Singer difference sets and the projective norm graph
aa r X i v : . [ m a t h . C O ] A ug Singer difference sets and the projective normgraph
Tamás Mészáros ∗∗†
Lajos Rónyai ‡ § Tibor Szabó ∗¶ August 16, 2019
Abstract
We demonstrate a close connection between the classic planar Singerdifference sets and certain norm equation systems arising from projectivenorm graphs. This, on the one hand leads to a novel description of planarSinger difference sets as a subset H of N , the group of elements of norm1 in the field extension F q / F q . H is given as the solution set of a simplepolynomial equation, and we obtain an explicit formula expressing eachnon-identity element of N as a product B · C − with B, C ∈ H . Thedescription and the definitions naturally carry over to the nonplanar andthe infinite setting. On the other hand, relying heavily on the differenceset properties, we also complete the proof that the projective norm graph
NG( q, does contain the complete bipartite graph K , for every primepower q ≥ . This complements the property, known for more than twodecades, that projective norm graphs do not contain K , (and henceprovide tight lower bounds for the Turán number ex( n, K , ) ). Let F be an arbitrary field, for t ≥ let K be a cyclic Galois extension of degree t − and let us denote by N the norm of this extension, i.e. for A ∈ K we have N( A ) = A · φ ( A ) · φ (2) ( A ) · · · φ ( t − ( A ) where the automorphism φ generatesthe Galois group of K / F and φ ( j ) denotes the j -fold iteration of φ . Then the projective norm-graph NG( F , K ) has vertex set K × F ∗ , where F ∗ denotes themultiplicative subgroup of F , and two vertices ( A, a ) and ( B, b ) are adjacent ifand only if N( A + B ) = ab . . ∗ Freie Universität Berlin † Position funded by the DRS Fellowship Program of Freie Universität Berlin ‡ Institute for Computer Science and Control, Hungarian Academy of Sciences and BME § Research supporten in part by NKFIH Grant No. K115288. ¶ Research supported in part by GIF grant No. G-1347-304.6/2016. For technical reasons we allow the two vertices to be the same, i.e. we allow loop edges. q = p k we use the standard notation F q for the finitefield with q elements. When F = F q , then K = F q t − and the automorphism φ above can be chosen to be the Frobenius automorphism φ ( X ) = X q and thegraph NG( F q , F q t − ) is denoted by NG( q, t ) . It is not difficult to show that NG( q, t ) has q t − ( q − q t − − ≈ n − t edges, where n = q t − ( q − isthe number of vertices. In [1] it was also proved that NG( q, t ) does not contain K t, ( t − as a subgraph, and hence has essentially the highest number of edgesamong graphs with this property, on the same number of vertices.Determining the largest number ex( n, H ) of edges a graph on n verticeswithout a subgraph isomorphic to H can have is one of the classic problemsof extremal graph theory, its history going back more than a century, to thetheorem of Mantel about triangle-free graphs. The value of ex( n, H ) is settledasymptotically when H is non-bipartite, but for bipartite graphs its order ofmagnitude is known only in a handful of cases. Even for the simplest bipartitegraphs, such as even cycles and complete bipartite graphs, the question is wideopen. The general upper bound of Kővári, Sós and Turán [18] states that ex( n, K t,s ) ≤ c s n − t for every t ≤ s . Matching lower bounds are known onlyfor t = 2 (Klein [10]), t = 3 (Brown [7]), and by the projective norm graphs forarbitrary t ≥ and s ≥ ( t − . The fundamental question of the order ofmagnitude of ex( n, K t,s ) is very much open for any ≤ t ≤ s ≤ ( t − .In general it is not known how large complete bipartite graphs the projectivenorm graph NG( q, t ) contains. For t ≥ it could even be the case that thereis an infinite sequence of prime powers q such that NG( q, t ) does not containa copy of K t,t and hence resolves the question of the order of magnitude of ex( n, K t,t ) for every t and s . With that, the determination of the largest integer s ( t ) , for which NG( q, t ) contains a K t,s ( t ) for every large enough prime power q , has potentially far reaching consequences. For t = 2 and t = 3 the projectivenorm graph NG( q, t ) does contain a K t, ( t − by combinatorial reasons (the KSTupper bound), so s (2) = 1 and s (3) = 2 . For t ≥ however, it was only knownthat t − ≤ s ( t ) ≤ ( t − .In this direction Grosu [14] has recently shown that NG( p, contains acopy of the complete bipartite graph K , for roughly -fraction of all primes p . In [2], among other things, this was extended for any prime power q if thecharacteristic p is not or . Furthermore, the proof also provided many copiesof K , . Computer calculations have also suggested that the same holds in thecase p ∈ { , } as well, but the arguments in [2] crucially used the restrictionon the characteristic. In this paper we provide different arguments to show theexistence of K , in N G ( q, for the cases when p and q ≥ ,and hence establish s (4) = 6 . In the process we uncover a close connectionbetween the norm equation systems arising from the projective norm graph andthe classic Singer difference sets. We consider this connection one of the maincontributions of our paper. In the next subsection we introduce the necessarybackground for the latter. 2 .2 Difference sets Given a multiplicative group G , a subset D ⊆ G is called a planar difference set if every non-identity element A ∈ G has a unique representationas a productof an element from D and an element from D − , where D − := { d − : d ∈ D} denotes the set of inverses of the elements of D . We refer to this representationas the mixed representation of A with respect to D .Difference sets in finite groups are central and diverse objects in designtheory with a rich history and numerous applications both inside and outsidemathematics. For a gentle introduction and survey the reader may consult e.g.[21]. If a finite group G admits a planar difference set of size m , then its order,by simple counting, must be of the form ℓ + ℓ + 1 , where ℓ = m − . Planardifference sets in Abelian groups are only known to exist if G is a cyclic group and ℓ is a prime power. In what follows we will simply write ’difference set’ insteadof planar difference set. The first construction was given by Singer [25], usingthe finite projective plane PG(2 , q ) . A collineation is a one-to-one mapping c onthe points of the plane carrying lines into lines. Singer proved that in PG(2 , q ) there is always a collineation c that cyclically permutes the q + q + 1 points.Then if we label the points P i , ≤ i ≤ q + q so that P i = c i ( P ) for every i ,then the indicies corresponding to points on the same line will form a differenceset of size q + 1 in the additive cyclic group Z q + q +1 . For the other directionhe remarks that such a difference set naturally induces a projective geometry oforder q . This strong connection motivates the name ’planar’ difference set.In general, for multiplicative groups G , G two difference sets D ⊂ G and D ⊂ G are called equivalent if there exists a group isomorphism ϕ : G → G and an element Γ ∈ G such that ϕ ( D ) = Γ · D . For example, in Abeliangroups any difference set D is equivalent to its inverse D − via the isomorphism Y → Y . In Singer’s construction from above, choosing different lines for thesame collineation also results in equivalent difference sets. Singer conjecturedthat every difference set in Z q + q +1 is equivalent to his construction. Thisconjecture is still very much open. Berman [3] and Halberstam and Laxton [15],verifying a related conjecture of Singer, determined the exact number of reduceddifference sets of Singer type in Z q + q +1 , where a(n additive) difference set iscalled reduced if it contains both and ∈ Z q + q +1 .We finish this subsection with an equivalent formulation of Singer’s construc-tion that will be useful later. We consider the cyclic group G = F ∗ q . F ∗ q . Theorder of this group is q + q + 1 and the cosets of those elements A ∈ F ∗ q forwhich the trace Tr( A ) = A + A q + A q = 0 form a difference set of size q + 1 which is equivalent to the Singer difference set (see e.g. [23]). Difference sets.
In order to treat infinite difference sets as well, we intro-duce our definitions and results for arbitrary fields, restricting to finite fieldsonly when necessary. This general approach also keeps the arguments more3ransparent.Let F be an arbitrary field and K a Galois extension of degree 3. Let φ bea nonidentity F -automorphism of K . Then the Galois group of the extension K over F is { φ, ψ, id } , where ψ = φ ◦ φ and, of course, id = φ ◦ φ ◦ φ . We denote by N K / F the norm of this extension: for A ∈ K we have N K / F ( A ) = A · φ ( A ) · ψ ( A ) ∈ F . When it causes no confusion, which will be the case most of the time, weomit writing the index K / F . Examples of such extensions are simplest cubicfields [24], which are important and well studied objects in algebraic numbertheory.We shall consider two functions h , h : K → K defined as h ( X ) = φ ( X ) · X + X + 1 and h ( X ) = φ ( X ) · X + φ ( X ) + 1 . For i = 1 , let H i denote the set of roots of h i in K and let N denote theset of elements in K ∗ with norm . It is easy to see that N is a subgroup ofthe multiplicative group K ∗ . In our first result we prove that H and H forma difference set in the cyclic group N with an explicit formula for the mixedrepresentation. Theorem 1.1.
The sets H and H are equivalent difference sets in the group N and H = H − . Furthermore, the unique mixed representation of an element A ∈ N \ { } (with respect to H i ) is given by the following explicit formulas: A = A · φ ( A ) − − φ ( A ) ∈ H and A = A − A · φ ( A ) A · φ ( A ) − ∈ H . (1)Infinite difference sets were earlier constructed by Hughes [16] using a greedy-like approach. Our construction is more explicit and so offers more possibilitiesto study these nice combinatorial structures.Next we spell out the statement of our theorem for finite fields. We shallshow that in this case our difference sets are of Singer type. Let F = F q , K = F q and take φ to be the Frobenius automorphism X → X q . Then h ( X ) and h ( X ) become polynomials over F q , namely h ( X ) = X q +1 + X + 1 and h ( X ) = X q +1 + X q + 1 , and the group N is the unique subgroup of order q + q + 1 in F ∗ q . Corollary 1.2.
The root set H ⊆ F q of the polynomial h ( X ) = X q +1 + X +1 ∈ F q [ X ] and the root set H ⊆ F q of the polynomial h ( X ) = X q +1 + X q +1 ∈ F q [ X ] are equivalent difference sets of size q + 1 in the order q + q + 1 cyclicgroup N of norm elements of F q with H = H − . Furthermore the uniquemixed representation of an element A ∈ N \ { } (with respect to H i ) is given bythe following explicit formulas: A = A q +1 − − A q ∈ H and A = A − A q +1 A q +1 − ∈ H . Moreover, both H and H are equivalent to the Singer difference set. Projective norm graphs.
As already mentioned earlier, one of the mainresults in [2] is that if p > then NG( q, contains K , as a subgraph. Relyingheavily on the properties of the Singer difference set from Corollary 1.2 andits connection to norm equation systems, here we settle the remaining cases ofcharacteristic and . Theorem 1.3.
Let q = p k > where p is a prime. Then NG( q, contains K , as a subgraph. In particular, NG(2 k , and NG(3 ℓ , contain K , as a subgraph for k ≥ and ℓ ≥ . Complementing Theorem 1.3 we remark that it is immediate that
NG(2 , does not contain K , , and in NG(3 , and NG(4 , we verified by computersearch that there is no K , either. H i We start by proving a few helpful properties of the sets H i , inlcuding theirunique mixed representation property from Theorem 1.1. Proposition 2.1. (i) For Y ∈ K ∗ we have h (cid:0) Y (cid:1) = h ( Y )Φ( Y ) · Y . In particular, H = H − . (ii) Every A ∈ N \ { } can be represented uniquely as a product A = A · A of an element A of H and an element A of H . This representation isgiven by A = A · φ ( A ) − − φ ( A ) and A = A − A · φ ( A ) A · φ ( A ) − . (2) (iii) H ∩ H = H ∩ F = H ∩ F = { Y ∈ F : Y + Y + 1 = 0 } , in particular, if F = F q , K = F q then H ∩ H = { } if q ≡ { α, α − } , where α = 1 , α = 1 if q ≡ ∅ if q ≡ . (iv) If Y ∈ K then φ ( h i ( Y )) = h i ( φ ( Y )) , in particular φ ( H i ) = H i . roof. (i) For Y ∈ K we have h (cid:18) Y (cid:19) = 1 Y · φ (cid:18) Y (cid:19) + φ (cid:18) Y (cid:19) + 1 = 1 Y · φ ( Y ) + 1 φ ( Y ) + 1 == 1 + Y + Y · φ ( Y ) Y · φ ( Y ) = h ( Y ) Y · φ ( Y ) . The rest of the statement follows by the definition of the H i ’s.(ii) First we show the uniqueness of the representation of the form (2). Supposethat A ∈ N \ { } and A = A · A with A ∈ H and A = A/A ∈ H . Then A · φ ( A ) + A + 1 = 0 ,AA · φ (cid:18) AA (cid:19) + φ (cid:18) AA (cid:19) + 1 = AA · φ ( A ) φ ( A ) + φ ( A ) φ ( A ) + 1 = 0 . By expressing φ ( A ) from the two equations we obtain A + 1 A = A · φ ( A ) A + φ ( A ) . Solving for A gives that the only possibility is A = A · φ ( A ) − − φ ( A ) . This gives uniqueness and the stated formula (1) for A = A/A as well. Itremains to verify that A i ∈ H i . We consider first A . Using N( A ) = A · φ ( A ) · ψ ( A ) = 1 we have h ( A ) = A · φ ( A ) − − φ ( A ) · φ (cid:18) A · φ ( A ) − − φ ( A ) (cid:19) + A · φ ( A ) − − φ ( A ) + 1 == A · φ ( A ) − − φ ( A ) · φ ( A ) · ψ ( A ) − − ψ ( A ) + φ ( A )( A − − φ ( A )= ψ ( A ) − − φ ( A ) · A − − ψ ( A ) + A · ψ ( A ) ( A − − φ ( A ) == 1 − AA · ψ ( A ) · (1 − φ ( A )) + A − A · ψ ( A ) · (1 − φ ( A )) = 0 , and hence A ∈ H . The verification of A ∈ H is a similar calculation that weleave to the reader.(iii) To see the desired equalties of sets note that for an element Y ∈ K we have h ( Y ) = h ( Y ) if and only Y = φ ( Y ) , which happens if and only if Y ∈ F .If furthermore F = F q , then for | q we have X + X + 1 = ( X − , whileotherwise the non-trivial third-roots of unity are in F if and only if | q − .(iv) The statement is a direct consequence of the fact that φ is an automorphismthat fixes F . 6 .2 Difference sets and a norm equation system After Proposition 2.1(ii), all we need in order to complete the proof of Theo-rem 1.1 is to show that H , H ⊆ N . In our next result we are not only provingthat the elements of H ∪ H indeed have norm , but we are also able tocharacterize them as the solutions in K of the norm equation system N( X ) = 1 , N( X + 1) = − , (3)which is closely related to projective norm graphs. Besides this connection, thesystem (3) also arises naturally in algebraic number theory. When K is a numberfield, then the algebraic integer solutions of (3) will be exceptional units in thesense of Nagell [22]. Exceptional units are interesting objects, in particular onecan show that their number is always finite. Proposition 2.2. Y ∈ K is a solution of (3) if and only if Y ∈ H ∪ H .Proof. The proof is adapted from [2] which handled the finite field setting. Wefirst write the equations of (3) in simple product form. X · φ ( X ) · ψ ( X ) = 1( X + 1) · φ ( X + 1) · ψ ( X + 1) = ( X + 1) · ( φ ( X ) + 1) · ( ψ ( X ) + 1) = − Expressing ψ ( X ) from both equations and clearing denominators gives ( X + 1) · ( φ ( X ) + 1) = − X · φ ( X ) − X · φ ( X ) · ( X + 1) · ( φ ( X ) + 1) . After expanding and simplifying we obtain X · ( φ ( X )) + X · ( φ ( X )) + X · φ ( X ) + 3 X · φ ( X ) + φ ( X ) + X + 1 = 0 . (4)It is easy to check that the left hand side of (4) is h ( X ) · h ( X ) = ( φ ( X ) · X + X + 1) · ( X · φ ( X ) + φ ( X ) + 1) . Clearly, by the above calculations, any solution Y of (3) will be a solutionof (4), hence h ( Y ) · h ( Y ) = 0 , which in turn is equivalent to Y ∈ H ∪ H .Conversely, let Y ∈ H ∪ H i.e. h ( Y ) · h ( Y ) = 0 . Then either h ( Y ) = 0 , in which case φ ( Y ) = − Y + 1 Y =: u ( Y ) , or h ( Y ) = 0 , in which case φ ( Y ) = − Y + 1 =: v ( Y ) . Note that the denominator is non-zero in both cases. Using part (iv) of Propo-sition 2.1, in the first case we obtain ψ ( Y ) = φ ◦ φ ( Y ) = u ( u ( Y )) = − − Y +1 Y + 1 − Y +1 Y = v ( Y ) , ψ ( Y ) = φ ◦ φ ( Y ) = v ( v ( Y )) = − − Y +1 + 1 = u ( Y ) . In both cases we have { φ ( Y ) , ψ ( Y ) } = { u ( Y ) , v ( Y ) } and therefore N( Y ) = Y · φ ( Y ) · ψ ( Y ) = Y · u ( Y ) · v ( Y ) = Y · (cid:18) − Y + 1 (cid:19) · (cid:18) − Y + 1 Y (cid:19) = 1 . Similarly, for the norm of Y + 1 we obtain N( Y + 1) = ( Y + 1) · φ ( Y + 1) · ψ ( Y + 1) = ( Y + 1) · ( φ ( Y ) + 1) · ( ψ ( Y ) + 1)= ( Y + 1) · ( u ( Y ) + 1) · ( v ( Y ) + 1)= ( Y + 1) · (cid:18) − Y + 1 Y + 1 (cid:19) · (cid:18) − Y + 1 + 1 (cid:19) = − . This finishes the proof.
Proof of Theorem 1.1.
Proposition 2.2 shows that N( X ) = 1 for every X ∈H ∪ H , hence H , H ⊆ N . The equality H = H − is just part (i) ofProposition 2.1, while the unique mixed representation of an element A ∈ N \{ } is provided by part (ii) of Proposition 2.1. Proof of Corollary 1.2.
Most of the statements follow directly by applying The-orem 1.1 to the setting F = F q , K = F q , and φ ( X ) = X q . Note that then N is an order q + q + 1 multiplicative group, hence any difference set in it, inparticular H and H , must have size q + 1 . In particular, this implies that both h and h have all their q + 1 roots in N ⊆ F q .It remains to prove the equivalence of H (and so of H ) to the Singerdifference set S , for which we use the description mentioned at the end of Sub-section 1.2, i.e. S = { A · F ∗ q : A ∈ F ∗ q , Tr( A ) = 0 } ⊆ F ∗ q . F ∗ q . Consider the group homomorphism
Φ : F ∗ q → F ∗ q , given by Φ( A ) = A q − .Since N ( φ ( A )) = A q − = 1 , the image of Φ is contained in N . The kernel is F ∗ and |N | = q + q + 1 = q − q − , so the quotient map ¯Φ provides an isomorphismbetween the groups F ∗ q . F ∗ q and N .We claim that ¯Φ maps S into H , i.e. if A ∈ F ∗ q is such that Tr( A ) = A q + A q + A = 0 , then A q − ∈ H . Indeed, h ( A q − ) = A ( q − q +1) + A q − + 1 = A q A + A q − + 1 == ( − A q − A ) A + A q − + 1 = − A q − − A q − + 1 = 0 . Using the injectivity of ¯Φ and |S| = |H | , we obtain that ¯Φ( S ) = H , showingthat ¯Φ gives the desired equivalence of the difference sets S and H .8 .3 Norm equation systems with the maximum numberof solutions Let F be an arbitrary field and K a Galois extension of degree . Finding a K , in NG( F , K ) , up to a couple of technicalities to be handled later, is essentiallyequivalent to finding distinct pairs ( A , a ) , ( A , a ) , ( A , a ) ∈ K × F ∗ such thatthe system N( X + A ) = a , NG( X + A ) = a , N( X + A ) = a , (5)has six solutions X ∈ K . Note that by the K , -freeness of the projective normgraph NG( F , K ) (see Subsection 5.1 in the Appendix), we also know that (5)can have at most six solutions for any values of the parameters.In what follows we will study this system for special sets of parameters andtry to find particular choices where the maximum possible six solutions areattained. The main result of this subsection is that in some cases this is indeedpossible. For a fixed element A ∈ N \ { } consider the following system of normequations: N( X ) = 1 , N( X + 1) = − , N( X + A ) = − . (6) Theorem 2.3.
Let q = p k > , where p is prime, and let F = F q , K = F q . Then there exists an element A ∈ N \ { } such that the system (6)has 6 solutions in F q . The starting point in the proof of Theorem 2.3 is an observation, valid overarbitary fields, that connects the solution set of the system (6) to the solutionset of (3) and hence to the difference sets H , H . Proposition 2.4.
An element Y ∈ K is a solution of (6) with parameter A ∈N \ { } if and only if Y and AY are both contained in H ∪ H .Proof. By Proposition 2.2 Y and AY are both contained in H ∪ H if and onlythey are both solutions of (3). We show that this is equivalent to Y being asolution of (6).Suppose first that Y ∈ K is a solution of (6). Then a fortiori Y is a solutionof (3) and Y = 0 . Also, N (cid:18) AY (cid:19) = N( A )N( Y ) = 11 = 1 , N (cid:18) AY + 1 (cid:19) = N (cid:18) A + YY (cid:19) = N( A + Y )N( Y ) = −
11 = − , hence AY is also a solution of (3).Conversely, assume that Y and AY are both solutions of (3). Then, in par-ticular, Y satisfies the first two equations from (6), as for the third one wehave N( Y + A ) = N (cid:18) Y (cid:18) AY (cid:19)(cid:19) = N( Y ) N (cid:18) AY + 1 (cid:19) = 1 · − − , Y is a solution of (6).By the previous propostion we will be looking for product representations Y · AY of an element A from the set H ∪ H . To prove Theorem 2.3 we actuallyneed to find an element A ∈ N \ { } that has three such product representations A = B C = B C = B C , such that the six elements B , C , B , C , B , C ∈H ∪ H are all distinct. For this we will crucially use that H and H = H − are difference sets in N and inverses of each other.Recall that if D is any difference set in some multiplicative group G then everyelement A ∈ G \ { } has a unique representation, called mixed representation, asa product B · C = A such that one of B and C is from D and the other is from D − . In the next propositions we summarize our knowledge about other productrepresentations. To this end we will call a product B · C = A a D -representation of the element A ∈ G if both B and C are from D . Proposition 2.5.
Let D be an arbitrary difference set in some multiplicativegroup G . Then every A ∈ G has at most one D -representation.Proof. Let us assume that D D = D D for some D , D , D , D ∈ D . Then D D − = D D − , and this, by the difference set property, is either or wehave D = D and D = D : in any case { D , D } = { D , D } .The explicit descriptions of our difference sets allow us to characterize whenan H i -representation with distinct factors exists. Proposition 2.6. (i) If char ( F ) = 2 then A ∈ N has an H -representation with different factorsif and only if ( A + 1 − Aφ ( A )) − A is a non-zero square in K .(ii) If char ( F ) = 2 then A ∈ N has an H -representation with different factorsif and only if ( Aφ ( A ) + φ ( A ) − − Aφ ( A ) is a non-zero square in K .Proof. In the case of finite fields most parts of the argument have already ap-peared in [2]. As the proof of the two parts is analogous, below we only presentthe one of (i).Suppose first that A ∈ N has an H -representation. This means that thereis an element Y ∈ K such that Y and AY are both roots of h , i.e. h ( Y ) = φ ( Y ) · Y + Y + 1 = 0 and h (cid:18) AY (cid:19) = φ (cid:18) AY (cid:19) · AY + AY + 1 = 0 . after expressing φ ( Y ) from both equations, letting them being equal and clearingdenominators we obtain Y + ( A + 1 − aφ ( A )) · Y + A = 0 . Clearly, the roleof Y and AY can be switched, which means that both Y and AY are roots of thequadratic equation X + ( A + 1 − Aφ ( A )) · X + A = 0 . (7)10f char( F ) = 2 , this is possible only if the discriminant D = ( A + 1 − Aφ ( A )) − A is a nonzero square in K .For the other direction suppose that D = ( A + 1 − Aφ ( A )) − A is a nonzerosquare in K , i.e. there is some element G ∈ K ∗ such that D = G . Thenwe know that the quadratic equation in (7) has two different roots, namely X ± = Aφ ( A ) − A − ± G . Clearly, X + · X − = A , so to finish the proof it is enoughto show that X ± ∈ H , i.e. h ( X ± ) = 0 .Using N ( A ) = Aφ ( A ) ψ ( A ) = 1 we have φ ( D ) = φ (cid:16) ( A + 1 − Aφ ( A )) − A (cid:17) = ( φ ( A ) + 1 − φ ( A ) ψ ( A )) − φ ( A ) == (cid:18) φ ( A ) + 1 − A (cid:19) − φ ( A ) = 1 A (cid:16) ( Aφ ( A ) + A − − A φ ( A ) (cid:17) == 1 A (cid:16) ( A + 1 − Aφ ( A )) − A (cid:17) = 1 A D. Then (cid:16) φ ( G ) G (cid:17) = φ ( G ) G = φ ( D ) D = A DD = A and hence φ ( G ) G = ± A . However N (cid:16) φ ( G ) G (cid:17) = φ ( G ) G · φ (cid:16) φ ( G ) G (cid:17) · ψ (cid:16) φ ( G ) G (cid:17) = φ ( G ) G · ψ ( G ) φ ( G ) · Gψ ( G ) = 1 which, aschar( F ) = 2 , excludes φ ( G ) G = − A . Therefore φ ( G ) = A G . As a consequancewe get φ ( X ± ) = φ (cid:18) Aφ ( A ) − A − ± G (cid:19) = φ ( A ) ψ ( A ) − φ ( A ) − ± φ ( G )2 == A − φ ( A ) − ± A G A X ± + 1 − Aφ ( A ) A .
Now we are ready to substitute X ± into h . h ( X ± ) = X ± · φ ( X ± ) + X ± + 1 = X ± · (cid:18) A X ± + 1 − Aφ ( A ) A (cid:19) + X ± + 1 == 1 A (cid:0) X ± + ( A + 1 − Aφ ( A )) · X ± + A (cid:1) = 0 , where at the last equality we just used that X ± are the roots of (7).For the elements of H i the existence of a H − i -representation follows directly. Proposition 2.7. If A ∈ H i then its unique H − i -representation is given by φ ( A ) · ψ ( A ) = A .Proof. On the one hand, as A ∈ H i ⊆ N , we have that A = N ( A ) φ ( A ) · ψ ( A ) = φ ( A ) · ψ ( A ) . On the other hand, by Proposition 2.1(i) we have A ∈ H − i , and so byProposition 2.1(iv) we have φ ( A ) = φ (cid:0) A (cid:1) ∈ H − i and ψ ( A ) = φ (cid:16) φ ( A ) (cid:17) ∈ H − i .The uniqueness follows from Proposition 2.5.11rom now on we will consider the special case F = F q (and K = F q ), andseparate the cases according to the characteristic modulo . In this subsection we settle the case of characteristic which was left open in [2].Our method extends to odd characteristic p ≡ , which was settled, ina much stronger form, already in [2].First we show the existence of six solutions when q is congruent to modulo (as opposed to p ). Proposition 2.8.
Let q > be a prime power such that q ≡ , and let F = F q and K = F q . Then there exists an element A ∈ N \ { } such that thesystem (6) has 6 solutions in F q .Proof. We will find an element A ∈ N \ { } that has an H i -representation A = B i · C i for both i = 1 and , such that B i = C i . These four elements,together with the two elements A ∈ H and A ∈ H from the unique mixedrepresentation of A (which exists by Proposition 2.1(ii)) give us six distinctsolutions of (6). Indeed, they are solutions of (6) by Proposition 2.4 and theirdistinctness follows immediately from the fact that the sets H and H aredisjoint by Proposition 2.1(iii).In order to find the appropriate element A , for a set H ⊆ N we define H ∗ := { B · C : B, C ∈ H , B = C } to be the set of its pairwise products from distinct factors, and show that H ∗ ∩H ∗ is not empty.First note that by Proposition 2.6(ii) the pairwise products of the elementsof H i are all distinct, hence the cardinality of H ∗ i is (cid:0) | H i | (cid:1) = q + q . Both H ∗ and H ∗ are subsets of the ( q + q ) -element set N \ { } . For this note that H i iscontained in N , which is closed under multiplication, and that
6∈ H ∗ i since H i is disjoint from its inverse H − i by our assumption on q and Proposition 2.1(iii).Hence the only way H ∗ and H ∗ could be disjoint is if their union is N \ { } . Inthis case however it would also hold that σ ( N \ { } ) = σ ( H ∗ ) + σ ( H ∗ ) , (8)where σ ( S ) denotes the sum of the elements of a subset S ⊆ F q . On the onehand the set N is the collection of all q + q + 1 roots in F q of the polyno-mial X q + q +1 − and so σ ( N ) is ( − times the coefficient of X q + q in thispolynomial, which is . From this we obtain that σ ( N \ { } ) = − . On theother hand H i is the set of all q + 1 roots in F q of the polynomial h i ( X ) , hence σ ( H ∗ i ) is the coefficient of X q − in h i ( X ) , which is 0 for q > . We arrived toa contradiction, as the left hand side of (8) is (-1), while the right hand side is . 12 roof of Theorem 2.3 for p ≡ . First note that F q is a cubic exten-sion of F q and the norm of an element B ∈ F q ⊆ F q is the same, irrespectivein which of the two fields we compute it: N F q / F q ( B ) = N F q / F q ( B ) . Thismeans that if for an element A ∈ F q with N F q / F q ( A ) = 1 the system (6) withthe norm function N F q /F q has six distinct solutions X , . . . , X ∈ F q , then thevery same six elements are also solutions of the the system (6) with the normfunction N F q /F q .Now let p ≡ be a prime and let q = p k = p ℓ m be an arbitrarypower where ℓ is odd. Then Proposition 2.8 gives the required six distinctsolutions when m = 0 and p ℓ > . By repeated application of the aboveobservation, the statement also follows for any positive integer m and p ℓ > .These include all the powers when p > .When p = 2 then only those powers are included where ℓ ≥ . So we are leftwith prime powers of the form m . To settle these last cases one first resolvesthe problem when the prime power is = 16 and then uses the above squaringtrick to deduce the case of arbitrary m = 16 m − .For q = 16 we have found the appropriate A ∈ F of norm , for whichthe system (6) has six distinct solution with the aid of a computer. To describethis example, let U be the primitive element of F ∗ whose minimal polynomialover F is X + X + X + X + X + X + 1 , and consider the system N( X ) = 1 , N( X + 1) = 1 , N( X + U ) = 1 . By Magma Calculator [6] it is easily verified that A = U is in N \ { } , andthat the system has indeed six solutions, namely U , U , U ∈ H and U , U , U ∈ H with A = U · U = U · U = U · U . Proof of Theorem 2.3 for p = 3 . Just like in the previous subsection, we willfind an element A ∈ N \{ } which has an H i -representation A = B i · C i for both i = 1 and , such that these four elements and the two elements A ∈ H and A ∈ H from the mixed representation of A (which exists by Proposition 2.1(ii))are pairwise distinct and hence provide six distinct solutions of (6).We do this in two steps. First we find an element that has both H - and H -representation, but in one of them the factors are not distinct. Lemma 2.9.
For i = 1 or there is an element C ∈ H i \ { } , such that C has an H − i -representation C = B · E with distinct factors B = E . Let us fix elements C ∈ H i \{ } and B, E ∈ H − i guaranteed by Lemma 2.9.We show that the element A := CE is the kind we are looking for. First observe13hat by Proposition 2.1(i) A = C · E = B · C provide a H i - and H − i -representation of A , respectively. Note furthermorethat as ( H i \ { } ) ∩ H − i = ∅ , we have C = E and hence A = 1 . Consequently,by Proposition 2.1(ii) there exists a unique mixed representation A = A · A with A i ∈ H i .Next we show that these six elements from the representations are all dis-tinct. Lemma 2.10.
The elements A i , C, E ∈ H i and A − i , C , B ∈ H − i are alldistinct. To finish the proof of Theorem 2.3 note that by Proposition 2.4 these ele-ments provide six distinct solutions of (6).We finish this subsection proving the two lemmas from the above proof.
Proof of Lemma 2.10.
For the distinctness first we establish that none of thesix elements is . This is certainly true for C and C by the choice of C inLemma 2.9. Now assume that B or E is , say B = 1 (the argument in thecase E = 1 is analogous). On the one hand, as E ∈ H − i , by Proposition 2.7 E has a unique H i -representation: E = E q · E q . On the other hand E = B · E = C · C is also a H i -representation of E , so by the uniqueness we musthave E q = C = E q = φ ( E q ) , and thus E q ∈ F q and E q = C ∈ H − i . Thatmeans φ ( E ) = E q ∈ F q ∩ H − i = H ∩ H = { } by Proposition 2.1(iii), whichis only possible if E = 1 . This contradicts C = 1 and implies B, E = 1 . Finallyassume that A or A is equal to , say A = 1 . Then A · A = A = B · C aretwo H -representations of A . By uniqueness either B or C should be , whichis a contradiction by the above.Since none of the six elements is and by part (iii) of Proposition 2.1 H ∩H = { } , we established that { A i , C, E } ∩ { A − i , B, C } = ∅ . We are left to show that (cid:12)(cid:12) { A i , C, E } (cid:12)(cid:12) = 3 and (cid:12)(cid:12) { A − i , B, C (cid:12)(cid:12) = 3 . Since theyare proved analogously we present just the first one.If A i = C , then A − i = E , which is a contradiction as A − i ∈ H − i and E ∈H i and none of them is . If A i = E , then A − i = C , which is a contradictionsimilarly as A − i ∈ H − i and C ∈ H i and none of them is . Finally, supposethat C = E . Then we have B = C · E = C , which is a contradiction as B ∈ H − i \ { } and C ∈ H i \ { } because the polynomial h i is defined over F ,hence if h i ( C ) = 0 , then h i ( C ) = 0 as well. Proof of Lemma 2.9.
We want to find a C ∈ H i \ { } for i = 1 or , such that C has an H − i -representation with different elements B and E . This happens14xactly if one of the formulas in Proposition 2.6 is a nonzero square in F q when we substitute A = C . It turns out that after simplifying the substitutedformula of (i) using h ( C ) = C q +1 + C q + 1 = 0 and clearing its square de-nominator we obtain the very same expression D ( C ) as after simplifying thesubstituted formula of (ii) using h ( C ) = C q +1 + C + 1 = 0 and clearing itssquare denominator: D = D ( C ) = (cid:0) C ( C + 1) + ( C + 1) − C (cid:1) − C + 1) C == (cid:0) C + 3 C + 1 (cid:1) · (cid:0) C + C + 1 (cid:1) · (cid:0) C + C − (cid:1) · (cid:0) C − C − (cid:1) . We aim to find an element C ∈ H ∪ H \ { } for which D is a square in F q ,or equivalently, N ( D ) is a square in F q .As it turns out the factors of N ( D ) can be conveniently expressed using thetrace T r ( C ) = C + C q + C q =: t of C : N ( C + 3 C + 1) = − t − t − ,N ( C + C + 1) = t + 3 t + 9 ,N ( C + C −
1) = t + 3 t + 1 ,N ( C − C −
1) = − t − t − , and so N ( D ) = ( − t − t − ( t + 3 t + 9)( t + 3 t + 1) . In characteristic this expression is a square if and only if t + 1 is a square.Using Theorem 5.18 from [20] we get that X y ∈ F q η ( y + 1) = − η (1) = − , where η is the quadratic character of F q . Therefore, as y + 1 = 0 has at mosttwo solutions, for at least q − elements y ∈ F q the expression y + 1 is asquare.This ensures the existence of many good “traces”, which we can use to con-struct many good C , as we now show that the trace function is a -to- functionon H ∪ H \ { } . That is, if T r ( C ) = T r ( C ) for C , C ∈ H ∪ H \ { } then C and C are conjugates of each other. For this we note that the minimalpolynomial of an element C of H ∪ H \ { } can be expressed just by the trace t of C : m t ( X ) = ( X − C )( X − C q )( X − C q ) = X − tX − ( t + 3) X − . Hence if C and C have the same trace, then they have the same minimalpolynomial.Consequently there are exactly | H ∪ H \{ }| = q elements in F ∗ q that aretraces of some element in H ∪ H \ { } .15n conclusion, there are at least (cid:0) q + q − (cid:1) − q = q − > elements t ∈ F q which are traces of an element C from H ∪ H \ { } , and for which t + 1 , andhence also D = D ( C ) , is a square. This completes the proof for q > .Otherwise, by our assumption on q , we are left with the case q = 9 . Then wecan directly find a t ∈ F such that it is the trace of an element C ∈ H ∪H \{ } and t + 1 is a nonzero square in F . In fact t = − will do. First note that t + 1 = 1 + 1 = 2 is in F and hence is a square in its quadratic extension F .Now, to finish the argument it is enough to show that m − ( X ) = X + X + X − is the minimal polynomial of some C ∈ H ∪ H \ { } over F , because then weautomatically have Tr( C ) = t = − . It is immediate that m − ( X ) is irreducibleover F , hence it can not have a root in F , and therefore it is irreducible over F . Next consider the polynomial h ( X ) for q = 3 . Then (when computed over F ) we have h ( X ) = X + X + 1 = ( X − X + X + X −
1) = ( X − m − ( X ) , which in view of Proposition 2.2 means that the roots of m − ( X ) solve thesystem (3) with norm function N F / F . But then, as we have seen earlier, theroots of m − ( X ) also solve the system (3) with norm function N F / F , andhence, again by Proposition 2.2 and the fact m − (1) = 0 , we have that any rootof m − is in H ∪ H \ { } , as desired. K , subgraphs in NG( q, In this subsection we prove Theorem 1.3. In the first lemma we connect solutionsof a norm equation systems with three equations to four adjacencies in theprojective norm-graph. For this let again F be an arbitrary field and K a Galoisextension of degree . Lemma 2.11.
Suppose that for A , A , A ∈ K ∗ and a , a , a ∈ F ∗ , the element Y ∈ K ∗ is a solution of (5). Then in the projective norm graph NG( F , K ) thevertex (cid:16) Y , Y ) (cid:17) is adjacent to all the vertices (cid:18) A , a N( A ) (cid:19) , (cid:18) A , a N( A ) (cid:19) , (cid:18) A , a N( A ) (cid:19) , (0 , . Proof.
We check the adjacenies from the statement. For the first three vertices,using N( Y + A i ) = a i , we have N (cid:18) A i + 1 Y (cid:19) = N (cid:18) Y + A i A i Y (cid:19) = N( Y + A i )N( A i Y ) = a i N( A i ) · Y ) . For the last vertex we have N (cid:18) Y (cid:19) = N (cid:18) Y (cid:19) = 1 · Y ) , as requested. 16he statement of Theorem 1.3 now follows easily. Proof of Theorem 1.3.
By Theorem 2.3 there exists an element A ∈ N \ { } such that the system (6) admits solutions in F q . We choose an element C = 0 , − , − A from F q such that N( C ) = 1 . This is clearly possible since q − − ( q + q + 1) > for q > . Let A = C , A = C + 1 , A = C + A , a = 1 , a = − , and a = − . The corresponding norm equation system of theform (5), as the translation of (6) by C , also has six distinct solutions, whichwe denote by Y , . . . , Y . Since by the choice of C the elements A , A , and A ,as well as the solutions Y , . . . , Y are non-zero, Lemma 2.11 is applicable. Asthe elements A , A , A are also distinct, so are the vertices (cid:18) A , a N ( A ) (cid:19) , (cid:18) A , a N ( A ) (cid:19) , (cid:18) A , a N ( A ) (cid:19) , (0 , . which we denote by ( B i , b i ) , i = 1 , , , , . By Lemma 2.11 each of them isadjacent to the six distinct vertices (cid:16) Y j , N ( Y j ) (cid:17) , j = 1 , . . . , .In order for these adjacencies to indeed give rise to a K , in N G ( q, ,we need to make sure that none of them represents a loop.If p = 2 , then NG( q, has no loop edges, so all the participating vertices are different, and they form a K , .Assume now that p > . Let us call an element D ∈ F q bad for the pairof indicies ( i, j ) , if B i + D = Y j − D . For every fixed pair ( i, j ) there existsexactly one D which is bad for ( i, j ) , namely D = (cid:16) Y j − B i (cid:17) . Therefore,together there are at most elements D which are bad for some pair ( i, j ) . As q ≥ > , there exists an element D ∈ F q which is not bad for any pair.For this D , the ten vertices ( B i + D, b i ) , i = 1 , , , , and (cid:16) Y j − D, N ( Y j ) (cid:17) , j = 1 , . . . , , are all distinct and hence give rise to a K , . In general an ( n, m, λ ) difference set is a set D of m elements in a multiplicativegroup G of order n , such that any element g ∈ G has exactly λ mixed productrepresentations with respect to D . Singer’s construction naturally generalizesto the case with parameters ( n, m, λ ) = ( q t − q − , q t − − q − , q t − − q − ) for any t ≥ ,which are called Singer parameters. Unlike for t = 3 , the planar case, for manyvalues t > difference sets having Singer parameters yet being inequivalent toSinger’s construction are known to exist (see e.g. [13]).In the case of planar difference sets we made use of a convenient description ofthe Singer difference set inside the multiplicative group F ∗ q . F ∗ q as the collectionof cosets of elements of trace . In what follows, we extend this description tothe general setting of degree t cyclic Galois extensions over arbitrary fields F ,that we encountered already in Section 1 in connection with projective normgraphs. To this end, let F be an arbitrary field and K a cyclic Galois extension17f degree t with t ≥ . Let Tr K / F be the trace function corresponding to thisextension, i.e. for A ∈ K we have Tr K / F ( A ) = A + φ ( A ) + φ (2) ( A ) + · · · + φ ( t − ( A ) , where φ ( i ) denotes the i -fold iteration of some generator φ of the Galois group.Now consider the subset S t = { Y | Y ∈ K ∗ , Tr K / F ( Y ) = 0 } of the multiplicative group K ∗ / F ∗ , where Y denotes the image of Y ∈ K ∗ underthe natural map K ∗ → K ∗ / F ∗ .It is known [23] that when F = F q is a finite field then S t is a Singer differenceset with parameters ( q t − q − , q t − − q − , q t − − q − ) . This can be extended for arbitrary F , using the natural ( t − -dimensional projective space structure on K ∗ / F ∗ (which is induced by the t -dimensional F -vector space structure of K ). It turnsout that for any non-identity element A , the set of S t -elements from the mixedrepresentations of A with respect to S t forms a subspace of projective dimension t − . Proposition 3.1.
Let L be a subspace of K ∗ / F ∗ of projective dimension t − .Then for every element A ∈ K ∗ / F ∗ \ (cid:8) (cid:9) the set RL (cid:0) A (cid:1) = { B ∈ L | ∃ C ∈ L such that A = BC − } = { B ∈ L | B/A ∈ L} forms a subspace of projective dimension t − .Proof. Let L be a ( t − -dimensional subspace of K over F such that L = L \ { } / F ∗ . Then for any element A ∈ K ∗ / F ∗ \ (cid:8) (cid:9) we have RL (cid:0) A (cid:1) = RL ( A ) \ { } / F ∗ , where A ∈ K ∗ \ F ∗ is such that A = A · F ∗ and RL ( A ) = { B ∈ L | ∃ C ∈ L such that A = BC − } . Observe that RL ( A ) = L ∩ A L , where A L = { AL | L ∈ L} .Since A F ∗ , the ( t − -dimensional subspaces L and A L are different,hence their intersection has dimension t − . Therefore the projective dimensionof RL (cid:0) A (cid:1) is indeed t − . Corollary 3.2.
Let L be a subspace of F ∗ q t / F ∗ of projective dimension t − .Then L is a difference set in the multiplicative group F ∗ q t / F ∗ with parameters ( q t − q − , q t − − q − , q t − − q − ) . In particular, so is S t .Proof. The set RL (cid:0) A (cid:1) is in a one-to-one correspondence with the collection ofmixed L -respresentations of A . Since a projective space of dimension t − i over F q has size q t − i +1 − q − , the parameters of the difference set follow.18inally note that as Tr K / F : K → F is a non-trivial F -linear function,Ker (Tr K / F ) is a ( t − -dimensional subspace of K . Hence the set S t = Ker (Tr K / F ) \ { } / F ∗ is a subspace of K ∗ / F ∗ of projective dimension t − ,and as such is a difference set with Singer parameters.In Corollary 1.2 we gave an alternative description of planar difference setsof Singer type over finite fields as the root set of a simple polynomial and gaveexplicit formulas of the product representation of each element. Here we extendthis result to the general setting. For this let N K / F denote the norm function ofthe extension K / F and N the group of elemens of norm in K . Furthermore,we shall consider the function d t ( X ) = 1 + X + Xφ ( X ) + Xφ ( X ) φ (2) ( X ) + · · · + Xφ ( X ) · · · φ ( t − ( X ) . We remark that the function d t ( X ) appears in a paper of Foster [12] in theformulation of the Murphy condition.In the next theorem we show that the set D t = { A ∈ K | d t ( A ) = 0 } of roots of d t ( Y ) in K is contained in the multiplicative group N and has thesame difference set property that the mixed representation of any element A ∈N \ { } with respect to D t form (in some sense) a projective space of dimension t − over F . In addition we will also be able to describe concisely these productrepresentations. Theorem 3.3.
There is a group isomorphism
Φ : K ∗ / F ∗ → N such that Φ( S t ) = D t . In particular, through Φ , the set D t inherits the difference setproperty of S t just like the projective space structure. Moreover, given an ele-ment A ∈ N \ { } the different mixed representations of A with respect to D t are exactly the products B · (cid:0) BA (cid:1) − , where B is a root in K of the function f t,A ( X ) = d t ( X ) − φ ( t − ( A ) · d t (cid:18) XA (cid:19) . Proof.
Consider the K ∗ → K ∗ map Φ defined by X → φ ( X ) X . On the one hand,one readily sees that the map Φ maps K ∗ into N . On the other hand, by Hilbert’sTheorem 90 [19] we know that for every A ∈ N there is an element Y ∈ K ∗ such that A = Φ( Y ) , which in turn shows that Φ is surjective. Therefore, asKer (Φ) = F ∗ , the quotient map Φ : K ∗ / F ∗ → N provides an isomorphismbetween the respective groups.Next we show that the image of S t under the map Φ is D t . For this let Y ∈ K ∗ . Then, on the one hand, we have d t (Φ( Y )) = d t (cid:18) φ ( Y ) Y (cid:19) = 1 + t − X j =0 j Y i =0 φ ( i +1) ( Y ) φ ( i ) ( Y ) == 1 + t − X j =0 φ ( j +1) ( Y ) Y = 1 Y Tr K / F ( Y ) . Y ∈ S t , then Φ (cid:0) Y (cid:1) ∈ D t . Finally, let A ∈ D t , i.e. A is a root of d t . Then N K / F ( A ) = 1 + A · φ ( d t ( A )) − d t ( A ) = 1 , and hence, again by Hilbert’sTheorem 90, there is an element Y ∈ K ∗ such that A = Φ( Y ) and so A = Φ (cid:0) Y (cid:1) .By the above calculations Tr K / F ( Y ) = Y · d t (Φ( Y )) = Y · d t ( A ) = 0 , meaningthat Y ∈ S t . This concludes the proof of Φ( S t ) = D t .Now let us turn to the second part of the theorem. Given an element A ∈N \ { A } first take a mixed representations A = B · C − with B, C ∈ D t . Then,in particular, we have C = BA ∈ D t and hence d t ( B ) = d t (cid:0) BA (cid:1) = 0 . Therefore,we have that B is also a root of the function f t,A ( X ) = d t ( X ) − φ ( t − ( A ) · d t (cid:18) XA (cid:19) . For the other direction suppose that B is a root of f t,A . Note that then nec-essarily B = 0 , as otherwise we would have f t,A (0) = 1 − φ ( t − ( A ) , whichis a contradiction as for A ∈ N \ { } we have − φ ( t − ( A ) = 0 . To finish theproof we need to show that B, BA ∈ D t , as then the product B · (cid:0) BA (cid:1) − is a validproduct representation of A with respect to D t . Using that N K / F ( A ) = 1 and φ ( t ) ≡ id, we have φ (0) = φ ( f t,A ( B )) = φ ( d t ( B )) − φ ( t ) ( A ) · φ (cid:18) d t (cid:18) BA (cid:19)(cid:19) = 1 B · (cid:0) d t ( B ) + N K / F ( B ) − (cid:1) − A · BA · (cid:18) d t (cid:18) BA (cid:19) + N K / F (cid:18) BA (cid:19) − (cid:19) = 1 B (cid:18) d t ( B ) − d t (cid:18) BA (cid:19)(cid:19) = ⇒ d t ( B ) = d t (cid:18) BA (cid:19) , and hence f t,A ( B ) = d t ( B ) (cid:16) − φ ( t − ( A ) (cid:17) . However, as remarked earlier,for A ∈ N \ { } we have − φ ( t − ( A ) = 0 , so this at once implies that d t ( B ) = d t (cid:0) BA (cid:1) = 0 , and so B, BA ∈ D t , as required.Next we spell out the special case of Theorem 3.3 when F = F q , K = F q t and φ is the Frobenius automorphism X → X q . This gives a description of theclassic Singer difference set inside N as the set of roots of a simple polynomialand describes the mixed representations of any element also using the roots ofa polynomial. Corollary 3.4.
Let q = p k be a prime power, t ≥ an integer, and let us defineover F q the polynomial d t ( Y ) = 1 + Y + Y q + Y q + q + · · · + Y q + ··· + q t − of degree q t − − q − . Then the set D t = { A ∈ F q t | d t ( A ) = 0 } f roots of d t ( Y ) forms a ( q t − q − , q t − − q − , q t − − q − ) -difference set in the cyclic group N of norm elements of F q t , which is equivalent to the Singer difference set S t . Moreover, given an element A ∈ N \ { } , the q t − − q − different mixed D t -representations of A are exactly the products B · (cid:0) BA (cid:1) − , where B is a root in F q t of the degree q t − − q − polynomial f t,A ( X ) = d t ( X ) − A q + ··· + q t − · d t (cid:18) XA (cid:19) In connection with Corollary 3.4 first note, that in particular it implies thatthe polynomials d t ( X ) and f t,A ( X ) always split over F q t . Also, in the specialcase t = 3 , we recover the difference set H from Corollary 1.2. In this casethe polynomial f t,A is linear and its unique root is exactly the elment A fromCorollary 1.2. It is widely conjectured [5, 9] that the KST upper bound, ex( n, K t,s ) ≤ c t,s n − /t , (9)is tight up to constant factor for every s ≥ t ≥ . Together with the results of[2], Theorem 1.3 establishes that for t = 4 , the projective norm graph NG( q, does not resolve this conjecture beyond the cases s ≥ .Several interesting questions remain open. Complete bipartite graphs in projective norm graphs.
Both in [2] andin Theorem 1.3 we could only find a special kind of copies of K , . The numberof these copies is only roughly q . If that was it, then a simple uniform randomsubgraph of NG( q, with a few deleted edges would prove the tightness of theKST-bound for t = 4 and s = 6 . We think however, also supported by computerexperiments, that the number of copies of K , in NG( q, should be the sameorder as their typical number in the random graph of the same edge density. Conjecture 1.
The number of copies of K , in NG( q, is Θ( q ) . The determination of s ( t ) is still widely open for t ≥ , when we do not evenknow whether there is a K t,t in NG( q, t ) for every large enough q . While it isprobably more realistic to expect that there are copies of K t, ( t − for every t ≥ and large enough q (besides numerology, i.e. that s ( t ) = ( t − for t = 2 , and , there are also algebro-geometric heuristics pointing towards this), we harboura slim hope that t = 4 was still a special case. At least the graph NG( q, seemsquite special, with a unique structure and symmetries, and maybe that alone isresponsible for the presence of K , subgraphs.21 nfinite projective norm graphs. The first constructions of dense K t,t -free graphs were motivated by simple facts from real Euclidean geometry: twolines of the plane intersect in at most one point; three unit spheres in -spaceintersect in at most two points. Consequently the point/line incidence graph ofthe Euclidean plane is K , -free, and the unit-distance graph of the Euclidean -space is K , -free. Furthermore these infinite K t,t -free graphs are “dense” interms of the dimension of the neighborhoods. So when defined over appropriatefinite fields, in a way that the algebra in the proof of their K t,t -freeness carriesover, their number of edges verifies the tightness of the KST-bound for t = 2 and t = 3 .The (projective) norm graphs were not constructed this way, yet one candefine them over an arbitrary field F , see Section 1. The key lemma from [17]holds over any field, which implies that the proof of the K t, ( t − -freeness of NG( q, t ) also extends to NG( F , K ) for arbitrary F and arbitrary Galois extension K of degree t − . (For completeness we include the argument in the Appendix.)In particular, if t = 4 then NG( F , K ) does not contain K , for any field F . Afterseeing that NG( q, does contain K , for any q > , it might seem plausiblethat the same is true for infinite fields. The tightness of the KST-bound.
The tightness of the order of magnitudeof the KST-bound is a central question of the area. This conjecture suggests thatwhatever density is not ruled out by simple double counting, should essentiallybe possible to realize with a construction. Here we speculate that this mightnot be the case and offer a counter-conjecture.In any graph with cn / edges, the number of common neighbors of anaverage -tuple is (at least) a constant c ′ depending on c . If this graph with cn / edges is random then this constant average is spread out over (cid:0) n (cid:1) distributionsthat are each approximately Poisson with mean c ′ . Consequently for any s , apositive constant proportion of -tuples have at least s neighbors. In contrast,in any K ,s -free construction with cn / edges (matching the KST-bound), no -tuple can have more than s − neighbors. So in such constructions each ofthe Poisson-tails has to be absorbed by the -tuples with at most s − commonneighbors. Should such graphs exist for some s , they must be extremely rare,their mere existence has to be a coincidence and should require quite a bit ofstructure.In all known constructions (including Klein [10], Brown [7], KRS [17], ARS [1]and Bukh [8]) this is realized using the algebro/geometric notion of dimensionand its strong correlation with the cardinality of the corresponding variety: an“everyday” d -dimensional variety over F q has roughly Θ( q d ) points. To achievethat the common neighborhood of four vertices is less than a constant s , one ap-peals to the geometric intuition that in the four-dimensional space the intersec-tion of four hypersurfaces, that are in general enough position, is -dimensional,and hence it is the union of constantly many points. A graph can be definedon a four-dimensional space of roughly q =: n vertices, and the neighbor-hood of each vertex can be chosen to be some hypersurface, which then have22oughly the desired size q = n / . For a K ,s -free graph the intersection ofany four of the neighborhood-hypersurfaces should have size < s . Now if theneighborhood-hypersurfaces are carefully chosen, so that any four of them arein general enough position, then their intersection is -dimensional and hencehas size Θ( q ) , a constant.How to choose the hypersurfaces and what is this constant? Even thoughchoosing randomly is a generally good strategy (witnessed by the random al-gebraic construction of Bukh [8]), finding good explicit choices, as it is oftenthe case, is not so straightforward. By the KST-bound the constant boundingthe neighborhoods of t -tuples in any graph with cn − /t edges is at least t − ,and the projective norm graph chooses neighborhoods where they are boundedby not more than ( t − . The current analysis of the random choice gives anupper bound of t O ( t ) .Now how small could this constant be? We believe that the presence ofsome notion of “dimension” in this problem is a necessity and this constant isjust going to be in the nature of the geometry of the hypersurface-neighborhoodswe have chosen. As such, it will not just be limited by the simple combinatorialrestrictions of the KST-bound but also by those of geometry/algebra. And thenits extrema should be delivered by a regular, rigid structure with distinctiveproperties. For t = 4 we have seen ample evidence that the projective normgraph NG( q, fits this bill, and tend to accept it as the limit of what algebracan offer in this realm. Since we know now that K , does occur in NG( q, ,we conjecture the following. Conjecture 2. ex( n, K , ) = o ( n / ) . We note that should this conjecture be true, it of course implies that theKST-bound is not tight for the symmetric case K , either. That further impliesthat ex( n, K t,t ) = o ( n − /t ) for every t ≥ ; this is the consequence of (anadaptation of) a theorem of Erdős and Simonovits [11].While we do believe Conjecture 2, at the same time we also think that it ismore likely that we see it disproved than proved. For a proof one might need todevelop a two step approach. Given a K , -free graph with cn / edges, buildup a significant-enough proportion of a pseudo-algebraic/geometric frameworkusing the neighborhoods as hypersurfaces, with surfaces having appropriate in-tersection sizes and structure. Then, provided the pseudo-algebra/geometrygives a structure rigid enough, establish the existence of a K , . Preliminaryresults in this direction were proven by Blagojevic, Bukh, and Karasev [4] andin this paper. References [1] N. Alon, L. Rónyai, T. Szabó. Norm-graphs: variations and applications.Journal of Combinatorial Theory, Series B :280–290 (1999).[2] T. Bayer, T. Mészáros, L. Rónyai, T. Szabó. 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K , -freeness of NG( F , K ) As before, let F be an arbitrary field and for t ≥ let K be a cyclic Galoisextension of degree t − , whose Galois group is generated by the automorphism φ . We aim to prove that the projective norm graph NG( F , K ) is K t, ( t − -free,i.e. that the common neighbourhood of t vertices has size at most ( t − . Theproof is exactly the same as in the case of finite fields, and it is included hereonly because we need some of the steps anyway.For ℓ ≥ let U = { ( B i , b i ) | ≤ i ≤ ℓ } be an ℓ -set of vertices. Without lossof generality we may assume that B i = B j for i = j , as otherwise the commonneighbourhood of U would be empty. For ≤ i ≤ ℓ − let A i = A i ( U ) := 1 B i − B ℓ ∈ K and a i = a i ( U ) := b i b ℓ · N( B i ) ∈ F ∗ , and consider the system N ( X + A i ) = a i ≤ i ≤ ℓ − (10)of norm equations. Simple substitutions show that a vertex ( X, x ) is in thecommon neighbourhood of U if and only if ξ (( X, x )) := X + B ℓ is a solution to(10). Note that ξ (( X, x )) is well-defined, as N( X + B ℓ ) = b ℓ · x = 0 . Therefore,we have that the size of the common neighbourhood of U is at most the numberof solutions to (10). We remark that it can be less (by one) exactly if is asolution to (10), which happens if a = a = · · · = a ℓ .Now set ℓ = t and let us recall the key lemma from the original proof due toKollár, Rónyai and Szabó [17]. 25 emma 5.1. Let K be a field and α i,j , β j ∈ K for ≤ i, j ≤ m such that α i ,j = α i ,j for i = i . Then the system of equations ( x − α i, )( x − α i, ) · · · ( x m − α i,m ) = β i i = 1 , , . . . , m has at most m ! solutions ( x , x , . . . , x m ) ∈ K m . To finish the proof of the K t, ( t − -freeness we just need to realize that theequations in (10) can be rewritten, namely for ≤ i ≤ t − we have N( X + A i ) = ( X + A i ) · φ ( X + A i ) · · · φ ( t − ( X + A i ) == ( X + A i ) ( φ ( X ) + φ ( A i )) · · · (cid:16) φ ( t − ( X ) + φ ( t − ( A i ) (cid:17) = a i . Now if for ≤ i, j ≤ t − we set α i,j = φ ( j − ( A i ) , x j = φ ( j − ( X ) and β i = a i then Lemma 5.1 applies with m = t − and gives that the system (10) has atmost ( t − solutions and hence any t -set of vertices in NG( F , K ) has at most ( t −1)!