Resolvability of a cyclic orbit of a subset of Z v and spread decomposition of Singer cycles of projective lines

Transcription

1 ACA2015, Augst 26-30, 2015, Michigan Tech. Univ. Resolvability of a cyclic orbit of a subset of Z v and spread decomposition of Singer cycles of projective lines Masakazu Jimbo (Chubu University) jimbo@isc.chubu.ac.jp & Miwako Mishima (Gifu University) & Koji Momihara (Kumamoto University)

2 Resolvability of a cyclic orbit v, k: positive integers such that k v Z v : a residue ring of integers modulo v. For a subset (block) B Z v of size k, Orb Zv (B) = {B + i i Z v } is called a cyclic orbit of B. Orb Zv (B) is said to be full if Orb Zv (B) = v, otherwize short ??? v-1 B B+1 B+2 Find a resolu on class from a cyclic orbit. A subfamily R Orb Zv (B) satisfying E R E = Z v is called a resolution class. If Orb Zv (B) is decomposed into disjoint resolution classes then Orb Zv (B) is said to be resolvable.

3 For any A, B Z v, let A + B = {a + b a A, b B}, which is considered as a mulitiset. If Orb Zv (B) has a resolution class, then there is a subset C Z v satisfying B B+1 B+2 B + C = Z v. (1) c 1 B+c 1 a resolu on class R={B, B+c 1, B+c 2, B+c 3 } If (1) holds, we say that (B, C) is a factorization of Z v. Note that if (B, C) is a factorization of Z v, then (B + x, C + y) is also a factorization for x, y Z v. c 2 c 3 v-1 B+c 2 B+c 3 Let C={0, c 1, c 2, c 3 }, then B+C=Z v

4 Lemma 1 If Orb Zv (B) has a resolution class then Orb Zv (B) is resolvable. Proof: A resolution class is represented as R = c C (B + c) with C Z v. Take R + b 1, R + b 2,... as resolution classes for b i B. What is the condition for B that Orb Zv (B) is resolvable?

5 A subset A Z v is said to be periodic if its stabilizer N = {g Z v A + g = A} is a nontrivial subgroup of Zv. Theorem 1 (see, Szabo and Sands (2009)) In the case when k = B is a power of a prime, if B + C = Z v, then B or C is periodic. Hereafter, we consider only the case when k = B is a power of a prime.

6 The structure of a factor (B, C) of Z v Theorem 2 For a factorization (B, C) of Z v, assume that k = B is a power of a prime and 0 B, 0 C. Then, B and C are represented by B = N 1 + +N t 1 +B t, C = M 0 +M 1 + +M t 1 +C t, where n 0 = v, N 0 = {0}, m i = n i N i, M i = {0 g < m i C + g(mod mi ) = C}, for i = 0, 1,... n i = m i 1 M i 1, N i = {0 g < n i B + g(mod ni ) = B}, for i = 1, 2,... and t = max i {i Ni {0}}, t = max i {i Mi {0}}, B t = N t (mod n t ), B t Z mt 1, M t = M t (mod m t ), M t Z nt.

7 Example 1 For v = 48, let B = {0, 9, 24, 33} and C = {0, 1, 2, 6, 7, 8, 12, 13, 14, 18, 19, 20}. Then, B+C = Z 48. B = N 1 + B 2 = {0, 24} + {0, 9}, C = M 1 + C 2 = {0, 6, 12, 18} + {0, 1, 2} Then {0, 24} is a subgroup of Z 48, {0, 6, 12, 18} is a subgroup of Z 24, {0, 9} {0, 3} is a subgroup of Z 6, and {0, 1, 2} is Z 3. Is this true for any k = B? Example 2 (Zsabo and Sands (2009)) For v = 72, let B = {0, 8, 16}+{0, 8} and C = ({0, 24, 66}+{0, 36}) ({0, 24, 48} + {0, 44} + {1}) Then, B + C = Z 72. But both of B and C are not periodic in Z 72. That is, a bad factorization!!

8 Minimum factorization theorem Theorem 3 For a prime p and integrs r, t, m satisfying r t and (m, p) = 1, let v = p t m. For a subset B Z v with B = k = p r, assume that there is a subset C Z v such that (B, C) is a factorization of Z v. Then e = min{n C, (B, C) is a factorization of Z n, n v}. is a divisor of p t. By this theorem, in order to check the resolvablity of a cyclic orbit Orb Zv (B) for B = p r, we have only to check whether Orb Zp (B) has a resolution class, or not i for each i t.

9 Application: Spread decomposition of a line orbit of PG(2n 1, q) For a prime power q, the lines of PG(2n 1, q) consist of a number of Singer cycles of length v = q2n 1 q 1 and a single short orbit. Among them, some Singer cycle of full length may be decomposed into spreads (resolution classes). We want to decide the number of Singer cycles which are decomposed into spreads. In this talk, we consider the cases of q = 3 and 4. In these cases the sizes of lines L are q + 1 = 4, 5, which are a prime or a prime power.

10 The case of q = 3: PG(2n 1, 3) Let g = GF(3 2n ) /GF(3). Then, g = v = 32n 1 2. A base line L of Orb g (L) is represented by L = {1 = q 0, x = g b 1, x + 1 = g b 2, x 1 = g b 3}, or simply by pow(l) = {0, b 1, b 2, b 3 } Z v.

11 A line orbit is decomposed into spreads. There is a subset C Z v such that pow(l) + C = Z v. There is a subset C Z 2 i such that pow(l) + C = Z 2 i for some i such that 2 i v. (by Theorem 3) There is a proper distribution of the elements of L = {1, x, x + 1, x 1} in cyclotomic cosets C (2i ) for some i such that 4 2 i v. j

12 What is the proper disribution Lemma 2 If a Singer cycle Orb g (L) of a line L = {1, x, x + 1, x 1} has a resolution then the four elements must fall in C (2i ) 0, C (2i ) j, C (2i ) 2 i 1, C(2i ) 2 i 1 +j, respectively. That is pow(l) (mod 2 i ) {0, j 1, j 2, j 3 } = {0, 2 i 1 } + {0, j} = {0, j, 2 i 1, 2 i 1 + j} for 1 j < 2 i 2. We call i the depth of cyclotomic cosets.

13 Counting the Singer cycles which is decomposed into spreads. Theorem 4 Let M i = {x x, (x + 1) 1 (x 1) C (2i ) 2 i 1 x + 1 / C(2i ) 0 }. Then, for any x M i, a Singer cycle of a line L = {1, x, x + 1, x 1} is decomposed into four spreads. And total # of spreads within Singer cyclces = 1 2 M i. We count M i by using a number theoretical technique.

14 Counting methods e: a divisor of v ζ e C: a primitive e-th root ψ(g h ) = ζe h: a multiplicative character of order e of F 3 2n Note that f (e)(z) = 1 C j e e 1 u=0 ζ uj e ψ u (z) = 1 if z C (e) j 0 otherwize M i ={x x, (x + 1) 1 (x 1) C (2i ) 2 i 1 x + 1 / C(2i 1 ) 0 } ={x x, (x + 1) 1 (x 1) C (2i ) 2 i 1} \ {x x, (x + 1) 1 (x 1) C (2i ) 2 i 1 x + 1 C(2i 1 ) 0 }

15 Representation of M i by Jacobi-like sum. Hence, where M i = f C (x)f (2i) 2 C (2 i ) ((x + 1) 1 (x 1)) x F 3 2n\F i 1 2 i 1 3 f C (x)f (2i) 2 C (2 i ) ((x + 1) 1 (x 1))f x F 3 2n\F i 1 2 C (2 i 1 ) i = 1 ( 1) u+v T 2 2i ψi (u, v, v) (u,v) Z 2 2 i 1 2 3i 1 T ψi (u, v, w) = (u,v,w) Z 2 2 i Z 2 i 1 ( 1) u+v T ψi (u, 2w v, v), x F 3 2n\F 3 ψ u i (x)ψ v i (x + 1)ψ w i (x 1). and ψ i is a multiplicative character of F q of order 2 i. (x + 1)

16 Evaluation of T ψi (u, v, w) for i 2 If i 2, then we can compute the exact values of T ψi (u, v, w) for any (u, v, w) Z 3 2 i since T ψi (u, v, w) can be reduced to computable Jacobi sums. Lemma 3 Let q be a power of 3. For any a, b Z m, q 3 if a = b = 0, T ψ (a, b, 0) = 1 ψ a ( 1) if a 0 and b = 0, ψ a ( 1)J(ψ a, ψ b ) ψ 2a+b ( 1) if a, b 0. Lemma 4 Let q be a power of 3, ψ i be a multiplicative character of F q of order 2 i and η be the quadratic character of F q. Then, for any u, v, w Z 4, T ψ2 (u, v, w) = { J(ψ3, η) + J(ψ3 5, η) if u, v and w are distinct, J(ψ3 i, ψj 2 ) + J(ψi+4 3, ψ j 2 ) if (u, v, w) {(i, j, j), (j, i, j), (j, j, i)}.

17 # of spreads for depth i 2 Theorem 5 The total number K of spreads obtained by decomposing every Singer cycle of lines in P G(2n 1, 3) depth at most i = 2 is { 1 K = n 2 3 n+1 + 2I(n) + 2 } if n is even, { 1 3 2n n + 6 I(n) + 1 } if n is odd, 64 where I(n) = (1 i 2) n + (1 + i 2) n, i = 1. If n is odd, v = 4m with m odd. Thus Theorem 5 gives the best decomposition in the sense that the total number of spreads within Singer cycles is maximum.

18 More decomposition for even n For even n, as for i 3, unlike the case i 2, not every T ψi (u, v, w) can be reduced to Jacobi sums, and thus we need to find another way to compute T ψi (u, v, w). Applying Theorem 5.39 in Finite Fields by Lidl and Neiderreiter, we have T ψi (u, v, w) = ψi u (x)ψi v (x + 1)ψi w (x 1) = ψ (l) i (f(x)) x F 3 2n\F 3 x F 3 2 i 1 l\f 3 = ω l 1 ω l 2 ψ (l) i (f(0)) ψ (l) (f(1)) ψ (l) (f( 1)), where (ω 1, ω 2 ) is a solution to the system of equations ψ i (c u (1 + c) v ( 1 + c) w ) = ω 1 ω 2, c F 3 2 i 1 ψ i (c u (1 + b + c) v (1 b + c) w ) = ω 1 ω 2. (b,c) F 2 3 2i 1 i i

19 # of speads for depth i 3 n Numbers spreads depth i in the column for i means the best decomposition in the sense that the total number of spreads is maximum.

20 The case of PG(2n 1, 4) Evaluation of M i leads us to a decomposition of the Singer cycles of lines in PG(2n 1, 4). Theorem 6 Let s be the highest power of 5 in n and r be an integer within 1 r s + 1. The Singer cycles in P G(2n 1, 4) can be decomposed into ( r ) K = 1 3 i=1 M i + δ spreads, where δ = 0 or 1 depending on if r = s + 1 or otherwise. To conut M i, instead of T ψi (u, v, w), we use Q ψi (u, v, w, z) = ψi u (x)ψi v (x + 1)ψi w (x + 2)ψi z (x + 4) x F 4 2n\F 4 where ψ i is a multiplicative character of order 5 i.

21 Size of M 1 for q = 4 By a similar manner to the case of q = 3, we obtain the following for M 1. M 1 =

22 Number of spreads by Theorem 6 for PG(2n 1, 4) Considering just M 1, the number of spreads obtained by our decomposition is calculated as follows. n Number of spreads in the column for n means the best decomposition in the sense that the total number of spreads is maximum.

23 Thank you.