Almost orthogonal subsets of vector spaces over finite fields
aa r X i v : . [ m a t h . C O ] J a n ALMOST ORTHOGONAL SUBSETS OF VECTOR SPACES OVERFINITE FIELDS
ALI MOHAMMADI AND GIORGIS PETRIDIS
Abstract.
We prove various results on the size and structure of subsets of vector spacesover finite fields which, in some sense, have too many mutually orthogonal pairs of vectors.In particular, we obtain sharp finite field variants of a theorem of Rosenfeld and an almostversion of a theorem of Berlekamp. Introduction
Background. An orthogonal set of vectors in R n , that is a set of non-zero vectors withthe property that every pair of distinct vectors is mutually orthogonal, is linearly independentand therefore contains at most n elements. Erd˝os asked the question of determining themaximum size of a set of almost orthogonal vectors: a set of non-zero vectors with theproperty that among any three distinct vectors, at least two are mutually orthogonal [14].The union of two disjoint orthogonal sets is an almost orthogonal set of size 2 n . Rosenfeld,confirming a belief of Erd˝os, proved that 2 n is the maximum size of an almost orthogonalsubset of R n [16]. Deaett gave a short and elegant proof of Rosenfeld’s theorem [6], whichhas similarities with an argument of Pudl´ak [15]. Deaett also proved that for dimension 4and lower every almost orthogonal set of maximum size is the union of two orthogonal sets;and provided examples in dimension 5 and higher of almost orthogonal sets of maximumsize that are not the union of two disjoint orthogonal sets.In C n , the existence of self-orthogonal vectors (like (1 , i ) ∈ C ) changes the answer toboth questions. Even in dimension two, the span of (1 , i ) is an uncountable set of orthogonalvectors. Deaett proved, however, that in C n the maximum number of almost orthogonalvectors none of which is self-orthogonal is 2 n , generalising Rosenfeld’s theorem [6].Both questions have also been investigated over finite fields. The size of the largest orthog-onal set in ( Z / (2 Z )) n was determined by Berlekamp [5] and the size of the largest orthogonalset in ( Z / ( p Z )) n for primes p was determined by Zame [23]. There are similarities in theirmethods. The question Berlekamp answered is equivalent to solving another question ofErd˝os: determining the size of the largest family of subsets of { , , . . . , n } with the prop-erty that every two distinct elements have even intersection. Erd˝os’ question was solvedindependently by Graver [5, 8].A key to Berlekamp’s and Zame’s arguments is the existence of self-orthogonal vectors.Self-orthogonal vectors exist over any finite field when the dimension is at least 3 or, indimension 2, when the order of the field is 2 or is congruent to 1 modulo 4. There are furtherintricacies when working in vector spaces over finite fields. For example, as is detailed in thenext subsection, in dimension 6, the dot product is equivalent to the symmetric bilinear form( x , y ) x y − x y + x y − x y + x y − x y when the order of the field is congruent to 1modulo 4, and to the symmetric bilinear form ( x , y ) x y − x y + x y − x y + x y + x y when the order of the field is congruent to 3 modulo 4. The difference points to the fact that the largest orthogonal subspace has dimension that depends on the order of the field [21].In addition to this, expressing the dot product in these equivalent ways has the advantagethat it makes clear the existence of self-orthogonal vectors.There does not seem to be a significant difference between studying the dot product andstudying any symmetric non-degenerate bilinear form and this is the approach taken in theliterature recently. Ahmadi and Mohammadian [1], using an argument similar to Berlekamp,determined the size of the largest orthogonal set with respect to any non-degenerate sym-metric bilinear form over fields of odd order (see also [10, 21]). Ahmadi and Mohammadianalso made progress on the question of determining the size of the largest almost orthogonalset with respect to any non-degenerate symmetric bilinear form.The main purpose of this paper is to determine the size of the largest almost orthogonalset with respect to any bilinear form in any vector space over any sufficiently large finite fieldof odd order; and also in ( Z / (2 Z )) n for sufficiently large n . It is worth recording here that,unlike Rosenfeld’s and Deaett’s theorems, it is not always the case that the size of the largestalmost orthogonal set equals twice the size of the largest orthogonal set. The ( Z / (2 Z )) n question has a set system formulation that can be thought of as an “almost” version ofBerlekamp’s theorem: determine the size of the largest family of subsets of { , , . . . , n } withthe property that among every three distinct elements, at least two have even intersection.We show that the size of the largest family almost doubles. We also investigate the finite fieldanalogue of another question that Erd˝os asked for Euclidean space: determine the maximumsize of subsets of R n with the property that among any k of their elements, at least two aremutually orthogonal [4, 7].1.2. Notation and definitions.
Throughout the paper, we use m and n to be positiveintegers, p a prime and q = p m . We also use F q to denote a finite field of order q and write F ∗ q = F q \ { } . A bilinear form over F nq is a mapping B : F nq × F nq → F q , which takes the form B ( x , y ) = x T A y , for all x , y ∈ F nq , for some n × n matrix A over F q . We say B is symmetric if A is a symmetric matrix and say B is degenerate if det( A ) = 0. We call two bilinear forms equivalent if their correspondingmatrices are equivalent ( A, B are equivalent if A = M T BM for an invertible matrix M ).Let γ ∈ F q be a non-square and η : F q → C denote the quadratic character of F q . Namely, η ( a ) = , if a = 0;1 , if a is a non-zero square; − , if a is not a square.For any bilinear form B , with associated matrix A , we define ε ( B ) ∈ { , , γ } as the elementsatisfying η (det( A )) = η (( − ⌊ n ⌋ ε ( B )). For odd q , by a result in [9, p. 79], which also appearsas [1, Theorem 1], any non-degenerate symmetric bilinear form, B , over F nq is equivalent tothe form(1) ( x , y ) x y − x y + · · · + x n − y n − − x n − y n − + ε ( B ) x n y n for odd n and is equivalent to the form(2) ( x , y ) x y − x y + · · · + x n − y n − − x n − y n − + x n − y n − − ε ( B ) x n y n for even n , where x = ( x , . . . , x n ) and y = ( y , . . . , y n ). LMOST ORTHOGONAL SUBSETS OF VECTOR SPACES OVER FINITE FIELDS 3
For q = 2, one can infer from (5) in [12, p. 7] that every non-degenerate symmetricbilinear form in odd dimension is equivalent to the dot product that arises from the n × n identity matrix I n . In even dimensions every non-degenerate symmetric bilinear form iseither equivalent to the dot product or to the hyperbolic form H that arises from the matrix H ⊕ · · · ⊕ H , where H = (cid:18) (cid:19) . Definition 1.1.
We refer to two vectors v , v ∈ F nq \ { } as mutually orthogonal if B ( v , v ) = 0. If v ∈ F nq \ { } , satisfies B ( v , v ) = 0, we refer to it as self-orthogonal .We call a subset S ⊂ F nq \ { } an orthogonal set if every distinct pair of elements of S aremutually orthogonal and say S ⊂ F nq \ { } is ( k, l ) -orthogonal if for any k vectors in S atleast l of them are pairwise mutually orthogonal. Furthermore, we call a subspace V ⊂ F nq an orthogonal subspace if V \ { } is an orthogonal set. We denote by S k,l = S k,l ( q, n, B ) themaximum size of any ( k, l )-orthogonal subset of F nq with respect to B .Given a set X ⊂ F nq , we use h X i to denote the subspace of F nq generated by X and write h v , · · · , v k i , instead of h{ v , · · · , v k }i . We also define the orthogonal complement of X by X ⊥ = { v ∈ F nq : B ( v , x ) = 0 for all x ∈ X } , which constitutes a subspace of F nq .1.3. Previous results for almost orthogonal sets.
In [1, Examples 12–15], explicit ex-amples of (3 , q , showing that(3) S , ( q, n, B ) ≥ q n − , if n is odd and ε ( B ) = 1;2 q n − − q + 1 , if n is odd and ε ( B ) = γ ;2 q n − q − , if n is even and ε ( B ) = 1;2 q n − + 2 , if n is even and ε ( B ) = γ. The examples also work for q = 2, showing that(4) S , (2 , n, · ) ≥ ( n +12 , if n is odd;2 n +1 − , if n is even . The authors further conjectured, in [1, Conjectures 11 and 16], that the inequalities in (3)and (4) could be replaced by equalities and outlined a proof, in [1, Theorem 17], that for all n and either choices (1) and (2) of bilinear forms, S , ≤ q ⌊ n ⌋ .1.4. Improved lower-bounds on S , . We proceed to present examples of (3 , F nq for odd q that have slightly more elements than the examples given in (3). The-orem 2.1, which is stated in the next section, shows that the examples described below areof maximum size. We denote by { e , . . . , e n } the standard basis of F nq . Example 1.2.
Let q be odd and n = 2 k + 1 ≥ and B satisfy ε ( B ) ∈ { , γ } . Consider thetwo mutually disjoint orthogonal sets S = (cid:0) { ( x , x , . . . , x k , x k ,
0) : x , . . . , x k ∈ F q } \ { } (cid:1) ⊔ { e n − + e n − + e n } and S = (cid:0) { ( x , − x , . . . , x k , − x k ,
0) : x , . . . , x k ∈ F q } \ { } (cid:1) ⊔ { e n − − e n − − ε ( B ) − e n } . Then, the set S = S ⊔ S ⊔ { e n } is a (3 , -orthogonal set of size q k + 1 = 2 q ( n − / + 1 . A. MOHAMMADI AND G. PETRIDIS
Example 1.3.
Let q be odd and n = 2 k ≥ and B satisfy ε ( B ) = 1 . Consider the twomutually disjoint orthogonal sets S = { ( x , x , . . . , x k , x k ) : x , . . . , x k ∈ F q } \ { } and S = { ( x , − x , . . . , x k , − x k ) : x , . . . , x k ∈ F q } \ { } . Then, the set S = S ⊔ S is a (3 , -orthogonal set of size q k − q n/ − . Example 1.4.
Let q be congruent to 3 modulo 4 and n = 2 k ≥ and B satisfy ε ( B ) = − ( − is not a square). Consider the three pairwise disjoint orthogonal sets S = (cid:0) { ( x , x , . . . , x k − , x k − , ,
0) : x , . . . , x k − ∈ F q } \ { } (cid:1) ⊔ { e n − + e n − + e n − + e n , e n − + e n − + e n − − e n } and S = (cid:0) { ( x , − x , . . . , x k − , − x k − , ,
0) : x , . . . , x k − ∈ F q } \ { } (cid:1) ⊔ { e n − − e n − − e n − − e n , e n − − e n − − e n − + e n } , and S = { e n − + e n , e n − − e n } . Then, the set S = S ⊔ S ⊔ S is a (3 , -orthogonal set of size q k − + 4 = 2 q n/ − + 4 . For n = 2 , { e , e , e , e } is a (3 , -orthogonal set of size 4. Next, we provide examples for q = 2. Theorem 2.2 shows they are of maximum size.The following example is obtained by adding a single element to the example given by [1,Example 12]. Example 1.5.
Let n = 2 k + 1 ≥ and B denote the dot product in F n . Consider the disjointorthogonal sets S = (cid:0) { (0 , x , x . . . , x k , x k ) : x , . . . , x k ∈ F } \ { } (cid:1) ⊔ { e } and S = (cid:0) { ( x , x , . . . , x k , x k ,
0) : x , . . . , x k ∈ F } \ { } (cid:1) ⊔ { e n } . Then, the set S = S ⊔ S ⊔ { e + e + · · · + e n } is a (3 , -orthogonal set of size k +1 + 1 =2 ( n +1) / + 1 . The example below is the same as [1, Example 14].
Example 1.6.
Let n = 2 k ≥ and B denote the dot product in F n . Consider the disjointorthogonal sets S = { ( x , x , . . . , x k , x k ) : x , . . . , x k ∈ F } \ { } and S = { ( x k , x , x , . . . , x k − , x k − , x k ) : x , . . . , x k ∈ F } \ { } . Then, noting (1 , , . . . , , ∈ S ∩ S , it follows that the set S = S ∪ S is a (3 , -orthogonalset of size k +1 − n/ − . Our final example concerns the hyperbolic form.
LMOST ORTHOGONAL SUBSETS OF VECTOR SPACES OVER FINITE FIELDS 5
Example 1.7.
Let n = 2 k ≥ and H denote the hyperbolic form in F n . Consider thedisjoint orthogonal sets S = h e , e , e , . . . , e k − i \ { } and S = h e , e , e , . . . , e k i \ { } . It follows that the set S = S ∪ S is a (3 , -orthogonal set of size k +1 − n/ − . Main results
Our first main result is an upper bound on the size of (3 , q . Theorem 2.1.
Let n ≥ be an integer and q ≥ be an odd prime power. If S ⊂ F nq is (3 , -orthogonal with respect to a non-degenerate symmetric bilinear form B , then | S | ≤ q n − + 1 , if n is odd ;2 q n − , if n is even and ε ( B ) = 1;2 q n − + 4 , if n ≥ is even and ε ( B ) = γ ;4 , if n = 2 and ε ( B ) = γ. Combined with Examples 1.2, 1.3, and 1.4, Theorem 2.1 establishes the value of S , forall n , sufficiently large q , and B : S , ( q, n, B ) = q n − + 1 , if n ≥ q n − , if n ≥ ε ( B ) = 1;2 q n − + 4 , if n ≥ ε ( B ) = γ ;4 , if n = 2 and ε ( B ) = γ. In contrast to Euclidean space R n , S , is sometimes larger than twice the size of the largestorthogonal set (specifically when n ≥ n ≥ ε ( B ) = γ ). See [1]or Lemma 3.3 below for the size of the largest orthogonal set for the various possibilities of n , q , and B .The proof of Theorem 2.1 relies on the framework developed by Ahmadi and Mohamma-dian [1]. It also has similarities with the work of Berlekamp [5] and the paper of Deaett [6].In Section 5 we present a different argument for even n with ε ( B ) = 1 that is based oncharacter sum estimates. The proof in Section 5 works for all odd q .We also answer the corresponding question for q = 2. Theorem 2.2.
Let n be an integer. If S ⊂ F n is a (3 , -orthogonal with respect to the dotproduct, then | S | ≤ ( n +12 + 1 , if n ≥ is odd ;2 n +1 − , if n ≥ is even . For even n ≥ and S ⊂ F n be (3 , -orthogonal with respect to the hyperbolic form H , | S | ≤ n +1 − . A. MOHAMMADI AND G. PETRIDIS
Combined with Examples 1.5, 1.6, and 1.7, Theorem 2.2 establishes the value of S , forall sufficiently large n : S , (2 , n, B ) = n +12 + 1 , if n ≥
21 is odd and B = · ;2 n +1 − , if n ≥
18 is even and B = · ;2 n +1 − , if n ≥ B = H . As highlighted in [1], the quantity S , (2 , n, · ) may be interpreted as the size of a maximallylarge family F of non-empty subsets of an n -element set such that among every three distinctelements of F , there is a pair of sets whose intersection is of even cardinality. The valuesabove confirm [1, Conjecture 16] of Ahmadi and Mohammadian for sufficiently large, even n . Although, for odd n , we have shown that the relevant value is larger by one than whatwas conjectured. The sufficiently large n assumption cannot be removed, see Remark 6.7.We also prove an upper bound for S k, ( q, n, B ). This is the finite field analogue of anotherquestion of Erd˝os [4, 7]. Theorem 2.3.
Let q be an odd prime power and k ≥ be an integer. Suppose that S ⊂ F nq \ { } is ( k, -orthogonal with respect to a non-degenerate symmetric bilinear form B .Then | S | ≤ (cid:18) k − k − q − k + 1 (cid:19) ( q n/ + 1) . Theorem 2.3 implies S k, ≤ ( k − o q →∞ (1)) q n/ , which is asymptotically sharp in some cases. For example for odd q , k = n = 4 and ε ( B ) = 1the union of the following three pairwise disjoint orthogonal sets has size 3( q − S = { ( x , x , x , x ) : x , x ∈ F q } \ { } and S = { ( x , − x , x , − x ) : x , x ∈ F q } \ { } and S = { ( x , x , x , − x ) : x , x ∈ F q } \ { } . Outline of the proofs of Theorems 2.1 and 2.2.
The proofs of Theorems 2.1and 2.2 are based on an inductive scheme developed by Ahmadi and Mohammadian [1]. Tooutline the argument let us denote by d n = d n ( q, B ) the dimension of the largest orthogonalsubspace of F nq with respect to a non-degenerate symmetric bilinear form B . Both theoremstake the form(5) S , ( q, n, B ) = 2 q d n + f ( q, n, B ) , where f ( q, n, B ) ∈ {− , − , , } . We show this by proving by induction a weaker statementof the form S , ( q, n, B ) ≤ (2 + o (1)) q d n . Note here that the o (1) term is for q → ∞ for odd q and n → ∞ for q = 2. A technical difficulty in carrying out the induction is that one mustensure that the restriction of B to the orthogonal complement considered is non-degenerate.The inductive argument that proves the weaker bound is based on the basic observationthat for every v ∈ S , the set of elements in S \ { v } not orthogonal to v constitute anorthogonal set. The structure of orthogonal sets has been determined by Berlekamp, andAhmadi and Mohammadian [5, 1]. A key feature is that they contain few elements not in asingle orthogonal subspace. We make repeated use of this fact. The second basic fact we use LMOST ORTHOGONAL SUBSETS OF VECTOR SPACES OVER FINITE FIELDS 7 to our advantage for large odd q is that a proper subspace of a vector space is significantlysmaller than the vector space.Once the weak bound has been established, it is used to determine f ( q, n, B ). It is at thispoint that different arguments must be used according to ( q, n, B ). The key observation,implicit in the literature, is that if an orthogonal set contains just a few elements that arenot self-orthogonal, then it is much smaller than q d n (for large q ).For q = 2, which is not large, slightly different arguments are utilised. The basic fact thatdrives the proof is that, unlike for odd q , the set of vectors not orthogonal to any v containsat most half of any orthogonal subspace.The proof of Theorems 2.1 probably yields a characterisation of nearly extremal sets: theyare mostly contained in two disjoint orthogonal subspaces of maximum dimension.3. Preparations
We begin with some basic facts about Ramsey numbers. Given positive integers s, t the
Ramsey number R ( s, t ) is the least integer with the property that every graph on R ( s, t )vertices either contains a K s or the complement of the graph contains a K t . We will use thefollowing bounds:(6) R (3 ,
3) = 6 , R (3 ,
4) = 9 , R ( s, t ) ≤ (cid:18) s + t − s − (cid:19) . The connection between Ramsey numbers and almost orthogonal sets goes back to at leastthe paper of Deaett [6]. A connection with other similar questions of Erd˝os is detailed in [14].We deduce the following elementary observation concerning graphs.
Lemma 3.1.
A triangle-free graph with the property that its complement is also triangle-freeis either the 5-cycle or has at most 4 vertices.Proof.
We denote by H the graph. Its order is at most R (3 , − H is 5. Note that H does not have a vertex of degree atleast 3. This is because if such a vertex existed, then either two of its neighbours would beconnected by an edge, giving rise to a triangle; or none of its neighbours would be connectedby an edge, giving a triangle in the complement. Similarly, the complement has no edgeof degree at least 3. Therefore all vertices of the graph have degree 2. Graphs of constantdegree 2 contain a cycle. The graph H has no 3-cycle. It does not contain a 4-cycle (the fifthvertex would be isolated). Therefore it contains a 5-cycle, which is the entire graph. (cid:3) We proceed with results concerning orthogonal sets. The most important is a structuralcharacterisation of orthogonal sets which we take from [1, Lemma 3] – see also [5] for q = 2. Lemma 3.2.
Let B denote a non-degenerate, symmetric bilinear form over F nq , where q isa prime power and n ≥ . Suppose that S ⊂ F nq is an orthogonal set. Then, there exists anorthogonal subspace V ⊂ { x ∈ F nq : B ( x , x ) = 0 } and a set T = { x ∈ S : B ( x , x ) = 0 } , suchthat S ⊂ V ⊔ T and V ) + | T | ≤ n. Next, we recall [1, Theorem 4], which relying mainly on Lemma 3.2, obtains the followingsharp bound on orthogonal sets. We note that in [1] S , ( q, n, B ) is denoted by S ( q, n ). A. MOHAMMADI AND G. PETRIDIS
Lemma 3.3.
For n ≥ and a prime power q ≥ , let B denote a non-degenerate symmetricbilinear form over F nq . Then S , ( q, n, B ) = q n − , if n is odd ; q n − , if n is even and ε ( B ) = 1; q n − + 1 , if n is even and ε ( B ) = γ. An analogue of Lemma 3.3, for q = 2, was earlier proved by Berlekamp [5]. Also see [8]. Lemma 3.4. S , (2 , n, · ) = n, if n ≤ n − , if n is odd and n ≥ n , if n is even and n ≥ . The following corollary is central to our considerations.
Lemma 3.5.
For s ∈ S , define (7) S s = { x ∈ S \ { s } : B ( x , s ) = 0 } . If | S s | ≥ , then S s is an orthogonal set. In particular, we may write for all s ∈ SS s = R s ⊔ T s , where T s = { x ∈ S s : B ( x , x ) = 0 } and h R s i = V s , an orthogonal subspace of F nq thatcontains only self-orthogonal vectors.Proof. Given two distinct vectors x , x ∈ S s , by the (3 , S , two of { x , x , s } must be mutually orthogonal. Thus, given the definition of S s , we must have B ( x , x ) = 0. The rest follows from Lemma 3.2 or is immediate when | S s | ≤ (cid:3) We also collect some basic facts about orthogonal subspaces as follows.
Lemma 3.6.
Let V ⊂ F nq denote an orthogonal subspace with at least three elements.(1) Every vector in V is self-orthogonal.(2) Suppose that V is of maximum dimension and V = h R i for some R ⊂ F nq . If z / ∈ V is a self-orthogonal vector, then z is not orthogonal to R .Proof. For the first statement, let x be any element of V and y some other element of V . Itfollows that x + y ∈ V \ { x } and so0 = B ( x , x + y ) = B ( x , x ) + B ( x , y ) = B ( x , x ) . For the second statement, we have B ( z , z ) = 0. Suppose for a contradiction that z ⊥ R .Then z ⊥ V . Note that by the first part, for all λ, µ ∈ F q and x , y ∈ V we have B ( λ z + x , µ z + y ) = λµ B ( z , z ) + λ B ( z , y ) + µ B ( z , x ) + B ( x , y ) = 0 . Hence h{ z } ∪ R i is an orthogonal subspace that strictly contains V , a contradiction. (cid:3) The next result forms the basis of the induction argument in the proof of Theorem 2.1.The key is to show that if we restrict B to a certain type of orthogonal complement, thenit remains non-degenerate; and that, under a further condition, the equivalence class of B isconserved. LMOST ORTHOGONAL SUBSETS OF VECTOR SPACES OVER FINITE FIELDS 9
Lemma 3.7.
Let n ≥ , B be a non-degenerate symmetric bilinear form over F nq , and { v , w } ⊂ F nq be linearly independent.(1) If B ( v , w ) = B ( v , v ) B ( w , w ) , then the restriction B ↾ { v , w } ⊥ of B to the orthogonal complement of { v , w } , is anon-degenerate symmetric bilinear form.(2) If q is odd, v and w are not mutually orthogonal, and w is self-orthogonal (that is B ( v , w ) = 0 and B ( w , w ) = 0 ), then ε ( B ↾ { v , w } ⊥ ) = ε ( B ) .(3) If q = 2 , n is even, v and w are not mutually orthogonal, and both v , w are self-orthogonal (that is B ( v , w ) = 0 and B ( v , v ) = B ( w , w ) = 0 ), then B is equivalent to H if and only if B ↾ { v , w } ⊥ is equivalent to H .Proof. Throughout the proof we write a = B ( v , v ) , b = B ( v , w ) , c = B ( w , w ) . For (1) we first show h v , w i ∩ { v , w } ⊥ = { } and therefore that F nq = h v , w i ⊕ { v , w } ⊥ .Suppose λ v + µ w ∈ { v , w } ⊥ . Applying B ( v , − ) and then B ( w , − ) to both sides gives thelinear system (cid:26) λa + µb = 0 λb + µc = 0 . It follows that λ = µ = 0 because ac = b .We write M for the matrix of B ↾ h v , w i with respect to the basis { v , w } , M for thematrix of B ↾ { v , w } ⊥ with respect to any basis, and M for the matrix of B with respect tothe union of these two bases, then M = (cid:18) M M (cid:19) . It follows immediately that if B is non-degenerate, then so is B ↾ { v , w } ⊥ .For (2) we have b = 0 and c = 0. We show M = (cid:18) a bb (cid:19) ∼ (cid:18) − (cid:19) , which proves ε ( B ↾ { v , w } ⊥ ) = ε ( B ).Let α, β, γ be solutions to α − γ = a and β ( α − γ ) = b . Such α, γ exist because everyelement of F q is the difference of two squares. The characteristic is not 2, so we can alwaystake α = γ (even when a = 0). Then there exists a suitable β . Now a simple calculationconfirms (cid:18) α βγ β (cid:19) T (cid:18) − (cid:19) (cid:18) α βγ β (cid:19) = (cid:18) α − γ β ( α − γ ) β ( α − γ ) 0 (cid:19) = (cid:18) a bb (cid:19) ;and det (cid:18) α βγ β (cid:19) = β ( α − γ ) = b = 0.For (3) we have b = 1 and a = c = 0. Therefore M = H . So B is equivalent to H if andonly if B ↾ { v , w } ⊥ is equivalent to H . (cid:3) Remark 3.8.
The condition in part (1) of the lemma is necessary. Take, for example, n = 4and B the bilinear form given by the diagonal matrix with diagonal entries (1 , − , , − B is the dot product when q = 2. Take v = (1 , , ,
0) and w = (1 , , , a, b, c defined in the proof of the lemma allequal to 1. Hence ac = b . It is not true that h v , w i trivially intersects { v , w } ⊥ because thespan of w − v = (0 , , ,
0) lies in both subspaces. Furthermore, B restricted to { v , w } ⊥ isdegenerate because w − v is orthogonal to both w − v and (0 , , , { v , w } ⊥ .The next step is to bound the number of vectors in any (3 , F nq that are not self-orthogonal. It may be true that, analogously to the results of Rosenfeldand Deaett [16, 6], there are at most 2 n such vectors. We prove a weaker result thatsuffices for our purposes. As part of the proof, we require a straightforward adaptation of [6,Proposition 4.4], which we state. The proof is nearly identical to that in [6]. Lemma 3.9.
Let n ≥ be a positive integer, F be a field, and S ⊂ F n be a (3 , -orthogonalset with respect to a symmetric bilinear form. If B ⊂ S is an orthogonal basis for F n , then S \ B is an orthogonal set. We state and prove another result that is implicit in [6, Section 4]. It is convenient tophrase many of the subsequent arguments in terms of the simple graph G with vertex set S and edges given by pairs of elements of S that are not mutually orthogonal ( xy is an edgeprecisely when B ( x , y ) = 0). Lemma 3.10.
Let n ≥ be a positive integer, F a field, and D ⊂ F n be a (3 , -orthogonalset with respect to a symmetric bilinear form. If D consists entirely of vectors that are notself-orthogonal, then | D | ≤ max { n, R (3 , n ) − } (6) = ( n, if ≤ n ≤ n ( n +1)2 − , if n ≥ . Proof.
We use the graph G described just above the statement of the lemma. The claim istrue for n = 0. For n ≥ F n and so is linearly independent (we need here that all vectors in D are not self-orthogonal). If G has an independent set B of size n , then that set is linearly independentand therefore is a basis for F n . By Lemma 3.9 we get that D \ B is orthogonal and hencecontains at most n elements. Hence | D | = | B | + | D \ B | ≤ n . If G , which is triangle-free,has no independent set of size n , then | D | < R (3 , n ), by the definition of R (3 , n ). (cid:3) Note that by work of Ajtai, Koml´os and Szemer´edi, and of Kim [2, 11] R (3 , n ) = (1 + o n →∞ (1)) n log n , with stronger explicit upper bounds in [17]. This means that | D | = o ( n ).We also extract this consequence of Lemma 3.2 and Lemma 3.9 from the proof of [1,Theorem 17]. Lemma 3.11.
Let S ⊂ F nq be a (3 , -orthogonal set with respect to a non-degenerate symmet-ric bilinear form B . If every pair of linearly independent vectors in S is mutually orthogonal(that is B ( x , y ) = 0 for every linearly independent { v , w } ⊂ S ), then | S | ≤ S , ( q, n, B ) + n . LMOST ORTHOGONAL SUBSETS OF VECTOR SPACES OVER FINITE FIELDS 11
As the final result of this section, we recall [1, Theorem 17], which is a quantitativelyweaker version of Theorem 2.1. We will use this result in Section 5 and so provide a proofwhich follows the same scheme as that introduced in [1], while paying special attention tocertain intricacies involved in carrying out the induction. In particular, the proof relies onLemma 3.7 to sidestep a potential issue that appears to have been overlooked in the originalproof of [1, Theorem 17]. It also serves as a prelude to the proof of Theorem 2.1.We employ for the first of many times a decomposition of a (3 , S thatappears in [1], and so detail it. Given two distinct elements x , y ∈ S , every element of S \ { x , y } is either not orthogonal to x , or not orthogonal to y , or orthogonal to both x and y . Using the notation of Lemma 3.5 we decompose S as follows(8) S = S x ∪ S y ∪ S xy ∪ { x , y } , where S x and S y are defined in (7) and S xy = S ∩ { x , y } ⊥ . Note that { x , y } can be leftout if B ( x , y ) = 0 because x ∈ S y and vice versa. When bounding | S xy | by induction it isessential that B ↾ { x , y } ⊥ is non-degenerate. Proposition 3.12.
Let q be odd and let B be a non-degenerate symmetric bilinear form over F nq . If S ⊂ F nq \ { } is (3 , -orthogonal, then | S | ≤ q ⌊ n ⌋ . Proof.
We proceed by induction on n . Note that the result is true for n ∈ { , } because | S | ≤ n .If every linearly independent pair of vectors in S is mutually orthogonal, by Lemma 3.11and Lemma 3.3, we have | S | ≤ S , ( q, n, B ) + n ≤ q ⌊ n ⌋ . Hence suppose there exists a linearly independent pair { x , y } ⊂ S , with B ( x , y ) = 0. If atleast one of these vectors is self-orthogonal, by Lemma 3.7 (1), B ↾ { x , y } ⊥ is non-degenerate.Recalling the decomposition (8) and noting that x ∈ S y and y ∈ S x , we have | S | ≤ | S x | + | S y | + | S xy | . For n ∈ { , } , we have | S xy | ≤ q and for n ≥ | S xy | ≤ q ⌊ n ⌋− by the inductionhypothesis. Furthermore, by Lemma 3.3 and Lemma 3.5, we have | S x | , | S y | ≤ q ⌊ n ⌋ . Addingthis all up, we obtain the required result in this case.Next, suppose that neither x nor y is self-orthogonal and note that in this case, we mayno longer assume B ↾ { x , y } ⊥ is non-degenerate (see Remark 3.8). If every pair of elementsof S xy is mutually orthogonal, by Lemma 3.3, we have | S xy | ≤ q ⌊ n ⌋ and the required resultfollows. Hence suppose there exist v , w ∈ S xy , with B ( v , w ) = 0. Again, if at least oneof { v , w } is self-orthogonal, we may repeat the arguments of the first case to obtain therequired result. Thus assume otherwise. Consider the decomposition(9) S = S x ∪ S v ∪ S xv ∪ { x , v } . By Lemma 3.7 (1), B ↾ { x , v } ⊥ is non-degenerate. Employing the notation of Lemma 3.5,note that y ∈ T x and w ∈ T v . Suppose there exists z ∈ R v , with B ( y , z ) = 0. Thenby Lemma 3.7 (1), B ↾ { y , z } ⊥ is non-degenerate and we may repeat the arguments of thefirst case, with z in place of x , to obtain the required result. Otherwise, if y is orthogonalto R v , it follows that R v ⊔ { y , w } is an orthogonal set, which by Lemma 3.2, implies thatdim( V v ) ≤ ⌊ n/ ⌋ −
1. By a similar argument, we may assume R x ⊔ { y , w } is an orthogonal set and that dim( V x ) ≤ ⌊ n/ ⌋ −
1. Furthermore note that w ∈ S xy and so w S x andsimilarly y S v .For n ∈ { , } , by Lemma 3.2 and the above observations, we have | S x ∪ S v | ≤ n − | S xv | ≤ q . Thus going back to (9), we get | S | ≤ n + q ≤ q as required. For n ≥
4, we mayagain use Lemma 3.2 to see | S x ∪ S v | ≤ q ⌊ n/ ⌋− + 3) −
2. Then | S | ≤ (2 q ⌊ n/ ⌋− + 4) + 3 q ⌊ n ⌋− + 2= 5 q ⌊ n/ ⌋− + 6 ≤ q ⌊ n/ ⌋ , for all n ≥ q ≥ (cid:3) Proof of Theorem 2.1
We set d n = n − , if n ≥ n , if n ≥ ε ( B ) = 1; n − , if n ≥ ε ( B ) = γ. It was proved in [21] that d n is the dimension of the largest orthogonal subspace of F nq (alsofollows from Lemma 3.2). Note that d n − = d n − n . For n = 0 and n = 1 the size of the largest (3 , | S | ≤ q d n + O ( q d n − ) or that S possesses certain properties thatmake proving the theorem a matter of case analysis. For sufficiently large q the former upperbound is smaller than the in the theorem.We phrase the argument in terms of the graph G with vertex set S and two vectors adjacentprecisely when they are not mutually orthogonal. The two properties of G we use is that itis triangle free (follows from S being (3 , G having size at most S , ( q, n, B ) (because an independent set is an orthogonal subset of F nq ). We will also use the fact that every orthogonal set in F nq has size at most S , ( q, n, B ),a quantity that is determined in Lemma 3.3.By Lemma 3.11 we may assume from now on the existence of linearly independent { v , w } ⊂ S with vw an edge (that is B ( v , w ) = 0). This is because n ≤ q d n − n ≥ q ≥
5. We decompose S in the neighbourhood S v of v , the neighbourhood S w of w , and theset of vertices S vw that are not adjacent to either v or w : S = S v ∪ S w ∪ S vw , where S vw ⊂ { v , w } ⊥ . We follow the set up of Lemma 3.5 and decompose S v = R v ⊔ T v with R v spanning the orthogonal vector space V v .When n = 2 we get that S vw is a subset of a zero dimensional vector space that doesnot include and so is empty. Since both S v and S w are orthogonal sets, we get | S | ≤ S , ( q, , B ). This proves the theorem for n = 2.For n ≥ S vw . We need the following tobe able to apply the second part of Lemma 3.7. Lemma 4.1.
For n ≥ , let S ⊂ F nq be a (3 , -orthogonal set with respect to a non-degeneratesymmetric bilinear form B . If every pair of linearly independent self-orthogonal vectors in S LMOST ORTHOGONAL SUBSETS OF VECTOR SPACES OVER FINITE FIELDS 13 is orthogonal to one another (that is B ( x , y ) = 0 for all linearly independent self-orthogonal x , y ∈ S ), then | S | ≤ ( S , ( q, n, B ) + 2 n, if ≤ n ≤ S , ( q, n, B ) + n ( n +1)2 − , if n ≥ . Proof.
Let D be the set of vectors in S , which are not self-orthogonal: D = { x ∈ S : B ( x , x ) = 0 } . By the hypothesis on S we have that S \ D is an orthogonal set. Hence | S \ D | ≤ S , ( q, n, B ).The claim follows by bounding | D | via Lemma 3.10. (cid:3) The upper bound on | S | in Lemma 4.1 is smaller than the bound in Theorem 2.1 for n ≥ q ≥
5. From now on we assume the existence of linearly independent vectors { v , w } that are self-orthogonal but are not mutually orthogonal: B ( v , v ) = B ( w , w ) = 0, but B ( v , w ) = 0 . Recalling the definition of f = f ( q, n, B ) inferred from Theorem 2.1 and (5), we get fromLemma 3.7 (2)(10) | S vw | ≤ q d n − + f = 2 q d n − + f. Therefore | S vw | is much smaller than the bound on | S | we are trying to prove. What drivesthe proof is that if either V v or V w is not of maximum dimension, then we are done. Tosee why, suppose V v is not of maximum dimension. Then using Lemma 3.2, Lemma 3.3,Lemma 3.5 and (10) we get | S | ≤ | S v | + | S w | + | S vw |≤ ( q d n − + 3) + ( q d n + 1) + (2 q d n − + f )= q d n + 3 q d n − + 4 + f. Since for q ≥ q d n − + 4 ≤ q d n , we assume from now on dim( V v ) = dim( V w ) = d n .One more property we need is that v ∈ V w and w ∈ V v . To confirm, say the latter,note that there is no edge from w to R v (because the graph is triangle-free). Therefore w ⊥ h R v i = V v . By Lemma 3.6 (2), and using the fact that w is self-orthogonal, we get w ∈ V v .We summarise all this in a proposition. Proposition 4.2.
Let q ≥ be an odd prime power, n ≥ and S ⊂ F nq be a (3 , -orthogonalset with respect to a non-degenerate symmetric bilinear form. Then | S | satisfies the upperbound of Theorem 2.1 unless there exist linearly independent self-orthogonal vectors v and w with dim( V v ) = dim( V w ) = d n ; and v ∈ V w and w ∈ V v . In this case S vw = S ∩ { v , w } ⊥ satisfies | S vw | ≤ q d n − + f ( q, n, B ) , with f ( q, n, B ) inferred from Theorem 2.1 and (5) . The final preparatory result is that for the remaining S described in Proposition 4.2, R z is considerably smaller than q d n for all z ∈ S vw . The proof is typical of forthcomingconsiderations. The key observation is that if a subspace of a vector space does not containa single element of the vector space, then it is considerably smaller. Lemma 4.3.
Let q be an odd prime power and let S ⊂ F nq be (3 , -orthogonal with respect toa non-degenerate symmetric bilinear form B . Suppose { v , w } ⊂ S is a linearly independent subset that consists of two self-orthogonal vectors that are not mutually orthogonal (that is B ( v , v ) = B ( w , w ) = 0 , but B ( v , w ) = 0 ). If z ∈ S vw = S ∩ { v , w } ⊥ , then | R z | ≤ q d n − − . Proof.
We have S ⊂ ( V v \ { } ) ∪ T v ∪ ( V w \ { } ) ∪ T w ∪ S vw . By Lemma 3.5 we know that R z contains only self-orthogonal vectors and so is disjoint from T v ∪ T w . Hence | R z | ≤ ( | V v ∩ V z | −
1) + ( | V w ∩ V z | −
1) + | R z ∩ S vw | .V v = V z because w ∈ V v \ V z . Hence V v ∩ V z is a proper subspace of V v and is therefore not ofmaximum dimension. This means | V z ∩ V v | ≤ q d n − . Similarly | V z ∩ V w | ≤ q d n − . Moreover,note that R z ∩ S vw is an orthogonal subset of { v , w } ⊥ , on which non-degeneracy and typeof B is preserved by Lemma 3.7 (2). Thus by Lemma 3.2, we have | R z ∩ S vw | ≤ q d n − − (cid:3) We begin the final stage of the proof of the theorem. We assume we are in the remainingcase detailed in Proposition 4.2. Let S ∗ vw = S vw \ ( V v ∪ V w ) . We distinguish between two different cases.
Case 1:
An edge exists between R v ∪ R w and S ∗ vw .Suppose uz is an edge with z ∈ S vw and, say, u ∈ R v . Our first claim is that { u , z } islinearly independent. Indeed if z = λ u , then we would have B ( z , u ) = λB ( u , u ) = 0, whichcontradicts uz being an edge. Furthermore, u is self-orthogonal and so by Lemma 3.7 (2)we get that B ↾ { u , z } ⊥ is non-degenerate and ε ( B ) is preserved.We have | S | ≤ | S u | + | R z | + | T z | + | S uz | . We have the following bounds: by Lemma 3.3 and Lemma 3.5 | S u | ≤ q d n + 1; by Lemma 3.2and Lemma 4.3 (and its proof), | R z | + | T z | ≤ q d n − + 1; and by induction, just like in (10), | S uz | ≤ q d n − + f . In total | S | ≤ q d n + 5 q d n − + 2 + f. We are done because for q ≥ n ≥
3, 5 q d n − + 2 ≤ q d n . Case 2:
No edge exists between R v ∪ R w and S ∗ vw .There is no edge from S ∗ vw to R v and therefore S ∗ vw is orthogonal to R v . It followsthat S ∗ vw is orthogonal to V v = h R v i . Similarly, S ∗ vw is orthogonal to V w . All vectors in T v ∪ T w ∪ S vw are not self-orthogonal by Lemma 3.6 (2) and dim( V v ) being maximum. Weuse the decomposition(11) S ⊂ ( V v \ { } ) ∪ ( V w \ { } ) ∪ ( T v ∪ T w ∪ S vw ) . We consider the three different possibilities separately.Even n ≥ ε ( B ) = 1. Our aim is to show that T v = T w = S ∗ vw = ∅ . Then by (11) | S | ≤ ( | V v | −
1) + ( | V w | − ≤ q d n −
1) = 2 q n/ − . We may assume T v = T w = ∅ else, by Lemma 3.2, V v or V w are not of maximum dimension,which is not allowed by Proposition 4.2. To show that S ∗ vw = ∅ , suppose for a contradiction LMOST ORTHOGONAL SUBSETS OF VECTOR SPACES OVER FINITE FIELDS 15 that z ∈ S ∗ vw , then V v ∪ { z } would be an orthogonal set, forcing, via Lemma 3.2, V v not tohave maximum dimension.Odd n ≥
3. We want to show | T v | + | T w | + | S ∗ vw | ≤
3. Then by (11) | S | ≤ ( | V v | −
1) + ( | V w | −
1) + 3 ≤ q d n + 1 = 2 q ( n − / + 1 . For any distinct x , y ∈ S ∗ vw , xy is an edge. This is because if xy were not an edge, then V v ∪ { x , y } would be an orthogonal set of size | V v | + 2, which, by Lemma 3.2, would force V v not to have maximum dimension. Therefore the induced subgraph on S ∗ vw is completeand triangle-free. Hence S ∗ vw must have at most two vertices. Moreover, by Lemma 3.2, | T v | , | T w | ≤ | S ∗ vw | = | T v ∪ T w | = 2. Suppose S ∗ vw = { x , y } with xy an edge, and T v = { u } . The graphis triangle-free and so one of ux , uy is not an edge. Suppose that ux is not an edge. Then V v ∪ { u , x } is an orthogonal set, forcing V v not to be of maximal dimension.Even n ≥ ε ( B ) = γ . We want to show | T v | + | T w | + | S ∗ vw | ≤
6. Then by (11) | S | ≤ ( | V v | −
1) + ( | V w | −
1) + 6 ≤ q d n + 4 = 2 q n/ − + 4 . In fact, by Lemma 3.2 and Proposition 4.2, | T w | ≤ | T v | + | S ∗ vw | ≤ T v ∪ S ∗ vw is orthogonal to R v and therefore is orthogonal to V v .Similarly, S ∗ vw is orthogonal to h V v ∪ V w i = V v + V w . Now consider the graph H induced on T v ∪ S ∗ vw . This is a triangle-free graph. Moreover, it has no independent set of size 3 becauseotherwise we could join this set to V v and obtain an orthogonal set of size | V v | + 3, whichwould force V v not to have maximum dimension. By Lemma 3.1 we get | T v ∪ S ∗ vw | ≤ H is a 5-cycle. Our final task is to rule out this possibility. Suppose fora contradiction that H is a 5-cycle.We set T v = { u , u } and S ∗ vw = { z , z , z } . u u is not an edge (because both u , u are incident to v ) and so H can be taken to be the 5-cycle z u z u z . The complementof H is the 5-cycle z z z u u . In the complement of H vertex adjacency is equivalent toorthogonality, and so { z , z } ⊂ ( V v + V w ) ⊥ is an orthogonal (and hence linearly independent)set in the orthogonal complement of V v + V w .We make a small digression to investigate the dimension of V v + V w . We may assume V v ∩ V w = { } . This is because if the intersection is non-trivial, then | S | ≤ | ( V v ∪ V w ) \ { }| + | T v | + | T w | + | S ∗ vw | ≤ (2 q d n − q ) + 7 ≤ q d n + 4 , and we are done. We may therefore assume that dim( V v + V w ) = 2 d n = n − F nq = ( V v + V w ) ⊕ h z i ⊕ h z i . To complete the argument we exploit the orthogonality relations encoded in the com-plement of H and show that z = , the contradiction we are after. To start note that z ∈ h z , z i . As z is orthogonal to z , we get z = λ z . We are left to show λ = 0.We next show u , u ∈ V v ⊕ h z , z i . Let’s start with, say, the decomposition u = α z + β z + x v + x w , for α, β ∈ F q , x v ∈ V v , and x w ∈ V w . Suppose for a contradiction that x w = . ByLemma 3.6 (2) and the maximality of dim( V v ) we get that the self-orthogonal vector x w isnot orthogonal to V v . Therefore there exists y ∈ V v \ { } such that B ( x w , y ) = 0. But then B ( u , y ) = B ( x w , y ) = 0 , a contradiction to u being orthogonal to the whole of V v . We therefore have u = α z + β z + x v and u = α ′ z + β ′ z + x ′ v . Now, u is orthogonal to z and so 0 = α ′ B ( z , z ), which gives α ′ = 0. Moreover u / ∈ V v ,which gives β ′ = 0. Next, u is orthogonal to u and so 0 = ββ ′ B ( z , z ). Hence β = 0 and,similarly to above, u = α z + x v for α ∈ F ∗ q . Finally, u is orthogonal to z = λ z . Hence0 = λα B ( z , z ), which implies the desired λ = 0. H is therefore not a 5-cycle and consequently | T v ∪ S ∗ vw | ≤
4. The proof of the theorem isconcluded.5.
Character sum proof of Theorem 2.1 for even n , ε ( B ) = 1 and all odd q First, we recall some basic facts from the theory of character sums, which we use to givean alternative proof of Theorem 2.1 for even n and ε ( B ) = 1 that holds for all odd q . See,for example, [13, Chapter 5] for more details. Lemma 5.1.
Let H be a subgroup of a finite abelian group G and χ a character of G , then X g ∈ H χ ( g ) = ( | H | if χ is trivial on H, otherwise , Let e p ( x ) = exp(2 πix/p ), Tr( x ) = x + x p + · · · + x p m − (recalling q = p m ) and ψ ( x ) = e p (Tr( x )). Then the functions { ψ ( λx ) : λ ∈ F q } determine all of the characters of F q . Lemma 5.2.
Let B denote a non-degenerate, symmetric bilinear form over F nq , where q isodd and n ≥ . Let V denote a subspace of F nq . Suppose that s V ⊥ . Then ψ ( B ( s , x )) is anontrivial character of V . The next result is a slight extension of Vinogradov’s bound on bilinear character sums,which appears, for example, in [22, p. 92]. Also see [18, Lemma 5] for the special case, where B is the dot product. Lemma 5.3.
Given
X, Y ⊂ F nq and λ ∈ F ∗ q , we have (cid:12)(cid:12)(cid:12)(cid:12) X x ∈ X X y ∈ Y ψ (cid:16) λ B ( x , y ) (cid:17)(cid:12)(cid:12)(cid:12)(cid:12) ≤ p | X || Y | q n . Proof.
We apply the triangle inequality and then the Cauchy-Schwarz inequality to get (cid:12)(cid:12)(cid:12)(cid:12) X x ∈ X X y ∈ Y ψ (cid:16) λ B ( x , y ) (cid:17)(cid:12)(cid:12)(cid:12)(cid:12) ≤ | X | X x ∈ X (cid:12)(cid:12)(cid:12)(cid:12) X y ∈ Y ψ (cid:16) B ( λ x , y ) (cid:17)(cid:12)(cid:12)(cid:12)(cid:12) ≤ | X | X x ∈ F nq (cid:12)(cid:12)(cid:12)(cid:12) X y ∈ Y ψ (cid:16) B ( x , y ) (cid:17)(cid:12)(cid:12)(cid:12)(cid:12) = | X | X x ∈ F nq X y , z ∈ Y ψ (cid:16) B ( x , y − z ) (cid:17) = | X | X y , z ∈ Y X x ∈ F nq ψ (cid:16) B ( x , y − z ) (cid:17) = | X || Y | q n . LMOST ORTHOGONAL SUBSETS OF VECTOR SPACES OVER FINITE FIELDS 17
To obtain the last equality, we used the fact that the inner sum in the penultimate line equals q n if y = z and zero otherwise. This, in turn, follows from the observation that F nq ⊥ = { } ,combined with Lemmas 5.1 and 5.2. (cid:3) Proof of Theorem 2.1 for even n and ε ( B ) = 1 . Firstly, replace S by S ⊔ { } . This makescalculations easier. We will take away at the end of the proof. For s ∈ S , write S ′ s = { x ∈ S : B ( x , s ) = 0 } . Also write D = { x ∈ S : B ( x , x ) = 0 } . Recalling (7), note that X s ∈ S | S ′ s | = X s ∈ S | S s | + | D | . Now | S | − X s ∈ S | S s | − | D | = | S | − X s ∈ S | S ′ s | (12) = (cid:12)(cid:12)(cid:12)(cid:12) X s ∈ S X s ∈ S \ S ′ s ψ ( B ( s , s )) (cid:12)(cid:12)(cid:12)(cid:12) . Here, we used just that for s ∈ S and s ∈ S \ S ′ s , we have ψ ( B ( s , s )) = 1. By thetriangle inequality, we also have (cid:12)(cid:12)(cid:12)(cid:12) X s ∈ S X s ∈ S \ S ′ s ψ ( B ( s , s )) (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12) X s ∈ S X s ∈ S ψ ( B ( s , s )) (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12) X s ∈ S X s ∈ S ′ s ψ ( B ( s , s )) (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12) X s ∈ S X s ∈ S ψ ( B ( s , s )) (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12) X s ∈ S X s ∈ S s ψ ( B ( s , s )) (cid:12)(cid:12)(cid:12)(cid:12) + | D | . Next, let W = { x ∈ S : | S x | ≥ } . Then, using Lemmas 3.2 and 3.5, we have (cid:12)(cid:12)(cid:12)(cid:12) X s ∈ W X s ∈ S s ψ ( B ( s , s )) (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12) X s ∈ W X s ∈ ( V s ⊔ T s ) ψ ( B ( s , s )) (cid:12)(cid:12)(cid:12)(cid:12) (13) + (cid:12)(cid:12)(cid:12)(cid:12) X s ∈ W X s ∈ ( V s ⊔ T s ) \ S s ψ ( B ( s , s )) (cid:12)(cid:12)(cid:12)(cid:12) . Using Lemmas 5.1 and 5.2, the first sum, on the RHS of (13), is bounded as follows. (cid:12)(cid:12)(cid:12)(cid:12) X s ∈ W X s ∈ ( V s ⊔ T s ) ψ ( B ( s , s )) (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12) X s ∈ W X s ∈ V s ψ ( B ( s , s )) (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12) X s ∈ W X s ∈ T s ψ ( B ( s , s )) (cid:12)(cid:12)(cid:12)(cid:12) ≤ X s ∈ W | T s | . The second sum, on the RHS of (13), is bounded trivially by X s ∈ W | V s | + | T s | − | S s | . Going back to (12), we have | S | ≤ | D | + (cid:12)(cid:12)(cid:12)(cid:12) X s ∈ S X s ∈ S ψ ( B ( s , s )) (cid:12)(cid:12)(cid:12)(cid:12) + X s ∈ W | V s | + 2 | T s | + X s ∈ S \ W | S s | . By Lemma 5.3, we know (cid:12)(cid:12)(cid:12)(cid:12) X s ∈ S X s ∈ S ψ ( B ( s , s )) (cid:12)(cid:12)(cid:12)(cid:12) ≤ | S | q n/ . For s ∈ W , write k s = dim( V s ). Then by Lemma 3.2, we know | V s | +2 | T s | ≤ q k s +2 n − k s ≤ q n/ . Furthermore, for s ∈ S \ W , we have 2 | S s | ≤ q n/ . So adding it all up, | S | ≤ | S | q n/ + 2 | D | . At this stage we go back to the original S that does not include . The above becomes(14) | S | ≤ q n/ + j | D || S | k − . Note that from (14), one can only deduce the bound | S | ≤ q n/ + 1. However, we proceedto sharpen this bound through an analysis of the set D . Some aspects of the remainingarguments can certainly be simplified if we are not aiming to prove the theorem for all q . Todeal with some small technicalities that follow, we require the bound(15) | S | ≤ q − , for n = 2, all odd q and either scenarios ε ( B ) ∈ { , γ } . This bound has already beenestablished as the base case of the induction in the proof of Theorem 2.1 . In particular,henceforth assume n ≥ | S | ≤ q n/ − , which follows from (14) if 2 | D | < | S | . So suppose otherwise. Using the bound on | D | provided by Lemma 3.10, we get | S | ≤
16 for n = 4 and | S | ≤ n ( n + 1) − q ≥ n ≥ | S | = 2 q n/ − | S | . In particular, the following observations will be useful.
Claim 5.4.
Let S ⊂ F nq denote a maximal (3 , -orthogonal set. Then | S \ D | ≡ q − .Proof. Each v ∈ S \ D is self-orthogonal and so S must contain the entire punctured line l v = { λ v : λ ∈ F ∗ q } , otherwise maximality of S is violated. The result follows noting that l v ∩ l w is empty if w l v and of size q − (cid:3) Claim 5.5.
At least one of the following statements holds.(1) | S | ≤ q n/ − ,(2) | D | = 2 ,(3) | T v | ≤ for each v ∈ S , where T v = D ∩ S v . In the proof of Theorem 2.1, when applying Lemma 3.11, to avoid a lengthy multi-case analysis, it isassumed that q >
3. However one may easily confirm that, for our purposes here, the argument remainsvalid when q = 3. LMOST ORTHOGONAL SUBSETS OF VECTOR SPACES OVER FINITE FIELDS 19
Proof.
We suppose that neither (2) nor (3) is true and prove (1). Thus, assume there exist v ∈ S , distinct elements w , w ∈ D \ { v } , with B ( v , w ) and B ( v , w ) both non-zero, andpotentially a fourth element w ∈ D , which need not be distinct from the previous ones. Weconsider two main cases as to whether v is self-orthogonal or not.First, assume B ( v , v ) = 0, which in particular implies v ∈ T w ∩ T w . By the (3 , S , we have B ( w , w ) = 0 and so firstly { w , w } is linearly independentand secondly by Lemma 3.7 (1), B ↾ { w , w } ⊥ is non-degenerate. Write S = S w ∪ S w ∪ S w w ∪ { w , w } . For ( q, n ) = (3 , | S | ≤ q n/ − + 1) − q −
1) + 2 = 13 <
16 = 2 q n/ − . For other admissible choices of ( q, n ), by Proposition 3.12, we have | S | ≤ q n/ − + 1) − q n/ − + 2 = 5 q n/ − + 3 < q n/ − . Next, assume B ( v , v ) = 0. In this case w
6∈ { v , w , w } . We split this case further byfirst assuming that w is orthogonal to R v (using the notation of Lemma 3.5). This implies R v ⊔{ w , w , w } is an orthogonal set. Further note that { w , v } is linearly independent andthat by parts (1) and (2) of Lemma 3.7, B ↾ { w , v } ⊥ is non-degenerate and its equivalenceclass is preserved. Write S = S w ∪ S v ∪ S w v . For ( q, n ) = (3 , | S | ≤ ( q n/ −
1) + ( q n/ − + 3) + 2 q n/ − − q n/ − . For the remaining combinations of ( q, n ), we use the bound (16) to get | S | ≤ ( q n/ −
1) + ( q n/ − + 3) + 2 q n/ − − ≤ q n/ − . Finally, suppose there exists some u ∈ R v , such that B ( w , u ) = 0. By definition B ( u , u ) = 0 and B ( u , v ) = 0, from which we may deduce { u , v } is linearly independentand that by parts (1) and (2) of Lemma 3.7, B ↾ { u , v } ⊥ is non-degenerate and its equiva-lence class is preserved. Write S = S u ∪ S v ∪ S uv and note that w , w ∈ T v and w ∈ T u .We use the bound (16) to get | S | ≤ q n/ − + 1) + 2 q n/ − − q n/ − + 1 ≤ q n/ − q and n ≥ (cid:3) Claim 5.6.
Suppose that S ⊂ { z } ⊥ , where z ∈ F nq is not self-orthogonal. Then | S | ≤ q n/ − ≤ q n/ − , for all n ≥ and odd q .Proof. Since z is not self-orthogonal, the restriction of B on { z } ⊥ remains non-degenerate.Now, using that S = S ∩{ z } ⊥ , we may use Proposition 3.12, to obtain the required result. (cid:3) Writing Q = { v ∈ S : v ∈ ( S \ { v } ) ⊥ } , note that S = [ v ∈ S S v ∪ Q. Now if z ∈ D ∩ Q , then S \ { z } ⊂ { z } ⊥ . Thus, by Claim 5.6, we have | S | ≤ q n/ − ≤ q n/ − , for all n ≥ q . In particular, we may assume(17) D = [ v ∈ S T v . Suppose | S | = 2 q n/ −
1. Note that, if | D | is even, by Claim 5.4 we must have that | S | is even leading to a contradiction. Recalling Claim 5.5, if statement (2) holds, we are doneand so assume statement (3) is true. Let w ∈ D and so, by (17), we have T v = { w } forsome v ∈ S . If v is self-orthogonal, then firstly { v , w } is linearly independent and secondlyby parts (1) and (2) of Lemma 3.7, B ↾ { v , w } ⊥ is non-degenerate and its equivalence class isunchanged. We write S = S v ∪ S w ∪ S vw and use Lemma 3.2 as before, being mindful of the crucial fact that S v contains exactly onenon-self-orthogonal element. For ( q, n ) = (3 , | S | ≤ q n/ − + ( q n/ −
1) + 2 q n/ − − q n/ − + q n/ − <
16 = 2 q n/ − . For other combinations of ( q, n ), by (16), we have | S | ≤ q n/ − + ( q n/ −
1) + 2 q n/ − − q n/ − + q n/ − ≤ q n/ − . Both bounds above contradict the presumed size of S . Then, we must have that v is notself-orthogonal. It follows that T w = { v } , which in turn implies that elements of D occur inpairs. As explained above, this contradicts the presumed parity of | S | , concluding the proof. (cid:3) Proof of Theorem 2.2
The proof of Theorem 2.2 is similar to the proof of Theorem 2.1. The differences arisefrom having characteristic 2 (the theory of bilinear forms is different) and not being able toassume that q is large enough. We are however free to assume that n is large enough. A factspecial to F n that we use is that every two distinct non-zero vectors are linearly independent.In particular the requirement for { v , w } to be linearly independent in Lemma 3.7 becomesredundant.From now on we use the notation in Lemma 3.2 and Lemma 3.5. The following simpleinequality will be useful. It is specific to F n , is true for all n , and is sharp. Lemma 6.1.
For n ≥ , let S ⊂ F n be a (3 , -orthogonal set with respect to a non-degeneratesymmetric bilinear form B . If v ∈ S , then in the notation of Lemma 3.5, | R v | ≤ | V v | / .Proof. Note that R v is disjoint from R v + R v . Indeed, if x , y ∈ R v , then B ( v , x ) = B ( v , y ) =1. Therefore B ( v , x + y ) = 1 + 1 = 0. This means that x + y / ∈ R v .Now, V v is a vector space containing R v . Therefore R v and R v + R v are two disjoint setscontained in V v . Hence 2 | R v | ≤ | R v | + | R v + R v | ≤ | V v | . (cid:3) LMOST ORTHOGONAL SUBSETS OF VECTOR SPACES OVER FINITE FIELDS 21
We derive a bound on | S v | . Lemma 6.2.
For n ≥ , let S ⊂ F n be a (3 , -orthogonal set with respect to a non-degeneratesymmetric bilinear form B . If v ∈ S , then in the notation of Lemma 3.5: • If B = · , then | S v | ≤ n, if n ≤ n − − , if n is odd and n ≥ n − , if n is even and n ≥ . • If B = H and n is even, then | S v | ≤ n − .Proof. We begin with B = · . By Lemma 3.5 and Lemma 6.1, we have | S v | ≤ | R v | + | T v | ≤ | V v | | T v | . By Lemma 3.2 we have dim( V v ) ≤ ⌊ ( n − | T v | ) / ⌋ . Setting t = | T v | we get | S v | ≤ ⌊ n − t ⌋− + t. A routine calculation confirms that, for n ≤
7, the right side is maximum when t = n .Otherwise, the maximum is achieved when t = 1 for odd n and when t = 0 for even n .If B = H , there are no non-self-orthogonal vectors and so, similarly to above, | S v | ≤| V v | / ≤ n − . (cid:3) We first prove the theorem for the hyperbolic form H . Proposition 6.3.
Let n ≥ be even and S ⊂ F n be a (3 , -orthogonal set with respect tothe hyperbolic form H . Then | S | ≤ n +1 − . Proof.
We prove the claim by induction. For n = 2, F \ { } is not (3 , n = 2.For the inductive step, we may assume there exist linearly independent v , w such that v · w = 1. If not, then S is an orthogonal set and by Lemma 3.2, we have the better bound | S | ≤ n −
1. By Lemma 3.7 (3) H ↾ { v , w } ⊥ is non-degenerate and is equivalent to H (inthis lower-dimensional vector space). By the induction hypothesis we have | S vw | ≤ n − . Hence by Lemma 6.2, | S | ≤ | S v | + | S w | + | S w | ≤ n − + 2 n − + (2 n −
2) = 2 n +1 − . (cid:3) From now on we mainly restrict our attention to the dot product, though we will useProposition 6.3 for even n because we sometimes use Lemma 3.7 (1) and the restrictionof the dot product may be equivalent to H . To prove the theorem we consider two casesseparately depending on whether S contains a vector that is not self-orthogonal or not. Wefirst prove the theorem when all vectors in S are self-orthogonal. The proof is similar to thatof Proposition 6.3. Proposition 6.4.
For n ≥ , let S ⊂ F n be a (3 , -orthogonal set with respect to the dotproduct. If S consists entirely of self-orthogonal vectors, then | S | ≤ ( n +12 − , if n is odd ;2 n +1 − , if n is even . Proof.
We prove the claim by induction. For n = 1, | S | = 0. For n = 2, we have S ⊂ { (1 , } and the claim follows.For the inductive step, we may assume there exist linearly independent v , w such that v · w = 1. If not, then S is an orthogonal set and by Lemma 3.4, we have a better boundon | S | than required. Furthermore, by Lemma 3.7 (3) the dot product restricted to { v , w } ⊥ is equivalent to the (lower dimensional) dot product and S vw contains only self-orthogonalvectors. For even n , by the induction hypothesis we have | S vw | ≤ n − . All vectors in S are self-orthogonal and so T v = T w = ∅ . Therefore S v = R v . Lemma 6.1gives | R v | ≤ n − . The same holds for w . Putting everything together gives | S | ≤ n − + 2 n − + (2 n −
3) = 2 n +1 − . For odd n , the induction hypothesis gives | S vw | ≤ n − − . Again T v = T w = ∅ and so by Lemma 6.2, we have | S v | , | S w | ≤ n − − . This gives | S | ≤ n − − + 2 n − − + (2 n − −
2) = 2 n +12 − , as required. (cid:3) The next step is to prove a bound for all S that is weaker than that in Theorem 2.2. Itwill be used to prove the theorem when S contains a vector that is not self-orthogonal. Lemma 6.5.
For n ≥ , let S ⊂ F n be a (3 , -orthogonal set with respect to a non-degeneratesymmetric bilinear form. Then | S | ≤ n +12 + 2 n − , if n = 1 , n +12 + n ( n +1)2 − , if n ≥ is odd ;2 n +1 + 2 n − , if n = 2 , n +1 + n ( n +1)2 − , if n ≥ is even . Proof.
If the bilinear form is equivalent to H , the result follows from Proposition 6.3.If the bilinear form is equivalent to the dot product, we let D ⊂ S be the collection ofvectors in S that are not self-orthogonal. The claim follows by applying Proposition 6.4 to S \ D and Lemma 3.10 to D . (cid:3) We continue with the case when there is a vector that is non-self-orthogonal. The proofis longer because we cannot initiate the induction (for example, Remark 6.7 on p. 24 shows S , (2 , , · ) ≥ > −
3) and because we can no longer assume, say, T v = ∅ . LMOST ORTHOGONAL SUBSETS OF VECTOR SPACES OVER FINITE FIELDS 23
Proposition 6.6.
Let n be an integer and S ⊂ F n be a (3 , -orthogonal with respect to thedot product. If S contains a vector that is not self-orthogonal, then | S | ≤ ( n +12 + 1 , if n ≥ is odd ;2 n +1 − , if n ≥ is even . Proof.
We begin with familiar notation. We let G be the simple graph with vertex set S andedges given by not mutually orthogonal pairs of vertices, and D = { v ∈ S : v · v = 1 } . Even n . Let z ∈ D . We consider two separate cases according to whether there exists anedge between z and S \ D or not.Suppose first that there is no edge between z and S \ D . Then S \ D ⊂ { z } ⊥ . Thedot product restricted to { z } ⊥ is non-degenerate (because z is not self-orthogonal). Thedimension of { z } ⊥ is odd and so the restriction is equivalent to the (lower dimensional) dotproduct. Moreover, all elements of S \ D are self-orthogonal. Therefore by Proposition 6.4we get | S \ D | ≤ n −
2. By Lemma 3.10 we have | D | ≤ n ( n +1)2 −
1. Hence (because n ≥ | S | ≤ (2 n −
2) + ( n ( n +1)2 − ≤ n +12 − . Next we suppose that there exists and edge vz with v ∈ S \ D . We have | S | ≤ | S v | + | S z | + | S vz | . By Lemma 3.7 (1) the dot product restricted to { v , z } ⊥ is non-degenerate. Lemma 6.5 gives | S vz | ≤ n + ( n − n − − . To bound | S v | note z ∈ T v . Writing t = | T v | , and applying Lemma 3.5 and Lemma 6.1 weget (using n ≥ | S v | ≤ ⌊ n − − t ⌋− + t ≤ n − + 1 . By Lemma 6.2 we get | S z | ≤ n − .Putting everything together gives (using n ≥ | S | ≤ n +1 − − (2 n − − ( n − n − ) ≤ n +1 − . Odd n . If D contains up to three elements, the required result follows from Proposition 6.4.Let x , y , z ∈ D denote three distinct elements and let H be the graph induced on { x , y , z } . H is not a triangle. We consider three cases based on the number of edges in H .Suppose H is the empty graph. First, assume a pair of the sets R x , R y , R z has non-emptyintersection. Namely, say v ∈ R x ∩ R y . Consider the decomposition(18) S = S x ∪ S v ∪ S xv . Note that x , y ∈ T v and that by Lemma 3.7 (1), we may apply Lemma 6.5 to obtain(19) | S | ≤ (2 n − − + 3) + (2 n − − + 1) + (2 n − + ( n − n − − ≤ n +12 + 1 , for n ≥
21. Thus suppose R x , R y , R z are pairwise disjoint. This means that R z ∪ { x , y } isorthogonal. Since z ∈ S xy and the dot product restricted to { x , y } ⊥ is non-degenerate (byLemma 3.7 (1)), Lemma 6.2 gives | S z | = | S z ∩ S xy | ≤ n − − . Considering the decomposition(20) S = S x ∪ S z ∪ S xz ∪ { x , z } , and using Lemma 3.7 (1), Lemma 6.2 (as well as its proof), and Lemma 6.5, we have(21) | S | ≤ (2 n − − + 1) + (2 n − − + 3) + (2 n − + ( n − n − −
3) + 2 ≤ n +12 + 1for n ≥
21. (We can do better but will refer later to (21)).Next, suppose H has exactly one edge. Without loss of generality take yx to be the edge.We split this case further. First suppose there exists an edge between z and R x ∪ R y . Say,an edge between z and v ∈ R x . We consider the decomposition (18) Then noting that x , z ∈ T v and that Lemma 3.7 (1) allows one to apply Lemma 6.5, one recovers the samebound on | S | as (19). Next, suppose there is no edge between z and R x ∪ R y . In particular, R x ∪ { y , z } is an orthogonal set. It follows that V x does not have the maximum dimensionthat is possible for orthogonal subspaces. Thus, using the decomposition (20) and usingLemma 3.7 (1), Lemma 3.2, Lemma 6.1 and Lemma 6.5, we may obtain the same bound on | S | as (21).Finally, suppose H has two edges. Without loss of generality let yxz be the path of length2. Here, we proceed to show that we may assume | D | = 3, which, as pointed out earlier,gives the required result.Suppose there exists a fourth vector w ∈ D . If w forms an edge with x , then there isno edge between any two of { y , z , w } and we are done by the arguments of the first case.If, on the other hand, w does not form an edge with x , by Lemma 3.7 (1), dot product isnon-degenerate on { x , w } ⊥ , so we use S = S x ∪ S w ∪ S xw ∪ { x , w } . Then, noting | T x | ≥ | S | ≤ (2 n − − + 3) + (2 n − − + 1) + (2 n − + ( n − n − −
3) + 2 ≤ n +12 + 1 , which is the same as (21). (cid:3) The proof of Theorem 2.2 is completed by combining Propositions 6.3, 6.4 and 6.6.
Remark 6.7.
Theorem 2.2 is false for small n . For n = 2, S , (2 , , · ) = 3 as we see bytaking S = F \ { } . For n = 4 the example below shows S , (2 , , · ) ≥ S = { (1 , , , , (1 , , , , (1 , , , , (0 , , , , (0 , , , , (0 , , , , (1 , , , } . The graph of S is indeed triangle-free: using the implicit order on the vertices, it is the unionof the 6-cycle 234567 with the edges 12 and 47.7. Proof of Theorem 2.3
The following is essentially the same as [10, Equation 2.4] and [18, Lemma 5]. Also see [3]or apply the point-hyperplane incidence bound in [20].
Lemma 7.1.
For
X, Y ⊂ F nq , define O ( X, Y ) = |{ ( x , y ) ∈ X × Y : B ( x , y ) = 0 }| . Then (cid:12)(cid:12)(cid:12)(cid:12) O ( X, Y ) − | X || Y | q (cid:12)(cid:12)(cid:12)(cid:12) ≤ p | X || Y | q n . LMOST ORTHOGONAL SUBSETS OF VECTOR SPACES OVER FINITE FIELDS 25
The following result is due to Tur´an [19].
Lemma 7.2.
Any graph of n vertices, which is K r +1 -free contains at most (1 − /r )( n / edges.Proof of Theorem 2.3. Let G = G ( S, E ) be the simple graph, where ( s , s ) ∈ S forms anedge in E if s = s and B ( s , s ) = . Then using the fact that S is ( k, G is K k -free and thus by Lemma 7.2, | E | ≤ k − k − | S | . Denoting G ′ = G ( S, E ) as the complement of G , we deduce that | E | ≥ | S | ( | S | − − | E | ≥ | S | k − − | S | . Now, clearly O ( S, S ) ≥ | E | . Hence, applying Lemma 7.1, we have | S | k − − | S | q − | S | ≤ | S | q n/ , which gives | S | ≤ (cid:18) q ( k − q − k + 1 (cid:19) ( q n/ + 1) . (cid:3) References [1] O. Ahmadi and A. Mohammadian, ‘Sets with many pairs of orthogonal vectors over finite fields’,
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Email address : [email protected] G.P.: Department of Mathematics, University of Georgia, Athens, GA, 30602 USA
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