Splitting Subspaces, Singer Cycles and Linear Recurrences

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1 Splitting Subspaces, Singer Cycles and Linear Recurrences Department of Mathematics Indian Institute of Technology Bombay Powai, Mumbai , India srg/ Séminaire de Théorie des Nombres Institut de Mathématiques de Bordeaux Bordeaux, France May 11, 2012

2 Wonderful world of finite fields Galois (1830): For every prime power q, there is a unique field with q elements, denoted by F q or by GF(q).

3 Wonderful world of finite fields Galois (1830): For every prime power q, there is a unique field with q elements, denoted by F q or by GF(q). 1 Galois Theory is easy: For every n 1, the field F q has a unique field extension of degree n, namely, F q n. This extension is Galois and its Galois group over F q is cyclic of order n.

4 Wonderful world of finite fields Galois (1830): For every prime power q, there is a unique field with q elements, denoted by F q or by GF(q). 1 Galois Theory is easy: For every n 1, the field F q has a unique field extension of degree n, namely, F q n. This extension is Galois and its Galois group over F q is cyclic of order n. 2 Primitive objects exist: The multiplicative group F q of nonzero elements of F q is cyclic [a generator of this is called a primitive element; the minimal polynomial over F q of a generator of F q n is called a primitive polynomial].

5 Wonderful world of finite fields Galois (1830): For every prime power q, there is a unique field with q elements, denoted by F q or by GF(q). 1 Galois Theory is easy: For every n 1, the field F q has a unique field extension of degree n, namely, F q n. This extension is Galois and its Galois group over F q is cyclic of order n. 2 Primitive objects exist: The multiplicative group F q of nonzero elements of F q is cyclic [a generator of this is called a primitive element; the minimal polynomial over F q of a generator of F q n is called a primitive polynomial]. 3 Vector spaces are interesting: Finite dimensional vector spaces over F q are naturally endowed with a multiplicative structure. Indeed, F n q F q n.

6 Wonderful world of finite fields Galois (1830): For every prime power q, there is a unique field with q elements, denoted by F q or by GF(q). 1 Galois Theory is easy: For every n 1, the field F q has a unique field extension of degree n, namely, F q n. This extension is Galois and its Galois group over F q is cyclic of order n. 2 Primitive objects exist: The multiplicative group F q of nonzero elements of F q is cyclic [a generator of this is called a primitive element; the minimal polynomial over F q of a generator of F q n is called a primitive polynomial]. 3 Vector spaces are interesting: Finite dimensional vector spaces over F q are naturally endowed with a multiplicative structure. Indeed, F n q F q n. 4 Counting counts: Objects of interest in algebra and geometry, such as polynomials of a given degree, matrices of a given size, affine and projective varieties, are finite sets in the setting of finite fields. Thus one can try to enumerate or estimate them.

7 Splitting Subspaces Let m, n be positive integers and σ F q mn. An m-dimensional subspace W of F q mn is σ-splitting if F q mn = W σw σ n 1 W.

8 Splitting Subspaces Let m, n be positive integers and σ F q mn. An m-dimensional subspace W of F q mn is σ-splitting if F q mn = W σw σ n 1 W. Question (Niederreiter, 1995) Given σ such that F q mn = F q (σ), what is the number of σ-splitting subspaces of F q mn of dimension m? Denote by S(σ, m, n; q) the number of m-dimensional σ-splitting subspaces of F q mn.

9 Splitting Subspaces Let m, n be positive integers and σ F q mn. An m-dimensional subspace W of F q mn is σ-splitting if F q mn = W σw σ n 1 W. Question (Niederreiter, 1995) Given σ such that F q mn = F q (σ), what is the number of σ-splitting subspaces of F q mn of dimension m? Denote by S(σ, m, n; q) the number of m-dimensional σ-splitting subspaces of F q mn. Note: For an arbitrary σ F q mn, there may not be any σ-splitting subspace; for example, this happens if σ F q and n > 1. However, if σ F q mn satisfies F q mn = F q (σ), then a σ-splitting subspace exists, e.g., W = span{1, σ n, σ 2n,..., σ (m 1)n }.

10 Connection with Cryptography (via LFSRs) A (homogeneous) linear recurrence or a LFSR of order n is given by s i+n = s i c 0 + s i+1 c s i+n 1 c n 1 (1) for i = 0, 1,..., where c 0, c 1,..., c n 1 F q and (s 0, s 1,, s n 1 ) F n q is the initial state. The sequence s = (s 0, s 1,... ) generated by the LFSR (1) is always (ultimately) periodic, i.e., r 1 and n 0 0 s.t. s i+r = s i i n 0. Moreover, we can take r q n 1 and s is periodic (i.e. can take n 0 = 0) if c 0 0. We say that (1) is primitive if s is periodic with maximum possible period for any choice of nonzero initial state. Easy Question: What is the number of primitive linear recurrence sequences of order n over F q?

11 Connection with Cryptography (via LFSRs) A (homogeneous) linear recurrence or a LFSR of order n is given by s i+n = s i c 0 + s i+1 c s i+n 1 c n 1 (1) for i = 0, 1,..., where c 0, c 1,..., c n 1 F q and (s 0, s 1,, s n 1 ) F n q is the initial state. The sequence s = (s 0, s 1,... ) generated by the LFSR (1) is always (ultimately) periodic, i.e., r 1 and n 0 0 s.t. s i+r = s i i n 0. Moreover, we can take r q n 1 and s is periodic (i.e. can take n 0 = 0) if c 0 0. We say that (1) is primitive if s is periodic with maximum possible period for any choice of nonzero initial state. Easy Question: What is the number of primitive linear recurrence sequences of order n over F q? φ(q n 1) Answer:, since the LFSR (1) is primitive if and only if n its characteristic polynomial, viz., f (X ) = X n c n 1 X n 1 c 1 X c 0 is primitive

12 Vector recurrences, aka σ-lfsrs Linear recurrences or LFSRs of order n over F q admit an important and useful generalization that goes back to Niederreiter ( ), where (1) is replaced by the vector recurrence or σ-lfsr: s i+n = s i C 0 + s i+1 C s i+n 1 C n 1 (2) where C 0, C 1,..., C n 1 M m (F q ) and the initial state (s 0,..., s n 1 ) is in (F m q ) n. As before, the resulting infinite sequence s = (s 0, s 1,... ) is always ultimately periodic, and is periodic if C 0 is nonsingular. One has similar notions of primitivity for these vector recurrences.

13 Vector recurrences, aka σ-lfsrs Linear recurrences or LFSRs of order n over F q admit an important and useful generalization that goes back to Niederreiter ( ), where (1) is replaced by the vector recurrence or σ-lfsr: s i+n = s i C 0 + s i+1 C s i+n 1 C n 1 (2) where C 0, C 1,..., C n 1 M m (F q ) and the initial state (s 0,..., s n 1 ) is in (F m q ) n. As before, the resulting infinite sequence s = (s 0, s 1,... ) is always ultimately periodic, and is periodic if C 0 is nonsingular. One has similar notions of primitivity for these vector recurrences. Problem: Determine the number of primitive vector recurrences of order n over F m q (or equivalently, over F q m).

14 Vector recurrences, aka σ-lfsrs Linear recurrences or LFSRs of order n over F q admit an important and useful generalization that goes back to Niederreiter ( ), where (1) is replaced by the vector recurrence or σ-lfsr: s i+n = s i C 0 + s i+1 C s i+n 1 C n 1 (2) where C 0, C 1,..., C n 1 M m (F q ) and the initial state (s 0,..., s n 1 ) is in (F m q ) n. As before, the resulting infinite sequence s = (s 0, s 1,... ) is always ultimately periodic, and is periodic if C 0 is nonsingular. One has similar notions of primitivity for these vector recurrences. Problem: Determine the number of primitive vector recurrences of order n over F m q (or equivalently, over F q m). This is open, in general. One difficulty is that the characteristic polynomial of (2) is X n C n 1 X n 1 C 1 X C 0 M m (F q )[X ] and this can not be realized as a minimal polynomial of any sort.

15 Connection with Group Theory, via Singer cycles Observe that GL n (F q ) is a finite group and GL n (F q ) =

16 Connection with Group Theory, via Singer cycles Observe that GL n (F q ) is a finite group and GL n (F q ) = (q n 1)(q n q) (q n q n 1). Question: What is the maximum possible order of an element of GL n (F q ) and how many elements are of the maximum order?

17 Connection with Group Theory, via Singer cycles Observe that GL n (F q ) is a finite group and GL n (F q ) = (q n 1)(q n q) (q n q n 1). Question: What is the maximum possible order of an element of GL n (F q ) and how many elements are of the maximum order? Answer. max {o(a) : A GL n (F q )} = q n 1. Definition: A Singer cycle in GL n (F q ) is an element of maximum possible order in GL n (F q ).

18 Connection with Group Theory, via Singer cycles Observe that GL n (F q ) is a finite group and GL n (F q ) = (q n 1)(q n q) (q n q n 1). Question: What is the maximum possible order of an element of GL n (F q ) and how many elements are of the maximum order? Answer. max {o(a) : A GL n (F q )} = q n 1. Definition: A Singer cycle in GL n (F q ) is an element of maximum possible order in GL n (F q ). Using results from group theory about the structure of Singer subgroups, we can show that the number of Singer cycles in GL n (F q ) is equal to φ(q n 1) (q n q)(q n q 2 ) (q n q n 1 ), n where φ denotes the Euler totient function.

19 Block Companion Singer Cycles By a (m, n)-block companion matrix over F q we mean T M mn (F q ) of the form C 0 I m C 1 T = , (3) I m 0 C n I m C n 1 where C 0, C 1,..., C n 1 M m (F q ) and I m denotes the m m identity matrix over F q, while 0 indicates the zero matrix in M m (F q ). If such a matrix T is a Singer cycle in GL mn (F q ), call it a (m, n)-block companion Singer cycle over F q. Problem: Determine the number of all (m, n)-block companion Singer cycles over F q. Remark: A solution to Niederreiter s splitting subspace question will also answer the last two problems.

20 Back to Splitting Subspaces Let m, n be positive integers and σ F q mn. Recall that an m-dimensional subspace W of F q mn is σ-splitting if F q mn = W σw σ n 1 W. and that S(σ, m, n; q) denotes the number of m-dimensional σ-splitting subspaces of F q mn. Easy Examples: Suppose σ F q mn is such that F q mn = F q (σ). Note that if n = 1, then S(σ, m, n; q) = 1, whereas if m = 1, then S(σ, m, n; q) = (q n 1)/(q 1). In other words, S(σ, m, n; q) = qmn 1 q m 1 if min{m, n} = 1. Exercise: Determine S(σ, m, n; q) when m = n = 2. In general, it is an open problem to determine S(σ, m, n; q) for arbitrary m and n.

21 Splitting Subspace Conjecture Motivated partly by the work of Zeng, Han and He (2008) on binary σ-lfsrs, we were lead to the following quantitative formulation of Niederreiter s Question: Splitting Subspace Conjecture: For any σ such that F q mn = F q (σ), S(σ, m, n; q) = qmn 1 q m 1 qm(m 1)(n 1). Note that this is true when min{m, n} = 1. The best known general result seems to be the following. Theorem The Splitting Subspace Conjecture holds in the affirmative if m = 2. Remark: q m(m 1) is the number of m m nilpotent matrices over F q, thanks to an old result of Fine and Herstein (1958).

22 Pointed Splitting Subspaces Definition By a pointed σ-splitting subspace of dimension m we shall mean a pair (W, x) where W is an m-dimensional σ-splitting subspace of F q mn and x W. The element x may be referred to as the base point of (W, x). Pointed Splitting Subspace Conjecture: For any x F q mn and any σ such that F q mn = F q (σ), the number of m-dimensional pointed σ-splitting subspaces of F q mn with base point x is q m(m 1)(n 1). We have: SSC PSSC, i.e., the two conjectures are equivalent. Remark: The PSSC is equivalent to the assertion that there is a one-to-one correspondence between pointed splitting subspaces and pointed n-tuples of m m nilpotent matrices over F q. One of the key ingredients in the proof in the m = 2 case is a result closely related to the question about the probability for a pair of polynomials in F q [X ] to be relatively prime.

23 Bounds and Asymptotics For any σ F q mn, define N(σ, m, n; q) := S(σ, m, n; q) m 1 i=0 (qm q i ). Theorem Let σ F q mn (q 2)q mn + 1 (q 1) Corollary be such that F q mn = F q (σ). m 1 q mn(m 1) N(σ, m, n; q) (q mn q i ). Let σ F q mn be such that F q mn = F q (σ) and let x F qmn. Then the number of m-dimensional pointed σ-splitting subspaces of F q mn with base point x is is asymptotically equivalent to q m(m 1)(n 1) as q. i=0

24 A Plethora of Conjectures and the Interrelations Besides the Splitting Subspace Conjecture (SSC) and the Pointed Splitting Subspace Conjecture (PSSC) stated earlier, one has also the following quantitative versions of the 2 problems stated earlier. 1. Primitive Vector Recurrence Conjecture (PVRC) The number of primitive vector recurrences of order n over F m q is φ(q mn 1) mn q m(m 1)(n 1) m 1 (q m q i ). (4) In the binary case (q = 2), this conjecture is due to Zeng, Han and He (2008). 2. Block Companion Singer Cycle Conjecture (BCSCC) The number of (m, n)-block companion Singer cycles over F q is also given by the formula (4) above. Analyzing the fibers of the characteristic map that sends a matrix to its characterisitic polynomials leads to the following refined versions of the last conjecture. i=1

25 A Plethora of Conjectures and the Interrelations 3. Primitive Fiber Conjecture (PFC) For any primitive polynomial f in F q [X ] of degree mn, the number of (m, n)-block companion Singer cycles over F q having f as its characteristic polynomial is q m(m 1)(n 1) m 1 (q m q i ). (5) 4. Irreducible Fiber Conjecture (IFC) For any monic irreducible polynomial f in F q [X ] of degree mn, the number of (m, n)-block companion Singer cycles over F q having f as its characteristic polynomial is given by (5) above. And here is a summary of the relation between these conjectures. i=1 PSSC SSC IFC = PFC = BCSCC PVRC

26 Coprime Polynomial Pairs Thanks to Knuth [ACP-2 ( 4.6.1, Ex. 5), 1969] and more recently, Corteel, Savage, Wilf, Zeilberger (and Zagier) [JCT-A, 1998], we know that the probability that two monic polynomials of degree n 1 over F q, chosen independently and uniformly at random, are relatively prime is 1 1 q. A bijective explanation was given by Reifegerste (2000) in the case q = 2. The case of arbitrary q was explained by Benjamin and Bennett (2007) by constructing an explicit surjective map {(f, g) : f, g F q [X ] monic, deg n, coprime} {(f, g) : f, g monic, deg n, non-coprime} such that the cardinality of each fiber is q 1. We remark that a consequence of the main result of Benjamin and Bennett and its proof was used in our original solution of the Splitting Subspace Conjecture in the case m = 2.

27 Side Remark: Connection with Riemann Zeta Function It is an elementary and well-known that the probability of two integers to be relatively prime is 1 ζ(2) = 6 π 2.

28 Side Remark: Connection with Riemann Zeta Function It is an elementary and well-known that the probability of two integers to be relatively prime is 1 ζ(2) = 6 π 2. To note that the result for polynomials fits in with this, recall that the Weil and Riemann zeta functions of any variety V over F q are ( ) Z V (T ) := exp V (F q n) T n ( and ζ V (s) := Z V q s ). n n=1

29 Side Remark: Connection with Riemann Zeta Function It is an elementary and well-known that the probability of two integers to be relatively prime is 1 ζ(2) = 6 π 2. To note that the result for polynomials fits in with this, recall that the Weil and Riemann zeta functions of any variety V over F q are ( ) Z V (T ) := exp V (F q n) T n ( and ζ V (s) := Z V q s ). n n=1 Now F q [X ] A 1 F q Z A 1(T ) = exp = ζ A 1(2) = and ( n=1 1 1 q(q 2 ) ) q n T n = exp ( log(1 qt ) 1) = n and 1 ζ A 1(2) = 1 1 q. 1 1 qt

30 Nonsingular Hankel Matrices Recall: M = (m ij ) is a Toeplitz matrix (resp: Hankel matrix) if m ij = m rs whenever i j = r s (resp: i + j = r + s). Daykin [Crelle, 1960] was perhaps the first to prove that the number of n n nonsingular Hankel matrices over F q is q 2n 2 (q 1). In particular, the probability that a Hankel matrix over F q is nonsingular is 1 1 q. [More generally, Daykin has formulas for the number of m n Hankel matrices over F q of a given rank r min{m, n}. ] Recently, Daykin s formula has been (re)proved [using subresultants and such] by Kaltofen and Lobo (1996) in the context of counting n n nonsingular Toeplitz matrices over F q. A simpler and more direct proof can be found in our paper in JCT-A 118 (2011), The case of Hankel matrices over F q of a given rank is also discussed there.

31 Coprime Polynomial Pairs and Nonsingular Hankel Matrices Curious Fact: The probability for a polynomial pair to be coprime and for a Hankel matrix to be nonsingular is the same! In fact, ( CPP n (F q ) = q 2n 1 1 ) = q 2n 1 (q 1), q and ( HGL n (F q ) = q 2n ) = q 2n 2 (q 1), q where CPP n (F q ) denotes the set of all ordered pairs of coprime monic polynomials over F q of degree n and HGL n (F q ) denotes the set of all n n nonsingular Hankel matrices with entries in F q.

32 Coprime Polynomial Pairs and Nonsingular Hankel Matrices Curious Fact: The probability for a polynomial pair to be coprime and for a Hankel matrix to be nonsingular is the same! In fact, ( CPP n (F q ) = q 2n 1 1 ) = q 2n 1 (q 1), q and ( HGL n (F q ) = q 2n ) = q 2n 2 (q 1), q where CPP n (F q ) denotes the set of all ordered pairs of coprime monic polynomials over F q of degree n and HGL n (F q ) denotes the set of all n n nonsingular Hankel matrices with entries in F q. Question: Can we explain this fact? In other words, can we give a combinatorial proof that CPP n (F q ) = q HGL n (F q )?

33 An explicit surjection Let F be any field and n be any positive integer. Theorem There is a surjective map σ : CPP n (F ) HGL n (F ) such that for any A HGL n (F ), the fiber σ 1 ({A}) is in one-to-one correspondence with F. In particular, CPP n (F q ) = q HGL n (F q ).

34 An explicit surjection Let F be any field and n be any positive integer. Theorem There is a surjective map σ : CPP n (F ) HGL n (F ) such that for any A HGL n (F ), the fiber σ 1 ({A}) is in one-to-one correspondence with F. In particular, CPP n (F q ) = q HGL n (F q ). Sketch of Proof: A key idea is to consider the Bezoutian. Recall that the Bezoutian (matrix) of u, v F [X ] of degree n is the n n matrix B n (u, v) = (b ij ) determined by the equation u(x )v(y ) v(x )u(y ) X Y n = b ij X i 1 Y j 1. i,j=1 Fact: Assume that deg u = n and deg v n. Then B n (u, v) is nonsingular u and v are coprime

35 Illustration of the Bezoutian Example: Suppose u(x ) = u 0 + u 1 X + + u n X n with u 0, u 1,..., u n F and v(x ) = 1. Then u(x ) u(y ) X Y = n k=1 u k X k Y k X Y = = n u k k=1 i=1 k X i 1 Y k i n u i+j 1 X i 1 Y j 1, i,j=1 where, by convention, u k := 0 for k > n. Thus u 1 u 2 u n 1 u n u 2 u 3 u n 0 B n (u, 1) =..... u n 1 u n 0 0 u n In particular, if deg u = n, i.e., if u n 0, then u and v are coprime, and moreover B n (u, v) is nonsingular.

36 Padé pairs and Hermite pairs Define the set of Padé pairs by P n (F ) := { (u, v) F [X ] 2 : u monic, deg u = n, and deg v < n }, and the set of Hermite pairs by HP n (F ) := {(u, v) P n (F ) : u and v are coprime}. Observe that CPP n (F ) is in bijection with HP n (F ) [Proof: The map given by (f, g) (f, g f ) does the job.]. Next, observe that for any (u, v) P n (F ), there are unique a i F, i 1, such that v(x ) u(x ) = a i X i. i=1 Define H n (u, v) to be the n n Hankel matrix (a i+j 1 ).

37 Completion of the proof Lemma (Barnett factorization) For any (u, v) P n (F ), B n (u, v) = B n (u, 1)H n (u, v)b n (u, 1). Thanks to the above results, (u, v) H n (u, v) gives a map HP n (F ) HGL n (F ). With some effort, we can show that it is surjective and that the fibre is in bijection with F. Combining this with the bijection CPP n (F ) HP n (F ), we obtain the desired result.

38 Completion of the proof Lemma (Barnett factorization) For any (u, v) P n (F ), B n (u, v) = B n (u, 1)H n (u, v)b n (u, 1). Thanks to the above results, (u, v) H n (u, v) gives a map HP n (F ) HGL n (F ). With some effort, we can show that it is surjective and that the fibre is in bijection with F. Combining this with the bijection CPP n (F ) HP n (F ), we obtain the desired result. Remark: One can replace HGL n (F ) by the set TGL n (F ) of n n nonsingular Toeplitz matrices over F q. Indeed if E denotes the n n matrix with 1 on the antidiagonal and 0 elsewhere, then E is nonsingular and the map given by A AE gives a bijection between TGL n (F ) and HGL n (F ).

39 References and Samrith Ram, Enumeration of splitting subspaces over finite fields, Contemp. Math., Vol. 574 (2012), Amer. Math. Society, Providence, to appear. Mario Garcia-Armas, and Samrith Ram. Relatively prime polynomials and nonsingular Hankel matrices over finite fields, Journal of Combinatorial Theory, Series A, Vol. 118, No. 3 (2011), pp and Samrith Ram, Block companion Singer cycles, primitive recursive vector sequences, and coprime polynomial pairs over finite fields, Finite Fields and Their Applications Vol. 17, No. 5 (2011), pp , Sartaj Ul Hasan and Meena Kumari, Primitive polynomials, Singer cycles, and word-oriented linear feedback shift registers, Designs, Codes and Cryptography, Vol. 58, No. 2 (2011), pp Available: srg/papers.html

40 Thank you for your attention!

Journal of Combinatorial Theory, Series A

Journal of Combinatorial Theory, Series A 118 (2011) 819 828 Contents lists available at ScienceDirect Journal of Combinatorial Theory, Series A wwwelseviercom/locate/jcta Relatively prime polynomials