TWO GEOMETRIC COMBINATORIAL PROBLEMS IN VECTOR SPACES OVER FINITE FIELDS

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1 TWO GEOMETRIC COMBINATORIAL PROBLEMS IN VECTOR SPACES OVER FINITE FIELDS EMMETT WYMAN Abstract. First, we show that the number of ordered right triangles with vertices in a subset E of the vector space F q over the finite field F q is equal to the expected value q 1 E 3 up to an error term O(q 3 E 3 ). Second, let E F d q. We give a lower bound for the maximum discrepancy between E \ H and the expected value q 1 E taken over all hyperplanes H in F d q. This result is easily extended to take H over all hyperplanes and spheres in F d q. Contents 1. Introduction 1. Fourier Analysis in F d q 3. Right Triangles Statement of Results Proof of Theorem 5 4. Discrepancies Statement of Results Proof of Theorem 3 9 References Introduction Geometric combinatorics in the Euclidean setting is a classical area of study. Recently, problems in the setting of vector spaces over finite fields have garnered much attention. Many of the results in the finite setting have continuous analogues. Consequentially, vector spaces over finite fields provide an excellent setting in which to study problems which may have similar results in Euclidean space [6]. In general, we are concerned with the asymptotic behavior of quantities dependent on a subset E of F d q as q!1, e.g. the number of distinct distances between points in E, or the number of distinct areas of triangles with vertices in E (see [10] and [11]). The following notation will be useful in our treatment of such quantities. Let and Y be quantities dependent on q. Wewrite. Y if apple CY for all large q, wherec is some constant. & Y means Y.. Wewrite Y if. Y and Y.. We write Y if /Y! 0 as q!1and we write Y if Y. In addition, we will use big O and little O notation: Y = O() if there exists a constant C such that Y <C for all large q, and = o(y )if/y! 0 as q!1. 1

2 EMMETT WYMAN Often, we will ask how large a subset E of F d q must be in order to ensure that it contains a certain kind of geometric structure, such as a simplex of particular volume or triangle with a particular angle. A k-simplex in F d q is a set of k + 1 points such that any n 1-dimension subspace of F d q contains no more than n of them. In [9], D. Hart and A. Iosevich prove that any subset E of F d q such that d> k+1 and E >Cq dk k+1 + k where C is a universal constant, then E contains a copy of every k-simplex up to a rotation (a transformation by a matrix in the orthogonal group) and translation. In [4], the authors prove that if E >Cq d+k,thene contains a positive proportion of non-congruent k-simplices. In [0], L. A. Vinh gives the following result regarding triangles with vertices in a subset E of the plane F q: Let s F q and suppose E q 3,thenthenumber A(E,s)= {(x, y, z) E 3 : s =(x y) (x z)? } of ordered triangles with vertices in E of area s is equal to (1 + o(1))q 1 E 3. This result demonstrates that if E is large enough, then the number of triangles of area s determined by E converges to the expected value q 1 E 3. Vinh goes on to prove that A(E,0) = (1 + o(1))q 1 E 3 if E q 5 3. We will prove an analogous result for angles using a similar method. Another result of interest is due to L. A. Vinh [19] regarding the proportion of right triangles determined by a subset E of the vector space F d q over the finite field F q of q elements for d, as follows. Theorem 1. Let E F d q and let D(E) ={(x, y, z) E E E :(x y) (x z) =0} denote the set of ordered right triangles with vertices in E. Then D(E) contains at least one element if E > q d+1 3. We will give a mild improvement on the constant for the case of d =. In 1954, K. F. Roth published On irregularities of distribution in Mathematika, which proves the following result. Let E be a collection of points in the square [0, 1] [0, 1]. The maximum discrepancy between E\([0, x] [0, y]) and the expected value xy E is bounded below by C log E where C is a universal constant (see [14]). We will prove a result in a similar spirit in the finite field setting. Let E F d q and consider the di erence D E (H) = E \ H q 1 E for each hyperplane H in F d q.we will show that if E is a su ciently small subset of F d q, then there exists a hyperplane H such that D E (H) is asymptotic to q 1 E 1.. Fourier Analysis in F d q Fourier analysis in F d q is a key technique applied to many problems in geometric combinatorics, and it will be an invaluable tool in the study of the problems presented herein. In the following section, we state and prove some of the tools of Fourier analysis in F d q. Let x =(x 1,...,x d ) and y =(y 1,...,y d ) F d q. Then we define the dot product x y := x 1 y x d y d.

3 TWO GEOMETRIC COMBINATORIAL PROBLEMS IN VECTOR SPACES OVER FINITE FIELDS3 We will also write kxk := x x = x 1 + x d. Let : F q! C be a nontrivial additive character with the following properties. 1. (x + y) = (x) (y). For each x F q, ( q if x =0, (mx) = 0 if m 6= 0. mf q It follows from the above properties that (0) = 1, (x) = ( x). Moreover, for any x F d q ( q d if x =0, (x m) = 0 if m 6= 0. mf d q For more information regarding additive characters, see [6]. Fourier transform on F d q and prove some useful properties. We now define the Definition 1. Let f : F d q! C. The Fourier transform ˆf of f is defined to be ˆf(m) :=q d xf d q f(x) ( x m). Note that, in particular, the Fourier transform of f at 0 evaluates to the average of f over F d q. ˆf(0) = q d xf d q f(x), Proposition 1. (Fourier Inversion Formula) Let f : F d q! C. Then Proof. mf d q f(x) = mf d q ˆf(m) (x m) =q d mf d q = q d yf d q = q d yf d q = f(x). ˆf(m) (x m). f(y) ( m y) (m x) yf d q f(y) yf d ( q d q if x = y 0 if x 6= y (m (x y))

4 4 EMMETT WYMAN Proposition. (Plancherel s Formula) Let f,g : F d q! C. Then ˆf(m)ĝ(m) =q d f(x)g(x). mf d q xf d q Proof. mf d q ˆf(m)ĝ(m) =q d mf d q = q d x,yf d q = q d x,yf d q f(y) ( xf d q x m) yf d q f(x)g(y) ((y x) m) mf d q ( q d if x = y f(x)g(y) 0 if x 6= y g(y) (y m) = q d xf d q f(x)g(x). We obtain Parseval s formula, ˆf(m) = q d f(x), mf d q xf d q a special case of Plancherel s formula, by setting f = g in Plancherel s formula. 3. Right Triangles 3.1. Statement of Results. Here we state the main theorem for the distribution of right triangles in the plane. Theorem. Let E F d q and D(E) ={(x, y, z) E E E :(x y) (x z) =0}. Then D(E) = q 1 E 3 + q 3 3 E, where < 1/, with the leading term dominating if E > 1 3 q 5 3. If we assume q 3 (mod 4), then < 1+o(1) and the leading term will dominate when E > (1 + o(1))q 5 3. A remark: The main term is the expected value of D(E) in the following sense. If we define D r (E) ={(x, y, z) E 3 :(x y) (x z) =r}, then E 3 can be expressed as the disjoint union S rf q D r (E). Hence E 3 = rf q D r (E), and so we expect to find E 3 /q elements in D 0 (E) =D(E).

5 TWO GEOMETRIC COMBINATORIAL PROBLEMS IN VECTOR SPACES OVER FINITE FIELDS5 3.. Proof of Theorem. We will make use of the following counting lemmas. Lemma 1. Let q be a prime power such that q 3 (mod 4). Then for each r F q, {(x, y) F q : x + y = r} apple q +1. Proof. We have that q = {(x, y) F q : x + y = r}. rf q Since q 3 (mod 4), the only solution to x +y = 0 is the trivial solution x = y = 0. Moreover, if r 1 and r are both nonzero quadratic residues in F q,thereexists F q such that r 1 = r. Hence {(x, y) F q : x + y = r 1 } = {( x, y) F q :( x) +( y) = r 1 } = { (x, y) F q :( x) +( y) = r } = {(x, y) F q : x + y = r }. This works similarly if r 1 and r are quadratic nonresidues. Note that since quadratic nonresidue, we have q 1= {(x, y) F q : x + y = r} and so rf q\{0} 1isa = q 1 {(x, y) F q : x + y =1} + q 1 {(x, y) F q : x + y = 1}, (q + 1) = {(x, y) F q : x + y =1} + {(x, y) F q : x + y = 1}. Now the number of solutions to x + y =1wherey 6= 0 is the same as the number of solutions to b a =1whereb 6= 0 (divide through by y and let a = x/y and b =1/y), which is equal to the number of solutions to b a =1ifb is allowed to be 0. This, however, is equal to the number of solutions to =1byletting = b a and = b + a. This has q 1 solutions, and so x + y = 1 has q +1 solutions, the extra two solutions coming from x = ±1, y = 0. It follows from the above equation that x + y = 1 also has q + 1 solutions. The lemma follows. Lemma. Let E F q. Then {(y, z, y 0,z 0 ) E 4 : y + z = y 0 + z 0 and y z = y 0 z 0 } ( (q + ) E if q 3 (mod 4) apple (q + 1) E otherwise. Proof. First consider the case where z = z 0.Theny = y 0 and the equation y z = y 0 z 0 is automatically satisfied. There are E options for both z and y, and so this case contributes E elements. Now consider the case where z 6= z 0. We may choose y and z in E di erent ways each. After we determine y 0, there is at most one value

6 6 EMMETT WYMAN for z 0 E. This yields at most E combinations of values for y, z, and z 0. Then we write z 0 = y + z y 0. Plugging into y z = y 0 z 0 yields () 0 = y z y 0 (y + z)+ky 0 k. Now write y 0 =(y1 0,y0 ), z =(z 1,z ), and y =(y 1,y ). There are at most q values which we could fix for y1 0, and then the above equation becomes the quadratic 0=(y z + y1 0 y1(y z 1 )) y(y z 1 )+y 0 as a function of y 0. Hence there are at most two values for which y0 can take, yielding q E for this case. Hence, we have (q + 1) E as an upper bound. Now suppose q 3 (mod 4). First note that q cannot be a power of, and so at () we complete the square to obtain ky + zk 4 y z = ky + zk 4 y 0 (y + z)+ky 0 k = y 0 y + z. There are, by Lemma 1, at most q + 1 possible values y 0 may take to satisfy this equation. Hence our upper bound becomes (q + ) E. With the above lemmas in hand, we proceed with the proof of Theorem. Proof. We write {(x, y, z) E 3 :(x y) (x z) =0} = q 1 sf q x,y,ze (s(x y) (x z)) and split the sum into two terms: one for s = 0 and one for s 6= 0, so that the above becomes = q 1 x,y,ze (0) + q 1 s6=0 x,y,ze (s(x y) (x z)).

7 TWO GEOMETRIC COMBINATORIAL PROBLEMS IN VECTOR SPACES OVER FINITE FIELDS7 The first term is evaluated to q 1 E 3, the main term. Now we estimate the second term. Applying the Cauchy-Schwarz inequality twice yields q 1 s6=0 x,y,ze apple q (q 1) s6=0 apple q (q 1) E s6=0 (s(x y) (x z)) x,y,ze xe apple q (q 1) E s6=0 xf q = q (q 1) E s6=0 xf q = q (q 1) E s6=0 (s(x y) (x z)) y,ze y,ze y,z,y 0,z 0 E xf q y,z,y 0,z 0 E Then by the properties of the character Then by Lemma, q(q =(q 1) E s6=0 (s(x y) (x z)) (s(x y) (x z)) (s(x y) (x z) s(x y 0 ) (x z 0 )) (s(y z y 0 z 0 )) (sx ( y z + y 0 + z 0 )). y,z,y 0,z 0 E y+z=y 0 +z 0 apple (q 1) E sf q = q(q 1) E 1) E y,z,y 0,z 0 E y+z=y 0 +z 0 y z=y 0 z 0 y,z,y 0,z 0 E y+z=y 0 +z 0 y,z,y 0,z 0 E y+z=y 0 +z 0 y z=y 0 z 0,we have 1. (s(y z y 0 z 0 )) (s(y z y 0 z 0 )) 1 apple q(q 1) E 3 (q + 1) apple q 3 E 3. Hence the second term is bounded above by 1 q 3 E 3, and so we write D(E) = q 1 E 3 + q 3 E 3.

8 8 EMMETT WYMAN where < 1/. It follows from direct computation that the first term exceeds the second if E > 1 3 q 5 3. Now if q 3 (mod 4), Lemma instead yields q(q 1) E 1 apple q(q 1) E 3 (q + ) = (q 3 + q q) E 3. y,z,y 0,z 0 E y+z=y 0 +z 0 y z=y 0 z 0 Hence the second term is bounded by q 3 E 3 (1+o(1)). Hence the first term exceeds the second when q 1 E 3 >q 3 E 3 (1 + o(1)), i.e. when E >q 5 3 (1 + o(1)). This concludes the proof of Theorem. 4. Discrepancies 4.1. Statement of Results. A hyperplane in F d q is a set of the form {x F d q : x m = t} for some t F q and some nonzero m F d q.asphereinf d q is a set in the form {x F d q : kxk + x m = t} for some t F q and m F d q. Let H (F d q) denote the set of all hyperplanes in F d q. We will write H = H (F d q) if it is clear that we are referring to the space F d q. Similarly, we define S = S (F d q) to be the set of all spheres and hyperplanes in F d q. We now state our primary result. Theorem 3. Let E F d q and let D E (H) = E \ H q 1 E for each H H. Then sup D E (H) q 1/ E 1/ (1 q d E ) 1/ (1 + O(q 1 )). It follows that if E q d, then the lower bound becomes q 1/ E 1/ (1 + O(q 1/ )). The above theorem tells us something about the distribution of points on each hyperplane. Suppose that we have H slots, and E V (F d q) points to assign randomly to each slot, the same number we get when we sum up E \ H over the H H. The square root principle tells us that each slot will have about E /q points, plus or minus something akin to q 1/ E 1/. Theorem 3, however, tells us that there will always be a plane with - roughly - a higher deviation than q 1/ E 1/ from the mean. In fact, it tells us if E \ H is very close to E /q for many planes, then there must be a plane with a high deviation from the mean. If we let D E (S) = E\S q 1 E for each S S, we have the following corollary. Corollary 1. Let E F d q. Then sup D E (S) q 1/ E 1/ (1 + o(1)). SS

9 TWO GEOMETRIC COMBINATORIAL PROBLEMS IN VECTOR SPACES OVER FINITE FIELDS9 The corollary follows trivially from Theorem 3 if E q d since H S. The corollary is nontrivial only if E & q d, in which case sup SS D E (S) q 1/ E 1/ (1 + o(1)) may be significantly larger than sup D E (H). 4.. Proof of Theorem 3. Definition. We say that V is a direction set of a space F d q if for each nonzero vector x F d q, there exists a unique v V such that v = x for some F q. For each q and d, we fix a direction set denoted by V (F d q). In the following proposition, we construct each V (F d q) inductively on d. In the process, we will determine the size of V (F d q). Proposition 3. There exists a direction set V of F d q for each q and d, and V = (q d 1)/(q 1). Proof. We prove the size of V first. Suppose that V is a direction set of F d q. Then since each nonzero element in F d q is uniquely expressed as v for some nonzero F q and v V,wehave(q 1) V = F d q \{0} = q d 1. We prove that a direction set V exists for each F d q by induction on d. Clearly, V = {1} is a direction set for F 1 q. Assume now that V is a direction set for F d q. Let V 0 =(F d q {1}) [ (V {0}). Suppose (x, x d+1 ) F d q F q is nonzero. Suppose x d+1 = 0 and x 6= 0, in which case by the inductive hypothesis there exists a unique v V and such that x = v, from which we obtain (v, 0) = (x, x d+1 ), where (v, 0) V 0. If x d+1 6= 0, then (x 1 d+1 x, 1) V 0 and (x, x d+1 )=x d+1 (x 1 1 d+1x, 1). Moreover, (xd+1x, 1) is the only point in F d q {1} which can be scaled to (x, x d+1 ). Hence, V 0 is a direction set for F d+1 q. Proposition 4. Each hyperplane H H (F d q) is uniquely expressible as H v,t {x F d q : x v =1} where v V (F d q) and t F q.itfollowsthat H (F d q) = q V (F d q). Proof. We will show that the map F q V (F q )! H (v, t) 7! H v,t is a bijection. To show that the map is one-to-one, let v 1,v V (F d q) and t 1,t F q such that H v1,t 1 = H v,t. Then by definition of the hyperplane, x v 1 = t 1 and x v = t have the same solution set. Now pick any x 0 in H v1,t 1,thenH v1,t 1 x 0 = H v1,0 and H v,t x 0 = H v,0. Thenx v 1 = 0 and x v = 0 have the same solution set, and so v 1 = v.butthenv 1 = v. Now t 1 = x 0 v 1 = x 0 v = t. Hence the map is one-to-one.

10 10 EMMETT WYMAN To show that the map is onto, we recall that every H H can be written as {x F d q : x m = s} for some s F q and nonzero m in F d q.thenthereexistunique v V (F d q) and F q \{0} such that m = v. Thenwewrite {x F d q : x m = s} = {x F d q : x v = s} = {x F d q : x v = 1 s} = H v, 1 s. The thrust of the proof of Theorem 3 comes from the following lemma, which is interesting in its own right. It states that the l norm of the size of the intersection E \ H over H H only depends on the size of E, as opposed to its structure. Hence, it is a very useful tool for when trying to find a lower bound on the maximum discrepancy sup D E (H) of E when nothing about the structure of E is assumed. Lemma 3. Let E F d q. Then E \ H = q d 1 E + q 1 ( V (F d q) 1) E. We conclude that D E (H) = q 1 E (q d E ).

11 TWO GEOMETRIC COMBINATORIAL PROBLEMS IN VECTOR SPACES OVER FINITE FIELDS 11 Proof. We have by the above proposition, E \ H = vv (F d q ) tf q = vv (F d q ) tf q = q E \ H v,t vv (F d q ) tf q = q 1 vv (F d q ) = q d 1 q 1 xf d q vv (F d q ) sf q = q d 1 vv (F d q ) sf q = q d 1 vv (F d q) sf q x,x 0 F d q s,s 0 F q x,x 0 F d q sf q E(x) sf q (s(t x v)) E(x)E(x 0 ) ((s s 0 )t) (( sx + s 0 x 0 ) v) E(x)E(x 0 ) (s( x + x 0 ) v) q d xf d q Ê(sv) E(x) ( sx v) Ê(sv) + q d 1 vv (F d q ) Ê(0). Since V (F d q) is a direction set, we have = q d 1 = q d 1 mf d q Then by Parseval s formula, mf d q \{0} Ê(m) + q 1 V (F d q) E = q d 1 xf d q Ê(m) q d 1 Ê(0) + q 1 V (F d q) E. E(x) q 1 E + q 1 V (F d q) E = q d 1 E + q 1 ( V (F d q) 1) E.

12 1 EMMETT WYMAN Hence P E \ H = q d 1 E + q 1 ( V (F d q) 1) E. Applying this formula to P D E(H) yields D E (H) = ( E \ H v,t q 1 E ) vd(f d q) tf q = vd(f d q ) tf q ( E \ H v,t q 1 E E \ H v,t + q E ) = q d 1 E + q 1 ( D(F d q) 1) E q 1 D(F d q) E + q 1 D(F d q) E = q d 1 E q 1 E = q 1 E (q d E ), as desired. We now proceed with the proof of Theorem 3 below. Proof. Suppose M 3, Hence Now so we have q 1 E (q d D E (H) for each H H. Then by Proposition 4 and Lemma E ) = 1 D E (H) apple M H = M q(q d 1)(q 1) 1. M q 1 E (1 q d E ) 1 q 1 1 q d. 1 q 1 1 q d = q d + q 1 1 q d apple q 1, M q 1 E 1 (1 q d E ) 1 (1 + O(q 1 )) 1 = q 1 E 1 (1 q d E ) 1 (1 + O(q 1 )). Hence as desired. sup D E (H) q 1 1 E (1 q d E ) 1 1 (1 + O(q )) We now prepare to prove the corollary to Theorem 3. Let P (F d+1 q ) denote the paraboloid {(x, kxk) F d q F q : x F d q}. As for hyperplanes and spheres, we may suppress the dependence on F d+1 q in the notation if it is clear from context.

13 TWO GEOMETRIC COMBINATORIAL PROBLEMS IN VECTOR SPACES OVER FINITE FIELDS 13 Proposition 5. Let : F d+1 q! F d q be the projection given by (x 1,...,x d+1 )= (x 1,...,x d ) for each (x 1,...,x d+1 ) F d+1 q. gives a natural bijective correspondence between H (F d+1 q ) and S (F d q) given by : {H \ P (F d+1 q ):H H (F d+1 q )}! S (F d q) {(x, kxk) F d+1 q :(x, kxk) (v, v d+1 )=t} 7! {x F d q : v d+1 kxk + v x = t}. Proof. Since the restriction P (F d+1 q ) : P (Fd+1 q )! F d q is a bijection, the map is injective. It remains to show that is surjective. Note that for each H m,s S (F d q), we have (H (m,0),s \ P (F d+1 q )) = H m,s. Now if we have a sphere Then {x F d q : kxk + m x = s} S (F d q), {H (m,1),s \ P (F d+1 q )} = {(x, kxk) F d+1 q :(x, kxk) (m, 1) = s} = { (x, kxk) :kxk + m x = s} We now prove Corollary 1. = {x F d q : kxk + m x = s} = H m,s. Proof. Let E F d q. Then consider the set E 0 F d+1 q given by E 0 =( P (F d+1 q ) ) 1 (E). Then since E 0 apple q d q d+1, by Theorem 3, sup D E 0(H) (1 + o(1))q 1/ E 0 1/ (F d+1 q ) =(1+o(1))q 1/ E 1/. Now if H H (F d+1 q ), let S = (H \ P (F d+1 q )). Then (E 0 \ H) =E \ S. Then since P (F d+1 q ) is bijective, E 0 \ H = E \ S, and so D E 0(H) = E 0 \ H q 1 E = E \ S q 1 E = D E (S). Then by Proposition 5, sup D E (S) = sup D E 0(H) (1 + o(1))q 1/ E 1/. SS

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