 0). If g is identically zero, then f = Lxn, with L a linear form. Otherwise, f would not be quadratic. In this case, let a = (0,..., 0, 1, 0). Then f(a) = 0 and the result holds for kn ≥ 1, since kak = 1. Therefore, we can assume that g is not identically zero.

60 0 Now suppose g has a nontrivial solution a = (a1, . . . , an−1). We know that kGk ≤ kFk, r(g) ≤ r(f), and s(g) ≤ s(f). Therefore, θ(g) ≤ θ(f) so we need kGkθ(g) ≤ kFkθ(f). Consider the case when f = g. Then all these inequalities become equalities θ(g) θ(f) and kn−1kGk = kn−1kFk . By the inductive hypothesis,

0 θ(g) 0 < max |aj| = ka k ≤ kn−1kGk . 1≤j≤n−1

Let a = (a1, . . . , an−1, 0). Then

0 max |aj| = ka k = kak = max |aj|. 1≤j≤n−1 1≤j≤n

Since g = f(x1, . . . , xn−1, 0), it follows that f(a) = 0. Therefore

θ(g) θ(f) 0 < kak ≤ kn−1kGk ≤ kn−1kFk .

Letting kn = kn−1 gives the result in this case. Therefore we can assume g is anisotropic. This, along with the fact that g has n − 1 ≥ 5 variables, gives that g is deﬁnite by the Hasse-Minkowski Theorem (see [21, p. 41]). Therefore one of r(g), s(g) is zero. Corollary 2.23 gives that rank(g) = r(g) + s(g). If r(g) + s(g) < n − 1, Corol- lary 2.12 gives that g is isotropic, a contradiction. Hence r(g) + s(g) = n − 1. Now since one of r(g) or s(g) is 0, the other is n − 1. Hence max( r(f), s(f)) ≥ max( r(g), s(g) ) = n − 1. Therefore n − 1 θ(f) ≥ 2 and therefore Theorem 4.3 applied to f completes the proof using the reasoning as (n−1)/2 3n(n+1) in the base case of n ≤ 5 and we take kn = 2 . Theorem 6.2. Deﬁne h = h(f) as the unique integer such that h min(r, s) ≤ max(r, s) < (h + 1) min(r, s).

max(r,s) That is, h = b min(r,s) c. Then the exponent θ in Theorem 6.1 cannot be replaced by h any number smaller than 2 . Proof (following [23]). When (r, s) = (1, n − 1), h = n − 1 and this result is the same as Kneser’s in Theorem 4.5. Now, we can assume r ≤ s, for if s < r we can multiply f by −1, which has no eﬀect on kFk or h. Let c be a large positive integer, to be determined later, and consider

r r X 2 X 2 2 2 f(x) = xi − (cxi − xi+r) + (cxi+r − xi+2r) + ··· + (cxi+hr−r − xi+hr) i=1 i=1 r+s X 2 − xi , i=hr+r+1

61 where the ﬁnal sum on the right is empty if s = hr. Then f has signature (r, s). When h = 1, we have

r r r+s X 2 X 2 X 2 f(x) = xi − (cxi − xi+r) − xi i=1 i=1 i=2r+1 2 2 2 2 2 2 2 = x1 + ··· + xr − c x1 + 2cx1x1+r − x1+r − · · · − c xr + 2cxrx2r 2 2 2 − x2r − x2r+1 − · · · − xr+s.

Therefore, choosing c ≥ 3 gives |1 − c2| > | − 2c| and we see that

kFk = |1 − c2| = |c2 − 1| = c2 − 1.

Now when h ≥ 2, we have the additional terms of

r X 2 2 − (cxi+r − xi+2r) + ··· + (cxi+hr−r − xi+hr) i=1 2 2 2 2 2 2 = −c xi+r + 2cxi+rxi+2r − xi+2r − · · · − c xi+hr−r + 2cxi+hr−rxi+hr − xi+hr,

2 which introduce a coeﬃcient of −1 − c on xi+r. So

kFk = | − 1 − c2| = c2 + 1.

To see that f(x) has nontrivial solutions, we can choose xi for 1 ≤ i ≤ r, not all zero, 2 2 and then let xi+r = (c − 1)xi for 1 ≤ i ≤ r, so that xi − (cxi − xi+r) = 0. Then let j−1 2 xi+jr = c (c − 1)xi so that (cxi+(j−1)r − xi+jr) = 0. Let xi = 0 for hr + r + 1 ≤ i ≤ r + s. Then (x1, . . . , xr+s) is a solution. Choose one of these nontrivial solutions. Since x1, . . . , xr appear in identical terms, we can choose the maximum value in the set {|x1|,..., |xr|} and permute the indices so that |xr| = max{|x1|,..., |xr|}. We can assume xr > 0 (for if it is not, we can choose −x, since f(x) = f(−x)). Then for such a solution,

r r+s 2 X 2 2 X 2 rxr ≥ (cxi − xi+r) + ··· + (cxi+hr−r − xi+hr) + xi i=1 i=hr+r+1 2 2 2 ≥ (cxr − x2r) + (cx2r − x3r) + ··· + (cxhr − xhr+r) .

Choose c such that (c − 1)2 ≥ r. Then

2 2 2 (cxr − xr) = (c − 1) xr 2 ≥ rxr 2 2 2 ≥ (cxr − x2r) + (cx2r − x3r) + ··· + (cxhr − xhr+r) . (6.1)

Now suppose xr > x2r. Then

cxr − x2r > cxr − xr = (c − 1)xr

62 and since xr, c − 1 > 0, both of these values are positive and therefore

2 2 (cxr − x2r) > (cxr − xr) which is a contradiction to eq. (6.1). So xr ≤ x2r.

Now suppose x2r > x3r. Then

cx2r − x3r > cx2r − x2r = (c − 1)x2r ≥ (c − 1)xr = cxr − xr and by similar reasoning as given above,

2 2 (cx2r − x3r) > (cxr − xr) which is again a contradiction to eq. (6.1). So x2r ≤ x3r. Repeating this line of reasoning, we can show xr ≤ x2r ≤ · · · ≤ xhr. Therefore, for 1 ≤ j ≤ h, we have

2 2 2 2 rxjr ≥ rxr ≥ (cxr − x2r) + ··· + (cxhr − xhr+r) which implies

2 2 rxjr ≥ (cxjr − xjr+r) 1/2 r xjr ≥ cxjr − xjr+r.

Choose c such that c > 2r1/2. Then, 1 1 x ≥ cx − r1/2x > cx − cx = cx . jr+r jr jr jr 2 jr 2 jr Therefore,

1 1 2 1 3 1 h kxk ≥ x > cx > c x > c x > ··· > c x . hr+r 2 hr 2 hr−r 2 hr−2r 2 r Case 1: When h = 1, kFk = c2 − 1. Then

h h h h 1 1 1 2 1 2 c = (kFk + 1)1/2 = (kFk + 1) > kFk . 2 2 4 4

Therefore for some nonzero a ∈ Zn with the form described such that f(a) = 0,

h h h 1 1 2 1 2 kak > c a > kFk a ≥ kFk 2 r 4 r 4 since ar is a positive integer. Hence

h 1 2 kFk < kak kFkθ 4 n

63 h and θ cannot be decreased to less than 2 . Case 2: When h ≥ 2, kFk = c2 + 1. Recall that c ≥ 3. Then kFk ≥ 10, so 1 kFk − 1 > 2 kFk. Then

h h 1/2!h h 1 1 1 1 1 2 c = (kFk − 1)1/2 > kFk = kFk . 2 2 2 2 8

Therefore for some nonzero a ∈ Zn with the form described such that f(a) = 0,

h h h 1 1 2 1 2 kak > c a > kFk a ≥ kFk 2 r 8 r 8 since ar is a positive integer. Hence

h 1 2 kFk < kak kFkθ 8 n

h and θ cannot be decreased to less than 2 . We note now that Theorem 6.2 is weak when r, s are nearly equal. To see this, assume r ≤ s and consider the difference between the upper bound of θ given by Cassels, s n−1 r+s−1 b r c 2 = 2 , and the lower bound of θ given in Theorem 6.2, 2 . As r tends to s, s r+s−1−b r c n the difference 2 approaches s − 1, which approaches 2 − 1. This difference is substantial, especially when n is large, indicating that Theorem 6.2 is weak in this case. Here we consider two specific cases where r, s are nearly equal: when (r, s) = (2, 2) and when (r, s) = (2, 3). We follow Watson’s proofs in [23].

We begin with the case (r, s) = (2, 2). Choose two large primes, p, q such that

q < p < 2q, p ≡ 5 (mod 8), and q ≡ 3 (mod 8).

Since p ≡ 5 (mod 8), we know p ≡ 1 (mod 4) and therefore we can ﬁnd integers α, β such that p = α2 + β2. Note that we can also ﬁnd integers γ, δ such that

q = γ2 + 2δ2 since q ≡ 3 (mod 8). Consider the quadratic form

f = p(x2 + y2) − q(z2 + 2t2), which has nontrivial zero x = (x, y, z, t) = (qα, qβ, pγ, pδ). Recalling that kFk = 4 max |fij|, we see that kFk = 2q. Now we see that for any nonzero x = (x, y, z, t) ∈ Z such that f(x) = 0 we have

p(x2 + y2) ≡ 0 (mod q)

64 and since q - p, it follows that x2 + y2 ≡ 0 (mod q)

Assume q - y. Then x2 ≡ −y2 (mod q) x2 ≡ −1 (mod q), y

−1 which is a contradiction since q ≡ 3 (mod 8) implies q = −1, that is −1 is not a quadratic residue modulo q, by [21, p. 7]. Hence

y ≡ 0 (mod q).

Therefore 0 ≡ x2 + y2 ≡ x2 (mod q) and x ≡ 0 (mod q). So we have q2 | x2 and q2 | y2 which gives

x2 + y2 ≡ 0 (mod q2).

Now we also see f(x) = 0 implies

q(z2 + 2t2) ≡ 0 (mod p).

Since p - q, z2 + 2t2 ≡ 0 (mod p). Assume p - t. Then z2 ≡ −2t2 (mod p) z 2 ≡ −2 (mod p). t −2 This is a contradiction since p ≡ 5 (mod 8) implies p = −1, that is −2 is not a quadratic residue modulo p, by [21, p. 7]. So

t ≡ 0 (mod p).

This then gives that 0 ≡ z2 + 2t2 ≡ z2 (mod p) and therefore z ≡ 0 (mod p).

65 Hence p2 | z2 and p2 | t2. So p2 | (z2 + 2t2). And since q(z2 + 2t2) = p(x2 + y2),

p2 | p(x2 + y2) which gives x2 + y2 ≡ 0 (mod p). Therefore x2 + y2 ≡ 0 (mod pq2). So a nontrivial solution x = (x, y, z, t) will have

1 3 x2 + y2 ≥ pq2 > q3 = kFk 2 since kFk = 2q. So we see that

1 3 kFk ≤ x2 + y2 ≤ x2 + y2 + z2 + t2 2

3/2 1 p kFk ≤ x2 + y2 + z2 + t2 ≤ 2 · max{|x|, |y|, |z|, |t|} = 2kxk. 2 Since this holds for any nontrivial integral solution of f, we can ﬁnd a solution such that 1 1 3/2 kFk ≤ kxk ≤ (3kFk)3/2 2 2 3 by Theorem 4.3. Hence, 2 is the smallest possible exponent on kFk. Note that this 1 is better than the exponent of 2 on kFk given by Theorem 6.2. Now we consider the case (r, s) = (2, 3). Choose p a large prime with

p ≡ −1 (mod 8) and m such that 4m < p < 4m+1. Consider the quadratic form

f = x2 + 4my2 − p(z2 + t2 + w2).

x2 + 4my2 − p(z2 + t2 + w2) = 0 Then kFk = p. Now, since this is an indeﬁnite quadratic form in r + s = 2 + 3 = 5 variables, the Hasse-Minkowski Theorem gives that f has a nontrivial solution (see [21, p. 41]). Now for any nonzero x = (x, y, z, t, w) ∈ Z5 such that f(x) = 0, we have x2 + 4my2 ≡ 0 (mod p) so that 4my2 ≡ −x2 (mod p).

66 Assume p - x. Then 2my 2 ≡ −1 (mod p) x

−1 which is a contradiction since p ≡ −1 (mod 8) gives p = −1, again by [21, p. 7]. So x ≡ 0 (mod p). Therefore 0 ≡ x2 + 4my2 ≡ 4my2 (mod p). Since p 6= 2, this implies y ≡ 0 (mod p). Assume x = 0. Write y = pu for some u ∈ Z. Then when f(x) = 0 we can rewrite x2 + 4my2 = p(z2 + t2 + w2) as

4mp2u2 = p(z2 + t2 + w2) 4mpu2 = z2 + t2 + w2.

Let u = 2rs where s is odd. Then

4mpu2 = 4m+rps2.

Let m0 = m + r. Since s is odd, s2 ≡ 1 (mod 8). Therefore

ps2 ≡ (−1)(1) ≡ −1 (mod 8) and we can write ps2 = 8v0 − 1 for v0 ∈ Z. So 4m0 (8v0 − 1) = z2 + t2 + w2, which is insoluble by Legendre’s Three-Square Theorem. Therefore x 6= 0.

Now assume x 6≡ 0 (mod 2m−1). Then x = 2rs, where r ≤ m − 2 and s is odd. Then we can rewrite x2 + 4my2 = p(z2 + t2 + w2) as

4rs2 + 4my2 = p(z2 + t2 + w2) 4r(s2 + 4m−ry2) = p(z2 + t2 + w2).

Since m − r ≥ 2, 8 | 4m−r. We know s2 ≡ 1 (mod 8) since s is odd. So we have

s2 + 4m−ry2 ≡ 1 (mod 8).

Now since p - 4r, p | s2 + 4m−ry2 we can write pj = s2 + 4m−ry2 for j ∈ Z and therefore z2 + t2 + w2 = 4rj.

67 Since p ≡ −1 (mod 8) and pj = s2 + 4m−ry2 ≡ 1 (mod 8), we must have

j ≡ −1 (mod 8).

This contradicts z2 + t2 + w2 = 4rj by Legendre’s Three-Square Theorem. So x ≡ 0 (mod 2m−1). This, combined with x ≡ 0 (mod p), gives that x ≡ 0 (mod 2m−1p). Hence

|x| ≥ 2m−1p.

Recalling that p < 4m+1 we have

p1/2 < 2m+1 1 p1/2 < 2m−1 4 1 p3/2 < 2m−1p. 4 Hence 1 1 |x| ≥ 2m−1p > p3/2 = kFk3/2 4 4 since kFk = p. Hence 1 kFk3/2 < |x| ≤ max{|x|, |y|, |z|, |t|, |w|} = kxk. 4

5−1 3 Therefore, Cassels’ exponent of 2 = 2 cannot be reduced below 2 . Note that this 1 is better than the exponent of 2 on kFk given by Theorem 6.2.

68 Chapter 7 Birch & Davenport 1958

This section contains an exposition of results from [1].

Theorem 7.1. Let n ≥ 3. Let Λ ⊂ Rn be an n-dimensional lattice of determinant ∆, and let n n X X f(x1, . . . , xn) = fijxixj i=1 j=1 be an indeﬁnite quadratic form which takes only integral values at the points x = (x1, . . . , xn) of Λ. Suppose f is isotropic over Λ. Then there is a nonzero point on Λ satisfying both f(x1, . . . , xn) = 0 and

1 2 n−1 2 2 (n−1) 2 0 < kxk2 ≤ γn−1 2kFk2 ∆ .

Proof (following [1]). Let a = (a1, . . . , an) 6= 0 be a point in Λ such that f(a) = 0 2 and kak2 is minimal. Note that this implies a is primitive. By Lemma 5.6, there is 0 0 0 0 a point z ∈ Λ such that the perpendicular distance, d = kz − projaz k, from z to a satisﬁes 1/(n−1) 1 2 ∆ d ≤ γn−1 . kak2 Since the z’s given in Lemma 5.6 and Lemma 5.7 are the same, Lemma 5.7 gives

1 X 0 0 2 21− n−1 2/(n−1) 0 < (aizj − ajzi) ≤ γn−1 kak2 ∆ . i,j

Now, we can shift z0 by an integral multiple of a to ﬁnd another point in Λ. That is, we consider z := z0 + ca ∈ Λ with c ∈ Z. Then 0 0 0 (z + ca) · a a · z projaz = proja(z + ca) = 2 a = 2 + c a. kak2 kak2

a·z0 1 Since c is an integer, we can choose c such that 2 + c ≤ . Then kak2 2 1 kproj zk = kproj (z0 + ca)k ≤ kak . a 2 a 2 2 2

Consider the triangle with sides z, projaz, and z − projaz. Note that

0 0 0 a · z 0 a · z 0 0 kz−projazk2 = z + ca − 2 + c a = z − 2 · a = kz −projaz k2 = d. kak2 2 kak2 2

69 Then

2 2 2 kzk2 = kprojazk2 + kz − projazk2 1 ≤ kak2 + d2 4 2 2/(n−1) 1 2 ∆ ≤ kak2 + γn−1 4 kak2 1 ≤ kak2 + γ ∆2/(n−1) kak2−1/(n−1) . (7.1) 4 2 n−1 2 Now we have two cases: Case 1: f(z) 6= 0. Let u = f(z) and v = Bf (a, z). Then u ∈ Z since z ∈ Λ. Note that v ∈ Z also, since f(a − z) = f(a) − 2Bf (a, z) + f(z) and f(a − z), f(a), and f(z) ∈ Z since a, z ∈ Λ. Deﬁne b = ua + vz. Then b ∈ Λ. Since a, z0 are linearly independent, it follows that a, z are linearly independent. Since u 6= 0, this implies b 6= 0. From here, this case follows exactly from the proof of Theorem 5.10. Note that in that proof, the assumption that f(z) 6= 0 was unnecessary since, under the assumptions of Theorem 5.10, b 6= 0 since f(b) = 0. Also note, since it has been shown that

0 0 d = kz − projazk = kz − projaz k, the result of Lemma 5.7 holds for z as well. 2 Case 2: f(z) = 0. Since z ∈ Λ and z 6= 0, the minimality of kak2 implies

2 2 kak2 ≤ kzk2.

Then by eq. (7.1) we have 1 kak2 ≤ kak2 + γ ∆2/(n−1) kak2−1/(n−1) 2 4 2 n−1 2 3 kak2 ≤ γ ∆2/(n−1) kak2−1/(n−1) 4 2 n−1 2 4 kak2 ≤ γ ∆2/(n−1) kak2−1/(n−1) 2 3 n−1 2 4 kak2n/(n−1) ≤ γ ∆2/(n−1) 2 3 n−1 4(n−1)/n kak2 ≤ γ(n−1)/n∆2/n. 2 3 n−1

Hence if we can show

4(n−1)/n γ(n−1)/n∆2/n ≤ γn−1 2kFk2(n−1)/2 ∆2, (7.2) 3 n−1 n−1 2

70 we have the desired result. Note that eq. (7.2) is true if and only if the following holds: (n−1)/n 4 2 γ(2n−n −1)/n∆2(1−n)/n ≤ 2kFk2(n−1)/2 . 3 n−1 2 2 Raising both sides to the power of n−1 , we have

42/n γ−2(n−1)/n∆−4/n ≤ 2kFk2. (7.3) 3 n−1 2

Let m1,..., mn be a basis of Λ. Then x = X1m1 + ··· + Xnmn, for X1,...,Xn ∈ Z. T Let X = (X1,...,Xn) and M = [m1 ··· mn], the n × n matrix whose columns are m1,..., mn with respect to the standard basis vectors. Then x = MX. Let A = (fij), the matrix associated to f. Then

T T T T f(x1, . . . , xn) = x Ax = (MX) AMX = X M AMX.

Since A is symmetric, (M T AM)T = M T AT M = M T AM. Let M T AM =: P = T (φij). Since P is symmetric, X P X = φ(X1,...,Xn) is a quadratic form. Since φ(X1,...,Xn) = f(x1, . . . , xn) ∈ Z, we see that φ has integral coeﬃcients. For φii = φ(ei) ∈ Z and φij = φ(0,..., 0, 1, 0,..., 0, 1, 0,..., 0)−φii −φjj ∈ Z. Let det(f) and det(φ) denote the determinant of the matrix associated to f, φ, respectively. Therefore

det(φ) = det(P ) = det(M T AM) = det(M)2 · det(A) = ∆2 · det(f).

So we have

The integral coeﬃcients of φ imply that all entries in 2P are integers. Hence

1 ≤ | det(2P )| = 2n| det(P )| = 2n| det(φ)|.

Therefore | det(f)| = ∆−2| det(φ)| ≥ 2−n∆−2. T Now, the columns of A, say fi, are of the form (f1i, . . . , fni) . Then Hadamard’s inequality (Proposition 2.32) gives

n n q Y Y 2 2 | det(f)| ≤ kfik2 = f1i + ··· + fni i=1 i=1

n 2 Y 2 2 2 2 2 2 | det(f)| ≤ (f1i + ··· + fni) = (f11 + ··· + fn1) ··· (f1n + ··· + fnn). i=1

71 And since we have shown ∆−22−n ≤ | det(f)|, we have

−4 −n 2 2 2 2 2 ∆ 4 ≤ | det(f)| ≤ (f11 + ··· + fn1) ··· (f1n + ··· + fnn). Taking the n-th root of both sides and using the inequality of arithmetic and geometric means, we see that q −4/n −1 n 2 2 2 2 ∆ 4 ≤ (f11 + ··· + fn1) ··· (f1n + ··· + fnn) f 2 + ··· + f 2 + ··· + f 2 + ··· + f 2 ≤ 11 n1 1n nn . n Then

−4/n −1 X 2 2 ∆ · 4 n ≤ fij = kFk2 i,j n ∆−4/n · ≤ 2kFk2. (7.4) 2 2 Note that for n ≥ 3, 4 2 n/2 ≤ γn−1 3 n n−1 2 4 n−1 since γ2 = 3 and γn−1 increases as n increases by Lemma 5.4. Then 42/n 2 ≤ γ2(n−1)/n 3 n n−1 42/n 2 γ−2(n−1)/n ≤ 1 3 n n−1 42/n 2 ∆−4/n γ−2(n−1)/n ≤ ∆−4/n 3 n n−1 42/n n ∆−4/n γ−2(n−1)/n ≤ ∆−4/n · . 3 n−1 2 Then by eq. (7.4), we have

42/n ∆−4/n γ−2(n−1)/n ≤ 2kFk2 3 n−1 2 which we have shown in eq. (7.3) is suﬃcient for the result. Theorem 7.1 can be reformulated as follows.

Theorem 7.2. Let h(y1, . . . , yn) be an indefinite quadratic form with integral coefficients, and let g(y1, . . . , yn) be a positive definite form with real coefficients. If the equation h = 0 is properly soluble in the integers, it has a solution satisfying

n−1 −1 2 (n−1)/2 0 < g(y1, . . . , yn) ≤ γn−1 (2Tr (hg ) ) (det(g)), where h and g denote the matrices (hij) and (gij) associated to h, g, respectively, and Tr denotes the trace of a matrix.

72 Theorem 7.1⇒Theorem 7.2: Following the method of Birch and Davenport in [1], to see how to deduce Theorem 7.2 from Theorem 7.1, ﬁrst consider that g positive deﬁnite implies g can be written

2 2 g(y1, . . . , yn) = x1 + ··· + xn where x1, . . . , xn are linearly independent linear forms in y1, . . . , yn. That is

xi = p1iy1 + ··· + pniyn where pij ∈ R. Then we can deﬁne a lattice Λ with basis p1,..., pn where pi = (pi1, . . . , pin). Let P = [p1 ··· pn] be the matrix whose columns are p1,..., pn with T respect to the standard basis vectors. Then x = (x1, . . . , xn) = P y when y = T (y1, . . . , yn) . Deﬁne ∆ := det(Λ) = | det(P )|. Then T T T T g(y1, . . . , yn) = x x = (P y) P y = y P P y. T Hence g = P P . Let det(g) := det(g) = det(gij). Then

det(g) = det(P T P ) = ∆2.

Let f(x1, . . . , xn) be as in Theorem 7.1. Then define h(y) := f(P y). Since f is indefinite, so is h. When y ∈ Zn, we have P y ∈ Λ and f(P y) ∈ Z. Then by Lemma 2.27, h(y1, . . . , yn) has integer coefficients. Since f has a nontrivial solution in Λ, h has a nontrivial integral solution. As in the proof of Theorem 7.1, let A = (fij), the matrix associated to f. Then

h(y) = f(P y) = (P y)T AP y = yT P T AP y.

Therefore h = P T AP . 2 n−1 2 1 (n−1) 2 2 Theorem 7.1 gives kxk2 ≤ γn−1 (2kFk2) 2 ∆ . We have shown ∆ = det(g) and 2 T 2 −1 2 kxk2 = x x = g(x1, . . . , xn). It remains to show that kFk2 = Tr((hg ) ). First, consider

hg−1 = P T AP (P T P )−1 = P T AP P −1(P T )−1 = P T A(P T )−1.

Then (hg−1)2 = P T A(P T )−1P T A(P T )−1 = P T A2(P T )−1. Now, since Tr(BC) =Tr(CB),

Tr((hg−1)2) = Tr(P T A2(P T )−1) = Tr(A2P T (P T )−1) = Tr(A2).

2 Pn Pn Let A = (αij). Then αij = k=1 fikfkj and, in particular, αii = k=1 fikfki = Pn 2 k=1 fik since A is symmetric. Therefore

n n n −1 2 2 X X X 2 2 Tr((hg ) ) = Tr(A ) = αii = fik = kFk2. i=1 i=1 k=1

73 Hence Theorem 7.1 implies Theorem 7.2. Theorem 7.2⇒Theorem 7.1: Now, to see how to deduce Theorem 7.1 from Theo- rem 7.2, ﬁrst consider that Λ has a basis p1,..., pn. If P = [p1 ··· pn] as before, this gives ∆ = det(Λ) = | det(P )|. Then for x ∈ Λ, we can write x = P y where y ∈ Zn. Let A be the n × n matrix associated to f. Then

T T T T f(x1, . . . , xn) = x Ax = (P y) AP y = y P AP y.

Since A is symmetric, it follows that P T AP is symmetric, since (P T AP )T = P T AT P = P T AP . Hence yT P T AP y is a quadratic form, say h(y), with h(y) = f(x) ∈ Z when x = P y ∈ Λ. Hence h has integer coeﬃcients and since f is isotropic over Λ, it follows h is isotropic over Zn. T T T T Now let g(y1, . . . , yn) = y P P y = (P y) P y = x x. Since g(y1, . . . , yn) = T 2 2 x x = x1 + ··· + xn, we have g positive deﬁnite. Then Theorem 7.2 gives that

n−1 n−1 −1 2 2 g(y1, . . . , yn) ≤ γn−1 2Tr(hg ) (det(g)).

T 2 −1 2 2 By deﬁnition, det(g) = det(P P ) = ∆ . The proof that Tr(hg ) = kFk2 is as T 2 above. Since g(y1, . . . , yn) = x x = kxk2, Theorem 7.1 holds.

74 Chapter 8 Davenport 1971

This section provides an exposition on the results in [6]. Details have been added. Pn Pn Theorem 8.1. Suppose the quadratic form f(x1, . . . , xn) = i=1 j=1 fijxixj as- sumes only integral values on the points of Λ and is isotropic over Λ. Then there exist two linearly independent points x, y ∈ Λ such that f(x) = f(y) = 0 and

n−1 2 kxk2kyk2 n kFk ∆ where ∆ = det(Λ).

Proof (following [6]). Let a ∈ Λ, a 6= 0, such that f(a) = 0 and kak2 is minimal. Let W = Ra. Then by Corollary 2.24 we can write Rn = W ⊕ W ⊥ with respect to the standard inner product. Let π : W ⊕ W ⊥ → W ⊥ be a projection map. Then Lemma 3.29 gives that Λ0 := π(Λ) is an (n−1)-dimensional lattice. Since a ∈ W ∩Λ, it follows from Lemma 3.28 that W ∩ Λ is a 1-dimensional lattice. Therefore, it has a basis v. Then there is c ∈ Z such that a = cb. Since kak2 is minimal, |c| = 1. Hence p T p 2 a is a basis of W ∩Λ and therefore det(W ∩Λ) = det(a a) = kak2 = kak2. Then Theorem 3.31 gives ∆ ∆0 = . kak2 0 ⊥ Let p1,..., pn−1 be a set of successive minimal points in Λ ⊂ W , i.e. kpik2 = 0 λi(Λ )2. Then, by deﬁnition,

kp1k2 ≤ · · · ≤ kpn−1k2. (8.1)

Also, Corollary 3.24 gives

0 −1 kp1k2 · · · kpn−1k2 n ∆ = ∆kak2 . (8.2)

⊥ Since π is surjective, each point pv is the projection on W of an inﬁnity of points of n R of the form pv +ra where r ∈ R. Now, there is some r0 ∈ R such that pv +r0a ∈ Λ so that pv + (r0 + t)a ∈ Λ 1 for t ∈ Z. Choose t such that |r0 + t| ≤ 2 and deﬁne

θv := r0 + t and zv := pv + θva. n ⊥ Now since p1,..., pn−1 are linearly independent and R = W ⊕ W , it follows that p1,..., pn−1, a is a linearly independent set. It is straightforward to show z1,..., zn−1, a are linearly independent as well.

75 ⊥f n n Now, consider W = {v ∈ R | Bf (v,W ) = 0} = {v ∈ R | Bf (v, a) = 0}. By Lemma 2.15, dim(W ⊥f ) = dim(Rn) − dim(W ) = n − 1. Note that a ∈ W ⊥f but a ∈/ W ⊥. Since dim(W ⊥) = dim(W ⊥f ) = n − 1, W ⊥ + W ⊥f = Rn. So dim(W ⊥ ∩ W ⊥f ) = dim(W ⊥) + dim(W ⊥f ) − dim(Rn) = n − 2. Therefore, not all of the points ⊥f p1,..., pn−1 can lie in W . Let pr be the ﬁrst that does not. Then

Bf (a, pr) = Bf (a, zr − θra) = Bf (a, zr) − θrf(a) = Bf (a, zr) 6= 0.

Now f(a + zr) = 2Bf (a, zr) + f(zr) by properties of bilinear forms. Since a and zr are in the lattice Λ, f(zr), f(a+zr) ∈ Z. Therefore 2Bf (a, zr) ∈ Z. Deﬁne b := ua + vzr, with u, v ∈ Z. Then b ∈ Λ and

2 2 2 f(b) = u f(a) + 2uvBf (a, zr) + v f(zr) = 2uvBf (a, zr) + v f(zr).

Hence if we let u = f(zr), v = −2Bf (a, zr) we have

2 f(b) = 2uvBf (a, zr) + v f(zr) 2 2 = −4f(zr)Bf (a, zr) + 4f(zr)Bf (a, zr) = 0.

Suppose c1a + c2b = 0. Then

c1a + c2(ua + vzr) = (c1 + c2u)a + c2vzr = 0.

Since a and zr are linearly independent, c1 + c2u = c2v = 0. We know v = −2Bf (a, zr) 6= 0 since Bf (a, zr) 6= 0. So c2 = 0. But this implies c1 = 0. Hence, a and b are linearly independent. Recalling that zr = pr + θra, we can write

b = ua + vzr = ua + v(pr + θra) = (u + vθr)a + vpr.

We can also write

f(pr) = f(zr − θra) 2 = f(zr) − 2Bf (zr, θra) + θr f(a)

= f(zr) − 2θrBf (zr, a)

= u + vθr and −2Bf (a, pr) = −2Bf (a, zr) = v.

76 We have now constructed, given a point a ∈ Λ with f(a) = 0, a point b ∈ Λ with f(b) = 0 such that a and b are linearly independent. Recalling that a and pr are perpendicular with respect to the standard inner product, we have

2 kbk2 = b · b

= [(u + vθr)a + vpr] · [(u + vθr)a + vpr] 2 2 2 2 = (u + vθr) kak2 + 2(u + vθr)v(pr · a) + v kprk2 2 2 2 2 = (u + vθr) kak2 + v kprk2.

Now, since f(pr) = u + vθr, we have

|u + vθr| = |f(pr)|

X = f p p ij ri rj i,j X ≤ |fijpri prj | i,j X ≤ kFk |pri ||prj | i,j n ! n ! X X = kFk |pri | |prj | i=1 j=1 2 ≤ kFk(n kprk) √ 2 ≤ kFk(n n kprk2) 2 n kFkkprk2. (8.3)

Similarly, since v = −2Bf (a, pr),

|v| n kFkkak2kprk2. Therefore,

2 2 2 2 2 kbk = (u + vθr) kak + v kprk 2 4 2 n 2kFk kprk2kak2 so that 2 kbk n kFkkprk2kak2. (8.4)

Let v < r. Recall that pr is the ﬁrst pi such that Bf (a, pr) 6= 0. Therefore, since v < r, we have that Bf (a, pv) = 0. Hence

f(zv) = f(pv + θva) = f(pv) + 2θvBf (pv, a) + f(a) = f(pv).

We have two cases for each pv with v < r: Case 1: f(pv) 6= 0. Since f takes on integer values on Λ and zv ∈ Λ,

|f(pv)| = |f(zv)| ≥ 1.

77 Therefore, by similar reasoning to that used in eq. (8.3),

2 1 ≤ |f(pv)| n kFk kpvk2.

Case 2: f(pv) = 0. Then f(zv) = 0. Now, zv, a ∈ Λ and are linearly independent. Hence we can assume n−1 2 kak2kzvk2 n kFk ∆ , n−1 2 for otherwise we have kak2kzvk2 n kFk ∆ , and we are done. Since zv = pv +θva 1 and |θv| ≤ 2 , 2 kzvk2 = zv · zv

= (pv + θva) · (pv + θva) 2 2 2 = kpvk2 + 2θv(a · pv) + θvkak2 2 2 2 = kpvk2 + θvkak2 1 ≤ kp k2 + kak2. v 2 4 2

Now, by the minimality of kak2 we have kak2 ≤ kzvk2 which, combined with the inequality above, gives 1 1 kz k2 ≤ kp k2 + kak2 ≤ kp k2 + kz k2. v v 4 v 4 v Hence 3 kz k2 ≤ kp k2 4 v 2 v 2 r3 kz k ≤ kp k 4 v 2 v 2

kzvk2 n kpvk2.

n−1 2 Therefore, by our assumption that kak2kzvk2 n kFk ∆ , we have

n−1 2 −1 kpvk2 n kzvk2 n kFk ∆ kak2 . (8.5)

2 We will now show that this implies kFkkpvk n 1, which is true in Case 1 as well. Note ﬁrst that Theorem 7.1 gives that since f is isotropic over Λ, there is some 2 n−1 2 nontrivial x ∈ Λ such that kxk2 n kFk ∆ . Since f(a) = 0 and kak2 is minimal, this gives

2 n−1 2 kak2 n kFk ∆ (n−1)/2 kak2 n kFk ∆ (n−1)/2 −1 1 n kFk ∆kak2 (n−1)/2 n−1 2 −1 kFk ∆ n kFk ∆ kak2 . Combining this with eq. (8.5), we have

n−1 2 −1 (n−1)/2 kpvk2 n kFk ∆ kak2 n kFk ∆.

78 −2/n Now, by eq. (7.4) in the proof of Theorem 7.1, kFk n ∆ . Therefore 2/n −1 ∆ n kFk −n/2 ∆ n kFk . Hence (n−1)/2 (n−1)/2 −n/2 −1/2 kpvk2 n kFk ∆ n kFk · kFk = kFk for all v < r. Therefore

1/2 kFk kpvk2 n 1 2 kFkkpvk2 n 1 as desired. Recall that eq. (8.2) gives

−1 ∆kak2 n kp1k2 · · · kpn−1k2 −1/2 and since we have shown kpvk2 n kFk for all v < r, we can write −1 −1/2 r−1 ∆kak2 n (kFk ) kprk2 · · · kpn−1k2 −1 −1/2 r−1 n−r ∆kak2 n (kFk ) kprk2 (by eq. (8.1)) 1 2 (r−1) −1 n−r kFk ∆kak2 n kprk2 . (8.6) Recall that eq. (8.4) gives

2 kbk2 n kFkkprk2kak2. Then

n−r n−r 2(n−r) n−r kbk2 n kFk kprk2 kak2 n−r n−r n−r 2(n−r) 2(n−r) kak2 kbk2 n kFk kprk2 kak2 n−r 2(n−r) 2 −2 r−1 n kFk kak2 ∆ kak2 kFk (by eq. (8.6)) n−1 2 2(n−r−1) n kFk ∆ kak2 n−1 2 n−1 2 n−r−1 n kFk ∆ (kFk ∆ ) (by Theorem 7.1) n−1 2 (n−1)(n−r−1) 2(n−r−1) n kFk ∆ kFk ∆ . Taking the (n − r)-th root of both sides, we have

n−1 2 (n−1)(n−r−1) 2(n−r−1) kak2kbk2 n kFk n−r ∆ n−r kFk n−r ∆ n−r n−1+(n−1)(n−r−1) 2+2(n−r−1) = kFk n−r ∆ n−r (n−1)(n−r) 2(n−r) = kFk n−r ∆ n−r = kFkn−1∆2. Thus we found a, b ∈ Λ such that a, b are linearly independent, f(a) = 0 = f(b), and n−1 2 kak2kbk2 n kFk ∆ , as desired.

79 Theorem 8.2. Let c1, . . . , cn ∈ Z, not all of the same sign, and suppose the equation

2 2 c1x1 + ··· + cnxn = 0 (8.7) has a nontrivial solution in Zn. Then there exist two linearly independent solutions x and y of eq. (8.7) such that

2 2 2 2 2 (|c1|x1 + ··· + |cn|xn)(|c1|y1 + ··· + |cn|yn) n |c1c2 ··· cn| .

1 1 Proof (following [6]). Let Λ be the lattice with basis (|c1| 2 , 0,..., 0),..., (0,..., 0, |cn| 2 ). Pn Pn Let f(x1, . . . , xn) = i=1 j=1 fijxixj be the quadratic form with ( sgn cj if i = j fij = 0 otherwise.

n Suppose x = (x1, . . . , xn) ∈ Z is a nontrivial solution of eq. (8.7). Then

1 1 1 2 1 2 f(|c1| 2 x1,..., |cn| 2 xn) = sgn c1(|c1| 2 x1) + ··· + sgn cn(|cn| 2 xn) 2 2 = sgn c1|c1|x1 + ··· + sgn cn|cn|xn 2 2 = c1x1 + ··· + cnxn = 0.

1 1 and therefore (|c1| 2 x1,..., |cn| 2 xn) is a solution of f. Now kFk = 1 and Deﬁnition 3.3 gives  1  |c1| 2 0 1  ..  2 ∆ := det(Λ) = det  .  = |c1 ··· cn| . 1 0 |cn| 2 Therefore, since f has a nontrivial solution in Λ and f(Λ) ⊆ Z, Theorem 8.1 gives that we can ﬁnd two linearly independent solutions x, y ∈ Λ of f, such that

n−1 2 kxk2kyk2 n kFk ∆ which is, in this case, q q 1 2 1 2 1 2 1 2 n−1 (|c1| 2 x1) + ··· + (|cn| 2 xn) · (|c1| 2 y1) + ··· + (|cn| 2 yn) n 1 · |c1 ··· cn|.

Squaring both sides and simplifying we have

2 2 2 2 2 (|c1|x1 + ··· + |cn|xn)(|c1|y1 + ··· + |cn|yn) n |c1 ··· cn| , as desired.

80 Chapter 9 Schulze-Pillot 1983

This section provides an exposition of the results in [20]. Many details have been added. n Let Λ ⊂ R be an n-dimensional lattice of determinant ∆. Let f ∈ R[x1, . . . , xn] be an isotropic quadratic form such that f(Λ) ⊆ Z. In Chapters 7 and 8 we showed there exist linearly independent x, y ∈ Λ such that f(x) = f(y) = 0 and

n−1 2 kxki n kFkj ∆ n−1 2 kxkikyki n kFkj ∆ , with i, j ∈ {1, 2, ∞}. In general, there is no such theorem for the product of the norms of n linearly independent isotropic vectors of f for n ≥ 3. That is, the following statement is false: Let Λ ⊂ Rn be an n-dimensional lattice, n ≥ 3, of determinant ∆. Let f ∈ R[x1, . . . , xn] be an isotropic quadratic form such that f(Λ) ⊆ Z. Then there exist constants C, θ, ω ∈ R>0, depending only on n, such that if x1,..., xn ∈ Λ are linearly independent isotropic vectors of f,

θ ω kx1ki · · · kxnki ≤ CkFkj ∆ , with i, j ∈ {1, 2, ∞}. In Theorem 9.1 we will show this statement is false for a speciﬁc lattice Λ, but we will show in Corollary 9.9 that it holds when Λ = Zn.

Theorem 9.1. Let n ≥ 3 and let c, θ, ω ∈ R>0 be constants depending only on n. Then there exists an n-dimensional lattice Λ ⊂ Rn of determinant ∆ and an isotropic Pn Pn quadratic form f(x1, . . . , xn) = i=1 j=1 fijxixj ∈ R[x1, . . . , xn] with f(Λ) ⊆ Z such that for any linearly independent isotropic vectors x1,..., xn ∈ Λ of f,

θ ω kx1k · · · kxnk > c · kFk ∆ . Proof (following [20]). Let

2 2 2 f(x) = x1x2 + x3 + x4 + ··· + xn.

Then kFk = 1. Let k ∈ Z with k > 0 and let Λk be the lattice with basis −1 n k e1, ke2, e3,..., en, where e1,..., en is the standard basis of Z . Then ∆ = −1 det(Λk) = 1. Let x ∈ Λk such that x is nonzero. Then x has the form (a1k , a2k, a3, . . . , an), where a1, . . . , an ∈ Z, and

−1 −1 kxk = max{|a1k |, |a2k|, |a3|,..., |an|} ≥ |a1k |.

−1 −1 If a1 6= 0, then |a1k | ≥ k . If a1 = 0, then some ai 6= 0 for 2 ≤ i ≤ n. Then −1 −1 −1 kxk ≥ |ai| ≥ k since ai ∈ Z and 0 < k ≤ 1. So for all nonzero x ∈ Λk, kxk ≥ k . −1 Now let y ∈ Λk satisfy f(y) = 0 with y 6= 0 and suppose y and k e1 are −1 linearly independent. Then y = (b1k , b2k, b3, . . . , bn) with b1, . . . , bn ∈ Z and not

81 2 2 all b2, . . . , bn equal 0. Suppose b2 = 0. Then f(y) = b3 + ··· + bn = 0, which implies −1 −1 b3 = ··· = bn = 0. Then y = (b1k , 0,..., 0), which contradicts that y and k e1 are linearly independent. Hence b2 6= 0. Then

−1 kyk = max{|k b1|, |b2k|, |b3|,..., |bn|} ≥ |b2k| ≥ k since b2 6= 0. Choose any n linearly independent solutions x1,..., xn of f. Then at most −1 one of these solutions is a multiple of k e1 (noting that such a vector with form −1 −1 (ck , 0,..., 0) is a solution of f = 0). So we may assume x2,..., xn and k e1 are linearly independent. Therefore kxik ≥ k for 2 ≤ i ≤ n. It was shown above that −1 −1 any nonzero lattice element has norm at least k , so kx1k ≥ k . Hence

−1 n−1 n−2 kx1k · · · kxnk ≥ k · k = k .

Then for given constant c we can choose a k ∈ Z such that

n−2 kx1k · · · kxnk ≥ k ≥ c.

Since kFk = ∆ = 1, this gives

θ ω kx1k · · · kxnk ≥ c = c · kFk ∆ .

Theorem 9.1 can be strengthened as follows:

Theorem 9.2. Let n ≥ 3 and let c, θ, ω ∈ R>0 be constants depending only on n. Then there exists an n-dimensional lattice Λ ⊂ Rn of determinant ∆ and an isotropic Pn Pn quadratic form f(x1, . . . , xn) = i=1 j=1 fijxixj ∈ R[x1, . . . , xn] with f(Λ) ⊆ Z such that for any linearly independent isotropic vectors x1,..., x` ∈ Λ of f where 3 ≤ ` ≤ n, θ ω kx1k · · · kx`k > c · kFk ∆ .

Proof. Let f and Λk be as in the proof of Theorem 9.1. It was shown in this proof −1 that for any nonzero x ∈ Λk, we have kxk ≥ k and that for solutions y of f = 0 −1 such that y, k e1 are linearly independent, we have kyk ≥ k. Then by the reasoning in the proof of Theorem 9.1, choosing ` linearly independent solutions x1,..., x` of f = 0 gives −1 `−1 `−2 kx1k · · · kx`k ≥ k · k = k . Then for the given constant c, we can choose a k ∈ Z such that

`−2 kx1k · · · kx`k ≥ k ≥ c, because ` − 2 ≥ 1. Since kFk = ∆ = 1, this gives

θ ω kx1k · · · kx`k > c = c · kFk ∆ .

82 Note that the results above hold for diagonal forms as well, as shown below.

Theorem 9.3. Let n ≥ 3 and let c, θ, ω ∈ R>0 be constants depending only on n. Then there exists an n-dimensional lattice Λ ⊂ Rn of determinant ∆ and a Pn Pn diagonal isotropic quadratic form f(x1, . . . , xn) = i=1 j=1 fijxixj ∈ R[x1, . . . , xn] with f(Λ) ⊆ Z such that for any linearly independent isotropic vectors x1,..., x` ∈ Λ of f where 3 ≤ ` ≤ n, θ ω kx1k · · · kx`k > c · kFk ∆ . Proof. Let 2 2 2 2 f(x) = x1 − x2 + x3 + ··· + xn. Then kFk = 1. Let k ∈ Z, k > 0, and e1,..., en be the standard basis vectors of n −1 −1 Z . Let v1 = k e1 + ke2 and v2 = k e1 − ke2 and let Λk be the lattice with basis v1, v2, e3,..., en. Then ∆ = 2. Let x ∈ Λk such that x is nonzero. Then x has the −1 form (k (a1 + a2), k(a1 − a2), a3, . . . , an) with a1, . . . , an ∈ Z. Then −1 kxk = max{|k (a1 + a2)|, |k(a1 − a2)|, |a3|,..., |an|}.

If a1 + a2 6= 0, then −1 −1 kxk ≥ |k (a1 + a2)| ≥ k .

If a1 + a2 = 0, then a1 = −a2 and hence k(a1 − a2) = 2ka1 = −2ka2. Since x is nonzero, one of a1 . . . , an is not zero, say ai. Then, since ai ∈ Z, −1 kxk ≥ |ai| ≥ k . −1 So for all nonzero x ∈ Λk, kxk ≥ k . Now let y ∈ Λk satisfy f(y) = 0 with y 6= 0 and y, v1 linearly independent. −1 Then y = b1v1 + b2v2 + b3e3 + ··· + bnen = (k (b1 + b2), k(b1 − b2), b3, . . . , bn) and if b1 6= 0, then at least one of b2, . . . , bn is nonzero. Suppose b1 = b2. Then −1 −1 2 2 2 y = (2b1k , 0, b3, . . . , bn) and Q(y) = (2b1k ) + b3 + ··· + bn = 0, which implies b1 = b3 = ··· = bn = 0. Since b1 = b2, this implies b2 = 0 which gives that y = 0, a contradiction. Hence, b1 6= b2. Then −1 kyk = max{|k (b1 + b2)|, |k(b1 − b2)|, |b3|,..., |bn|} ≥ |k(b1 − b2)| ≥ k since b1 − b2 6= 0. Now, choose ` linearly independent solutions x1,..., x` of f(x) = 0. Then at most one of these solutions is a multiple of v1. So we may assume x2,..., x`, v1. Therefore kxik ≥ k for 2 ≤ i ≤ `. It was shown above that any nonzero vector in Λk −1 −1 has norm at least k , so kx1k ≥ k and therefore −1 `−1 `−2 kx1k · · · kx`k ≥ k · k = k . Then for the given constants c, ω, we can choose a k ∈ Z such that `−2 ω kx1k · · · kx`k ≥ k ≥ c · 2 . Since kFk = 1 and ∆ = 2, this gives θ ω kx1k · · · kx`k > c · kFk ∆ .

83 Theorem 9.4. Let L be a lattice in Rn of determinant ∆ and Q(x) = xT Ax a nondegenerate quadratic form on Rn with Q(L) ⊆ Z having a nontrivial solution of Q(x) = 0 in L. Then there exist linearly independent vectors x1,..., xn in L with Q(xi) = 0, for i = 1, . . . , n, and

(n−1)/2 n | det(x1,..., xn)| n | det(A)| ∆ .

Proof (following [20]). Without loss of generality, it can be assumed that A = (aij) is a diagonal matrix with entries ±1. To see this, assume the result in this case, that is, when the matrix associated to the quadratic form has diagonal entries ±1 and hence determinant ± 1. Now, assume the associated matrix, A, is not of this form. Since A is symmetric with entries in R, there exists a diagonal matrix D with entries ±1 and a non-singular matrix F with A = F T DF . It is straightforward to show that FL = {F x | x ∈ L} is a lattice. The form yT Dy with y ∈ FL is isotropic because xT Ax with x ∈ L is isotropic. The form yT Dy is nondegenerate by Corollary 2.23 because A has rank n and D = (F T )−1AF −1 and therefore D has rank n. By Corollary C.3, there exist T y1,..., yn linearly independent vectors in FL such that yi Dyi = 0. Note that

−1 by assumption. Deﬁne xi := F yi for 1 ≤ i ≤ n. It is again straightforward to show x1,..., xn are linearly independent. Consider that

T T −1 T −1 T Q(xi) = xi Axi = yi (F ) AF yi = yi Dyi = 0 and

−1 −1 −1 | det(x1,..., xn)| = | det(F y1,...,F yn)| = | det(F )| | det(y1,..., yn)| −1 n n n | det(F )| · | det(F )| ∆ = | det(F )|n−1∆n.

Now since A = F T DF , det(A) = det(F )2 det(D) and | det(A)| = | det(F )|2. Then

n−1 n (n−1)/2 n | det(x1,..., xn)| n | det(F )| ∆ = | det(A)| ∆ .

Hence, we can assume that A is a diagonal matrix with entries ±1. T Now, let B(x, y) := BQ(x, y) = x Ay. Let ˜x1,..., ˜xs be a maximal set of linearly independent vectors in L such that:

n 1. Q vanishes identically on the subspace of R spanned by ˜x1,..., ˜xs

2. k˜x1k = min{kxk | x ∈ L, x 6= 0,Q(x) = 0}, and

k˜xik = min{kxk | x ∈ L such that ˜x1,..., ˜xi−1, x are linearly independent,

Q(x) = B(x, ˜x1) = ··· = B(x, ˜xi−1) = 0} (i ≥ 2)

84 3. k˜xik < max((i − 1)k˜xi−1k, 1) for i ≥ 2. Because Q is nondegenerate, property 1 gives s ≤ n/2 by Theorem 2.16. If A is written A = (aij), deﬁne |A| := max{|aij|}. Then, by our assumptions on A, |A| = 1 and | det(A)| = 1. Suppose v1,..., vn is a basis of L. Deﬁne the matrix V := [v1 ··· vn], whose columns are these basis vectors with respect to the standard basis vectors. Since Q(L) ⊆ Z, 1 V T AV ⊆ n×n 2Z giving that 1n det(V T AV ) = det(A) det(V )2 ≥ 2 1n 1 det(V )2 ≥ . 2 det(A)

Since v1,..., vn is a basis for L, | det(V )| = ∆ and

1n/2 1 1n/2 ∆ ≥ = 2 | det(A)|1/2 2 so that ∆ n 1. (9.1) By Theorem 7.1 and property 2, we have

(n−1)/2 k˜x1k n ·|A| ∆ = ∆. (9.2)

Now, by way of induction, assume k˜xik n ∆ for 2 ≤ i ≤ s − 1. By property 3,

k˜xi+1k < max(ik˜xik, 1).

If max(ik˜xik, 1) = 1, then k˜xi+1k < 1 n ∆. If max(ik˜xik, 1) = ik˜xik, then

k˜xi+1k < ik˜xik < nk˜xik n n∆ n ∆.

Hence for 1 ≤ i ≤ s, k˜xik n ∆. (9.3)

Now, consider the case when s > 1. By property 3, k˜x2k < max(k˜x1k, 1). If max(k˜x1k, 1) = k˜x1k, then k˜x2k < k˜x1k, which contradicts property 2. Therefore,

max(k˜x1k, 1) = 1 (9.4) and k˜x1k ≤ k˜x2k < 1. n Let W be the subspace of R generated by ˜x1,..., ˜xs. Then by Lemma 3.28, W ∩L is an s-dimensional lattice. Now we show, using property 2, that ˜x1,..., ˜xs are

85 successive minimum vectors of W ∩ L. Let y1,..., ys be successive minimum vectors of W ∩L. It can be assumed that they are linearly independent. Since y1,..., ys ∈ W , it follows that Q(yi) = 0 for 1 ≤ i ≤ s. So, by property 2, ky1k = k˜x1k and for 2 ≤ i ≤ s, Lemma 3.16 gives kyik ≤ k˜xik. Suppose kyik < k˜xik. Since Q(yi) = 0 and B(yi, ˜xj) = 0 for 1 ≤ j ≤ i − 1 (because yi, ˜xj ∈ W ), this contradicts property 2. So kyik = k˜xik and ˜x1,..., ˜xs are successive minimum vectors of W ∩ L. By Theorem 3.26, there is a basis x1,..., xs of W ∩L such that kx1k ≤ · · · ≤ kxsk and k˜xik n kxik n k˜xik (9.5) and | det(W ∩ L)| n kx1k · · · kxsk n | det(W ∩ L)|. (9.6) Let W ⊥ be the orthogonal complement of W with respect to the standard inner product. By Corollary 2.24, Rn = W ⊕ W ⊥. Let π : Rn → W ⊥ be the projection map onto W ⊥. Then by Lemma 3.29 π(L) is a lattice and by Theorem 3.31, | det π(L)| = ∆| det(W ∩ L)|−1.

Again by Theorem 3.26, there exists a basis ps+1,..., pn of π(L) with kps+1k ≤ · · · ≤ kpnk and | det π(L)| n kps+1k · · · kpnk n | det π(L)|. (9.7)

Let zi ∈ L, with s + 1 ≤ i ≤ n, be vectors such that π(zi) = pi. Then zi is of the Ps form zi = j=1 rijxj + pi, with rij ∈ R. Then for s + 1 ≤ i ≤ n and 1 ≤ ` ≤ s

s ! s X X h i B(zi, x`) = B rijxj + pi, x` = rijB(xj, x`) + B(pi, x`) = B(pi, x`). j=1 j=1 (9.8) Case 1: Assume s = 1 and B(pj, x1) 6= 0 for 2 ≤ j ≤ n. Then x1 is as given above (i.e. x1 is the basis of W ∩ L). Let

x2 = Q(z2)x1 − 2B(x1, z2)z2 . .

xn = Q(zn)x1 − 2B(x1, zn)zn.

Then x1,..., xn are the desired solutions. The proof of this is similar to, but simpler than, the proof of Case 2. See the remark at the end of the proof for this case. Case 2: Assume s ≥ 1. Since Q is nondegenerate and vanishes on W , there is some n ⊥ ⊥ vector v ∈ R \W = W \{0} such that B(v,W ) 6= 0. Since ps+1,..., pn span W , ⊥ there is some pj ∈ W such that B(pj,W ) 6= 0 with s + 1 ≤ j ≤ n. Deﬁne the following: t1 := min{s + 1 ≤ j ≤ n | B(pj,W ) 6= 0}

u1 := min{1 ≤ j ≤ s | B(xj, pt1 ) 6= 0}.

Note that u1 exists since x1,..., xs span W . Now for i ≥ 2 and ui−1 > 1 deﬁne:

Wi := Rx1 + ··· + Rxui−1−1,

86 ti := min{ti−1 + 1 ≤ j ≤ n | B(pj,Wi) 6= 0},

ui := min{1 ≤ j ≤ ui−1 − 1 | B(pti , xj) 6= 0}.

First note that the assumption ui−1 > 1 implies W2 = Rx1 + ··· + Rxu1−1 exists. To see why the ti’s exist, first assume ti−1, ui−1, and Wi−1 exist. Since ui−1 exists and ui−1 > 1, Wi is defined. Consider that Wi ⊆ W implies that B(W, Wi) = 0. Hence ⊥ there is some nonzero v ∈ W such that B(v,Wi) 6= 0. However, B(pj,Wi) = 0 for s + 1 ≤ j ≤ ti−1 by definition of t1, . . . , ti−1. Therefore, there is some j such that ti−1 + 1 ≤ j ≤ n and B(pj,Wi) 6= 0. The ui’s exist because x1,..., xui−1−1 span Wi. Finally, define: k := max{i | ui−1 > 1},

tk+1 := n + 1. Note that u1 > u2 > ··· > uk,

t1 < t2 < ··· < tk+1, and W2 ⊃ W3 ⊃ · · · ⊃ Wk.

One sees also that uk = 1, for by deﬁnition of the ui’s, 1 ≤ uk ≤ uk−1 − 1. By deﬁnition of k, uk ≤ 1 and hence uk = 1. Now, by the results in Appendix D, the ti’s satisfy the following:

t1 ≤ n − s + 1, t2 ≤ n − u1 + 2, . . . , tk ≤ n − uk−1 + 2. (9.9)

For s + 1 ≤ i < t1 deﬁne 0 pi := pi + pt1 , and for tj ≤ i < tj+1 deﬁne ( 0 pi if B(pi, xuj ) 6= 0 pi := pi + ptj otherwise.

In the case of s + 1 ≤ i < t1 this gives

0 kpik = kpi + pt1 k ≤ kpik + kpt1 k ≤ kpt1 k + kpt1 k = 2kpt1 k.

0 In the case tj ≤ i < tj+1 and pi = pi this gives 0 kpik = kpik ≤ 2kpik,

0 and in the case tj ≤ i < tj+1 and pi = pi + ptj this gives 0 kpik = kpi + ptj k ≤ kpik + kptj k ≤ kpik + kpik = 2kpik. Hence, in summary, ( 0 2kpt1 k for s + 1 ≤ i < t1 kpik ≤ (9.10) 2kpik for t1 ≤ i ≤ n.

87 0 0 0 0 0 Since each pi ∈ π(L), there is some zi ∈ L such that π(zi) = pi and zi has the form

s 0 X 0 zi = λijxj + pi, j=1

0 0 0 xi = Q(zi)xu1 − 2B(xu1 , zi)zi and for tj ≤ i < tj+1 put

0 0 0 xi = Q(zi)xuj − 2B(xuj , zi)zi.

It remains to show that these xs+1,..., xn, along with the basis x1,..., xs of W ∩ L chosen above, are the desired n linearly independent zeros of Q in L. First note 0 x1,..., xs ∈ L by deﬁnition. For 1 ≤ j ≤ k and s + 1 ≤ i ≤ n, Q(xuj + zi) = 0 0 0 0 0 Q(xuj ) + 2B(xuj , zi) + Q(zi). Since xuj , zi ∈ L, Q(xuj ),Q(zi),Q(xuj + zi) ∈ Z. Hence 0 2B(xuj , zi) ∈ Z and each xi with s + 1 ≤ i ≤ n is an integer linear combination of vectors in L and is therefore itself in L. For s + 1 ≤ i < t1,

0 0 0 Q(xi) = Q Q(zi)xu1 − 2B(xu1 , zi)zi

0 0 0 0 0 0 = Q Q(zi)xu1 + 2B Q(zi)xu1 , −2B(xu1 , zi)zi + Q − 2B(xu1 , zi)zi 0 2 0 0 0 0 2 0 = Q(zi) Q(xu1 ) − 4Q(zi)B(xu1 , zi)B(xu1 , zi) + 4B(xu1 , zi) Q(zi) 0 0 2 0 0 2 = 0 − 4Q(zi)B(xu1 , zi) + 4Q(zi)B(xu1 , zi) = 0.

Similarly, when tj ≤ i < tj+1, Q(xi) = 0. Since x1,..., xs ∈ W , Q(x1) = ··· = Q(xs) = 0. Therefore, Q(xi) = 0 for 1 ≤ i ≤ n. To continue further, consider that xs+1,..., xn can be rewritten as follows, by 0 Ps 0 replacing zi with `=1 λi`x` + pi. For s + 1 ≤ i < t1:

0 0 0 xi = Q(zi)xu1 − 2B(xu1 , zi)zi s ! s !" s # X 0 X 0 X 0 = Q λi`x` + pi xu1 − 2B xu1 , λi`x` + pi λi`x` + pi `=1 `=1 `=1 s ! s ! X X 0 0 = Q λi`x` xu1 + 2B λi`x`, pi xu1 + Q(pi)xu1 `=1 `=1 s !" s # " s # X X 0 0 X 0 − 2B xu1 , λi`x` λi`x` + pi − 2B(xu1 , pi) λi`x` + pi `=1 `=1 `=1 s ! " s # X 0 0 0 X 0 = 0 + 2B λi`x`, pi xu1 + Q(pi)xu1 + 0 − 2B(xu1 , pi) λi`x` + pi `=1 `=1

88 s ! s X 0 0 0 X 0 0 = 2B λi`x`, pi xu1 + Q(pi)xu1 − 2B(xu1 , pi) λi`x` − 2B(xu1 , pi)pi. `=1 `=1 (9.11)

Similarly, for tj ≤ i < tj+1 and 1 ≤ j ≤ k,

s ! s X 0 0 0 X 0 0 xi = 2B λi`x`, pi xuj +Q(pi)xuj −2B(xuj , pi) λi`x` −2B(xuj , pi)pi. (9.12) `=1 `=1

Recall that tk+1 = n + 1 so that this covers exactly up through i = tj+1 − 1 = n + 1 − 1 = n. Rewriting the xi’s in this way makes it easier to see why they are linearly independent. Consider r1, . . . , rn ∈ R. Then if r1x1 + ··· + rnxn = 0, one sees 0 0 0 0 r1x1 + ··· + rsxs + rs+1 xs+1 + 2B(xu1 , ps+1)ps+1 + ··· + rn (xn + 2B(xuk , pn)pn) 0 0 0 0 = rs+1 · 2B(xu1 , ps+1)ps+1 + ··· + rn · 2B(xuk , pn)pn. Now the left-hand side is in W and the right-hand side is in W ⊥. Therefore, since W ∩ W ⊥ = {0}, both sides must be zero. It is straightforward to show, by deﬁnition 0 0 of pi, that in each term on the right-hand side, 2B(xuj , pi) 6= 0. Therefore rs+1 = ··· = rn = 0. Hence the equality is now

r1x1 + ··· + rsxs = 0 and since x1,..., xs are linearly independent, r1 = ··· = rs = 0 and hence x1,..., xn are linearly independent. Consider the matrix (x1,..., xn) whose columns are x1,..., xn with respect to the standard basis vectors. Since each of xs+1,..., xn is a linear combination of x1,..., xs 0 0 and ps+1,..., pn by the work above, elementary column operations give 0 0 0 0 | det(x1,..., xn)| = | det(x1,..., xs, −2B(xu1 , ps+1)ps+1,..., −2B(xuk , pn)pn)|.

For s + 1 ≤ i < t1, 0 0 0 0 −2B(xu1 , pi)pi = −2B(xu1 , pi)pi − 2B(xu1 , pi)pt1 0 = −2B(xu1 , pi)pi − 2B(xu1 , pi)pt1 − 2B(xu1 , pt1 )pt1 0 = −2B(xu1 , pi)pi − 2B(xu1 , pt1 )pt1 , where the last equality uses B(xu1 , pi) = 0 since i < t1. For tj ≤ i < tj+1 there are 0 two cases. If B(pi, xuj ) 6= 0, then pi = pi and therefore 0 0 0 −2B(xuj , pi)pi = −2B(xuj , pi)pi = −2B(xuj , pi)pi.

In particular, this is true when i = tj since B(ptj , xuj ) 6= 0 by deﬁnition of uj and tj. 0 When B(pi, xuj ) = 0, pi = pi + ptj and therefore 0 0 0 0 −2B(xuj , pi)pi = −2B(xuj , pi)pi − 2B(xuj , pi)ptj 0 = −2B(xuj , pi)pi − 2B(xuj , pi)ptj − 2B(xuj , ptj )ptj 0 = −2B(xuj , pi)pi − 2B(xuj , ptj )ptj ,

89 where the last equality uses that B(xuj , pi) = 0 by assumption. Now, because −2B(x , p0 )p0 = −2B(x , p )p (since B(p , x ) 6= 0 and hence p0 = p ) uj tj tj uj tj tj tj uj tj tj for 1 ≤ j ≤ k as described above, this replacement can be made in the matrix. Hence, adding a multiple of this column to any other column is an elementary col- 0 0 0 umn operation. Therefore, columns of the form −2B(xu1 , pi)pi = −2B(xu1 , pi)pi − 0 0 0 2B(xu1 , pt1 )pt1 can be replaced with −2B(xu1 , pi)pi and columns of the form −2B(xuj , pi)pi = 0 0 −2B(xuj , pi)pi − 2B(xuj , ptj )ptj can be replaced with −2B(xuj , pi)pi. Hence,

0 0 | det(x1,..., xn)| = | det(x1,..., xs, −2B(xu1 , ps+1)ps+1,..., −2B(xuk , pn)pn)|.

Consider the matrix N := (x1,..., xs, ps+1,..., pn). Put X := (x1,..., xs) and p T P := (ps+1,..., pn). Then det(N) = det(N N). Since x1,..., xs ∈ W and ⊥ ps+1,..., pn ∈ W ,

T T xi · xj 0 X X 0 N N = = T 0 p` · pm 0 P P for 1 ≤ i, j ≤ s and s + 1 ≤ `, m, ≤ n. Therefore

det(N) = pdet(N T N) = pdet(XT X) det(P T P ) = pdet(XT X) · pdet(P T P )

Since x1,..., xs is a basis of W ∩ L and ps+1,..., pn is a basis of π(L), this gives

det(N) = det(W ∩ L) · det(π(L)) and therefore

0 0 | det(x1,..., xn)| = | det(W ∩ L) det(π(L)) · 2B(xu1 , ps+1) ··· 2B(xuk , pn)| 0 0 = |2B(xu1 , ps+1) ··· 2B(xuk , pn)| det(W ∩ L) · det(π(L)).

Then Theorem 3.31 gives det(π(L)) = ∆ · det(W ∩ L)−1 so that

0 0 | det(x1,..., xn) = |2B(xu1 , ps+1) ··· 2B(xuk , pn)| · ∆. (9.13)

Now for s + 1 ≤ i < t1,

0 0 0 Q(xu1 + zi) = Q(xu1 ) + 2B(xu1 , zi) + Q(zi) s ! X 0 0 = Q(xu1 ) + 2B xu1 , λi`x` + pi + Q(zi) `=1 s ! X 0 0 = Q(xu1 ) + 2B xu1 , λi`x` + 2B(xu1 , pi) + Q(zi) `=1 0 0 = 0 + 0 + 2B(xu1 , pi) + Q(zi).

0 0 0 0 Since xu1 , zi ∈ L, xu1 + zi ∈ L and hence Q(zi),Q(xu1 + zi) ∈ Z. Therefore 0 0 0 2B(xu1 , pi) ∈ Z. Here pi = pi +pt1 so that 2B(xu1 , pi) = 2B(xu1 , pi)+2B(xu1 , pt1 ).

90 Since i < t1, the ﬁrst term is 0 but by deﬁnition of t1, the second term is not. There- 0 fore 2B(xu1 , pi) 6= 0. Hence

0 1 ≤ |2B(xu1 , pi)| n n X X 0 = 2 am`xu p 1m i` m=1 `=1 n n X X ≤ 2 |a | |x | |p0 | m` u1m i` m=1 `=1 n n X X 0 ≤ 2 |A|kxu1 kkpik m=1 `=1 n n X X 0 = 2 kxu1 kkpik m=1 `=1 2 0 ≤ 2n · kxu1 kkpik 0 n kxu1 kkpik.

0 Similarly one can show for tj ≤ i < tj+1 and 1 ≤ j ≤ k, 2B(xuj , pi) ∈ Z and 0 2B(xuj , pi) 6= 0. Hence

0 0 1 ≤ |2B(xuj , pi)| n kxuj kkpik.

Applying these to eq. (9.13) gives

0 0 | det(x1,..., xn)| = |2B(xu1 , ps+1) ··· 2B(xuk , pn)| · ∆ 0 0 0 0 n kxu1 kkps+1k · · · kxu1 kkpt1−1k kxu1 kkpt1 k · · · kxu1 kkpt2−1k kx kkp0 k · · · kx kkp0 k ··· kx kkp0 k · · · kx kkp0 k · ∆ u2 t2 u2 t3−1 uk tk uk n " t1−1 # "t2−1 # t1−1−s Y 0 t2−1−(t1−1) Y 0 = kxu1 k kpik · kxu1 k kpik i=s+1 i=t1

"t3−1 # " n # t3−1−(t2−1) Y 0 n−(tk−1) Y 0 · kxu2 k kpik · · · kxuk k kpik · ∆ i=t2 i=tk

" t1−1 #" n #" k # t1−s−1 Y 0 Y 0 Y tj+1−tj = ∆kxu1 k kpik kpik kxuj k . i=s+1 i=t1 j=1

91 By eq. (9.10),

" t1−1 #" n #" k # t1−s−1 Y Y Y tj+1−tj | det(x1,..., xn)| n ∆kxu1 k 2kpt1 k 2kpik kxuj k i=s+1 i=t1 j=1 " n #" k # n−s t1−s−1 t1−s−1 Y Y tj+1−tj n 2 · ∆kxu1 k kpt1 k kpik kxuj k i=t1 j=1 " n #" k # n t1−s−1 t1−s−1 Y Y tj+1−tj ≤ 2 · ∆kxu1 k kpt1 k kpik kxuj k i=t1 j=1 " n #" k # t1−s−1 t1−s−1 Y Y tj+1−tj n ∆kxu1 k kpt1 k kpik kxuj k . i=t1 j=1

Now, there are two cases: t1 > s + 1 and t1 = s + 1. Case 2a: t1 > s + 1. For this case two lemmas are needed. To preserve the ﬂow of the proof, they will be stated here and proved later.

Lemma 9.5. If t1 > s + 1 holds, then

c1kpik ≥ 1 for s + 1 ≤ i ≤ n where c1 is a positive constant depending on n and |A|. s−1 −1 Lemma 9.6. If t1 > s + 1, then kpt1 k ≤ c2|A| ∆kxu1 k , where c2 is a positive constant depending on n and |A|.

By the proofs of Lemmas 9.5 and 9.6, the constants c1, c2 depend only on n when |A| = 1, as it does here. Thus we can use 1 n kpik for s + 1 ≤ i ≤ n and s−1 −1 −1 kpt1 k n |A| ∆kxu1 k = ∆kxu1 k . Then

" n #" k # t1−s−1 −1t1−s−1 Y Y tj+1−tj | det(x1,..., xn)| n ∆kxu1 k ∆kxu1 k kpik kxuj k i=t1 j=1 " n #" k # t1−s Y Y tj+1−tj = ∆ kpik kxuj k . i=t1 j=1

Now since t1 > s + 1, Lemma 9.5 gives

n t −1 n n Y Y1 Y Y kpik = 1 · kpik n kpik. (9.14)

i=t1 i=s+1 i=t1 i=s+1

Combining Theorem 3.31 and eqs. (9.6) and (9.7) gives

n s Y −1 Y −1 kpik n | det π(L)| = ∆| det(W ∩ L)| n ∆ kxik . i=s+1 i=1

92 Then " s #" k # t1−s Y −1 Y tj+1−tj | det(x1,..., xn)| ≤ ∆ · ∆ kxik kxuj k . i=1 j=1 −1 −1 Since kx1k ≤ · · · ≤ kxsk, it is true that kxsk ≤ · · · ≤ kx1k . Therefore " s #" k # " s #" k # Y −1 Y tj+1−tj Y −1 Y tj+1−tj kxik kxuj k ≤ kx1k kxuj k i=1 j=1 i=1 j=1 k −1s Y tj+1−tj = kx1k kxuj k . j=1

Recalling that uk = 1 and tk+1 = n + 1, this gives k k−1 s −1 Y tj+1−tj n+1−tk−s Y tj+1−tj kx1k kxuj k = kxuk k kxuj k . j=1 j=1

Since uk < uk−1, kxuk k ≤ kxuk−1 k and k−1 k−1 n+1−tk−s Y tj+1−tj n+1−tk−s Y tj+1−tj kxuk k kxuj k ≤ kxuk−1 k kxuj k . j=1 j=1

Since kxuk−1 k occurs tk − tk−1 times in the product, k−1 k−2 n+1−tk−s Y tj+1−tj n−s+1−tk−1 Y tj+1−tj kxuk−1 k kxuj k ≤ kxuk−1 k kxuj k . j=1 j=1 This process continues and ultimately gives

t1−s+1 n−s+1−t2 t2−t1 | det(x1,..., xn)| n ∆ kxu2 k kxu1 k t1−s+1 n−s+1−t2 t2−t1 ≤ ∆ kxu1 k kxu1 k t1−s+1 n−s+1−t1 = ∆ kxu1 k .

Note that the exponent on kxu1 k is positive since t1 < n − s + 1 by eq. (9.9). By eq. (9.5) and eq. (9.3), kxu1 k n k˜xu1 k n ∆. Thus,

t1−s+1 n−s+1−t1 n−2s+2 | det(x1,..., xn)| n ∆ ∆ = ∆ Since 1 ≤ s, the inequality n+2 ≤ n+2s holds and hence n−2s+2 ≤ n. Equation (9.1) n−2s+2 n gives 1 n ∆ and therefore ∆ ≤ ∆ . Hence n | det(x1,..., xn)| n ∆ .

Case 2b: t1 = s + 1. This eliminates two terms of the inequality, giving " n #" k # Y Y tj+1−tj | det(x1,..., xs)| n ∆ kpik kxuj k i=t1 j=1 " n #" k # Y Y tj+1−tj = ∆ kpik kxuj k . i=s+1 j=1

93 Again combining Theorem 3.31 and eqs. (9.6) and (9.7) as in case 2a, one sees

" s #" k # 2 Y −1 Y tj+1−tj | det(x1,..., xs)| n ∆ kxik kxuj k . i=1 j=1 Using the procedure of case 2a results in

2 n−s+1−t1 | det(x1,..., xs)| n ∆ kxu1 k 2 n−2s = ∆ kxu1 k .

Again by eq. (9.5) and eq. (9.3), kxu1 k n k˜xu1 k n ∆, so that 2 n−2s n−2s+2 | det(x1,..., xs)| n ∆ · ∆ = ∆ . Following the logic of case 2a, this gives

n | det(x1,..., xs)| n ∆ .

Remark: Here we consider Case 1 of Theorem 9.4. Recall that in this case s = 1 and B(pj, x1) 6= 0 for j = 2, . . . , n and the claim is that

x1,

x2 = Q(z2)x1 − 2B(x1, z2)z2, . .

xn = Q(zn)x1 − 2B(x1, zn)zn are the desired vectors in L. First note x1 ∈ L by deﬁnition, as it is a basis of W ∩ L. For 2 ≤ j ≤ n, Q(x1 + zj) = Q(x1) + 2B(x1, zj) + Q(zj). Since x1, zj ∈ L, Q(x1),Q(zj),Q(x1 + zj) ∈ Z. Hence 2B(x1, zj) ∈ Z. Therefore x2,..., xn are each integer linear combinations of vectors in L and therefore in L also. Since x1 is a basis of W ∩ L, Q(x1) = 0. For 2 ≤ i ≤ n, Q(xi) = Q Q(zi)x1 − 2B(x1, zi)zi = Q Q(zi)x1 + 2B Q(zi)x1, −2B(x1, zi)zi + Q − 2B(x1, zi)zi 2 2 = Q(zi) Q(x1) − 4Q(zi)B(x1, zi)B(x1, zi) + 4B(x1, zi) Q(zi) 2 2 = 0 − 4Q(zi)B(x1, zi) + 4Q(zi)B(x1, zi) = 0.

To see that x1,..., xn are linearly independent, consider that 2B(x1, zi) = 2B(x1, pi) for 2 ≤ i ≤ n by eq. (9.8) and zi = ri1x1 + pi so that xi can be rewritten

xi = Q(zi)x1 − 2B(x1, zi)zi

= Q(ri1x1 + pi)x1 − 2B(x1, pi)[ri1x1 + pi]

= Q(ri1x1)x1 + 2B(ri1x1, pi)x1 + Q(pi)x1 − 2B(x1, pi)ri1x1 − 2B(x1, pi)pi

= 2B(ri1x1, pi)x1 + Q(pi)x1 − 2B(x1, pi)ri1x1 − 2B(x1, pi)pi

= Q(pi)x1 − 2B(x1, pi)pi. (9.15)

94 Suppose r1x1 + ... + rnxn = 0 for r1, . . . , rn ∈ R. Then

r1x1 + r2 (x2 + 2B(x1, p2)p2) + ··· + rn (xn + 2B(x1, pn)pn)

= r2 · 2B(x1, p2)p2 + ··· + rn · 2B(x1, pn)pn. Now the left-hand side is in W and the right-hand side is in W ⊥. Therefore, since W ∩ W ⊥ = {0}, both sides must be zero. In each term on the right-hand side, 2B(x1, pi) 6= 0. Therefore r2 = ··· = rn = 0. Hence the equality is now

r1x1 = 0 and since x1 6= 0, r1 = 0 and hence x1,..., xn are linearly independent. Consider the matrix (x1,..., xn) whose columns are x1,..., xn with respect to the standard basis vectors. Since each of x2,..., xn is a linear combination of x1 and p2,..., pn by the work above, elementary column operations give

| det(x1,..., xn)| = | det(x1, −2B(x1, p2)p2,..., −2B(x1, pn)pn)|.

Deﬁne the matrix N := (x1, p2,..., pn). Put X := (x1) and P := (p2,..., pn). p T ⊥ Then | det(N)| = det(N N). Since x1 ∈ W and p2,..., pn ∈ W , T T x1 · x1 0 X X 0 N N = = T 0 p` · pm 0 P P for 2 ≤ `, m, ≤ n. Therefore | det(N)| = pdet(N T N) = pdet(XT X) det(P T P ) = pdet(XT X) · pdet(P T P ).

Since p2,..., pn is a basis of π(L) and x1 is a basis of W ∩ L, this gives

| det(x1,..., xn)| = | det(W ∩ L) det π(L) · 2B(x1, p2) ··· 2B(x1, pn)|

= |2B(x1, p2) ··· 2B(x1, pn)| · ∆.

Since B(x1, pi) 6= 0 for 2 ≤ i ≤ n,

1 ≤ |2B(x1, pi)| n n X X = 2 a x p m` 1m i` m=1 `=1 n n X X ≤ 2 |am`| |x1m | |pi` | m=1 `=1 n n X X ≤ 2 |A|kx1kkpik m=1 `=1 n n X X = 2 kx1kkpik m=1 `=1 2 ≤ 2n · kx1kkpik.

95 Then

Equation (9.5) gives kx1k n k˜x1k and eq. (9.3) gives k˜x1k n ∆ so that

2 n−2 | det(x1,..., xn)| n ∆ (∆) = ∆n. The proofs of Lemma 9.5 and Lemma 9.6 are below. In order to easily use these results in later proofs, it will not be assumed that A is a diagonal matrix with entries ±1 along the diagonal. Proof of Lemma 9.5 (following [20]). There is a vector s X zs+1 = λ`˜x` + ps+1 `=1 1 with |λ`| ≤ 2 . To see why the vector zs+1 has this form, consider that π(zs+1) = ps+1. ⊥ So zs+1 must have a ps+1 ∈ π(L) ⊂ W term and the remaining terms must live in W . Since ˜x1,..., ˜xs generate W , we can write any element in W as a linear 1 combination of these vectors. Also, |λ`| ≤ 2 because there are γ` ∈ R such that Ps `=1 γ`˜x` + ps+1 ∈ L. Adding integer multiples of ˜x1,..., ˜xs to this vector result in new vectors in L that project onto ps+1 and hence we can choose d1, . . . , ds ∈ Z such 1 that λ` := γ` + d` and |λ`| ≤ 2 . Since ˜x1,..., ˜xs ∈ W and s+1 < t1, the deﬁnition of t1 gives that B(ps+1, ˜x`) = 0 for 1 ≤ ` ≤ s. Therefore s ! X Q(zs+1) = Q λ`˜x` + ps+1 `=1 s ! s ! X X = Q λ`˜x` + 2B λ`˜x`, ps+1 + Q(ps+1) `=1 `=1

= Q(ps+1).

96 Case 1: Assume Q(ps+1) 6= 0. Since zs+1 ∈ L, Q(zs+1) = Q(ps+1) ∈ Z. Therefore

1 ≤ |Q(ps+1)|

= |B(ps+1, ps+1)| n n X X = a p p ij s+1i s+1j i=1 j=1 n n X X ≤ |aij| |ps+1i | |ps+1j | i=1 j=1 n n X X 2 ≤ |A| kps+1k i=1 j=1 2 2 = n |A|kps+1k , so that

2 2 1 ≤ n · |A| kps+1k 1/2 1 ≤ n · |A| kps+1k 1/2 1/2 1 ≤ n · |A| kps+1k ≤ · · · ≤ n · |A| kpnk.

Case 2: Assume Q(ps+1) = 0. Let S be the set of vectors z ∈ L such that

(i) ˜x1,... ˜xs, z satisfy properties 1 and 2 from the proof of Theorem 9.4, except that z need not be the element of minimal norm in the set in property 2

(ii) kzk ≤ kzs+1k.

To see that zs+1 ∈ S, consider that Q(zs+1) = Q(ps+1) = 0 and

s ! X B(zs+1, ˜xi) = B λ`˜x` + ps+1, ˜xi `=1 s X = B(λ`˜x`, ˜xi) + B(ps+1, ˜xi) `=1

= B(ps+1, ˜xi) = 0 for 1 ≤ i ≤ s since ˜xi ∈ W and s + 1 < t1. Also, ˜x1,..., ˜xs, zs+1 must be linearly independent. To see this, assume r1˜x1+···+rs˜xs+rs+1zs+1 = 0 with r1, . . . , rs+1 ∈ R. Then s ! X r1˜x1 + ··· + rs˜xs + rs+1 λ`˜x` + ps+1 = 0 `=1 s ! X r1˜x1 + ··· + rs˜xs + rs+1 λ`˜x` = −rs+1ps+1. `=1

97 Now the left-hand side is in W and the right-hand side is in W ⊥. Since W ∩W ⊥ = {0}, both sides must be 0. Hence rs+1 = 0. Then the equation becomes r1˜x1+···+rs˜xs = 0 and since ˜x1,..., ˜xs are linearly independent, r1 = ··· = rs = 0. So ˜x1,... ˜xs, zs+1 are linearly independent. Therefore (i) holds. It is clear that (ii) holds. Therefore S 6= ∅. Since L is a lattice (and therefore discrete) and the norms of elements of S are bounded above, |S| < ∞. Therefore one can ﬁnd z0 ∈ S such that kz0k is 0 0 minimal. Then ˜x1,... ˜xs, z satisfy properties 1 and 2. By deﬁnition of s, z cannot 0 0 satisfy property 3 and therefore kz k ≥ max(sk˜xsk, 1). But kzs+1k ≥ kz k by (ii) and therefore kzs+1k ≥ max(sk˜xsk, 1). So

s X kzs+1k = λ`˜x` + ps+1 `=1 s X ≤ |λ`| k˜x`k + kps+1k `=1 s ≤ k˜x k + kp k 2 s s+1 kz k ≤ s+1 + kp k. 2 s+1 Therefore kz k s+1 ≤ kp k 2 s+1 kzs+1k ≤ 2kps+1k.

Combining this with 1 ≤ max(k˜xsk, 1) ≤ kzs+1k gives

1 ≤ kzs+1k ≤ 2kps+1k

1 ≤ 2kps+1k ≤ 2kps+2k ≤ · · · ≤ 2kpnk. Hence 1 ≤ c1kpik for s + 1 ≤ i ≤ n and

( 1/2 n · |A| if Q(ps+1) 6= 0 c1 = . 2 if Q(ps+1) = 0

Proof of Lemma 9.6 (following [20]): Since t1 > s + 1, eq. (9.14) holds. Combining this, as in the proof of Theorem 9.4, with eq. (9.7) and Theorem 3.31, gives

n Y t1−1−s −1 kp`k n c1 · ∆| det(W ∩ L)| . `=t1

98 −1 c8 Qs −1 By eq. (9.6), | det(W ∩ L)| n kxik so that c7 i=1 n s Y t1−1−s Y −1 kp`k n c1 · ∆ kxik (9.16) `=t1 i=1 s n t1−1−s Y −1 Y −1 kpt1 k n c1 · ∆ kxik kp`k . i=1 `=t1+1

Recall that eq. (9.8) gives B(z`, xi) = B(p`, xi) for 1 ≤ i ≤ s and s + 1 ≤ ` ≤ n. Therefore

Q(z` + xi) = Q(z`) + 2B(z`, xi) + Q(xi)

= Q(z`) + 2B(p`, xi) + 0.

Since z` and xi ∈ L, Q(z` + xi),Q(z`) ∈ Z and therefore 2B(p`, xi) ∈ Z. Since

B(ptj , xuj ) 6= 0 for 1 ≤ j ≤ k by deﬁnition of tj, uj,

1 ≤ |2B(ptj , xuj )| n n X X = 2 amrxu pt j m j r m=1 r=1 n n X X ≤ 2 |a | |x | |p | mr uj m tj r m=1 r=1 n n X X ≤ 2 |A| kxuj k kptj k m=1 r=1 2 = 2n · |A| kxuj k kptj k.

Because kx1k ≤ · · · ≤ kxsk and kps+1k ≤ · · · ≤ kpnk, this gives 2 1 ≤ 2n |A| kxik kp`k or −1 −1 2 kxik kp`k ≤ 2n |A| (9.17) for uj ≤ i ≤ s and tj ≤ ` ≤ n and 1 ≤ j ≤ k. Write s n t1−1−s Y −1 Y −1 kpt1 k n c1 · ∆ kxik kp`k i=1 `=t1+1 u1 s n t1−1−s Y −1 Y −1 Y −1 = c1 · ∆ kxik kxik kp`k . i=1 i=u1+1 `=t1+1

Note that the center product has s − u1 terms while the rightmost product has n − t1 terms. Now s − u1 ≤ n − t1 because 1 ≤ u1 implies s − u1 ≤ s − 1 and s − 1 ≤ n − t1 by eq. (9.9). Applying eq. (9.17) gives

u1 n t1−1−s 2 s−u1 s−u1 Y −1 Y −1 kpt1 k n c1 · (2n ) ∆|A| kxik kp`k . i=1 `=t1+1+s−u1

99 Now t1 ≤ n−s+1 from eq. (9.9) implies t1+1+s−u1 ≤ n−s+1+1+s−u1 = n−u1+2. −1 Lemma 9.5 gives kpik ≤ c1 so that

u1 n t1−1−s 2 s−u1 s−u1 Y −1 Y −1 kpt1 k n c1 · (2n ) · ∆|A| kxik kp`k i=1 `=t1+1+s−u1

" u1 # " n # t1−1−s 2 s−u1 s−u1 Y −1 n+1−t1−s Y −1 ≤ c1 · (2n ) · ∆|A| kxik c1 kp`k i=1 `=n−u1+2

" u1 #" n # n−2s 2 s−u1 s−u1 Y −1 Y −1 = c1 · (2n ) · ∆|A| kxik kp`k . i=1 `=n−u1+2 Rewriting the ﬁrst product gives

"u2−1 #"u1−1 # n−2s 2 s−u1 s−u1 Y −1 Y −1 kpt1 k n c1 · (2n ) · ∆|A| kxik kxik i=1 i=u2 " n # −1 Y −1 kxu1 k kp`k .

`=n−u1+2

Now the indices on the factors of the second product are all at least u2 and, since eq. (9.9) gives t2 ≤ n − u1 + 2, all the indices on the factors in the third product are at least t2. Therefore eq. (9.17) can be applied. There are u1 − 1 − u2 + 1 = u1 − u2 factors in the second product and n − n + u1 − 2 + 1 = u1 − 1 factors in the third product. Since u1 − 1 ≥ u1 − u2 (since 1 ≤ u2), applying eq. (9.17) gives

"u2−1 # n−2s 2 s−u1+u1−u2 s−u1+u1−u2 −1 Y −1 kpt1 k n c1 · (2n ) · ∆|A| kxu1 k kxik i=1 " n # Y −1 kp`k .

`=n−u1+2+u1−u2 Repeating this procedure using the inequalities in eq. (9.9) gives   "uk−1−1 # n n−2s 2 s−uk−1 s−uk−1 −1 Y −1 Y −1 kpt1 k n c1 ·(2n ) ·∆|A| kxu1 k kxik  kp`k  .

i=1 `=n+2−uk−1

Recall that uk = 1. Then the terms of the ﬁrst product all have index at least uk and the terms of the second product all have index at least tk ≤ n + 2 − uk−1. The ﬁrst product has uk−1 − 1 terms and the second has n − n − 2 + uk−1 + 1 = uk−1 − 1 terms. Applying eq. (9.17) gives

n−2s 2 s−1 s−1 −1 kpt1 k n c1 · (2n ) · ∆|A| kxu1 k n−2s 2 n−1 s−1 −1 ≤ c1 · (2n ) · |A| ∆kxu1 k n−2s s−1 −1 n c1 · |A| ∆kxu1 k ,

100 so that s−1 −1 kpt1 k n c2|A| ∆kxu1 k where n c2 = max (c1 , 1) . Theorem 9.7. Let L and Q be as in Theorem 9.4. Then there exist linearly independent vectors x1,..., xn in L with Q(xi) = 0 and

(n−1)/2 n−2 n(n−1) n |A| ∆ kx1k · · · kxnk n |A| 2 ∆ . kx1k Proof (following [20]). Without loss of generality, it can be assumed that |A| = 1. To see this, assume the theorem in this case. Now, consider the case in which |A|= 6 1. −1 1/2 1/2 Deﬁne B := |A| · A and consider the lattice |A| L. Let ∆B = det(|A| L) = |A|n/2∆. Then y ∈ |A|1/2L has the form y = |A|1/2x for x ∈ L. Then

yT By = (|A|1/2x)T |A|−1A|A|1/2x = xT |A|1/2|A|−1A|A|1/2x T = x Ax ∈ Z. Now yT By is nondegenerate because xT Ax is, and yT By has a nontrivial solution because xT Ax does. Since |B| = 1, the theorem holds for B and the lattice |A|1/2L. 1/2 Thus there exist linearly independent isotropic vectors y1,... yn ∈ |A| L such that T yi Byi = 0 with 1 ≤ i ≤ n such that

(n−1)/2 n−2 n(n−1) |B| ∆B 2 n ky1k · · · kynk n |B| ∆B , ky1k

1/2 for some positive constant d. Since yi = |A| xi for xi ∈ L, the work above implies each xi is an isotropic vector of Q. Hence

−1/2 −1/2 kx1k · · · kxnk = |A| ky1k · · · |A| kynk −n/2 = |A| ky1k · · · kynk (n−1)/2 n−2 n(n−1) |B| · ∆B −n/2 2 n n |A| · |B| ∆B ky1k n/2 n−2 −n/2 n/2 n |A| ∆ = |A| (|A| ∆) 1/2 |A| kx1k (n−1)/2 n−2 n(n−1) n |A| ∆ = |A| 2 · ∆ , kx1k as desired. Hence, one can assume |A| = 1. Now, let ˜x1,..., ˜xs be as in the proof of Theorem 9.4. Set xi = ˜xi for 1 ≤ i ≤ s. Then x1,..., xs are linearly independent vectors in W ∩ L such that Q(xi) = 0

101 and det(W ∩ L) n kx1k · · · kxsk n det(W ∩ L) as before (by Corollary 3.24 and Theorem 3.26). Deﬁne xs+1,..., xn as in the proof of Theorem 9.4. Recall that in that proof there were two cases. Here, we assume the hypotheses of Case 2. We address Case 1, which is much simpler, in a remark at the end of this proof. In Case 2 of the proof of Theorem 9.4 when s + 1 ≤ i < t1,

2 2 2 2 2 kxik ≤ n |A|(2kpt1 k) kxu1 k + 2n |A| kxu1 k(2kpt1 k) + 2sn |A| kpt1 k kxsk kxu1 k s + 4n2|A|kx k kp k · kx k u1 t1 2 s 2 2 2 = 12n |A|kxu1 k kpt1 k + 4sn |A|kxsk kxu1 k kpt1 k 2 2 3 ≤ 12n |A|kxu1 k kpt1 k + 4n |A|kxsk kxu1 k kpt1 k 3 2 ≤ 12n |A|(kxu1 k kpt1 k + kxsk kxu1 k kpt1 k) 2 n |A|(kxu1 k kpt1 k + kxsk kxu1 k kpt1 k). (9.18)

When tj ≤ i < tj+1 and 1 ≤ j ≤ k,

s ! s 0 0 0 0 X 0 X xi = Q(pi)xuj − 2B(xuj , pi)pi + 2B pi, λi`x` xuj − 2B(xuj , pi) λi`x` `=1 `=1 and similarly 2 kxik n |A|(kxuj k kpik + kxsk kxuj k kpik). (9.19) To continue, the following lemma is necessary. To preserve the ﬂow of this proof, it will be proven later. Lemma 9.8.

2 kxik ≤ c5|A|kxu1 k kpt1 k if s + 1 ≤ i < t1 2 kxik ≤ c6|A|kxuj k kpik if tj ≤ i < tj+1 where c5, c6 are positive constants depending only on n and |A|.

102 By the proof of Lemma 9.8, the constants c5 and c6 depend only on n when |A| = 1, as it does here. Therefore Lemma 9.8 gives

" s #" t −1 #"t −1 # " n # Y Y1 Y2 Y kx1k · · · kxnk = kxik kxik kxik ··· kxik

i=1 i=s+1 i=t1 i=tk

" s #" t1−1 #"t2−1 # Y Y 2 Y 2 n kxik kxu1 k kpt1 k kxu1 k kpik

i=1 i=s+1 i=t1 " n # Y 2 ··· kxuk k kpik

i=tk " s # " k #" n # Y 2 t1−1−s Y tj+1−tj Y 2 = kxik (kxu1 k kpt1 k ) kxuj k kp`k .

i=1 j=1 `=t1

Case 1: Assume t1 > s + 1. Recalling the assumption that |A| = 1, Lemma 9.6 gives kxu1 k kpt1 k n ∆. Therefore

" s # " k #" n # Y t1−1−s t1−1−s Y tj+1−tj Y 2 kx1k · · · kxnk n kxik ∆ kpt1 k kxuj k kp`k .

i=1 j=1 `=t1 By eq. (9.16),

" s # Y t1−1−s t1−1−s kx1k · · · kxnk n kxik ∆ kpt1 k i=1 " k # " s # Y tj+1−tj 2 Y −2 kxuj k ∆ kxik j=1 i=1 " k #" s # t1−s+1 t1−1−s Y tj+1−tj Y −1 = ∆ kpt1 k kxuj k kxik . j=1 i=1 Using the process seen in the proof of Theorem 9.4, one sees

t1−s+1 t1−1−s n+1−s−t1 kx1k · · · kxnk n ∆ kpt1 k kxu1 k . Applying Lemma 9.6 gives

t1−s+1 −1 t1−s−1 n+1−s−t1 kx1k · · · kxnk n ∆ (∆kxu1 k ) kxu1 k t1−s+1 −1 t1−s−1 −1 s+t1−n−1 = ∆ (∆kxu1 k ) (kxu1 k ) n−2s+2 −1 t1−s−1 −1 s+t1−n−1 = ∆ (∆kxu1 k ) (∆kxu1 k ) n−2s+2 −1 2t1−n−2 = ∆ (∆kxu1 k ) .

−1 −1 Recalling that 1 ≤ u1 implies kxu1 k ≤ kx1k , gives

n−2s+2 −1 2t1−n−2 kx1k · · · kxnk n ∆ (∆kx1k ) .

103 Since x1 = ˜x1 and ˜x1 is the vector of minimal norm in L, Theorem 7.1 gives 1 n −1 −1 2t1−n−2 −1 n−2 ∆kx1k (see eq. (9.2)). Therefore, (∆kx1k ) ≤ (∆kx1k ) . By eq. (9.1), n+2−2s n 1 n ∆, so ∆ ≤ ∆ . Therefore

n −1 n−2 kx1k · · · kxnk n ∆ (∆kx1k ) , as desired. Case 2: Assume t1 = s + 1. Then

" s #" k #" n # Y Y tj+1−tj Y 2 kx1k · · · kxnk n kxik kxuj k kp`k

i=1 j=1 `=t1 " s #" k #" n # Y Y tj+1−tj Y 2 = kxik kxuj k kp`k . i=1 j=1 `=s+1 Using the process of case 1,

" k #" s # 2 Y tj+1−tj Y −1 kx1k · · · kxnk n ∆ kxuj k kxik . j=1 i=1 Following the steps in the proof of Theorem 9.4 then gives

2 n+1−s−t1 kx1k · · · kxnk n ∆ kxu1 k 2 n+1−s−(s+1) = ∆ kxu1 k 2 n−2s = ∆ kxu1 k 2 −1 2s−n = ∆ (kxu1 k ) 2 −1 2s−n ≤ ∆ (kx1k ) 2−2s+n −1 2s−n = ∆ (∆kx1k ) .

−1 Again using 1 n ∆ and 1 n ∆kx1k ,

n −1 n−2 kx1k · · · kxnk n ∆ (∆kx1k ) .

Remark: Now we address Case 1 from the proof of Theorem 9.4, when s = 1 and B(pj, x1) 6= 0 for all j with 2 ≤ j ≤ n. Recall from eq. (9.15), that for 2 ≤ i ≤ n,

xi = Q(pi)x1 − 2B(x1, pi)pi. Therefore,

kxik ≤ |Q(pi)| kx1k + |2B(x1, pi)| kpik 2 2 2 2 = n |A|kpik kx1k + 2n |A|kx1k kpik 2 2 = 3n kx1k kpik 2 n kx1k kpik .

104 Now combining this with Theorem 3.31 and eqs. (9.6) and (9.7) gives

n Y 2 kx1k · · · kxnk n kx1k kx1k kpik i=2 n 2 n kx1k det(π(L)) n −12 = kx1k ∆ det(W ∩ L) n −12 n kx1k ∆kx1k 2 −12−n = ∆ kx1k n −12−n = ∆ ∆kx1k .

−1 Since 1 n ∆kx1k by eq. (9.2), we have

n −1n−2 kx1k · · · kxnk n ∆ ∆kx1k .

Proof of lemma 9.8 (following [20]): Consider ﬁrst when s = 1. Then u1 = ··· = uk = 1. For s + 1 ≤ i < t1 using eq. (9.11),

0 0 0 0 0 xi = Q(pi)xu1 − 2B(xu1 , pi)pi + 2B (pi, λi1x1) xu1 − 2λi1B(xu1 , pi)x1 0 0 0 = Q(pi)xu1 − 2B(xu1 , pi)pi. So

0 0 0 kxik ≤ |Q(pi)| kxu1 k + |2B(xu1 , pi)| kpik 2 0 2 2 0 2 ≤ n |A|kpik kxu1 k + 2n |A| kxu1 k kpik 2 2 2 2 ≤ 4n |A|kpt1 k kxu1 k + 8n |A|kxu1 k kpt1 k 2 2 = 12n |A| kxu1 k kpt1 k 2 n |A| kxu1 k kpt1 k . (9.20)

Similarly when tj ≤ i < tj+1 for 1 ≤ j ≤ k, using eq. (9.12) gives

2 kxik n |A| kxuj k kpik . (9.21)

Now consider when s > 1. Since xs = ˜xs, property 3 gives

kxsk = k˜xsk < max ((s − 1)k˜xs−1k, 1) .

By Lemma 3.33, this gives k˜xsk < max ((s − 1)!k˜x1k, (s − 1)!). Since eq. (9.4) gives that k˜x1k < 1, we have kxsk = k˜xsk < (s − 1)! < n!.

Applying this to eq. (9.18) when s + 1 ≤ i < t1 gives

2 kxik n |A|(kxu1 k kpt1 k + n! · kxu1 k kpt1 k) and to eq. (9.19) when tj ≤ i < tj+1 gives

2 kxik n |A|(kxuj k kpik + n! · kxuj k kpik).

105 Case 1: Assume t1 > s + 1. Then Lemma 9.5 gives 1 ≤ c1kpik for s + 1 ≤ i ≤ n. Therefore,

2 2 c1kpik ≤ c1kpik 2 kpik ≤ c1kpik .

Then for s + 1 ≤ i < t1,

2 2 kxik n |A|(kxu1 k kpt1 k + n! · kxu1 k c11kpt1 k ) 2 = (n! · c1 + 1)|A|kxu1 k kpt1 k .

And similarly, when tj ≤ i < tj+1 for 1 ≤ j ≤ k,

2 kxik n (n! · c1 + 1)|A|kxuj k kpik .

Case 2: Assume t1 = s + 1. Then

2 kxs+1k n |A|(kxu1 k kps+1k + n!kxu1 k kps+1k) 2 ≤ n!|A|(kxu1 k kps+1k + kxu1 k kps+1k) 2 n |A|kxu1 k(kps+1k + kps+1k). (9.22)

Let c3 be the constant implied in eq. (9.22). Assume, by way of contradiction, that

1 kps+1k < min , 1 . 2c3|A|

1 If c3|A| > 2 , then 4c3|A| > 2 and ! 1 2 1 kxs+1k < c3|A| kxu1 k + 2c3|A| 2c3|A| 1 1 = kxu1 k + 4c3|A| 2 1 1 < kx k + u1 2 2

= kxu1 k.

1 If c3|A| ≤ 2 ,

2 kxs+1k < c3|A| kxu1 k(1 + 1) 1 ≤ kx k(2) 2 u1

= kxu1 k.

Hence, in either case, kxs+1k < kxu1 k.

106 Now, Q(xs+1) = 0 has been shown. Consider B(xs+1, ˜xi) for 1 ≤ i ≤ u1 − 1. Recall, by assumption, ˜xi = xi for 1 ≤ i ≤ s. By deﬁnition of xs+1 and properties of bilinear forms,

0 0 0 B(xs+1, ˜xi) = B Q(ps+1)xu1 − 2B(xu1 , ps+1)ps+1, ˜xi 0 0 0 = Q(ps+1)B(xu1 , ˜xi) − 2B(xu1 , ps+1)B(ps+1, ˜xi) 0 = −2B(xu1 , ps+1)B(ps+1, ˜xi) = 0

since i < u1 and therefore B(ps+1, ˜xi) = 0. By deﬁnition of the ˜xi’s, kxu1 k < max ((u1 − 1)kxu1 k, 1), so kxs+1k < max((u1 −1)kxu1 k, 1). Then x1,..., xu1−1, xs+1, xu1+1,..., xs satisfy properties 1, 2, and 3, which contradicts the minimality of kxu1 k. Therefore 1 min , 1 ≤ kps+1k. 2c3|A|

Let c4 := max(2c3|A|, 1). Then

1 ≤ c4kps+1k ≤ · · · ≤ c4kpnk.

Therefore

2 2 c4kpik ≤ c4kpik 2 kpik ≤ c4kpik for s+1 ≤ i ≤ n. Since t1 = s+1, apply this to the case tj ≤ i < tj+1 with 1 ≤ j ≤ k where 2 kxik n |A|(kxuj k kpik + n! · kxuj k kpik) to get

2 2 kxik n |A|(kxuj k kpik + n! · kxuj kc4kpik ) 2 ≤ |A|(1 + c4n!)kxuj k kpik .

Hence

2 kxik ≤ c5|A|kxu1 k kpt1 k for s + 1 ≤ i < t1 2 kxik ≤ c6|A|kxuj k kpik for tj ≤ i < tj+1 where c5 = (1 + n! · c1) ( (1 + n! · c1) when t1 > s + 1 c6 = (1 + n! · c4) when t1 = s + 1.

107 Remark: Here it is shown, following the methods of Schulze-Pillot, that the exponents in Theorem 9.7 are best possible. First, consider the lattice Λk and f from the proof of Theorem 9.1 where ∆ = 1 and |A| = 1. It is clear that f is nondegenerate and isotropic. Then Theorem 9.7 gives, for x1,..., xn nontrivial solutions of f,

−1 n−2 kx1k · · · kxnk n (kx1k )

−1 where d is a positive constant depending only on n. Let x1 = (−k , 0,..., 0). Then −1 f(x1) = 0 and kx1k = k . So

n−2 kx1k · · · kxnk n k

m holds for every k ∈ Z>0. Suppose kx1k · · · kxnk n k with m < n−2. It was shown n−2 in the proof of Theorem 9.1 that kx1k · · · kxnk ≥ k . Therefore

n−2 m k ≤ kx1k · · · kxnk n k for all k ∈ Z>0. This is a contradiction, so n − 2 is the smallest exponent such that n−2 (n−1)/2 −1 kx1k · · · kxnk n k and therefore the smallest possible exponent on |A| ∆kx1k . 0 Now, consider the lattice Λk with basis ke1, e2,..., en, where e1,..., en are the standard basis vectors and k ∈ Z>0. Consider also the quadratic form Q(x) = 2 Pn 2 0 0 −x1 + j=2 xj . Then ∆ = det(Λk) = k, |A| = 1, and Q(Λk) ⊆ Z. It is clear that Q is nondegenerate and isotropic. Then Theorem 9.7 gives n−2 n k kx1k · · · kxnk n k , kx1k where d is a positive constant depending only on n and x1,... xn are nontrivial solutions of Q. Let x1 = (k, k, 0,..., 0). Then Q(x1) = 0, kx1k = k and kx1k · · · kxnk n kn. Suppose there is m < n such that

m kx1k · · · kxnk n k .

0 Now, any x ∈ Λk has the form x = (a1 · k, a2, . . . , an) where ai ∈ Z for 1 ≤ i ≤ n. 0 Suppose x ∈ Λk is a nontrivial solution of Q = 0. Then a1 6= 0 for if a1 = 0, 2 2 Q(x) = a2 + ··· an = 0 implies a2 = ··· = an = 0, contradicting that x is a nontrivial solution. So a1 6= 0. Then

kxk = max{|a1 · k|, |a2|,..., |an|} ≥ |a1 · k| ≥ k.

Since x1,..., xn are nontrivial solutions,

n kx1k · · · kxnk ≥ k . Hence n m k ≤ kx1k · · · kxnk n k for every k ∈ Z>0, which is a contradiction. Hence n is the smallest exponent such n that kx1k · · · kxnk n k and therefore the smallest possible exponent on ∆. To show that the exponent on |A| is determined, one can show its exponent depends only on the exponent of ∆. To do this, assume the following:

108 (i) the inequality of Theorem 9.7 is

(n−1)/2 n−2 m n |A| ∆ kx1k · · · kxnk n ·|A| ∆ , kx1k

(ii) and if A is replaced by B = cA for some positive constant c and Λ is replaced −1/2 by c Λ, then if x1,..., xn are isotropic vectors satisfying (i), then y1 ..., yn, −1/2 −1/2 deﬁned yi = c xi, satisfy the theorem for cA and c Λ.

−1/2 Now, consider B = cA for some positive constant c. Deﬁning ∆B := det(c Λ) −n/2 gives ∆B = c ∆. Then for x1,..., xn linearly independent nontrivial solutions of xT Ax satisfying (i),

n/2 kx1k · · · kxnk = c ky1k · · · kynk (n−1)/2 n−2 n/2 m n |B| ∆B n c · |B| ∆B by assumption (ii) ky1k (n−1)/2 −n/2 n−2 n/2 m −n/2 n (c|A|) · c ∆ = c · (c|A|) (c ∆) −1/2 c kx1k (n−1)/2 n−2 m− n(n−1) m n |A| ∆ = c 2 |A| ∆ . kx1k

n(n−1) n(n−1) m− 2 Now, if m > 2 , then limc→0 c = 0. Since the left hand side is ﬁxed n(n−1) and the above holds for all c > 0, this is a contradiction. If m < 2 , then n(n−1) n(n−1) m− 2 limc→∞ c = 0, which is again a contradiction. Therefore, m = 2 . Corollary 9.9. Let L = Zn and the other notations be as before. Then there exist n linearly independent vectors x1,..., xn ∈ Z with Q(xi) = 0 and

2 n −1 kx1k · · · kxnk n |A| 2 .

Proof (following [20]). First note that here it is not assumed that |A| = 1. Since n L = Z , ∆ = 1. Let x1,..., xn be deﬁned as in Theorem 9.7, that is xi = ˜xi for 1 ≤ i ≤ s and xs+1,..., xn deﬁned as in Theorem 9.4. Note that x1,..., xs have the same properties here as in the proof of Theorem 9.7, namely that

det(W ∩ L) n kx1k · · · kxsk n det(W ∩ L).

Corollary 3.24 gives the right-hand inequality directly, since xi = ˜xi. For the left- 0 0 hand side, Theorem 3.26 gives that there is a basis x1,..., xs of W ∩ L such that 0 0 0 kxik n kxik for 1 ≤ i ≤ s and det(W ∩ L) n kx1k · · · kxsk. Therefore det(W ∩ L) n kx1k · · · kxsk. (n−2)/2 Case 1: Assume kx1k > |A| . Then Theorem 9.7 gives

(n−1)/2 n−2 n(n−1) |A| kx1k · · · kxnk n |A| 2 . kx1k

109 Then by assumption,

(n−1)/2 n−2 n(n−1) |A| kx k · · · kx k |A| 2 1 n n |A|(n−2)/2 n(n−1) n−2 = |A| 2 |A| 2 2 n −1 = |A| 2 .

(n−2)/2 n Case 2: Assume kx1k ≤ |A| . Note that since x1 ∈ Z , 1 ≤ kx1k. Hence 1 ≤ |A|(n−2)/2 and therefore 1 ≤ |A|. Also, recall from the proof of Theorem 9.4 that when s > 1, k˜x1k ≤ k˜x2k < 1. Since xi = ˜xi for 1 ≤ i ≤ s, this contradicts 1 ≤ kx1k. Therefore, s = 1 when L = Zn (this was not originally noted in [20]). Since s = 1, eq. (9.20) and eq. (9.21) give

" s #" t −1 #"t −1 # " n # Y Y1 Y2 Y kx1k · · · kxnk = kxik kxik kxik ··· kxik

i=1 i=s+1 i=t1 i=tk

" s #" t1−1 #"t2−1 # Y Y 2 Y 2 n kxik |A| kxu1 k kpt1 k |A| kxu1 k kpik ···

i=1 i=s+1 i=t1 " n # Y 2 ··· |A| kxuk k kpik

i=tk " s # " k # n−s Y 2 t1−1−s Y tj+1−tj = |A| kxik (kxu1 k kpt1 k ) kxuj k i=1 j=1 " n # Y 2 kp`k .

`=t1

Case 2a: Assume t1 > s + 1. Now since 1 ≤ |A| and n ≥ 2 (if n = 1, Q has no 1/2 nontrivial solutions), for the constant given in Lemma 9.5, c1, we have n · |A| ≥ c1. 1/2 Then Lemma 9.5 gives 1 ≤ c1kpik ≤ n · |A| kpik for s + 1 ≤ i ≤ n. Then

n " t1−1 #" n # Y 2 Y 2 Y 2 kp`k = 1 kp`k

`=t1 `=s+1 `=t1

" t1−1 #" n # Y 2 2 Y 2 ≤ n · |A|kp`k kp`k

`=s+1 `=t1 n 2(t1−1−s) t1−1−s Y 2 = n · |A| kp`k `=s+1 n 2n t1−1−s Y 2 ≤ n · |A| kp`k `=s+1 n t1−1−s Y 2 n |A| kp`k . `=s+1

110 By Theorem 3.31 and eqs. (9.6) and (9.7),

n Y 2 t1−1−s 2 kp`k n |A| det(π(L))

`=t1 2 = |A|t1−1−s ∆ det(W ∩ L)−1 s t1−1−s Y −2 n |A| kxik . i=1 Hence

" s # n−s t1−1−s Y 2 t1−1−s kx1k · · · kxnk n |A| · |A| kxik (kxu1 k kpt1 k ) i=1 " k #" s # Y tj+1−tj Y −2 kxuj k kxik j=1 i=1 " s # " k # n+t1−2s−1 Y −1 2 t1−1−s Y tj+1−tj = |A| kxik (kxu1 k kpt1 k ) kxuj k . i=1 j=1

Now, since s = 1, u1 = ··· = uk = 1. By Proposition A.2, we can choose the basis ps+1,..., pn of π(L) such that kp`k n 1 for s + 1 ≤ ` ≤ n. Recall −1 −1 kxsk ≤ · · · ≤ kx1k and tk+1 = n + 1. Hence

" s #" k # n+t1−2s−1 t1−1−s Y −1 Y tj+1−tj kx1k · · · kxnk n |A| kx1k kx1k kx1k i=1 j=1

n+t1−2s−1 t1−1−s −s tk+1−t1 = |A| kx1k · kx1k · kx1k n+t1−2s−1 n−2s = |A| kx1k .

(n−2)/2 By assumption, kxu1 k = kx1k ≤ |A| , therefore

n+t1−2s−1 (n−2)/2 n−2s kx1k · · · kxnk n |A| · (|A| ) .

Using t1 ≤ n − s + 1 and 1 ≤ s gives

n+n−s+1−2s−1 (n−2)/2 n−2s kx1k · · · kxnk n |A| (|A| ) ≤ |A|2n−3(|A|(n−2)/2)n−2 2 (n−2) +2n−3 = |A| 2 2 n −1 = |A| 2 .

Case 2b: Assume t1 = s + 1. Then

" s #" k #" n # n−s Y Y tj+1−tj Y 2 kx1k · · · kxnk n |A| kxik kxuj k kp`k . i=1 j=1 `=s+1

111 Applying Theorem 3.31 and eqs. (9.6) and (9.7) as in case 2a gives

" s #" k # n−s Y Y tj+1−tj 2 kx1k · · · kxnk n |A| kxik kxuj k det(π(L)) i=1 j=1 " s #" k # n−s Y Y tj+1−tj −12 = |A| kxik kxuj k ∆ det(W ∩ L) i=1 j=1 " s #" k # s n−s Y Y tj+1−tj Y −2 ≤ |A| kxik kxuj k kxik i=1 j=1 i=1 " s #" k # n−s Y −1 Y tj+1−tj = |A| kxik kxuj k . i=1 j=1

−1 −1 Again, s = 1 implies u1 = ··· = uk = 1. And kxsk ≤ · · · kx1k so that

" s #" k # n−s Y −1 Y tj+1−tj kx1k · · · kxnk n |A| kx1k kx1k i=1 j=1

n−s −s tk+1−t1 = |A| kx1k kx1k n−s −s n+1−(s+1) = |A| kx1k kx1k n−s n−2s = |A| kx1k .

(n−2)/2 By assumption, kx1k ≤ |A| so that

n−s (n−2)/2n−2s kx1k · · · kxnk n |A| |A| n−2 ≤ |A|n−1 |A|(n−2)/2 2 n −n+1 = |A| 2 2 n −n+1 ≤ |A| 2

Since 2 ≤ n, the inequality −n + 1 ≤ −1 holds and therefore

2 n −1 kx1k · · · kxnk n |A| 2 .

112 Chapter 10 Schlickewei 1985

Theorem 10.1. Let Λ be an n-dimensional lattice in Rn with determinant ∆. Let Pn Pn f(x) = i=1 j=1 fijxixj be a nonzero quadratic form with fij = fji. Assume f(Λ) ⊆ Z. Let d ≥ 1 be maximal such that there exist d linearly independent lattice points x1,..., xd ∈ Λ such that f vanishes on the linear space generated by x1,..., xd. Then there exist d such points such that

n−d 2 kx1k · · · kxdk n kFk2 ∆. (10.1) Proof (following [14]). Let hx, yi = x·y be the standard Euclidean inner product. By assumption, there are w1,..., wd ∈ Λ such that f vanishes on the subspace generated by w1,..., wd. Let Ld be the lattice generated by w1,..., wd. By Theorem 3.27, there are only ﬁnitely many d-dimensional lattices with determinant less than or equal to det(Ld). Hence there is a lattice of minimal determinant such that f vanishes on the lattice and the subspace it generates. Let x1,..., xd ∈ Λ be a lattice basis of the space where f vanishes such that det (hxi, xji) is minimal. Note that by Proposition 3.4, det(hxi, xji) > 0. Deﬁne q A := det(hxi, xji).

Let V be the d-dimensional subspace generated by x1,..., xd ∈ Λ and let Λd be the lattice generated by x1,..., xd. Then det(Λd) = A. By Theorem 3.26, there is a 0 0 basis x1 ,..., xd of Λd such that

0 0 A n kx1 k · · · kxd k n A. (10.2)

⊥ n Let V := {c ∈ R |hxi, ci = 0 for all i with 1 ≤ i ≤ d}. Then by Corollary 2.24, we can write Rn = V ⊕ V ⊥ and see that dim(V ) = d implies that dim(V ⊥) = n − d. Consider the projection map π : Rn → V ⊥ with respect to the standard inner product. Then Lemma 3.29 gives that π(Λ) ⊆ V ⊥ is an (n − d)-dimensional lattice. We will call it Λn−d. Then by Theorem 3.31,

∆ = det(V ∩ Λ) · det(Λn−d).

By Lemma 3.28, V ∩ Λ is a d-dimensional lattice and since Λd ⊆ V ∩ Λ, Lemma 3.5 gives that det(Λd) = c · det(V ∩ Λ) where c is a positive integer. Now, f vanishes on V ∩ Λ, so by minimality of det(Λd), c = 1 and A = det(Λd) = det(V ∩ Λ). Hence ∆ det(Λ ) = . (10.3) n−d A

Let λi(Λn−d) with 1 ≤ i ≤ n − d be the successive minima of Λn−d. Then by Corollary 3.24

1 n−d ! n−d Y √ 1 λi(Λn−d) ≤ n − d · (det(Λn−d)) n−d . i=1

113 Therefore

1 √ 1 n−d n−d λ1(Λn−d) ≤ n − d · (det(Λn−d)) n−d 1 √ ∆ n−d λ (Λ ) ≤ n − d . 1 n−d A

⊥ By Proposition 3.15, there is some nonzero z ∈ Λn−d ⊆ V such that kzk = λ1(Λn−d), so that 1 1 1 √ ∆ n−d √ ∆ n−d ∆ n−d kzk ≤ n − d ≤ n . (10.4) A A n A

Since π surjects onto Λn−d, there is some b ∈ Λ such that π(b) = z. Now, since b ∈ Rn = V ⊕ V ⊥, we write

b = γ1x1 + ··· + γdxd + z, with γ1, . . . , γd ∈ R. Let F be the n × n matrix F = (fij). It is straightforward to show hx,F yi = Bf (x, y). Then by hypothesis, for x, y ∈ V , hx,F yi = 0. Deﬁne Vd+1 := V ⊕ Rb. Now, suppose for every nonzero x ∈ Vd+1, hx,F bi = 0. Then, in particular, hb,F bi = 0 and hxi,F bi = 0 with 1 ≤ i ≤ d. This implies that f vanishes on Vd+1, a (d + 1)-dimensional subspace, which is a contradiction since d is maximal. Hence, there is some nonzero x ∈ Vd+1 such that hx,F bi= 6 0. Now, let Λd+1 be the lattice generated by x1,..., xd, b. Let M := [x1 ··· xd b], the matrix whose columns are x1,..., xd, b with respect to the standard basis vectors. p T Then det(Λd+1) = det(M M) and   hx1, x1i · · · hx1, xdi hx1, bi  . .. . .  T  . . . .  M M =   , hxd, x1i · · · hxd, xdi hxd, bi hb, x1i · · · hb, xdi hb, bi with 1 ≤ i, j ≤ d. Now,

hxi, bi = hxi, γ1x1 + ··· + γdxd + zi = γ1hxi, x1i + ··· + γdhxi, xdi + hxi, zi and

hb, zi = hγ1x1 + ··· + γdxd + z, zi = hγ1x1, zi + ··· + hγdxd, zi + hz, zi = hz, zi.

T Let m1,..., md+1 be the columns of M M. Then

T T md+1 − γ1m1 − · · · − γdmd = (hx1, zi,..., hxd, zi, hb, zi) = (0,..., 0, hz, zi) .

T T Replacing md+1 with (0,..., 0, hz, zi) gives a new matrix N with det(N) = det(M M). Let n1,..., nd+1 be the rows of N. Then

nd+1 − γ1n1 − · · · − γdnd = (hz, x1i,..., hz, xdi, hz, zi) = (0,..., 0, hz, zi).

114 0 Replacing nd+1 with (0,..., 0, hz, zi) gives a new matrix N , with upper left-hand 0 d×d matrix (hxi, xji), last column and row (0,..., 0, hz, zi), and det(N ) = det(N) = det(M T M). Therefore,

p T det(Λd+1) = det(M M) = pdet(N 0) q = hz, zi · det(hxi, xji)

= kzk2 A

n kzkA (by Lemma 2.33) 1 ∆ n−d A (by eq. (10.4)) n A 1− 1 1 = A n−d ∆ n−d .

0 We claim that if we can construct a d-dimensional sublattice Λd in Λd+1 such that f 2 2 0 0 1− n−d n−d vanishes on Λd and det(Λd) n A ∆ kFk2, we have the result. Indeed, if we 0 ﬁnd such a lattice Λd, then by minimality of A = det(Λd), we have

2 2 0 1− n−d n−d A ≤ det(Λd) n A ∆ kFk2 2 2 A n−d n ∆ n−d kFk2 n−d 2 A n ∆kFk2 .

Then eq. (10.2) gives that

n−d 0 0 2 kx1k · · · kxdk n ∆kFk2 , as desired. 0 Now we construct the lattice Λd ⊂ Λd+1. First, recall that f(Λ) ⊆ Z. Then for x, y ∈ Λ, f(x), f(y), f(x + y) ∈ Z and since f(x + y) = f(x) + 2hx,F yi + f(y), it follows that 2hx,F yi ∈ Z. ˜ Deﬁne ˜xi := 4xi for 1 ≤ i ≤ d and b := 4γ1x1 + ··· + 4γdxd + 2z = γ1˜x1 + ˜ ··· + γd˜xd + 2z and let Md+1 be the lattice generated by ˜x1,..., ˜xd, b. Consider ˜ x = a1˜x1 + ··· + ad˜xd + ab ∈ Md+1 where a1, . . . , ad, a ∈ Z. Then

x = a1(4x1) + ··· + ad(4xd) + a(4γ1x1 + ··· + 4γdxd + 2z)

= (4a1 + 4aγ1)x1 + ··· + (4ad + 4aγd)xd + 2az.

Now, consider the vector w = 2a1x1 + ··· + 2adxd + ab ∈ Λ. Plugging in for b gives w = (2a1 + aγ1)x1 + ··· + (2ad + aγd)xd + az. Since w, b ∈ Λ, it follows that

115 2hw,F bi ∈ Z. We see that

* d ! + X 2hw,F bi = 2 (2ai + aγi)xi + az,F b i=1 * d ! d ! !+ X X = 2 (2ai + aγi)xi + az,F γixi + z i=1 i=1 * d + * d !+ X X = 2 (2ai + aγi)xi,F z + 2 az,F γixi + 2haz,F zi i=1 i=1 * d + * d ! + X X = 2 (2ai + aγi)xi,F z + 2 aγixi ,F z + 2haz,F zi i=1 i=1 * d + X = 2 (2ai + 2aγi)xi,F z + 2haz,F zi i=1 * d ! + X = (4ai + 4aγi)xi + 2az,F z i=1 = hx,F zi, with x ∈ Md+1. Since 2hw,F bi ∈ Z, it follows that hx,F zi ∈ Z when x ∈ Md+1. ˜ Suppose for every u ∈ Md+1, we have hu,F zi = 0. Then h˜xi,F zi = hb,F zi = 0 for 1 ≤ i ≤ d. Then ˜ h˜xi,F bi = h˜xi,F (4γ1x1 + ··· + 4γdxd + 2z)i

= h˜xi,F (2z)i

= 2h˜xi,F zi = 0 and ˜ ˜ ˜ hb,F bi = h4γ1x1 + ··· + 4γdxd + 2z,F bi ˜ ˜ ˜ = γ1h˜x1,F bi + ··· + γdh˜xd,F bi + h2z,F bi = h2z,F b˜i = 2hz,F b˜i = 0.

It is clear that f vanishes on ˜x1,..., ˜xd by deﬁnition of ˜x1,... ˜xd. We have just shown ˜ it vanishes on b. It follows that f vanishes on Md+1, which is a (d + 1)-dimensional lattice, and therefore a contradiction. Hence there is some u0 ∈ Md+1 such that hu0,F zi= 6 0. Consider the linear map Md+1 → Z such that x 7→ hx,F zi. Let α ∈ Z be the smallest positive value of hx,F zi. Suppose there is v ∈ Md+1 such that hv,F zi = β and α - β. There are u, v ∈ Z such that uα + vβ = gcd(α, β). Then for u ∈ Md+1

116 with hu,F zi = α, uu + vv ∈ Md+1, and huu + vv,F zi = uhu,F zi + vhv,F zi = uα + vβ = gcd(α, β) < α since α - β, which is a contradiction. Therefore, α | hx,F zi for x ∈ Md+1 and there 1 1 is a linear map Md+1 → Z where x 7→ x, α F z . In particular, u 7→ hu, α F zi = 1. Let K be the kernel of this map and consider K ∩ Md+1. By Lemma 3.34, K contains d linearly independent elements of Md+1. Hence by Lemma 3.28, K ∩ Md+1 is a d-dimensional lattice, which we will call Md. Let W be the d-dimensional subspace 1 0 spanned by Md. Then for x ∈ W , x, α F z = 0. Let W be the (d + 1)-dimensional 0 ∼ d+1 space spanned by Md+1. Then W = R and by Corollary 2.24 we can write 0 ⊥ 0 W = W ⊕ W . Since u ∈ Md+1 ⊂ W , it can be written u = u1 + u2, with u1 ∈ W ⊥ ⊥ ⊥ and u2 ∈ W . Let π : W ⊕ W → W be the projection map with respect to the standard inner product h , i. Then π(u) = u2. Now by Corollary 3.30, π(Md+1) has dimension (d + 1) − d = 1. Let m be a basis of π(Md+1). Then there is some n ∈ W such that n + m ∈ Md+1 and π(n + m) = m. Then 1 1 1 1 n + m, F z = n, F z + m, F z = m, F z ∈ . α α α α Z

0 1 Now consider the linear map ϕ : W → R where x 7→ x, α F z . Then W ⊆ ker(ϕ) and by the Rank-Nullity Theorem, dim(ker(ϕ)) = dim(W 0)−dim(im(ϕ)) = d+1−1 = d. Since dim(W ) = d, it follows that W = ker(ϕ). Then, since u1 ∈ W , 1 1 1 1 1 1 = u, F z = u + u , F z = u , F z + u , F z = u , F z . α 1 2 α 1 α 2 α 2 α

1 Let r = m, α F z ∈ Z. Consider 1 1 1 m − ru , F z = m, F z − r u , F z = r − r = 0. 2 α α 2 α

⊥ ⊥ Hence m − ru2 ∈ ker(ϕ) = W . Since m, u2 ∈ W , it follows that m − ru2 ∈ W ⊥ and therefore m − ru2 ∈ W ∩ W . Hence m − ru2 = 0 and m = ru2. Now, since m is a basis of π(Md+1), u2 = sm with s ∈ Z. Therefore u2 = rsu2 and rs = 1. This p gives r = s = ±1. Hence u2 is a basis of π(Md+1) and det(π(Md+1)) = hu2, u2i = det(Md+1) ku2k2. Now by Corollary 3.32, det(π(Md+1)) = . We now show that det(Md+1∩W ) Md+1 ∩ W = Md. First, since Md spans W and Md ⊂ Md+1, Md ⊆ Md+1 ∩ W . Now, consider some v ∈ Md+1 ∩ W . Since v ∈ Md+1, to show v ∈ Md, it suﬃces to show 1 that hv, α F zi = 0, which holds since v ∈ W . Hence, v ∈ Md so Md = Md+1 ∩ W . det(Md+1) 1 Hence ku2k2 = det(π(Md+1)) = . Recall that u, F z = 1. Then, by det(Md) α Corollary 2.29, 1 1 det(Md+1) 1 p 1 = u2, F z ≤ ku2k2 · F z = · hF z,F zi α α 2 det(Md) α

117 so that 1 det(M ) ≤ det(M ) · phF z,F zi. (10.5) d d+1 α ˜ ˜ Recall that ˜x1,..., ˜xd, b is a basis of Md+1. Let B = [˜x1 ··· ˜xd b]. Then we have p T det(Md+1) = det(B B). Now  ˜  h˜x1, ˜x1i · · · h˜x1, ˜xdi h˜x1, bi . . . .  . .. . .  BT B =  . . .  .  ˜  h˜xd, ˜x1i · · · h˜xd, ˜xdi h˜xd, bi ˜ ˜ ˜ ˜ hb, ˜x1i · · · hb, ˜xdi hb, bi

For 1 ≤ i, j ≤ d, h˜xi, ˜xji = h4xi, 4xji = 16hxi, xji. Also ˜ h˜xi, bi = h˜xi, 4γ1x1 + ··· + 4γdxd + 2zi

= 4γ1h˜xi, x1i + ··· + 4γdh˜xi, xdi + 2h˜xi, zi

= 16γ1hxi, x1i + ··· + 16γdhxi, xdi + 2h˜xi, zi

= γ1h˜xi, ˜x1i + ··· + γdh˜xi, ˜xdi + 2h˜xi, zi, and ˜ hb, zi = h4γ1x1 + ··· + 4γdxd + 2z, zi

= 4γ1hx1, zi + ··· 4γdhxd, zi + 2hz, zi = 2hz, zi.

T Let p1,..., pd+1 be the columns of B B. Then ˜ T T pd+1 − γ1p1 − · · · γdpd = (2h˜x1, zi,..., 2h˜xd, zi, 2hb, zi) = (0,..., 0, 4hz, zi) .

T T Replacing pd+1 with (0,..., 0, 4hz, zi) gives a new matrix P with det(B B) = det(P ). Let q1,..., qd+1 be the rows of P . Then

qd+1 − γ1q1 − · · · γdqd = (2hz, ˜x1i,..., 2hz, ˜xdi, 4hz, zi) = (0,..., 0, 4hz, zi).

0 Replacing qd+1 with (0,..., 0, 4hz, zi) gives a matrix P whose upper-left d×d matrix is (h˜xi, ˜xji) = (16hxi, xji) and whose ﬁnal column and row is (0,..., 0, 4hz, zi). Also, det(P 0) = det(P ) = det(BT B). Therefore

p T det(Md+1) = det(B B) = pdet(P 0) q = 4hz, zi · det(16hxi, xji) q d = 4hz, zi · 16 det(hxi, xji) 2d+1 = 2 kzk2 · A 2n+1 ≤ 2 kzk2 · A

n kzk · A.

118 Combining this with eq. (10.4) gives

1 n−d ∆ 1− 1 1 det(M ) · A = A n−d ∆ n−d . d+1 n A Combining this with eq. (10.5) gives

1− 1 1 1 p det(M ) A n−d ∆ n−d hF z,F zi d n α 1− 1 1 1 = A n−d ∆ n−d kF zk α 2 1− 1 1 1 A n−d ∆ n−d kF zk. n α T Now F z = (f11z1 + ··· + f1nzn, . . . , fn1z1 + ··· + fnnzn) so

kF zk = max{|f11z1 + ··· + f1nzn|,..., |fn1z1 + ··· + fnnzn|}

≤ max{|f11||z1| + ··· + |f1n||zn|,..., |fn1|z1| + ··· + |fnn||zn|} ≤ nkFk kzk

n kFk2 kzk. Therefore 1− 1 1 1 det(M ) A n−d ∆ n−d kFk kzk d n α 2 1 and combining this with eq. (10.4) and α ≤ 1, since α is a positive integer, gives

1 n−d 1− 1 1 1 ∆ det(M ) A n−d ∆ n−d kFk d n α 2 A 1− 2 2 ≤ A n−d ∆ n−d kFk2.

0 0 0 Now let y1,..., yd be the successive minimum vectors of Md. Since Md ⊂ Md+1, yi with 1 ≤ i ≤ d can be written 0 ˜ yi = yi1˜x1 + ··· + yid˜xd + yib, where yij, yi ∈ Z for 1 ≤ i, j ≤ d. 0 0 Deﬁne yi := yi1x1 + ··· + yidxd + yib for 1 ≤ i ≤ d. Then, since y1,..., yd are linearly independent by deﬁnition, it follows that y1,..., yd are linearly independent 0 also. Now, we can rewrite both yi and yi as follows: 0 ˜ yi = yi1˜x1 + ··· + yid˜xd + yib

= yi1(4x1) + ··· + yid(4xd) + yi(4γ1x1 + ··· + 4γdxd + 2z)

= 4(yi1 + γ1yi)x1 + ··· + 4(yid + γdyi)xd + 2yiz and

yi = yi1x1 + ··· + yidxd + yib

= yi1x1 + ··· + yidxd + yi(γ1x1 + ··· + γdxd + z)

= (yi1 + γ1yi)x1 + ··· + (yid + γdyi)xd + yiz.

119 0 Thus, it is straightforward to show that kyik ≤ kyik for 1 ≤ i ≤ d. Now, by 0 0 Corollary 3.24, ky1k · · · kydk n det(Md). Thus

2 2 0 0 1− n−d n−d ky1k · · · kydk ≤ ky1k · · · kydk n det(Md) n A ∆ kFk2.

Let Γd ⊂ Λd+1 be the lattice generated by y1,..., yd. Then part 3 of Theorem 3.26 gives det(Γd) n ky1k · · · kydk so that

1− 2 2 det(Γd) n A n−d ∆ n−d kFk2.

0 Hence, Γd is a candidate for the lattice Λd. It remains only to show that f vanishes on Γd. Consider that, for 1 ≤ i, j ≤ d,

* d ! d ! !+ X X hyi,F yji = (yik + γkyi)xk + yiz,F (yjk + γkyj)xk + yjz k=1 k=1 * d + * d !+ X X = (yik + γkyi)xk,F (yjz) + yiz,F (yjk + γkyj)xk k=1 k=1

+ hyiz,F (yjz)i * d + * d !+ X X = yj (yik + γkyi)xk,F z + yi z,F (yjk + γkyj)xk k=1 k=1

+ yiyjhz,F zi * d + X Dyi E = y (y + γ y )x ,F z + y z,F z j ik k i k j 2 k=1 * d + X Dyj E + y (y + γ y )x ,F z + y z,F z i jk k j k i 2 k=1 y y = j hy0,F zi + i hy0,F zi. 4 i 4 j

0 0 0 Since y1,..., yd ∈ Md, hyi,F zi = 0 for 1 ≤ i ≤ d. Hence hyi,F yji = 0 and Γd 0 vanishes on f. Then Γd = Λd, as desired. The following theorem is equivalent to Theorem 10.1:

Theorem 10.2. Let f be a nonzero quadratic form with integer coefficients and associated n × n matrix F = (fij) and g a positive definite quadratic form with real coefficients and associated n×n matrix G = (gij). Consider d ≥ 1 maximal such that f vanishes on a rational linear subspace of dimension d. Then there exist d linearly n independent x1,..., xd ∈ Z such that f vanishes on the linear subspace generated by x1,..., xd and

−1 2 n−d 0 < g(x1) ··· g(xd) n (Tr (FG ) ) 2 · det(G). (10.6)

120 Theorem 10.1⇒Theorem 10.2: The same process used to show Theorem 7.1 2 2 implies Theorem 7.2 is used here to show g(xi) = kxik2 for 1 ≤ i ≤ d, ∆ = 2 −1 2 det(G), and kFk2 = Tr(FG ) , where F corresponds to the quadratic form f in Theorem 10.1. Then combining Theorem 10.1 with Lemma 2.33 gives

n−d 2 kx1k2 · · · kxdk2 n kx1k · · · kxdk n kFk2 ∆.

Then

2 2 g(x1) ··· g(xd) = kx1k2 · · · kxdk2 n−d 2 2 2 n kFk2 ∆ n−d = Tr(FG−1)2 2 det(G).

Theorem 10.2⇒Theorem 10.1: The same process used to show Theorem 7.2 2 implies Theorem 7.1 is used to show, again, that g(xi) = kxik2 for 1 ≤ i ≤ d, 2 2 −1 2 ∆ = det(G), and kFk2 = Tr(FG ) , where F corresponds to the quadratic form f in Theorem 10.1. Then

n−d n−d 2 2 −1 2 2 2 2 2 kx1k2 · · · kxdk2 = g(x1) ··· g(xd) n Tr(FG ) det(G) = kFk2 ∆ .

Combining this with Lemma 2.33 gives

n−d 2 kx1k · · · kxdk n kx1k2 · · · kxdk2 n kFk2 ∆.

Thus Theorem 10.1 and Theorem 10.2 are equivalent. Consider the following hypotheses:

(a) f is nondegenerate,

(b) there exists c ∈ R such that c 6= 0 and Λ = cZn,

Corollary 10.3. Let Λ, ∆, and f be as in Theorem 10.1. Suppose one of (a), (b), (c) 0 0 holds. Let d ≥ 1 so that there exist d linearly independent points x1,..., xd0 such that f vanishes on the subspace generated by x1,..., xd0 . Then there exist x1,..., xd0 such that n−d0 2 kx1k · · · kxd0 k n kFk2 ∆. (10.7) Proof (following [14]). Let d be maximal such that f vanishes on a d-dimensional sublattice of Λ. Let x1,..., xd be as in the statement of Theorem 10.1. Without loss 0 of generality, assume kx1k ≤ kx2k ≤ · · · ≤ kxdk. For 1 ≤ d ≤ d, this implies

d d0 (kx1k · · · kxd0 k) ≤ (kx1k · · · kxdk) .

121 Combining this with Theorem 10.1 gives

0 0 d n−d · d d0 d 2 d d kx1k · · · kxd0 k ≤ (kx1k · · · kxdk) n kFk2 ∆ . (10.8)

− 2 Now, we show that if one of (a), (b), or (c) holds, kFk2 n ∆ n . First, suppose that (a) holds, i.e. that f is nondegenerate. Then eq. (7.4) gives

− 4 n 2 ∆ n · ≤ 2kFk 2 2 so that r 2 4 1 − 2 2 ∆ n ≤ 2kFk n 2 and hence − 2 ∆ n n kFk2 as desired. Now suppose (b) holds, i.e. there is some c ∈ R such that c 6= 0 and Λ = cZn. Then f(Λ) = f(c2Zn) = c2f(Zn) and since f(Λ) ⊆ Z, it follows that c2f(Zn) ⊆ Z. Then the quadratic form c2f has integer coeﬃcients by Lemma 2.27. Then when 2 2 i = j, we have |c fii| ≥ 1 and when i 6= j, since fij = fji, |2c fij| ≥ 1. Then 2 1 n n c kFk2 ≥ 2 . Also, ∆ = det(cZ ) = c . Therefore 1 kFk ≥ 2 2c2 1 = 2 2∆ n − 2 n ∆ n . Finally, assume (c) holds, i.e. there is some x ∈ Λ such that f(x) = 0 and kxk n λ1, where λ1 is the ﬁrst successive minimum of Λ. Since f(Λ) ⊆ Z, |f(x)| ≥ 1. Also

n n X X |f(x)| = fijxixj i=1 j=1 n n X X ≤ |fij||xi||xj| i=1 j=1 ≤ n2kFkkxk2 2 n kFk2λ1. 2 Hence 1 n kFk2λ1. Now, let λ1, λ2, . . . , λn be the successive minima of Λ. Then 2 2 2 λ1 ≤ λ2 ≤ ... ≤ λn and therefore 1 n kFk2λ1 ≤ kFk2λ2 ≤ · · · ≤ kFk2λn. It follows n 2 2 2 that 1 n kFk2 λ1λ2 ··· λn. Now, Theorem 3.23 gives λ1λ2 ··· λn n ∆ so that n 2 1 n kFk2 ∆ −2 n ∆ n kFk2 − 2 ∆ n n kFk2.

122 − 2 We have now shown that ∆ n n kFk2 holds when (a), (b), or (c) holds. This n − 2 implies kFk2 n ∆. Combining this with eq. (10.8) gives

0 n−d · d d0 2 d d kx1k · · · kxd0 k n kFk2 ∆ 0 0 n−d − n ( d−d ) d0 2 2 d d = kFk2 ∆ 0 n−d d−d0 d0 2 d d n kFk2 ∆ ∆ n−d0 2 = kFk2 ∆, as desired. We predict there is a theorem such as the one below that is equivalent to Corol- lary 10.3. However, the details have not yet been worked out. In addition, we predict there will be one more hypothesis, to match the three hypotheses of Corollary 10.3. Consider the following hypotheses:

(a) f is nondegenerate

2 2 2 (b) g(x) = c (x1 + ··· + xn) where c ∈ R. Conjecture 10.4. Suppose one of (a) or (b) holds. Let f be a nonzero quadratic form with integer coefficients and associated n × n matrix F = (fij) and g a positive definite quadratic form with real coefficients and associated n × n matrix G = (gij). Consider d0 ≥ 1 such that f vanishes on a rational linear subspace of dimension d0. 0 n Then there exist d linearly independent x1,..., xd0 ∈ Z such that f vanishes on the linear subspace generated by x1,..., xd0 and

n−d0 −1 2 2 0 < g(x1) ··· g(xd0 ) n Tr(FG ) det(G).

Corollary 10.5. Let Λ, ∆, f, and d be as in Theorem 10.1. Then f has a nontrivial solution x ∈ Λ such that n−d 1 2d d kxk n kFk2 ∆ . (10.9)

Proof. By Theorem 10.1, there exist nonzero x1,..., xd ∈ Λ such that f(xi) = 0 for n−d 2 1 ≤ i ≤ d and kx1k · · · kxdk n kFk2 ∆. Without loss of generality, assume that kx1k ≤ kx2k ≤ · · · ≤ kxdk. Then

n−d d 2 kx1k ≤ kx1k · · · kxdk n kFk2 ∆ so that n−d 1 2d d kx1k n kFk2 ∆ .

123 Lemma 10.6. Let f be a quadratic form in Z[x1, . . . , xn]. Let n = r + s + t ≥ 3, r ≥ s ≥ 1 with sgn(f) = (r, s) and dim(rad(f)) = t. Put  s + t if r ≥ s + 3  d0 = s + t − 1 if r = s + 2 or r = s + 1 s + t − 2 if r = s.

Then f vanishes on a rational linear subspace of dimension ≥ d0.

In order to prove Lemma 10.6, we need the following two facts: Fact 1: An application of the Hasse-Minkowksi Theorem gives that an indefinite quadratic form f ∈ Q[x1, . . . , xn] with n ≥ 5 has a nontrivial zero. Fact 2: Given a field k with char(k) 6= 2 and f ∈ k[x1, . . . , xn] an isotropic quadratic 2 2 form, Corollary C.5 gives that f can be written f = x1 − x2 + R(x3, . . . , xn) where R 2 2 is a quadratic form in k[x3, . . . , xn]. Note that x1 −x2 has signature (1, 1), is isotropic, and describes a hyperbolic plane. Proof (following [14]). Without loss of generality, assume t = 0. Case 1: Suppose r ≥ s + 3. Since r ≥ s ≥ 1, f is indefinite and n = r + s ≥ 5. Then Fact 1 gives that f has a nontrivial zero and hence Fact 2 gives that we can split off a hyperbolic plane. Now, assume we can split off s − 1 hyperbolic planes. Then the new quadratic form, f 0, has signature (r − (s − 1), s − (s − 1)) = (r − s + 1, 1). Since r − s ≥ 3, it follows that f 0 is a form in r − s + 2 ≥ 5 variables. Therefore, Facts 1 and 2 give that we can split off another hyperbolic plane, resulting in the quadratic form f 00 with signature (r − s + 1 − 1, 1 − 1) = (r − s, 0). Hence f 00 is positive definite and there are no more hyperbolic planes to split off. In total, s hyperbolic planes have been removed so that f vanishes on a rational linear subspace of dimension at least s. Case 2: Suppose r = s+2. When s = 1, this gives r = 3 and therefore n = 4. In this case, there is no guarantee of splitting off a hyperbolic plane. Therefore we can say that f vanishes on a rational linear subspace of dimension at least s − 1 = 0. Now, assume s ≥ 2. Then r ≥ 4 and n ≥ 6. Therefore Facts 1 and 2 give that we can split off a hyperbolic plane. Now, assume we can split off s−2 hyperbolic planes, resulting in a quadratic form f 0 with signature (r − (s − 2), s − (s − 2)) = (r − s + 2, 2) = (4, 2). Combining Facts 1 and 2 gives that there is another hyperbolic plane to split off, resulting in a quadratic form f 00 with signature (4 − 1, 2 − 1) = (3, 1). Here we can cannot apply Fact 1 and therefore we have split off a total of s − 2 + 1 = s − 1 hyperbolic planes. So s − 1 ≥ d0. Case 3: Suppose r = s + 1. When s = 1, this gives r = 2 and n = 3. In this case, there is no guarantee of splitting off a hyperbolic plane. Therefore we can say that f vanishes on a rational linear subspace of dimension at least s − 1 = 0. Now, assume s ≥ 2. Then r ≥ 3 and n ≥ 5. Facts 1 and 2 give that we can split off a hyperbolic plane. Now, assume we can split off s − 2 hyperbolic planes, resulting in a quadratic form f 0 with signature (r − (s − 2), s − (s − 2)) = (r − s + 2, 2) = (3, 2). By Facts 1 and 2, we can split off another hyperbolic plane, resulting in a quadratic form f 00

124 with signature (2, 1). Here we cannot apply Fact 1 so in total, s−1 hyperbolic planes were removed so that s − 1 ≥ d0. Case 4: Suppose r = s. When s = 2, we have r = 2 and n = 4. In this case, there is no guarantee of splitting off a hyperbolic plane. Therefore we can say that f vanishes on a rational linear subspace of dimension at least s − 2 = 0. When s ≥ 3, we have r ≥ 3 and n ≥ 6. Then Facts 1 and 2 give that we can split off a hyperbolic plane. Now, assume we can split off s − 3 hyperbolic planes, resulting in a quadratic form f 0 with signature (r −(s−3), s−(s−3)) = (r −s+3, 3) = (3, 3). Since f 0 is a quadratic form in 6 variables, Facts 1 and 2 imply that we can remove one more hyperbolic plane, giving a quadratic form f 00 with signature (2, 2). Fact 1 does not apply here. 0 Hence, in total we remove s − 2 hyperbolic planes so that s − 2 ≥ d .

Corollary 10.7. Let f(x1, . . . , xn) be a quadratic form with integer coeﬃcients such that sgn(f) = (r, s) and n = r + s ≥ 5 and r ≥ s > 0. Then there exists a nonzero n θ x ∈ Z such that f(x) = 0 and kxk n kFk2 where

 1 r  2 · s when r ≥ s + 3  1 s+2 θ = 2 · s−1 when r = s + 2 or r = s + 1  1 s+1 2 · s−2 when r = s. Proof (following [14]). Let d ≥ 1 be maximal such that f vanishes on a d-dimensional n subspace with basis elements in Z . Let t = dim(rad(f)). Now we show that kFk2 ≥ 1 Pn Pn since n ≥ 5. Let f(x) = i=1 j=1 fijxixj with fij = fji. If one of fii 6= 0 for 1 Pn Pn 2 2 1 ≤ i ≤ n, then 1 ≤ |fii| and 1 ≤ i=1 j=1 fij = kFk2. If fii = 0 for all 1 ≤ i ≤ n, f nondegenerate with n ≥ 5 implies that f has at least three nonzero cross-terms fijxixj with i < j (with two nonzero cross-terms, there are at most four 1 variables). Since f has integer coeﬃcients and fij = fji, it follows that ≤ |fij| and q 2 1 2 therefore kFk2 ≥ 6 2 ≥ 1. Case 1: Suppose r ≥ s + 3. Since Λ = Zn and therefore ∆ = 1, Corollary 10.5 gives that there is some nonzero x ∈ Zn such that f(x) = 0 and

n−d 2d kxk n kFk2 .

Since d is maximal, Lemma 10.6 gives that d ≥ s + t. Since n = r + s and f has signature (r, s) and t = 0 so that d ≥ s. Then

n−d n−s r+s−s r 2d 2s 2s 2s kxk n kFk2 ≤ kFk2 = kFk2 = kFk2 .

Case 2: Suppose r = s + 2. Since by assumption r + s ≥ 5, it follows s ≥ 2. Therefore n = r + s ≥ 6. Consider g(x1, . . . , xn−1) := f(x1, . . . , xn−1, 0). Then by Lemma 2.25 and Proposition 2.26, g has one of the following three signatures: (r − 1, s, 0), (r, s − 1, 0), or (r − 1, s − 1, 1). In any case, g is a quadratic form in n − 1 ≥ 5 variables and g is indeﬁnite (since r − 1 > 1 and s − 1 ≥ 1) and therefore g has a nontrivial solution by the Hasse-Minkowski Theorem. Let D ≥ 1 be maximal

125 such that g vanishes on a rational linear subspace of dimension D. Case 2a: Suppose g has signature (r0, s0, t0) = (r − 1, s, 0). Then r0 = r − 1 = s + 2 − 1 = s + 1 = s0 + 1. Then Lemma 10.6 gives that g vanishes on a rational linear subspace of dimension ≥ s0 + t0 − 1 = s + 0 − 1 = s − 1. Case 2b: Suppose g has signature (r0, s0, t0) = (r, s−1, 0). Then r0 = r = s+2 = s0+3. Then Lemma 10.6 gives that g vanishes on a rational linear subspace of dimension ≥ s0 + t0 = s − 1 + 0 = s − 1. Case 2c: Suppose g has signature (r0, s0, t0) = (r − 1, s − 1, 1). Then r0 = r − 1 = s + 1 = s0 + 2. Then Lemma 10.6 gives that g vanishes on a rational linear subspace of dimension ≥ s0 + t0 − 1 = s − 1 + 1 − 1 = s − 1. Hence in every case g vanishes on a space of dimension at least s−1, and therefore n−1 D ≥ s−1. Now, by Corollary 10.5, there is some nonzero x0 = (x1, . . . , xn−1) ∈ Z such that g(x0) = 0 and n−1−D 2D kx0k n kGk2 .

Then x = (x1, . . . , xn−1, 0) is a solution of f and kx0k = kxk. Since kGk2 ≤ kFk2 and D ≥ s − 1,

kxk = kx0k n−1−D 2D n kGk2 n−1−(s−1) 2(s−1) n kFk2 n−s 2(s−1) ≤ kFk2 r+s−s 2(s−1) = kFk2 r 2(s−1) = kFk2 s+2 2(s−1) = kFk2 . Case 3: Suppose r = s + 1. Since Λ = Zn and therefore ∆ = 1, Corollary 10.5 gives that there is some nonzero x ∈ Zn such that f(x) = 0 and

n−d 2d kxk n kFk2 . Since d is maximal, Lemma 10.6 gives that d ≥ s + t − 1. Since n = r + s, it follows that f is nondegenerate and therefore t = 0. Hence d ≥ s − 1 and

n−d n−(s−1) r+s−(s−1) r+1 s+2 2d 2(s−1) 2(s−1) 2(s−1) 2(s−1) kxk n kFk2 ≤ kFk2 = kFk2 ≤ kFk2 = kFk2 . Case 4: Suppose r = s. Since r + s ≥ 5 by assumption, it follows r = s ≥ 3 and hence n = r + s ≥ 6. Again consider g(x1, . . . , xn−1) = f(x1, . . . , xn−1, 0). As in case 2, g has one of the following signatures: (r − 1, s, 0), (r, s − 1, 0), or (r − 1, s − 1, 1). In any case, g is a quadratic form in n − 1 ≥ 5 variables and g is indeﬁnite (since r − 1 > 1 and s − 1 > 1) and therefore g has a nontrivial solution by the Hasse- Minkowski Theorem. Let D ≥ 1 be maximal such that g vanishes on a rational linear subspace of dimension D.

126 Case 4a: Suppose sgn(g) = (r0, s0, t0) = (r −1, s, 0). Then r0 = r −1 = s−1 = s0 −1. Here, r0 < s0 and s0 = r0 + 1. Then Lemma 10.6 gives that g vanishes on a rational linear subspace of dimension r0 + t0 − 1 = r − 1 + 0 − 1 = r − 2 = s − 2. Case 4b: Suppose sgn(g) = (r0, s0, t0) = (r, s − 1, 0). Then r0 = r = s = s0 + 1. Then lemma 10.6 gives that g vanishes on a rational linear subspace of dimension s0 + t0 − 1 = s − 1 + 0 − 1 = s − 2. Case 4c: Suppose sgn(g) = (r0, s0, t0) = (r − 1, s − 1, 1). Then r0 = r − 1 = s − 1 = s0. Then Lemma 10.6 gives that g vanishes on a rational linear subspace of dimension s0 + t0 − 2 = s − 1 + 1 − 2 = s − 2. Hence in every case g vanishes on a space of dimension at least s−2, and therefore n−1 D ≥ s−2. Now, by Corollary 10.5, there is some nonzero x0 = (x1, . . . , xn−1) ∈ Z such that g(x0) = 0 and n−1−D 2D kx0k n kGk2 .

Then x = (x1, . . . , xn−1, 0) is a solution of f and kx0k = kxk. Since kGk2 ≤ kFk2 and D ≥ s − 2,

kxk = kx0k n−1−D 2D n kGk2 n−1−(s−2) 2(s−2) n kFk2 n−s+1 2(s−2) ≤ kFk2 r+s−s+1 2(s−2) = kFk2 r+1 2(s−2) = kFk2 s+1 2(s−2) = kFk2 .

127 Chapter 11 Schmidt 1985

For a linear form L ∈ R[x1, . . . , xn], let |L| denote the maximum absolute value of the kxk coeﬃcients of L. Recall that for x = (x1, . . . , xn), hxi = minxi6=0(|xi|) and q(x) = hxi . Schmidt proved the following in Proposition 1 of [19], beginning on page 90. Proposition 11.1. Suppose 0 < l ≤ s. Given D ≥ 1, for i = 1, . . . , l there are linear forms Li(Y) = L(Y1,...,Ys) with integer coeﬃcients and with |Li| ≤ D for i = 1, . . . , l such that every nonzero complex vector y = (y1, . . . , ys) has

l/s max(kyk, |L1(y)|,..., |Ll(y)|) s D hyi.

Note: The proof of Proposition 11.1 in [19] shows that the Li(Y)’s also depend on s. Corollary 11.2. Suppose n = r + s, r ≥ s > 0. Suppose 0 < l ≤ s. Given D ≥ 1, for i = 1, . . . , l, there are linear forms

Li(Y) = L(Y1,...,Ys) with integer coeﬃcients and with |Li| ≤ D for i = 1, . . . , l such that every nonzero complex vector y = (y1, . . . , ys) has

l/s max(kyk, |L1(y)|,..., |Ll(y)|) ≥ τ(n)D hyi, where τ(n) is a positive constant depending on n.

n Proof. Note ﬁrst that there are b 2 c possible values for s. For each s, Proposition 11.1 gives that there is some constant τ(s) depending only on s such that

l/s max(kyk, |L1(y)|,..., |Ll(y)|) ≥ τ(s)D hyi.

Deﬁne τ(n) := min n τ(s). Then 1≤s≤b 2 c

l/s max(kyk, |L1(y)|,..., |Ll(y)|) ≥ τ(n)D hyi.

n Since τ(n) depends only on 1,..., b 2 c, it depends only on n. Theorem 11.3. Suppose n = r+s, r ≥ s > 0. Given C ≥ 1 there is a quadratic form Pn Pn f(X) = i=1 j=1 fijXiXj ∈ Z[X1,...,Xn] with fij = fji such that the following properties hold: (i) sgn(f) = (r, s) (ii) kFk ≤ C

128 n r (iii) for x ∈ R \{0} with f(x) ≤ 0, q(x) n C 2s . 2 2 2 2 Proof (following [19]). Case 1: Suppose C < 8n . Let f = X1 + ··· + Xr − Xr+1 − 2 r · · · − Xr+s. Then kFk = 1 and therefore kFk ≤ C. Since 2s < r < n, one has r n 2 n r n C 2s < C < (8n ) , so that C 2s n 1. For any nonzero x ∈ , q(x) ≥ 1, so r R C 2s n 1 ≤ q(x) and the theorem holds. 2 1/2 4n2 2 1 Case 2: Now suppose C ≥ 8n . Deﬁne D = bρ(n)C c where C ≤ ρ(n) ≤ 2 . Then 4n2 ≤ ρ(n)2C so that 2n ≤ ρ(n)C1/2 and D ≥ 2n > 1. Using the division algorithm, there exist h, l ∈ Z such that r = sh + l with 0 ≤ l < s. Case 2a: Suppose l > 0. Since n = r + s = (sh + l) + s = (h + 1)s + l, relabel X1,...,Xn as Yij with 1 ≤ i ≤ s, 0 ≤ j ≤ h and Y1,...,Yl. With L1,...,Ll the linear forms from Proposition 11.1, set

Zij = Yij − DYi,j−1 (1 ≤ i ≤ s, 1 ≤ j ≤ h) (11.1a)

Zm = Ym − Lm(Y1h,...,Ysh) (1 ≤ m ≤ l). (11.1b)

We will show that Zij, Zm, Yi0 with 1 ≤ i ≤ s, 1 ≤ j ≤ h, and 1 ≤ m ≤ l are linearly independent linear forms in Z[X1,...,Xn]. To see this, let Lm(Y1h,...,Ysh) = bm1Y1h + ··· + bmsYsh with b1m, . . . , bsm ∈ Z and, for ease of notation, let Zi0 = Yi0 for 1 ≤ i ≤ s. Let  1 0 ········· 0 −D 1 0 ······ 0    0 −D 1 0 ··· 0   (Mi)(h+1)×(h+1) :=  ......   . . . .    −D 1 0 0 −D 1 for 1 ≤ i ≤ s. Let  0 ··· 0 ∗ 0 ··· 0 ∗ · · ·  N :=  ......  l×n  ...... Il  0 ··· 0 ∗ 0 ··· 0 ∗ · · · where the ﬁrst h columns of N contain all zeros, the nonzero columns are columns h + 1, 2(h + 1), ··· , s(h + 1), and the right-most l × l block is the l × l identity matrix. The entries of column i(h + 1) for 1 ≤ i ≤ s are −b1i,..., −bli. Then     Y10 Z10    .   .  M1 0 ······ 0 0 ··· 0  .   .       0 M2 0 ··· 0 0 ··· 0  Y1h Z1h    .   .   ......   .   .   . . . . .   .   .    Y  Z   0 ··· 0 Ms−1 0 0 ··· 0   s0  s0   ·  .  =  .  .  0 ······ 0 Ms 0 ··· 0   .   .          Y  Z     sh  sh  N   Y   Z   1   1   .   .   .   .  Yl Zl

129 Since the n × n matrix on the left-hand side is a lower triangular matrix with ones along the diagonal, it has determinant one. Therefore, Zij, Zm, and Yi0 with 1 ≤ i ≤ s, 1 ≤ j ≤ h, and 1 ≤ m ≤ l are linearly independent. Let

s h l s X X 2 X 2 X 2 f = Zij + Zm − Yi0. i=1 j=1 m=1 i=1

Then f can be considered the result of substituting n linearly independent linear forms Ps Ph 2 Pl 2 Ps 2 into the rank n quadratic form i=1 j=1 uij + m=1 um − k=1 uk and therefore is nondegenerate. Also f has signature (sh+l, s) = (r, s). Then f has integer coefficients 2 since D is an integer and each Lm has integer coefficients. Now, expanding the Zij 2 2 2 2 and Zm terms gives that kFk ≤ 2D since D ≥ 1 implies D ≤ D + 1 ≤ 2D . Recall 1/2 2 2 4n2 2 1 that D = bρ(n)C c, so that D ≤ ρ(n) C. Recalling that C ≤ ρ(n) ≤ 2 , one has 2ρ(n)2 ≤ 1 so that kFk ≤ 2D2 ≤ 2ρ(n)2C ≤ C. n r It remains to show that for nonzero x ∈ R with f(x) ≤ 0, q(x) n C 2s . Suppose n x ∈ R , x 6= 0, and f(x) ≤ 0. Relabel x1, . . . , xn to yij and ym as before, with 1 ≤ i ≤ s, 0 ≤ j ≤ h, and 1 ≤ m ≤ l. Also, define zij and zm, with 1 ≤ i ≤ s, 1 ≤ j ≤ h, 1 ≤ m ≤ l, as in eqs. (11.1a) and (11.1b) with yij and ym. Put Ps Ph 2 Pl 2 y = (y10, . . . , ys0). Suppose y = 0. Then f(x) = i=1 j=1 zij + m=1 zm ≤ 0, which implies that zij = zm = 0 for all i, j, m. By definition of zij and zm, this gives that x = 0, a contradiction. Thus y 6= 0. Then f(x) ≤ 0 implies

s h l s X X 2 X 2 X 2 zij + zm ≤ yi0 i=1 j=1 m=1 i=1 ≤ skyk2.

2 2 √ √ Hence for each zij, zij ≤ skyk so that |zij| ≤ skyk. Similarly, |zm| ≤ skyk. Recalling that zij = yij − Dyi,j−1 gives

yi1 = Dyi0 + zi1 2 yi2 = Dyi1 + zi2 = D(Dyi0 + zi1) + zi2 = D yi0 + Dzi1 + zi2 2 3 2 yi3 = Dyi2 + zi3 = D(D yi0 + Dzi1 + zi2) + zi3 = D yi0 + D zi1 + Dzi2 + zi3

Iterating this process gives

h h−1 yih = D yi0 + D zi1 + ··· + Dzi,h−1 + zih (11.2)

130 Let zi0 = yi0 for 1 ≤ i ≤ s. Combining this with eq. (11.2) gives, for 1 ≤ m ≤ l,

ym = Lm(y1h, . . . , ysh) + zm h h ! X k X k = Lm D z1,h−k,..., D zs,h−k + zm k=0 k=0 h ! X k = D Lm(z1,h−k, . . . , zs,h−k) + zm k=0 h−1 ! h X k = D Lm(y) + D Lm(z1,h−k, . . . , zs,h−k) + zm (11.3) k=0 Now, consider the following cases. Case 2a(i): Suppose kyk ≤ δ(n)Dl/shyi where δ(n) is a constant depending only on 1 l/s n such that δ(n) < n2 . Then kyk ≤ D hyi and Corollary 11.2 gives that there is some Lm, with 1 ≤ m ≤ l, such that l/s |Lm(y)| n D hyi. (11.4)

Recalling that |Lm| ≤ D and using eq. (11.3), h h |D Lm(y)| = |D Lm(y) − ym + ym| h ≤ |D Lm(y) − ym| + |ym| h = |ym − D Lm(y)| + |ym| h−1 !

X k = D Lm(z1,h−k, . . . , zs,h−k) + zm + |ym| k=0 h−1 ! X k ≤ D |Lm(z1,h−k, . . . , zs,h−k)| + |zm| + |ym| k=0 h−1 s ! X k X ≤ D D|zi,h−k| + |zm| + |ym| k=0 i=1 h−1 s ! X X k+1 ≤ D |zi,h−k| + |zm| + |ym| k=0 i=1 h√ √ ≤ hs · D skyk + skyk + |ym| h√ ≤ (h + 1)s · D skyk + |ym|. Therefore h h√ |ym| ≥ |D Lm(y)| − (h + 1)s · D skyk l and by the assumption kyk ≤ δ(n)D s hyi and eq. (11.4),

h h√ l |ym| ≥ |D Lm(y)| − (h + 1)s · D s · δ(n)D s hyi h l h√ l n D · D s hyi − (h + 1)s · D s · δ(n)D s hyi h+ l √ = D s hyi 1 − (h + 1)s s · δ(n) .

131 √ Since n = r + s = s(h + 1) + l with l > 0, (h + 1)s < n and s < n. So

h+ l 2 |ym| n D s hyi 1 − n δ(n) .

1 2 Since δ(n) < n2 , 1 − n δ(n) > 0 and therefore

h+ l r |ym| n D s hyi = D s hyi.

Therefore |ym| r q(x) ≥ D s . hyi n Now since D = bρ(n)C1/2c and D ≥ 1, D ≤ ρ(n)C1/2 < D + 1 ≤ 2D. This gives 1/2 C n D so that r r q(x) n D s n C 2s as desired. l/s Case 2a(ii): Suppose kyk > δ(n)D hyi. Let 1 ≤ i ≤ s such that kyk = |yi0|. By eq. (11.2)

h h h |D yi0| = |D yi0 − yih + yih| ≤ |D yi0 − yih| + |yih| h = |yih − D yi0| + |yih| h−1 = |D zi1 + ··· + Dzi,h−1 + zih| + |yih| h−1 ≤ D |zi1| + ··· + D|zi,h−1| + |zih| + |yih| h−1√ ≤ h · D skyk + |yih|. √ Note that since n = r + s = sh + l + s = s(h + 1) + l, n ≥ sh ≥ s · h. Therefore

h h−1√ |yih| ≥ |D yi0| − h · D skyk √ = Dhkyk − h · Dh−1 skyk √ = Dh−1kyk(D − h s) ≥ Dh−1kyk(D − n)

1 Recall that D ≥ 2n, so that 2D ≥ 2n + D and hence D − n ≥ 2 D. Hence

h−1 h |yih| ≥ D kyk(D − n) n D kyk.

Then by assumption,

h h+ l r |yih| n D kyk > δ(n)D s hyi = δ(n)D s hyi so that |yih| r q(x) ≥ D s . hyi n 1/2 It has been shown that C n D, so

r r q(x) n D s n C 2s

132 as desired. Case 2b: Suppose l = 0. Then r = sh and n = r +s = s(h+1). Rename X1,...,Xn as Yij with 1 ≤ i ≤ s and 0 ≤ j ≤ h and again set

Zij = Yij − DYi,j−1 (1 ≤ i ≤ s, 1 ≤ j ≤ h). (11.5)

Let s h s X X 2 X 2 f(X) = Zij − Yi0. (11.6) i=1 j=1 i=1

Deﬁning the (h + 1) × (h + 1) matrices Mi for 1 ≤ i ≤ s as in case 2a and letting Zi0 = Yi0 as before, we have     Y10 Z10    .   .  M1 0 ······ 0  .   .       0 M2 0 ··· 0  Y1h Z1h  . . .   .   .   . .. .  ·  .  =  .  .        0 ··· 0 M 0  Y  Z   s−1   s0  s0 0 ······ 0 M  .   .  s  .   .  Ysh Zsh

Since the left-most matrix is a lower triangular matrix with ones on the diagonal, it has a determinant of one. Therefore the Zij’s, with 1 ≤ i ≤ j and 1 ≤ j ≤ h, and the Yi0’s, with 1 ≤ i ≤ s, are linearly independent. Hence f can be considered as the result of substituting n linearly independent linear forms into the rank n diagonal Ps Ph 2 Ps 2 quadratic form i=1 j=1 uij − k=1 uk and is therefore nondegenerate. Then f is 2 of the form (sh, s) = (r, s). Also, f has integer coefficients. Expanding the Zij terms 2 2 2 gives that kFk = D + 1 because D + 1 is the coefficient on Yij for 1 ≤ i ≤ s and 1 ≤ j ≤ h − 1 and D ≥ 1 implies 2D ≤ D2 + 1. So kFk = D2 + 1 ≤ 2D2, again since D ≥ 1. Hence kFk ≤ 2D2 ≤ 2ρ(n)2C ≤ C n r as before. It remains to show that for nonzero x ∈ R with f(x) ≤ 0, q(x) n C 2s . n Let x ∈ R \{0} such that f(x) ≤ 0. Rename x1, . . . , xn as yij with 1 ≤ i ≤ s and 0 ≤ j ≤ h and zij be defined as in eq. (11.5). Then f(x) ≤ 0 implies

s h s X X 2 X 2 zij ≤ yi0 i=1 j=1 i=1 ≤ skyk2.

2 2 In particular, this gives zij ≤ skyk and therefore √ |zij| ≤ skyk s kyk (11.7) for 1 ≤ i ≤ s and 1 ≤ j ≤ h.

133 Using the same process as in the l > 0 case, for 1 ≤ i ≤ s such that kyk = |yi0|, one uses eq. (11.2) to show h |yih| n D kyk. r Since l = 0, h = s and r r |yih| n D s kyk n D s hyi.

1 It has been shown that C 2 n D, so

r r |yih| n D s hyi n C 2s hyi.

Therefore, |yih| r q(x) ≥ C 2s hyi n as desired. Now, consider the consequences of this theorem. If x ∈ Zn and x is nonzero, kxk hxi ≥ 1 so that kxk ≥ hxi = q(x). Then if the quadratic form f in Theorem 11.3 has a nonzero integral solution x,

r r kxk ≥ q(x) n C 2s ≥ kFk 2s .

This has two major consequences. First, it is an improvement on Theorem 6.2, which 1 r r r shows the exponent on kFk cannot be reduced below 2 b s c. Since b s c ≤ s , the result above provides a slightly better lower bound on the exponent. Secondly, when r combined with Corollary 10.7, it shows that when r ≥ s + 3, the exponent of 2s is best possible. If focusing solely on nonzero integral solutions of f, the proof of this theorem could 1 b r c have appealed to Theorem 6.2 in the l = 0 case, which would give kFk 2 s n kxk. r r In this case since r = sh, it follows that b s c = bhc = h = s . The following theorem, found in [16, Thm. 1], follows from Theorem 11.3.

Theorem 11.4. Suppose n ≥ 2d > 0. Given C ≥ 1, there is a quadratic from Pn Pn f(X) = i=1 j=1 fijXiXj ∈ Z[X1,...,Xn] with fij = fji, such that the following properties hold:

(i) f(X) vanishes on a d-dimensional rational subspace S

(ii) kFk ≤ C

(iii) for x ∈ S\{0} and f(x) ≤ 0,

n−d q(x) n C 2d

(iv) for x ∈ Rn\S and f(x) ≤ 0,

n−d q(x) n C 2d .

134 Proof (following [16]). Put r = n − d and s = d. Then r + s = n and r ≥ s > 0. By Theorem 11.3, there is a quadratic form f(X) ∈ Z[X1,...,Xn] with signature (r, s) = (n − d, d). In the proof of Theorem 11.3, two diﬀerent forms were considered. 2 2 2 2 2 When C < 8n , f(X) = X1 + ··· + Xn−d − Xn−d+1 − · · · − Xn is a form with signature (n − d, d) and f vanishes on the rational subspace S spanned by e1 + en−d+1, e2 + en−d+2,..., ed + en, where e1,..., en are the standard basis vectors of Rn. Hence dim(S) = d and f vanishes on a d-dimensional rational subspace. In addition, kFk = 1 so kFk ≤ C. When C ≥ 8n2, we write r = sh + l with 0 ≤ l < s in the proof of Theorem 11.3. Then here n − d = dh + l with 0 ≤ l < d. Then the quadratic form is f(X) = Pd Ph 2 Pl 2 Pd 2 i=1 j=1 Zij + m=1 Zm − i=1 Yi0, where Zij and Zm are deﬁned as in eqs. (11.1a) and (11.1b). Letting Zi1 = Yi0 for 1 ≤ i ≤ d, Zij = 0 for 1 ≤ i ≤ d, 2 ≤ j ≤ h, and Zm = 0 for 1 ≤ m ≤ l gives a d-dimensional rational subspace S where f vanishes. Thus (i) holds. The work done in the proof of Theorem 11.3 shows kFk ≤ C thus (ii) holds for all C ≥ 1. Theorem 11.3 gives, in both of these cases, that for any nonzero x ∈ n such that r n−d R f(x) ≤ 0, q(x) n C 2s = C 2d . Thus (iii) and (iv) hold. In Theorem 11.4, we separate statements (iii) and (iv) in order to make a direct comparison between Theorem 11.4 and Theorem 12.3 more straightforward. More details will be provided following the proof of Theorem 12.3.

135 Chapter 12 Schlickewei & Schmidt 1988

Schlickewei and Schmidt proved the following proposition on page 300 of [16]. It is similar to, but stronger than, Proposition 11.1.

Proposition 12.1. Suppose 0 < l ≤ d. Given B ≥ 1 there are linearly independent linear forms Li(Y) = Li(Y1,...,Yd)(i = 1, . . . , d)

1− l with integer coeﬃcients and with |Li| ≤ B d for i = 1, . . . , d satisfying the following conditions:

(i) Every nonzero complex vector y = (y1, . . . , yn) has

l/d l/d l/d max{kyk,B |L1(y)|....,B |Ll(y)|} d B hyi.

(ii) Every nonzero complex vector y with

max{|Ll+1(y)|,..., |Ld(y)|} = max{|L1(y)|,..., |Ld(y)|}

satisﬁes l/d kyk d B max{|L1(y)|,..., |Ld(y)|}.

Note: The proof of Proposition 12.1 in [16] shows that the Li(Y)’s also depend on d.

Corollary 12.2. Suppose n ≥ 2d > 0. Given B ≥ 1 there are linearly independent linear forms Li(Y) = Li(Y1,...,Yd),

1− l where 1 ≤ i ≤ d, with integer coeﬃcients and with |Li| ≤ B d for 1 ≤ i ≤ d satisfying the following conditions:

(i) Every nonzero complex vector y = (y1, . . . , yn) has

l/d l/d l/d max{kyk,B |L1(y)|....,B |Ll(y)|} ≥ τ(n)B hyi,

where τ(n) is a positive constant depending on n.

(ii) Every nonzero complex vector y with

max{|Ll+1(y)|,..., |Ld(y)|} = max{|L1(y)|,..., |Ld(y)|}

satisﬁes l/d kyk ≥ ν(n)B max{|L1(y)|,..., |Ld(y)|}, where ν(n) is a positive constant depending on n.

136 n Proof. Note ﬁrst that there are b 2 c possible values for d. For each d, Proposition 12.1 gives that there is some constant τ(d) depending only on d such that every nonzero complex vector y = (y1, . . . , yn) has

l/d l/d l/d max{kyk,B |L1(y)|,...,B |Ll(y)|} ≥ τ(d)B hyi.

Deﬁne τ(n) := min n τ(d). Then 1≤d≤b 2 c

l/d l/d l/d max{kyk,B |L1(y)|,...,B |Ll(y)|} ≥ τ(n)B hyi.

n Since τ(n) depends only on 1,..., b 2 c, it depends only on n and thus (i) holds. Now, Proposition 12.1 also gives that for every nonzero complex vector y = (y1, . . . , yn) with max{|Ll+1(y)|,..., |Ld(y)|} = max{|L1(y)|,..., |Ld(y)|} there is some constant ν(d) depending only on d such that

l/d kyk ≥ ν(d)B max{|L1(y)|,..., |Ld(y)|}.

Deﬁne ν(n) := min n ν(d). Then 1≤d≤b 2 c

l/d kyk ≥ ν(n)B max{|L1(y)|,..., |Ld(y)|}.

n Since ν(n) depends only on 1,..., b 2 c, it depends only on n. Thus (ii) holds. Theorem 12.3. Suppose n ≥ 2d > 0. Given C ≥ 1, there is a nondegenerate quadratic form f with integral coeﬃcients in n variables with the following properties:

(i) There is a d-dimensional rational subspace S on which f vanishes.

(ii) kFk ≤ C.

(iii) Any point x ∈ S\{0} has n−2d q(x) n C 2d .

(iv) Any point x ∈ Rn\S with f(x) ≤ 0 satisﬁes

n q(x) n C 2d . √ 4 n 2 2 Proof (following [16]). Case 1: Suppose C ≤ ( 2n ) . Let f(X) = X1 + ··· + Xd − X2 − · · · − X2. Then kFk = 1 so that kFk ≤ C. Since 0 < 2d ≤ n, n < n so d+1 n √ 2d n n 4 nn n that C 2d < C ≤ ( 2n ) . Therefore C 2d n 1. Since 1 ≤ q(x) for all nonzero x ∈ n, R n q(x) n C 2d .

n n−2d Since C 2d ≥ C 2d , this gives√ both (ii) and (iii). Case 2: Now, assume C > ( 2n4)n.

137 1 4 1 d n 1 2 Case 2a: Suppose d | n. Put B = bη(n)C 2d c where 1 < η(n) ≤ (which is √ C n 2 4 n n4 n4 possible since C > ( 2n ) ). Since C ≥ 1 and 2d ≤ n, it follows that 1 ≤ 1 . C 2d C n 1 1 1 2 1 2d Also, since 1 ≤ d, η(n) ≤ 2 ≤ 2 . Therefore

1 n4 1 2d 1 < η(n) ≤ . (12.1) C 2d 2

4 1 4 Then n < η(n)C 2d . Because n is an integer, it follows that

4 1 1 d n ≤ bη(n)C 2d c ≤ bη(n)C 2d c = B.

It follows that B ≥ 1. Since d | n, n = hd for some h ∈ Z. Since n ≥ 2d, h ≥ 2. Rename X1,...,Xn as Yij with 1 ≤ i ≤ h and 1 ≤ j ≤ d. Suppose h ≥ 3 and set

Zij = Xij − BXi−1,j (3 ≤ i ≤ h, 1 ≤ j ≤ d). (12.2)

Deﬁne d d h d X 2 2 X X X 2 f(X) = B X1j − X1jX2j + Zij. j=1 j=1 i=3 j=1

Note that, by substituting for Zij and completing the square, f can be re-written as

d 2 d h d X 1 X 1 X X f(X) = BX − X − X2 + (X − BX )2, 1j 2B 2j 4B2 2j ij i−1,j j=1 j=1 i=3 j=1 that is as the sum of squares of linear forms. Since these forms are are linearly independent and there are d + d + d(h − 2) = dh = n terms, it follows that f is nondegenerate. Now f has integer coefficients and kFk = B2 + 1. Since B ≥ 1, 2 2 kFk ≤ 2B so that kFk n B . 1 1 2d d 2 2d 2d Now, since B = bη(n)C c , B ≤ η(n) C. By eq. (12.1), η(n) ≤ 2 so that 2η(n)2d ≤ 1 gives kFk ≤ 2B2 ≤ 2η(n)2dC ≤ C so that kFk ≤ C. Let S be the subspace defined by X1j = Z3j = ··· = Zhj = 0 for 1 ≤ j ≤ d. i−2 Then Xij = B X2j so that X2j for 1 ≤ j ≤ d are the free variables. Then S is d-dimensional and has an integral basis. Also, f(S) = 0. Choose some nonzero x ∈ S. Rename x1, . . . , xn as xij as above and define zij as in eq. (12.2). Let x2 = (x21, . . . , x2d). Suppose x2 = 0. Then z3j = x3j − Bx2j = 0 implies x3j = 0, which implies x4j = 0 and so on so that xij = 0 for 2 ≤ i ≤ h and 1 ≤ j ≤ d. Since x ∈ S, x1j = 0 for 1 ≤ j ≤ d. Hence x = 0, a contradiction. Therefore x2 6= 0. Without loss of generality, assume x21 6= 0. Then z31 = ··· = zh1 = 0 implies

z31 = x31 − Bx21 = 0

x31 = Bx21

138 and

z41 = x41 − Bx31 = 0

x41 = Bx31 2 = B x21

h−2 and so on so that xh1 = B x21. Then |x | q(x) ≥ h1 = Bh−2. |x21|

1 d Now since B = bη(n)C 2d c

1 1 1 B d ≤ η(n)C 2d < B d + 1 d 1 1 d B ≤ η(n) C 2 < (B d + 1) 1 d ≤ (2B d ) = 2dB.

d 1 1 2 1 2 2 Then C < η(n) B. Since η(n) was chosen such that η(n) ≤ 2 , it follows that √ d n 2 2 2 1 2 2 2 ≤ η(n) . Hence η(n) ≤ η(n) and C n B. Then

h−2 h−2 hd−2d n−2d q(x) ≥ B n C 2 = C 2d = C 2d . Thus (ii) holds. n Now, choose x ∈ R \S with f(x) ≤ 0 and let x1 = (x11, . . . , x1d). Then f(x) ≤ 0 implies

d d h d d X 2 2 X X 2 X X B x1j + zij ≤ x1jx2j ≤ kx1k kx2k = d · kx1k kx2k j=1 j=1 i=3 j=1 j=1

Pd 2 2 2 2 Pd 2 2 so that j=1 B x1j ≤ d·kx1k kx2k. Recall that by Lemma 2.33 B kx1k ≤ d j=1 B x1j, so 2 2 2 B kx1k ≤ d kx1k kx2k. Similarly, 2 |zij| ≤ dkx1k kx2k. Pd Ph 2 Note that x1 6= 0 for if it were, f(x) = j=1 i=3 zij ≤ 0 would hold true, implying zij = 0 for 3 ≤ i ≤ h and 1 ≤ j ≤ d. But then x ∈ S, which is a contradiction. Therefore

2 −2 kx1k ≤ d B kx2k (12.3)

2 −2 2 2 3 −2 2 so that kx1k kx2k ≤ d B kx2k . Hence |zij| ≤ dkx1k kx2k ≤ d B kx2k . Then

3 −1 |zij| ≤ d 2 B kx2k

139 when 3 ≤ i ≤ h, 1 ≤ j ≤ d. Now by deﬁnition of the zij’s

x3j = Bx2j + z3j

x4j = Bx3j + z4j

= B(Bx2j + z3j) + z4j 2 = B x2j + Bz3j + z4j, and so on. Iterating this process gives

h−2 h−3 xhj = B x2j + B z3j + ··· + Bzh−1,j + zhj (12.4) for 1 ≤ j ≤ d. Note that x2 6= 0. For if it were, f(x) ≤ 0 would imply zij = 0 for 3 ≤ i ≤ h and 1 ≤ j ≤ d and x1j = 0 for 1 ≤ j ≤ d. Then x ∈ S, a contradiction. Assume, without loss of generality, that kx2k = |x21|. Now by eq. (12.4)

h−2 h−2 |B x21| = |B x21 − xh1 + xh1| h−2 ≤ |xh1 − B x21| + |xh1| h−3 = |B z31 + ··· + Bzh−1,1 + zh1| + |xh1| h−3 ≤ B |z31| + ··· + B|zh−1,1| + |zh1| + |xh1| h−3 3 −1 3 −1 3 −1 ≤ B (d 2 B kx2k) + ··· B(d 2 B kx2k) + d 2 B kx2k + |xh1| h−4 3 ≤ (h − 2)B d 2 kx2k + |xh1|.

Then, using eq. (12.3)

h−2 h−4 3 |xh1| ≥ B |x21| − (h − 2)B d 2 kx2k h−2 h−4 3 = B kx2k − (h − 2)B d 2 kx2k h−2 −2 3 = B kx2k 1 − (h − 2)B d 2

h −2 −2 3 ≥ B d kx1k 1 − (h − 2)B d 2

h −2 −2 − 1 = B kx1k d − (h − 2)B d 2 .

Now, since n = hd and h ≥ 3, h − 2 < n and d < n. Combining this with d ≥ 1.

h −2 −2 |xh1| > B kx1k n − nB .

Recall that B ≥ n4, which implies B > n3 since n > 1. It follows that n−2 −nB−2 > 0 and therefore h |xh1| n B kx1k. 1 It has been shown that C 2 n B so that

h h hd n n |xh1| n B kx1k n C 2 kx1k = C 2d kx1k = C 2d kx1k ≥ C 2d hx1i.

140 Therefore |xh1| n q(x) ≥ n C 2d , (12.5) hx1i as desired. Now consider when h = 2. Let

d d X 2 2 X f(X) = B X1j − X1jX2j. j=1 j=1

Then, by similar a argument to that given in the h ≥ 3 case, f is nondegenerate. It has integer coeﬃcients and kFk = B2. Therefore, by choice of η(n),

kFk ≤ 2B2 ≤ 2η(n)2dC ≤ C.

Let S be the subspace given by X1j = 0 for 1 ≤ j ≤ d. Then X2j are free variables for 1 ≤ j ≤ d and S is d-dimensional. Also f(S) = 0. Choose x ∈ S such that x 6= 0. Since x has the form (x11, . . . , x1d, x21, . . . , x2d) and x1j = 0 for 1 ≤ j ≤ d, there is some j with 1 ≤ j ≤ d such that x2j 6= 0. Then since h = 2 and therefore n = 2d,

kxk kx2k 0 n−2d q(x) = = ≥ 1 = C = C 2d . hxi hx2i

n Now, let x ∈ R \S with f(x) ≤ 0. Then one of x1j with 1 ≤ j ≤ d must be nonzero. Otherwise, x ∈ S, which is a contradiction. Therefore, kx1k 6= 0. Since f(x) ≤ 0, it follows that d d X 2 2 X B x1j ≤ x1jx2j. j=1 j=1

If x21 = ··· = x2d = 0, then x11 = ··· = x1d = 0 and x = 0, a contradiction. Therefore, x2 6= 0 and kx2k 6= 0. Hence, eq. (12.3) from the h ≥ 3 case still holds. Therefore 2 −2 2 −2 kx1k ≤ d B kx2k ≤ n B kx2k and hence 2 kx2k B n 2 2d q(x) ≥ ≥ 2 n B n C = C kx1k n since n = 2d. Case 2b: Now suppose d - n. Then, by the division algorithm, n = hd+l for h, l ∈ Z and 0 < l < d. Relabel X1,...,Xn as Yij with 1 ≤ i ≤ h, 1 ≤ j ≤ d and and Ym with 1 ≤ m ≤ l. Let L1(W1,...,Wd),...,Ld(W1,...,Wd) be the forms of Corollary 12.2. As before, h ≥ 2. When h ≥ 3, put

Zij = Yij − BYi−1,j (3 ≤ i ≤ h, 1 ≤ j ≤ d) (12.6a) l/d Zm = Ym − B Lm(Yh1,...,Yhd) (1 ≤ m ≤ l). (12.6b)

141 Deﬁne

d d h j l X 2 2 X X X 2 X 2 f(X) = B Y1j − Y1jLj(Y21,...,Y2d) + Zij + Zm. j=1 j=1 i=3 j=1 m=1

Since B is an integer and L1,...,Ld have integral coeﬃcients, f has integral co- 1−l/d l/d eﬃcients. Since |Lm| ≤ B , |B Lm| ≤ B. In addition, B ≥ 1 gives that 1 l 1− d 2 2 2 2 2 B ≤ B ≤ B and B − 1 ≤ B so kFk ≤ 2B . In case 2a it was shown that this, along with the right choice of η(n), gives kFk ≤ C. Note that, by substituting for Zij and Zm and completing the square, f can be re-written as

d 2 d X 1 X 1 f(X) = BY − Y L (Y ,...,Y ) − Y L (Y ,...,Y )2 1j 2B 1j j 21 2d 4B2 1j j 21 2d j=1 j=1 h d l X X 2 X l/d 2 + (Yij − BYi−1,j) + Ym − B Lm(Yh1,...,Yhd) . i=3 j=1 m=1

Since the Li’s are linearly independent, the square terms are linearly independent. As there are d + d + d(h − 2) + l = dh + l = n square terms, this gives that f is nondegenerate. Let S be the rational subspace given by Y1j = Z3j = ··· = Zhj = 0 with 1 ≤ j ≤ d and Z1 = ··· = Zl = 0. Then it is straightforward to show that Y2j with 1 ≤ j ≤ d are the free variables and thus dim(S) = d. Also, f vanishes on S. Let x ∈ S with x 6= 0. Let y2 = (y21, . . . , y2d). By way of contradiction, assume y2 = 0. Then 0 = z3j = y3j − By2j = y3j and so on until yij = 0 for 1 ≤ i ≤ h, 1 ≤ j ≤ d. In particular, yh1, . . . , yhd = 0. Combining this with zm = 0 for 1 ≤ m ≤ l gives ym = 0 for 1 ≤ m ≤ l. Hence x = 0, a contradiction. So y2 6= 0. Then z3j = ··· = zhj = 0 for 1 ≤ j ≤ d implies

y3j = By2j

y4j = By3j 2 = B(By2j) = B y2j

y5j = By4j 2 3 = B(B y2j) = B y2j. Iterating this process gives h−2 yhj = B y2j (12.7) for 1 ≤ j ≤ d. Now, zm = 0 for 1 ≤ m ≤ l along with eq. (12.15) give

l ym = B d Lm(yh1, . . . , yhd) l h−2 h−2 = B d Lm(B y21,...,B y2d) l +h−2 = B d Lm(y21, . . . , y2d). (12.8) Now, Corollary 12.2 gives that

l l l max{ky2k,B d |L1(y2)|,...,B d |Ll(y2)|} n B d hy2i. (12.9)

142 Suppose this maximum is ky2k. Consider that

h−2 h−2 max |yhj| = B max |y2j| = B ky2k. 1≤j≤d 1≤j≤d

Then h−2 h−2+ l max |yhj| = B ky2k n B d hy2i. 1≤j≤d

Since kxk ≥ max1≤j≤d |yhj| and hy2i ≥ hxi, this gives

h−2+ l kxk n B d hxi.

l Now suppose the maximum in eq. (12.9) is one of the B d |Lm(y2)| where 1 ≤ m ≤ l. Consider that, by eq. (12.8),

h−2 l h−2+ l max |ym| = B max B d |Lm(y2)| n B d hy2i. 1≤m≤l 1≤m≤l

Again, this gives h−2+ l kxk n B d hxi.

1 Therefore in either case, since C 2 n B,

h−2+ l dh−2d+l dh−2d+l n−2d kxk n B d hxi = B d hxi n C 2d hxi = C 2d hxi, so that kxk n−2d q(x) = C 2d hxi n for all nonzero x ∈ S. n Now, consider x ∈ R \S with f(x) ≤ 0. Let y1 = (y11, . . . , y1d) and deﬁne M := max{|L1(y2)|,..., |Ld(y2)|}. Since f(x) ≤ 0,

d d h l d X 2 2 X X 2 X 2 X B y1j + zij + zm ≤ y1jLj(y2) i=1 j=1 i=3 m=1 j=1

≤ dky1kM. (12.10)

2 2 Pd 2 2 2 2 2 Lemma 2.33 gives that B ky1k ≤ d i=1 B y1j so that B ky1k ≤ d ky1kM. Now, we will show that y1 6= 0. Assume y1 = 0. Then since f(x) ≤ 0, it follows that zij = 0 for 1 ≤ j ≤ d and 3 ≤ i ≤ h and zm = 0 for 1 ≤ m ≤ l. Then x ∈ S, which is a contradiction. So y1 6= 0 and ky1k= 6 0 and therefore

2 2 B ky1k ≤ d M and 2 −2 ky1k ≤ d B M. (12.11) Then 2 −2 2 ky1kM ≤ d B M . (12.12)

143 2 3 −2 2 By eqs. (12.10) and (12.12), zij ≤ d B M so that

3 −1 |zij| ≤ d 2 B M. (12.13)

Similarly, 3 −1 |zm| ≤ d 2 B M. (12.14)

Case 2b(i): Suppose M = max{|L1(y2)|,..., |Ll(y2)|}. Without loss of generality, say M = |Ll(y2)|. Suppose y2 = 0. Then f(x) ≤ 0 implies zm = 0 for 1 ≤ m ≤ l and zij = 0 for 3 ≤ i ≤ h and 1 ≤ j ≤ d. This then implies ym = 0 and yij = 0 and hence x = 0, a contradiction. Therefore y2 6= 0. By deﬁnition of zij, we have

y3j = z3j + By2j

y4j = z4j + By3j

= z4j + B(z3j + By2j) 2 = z4j + Bz3j + B y2j.

Iterating this process gives

h−3 h−2 yhj = zhj + Bzh−1,j + ··· + B z3j + B y2j. (12.15)

By deﬁnition of ym, eq. (12.15) gives

l ym = zm + B d Lm(yh1, . . . , yhd) h−3 h−3 ! l X k h−2 X k h−2 = zm + B d Lm B zh−k,1 + B y21,..., B zh−k,d + B y2d k=0 k=0 h−3 ! X l +k l +h−2 = zm + B d Lm(zh−k,1, . . . , zh−k,d) + B d Lm(y2). k=0 Therefore by eqs. (12.13) and (12.14)

l +h−2 l +h−2 |B d Ll(y2)| = |B d Ll(y2) − yl + yl| l +h−2 ≤ |yl − B d Ll(y2)| + |yl| h−3 ! X l +k d = zl + B Ll(zh−k,1, . . . , zh−k,d) + |yl| k=0 h−3 ! X l +k ≤ |zl| + B d |Ll(zh−k,1, . . . , zh−k,d)| + |yl| k=0 h−3 d ! X l +k X 1− l ≤ |zl| + B d B d |zhj| + |yl| k=0 j=1

144 h−3 d ! X k+1 X = |zl| + B |zhj| + |yl| k=0 j=1 h−3 ! 3 −1 X k+1 3 −1 ≤ d 2 B M + B d(d 2 B M) + |yl| k=0 h−3 ! 5 −1 X k 5 ≤ d 2 B M + B d 2 M + |yl| k=0 5 h−3 ≤ (h − 1)d 2 B M + |yl|.

Therefore by eq. (12.11)

l +h−2 5 h−3 |yl| ≥ B d |Ll(y2)| − (h − 1)d 2 B M l +h−2 5 h−3 = B d M − (h − 1)d 2 B M

l +h−2 5 −1− l = B d M 1 − (h − 1)d 2 B d

l +h −2 5 −1− l ≥ B d d ky1k 1 − (h − 1)d 2 B d

l +h −2 1 −1− l = B d ky1k d − (h − 1)d 2 B d .

l Since n = dh + l, it follows that d, h − 1 < n. Also since l < d, it follows that d < 1. Therefore l +h −2 1 −1 |yl| ≥ B d ky1k n − n · n 2 B .

4 7 −2 1 −1 Again, since B ≥ n and n > 1, it follows that B > n 2 . Therefore n −n·n 2 B > 0 and thus l +h |yl| n B d ky1k. (12.16)

1 It has been shown that C 2 n B, so

l +h dh+l n n n |yl| n B d ky1k = B d ky1k = B d ky1k n C 2d ky1k ≥ C 2d hy1i.

Therefore |yl| n q(x) ≥ n C 2d , hy1i as desired. Case 2b(ii): Suppose M = max{|Ll+1(y2)|,..., |Ld(y2)|}. Then part (b) of Corol- lary 12.2 gives

l l ky2k n B d max{|L1(y2)|,..., |Ld(y2)|} = B d M. (12.17)

145 Then, by combining eqs. (12.13) to (12.15), we have

h−2 h−2 |B y2j| = |B y2j − yhj + yhj| h−2 ≤ |B y2j − yhj| + |yhj| h−3 = |zhj + Bzh−1,j + ··· + B z3j| + |yhj| 3 −1 3 −1 h−3 3 −1 ≤ d 2 B M + B(d 2 B M) + ··· + B (d 2 B M) + |yhj| 3 h−4 ≤ (h − 2)d 2 B M + |yhj|.

By eqs. (12.12) and (12.17) this gives

h−2 3 h−4 |yhj| ≥ B |y2j| − (h − 2)d 2 B M h−2 3 h−4 = B ky2k − (h − 2)d 2 B M h−2+ l 3 h−4 ≥ B d M − (h − 2)d 2 B M

h−2+ l 3 −2− l = B d M 1 − (h − 2)d 2 B d

h+ l −2 3 −2− l ≥ B d d ky1k 1 − (h − 2)d 2 B d

h+ l −2 − 1 −2− l = B d ky1k d − (h − 2)d 2 B d .

l Since n = hd + l, we have h − 2, d < n. Recall that 1 ≤ d and l < d so d < 1. Therefore h+ l −2 −2 |yhj| ≥ B d ky1k n − nB .

4 3 −2 −2 Since B ≥ n > n 2 , it follows that n − nB > 0 so that

h+ l |yhj| n B d ky1k.

1 Since C 2 n D,

h+ l dh+l n n n |yhj| n B d ky1k = B d ky1k = B d ky1k n C 2d ky1k ≥ C 2d hy1i.

Hence |yhj| n q(x) ≥ n C 2d . hy1i Now consider when h = 2. Let

d d l X 2 2 X X 2 f(X) = B Y1j − Y1jLj(Y21,...,Y2d) + Zm. j=1 j=1 m=1

Just as before, kFk ≤ 2B2 which implies kFk ≤ C. Let S be the s-dimensional rational subspace deﬁned by Y1j = 0 for 1 ≤ j ≤ d and Zm = 0 for 1 ≤ m ≤ l. Then nonzero x = (y11, . . . , y1d, y21, . . . , y2d, y1, . . . , yl) ∈ S has y1j = 0 for 1 ≤ j ≤ d. Let y2 = (y21, . . . , y2d). Suppose y2 = 0. Then, since zm = 0 for 1 ≤ m ≤ l, it follows

146 that ym = 0 for 1 ≤ m ≤ l. Then x = 0, which is a contradiction. So y2 6= 0 and ky2k, hy2i= 6 0. Corollary 12.2 gives that

l l l max{ky2k,B d |L1(y2)|,...,B d |Ll(y2)|} n B d hy2i.

Suppose the maximum of this set is ky2k. Then, since h = 2 and n = hd + l,

ky2k l l +h−2 l+hd−2d l+hd−2d n−2d q(x) ≥ n B d = B d n B d n C 2d = C 2d . hy2i

l Now suppose, without loss of generality, the maximum of this set is B d |Ll(y2)|. Since l zm = 0 for 1 ≤ m ≤ l, it follows that ym = B d Lm(y21, . . . , y2d). Then

l l l max |ym| = B d max |Lm(y2)| = B d |Ll(y2)| n B d hy2i. 1≤m≤l 1≤m≤l

Then following a similar process to the one above gives

max1≤m≤l |ym| l n−2d q(x) ≥ n B d n C 2d . hy2i

Now consider nonzero x ∈/ S with f(x) ≤ 0. We can show y1, y2 6= 0 as in the h ≥ 3 case (see case 2b). Recall M = max{|L1(y2)|,..., |Ld(y2)|} and suppose M = max{|L1(y2)|,..., |Ll(y2)|}. Without loss of generality, we can assume M = |Ll(y2)|. l Now, yl = zl + B d Ll(y2) and by eq. (12.14)

l l |B d Ll(y2)| = |B d Ll(y2) − yl + yl| l ≤ |yl − B d Ll(y2)| + |yl|

= |zl| + |yl| 3 −1 ≤ d 2 B M + |yl|.

Therefore by eq. (12.11)

l 3 −1 |yl| ≥ B d |Ll(y2)| − d 2 B M l 3 −1 = B d M − d 2 B M l 3 −1− l = B d M(1 − d 2 B d ) l 2 −2 3 −1− l ≥ B d B d ky1k(1 − d 2 B d ) l +2 −2 − 1 −1− l = B d ky1k(d − d 2 B d ).

−2 − 1 −1− l −2 −1 Since 1 < d < n, it follows that d − d 2 B d ≥ n − B . Therefore |yl| ≥ l +2 −2 −1 4 2 −2 −1 B d ky1k(n −B ). Since B ≥ n , it follows that B > n and therefore n −B > 0 and l +2 |yl| n B d ky1k.

147 Since h = 2 this gives

|yl| l +h l+hd n n q(x) ≥ n B d = B d = B d n C 2d . ky1k

Now suppose M = max{Ll+1(y2),...,Ld(y2)|}. By Corollary 12.2 and eq. (12.11), we have

l ky2k n B d max{|L1(y2)|,..., |Ld(y2)|} l = B d M l 2 −2 ≥ B d B d ky1k l +2 −2 = B d d ky1k l +2 −2 ≥ B d n ky1k.

l +h Since h = 2, this gives ky2k n B d ky1k so that

ky2k l +h l+hd n n q(x) ≥ n B d = B d + B d n C 2d . ky1k

Comparing Theorem 11.4 with Theorem 12.3, we can see that statement (iv) of Theorem 12.3 is an improvement on statement (iv) of Theorem 11.4. However, we cannot make a direct comparison between part (iii) of each statement, as part (iii) of Theorem 11.4 has the extra hypothesis that f(x) ≤ 0 and a stronger result since n−d n−2d 2d ≥ 2d . Now, we consider the implications of this theorem. Let C ≥ 1 and suppose Pn Pn n ≥ 2d > 0. Suppose f(x1, . . . , xn) = i=1 j=1 fijxixj is nondegenerate, isotropic n n over Z and f(Z ) ⊆ Z. Then by Lemma 2.27, f ∈ Z[x1, . . . , xn]. Suppose f vanishes on a sublattice Γ ⊂ Zn with dim(Γ) = d. Then Γ generates a rational n linear subspace S of dimension d where f vanishes. Given x1,..., xn ∈ Z linearly n independent nontrivial solutions of f, at most d of them lie in S. Since xi ∈ Z implies kxik ≥ q(xi), Theorem 12.3 gives

kx1k · · · kxnk ≥ q(x1) ··· q(xn) n−2d d n n−d n (C 2d ) · (C 2d ) 2 n −d = C 2d 2 n −d ≥ kFk 2d .

Therefore the exponent on kFk in [17, Cor. 1] (see eq. (1.16)) is best possible.

148 Appendices

Appendix A: Basis of a Projection Lattice

Let Z denote the ring of integers. Let A = (aij) be an m × n matrix with entries in n Z. Thus each aij ∈ Z. Let M be the Z-submodule of Z generated by the rows of A. Thus M is the set of all Z-linear combinations of the vectors wi := (ai1, ai2, . . . , ain), 1 ≤ i ≤ m.

Proposition A.1. With the notation above, M has a Z-basis v1,..., vr, r ≤ min{m, n}, where these vectors satisfy the following properties.

1. vi = (0,..., 0, bii, . . . , bin), bij ∈ Z, i ≤ j ≤ n, 1 ≤ i ≤ r.

2. For each j with 1 ≤ j ≤ r, we have |bij| ≤ |bjj| for 1 ≤ i ≤ j. Proof. (1) The proof is by induction on n ≥ 1. First assume that n = 1. (Thus A is an m×1 column matrix.) As Z is a PID, it follows that the ideal (a11, a21, . . . , am1) = (b) for some nonnegative integer b. The inclusion “⊆” implies that b | ai1 for 1 ≤ i ≤ m, Pm and the inclusion “⊇” implies that b ∈ i=1 Z · ai. It follows that b is a Z-basis for M, the Z-submodule of Z generated by the rows of the column matrix A. Now assume that n ≥ 2 and that the proposition has been proved for matrices n as above with fewer than n columns. Let e1,..., en be the standard basis of Z . n Define the projection map π : Z → Z by π(x1, . . . , xn) = x1. Let K = ker(π). Thus K = Ze2 + ··· + Zen. Since im(π) ⊆ Z, we have im(M) = (b11) for some nonnegative integer b11. Suppose first that b11 = 0. Then ∼ n−1 M ⊆ ker(π) = Ze2 + ··· + Zen = Z . By induction on n, there is a Z-basis of M that satisfies (1). Now suppose that b11 6= 0. Choose v1 ∈ M such that π(v1) = b11. Then v1 = b11e1 +b12e2 +···+b1nen for some b1j ∈ Z, 2 ≤ j ≤ n. Then M = Z·v1 ⊕(M ∩ker(π)). (If x ∈ M and π(x) = tb11, then x − tv1 ∈ M ∩ ker(π). The rest of the argument ∼ n−1 is straightforward.) Since M ∩ ker(π) ⊆ K = Z , by induction on n, there is a Z- basis v2,..., vr of M ∩ ker(π), for some r, that satisfies (1). (2) Assume that we have a Z-basis of M that satisfies (1). We prove by induction on j ≥ 1 that we can arrange this basis to have the additional property given in (2). The statement is trivial for j = 1. Now suppose that j ≥ 2 and that the statement has been proven for smaller values of j. For each i, 1 ≤ i < j, choose ci ∈ Z so that when we add ci times row j to row i we have |cbjj + bij| < |bjj|. This operation does not affect any entries in columns 1, . . . , j − 1. The new rows of A are still a Z-basis of M, and now (2) holds for the first j rows. The proof finishes by induction on j. Proposition A.2. Let L = Zn. Choose v ∈ L such that v 6= 0 and define W = Rv. Let π : Rn → W ⊥ be the projection map with respect to the standard inner product n on R . Then π(L) has a basis p2,..., pn such that kpik n 1 for 2 ≤ i ≤ n.

149 Proof. Recall π(L) is an (n − 1)-dimensional lattice by Lemma 3.29. Write v = n (a1, . . . , an) with ai ∈ Z. Let e1,..., en be the standard basis vectors of Z . Since e1,..., en generate L, π(e1), . . . , π(en) generate π(L). Now v = a1e1 +···+anen ∈ W Pn so π(v) = i=1 aiπ(ei) = 0. Since v 6= 0, without loss of generality we can re-index so that an = kvk and, if necessary, multiply v by −1 so that an > 0. Therefore we have the dependence relation

n−1 X ai π(e ) = − π(e ), n a i i=1 n and π(e ), . . . , π(e ), − Pn−1 ai π(e ) generate π(L). 1 n−1 i=1 an i Deﬁne  1 0 ······ 0   0 1 0 ··· 0     . .. .   . . .  A :=   .  0 ······ 1 0     0 ······ 0 1  − a1 − a2 · · · · · · − an−1 an an an

Thus A is an n × (n − 1) matrix whose rows generate π(L) with respect to π(e1),..., n π(en−1). Define B := anA and let M be the Z-module of Z generated by the rows of B. Then by Proposition A.1, there is a matrix B0 whose first n − 1 rows are the Z-basis v1,..., vn−1 of M given in the proposition and whose last row is the zero vector. Define A0 := 1 B0. Then the rows of A0 form a basis for the same -module an Z as 1 B = A. an Now consider the first row of A, [1, 0,..., 0], which must be a Z-linear combination of the rows of A0. Only the first row of A0 has a nonzero element in the first column, 0 0 and therefore there is some nonzero integer z such that z · a11 = 1. Hence |a11| = 1 | z | ≤ 1. The second row of A is [0, 1, 0,..., 0] is a Z-linear combination of the rows of A0. Only the first two rows of A0 have nonzero entries in the second column, so there 0 0 0 are nonzero integers x, y such that xa12 + ya22 = 1. This also implies xa11 = 0 and 0 0 1 hence x = 0. Therefore ya22 = 1 and |a22| = | y | ≤ 1. Continuing this process gives 0 0 0 |ajj| ≤ 1 for 1 ≤ j ≤ n − 1. Now, since |bij| ≤ |bjj| for 1 ≤ i ≤ j and 1 ≤ j ≤ n − 1, 0 0 0 0 |aij| ≤ |ajj| for 1 ≤ i ≤ j and 1 ≤ j ≤ n − 1. Thus, |aij| ≤ 1 for all entries of A . T 0 T Define [p2 p3 ... pn 0] := A [π(e1) . . . π(en−1)] . Then p2,..., pn form a basis of π(L). Then

0 0 kp2k2 = ka11π(e1) + ··· + a1,n−1π(en−1)k2 0 0 ≤ |a11| kπ(e1)k2 + ··· + |a1,n−1| kπ(en−1)k2. n ⊥ Now since R = W ⊕ W by Corollary 2.24, ei can be written ei = yi + zi where ⊥ 2 2 2 yi ∈ W and zi ∈ W . Then 1 = keik2 = kyik2 + kzik2, which implies kzik2 ≤ 1. Since π(ei) = zi, kπ(ei)k2 ≤ 1. Hence 0 0 kp2k2 ≤ |a11| + ··· + |a1,n−1| ≤ n − 1.

150 Similarly we show kp3k2 ≤ n − 2,..., kpnk2 ≤ 1. We know from Lemma 2.33 that kpik ≤ kpik2, and therefore kpik n 1 for 2 ≤ i ≤ n. Lemma A.3. Let L = Zn. Choose v ∈ L such that v 6= 0 and deﬁne W = Rv. Let π : Rn → W ⊥ be the projection map with respect to the standard inner product n on R . Let ˜p2,..., ˜pn be the successive minimum vectors of π(L). Then π(L) has a basis p2,..., pn such that for 2 ≤ i ≤ n

(1) k˜pik n kpik n k˜pik

(2) | det π(L)| n kp2k · · · kpnk n | det π(L)|

(3) kpik n 1. Proof. Recall π(L) is an (n − 1)-dimensional lattice by Lemma 3.29. Statements (1) 0 0 and (2) are given by Theorem 3.26. To show (3), consider the basis p2,..., pn of 0 π(L) given by Proposition A.2 so that kpik n 1 for 2 ≤ i ≤ n. Re-index so that 0 0 kp2k ≤ · · · ≤ kpnk. Then for 2 ≤ i ≤ n

kpik n k˜pik (from (1)) 0 ≤ kpik (by Lemma 3.16)

n 1 (by Proposition A.2).

Hence (3) holds.

Appendix B: Hermite’s Constant

Proposition B.1. Let f(x1, . . . , xn) ∈ R[x1, . . . , xn] be a positive deﬁnite quadratic form. Then f attains a minimum on Zn − {0}.

Proof. By Proposition 2.19, there exist L1,...,Ln ∈ R[x1, . . . , xn] linearly independent forms such that 2 2 f = L1 + ··· + Ln Pn and Li = j=1 cijxj, for 1 ≤ i ≤ n. Let C = (cij), which is an n × n invertible −1 matrix. Denote its inverse C = (dij). Since there are a ﬁnite number of dij, we can ﬁnd a bound such that |dij| ≤ M, for all i, j. n Now, let b = (b1, . . . , bn) ∈ Z − {0}. Then   L1(b)  .  Cb =  .  . Ln(b) And so   L1(b) −1  .  b = C  .  . Ln(b)

151 Then n n X X |bi| ≤ M|Lj(b)| = M |Lj(b)|. j=1 j=1 And so n 2 2 2 X 2 2 2 bi ≤ M |Lj(b)| ≤ M n(|L1(b)| + ··· + |Ln(b)| ) j=1 by Lemma 2.30. Now, let K ∈ R such that K > 0. Suppose b ∈ Zn − {0} such that

n X 2 f(b) = Li(b) ≤ K. i=1 Then 2 2 2 2 2 bi ≤ M n(|L1(b)| + ··· + |Ln(b)| ) ≤ M nK, which implies √ |bi| ≤ M nK, for 1 ≤ i ≤ n. However, there are only finitely many integer values that satisfy this inequality. Therefore, there are finitely many b such that f(b) ≤ K and so we need only check finitely many values, one of which will attain the minimum of f.

Lemma B.2. Let S be an n × n matrix with entries in R and S = (sij). Assume that S is symmetric, i.e. sij = sji, and s11 6= 0. Then there exists A ∈ GLn(R) such that   s11 0 ··· 0  0  T   A SA =  .   . T  0 and for all k, 1 ≤ k ≤ n, corresponding k × k submatrices of S and AT SA that contain the ﬁrst row and column have the same determinant.

Proof. Let 1 − s12 · · · − s1n  s11 snn 0  A =   . .  . In−1  0

s1j Let s1,..., sn be the columns of S. Then multiplying S by A on the right adds − s1 s11 0 0 to sj for 2 ≤ j ≤ n. Now, let s1,..., sn be the rows of SA. Then multiplying SA on the left by AT adds − sj1 s0 to s0 for 2 ≤ j ≤ n. These row and column operations s11 1 j preserve the determinant of any principal k × k submatrix of S, 1 ≤ k ≤ n, that contains the ﬁrst row and column.

152 Definition B.3. An n × n matrix P with entries in R is positive definite if for every nonzero x ∈ Rn, xT P x > 0. Lemma B.4. Let A be a non-singular matrix with real entries. Then S is positive definite if and only if AT SA is positive definite.

Proof. First assume S is positive deﬁnite. Let x ∈ Rn, with x 6= 0. Then, since A is invertible, Ax 6= 0. Since S is positive deﬁnite, we have that

T n y Sy > 0 ∀y ∈ R , y 6= 0. In particular, this gives

(Ax)T SAx = xT AT SAx > 0.

Therefore AT SA is positive deﬁnite. Now assume AT SA is positive deﬁnite. Then for nonzero x ∈ Rn, we have 0 < xT AT SAx = (Ax)T SAx.

Since A is invertible, every nonzero y ∈ Rn has the form y = Ax, with x 6= 0, and we can write x = A−1y. Therefore

0 < (Ax)T SAx = (AA−1y)T S(AA−1)y = yT Sy and thus S is positive definite. Theorem B.5. Let S be the space of n × n symmetric matrices with entries in R. Let Sk be the upper-lefthand k × k submatrix of an n × n matrix S. Then for S ∈ S, S is positive definite if and only if det(Sk) > 0 for 1 ≤ k ≤ n. Proof. We proceed by induction on n. If n = 1, then the result is obvious. Assume that n ≥ 2 and that the result is true for k ×k matrices where k < n. Let S = (sij) ∈ T S. If S is positive definite, then (1, 0,..., 0)S(1, 0,..., 0) = s11 > 0. If det(Sk) > 0 for 1 ≤ k ≤ n, then det(S1) = s11 > 0. Thus in either direction, we can assume s11 > 0. By Lemma B.2, there exists A ∈ GLn(R) such that   s11 0 ··· 0  0  T   A SA =  .   . T  0

T and det(Sk) = det((A SA)k) for 1 ≤ k ≤ n (note that Lemma B.2 actually gives a stronger result, but this is all we need). Since

(AT SA)T = AT ST A = AT SA

153 we have that AT SA is symmetric and therefore we know that T is symmetric. Then

S is positive deﬁnite ⇐⇒ AT SA is positive deﬁnite, by Lemma B.4,

⇐⇒ T is positive deﬁnite (to see this let x = (0, x2, . . . , xn)),

⇐⇒ det(Tk) > 0, for 1 ≤ k ≤ n − 1, by the inductive hypothesis, T ⇐⇒ det((A SA)k) > 0, for 1 ≤ k ≤ n,

⇐⇒ det(Sk) > 0 for 1 ≤ k ≤ n, by Lemma B.2.

Deﬁnition B.6. For an n × n matrix A with n ≥ 2, its adjoint matrix is

∗ ∗ A := (aij)

∗ i+j where aij = (−1) Mji and Mji is the determinant of the (n − 1) × (n − 1) matrix resulting from the removal of the jth row and ith column of the matrix A. If n = 1, ∗ ∗ ∗ then A := [1]. It is well known that AA = A A = det(A)In and so if A is invertible, A∗ = det(A)A−1.

Note: The following ﬁve lemmas can be proved for all n × n matrices with entries in some commutative ring R. However, we are only concerned with the case in which A ∈ GLn(R), so these are the only cases for which we will prove the results.

∗ ∗ ∗ Lemma B.7. If A, B ∈ GLn(R), then (AB) = B A . Proof. We have

(AB)∗ = det(AB)(AB)−1 = det(A) det(B)B−1A−1 = det(B)B−1 det(A)A−1 = B∗A∗.

∗ T T ∗ Lemma B.8. For A ∈ GLn(R), (A ) = (A ) .

∗ ∗ Proof. Since A A = det(A)In, it follows that A A is symmetric. Therefore we have

T T ∗ T ∗ ∗ T T ∗ T A (A ) = det(A )In = det(A)In = A A = (A A) = A (A ) .

T ∗ T T ∗ Since A ∈ GLn(R), A ∈ GLn(R) and therefore (A ) = (A ) . ∗ Corollary B.9. If A ∈ GLn(R) is symmetric, then A is symmetric as well. Proof. By Lemma B.8, (A∗)T = (AT )∗ = A∗.

154 ∗ n−1 Lemma B.10. For A ∈ GLn(R), det(A ) = det(A) . Proof. We have

∗ AA = det(A)In ∗ n det(AA ) = det(A) det(In) det(A) det(A∗) = det(A)n det(A∗) = det(A)n−1.

∗ ∗ n−2 Lemma B.11. For A ∈ GLn(R) and n ≥ 2, (A ) = det(A) A. Proof. We have

(A∗)∗ = det(A∗)(A∗)−1 = det(A)n−1(A∗)−1 (by Lemma B.10) = det(A)n−1(det(A)A−1)−1 1 = det(A)n−1 · · A det(A) = det(A)n−2A.

Lemma B.12. Let A ∈ GLn(R) be symmetric. Let B ∈ GLn(R) be a matrix such that BT AB = D, where D is a diagonal matrix. Then

(i) A is positive deﬁnite if and only if BT AB is positive deﬁnite,

(ii) D is positive deﬁnite if and only if D−1 is positive deﬁnite,

(iii) A is positive deﬁnite if and only if A−1 is positive deﬁnite,

(iv) If A is positive deﬁnite, then A∗ is positive deﬁnite.

(v) If A∗ is positive definite, then either A is positive definite or A is negative definite.

Proof. Statement (i) is true by Lemma B.4. Now, let d1, . . . , dn be the diagonal T entries of D. Let e1,..., en be the standard basis vectors. Then ei Dei = di. Hence if D is positive definite, di > 0 for 1 ≤ i ≤ n. Now, if d1, . . . , dn > 0, then T 2 2 n x Dx = d1x1 + ··· + dnxn > 0 for all nonzero x ∈ R . Hence D is positive definite. 1 1 1 1 Now d1, . . . , dn > 0 if and only if ,..., > 0. Since ,..., are the diagonal d1 dn d1 dn entries of D−1, this holds if and only if D−1 is positive definite. Thus (ii) holds. Now A is positive definite if and only if BT AB = D is positive definite if and only if D−1 = (BT AB)−1 = B−1A−1(B−1)T is positive definite if and only if A−1 is positive definite (by part (i)). Hence (iii) holds.

155 To show statement (iv), assume A is positive deﬁnite. Then

xT A∗x = xT det(A)A−1 x = (det(A))(xT A−1x).

Theorem B.5 gives that det(A) > 0 and part (iii) gives that A−1 is positive definite. Therefore xT A∗x > 0 for all nonzero x ∈ Rn and A∗ is positive definite. Thus (iv) holds. Assume A∗ is positive definite. Then by part (iv)(A∗)∗ is positive definite and by Lemma B.11 (A∗)∗ = det(A)n−2A. Then for all nonzero x ∈ Rn, xT det(A)n−2Ax > 0. Suppose det(A) > 0. Then det(A)n−2 > 0 and thus xT Ax > 0 for all nonzero x and hence A is positive definite. Now, suppose det(A) < 0. If n is even, n−1 is odd. Then det(A∗) = det(A)n−1 < 0, which is a contradiction by Lemma B.10. Therefore, n is odd. By part (iv), (A∗)∗ is positive definite and by Lemma B.11, (A∗)∗ = det(A)n−2A. Then for all nonzero x ∈ Rn, xT det(A)n−2Ax > 0. Now det(A)n−2 < 0, and therefore xT Ax < 0 for all n nonzero x ∈ R . Hence A is negative definite. Theorem B.13. Let a, b, c ∈ R and f = ax2+bxy+cy2 be a positive definite quadratic form with associated matrix 1 a 2 b M = 1 . 2 b c 1 2 2 Let ∆(f) = det(M) = ac − 4 b . Then the minimum of f on Z − {0}, denoted m, is q 4∆ q 4∆ less than or equal to 3 . The minimum equals 3 if and only if f is equivalent to the form m(x2 + xy + y2).

Proof (following [22], page 91). First we note that, since f is positive definite, f(x) = xT Mx > 0 for all nonzero x = (x, y)T ∈ Rn. Then Theorem B.5 implies ∆(f) = 1 2 det(M) = ac − 4 b > 0 and so 4∆(f) > 0. Now by Proposition B.1, f attains a minimum value m over Z2 − {0}. Then there is some (p, r) ∈ Z2 − {0} such that f(p, r) = m. Since m is a minimum, gcd(p, r) = 1 and therefore, by the unimodular row lemma (see [13]), the vector (p, r) can be extended to an integral basis whose matrix, U, is unimodular and has (p, r) as the first column. Since (U T MU)T = U T M T U = U T MU, we can define a quadratic T T T T form g(x) := x U MUx. Then g(e1) = e1 U MUe1 = f(p, r) = m. Also, ∆(g) = det(U T MU) = det(U)2 det(M) = det(M) = ∆(f). Write g(x) = a0x2 + b0xy + c0y2. Then g(1, 0) = a0. Therefore m = a0 > 0. Now g(0, 1) = c0 and therefore m = a0 ≤ c0. Now let 1 ` V = . 0 1 Since det(V ) = 1, the map v 7→ V v is a bijection and therefore there is some T x1 = (x1, y1) such that V x1 = x. Then

T T T T T T T g(x) = x U MUx = (V x1) U MUV x1 = x1 V U MUV x1 =: g1(x1).

156 T T 2 2 Therefore, ∆(g1) = det(V U MUV ) = det(U) det(V ) det(M) = det(M) = ∆(g). Since g(x) = a0x2 + b0xy + c0y2,

0 b0 T a 2 U MU = b0 0 . 2 c

Then the matrix associated to g1 is

0 0 b0 a a ` + 2 0 b0 0 2 0 0 , a ` + 2 a ` + b ` + c

0 2 0 0 0 2 0 0 2 0 0 so that g1(x1) = a x1 + (2a ` + b )x1y1 + (a ` + b ` + c )y1. Now since [−a , a ] has length 2a0, we can always ﬁnd some integer ` such that −a0 ≤ b0 + 2a0` < a0. If 0 0 b + 2a ` is negative, then we work with g1(x1, −y1) to ensure that the coeﬃcient on 00 00 0 x1y1, which we will call b , is nonnegative. Then 0 ≤ b ≤ a . 00 0 2 0 0 00 Let c = a ` + b ` + c . Then g1(0, 1) = c . Since the unimodular transformation corresponding to V represents a bijective linear map, it follows that m = a0 is the 0 00 minimum of g1 as well and therefore a ≤ c . Now we have 0 ≤ b00 ≤ a0 ≤ c00, thus

0 00 00 2 0 2 0 2 0 2 2 4∆(g1) = 4a c − (b ) ≥ 4(a ) − (a ) = 3(a ) = 3m .

Therefore

2 4∆(f) = 4∆(g1) ≥ 3m 4∆ ≥ m2 3 r 4∆ ≥ m 3

00 00 0 Equality is achieved if and only if c = b = a = m, in which case g1(x, y) = 2 2 m(x + xy + y ). P Theorem B.14. Let fn = fijxixj where fij ∈ R, be a positive deﬁnite quadratic form in x1, . . . , xn and let M be the associated matrix. Let ∆(fn) = det(M) and let L(fn) be the minimum of fn when (x1, . . . , xn) ∈ Z − {0}.

1. Then there exists some Λn such that

n L(fn) ≤ Λn∆(fn), for all such fn

where Λn depends only on n. We deﬁne

n λn := inf{Λn|L(fn) ≤ Λn∆(fn) for all such fn}

so that n L(fn) ≤ λn∆(fn).

157 n−2 n 2. λn ≤ λn−1. Proof (following [11]). First, note that by Proposition B.1, we know this minimum T n is attained by fn. So there exists c = (c1, . . . , cn) ∈ Z such that fn(c) = L(fn). Therefore, gcd(c1, . . . , cn) = 1 and the unimodular row lemma (see [13]) gives that there exists some matrix U ∈ SLn(Z) such that the ﬁrst column of U is c. Deﬁne N := U T MU and note that since M is symmetric, N is symmetric as well. Let gn(y1, . . . , yn) be T the quadratic form associated to N. Let y = (y1, . . . , yn) Then

n n X X T gn(y1, . . . , yn) = nrsyrys = y Ny. r=1 s=1

So for e1 = (1, 0,..., 0),

T T T T n11 = gn(e1) = e1 Ne1 = e1 U MUe1 = c Mc = fn(c) = L(fn). (18)

T Now, by deﬁnition, if x = (x1, . . . , xn) , then

T T T x Mx = fn(x) and x U MUx = gn(x).

Since U is an invertible matrix, it represents a bijective linear map, and thus Ux = 0 if and only if x = 0. Therefore, fn(x) and gn(x) take on the same values and so L(fn) = L(gn). ∗ ∗ By deﬁnition, M is symmetric and therefore by Corollary B.9, M = (mij) is ∗ symmetric as well. So we let Fn(X1,...,Xn) be the quadratic form associated to M . Now consider

N ∗ = (U T MU)∗ = U ∗M ∗(U T )∗ = U ∗M ∗(U ∗)T by Lemmas B.7 and B.8. Let Gn(Y1,...,Yn) be the quadratic form associated to ∗ ∗ ∗ N . Since U ∈ SLn(Z), U has integer entries and by B.10, U ∈ SLn(Z). Then L(Fn) = L(Gn). 1/n Assume ∆(fn) 6= 1. Then we can divide every entry of M by ∆ (fn). To this matrix we associate the quadratic form

fn hn = 1/n . ∆ (fn) Then we see that

L(fn) ∆(fn) L(hn) = 1/n and ∆(hn) = = 1. ∆ (fn) ∆(fn) Therefore Ln(h ) Ln(f )/∆(f ) Ln(f ) n = n n = n ∆(hn) 1 ∆(fn)

158 n n and hence L (hn) ≤ λn∆(hn) if and only if L (fn) ≤ λn∆(fn). Therefore we may assume ∆(fn) = 1. ∗ n−1 This assumption and Lemma B.10 give that ∆(Fn) = det(M ) = det(M) = n−1 n−1 ∆ (fn) = 1 = 1. Now, let Y1 = 0. Then Gn becomes an (n − 1)-ary form, which we call Gn−1, whose minimum is at least L(Gn) = L(Fn), since we are only considering a subset of the initial values. Now, we know by Lemma B.11 that

n−2 (N ∗)∗ = det(N)n−2 · N = det(U T ) det(M) det(U) · N = 1n−2 · N = N.

∗ Therefore, since the adjoint of N is N, n11, which is an entry in N, satisﬁes

1+1 n11 = (−1) ∆(Gn−1) and so ∆(Gn−1) = n11. We now proceed by induction on n, with n ≥ 2. By Theorem B.13 we can take 4 Λ = 2 3 to satisfy this theorem, since we have shown that 4∆(f) L2(f) ≤ 3 for all positive deﬁnite f(x1, x2). We also showed in Theorem B.13 that there is a particular case in which 4∆(f) L2(f) = 3 4 and therefore we can conclude that λ2 = 3 . Assume that λ2, . . . , λn−1 have been found satisfying statement (1) of the theorem:

n−1 L (Gn−1) ≤ λn−1∆(Gn−1). We know that L(Fn) = L(Gn) ≤ L(Gn−1) and ∆(Gn−1) = n11 = L(fn) by eq. (18). Therefore we can conclude that

n−1 n−1 L (Fn) ≤ L (Gn−1) ≤ λn−1∆(Gn−1) = λn−1L(fn). (19)

We have just shown that for a positive deﬁnite quadratic form fn with associated ∗ n−1 matrix M and Fn the quadratic form associated to M , L (Fn) ≤ λn−1L(fn). Now note that, by Lemma B.11,

(M ∗)∗ = det(M)n−2 · M = 1n−2 · M = M.

159 ∗ Therefore, since Fn corresponds to M , which is positive deﬁnite by part (iv) of ∗ ∗ Lemma B.12, and fn corresponds to M = (M ) , we can conclude

n−1 L (fn) ≤ λn−1L(Fn). (20)

Then combining equations 19 and 20, we have

(n−1)2 n−1 n−1 n−1 L (fn) ≤ λn−1L (Fn) ≤ λn−1 · λn−1L(fn) so that

(n−1)2−1 n L (fn) ≤ λn−1 n n/(n−2) L (fn) ≤ λn−1 n n/(n−2) L (fn) ≤ λn−1 ∆(fn),

n/(n−2) n since ∆(fn) = 1. So we have found some Λn = λn−1 such that L (fn) ≤ Λn∆(fn). Deﬁne n λn := inf{Λn|L (fn) ≤ Λn∆(fn)}. Then n L (fn) ≤ λn∆(fn) as desired. n/(n−2) Now since Λn = λn−1 , the deﬁnition of λn gives

n/(n−2) λn ≤ λn−1 . So n−2 n λn ≤ λn−1.

Appendix C: Solutions of Nondegenerate Quadratic Forms

n Lemma C.1. Let k be a ﬁeld and let W be a subspace of k with dimk(W ) = n − 1. Then there exists a nonzero linear form

L = a1x1 + ··· + anxn ∈ k[x1, . . . , xn] such that W is the set of zeros of L.

Proof. Let {v1,..., vn−1} be a basis of W and write vi = (di1, . . . , din), 1 ≤ i ≤ n−1. n Let A be the (n − 1) × n matrix whose (i, j)-entry is dij. The function fA : k → n−1 k given by fA(v) = Av is a linear map. The dimension theorem implies that T dimk(ker(fA)) ≥ 1. Let w = (a1, . . . , an) be a nonzero element in ker(fA). Then vi · w = 0, 1 ≤ i ≤ n − 1. Let L = a1x1 + ··· + anxn ∈ k[x1, . . . , xn]. Thus W is contained in the set of zeros of L = 0. Since L doesn’t vanish on all of kn, it follows that W equals the set of zeros of L.

160 Proposition C.2. Let k be a ﬁeld, and let f ∈ k[x1, . . . , xn] be a nondegenerate quadratic form. Assume that f is isotropic over k. Let W be a subspace of kn with dimk(W ) = n − 1. Then f has an isotropic vector deﬁned over k that does not lie in W .

Proof. An invertible linear change of variables allows us to assume that the isotropic vector of f is e1 = (1, 0,..., 0). By Lemma C.1, there exists a nonzero linear form L = a1x1 + ··· + anxn ∈ k[x1, . . . , xn] such that W is the set of zeros of L. Then f = x1(b2x2 + ··· + bnxn) + Q(x2, . . . , xn), where Q is a quadratic form. We have bi 6= 0 for some i because f is nondegenerate (if b2 = ··· = bn = 0, then we could write f as a quadratic form in n − 1 variables). If a1 6= 0, then we let c = (1, 0,..., 0), and this would give f(c) = 0 and L(c) 6= 0. Thus c is an isotropic vector of f with c ∈/ W . Suppose that a1 = 0. Let M(x2, . . . , xn) = b2x2 + ··· + bnxn. Since there is at least one bi 6= 0, and L 6= 0, we have LM 6= 0. −Q(c2,...,cn) Choose c2, . . . , cn ∈ k such that L(c2, . . . , cn)M(c2, . . . , cn) 6= 0. Let c1 = . M(c2,...,cn) Then f(c) = 0 and L(c) 6= 0, as desired.

Corollary C.3. Let k be a ﬁeld, and let f ∈ k[x1, . . . , xn] be a nondegenerate quadratic form. Assume that f is isotropic over k. Then f has n linearly independent isotropic vectors deﬁned over k.

n Proof. Let v1,..., vm ∈ k be a maximal set of linearly independent isotropic vectors of f in kn. We have m ≥ 1 by hypothesis, and we clearly have m ≤ n. Suppose that n m ≤ n − 1. Let W be a subspace of k containing v1,..., vm such that dimk(W ) = n − 1. Then there exists an isotropic vector w ∈ kn of f such that w ∈/ W by Proposition C.2. Then {v1,..., vm, w} is a linearly independent set of m+1 isotropic n vectors of f in k , which is a contradiction. Thus m = n. The remaining results give another approach to proving Corollary C.3.

Lemma C.4. Let k be a ﬁeld, and let f ∈ k[x1, . . . , xn] be a nondegenerate quadratic form. Assume that f is isotropic over k. Then there exists an invertible linear change of variables that lets us write f = x1x2 + R(x3, . . . , xn). Proof. An invertible linear change of variables allows us to assume that the isotropic vector of f is v = (1, 0,..., 0). Then f = x1(b2x2 + ··· + bnxn) + Q(x2, . . . , xn), where Q is a quadratic form. We have bi 6= 0 for some i because f is nondegenerate (if b2 = ··· = bn = 0, then we could write f as a quadratic form in n − 1 variables). Another invertible linear change of variables, involving only x2, . . . , xn, allows to assume that f = x1x2 + Q(x2, . . . , xn) and therefore

f = x1x2 + Q(x2, . . . , xn)

= x1x2 + x2L(x2, . . . , xn) + R(x3, . . . , xn)

= x2(x1 + L(x2, . . . , xn)) + R(x3, . . . , xn),

161 where L is a linear form and R is a quadratic form. A last invertible linear change of variables now allows us to assume that

f = x1x2 + R(x3, . . . , xn).

Corollary C.5. Let k be a ﬁeld with char(k) 6= 2, and let f ∈ k[x1, . . . , xn] be a nondegenerate quadratic form. Assume that f is isotropic over k. Then there exists 2 2 an invertible linear change of variables that lets us write f = x1 − x2 + R(x3, . . . , xn).

Proof. By Lemma C.4, we can write f = x1x2 +R(x3, . . . , xn). Consider an invertible linear change of variables such that x1 7→ x1 + x2, x2 7→ x1 − x2 and xi 7→ xi for 3 ≤ i ≤ n. Then 1 1 x x + x · 1 = 1 2 1 −1 x2 x1 − x2 and the determinant of this change of variables matrix is −2. Hence, this is an 2 2 invertible linear change of variables where x1x2 7→ (x1 + x2)(x1 − x2) = x1 − x2. 2 2 Therefore we can write f = x1 − x2 + R(x3, . . . , xn).

Proof #2 of Corollary C.3. By Lemma C.4, we can assume that f = x1x2+R(x3, . . . , xn). Let v1 = (1, 0,..., 0) and let v2 = (0, 1, 0,..., 0). Then v1 and v2 are isotropic vectors of f. For 3 ≤ i ≤ n, let

vi = (−R(0,..., 0, 1, 0,..., 0), 1, 0,..., 0, 1, 0,..., 0),

th where the second entry and the i entry of vi are equal to 1. That is, x2 = xi = 1, x1 = −R(0,..., 0, 1, 0,..., 0), and xj = 0 where j 6= 1, 2, i. Then v3,..., vn are isotropic vectors of f. It is straightforward to check that v1,..., vn are linearly independent isotropic vectors of f.

Appendix D: Supplement to Chapter 9

Let V be ﬁnite dimensional vector space over k, with char(k) 6= 2. Let f : V → k be a nondegenerate quadratic map and suppose f vanishes on a subspace W ⊆ V . Let ⊥f W = {v ∈ V | Bf (v,W ) = 0}.

⊥f ⊥ ⊥f Lemma D.1. Assume that Bf (W, W ) = {0}. Let Y = W ∩ W . Then W = W L Y .

Proof. The hypothesis implies that W ⊆ W ⊥f . Thus W + Y ⊆ W ⊥f . We have W ∩ Y = {0} because Y ⊆ W ⊥ and W ∩ W ⊥ = {0} by Corollary 2.24. Let v ∈ W ⊥f . Then v = x + z where x ∈ W and z ∈ W ⊥. This gives z = v − x ∈ W ⊥f ∩ W ⊥ = Y . Thus v = x + z ∈ W + Y . It follows that W ⊥f = W + Y , and thus ⊥ L W f = W Y because W ∩ Y = {0}.

Lemma D.2. Suppose dimk(V ) = n and W ⊆ V is s-dimensional over k. Then ⊥ there are at least s linearly independent pi’s in W such that Bf (pi,W ) 6= 0.

162 ⊥ Proof. Suppose that there are at most s − 1 linearly independent pi’s in W such that Bf (pi,W ) 6= 0. Then there are at least (n − s) − (s − 1) = n − 2s + 1 linearly ⊥ independent pi’s in W such that Bf (pi,W ) = 0. Let Z be the span of the pi’s such ⊥f ⊥ that Bf (pi,W ) = 0. Then Z ⊆ W ∩ W =: Y . We have dimk(Z) ≥ n − 2s + 1 ⊥f because the pi’s are linearly independent. Now dimk(W ) = n − s by Lemma 2.15. ⊥f Then, by Lemma D.1, dimk(Y ) = dimk(W ) − dimk(W ) = (n − s) − s = n − 2s, which is a contradiction. Now, let f := Q where Q is the quadratic form in Chapter 9 and let W be as in Chapter 9. Let t1, u1, ti, ui, and Wi be as in Chapter 9, that is let Wi ⊆ W with dimk(Wi) = ui−1 − 1 for 2 ≤ i ≤ k.

Lemma D.3. t1 ≤ n − s + 1.

⊥ Proof. By Lemma D.2, there are at least s linearly independent pi’s in W such that Bf (pi,W ) 6= 0. Suppose that t1 > n − s + 1. Then Bf (pj,W ) = 0 for s + 1 ≤ j ≤ n − s + 1. Since |{pn−s+2,..., pn}| = s − 1, we obtain a contradiction to Lemma D.2.

Lemma D.4. There are at least ui−1−1 linearly independent pj’s such that Bf (pj,Wi) 6= 0.

⊥f ⊥ Proof. Since Wi ⊆ W , Bf (Wi,Wi) = {0}. Let Yi = Wi ∩ Wi . By Lemma D.1, ⊥f L ⊥f Wi = Wi Yi. By Lemma 2.15, dimk(Wi ) = n − (ui−1 − 1) = n − ui−1 + 1 and therefore dimk(Yi) = n − ui−1 + 1 − (ui−1 − 1) = n − 2ui−1 + 2. ⊥ ⊥ Note that pj ∈ W ⊆ Wi for s + 1 ≤ j ≤ n. Suppose there are at most ui−1 − 2 pj’s such that Bf (pj,Wi) 6= 0. Then there are at least (n − ui−1 + 1) − (ui−1 − 2) = n − 2ui−1 + 3 pj’s such that Bf (pj,Wi) = 0. Let Zi be the span of the pj’s such that ⊥ ⊥f Bf (pj,Wi) = 0. Then Zi ⊆ Wi ∩ Wi = Yi. We have dim(Zi) ≥ n − 2ui−1 + 3 since the pj’s are linearly independent. But dim(Yi) = n − 2ui−1 + 2, which is a contradiction.

Lemma D.5. ti ≤ n − ui−1 + 2 for 2 ≤ i ≤ k.

Proof. By Lemma D.4, there are at least ui−1 − 1 linearly independent pj’s such that Bf (pj,Wi) 6= 0. Suppose that ti > n − ui−1 + 2. Then Bf (pj,Wi) = 0 for ti−1 + 1 ≤ j ≤ n − ui−1 + 2. Since |{pn−ui−1+3,..., pn}| = ui−1 − 2, we obtain a contradiction to Lemma D.4.

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165 Vita

Deborah H. Blevins

Place of Birth:

• Lexington, KY Education:

• University of Kentucky, Lexington, KY M.A. in Mathematics, May 2018

• University of the Cumberlands, Williamsburg, KY B.A. in Mathematics, May 2015 summa cum laude

Professional Positions:

• Graduate Teaching Assistant, University of Kentucky Fall 2015–Spring 2021 Honors

• Enochs Scholarship in Algebra, University of Kentucky

• NSF Mathematical Sciences Graduate Internship, Oak Ridge National Labora- tory

• College of Arts and Sciences Outstanding Teaching Assistant Certiﬁcate, Uni- versity of Kentucky

• Wimberly C. Royster Oustanding Teaching Assistant Award, University of Ken- tucky

• Graduate Scholars in Mathematics Fellowship, University of Kentucky Publications & Preprints:

• Time-Based CAN Intrusion Detection Benchmark with P. Moriano, R. Bridges, M. Verma, S. Holliﬁeld, and M. Iannacone; pre-print arXiv: 2101.05781

166