Bent and Z 2 k -bent functions from spread-like partitions
aa r X i v : . [ m a t h . N T ] S e p Bent and Z k -bent functions from spread-likepartitions Wilfried Meidl, Isabel PirsicJohann Radon Institute for Computational and Applied Mathematics,Austrian Academy of Sciences, Altenbergerstrasse 69, 4040-Linz, Austria.e-mail: [email protected]; [email protected]
Abstract
Bent functions from a vector space V n over F of even dimension n = 2 m into the cyclic group Z k , or equivalently, relative differencesets in V n × Z k with forbidden subgroup Z k , can be obtained fromspreads of V n for any k ≤ n/
2. In this article, existence and construc-tion of bent functions from V n to Z k , which do not come from thespread construction is investigated. A construction of bent functionsfrom V n into Z k , k ≤ n/
6, (and more generally, into any abeliangroup of order 2 k ) is obtained from partitions of F m × F m , whichcan be seen as a generalization of the Desarguesian spread. As for thespreads, the union of a certain fixed number of sets of these partitionsis always the support of a Boolean bent function. Keywords
Relative difference set, bent function, partial spread, vectorialbent function, Z k -bent, partitions Mathematics Subject Classification
Let ( A, + A ), ( B, + B ) be finite abelian groups. A function f from A to B iscalled a bent function if | X x ∈ A χ ( x, f ( x )) | = p | A | (1)for every character χ of A × B which is nontrivial on B . Alternatively, f : A → B is bent if and only if for all nonzero a ∈ A the function D a f ( x ) = f ( x + A a ) − B f ( x ) is balanced, i.e., every value in B is taken on the samenumber | A | / | B | times. The graph of f , G = { ( x, f ( x )) : x ∈ A } , is then arelative difference set in A × B relative to B , see [15]. For background onrelative difference sets we refer to [16].1n the classical case, A = V n and B = V m are elementary abelian 2-groups, i.e., they are vector spaces of dimension n and m respectively overthe prime field F . In this case the character sum in (1), called Walshtransform of f at ( a, b ) ∈ V ∗ m × V n , is of the form W f ( a, b ) = X x ∈ V n ( − h a,f ( x ) i m ⊕ + h b,x i n , where h , i k denotes an inner product in V k . If V k = F k we may use theconventional dot product, the standard inner product in F k , the finite fieldof order 2 k , is h u, v i k = Tr k ( uv ), the absolute trace of uv . A function f : V n → V m is bent, if m > vectorial bent , if and only if |W f ( a, b ) | = 2 n/ for all nonzero a ∈ V m and b ∈ V n . As is well known, n must then be even and m can be at most n/
2. Throughout the article, n = 2 m shall always be an even integer.For Boolean bent functions, i.e., bent functions f from V n to F , the dual f ∗ is the Boolean function defined by W f (1 , b ) = W f ( b ) = 2 n/ ( − f ∗ ( b ) ,which is always a Boolean bent function as well. There are many examplesand constructions of Boolean bent functions in the literature. Even severalclasses of bent functions from V n to V n/ are known, such as Maiorana-McFarland functions, spread bent functions, Dillons H -class, see [4], andKasami bent functions, cf.[2]. For background on Boolean and vectorial(Boolean) bent functions we also refer to [13].In this article we are particularly interested in bent functions f from V n to the cyclic group Z k , hence by (1), in functions f for which H f ( a, b ) = X x ∈ V n ζ af ( x )2 k ( − h b,x i , (2)where ζ k is a complex primitive 2 k th root of unity, has absolute value 2 n/ for all nonzero a ∈ Z k and b ∈ V n . Again such functions can only exist for m ≤ n/
2, [14, 17].We remark that functions f : V n → Z k satisfying the weaker conditionthat |H f (1 , b ) | = 2 n/ for all b ∈ V n are referred to as generalized bentfunctions. They have been intensively studied in many papers, see [5, 7, 8,9, 10, 11, 18]. If not also bent, generalized bent functions do not correspondto relative difference sets. However, as easily observed, f : V n → Z k is bentif and only if 2 t f is a generalized bent function for every t , 0 ≤ t ≤ k − V n to any abelian group of order 2 n/ , inparticular also bent functions from V n to Z n/ (and as their projections,2ent functions into cyclic groups of smaller order). Unlike in the case ofbent functions between elementary abelian groups, it seems difficult to findother examples for bent functions from V n to Z k when k ≥
3. In [9]it is observed that one example of a bent function from V n to Z , whichdoes not come from (partial) spreads, can be obtained from a secondaryconstruction of Boolean bent functions in [12]. As also pointed out in [10],finding bent functions into the cyclic group which are not related to spreadsis an interesting problem.In Section 2 we revisit the construction of bent functions via spreads, inparticular we point to constructing vectorial bent functions from V n to V m ,and more general of bent functions from V n into arbitrary abelian groups G , | G | = 2 k , obtained from partial, but not complete spreads. In Section 3,a construction of bent functions from V n = F m × F m into Z k is given inpolynomial form. With an argument using algebraic degree we show thatthis construction yields bent functions into Z k , k ≤ m/
3, which cannotbe obtained from (partial) spreads. In Section 4, analysing partitions of V n coming with the functions of Section 3, we obtain large classes of bentfunctions (Boolean, vectorial, and into the cyclic group), which have similarproperties as spread functions, in fact can be seen as a generalization ofspread functions obtained from the Desarguesian spread. In particular, theunion of a certain fixed number of sets of these partitions is always thesupport of a Boolean bent function. Recall that a partial spread S of V n , n = 2 m , is a set of m -dimensionalsubspaces of V n which pairwise intersect trivially. If |S| = 2 m + 1, henceevery nonzero element of V n is in exactly one of those subspaces, then S is called a (complete) spread. The standard example is the Desarguesianspread, which has for V n = F m × F m the representation S = { U, U s : s ∈ F m } , with U = { (0 , y ) : y ∈ F m } and for s ∈ F m , U s = { ( x, sx ) : x ∈ F m } .As Dillon showed in his thesis [4], a Boolean bent function f is obtainedby choosing the nonzero elements of 2 m − subspaces of S as the support of f (PS − bent functions), or the elements of 2 m − + 1 subspaces of S for thesupport of f (PS + bent functions). Clearly, every Boolean bent functionwhich is constant on the subspaces of a (partial) spread is of this form.As is also well known, with a complete spread S one gets a vectorial bentfunction from V n to V m by mapping to every nonzero z ∈ V m the nonzero3lements of exactly one subspace of S , the remaining two subspaces aremapped to 0. From projections of a bent function from V n to V m one canget vectorial bent functions from V n to V k for any 1 ≤ k ≤ n/
2. We remarkthat the spread construction can also be applied to elementary abelian p groups, p odd, see for instance [6] for PS − and PS + bent functions from F np to F p . The above extreme cases, Boolean bent functions for which solely2 m − , respectively 2 m − + 1 subspaces are required, and the construction ofvectorial bent functions from V n to V m with complete spreads are mostlyconsidered in the literature. However, the spread construction can also beapplied with not complete spreads (with more elements) to obtain vectorialbent functions. The resulting functions are then in general not projectionsof a vectorial spread bent function into V m . The procedure is standard,but as far as we know, the argument has not been given explicitly in theliterature on bent functions, hence we give it below. We consider the mostgeneral situation, i.e., functions from V n to B , where B is any abelian groupof order 2 k , 1 ≤ k ≤ m . The bentness condition is then |T f ( b ) | = 2 m for all b ∈ V n and all nontrivial characters χ of B , where, for short, denoting theinner product h u, v i in V n by h u, v i = u · v , T f ( b ) = X x ∈ V n χ ( f ( x ))( − b · x . Let S = { U j , ≤ j ≤ (2 k − m − k } be a partial spread of V n and B anabelian group of order 2 k . Define f : V n → B as follows:Construction I.- Every nonzero element γ of B has as preimage the union of exactly2 m − k elements of S except from 0 ∈ V n , i.e., f − ( γ ) = S m − k i =1 U ∗ γ,i ,where U ∗ = U \ { } .- All other elements are mapped to 0 ∈ B , i.e., f − (0) = V n \ S j U ∗ j .First observe that T f (0) = X γ ∈ B ∗ X x ∈ U ∗ γ,i ≤ i ≤ m − k χ ( γ ) + 2 m − X x ∈ S j U ∗ j
1= ( − m − k (2 m −
1) + 2 m − (2 k − m − k (2 m −
1) = 2 m . If b = 0, then T f ( b ) = X γ ∈ B ∗ X x ∈ U ∗ γ,i χ ( γ )( − b · x − X x ∈ S j U ∗ j ( − b · x . (3)4e use that for every nonzero b ∈ V n we have at most one U j ∈ S for which b · x = 0 for all x ∈ U j . Case 1:
There is no U j ∈ S such that b · x = 0 for all x ∈ U j .Then we have T f ( b ) = X γ ∈ B ∗ χ ( γ )2 m − k ( − − (2 k − m − k ( −
1) = 2 m . Case 2: b · x = 0 for all x ∈ U γ ,i for (exactly) one γ ∈ B ∗ and one i ∈ { , . . . , m − k } .In this case, T f ( b ) = X γ ∈ B ∗ χ ( γ )2 m − k ( −
1) + 2 m χ ( γ ) − ((2 k − − m − k + 1 + 2 m − m − k + 2 m χ ( γ ) + (2 k − m − k − m = 2 m χ ( γ ) . Similarly, one shows that for a partial spread with (at least) 2 m − m − k +1subspaces, the function g : V n → B defined as follows is a bent function:Construction II.- For an element ˜ γ ∈ B ∗ we have g − (˜ γ ) = S m − k +1 i =1 U ˜ γ,i , i.e., ˜ γ has theunion of 2 m − k + 1 elements of a partial spread as preimage (note thatalso f (0) = ˜ γ ),- if γ ∈ B ∗ , γ = ˜ γ , then g − ( γ ) = S m − k i =1 U ∗ γ,i , i.e., the preimage of γ consists of the nonzero elements of 2 m − k elements of a partial spread,- the remaining elements are mapped to 0.If k = 1, hence B = F , then the functions in Construction I and II areconventional PS − and PS + bent functions, respectively. Hence one may seethe vectorial partial spread functions from V n to V k obtained by Construc-tion I and II with B = V k as vectorial PS − and PS + bent functions . Notethat for Construction I respectively II one needs partial spreads S with atleast 2 m − m − k respectively 2 m − m − k + 1 elements.In this article we are interested in bent functions from V n to the cyclicgroup Z k . Canonical examples one obtains with the partial spread con-struction with B = Z k for all k ≤ n/
2, which we will also call partialspread bent functions from V n to Z k . In the following sections we investi-gate existence and construction of bent functions from V n to Z k which donot come from partial spreads. 5 Z k -bent functions not obtained from spreads As we have to distinguish addition in different structures, we denote theaddition in the complex numbers and in the ring Z k by +, the addition inthe elementary abelian groups F , V n and F m is denoted by ⊕ .Let f be a function from V n to Z k , then we can write f as f ( x ) = a ( x ) + 2 a ( x ) + · · · + 2 k − a k − ( x ) (4)for uniquely determined Boolean functions a j , 0 ≤ j ≤ k −
1, from V n to F .Recall that a function f : V n → Z k is called generalized bent if for H f given as in (2) we have |H f (1 , u ) | = 2 n/ for all u ∈ V n . Naturally, bentfunctions from V n to Z k are generalized bent. The converse does not hold,but we have the following obvious lemma, see [5]. Lemma 1.
A function f : V n → Z k is bent if and only if t f is generalizedbent for all t , ≤ t ≤ k − . Generalized bent functions are intensively studied in the literature, see[5, 7, 8, 9, 10, 11, 18]. A comprehensive characterization of generalized bentfunctions via partitions has been given in [11]. In [5], another character-ization of generalized bent functions has been given via properties of theBoolean functions a i in (4). Proposition 1.
Let f : V n → Z k be given as f ( x ) = a ( x ) + 2 a ( x ) + · · · +2 k − a k − ( x ) for some Boolean functions a j , ≤ j ≤ k − , and let A be theaffine space of Boolean functions, A = a k − ⊕ h a k − , . . . , a i . Then f is generalized bent if and only if all functions in A are Boolean bentfunctions, and for any three functions b , b , b ∈ A we have ( b ⊕ b ⊕ b ) ∗ = b ∗ ⊕ b ∗ ⊕ b ∗ . (5)One of the main objectives in this article is to construct and to under-stand infinite classes of bent functions from V n to the cyclic group Z k , orequivalently of relative difference sets in V n × Z k relative to Z k , whichdo not come from partial spreads. We therefore need a tool to distinguishsuch bent functions from partial spread bent functions. As in our analysisof bent functions from V n to Z k , Boolean bent functions play a major role,for better understanding, in the remainder of the article we will denote bent6unctions from V n to Z k as Z k -bent functions. Bent functions from V n to F will be referred to as Boolean bent functions or simply as bent functions.Recall that two Boolean functions f, g from V n to F are called extendedaffine equivalent (EA-equivalent) if g ( x ) = f ( L ( x ) + b ) + a ( x ) for some linearcoordinate transformation L on V n , an element b ∈ V n and an affine map a ( x ) from V n to F . If a ( x ) is the zero-map, then f and g are called afflneequivalent , if additionally b = 0, then f and g are called linear equivalent . Lemma 2.
Let f : V n → Z k with f ( x ) = a ( x )+ 2 a ( x )+ · · · + 2 k − a k − ( x ) be a partial spread Z k -bent function. Then all Boolean functions a i , ≤ i ≤ k − , are Boolean partial spread bent functions, all of algebraic degree m = n/ .Proof. Since f is Z k -bent, by Lemma 1, for every t , 0 ≤ t ≤ k −
1, everyBoolean function in A t = a k − t − ⊕ h a k − t − , . . . , a i is bent. In particular, a i , 0 ≤ i ≤ k −
1, is a Boolean bent function. Let U ∈ S , and x , x ∈ U ∗ .If a j ( x ) = a j ( x ), then f : V n → Z k cannot be constant on U ∗ . Hence a j is a Boolean bent function which is constant on the nonzero elements of U for every U ∈ S , consequently a Boolean partial spread bent function. By[4, p.96], every Boolean partial spread bent function on a spread with morethan 2 m − subspaces has algebraic degree m . ✷ Remark 1.
With the same argument, all bent functions in A t , ≤ t ≤ k − ,are partial spread bent functions of algebraic degree m . In [12], Mesnager presented several examples of Boolean bent functions b , b , b satisfying ( b ⊕ b ⊕ b ) ∗ = b ∗ ⊕ b ∗ ⊕ b ∗ . The objective in [12] is to usethose functions in a secondary construction of Boolean bent functions dueto Carlet [1]. As shown in [9] this secondary construction is equivalent toconstructing generalized bent functions to Z . As observed in [9, 10], one ofthe examples, using Maiorana-McFarland functions, potentially yields Z -bent functions from V n = F m × F m to Z . Recall that for a permutation π of F m , the Boolean function h ( x, y ) = Tr m ( xπ ( y )) from F m × F m to F is a bent function belonging to the Maiorana-McFarland class. Noting thatfor the Maiorana-McFarland bent function b ( x, y ) = Tr m ( βxy d ), gcd(2 m − , d ) = 1, from F m × F m to F , we have b ∗ ( x, y ) = Tr m ( β − e x e y ), ed ≡ m −
1, we arrive at the following observation which we here state asa lemma. By convention, all powers of 0 are equal to 0 (including powerswith negative exponents).
Lemma 3. [12] Let d, e be integers such that gcd(2 m − , d ) = 1 and ed ≡ m − , and suppose that β , β , β ∈ F m satisfy ( β ⊕ β ⊕ β ) − e = β − e ⊕ β − e ⊕ β − e . (6)7 hen the Boolean bent functions b i ( x ) = Tr m ( β i xy d ) , i = 0 , , , satisfy ( b ⊕ b ⊕ b ) ∗ = b ∗ ⊕ b ∗ ⊕ b ∗ . As easily observed, f ( x ) = b ( x ) + 2( b ⊕ b )( x ) + 4( b ⊕ b )( x ) is thengeneralized bent, and since b i ⊕ b j , 0 ≤ i < j ≤
2, is bent, f is even Z -bent.For the details we refer to [9].Trivially, with − e = −
1, and more general − e ≡ v mod 2 m −
1, equation(6) is satisfed for all choices of β , β , β ∈ F m . In [12, Table 1], for 4 ≤ n ≤ e for which there exist β , β , β such that (6) is satisfiedare listed.In fact we have recalculated the values and slightly expanded the table,also completing the table with some additional entries not present in theoriginal. We find that the examples are so prolific that it is in all but twocases more efficient to list the complements, i.e., the coset leaders of thecyclotomic classes that do not fulfill the condition (also we omit the classled by 1). n n − e not fulfilling (6)With the next lemma we obtain large sets of elements of which everythree satisfy the condition (6) for some fixed (nontrivial) e . In fact evenstronger conditions are satisfied, which we will require to construct Z k -bentfunctions for k > Lemma 4.
Let m, j be integers such that and gcd(2 m − , j −
1) = 2 k − ,and let e = 2 m − j − . Then for any sum P i β i of elements of F k wehave P β − ei = ( P β i ) − e . In particular, for any β , β , β ∈ F k we have ( β ⊕ β ⊕ β ) − e = β − e ⊕ β − e ⊕ β − e .Proof. As k divides j , for β ∈ F k we have β − e = β j +1 = β , hence P β − ei = ( P β i ) − e if β i ∈ F k . In particular we have ( β ⊕ β ⊕ β ) − e =8 − e ⊕ β − e ⊕ β − e for β , β , β ∈ F k . ✷ Before we show the main results of this section, we apply Lemma 4 to thesecondary construction of bent functions in [1], and obtain with Theorem14 in [12] the following
Corollary 1.
Let m, j be integers such that gcd(2 m − , j + 1) = 1 and gcd(2 m − , j −
1) = 2 k − , let e = 2 m − j − , and let d be the inverse of e modulo m − . Then for any β , β , β ∈ F ∗ k the Boolean function g ( x ) = Tr m ( β − e x e y )Tr m ( β − e x e y ) ⊕ Tr m ( β − e x e y )Tr m ( β − e x e y ) ⊕ Tr m ( β − e x e y )Tr m ( β − e x e y ) is bent. Its dual is g ∗ ( x ) = Tr m ( β xy d )Tr m ( β xy d ) ⊕ Tr m ( β xy d )Tr m ( β xy d ) ⊕ Tr m ( β xy d )Tr m ( β xy d ) . We now come back to our primary objective, the construction of Z k -bentfunctions, and state the main results of this section. Theorem 1.
Let m, j be integers such that gcd(2 m − , j + 1) = 1 and gcd(2 m − , j −
1) = 2 k − , let e = 2 m − j − , and let d be the inverseof e modulo m − . Then for a basis { α , α , . . . , α k − } of F k over F , thefunctions f and f given as f ( x ) = k − X i =0 Tr m ( α i xy d )2 i , f ( x ) = k − X i =0 Tr m ( α − ei x e y )2 i (7) are Z k -bent functions from F m × F m to Z k .Proof. By Lemma 1, a function f : V n → Z k given as f ( x ) = a k − ( x )2 k − + a k − ( x )2 k − + · · · + a ( x ) is Z k -bent if and only if 2 t f ( x ) = a k − t − ( x )2 k − + · · · + a ( x )2 t is generalized bent for all 0 ≤ t ≤ k −
1. With Proposition 1,the function 2 t f ( x ) is generalized bent if and only if every Boolean functionin the affine space A t = a k − t − ⊕ h a k − t − , . . . , a i is bent, and for each threeBoolean bent functions b , b , b ∈ A t we have ( b ⊕ b ⊕ b ) ∗ = b ∗ ⊕ b ∗ ⊕ b ∗ .(Note that then also b ⊕ b ⊕ b is a bent function in A t .)Observe that for the function f , every Boolean function in A t , 0 ≤ t ≤ k −
1, is of the form b ( x ) = Tr m ( βxy d ), where β is an element in F ∗ k ,thus b ( x ) is bent. Let now b , b , b ∈ A t be given as b i ( x ) = Tr m ( β i xy d ),9 = 0 , ,
2. By Lemma 4, equation (6) is satisfied, and by Lemma 3 thefunction f is Z k -bent.For the function f , the Boolean functions in A t , 0 ≤ t ≤ k −
1, are of theform b ( x ) = Tr m ( ˜ βx e y ), ˜ β ∈ F ∗ k . Since gcd( − e, k −
1) = 1, hence ˜ β = β e forsome β ∈ F k , for three functions b , b , b ∈ A t we can assume that b i ( x ) =Tr m ( β − ei x e y ), β i ∈ F ∗ m , i = 0 , ,
2. Then b ⊕ b ⊕ b = Tr m (( β − e ⊕ β − e ⊕ β − e ) x e y ) = Tr m (( β ⊕ β ⊕ β ) − e x e y ). For the duals, b ∗ i ( x ) = Tr m ( β i xy d ), i = 0 , ,
2, and ( b ⊕ b ⊕ b ) ∗ ( x ) = Tr m (( β ⊕ β ⊕ β ) xy d ), the condition(5) is satisfied. Consequently, by Proposition 1, f is Z k -bent. ✷ Remark 2.
For j = 0 the conditions of Theorem 1 are satisfied. In thiscase, k = m and one obtains a bent function from V n to Z m , n = 2 m . Aseasily observed, e = 2 m − (hence − e ≡ m − trivially satisfiescondition (6) ), its inverse d and m − belong to the same cyclotomic cosetof modulo m − , i.e., Tr m ( αxy d ) = Tr m ( αx v (2 m − y ) for some integer v .Hence Tr m ( αxy d ) is obtained form the classical Boolean partial spread bentfunction Tr m ( αx m − y ) by applying a linear coordinate transformation to thevariable x . Therefore, Tr m ( αxy d ) and Tr m ( αx m − y ) are affine equivalent.The functions f , f in Theorem 1 are then partial spread Z k -bent functionsfrom F m × F m to Z m as constructed in Section 2.Clearly, with e = d = 2 m − , the functions f , f in (7) are partial spread Z k -bent functions, see also [12, Corollary 17]. With Theorem 1, for j > Z k -bent functions from F m × F m to Z k which provably do notcome from partial spreads. Corollary 2.
Let m and j > be integers such that gcd(2 m − , j + 1) = 1 and gcd(2 m − , j −
1) = 2 k − , and let e, d , α i , ≤ i ≤ k − , be asin Theorem 1. Then the functions f , f in (7) are Z k -bent functions from F m × F m to Z k , which do not come from partial spreads.Proof. If j >
0, then the binary weight of e = 2 m − j − m −
2. Hencethe algebraic degree of Tr m ( α − ei x e y ) in f is m −
1. By Lemma 2, f is not apartial spread Z k -bent function. The bent functions Tr m ( α i xy d ) in f arethe duals of the bent functions Tr m ( α − ei x e y ) in f , hence also not partialspread bent functions. Again with Lemma 2, f is not a partial spread Z k -bent function. ✷ The largest possible k in the construction of Theorem 1, besides from k = m , which as explained above solely yields the known spread functions, is k = m/
3. As easily seen, for every integer m divisible by 3 and k = m/ k + 1 , m −
1) = 1. Hence the conditions of Theorem 1 aresatisfied and we have the following corollary.
Corollary 3.
Let n = 2 m be divisible by , then there exists a bent functionfrom V n to Z m/ which is not obtained from a (partial) spread. We expect that Z m -bent functions from V m to Z m one only can obtainfrom (complete) spreads.We finish this section with a remark on the vectorial bent function F ( x ) = (Tr m ( α xy d ) , Tr m ( α xy d ) , . . . , Tr m ( α k − xy d )) from F m × F m to F k associated with f , a projection onto F k of the vectorial Maiorana-McFarland bent function ˜ F ( x ) = Tr nm ( xy d ) from F m × F m to F m . Asshown in [3, Theorem 2], ˜ F is a vectorial dual-bent function , i.e., a vectorialbent function for which the set of the duals of all component functions againforms a vectorial bent function (of the same dimension). The projection F attached to our Z k -bent function satisfies the even stronger condition that( b ⊕ ¯ b ) ∗ = b ∗ ⊕ ¯ b ∗ for all component functions b, ¯ b of F . Z k -bent functions from the par-titions In this section we look at the partitions of F m × F m which we obtain fromthe sets of the preimages of the Z k -bent functions f and f in Theorem1. As we will see, these partitions share some properties with spreads, theycan be seen as a generalization of the Desarguesian spread. For simplicitywe take j = k a divisor of m .For integers m, e with gcd(2 m − , e ) = 1 and an element s ∈ F m define U s := { ( x, sx − e ) : x ∈ F m } , U ∗ s = U s \ { } , and U = { (0 , y ) : y ∈ F m } . Then U , U ∗ s , s ∈ F m , form a partition of F m × F m . Note that U , U s , s ∈ F m , are the subspaces of the Desarguesian spread if − e ≡ m − − e ≡ v mod 2 m − U s is not a subspaceif we do not have − e ≡ v mod 2 m − v .Similarly, for integers m, d with gcd(2 m − , d ) = 1 and an element s ∈ F m define V s := { ( x − d s, x ) : x ∈ F m } , V ∗ s = V s \ { } , and V = { ( x,
0) : x ∈ F m } . Note that as above for the sets U and U s , if − d ≡ v mod 2 m −
1, then V s and V are the subspaces of the Desarguesian spread.11or a divisor k of m and an element γ of F k let A ( γ ) = [ s ∈ F m Tr mk ( s )= γ U ∗ s and B ( γ ) = [ s ∈ F m Tr mk ( s )= γ V ∗ s . (8) Lemma 5.
Let k be a divisor of m , gcd(2 m − , k + 1) = 1 and let e =2 m − k − and d be the integer such that de ≡ m − . For the Z k -bent functions f and f in Theorem 1 let Γ = { A ( i ) = { ( x, y ) ∈ F m × F m : f ( x, y ) = i } ; i = 0 , . . . k − } be the partition of F m × F m obtained with the set of the preimages for f ,and let Γ = { B ( i ) = { ( x, y ) ∈ F m × F m : f ( x, y ) = i } ; i = 0 , . . . k − } be the partition of F m × F m obtained with the set of the preimages for f .Then Γ = {A (0) ∪ U, A ( γ ); γ ∈ F ∗ m } Γ = {B (0) ∪ V, B ( γ ); γ ∈ F ∗ m } , where A (0) ∪ U = A (0) and B (0) ∪ V = B (0) .Proof. First observe that on U ∗ s = { ( x, sx k +1 ) : x ∈ F ∗ m } the bentfunction Tr m ( α − e x e y ) = Tr m ( α − e s ) is constant, hence f is constant on U ∗ s .Let s , s ∈ F m , then f takes on the same value on U s and U s if and onlyif Tr m ( α − ei s ) = Tr m ( α − ei s ) for all α i of the basis { α , . . . , α k − } of F k .Hence we requireTr m ( α − ei s ⊕ α − ei s ) = Tr k ( α − ei Tr mk ( s ⊕ s )) = 0 for all i = 0 , . . . , k − . (9)Using that P j α − ei j = ( P j α i j ) − e , we infer that { α − ei ; i = 0 , . . . , k − } isa basis of F k as well. Hence (9) holds if and only if Tr mk ( s ⊕ s ) = 0.Consequently, f is constant on A ( γ ) and A ( γ ), A ( γ ) with γ = γ aremapped to different constants. Additionally, the elements of U , as well asthe elements of A (0) are mapped to 0. The same argument applies for Γ with f . ✷ Note that for ( x, y ) ∈ U ∗ s we have Tr m ( αxy d ) = Tr m ( αxs d x − ed ) = Tr m ( αs d ).Hence also f is constant on U ∗ s and certainly on U . In fact we can write Γ also in terms of the sets U and S s ∈ F m Tr mk ( sd )= γ U ∗ s . Observe that the partitionsΓ and Γ , though looking similar, are different. In the further we will usethe representation of Γ as in Lemma 5.12 emark 3. In the special case k = m , the partition { U, A ( γ ); γ ∈ F m } = { U, U ∗ s ; s ∈ F m } with U s = { ( x, sx ) : x ∈ F m } , reduces to (a repre-sentation of ) the Desarguesian spread partition of F m × F m . The sameapplies to the partition { V, B ( γ ); γ ∈ F m } . As already observed in Remark2, f and f are then spread Z m -bent functions. As pointed out in Section2, many more can be obtained with the partition, by taking as preimage ofevery element in Z m exactly one of these sets, except for one (w.l.o.g. ),which has the union of two elements of the partition as preimage. To deduce a comparable result for the case k < m we will need thefollowing lemma.
Lemma 6.
Let m, k be integers such that k divides m and gcd(2 m − , k +1) = 1 , let e = 2 m − k − and d be the inverse of e modulo m − , i.e., de ≡ m − . Let u, v ∈ F m and γ ∈ F k .(i) If v = 0 , then Ω γ = X s :Tr mk ( s )= γ X x ∈ F ∗ m ( − Tr m ( ux ) ⊕ Tr m ( vsx k +1 ) = (cid:26) m − m − k : γ = Tr mk ( uv d ) , − m − k : otherwise.(ii) If u = 0 , then Υ γ = X s :Tr mk ( s )= γ X x ∈ F ∗ m ( − Tr m ( usx − d ) ⊕ Tr m ( vx ) = (cid:26) m − m − k : γ = Tr mk ( vu e ) , − m − k : otherwise.Proof. (i) Let s γ ∈ F m such that Tr mk ( s γ ) = γ . Then { s ∈ F m :Tr mk ( s ) = γ } = { s γ ⊕ s : Tr mk ( s ) = 0 } . Denoting the subspace { s ∈ F m :Tr mk ( s ) = 0 } by Λ we obtainΩ γ = X s ∈ Λ X x ∈ F ∗ m ( − Tr m ( ux ) ⊕ Tr m ( vs γ x k +1 ⊕ vsx k +1 ) = X x ∈ F ∗ m ( − Tr m ( ux ⊕ vs γ x k +1 ) X s ∈ Λ ( − Tr m ( vsx k +1 ) . P s ∈ Λ ( − Tr m ( vsx k +1 ) = 0 if vx k +1 Λ ⊥ , the orthogonalcomplement of Λ , and P s ∈ Λ ( − Tr m ( vsx k +1 ) = 2 m − k if vx k +1 ∈ Λ ⊥ .Since for z ∈ F k and s ∈ Λ we have Tr m ( zs ) = Tr k ( z Tr mk ( s )) = 0, andsince Λ ⊥ has dimension k , we have Λ ⊥ = F k . Therefore,Ω γ = 2 m − k X x ∈ F ∗ m : vx k +1 ∈ F ∗ k ( − Tr m ( ux ⊕ s γ vx k +1 ) . (10)As gcd(2 m − , k + 1) = 1, there exists a unique element µ ∈ F m with µ k +1 = v − , i.e., µ e = v , hence µ = v d . Then vx k +1 ∈ F k if and only if x k +1 ∈ µ k +1 F k = µ k +1 F k +12 k . Hence (10) reduces toΩ γ = 2 m − k X z ∈ F ∗ k ( − Tr m ( uµz ⊕ s γ vµ k +1 z k +1 ) = 2 m − k X z ∈ F ∗ k ( − Tr m ( uµz ⊕ s γ z ) . Let ν = Tr mk ( uµ ) = Tr mk ( uv d ), thenTr m ( uµz ⊕ s γ z ) = Tr k ( z Tr mk ( uµ )) ⊕ Tr k ( z Tr mk ( s γ )) = Tr k ( ν z ) ⊕ Tr k ( γz )= Tr k (( ν ⊕ γ ) z ) . Consequently,Ω γ = 2 m − k X z ∈ F ∗ k ( − Tr k (( ν ⊕ γ ) z ) = (cid:26) m − m − k : ν ⊕ γ = 0 , − m − k : otherwise.(ii) For Υ γ with the same arguments we obtainΥ γ = 2 m − k X x ∈ F ∗ m : ux − d ∈ F ∗ k ( − Tr m ( us γ x − d ⊕ vx ) . If µ ∈ F m is the unique element such that µ d = u , i.e., µ = u e , we getΥ γ = 2 m − k X z ∈ F ∗ k ( − Tr m ( s γ z − d ⊕ vµz ) = 2 m − k X z ∈ F ∗ k ( − Tr k ( γz − d ) ⊕ Tr k ( z Tr mk ( vµ )) . Changing the order of summation with z → z k +1 = z − e we haveΥ γ = 2 m − k X z ∈ F ∗ k ( − Tr k ( γ ( z − e ) − d ) ⊕ Tr k ( z k +1 Tr mk ( vµ )) = 2 m − k X z ∈ F ∗ k ( − Tr k ( γz ) ⊕ Tr k ( z Tr mk ( vµ )) = 2 m − k X z ∈ F ∗ k ( − Tr k (( γ ⊕ Tr mk ( vµ )) z ) , γ . ✷ Theorem 2.
Let m, k be integers such that k divides m and gcd(2 m − , k +1) = 1 , and let π ( i ) = γ i be a one-to-one map from Z k to F k . Definefunctions f A , f B : F m × F m → Z k as follows:- If ( x, y ) ∈ A ( γ i ) then f A ( x, y ) = i , and, w.l.o.g., f A (0 , y ) = 0 for all y ∈ F m ;- If ( x, y ) ∈ B ( γ i ) then f B ( x, y ) = i , and, w.l.o.g., f ( x,
0) = 0 for all x ∈ F m .Then f A , f B are Z k -bent functions.Proof. For u, v ∈ F m and 0 ≤ t ≤ k −
1, for the function f A we have H f A (2 t , ( u, v )) = X x,y ∈ F m ζ t f A ( x,y )2 k ( − Tr m ( uv ⊕ vy ) = k − X i =0 X ( x,y ) ∈A ( γ i ) ζ t i k ( − Tr m ( ux ⊕ vy ) + X y ∈ F m ( − Tr m ( vy ) = k − X i =0 ζ t i k X Tr mk ( s )= γ i X x ∈ F ∗ m ( − Tr m ( ux ⊕ vsx k +1 ) + X y ∈ F m ( − Tr m ( vy ) . If v = 0, then H f A (2 t , ( u, v )) = 2 m − k X x ∈ F ∗ m ( − Tr m ( ux ) 2 k − X i =0 ζ t i k + X y ∈ F m m . Let v = 0. Then with P ( x,y ) ∈A ( γ j ) ( − Tr m ( ux ⊕ vy ) = Ω γ , and applyingLemma 6, we obtain H f A (2 t , ( u, v )) = k − X i =0 ζ t i k Ω γ i + X y ∈ F m ( − Tr m ( vy ) = − m − k k − X ci =0 ζ t i k + 2 m ζ t ι k = 2 m ζ t ι k , (11)if γ ι = Tr mk ( uµ ) , where µ k +1 = v − .In the same way, for f B we obtain H f B (2 t , (0 , v )) = 2 m (as w.l.o.g. f B ( x,
0) =0), and H f B (2 t , ( u, v )) = 2 m ζ t ι k (12)15f Tr mk ( u e v ) = γ ι . ✷ From Theorem 2 we follow the next theorem on the related Boolean bentfunctions, which, just as the spread bent functions, are exactly those bentfunctions which are constant on the elements of a certain partition of V n . Theorem 3.
Let m, k be integers such that k divides m and gcd(2 m − , k +1) = 1 , let e = 2 m − k − and d such that de ≡ m − . For γ ∈ F k let A ( γ ) = [ s ∈ F m :Tr mk ( s )= γ { ( x, sx k +1 ) : x ∈ F ∗ m } , U = { (0 , y ) : y ∈ F m } , and B ( γ ) = [ s ∈ F m :Tr mk ( s )= γ { ( x − d s, x ) : x ∈ F ∗ m } , V = { ( x,
0) : x ∈ F m } . I. Every Boolean function whose support is the union of k − of the sets A ( γ ) is a bent function. Likewise, their complements, i.e., the Booleanfunctions with U and k − of the sets A ( γ ) as their support, are bent.II. Every Boolean function whose support is the union of k − of the sets B ( γ ) is a bent function. Likewise the Boolean functions with V and k − of the sets B ( γ ) as their support, are bent.The duals of the bent functions of the class in I are in the class in II (andvice versa).Proof. The bentness of the Boolean functions in I and II follows imme-diately with (11) and (12) for t = k −
1. Note that the Boolean function2 k − f has as support all A ( γ ) respectively B ( γ ) which f maps to an odd i .Moreover, we explicitly see the dual function of a bent function of I from(11) (for v = 0) H f (2 k − , ( u, v )) = 2 m ( − ι if γ ι = Tr mk ( uv d ) . Again observing that Tr mk ( uv d ) is constant γ for all( u, v ) ∈ B ( γ ), we infer that the dual is in class II. Note that clearly, as thedual is bent, we must have exactly 2 k − of the sets B ( γ ) in the support(additionally V may be in the support). ✷ Remark 4.
The Maiorana-McFarland functions Tr m ( α − e x e y ) , α ∈ F ∗ k ,and the functions g ( x ) in Corollary 1 certainly belong to the functions ofclass I, their duals Tr m ( αxy d ) and g ∗ in Corollary 1 belong to class II. s easily seen, in general these bent functions and their duals have differ-ent algebraic degree, hence in general they are not EA-equivalent. Thesespread-like partitions, which also give rise to Z k -bent functions and are themain object of our interest, yield a large quantity of bent functions of classI respectively class II. If some given function in class I or class II can beobtained from any of the numerous known primary or secondary bent func-tion constructions is as usual very difficult to answer. An arbitrary functionin class I or class II does not necessarily belong to the completed Maiorana-McFarland class as Example 6.3.16 in [4] shows (special case k = m , whereour partition reduces to the Desarguesian spread - see the remark below). Remark 5. (i) In the special case k = m , the set S = { U, A ( γ ) : γ ∈ F m } reduces to the Desarguesian spread, and f in Theorem 2 is aspread function on the complete Desarguesian spread, described as inSection 2. Corollary 3 describes then the PS − ap and PS + ap functions,cf. [4]. Hence we may see the Z k -bent functions in Theorem 2, andthe Boolean bent functions in Corollary 3 as generalizations of theDesarguesian spread bent functions.(ii) As for the spread functions in Section 2, also the proof of Theorem 2,holds not only for functions from F m × F m to Z k , but for functionsfrom F m × F m into any abelian group of order k . The bentness isa property of the partition of F m × F m . In particular, also manymore vectorial bent functions in dimension k are obtained. As easilyobserved, these vectorial bent functions from F m × F m to F k belongto the class of vectorial dual-bent functions, where the duals for thefunctions that are constant on Γ are functions that are constant on Γ . For a spread S = { U j : j = 0 , , . . . , m + 1 } of V n denote (for a fixedinner product) by S ⊥ the spread S ⊥ = { U ⊥ j : j = 0 , , . . . , m + 1 } .As is well known, the duals of the Boolean partial spread bent functionsfrom the spread S are the Boolean spread bent functions from S ⊥ . Moreover,a vectorial spread bent function from S is vectorial dual-bent, with a dualobtained from S ⊥ (see [3, Theorem 3]). Hence also with this respect thefunctions obtained from the partitions Γ and Γ behave like spread bentfunctions. In fact, for the special case that e ≡ v mod 2 m −
1, in which caseΓ , Γ are spreads, we see that Γ ⊥ = Γ . For the Desarguesian spread (moregeneral for every symplectic spread), the Boolean spread bent functions andtheir duals are EA-equivalent, see the discussion on weak self-duality in [3].17s observed above, with this respect the (vectorial) bent functions obtainedfrom Γ , Γ in general behave different.We close this section with some more remarks on equivalence of parti-tions. We here call two partitions of V n equivalent, if one is obtained fromthe other with a linear coordinate transformation on V n . Clearly, as for thespreads, the partitions Γ and Γ represent then a whole equivalence classof partitions. Let L be a linear permutation of F m × F m and let L (Γ i ), i = 1 ,
2, be the partition of F m × F m obtained via the coordinate trans-formation described by L . Then every Boolean function g which is constanton the elements of L (Γ i ) and has a support of cardinality 2 m − − m − or 2 m − + 2 m − is of the form g ( x ) = f ( L − ( x )) for a bent function inthe class I respectively II. Hence g is a bent function affine equivalent to f .Obviously two such partitions which are equivalent via a coordinate transfor-mation yield the same sets of bent functions (in the sense of EA-equivalencefor Boolean functions).Though we observed that we have functions f from class I and functions f from class II which are not EA-equivalent, the partitions Γ and Γ maystill be equivalent (note that in general also the set of bent functions for agiven spread contains several EA-equivalence classes). We remark that forthe Desarguesian spread (and in general for symplectic spreads) S and S ⊥ are equivalent. Numerically we confirmed that in general Γ and Γ are notequivalent. Hence with respect to equivalence, the partitions Γ , Γ behavedifferent than S and S ⊥ for a Desarguesian spread, the special case when e ≡ v mod 2 m − Acknowledgement.
The authors are supported by the FWF Project P 30966.
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