The largest complete bipartite subgraph in point-hyperplane incidence graphs

Given m points and n hyperplanes in Rd (d > 3), if there are many incidences, we expect to find a big cluster Kr,s in their incidence graph. Apfelbaum and Sharir [1] found lower and upper bounds for the largest size of rs, which match (up to a constant) only in three dimensions. In this paper we close the gap in four and five dimensions, up to some polylogarithmic factors. Mathematics Subject Classifications: 05C35, 52C10, 05D10


Introduction
Throughout this paper, let d denote an integer at least three.Given a set P of m points and a set Q of n hyperplanes in R d , their incidence graph G(P, Q) is a bipartite graph with vertex set P ∪ Q and (p, q) ∈ P × Q forms an edge iff p ∈ q.It is proved in [1] that if this graph does not contain K r,s (a complete bipartie graph with two parts A, B, |A| = r, |B| = s and (a, b) is an edge for every a ∈ A, b ∈ B) as a subgraph for some fixed integers r, s 2, then it can have at most O d ((mn) d/(d+1) + m + n) edges.Here the notation f = O d (g) means there exists some constant C that depends on d such that f Cg.The number of incidences between P and Q, denoted by I(P, Q) is the number of edges of G(P, Q).
Conversely, when the incidence graph has many edges, we expect to find a big subgraph isomorphic to K r,s .How big can rs (the number of edges of K r,s ) be in terms of m, n and the number of edges of the incidence graph?To make the question precise, we use the following definition: Definition 1.Given a set P of points and Q of hyperplanes in R d , let rs(P, Q) be the maximum size of a complete bipartite subgraph of its incidence graph, and rs d (m, n, I) the electronic journal of combinatorics 27(1) (2020), #P1.12 be the minimum of this quantity over all choices of m points and n hyperplanes in R d with I incidences.To be precise: We are interested in how big rs d (m, n, I) can be in terms of m, n and I. Apfelbaum and Sharir [1] gave a satisfactory answer to this question when d = 3.
2. If m n, I = O(nm 1/2 ) and I = Ω(m 3/4 n 3/4 ), then rs 3 (m, n, I) = Θ( However, much less is known in higher dimensions.One thing we know is that rs d (m, n, I) max{ I m , I n } by looking at the star subgraphs centering at the point and the hyperplane with maximum degrees in G(P, Q).As noted in [1], when the dimension d increases beyond 3, there are progressively more ranges of I (as a function of m and n) where the bounds for rs d change qualitatively.At one extreme, when I is small enough, we expect the graph not to contain any big complete bipartite subgraph.Indeed, if I = O d,ε ((mn) 1− 2 d+2 −ε ) for some ε > 0 and d is odd, Brass and Knauer [3] constructed an example of point-hyperplane incidence graph in R d with I incidences and no K t,t for some fixed integer t 2. A similar result holds if d is even, and These bounds have been slightly improved in [2].At another extreme, when I is very large, we expect to find a large K r,s .
Theorem 3 (Apfelbaum and Sharir [1]).For any d 3, if Morever, if the electronic journal of combinatorics 27(1) (2020), #P1.12 These lower and upper bounds only match (up to a constant) when d = 3 (which is why we have a tight bound in part 1 of Theorem 2).In this paper we close the gap (up to polylogarithmic factors) in four and five dimensions for this range of I.More specifically we prove the following two results.The main tool used to prove Theorem 4 and Theorem 5 is an incidence bound between points and nondegenerate hyperplanes by Elekes and Tóth [5], which is reviewed in the next section.We then present the proof of Theorem 4 and sketch the proof of Theorem 5 in the subsequent sections.At the end we explain why our method does not work in six dimensions.
We use the following notation.Let A and B be two sets of geometric objects in R d .Their incidence graph G(A, B) is a bipartite graph on A × B, where (a, b) forms an edge iff a ⊂ b.The number of incidences between A and B, denoted by I(A, B), is the number of edges of this graph.In this paper, A is either a set of points or a set of lines, and B is a set of higher dimensional flats.An affine d -dimensional flat, or a d -flat is a subset of R d that is congruent to R d for some integer 0 d d.Points, lines, planes and hyperplanes are flats of dimensions 0, 1, 2 and d − 1 respectively.Given a flat Given a set S of m points in R d and some β ∈ (0, 1), an affine hyperplane H is βdegenerate with respect to (w.r.t.) S if there exists a proper subflat F ⊂ H that contains more than β fraction of the number of points of S in H, i.e. |F ∩S| > β|H ∩S|.Otherwise, H is β-nondegenerate.Elekes and Tóth proved the following incidence bound.
Theorem 6 (Elekes-Tóth [5]).If S is a set of m points and H is a set of n β-nondegenerate hyperplanes w.r.t.S (for any 0 < β < 1)1 in R d (for any integer d 2), then there exists a constant C β,d > 0 such that This implies the maximum number of β-nondegenerate, k-rich (i.e. containing at least . Elekes-Tóth in fact proved the second statement; it is shown to be equivalent to (3) in [1].When d = 2, it reduces to the well known Szemerédi-Trotter point-line incidence bound [7].
Since points and hyperplanes are dual to each other, we also have a dual version of the above result.Given a set H of n hyperplanes in R d , a point p is β-nondengenerate with respect to (w.r.t.) H if there does not exist a line such that #{H ∈ H : ⊂ H} β#{H ∈ H : p ∈ H}.

Proof in four dimensions
We first outline our strategy.Let S be a set of m points, and H be a set of n hyperplanes in R 4 .There are two ways to form a big K r,s in the incidence graph G(H, S): either a plane contains many points of S and belongs to many hyperplanes of H, or a line does.By an averaging argument, we can assume that each hyperplane is Ω( I m )-rich (i.e.contains at least Ω( I m ) points of S).By Theorem 6, the contribution from β-nondegenerate hyperplanes is negligible, so we can assume that each hyperplane is β-degenerate, i.e. it contains some plane with at least β portion of the total number of points in that plane, hence the plane is Ω(β I m )-rich.In this case, we say each hyperplane degenerates to a rich plane.Either one of those planes belongs to many hyperplanes, which would form a big K r,s , or we can find a subset P i of planes such that I(S, P i ) is large.We then repeat our argument: using the averaging argument and Corollary 7, we can assume that each point in S belongs to many planes in P i and degenerates to a line.Either one of those lines contains many points, which then form a big K r,s , or we can find a subset L j of lines such that I(L j , P i ) is large.But after some transformation, this number is the same as the number of incidences between points and lines in R 2 and hence cannot be too large by Theorem 6 for d = 2, or equivalently, Szemerédi and Trotter's theorem in [7].
We now give a detailed proof.
Proof of Theorem 4. Assume for contradiction that there exist a set S of m points and a set H of n hyperplanes in R 4 with I incidences where I C 4 (mn 2/3 + nm 3/5 ) and the incidence graph G(S, H) contains no K r,s where rs C 4 I mn 5/2 mn(log mn) −4 .We shall choose the suitable positive constants C 4 and C 4 to derive a contradiction.
Indeed, remove all the hyperplanes that contain fewer than I 4n points and the hyperplanes that are β-nondegenerate.The number of incidences from the non-rich hyperplanes is at most n I 4n = I 4 .By Theorem 6, the number of incidences from the β-nondegenerate hyperplanes is at most C β,4 ((mn) 4/5 +mn 2/3 ) < C 4 4 (mn 2/3 +nm 3/5 ) if we choose C 4 > 8C β,4 .Indeed, this only fails if (mn) 4/5 mn 2/3 and (mn) 4/5 nm 3/5 , which are equivalent to m n 2/3 and n m, but those two inequalities cannot hold at the same time.Therefore, after the removal, there remain at least I 2 incidences.Assume that there remain n 1 hyperplanes for some n 1 n.Throughout the proof, there are many inequalities that involve n 1 , but we can always use n to upper bound n 1 in the correct direction.Therefore, without loss of generality, we can assume that n 1 = n.
Step 2: For each I 4n -rich β-degerenate hyperplane H, we can find a proper subflat P ⊂ H so that |P | β|H| βI 4n .Since dim(H) = 3, we can assume that P is a plane.Let P denote the set of these planes.We claim that no plane in P belongs to more than s 0 hyperplanes in H, where where c 1 is a sufficiently small constant to be specified later.Indeed, assume that there are at least s 0 hyperplanes that degenerate to a same plane, then we have a configuration of K r,s with r = βI 4n and s = s 0 .This leads to a contradiction if we choose C 4 < βc 1 4 : mn(log mn) −4 .
Step 3: We use a dyadic decomposition to find a subset of planes with many incidences with S. Let P j denote the set of all planes that are assigned to at least 2 j and fewer than 2 j+1 hyperplanes, for 1 j < log s 0 < log n (here the logarithm is in base 2).The contribution to incidences from the planes must be at least β fraction of the number of incidences from the β-degenerate hyperplanes, which implies log s 0 j=0 2 j+1 I(S, P j ) β 4 I. Hence there must exist some i such that where c 2 = β/8.We claim that the following holds where β is the same as before, and C β,3 is defined in Theorem 6: the electronic journal of combinatorics 27(1) (2020), #P1.12Indeed, assume otherwise.By (6), , where c 3 = c −1 2 4C β,3 .We shall now derive a contradiction using two facts: |P i | n 2 i and 2 i s 0 .The first fact follows from log s 0 j=0 2 j |P j | n, since each hyperplane is assigned to exactly one plane.Using the formula for s 0 in (5), we have: By our assumption, I C 4 nm 3/5 , thus the second term on the right hand side of (8), c 3 (log n)nm 1/2 , is less than I 2 for large values of m, n.This implies the first term of (8) must be at least I 2 Rearranging we get 1 m 3/8 n 5/8 log n log mn (c 4 nm 3/5 ) 5/8   where c 4 = (2c 3 c 1/4 1 ) 8/5 .However, we can choose c 1 small enough and C 4 big enough so that c 4 < C 4 and hence this contradicts I C 4 nm 3/5 .So (7) must hold.
Step 4: Let us consider G(S, P i ), the incidence graph between the points S and the set of planes P i from Step 3. We can use an averaging argument similar to Step 1 to assume that each point in S is I 4m -rich (i.e.belongs to at least I 4m planes in P i ).Since the bound in (7) is the same as that in Corollary 7, we can also assume each point in S is β-degenerate w.r.t.P i (in the sense defined before Corollary 7).Each such point degenerates to a line that is β I 4m -rich.Let L denote the set of all these lines.We claim that no line in L contains more than r 0 points, where for some small enough positive constant c 5 to be chosen later.Indeed, since each plane in P i belongs to at least 2 i hyperplanes in H by definition, each line in L belongs to at least This contradicts our assumption if we choose C 4 < c 2 c 5 β/4.
Step 5: Similar to Step 3, we use a dyadic decomposition to find a subset of lines in L that form many incidences with P i .Here we say a line is incident to a plane P if ⊂ P .Let L k denote the set of all lines that contain at least 2 k and fewer than 2 k+1 points for 1 k < log r 0 < log m.The contribution to I from the lines must be at least β fraction, which implies log r 0 k=0 2 k+1 I(L k , P i ) β 4 I .Hence there must exist some j such that where c 6 = β 2 /64.claim that the following holds where C ST is the constant in Szemerédi and Trotter's theorem [7]: Indeed, assume otherwise.By (10): We make use of the following four facts: 2 j , 2 i s 0 and 2 j r 0 .We already showed |P i | n 2 i .The second fact holds for a similar reason: since each point is assigned to exactly one line we have log r 0 k=0 2 k |L k | m, and thus |L j | m 2 j .We now write: Using the formula for s 0 and r 0 in ( 5) and ( 9), we have: the electronic journal of combinatorics 27(1) (2020), #P1.12 We can choose c 1 and c 5 small enough so that (c 1 c 5 ) 1/3 c 7 log m log n (log mn) 7/3 < 1 2 and I mn for large values of m, n.In this case, the right hand side of the last inequality is strictly less than I, a contradiction.So (11) must hold.
Step 6: Project the set of planes P i and the set of lines L j to a generic three dimensional subspace, then intersect them with a generic plane Pi within this subspace.After this transformation, P i becomes a set of lines P * and L j becomes a set of points L * in Pi.By (11), , which violates the Szemerédi-Trotter theorem [7].This gives the desired contradiction and finishes our proof.

Sketch of the proof in five dimensions
The proof method is the same as that in four dimensions, but the exponents are different and the method is repeated one more time.In particular, in the previous section, we unwrap a point-hyperplane configuration in R 4 with many incidences in two layers: hyperplanes degenerate to planes and points degenerate to lines.At each layer, either we can find a big K r,s subgraph, or the number of incidences remain larger than the nondegenerate bound in Theorem 6, and we can keep unwrapping.In R 5 , we unwrap in three layers: hyperplanes degenerate to 3-flats, points degenerate to lines, and 3-flats degenerate to planes.The detailed proof is quite similar to that in the four dimensions case, so we only give an outline here.For simplicity, we ignore the polylogarithmic factors and write f g (or f g) to indicate there exists some constants a, b > 0 such that f a(log mn) b g (or f a(log mn) b g.Proof sketch of Theorem 5. Prove by contradiction.Let S denote the set of m points and H denote the set of n hyperplanes in R 5 .Assume that I(S, H) (mn 3/4 + nm 2/3 ) but their incidence graph does not contain any K r,s subgraph where rs I mn 3 mn.
Step 1: We can assume that every hyperplane is I n -rich, and β-degenerate with respect to S for some β ∈ (0, 1).The choice of β is quite flexible, so we can assume β = 1/2.
Step 2: For each such hyperplane H, we can find a 3-dimensional flat (or a 3-flat) F such that F ⊂ H and |F ∩ S| β|H ∩ S| βI n .Let F denote the set of these 3-flats.Using our assumption on rs, no flat in F belongs to more than s 0 hyperplanes where s 0 c 1 I 2 m 2 n(log mn) 10 for some sufficiently small positive constant c 1 to be chosen later.
the electronic journal of combinatorics 27(1) (2020), #P1.12 Step 3: Let F j denote the set of all 3-flats in F that are assigned to at least 2 j and fewer than 2 j+1 hyperplanes where j log s 0 < log n.Then there exists an i such that I(F i , S) I 2 i .We show that + nm 2/3 , which cannot happen given our condition I mn 3/4 + nm 2/3 .
Step 4: Since I is large, using Corollary 7, we can assume that each point in S is I m -rich (i.e.belongs to at least I m flats in F i , and is β-degenerate w.r.t.F i .Each such point degenerates to a βI m -rich line.Let L denote that set of these lines.Then no line in L can contain more than r 0 points where r 0 c 2 mn 2 (log mn) 10 for some sufficiently small positive constant c 2 to be chosen later Step 5: We use a dyadic decomposition to find a subset of lines with many incidences with F i .Let L k denote the set of all lines in L that contain more than 2 k and fewer than 2 k+1 points.Then there exists a j such that I(F i , L j ) Indeed, assume otherwise.Using I and 2 j r 0 I 2 mn 2 , we have This cannot happen with an appropriate choice of c 1 , c 2 and logarithmic factors.
Step 6: Project the set of 3-dim flats F i and the set of lines L j into a generic four dimensional subspace, then intersect them with a generic 3-dimensional flat in this subspace.After this transformation, F i becomes a set of planes and L j becomes a set of points in R 3 .Because of the inequality in the previous step, we can assume that each 3-flats in F i degenerate to a plane.Let P denote the set of all such planes.Then no plane belongs to more than t 0 flats in F i where t 0 c 3 I 2 m 2 n(log mn) 10 for some sufficiently small positive constant c 3 to be chosen later.
the electronic journal of combinatorics 27(1) (2020), #P1.12 Step 7: Using a dyadic decomposition, there exists some subset P k of planes, each belongs to at least 2 k and fewer than 2 k+1 3-flats in F i such that I : Step 8: Project the set of planes P k and the set of lines L j into a generic three dimensional subspace, then intersect them with a generic plane in this subspace.After this transformation, P k becomes a set of lines and L j becomes a set of points in R 2 .Hence, the inequality on I(L j , P k ) in Step 7 violates Szemerédi and Trotter's theorem for appropriate choices of c 1 , c 2 , c 3 and logarithmic factors.This gives the desired contradiction and finishes our proof.

Discussion
We first compare our approach with that of Apfelbaum and Sharir in [1].Their proof of (1) relies on the incidence bound (3) to unwrap the point-hyperplane configuration with many incidences: for each k = d − 1, d − 2, . . ., 2, either we can find a big K r,s subgraph involving a (k − 1)-flat (i.e. a flat that contains at least r points and belongs to at least s hyperplanes) or the rich k-flats degenerate to rich (k − 1)-flats.In this paper we obtain a stronger result by combining (3) with its dual incidence bound (4) to unwrap the point-hyperplane configuration from both directions.
Our approach can be used to obtain an improved lower bound in six and higher dimensions, but is not good enough to match the upper bound (2).To understand why that is the case, let us revisit the construction that attains this upper bound.We start with a (d − 2)-dimensional rectangular integer grid, which we will denote by G, with many rich 'hyperplanes' in the first d − 2 coordinates of R d ; note that these rich hyperplanes are of dimension d − 3. Extend the configuration to the (d − 1) st coordinate of R d so that the points of G become parallel lines L and the hyperplanes of G become parallel (d − 2)-dimensional flats F. Our point set is obtained by putting an equal number of points in each line in L and our hyperplane set is obtained by extending each flat in F to an equal number of hyperplanes in an arbitrary direction.
In this case, every hyperplane is 1-degenerate (since it has a (d − 2)-dimensional subflat which contains the same number of points), and so is every point.Therefore, our argument in the first two layers is not wasteful.However, the further layers are no longer β-degenerate for any fixed constant β ∈ (0, 1).As shown in [5], the second term of (3) -which dominates the estimate in dimensions at least 4 -is no longer tight when β is not fixed.One potential fix is to obtain an incidence bound with explicit dependence of β that is stronger than (6) when β = o(1).