Large Sets of q -Analogs of Designs Michael Braun, Michael Kiermaier, - PDF document

Large Sets of q -Analogs of Designs Michael Braun, Michael Kiermaier, Axel Kohnert ∗ , Reinhard Laue Universit¨ at Bayreuth, laue@uni-bayreuth.de Abstract Joining small Large Sets of t -designs to form large Large Sets of t -designs allows to recursively construct infinite series of t -designs. This concept is generalized from ordinary designs over sets to designs over finite vector spaces, i.e. designs over GF ( q ), using three types of joins. While there are only very few general constructions of such q -designs known so far, from only one large set in the literature and two new ones in this paper this way many infinite series of Large Sets of q -designs with constant block sizes are derived. Keywords: q -analog, t -design, Large Set, subspace design AMS classifications : Primary 51E20; Secondary 05B05, 05B25, 11Txx ∗ † 11.12.2013 1

1 Classic and Subspace t -designs t -( v, k, λ ) design D = ( V, B ) t -( v, k, λ ) q design D = ( V, B ) V point set size v V GF ( q )- vector space dim v � V � V � � B ⊆ B ⊆ k k q � V � V � � each T ∈ in λ B ∈ B each T ∈ q in λ B ∈ B t t Subspace designs Cameron 1974 [10]: Infinite series for t = 2: Thomas 1987 q = 2 [18] Suzuki q > 2 [14], [15], Itoh 1997 [12]. ∀ t ∃ simple t -subspace design: Fazelli,Lovett,Vardy 2013 [11], q -analog to Teirlinck’s theorem [16]: Computer search: t = 2 , 3 M.,S. Braun et al. 2005-2011 [4, 5, 9, 6], 2-(13 , 3 , 1) 2 Braun,Etzion, ¨ O sterg ˚ a rd,Wassermann,Vardy 2013 [7], Large Set: � V � LS q [ N ]( t, k, v ) partition of q into N disjoint t -( v, k, λ ) q k with λ = [ v − t k − t ] q N . LS 2 [3](2 , 3 , 8) Braun,Kohnert, ¨ O sterg ˚ a rd,Wassermann 2013[8] � V � Three disjoint 2-(8 , 3 , 21) 2 that partition 3 2 2

Table 1: Table of LS 2 [3](2 , k, v ) v S t a r t : ? 8 3 - ? 9 - - ? 10 - - - 11 - - - - 12 - - - - 13 3 ? 5 - 14 - ? ? - - 15 - - 5 - - - 16 - - - - - - 17 - - - - - - - 18 - - - - - - - 19 3 ? 5 ? ? ? 9 ? 20 - ? ? ? ? ? ? ? 21 - - ? ? ? ? ? ? ? 22 - - - ? ? ? ? ? ? 23 - - - - ? ? ? ? ? ? 24 - - - - - ? ? ? ? ? 25 3 ? 5 - - - ? ? 11 ? ? 26 - ? ? - - - - ? ? ? ? 27 - - 5 - - - - - 11 ? ? ? 28 - - - - - - - - - ? ? ? 29 - - - - - - - - - - ? ? ? 30 - - - - - - - - - - - ? ? 31 3 ? 5 - - - ? ? 11 - - - 15 ? 32 - ? ? - - - - ? ? - - - - ? 33 - - ? - - - - - ? - - - - - ? 34 - - - - - - - - - - - - - - - 35 - - - - - - - - - - - - - - - 36 - - - - - - - - - - - - - - - - 37 3 ? 5 ? ? ? 9 ? 11 ? ? ? 15 ? 17 - - 38 - ? ? ? ? ? ? ? ? ? ? ? ? ? ? - - 39 - - 5 ? ? ? ? ? 11 ? ? ? ? ? 17 - - - 40 3

Table 2: Table of LS q [2](2 , k, v ) , q = 3 , 5 v S t a r t ( N e w ) : 3 6 - 7 - - 8 - - 9 3 ? ? 10 - ? ? 11 - - ? ? 12 - - - ? 13 3 - - - 7 14 - - - - - 15 - - - - - - 16 - - - - - - 17 3 ? ? ? 7 ? ? 18 - ? ? ? ? ? ? 19 - - ? ? ? ? ? ? 20 - - - ? ? ? ? ? 21 3 - - - 7 ? ? ? ? 22 - - - - - ? ? ? ? 23 - - - - - - ? ? ? ? 24 - - - - - - - ? ? ? 25 3 ? ? ? 7 - - - ? ? ? 26 - ? ? ? ? - - - - ? ? 27 - - ? ? ? - - - - - ? ? 28 - - - ? ? - - - - - - ? 29 3 - - - 7 - - - ? - - - 15 30 - - - - - - - - - - - - - 31 - - - - - - - - - - - - - - 32 - - - - - - - - - - - - - - 33 3 ? ? ? 7 ? ? ? ? ? ? ? 15 ? ? 34 - ? ? ? ? ? ? ? ? ? ? ? ? ? ? 35 - - ? ? ? ? ? ? ? ? ? ? ? ? ? ? 36 - - - ? ? ? ? ? ? ? ? ? ? ? ? ? 37 3 - - - 7 ? ? ? ? ? ? ? 15 ? ? ? ? 38 - - - - - ? ? ? ? ? ? ? ? ? ? ? ? 39 - - - - - ? ? ? ? ? ? ? ? ? ? ? ? 40 4

Recursion : (classical strategy, Khosrovshahi, Ajoodani- Namini [1, 2, 3] Teirlinck [17]) Partition the problem, insert small Large Sets into parts to obtain large Large Sets. Notation: Part ( S ) set of partitions of set S , First partition � V � {B 1 , . . . , B m } ∈ Part ( ) k q Second partition Decompose each B i into a join of two components: 5

• Ordinary join at U ≤ V : k 2 V ✭ ✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭ U + K r r U r K ✭ ✭✭✭✭✭✭✭✭✭✭✭ r K 1 r k 1 ( U + K ) /U = K 2 { 0 } r K 1 ∗ U K 2 = { K : U ∩ K = K 1 , ( U + K ) /U = K 2 } � U � V/U � � For K 1 ⊆ q , K 2 ⊆ k 1 k 2 q K 1 ∗ U K 2 = ∪{ ( K 1 ∗ U K 2 ) : K 1 ∈ K 1 , K 2 ∈ K 2 } Partition by ordinary joins: q -Vandermonde: k 1 = k � v 1 + v 2 � � v 1 � � v 2 � � q ( v 1 − k 1 )( k − k 1 ) = · k k 1 k − k 1 q q q k 1 =0 Example 1.1. For v = 10 , k = 3 , v 1 = 6 , v 2 = 4 the formula reads as � 10 � � 6 � � 4 � � 6 � � 4 � � 6 � � 4 � � 6 � � 4 � = q 18 + q 10 + q 6 + q 0 . 3 0 3 1 2 2 1 3 0 q q q q q q q q q 6

• Avoiding join and Covering join V ✭✭✭✭✭✭✭✭✭✭✭✭✭✭ ✭ W r U 2 r � r � � W/U 2 = K 2 F � � ✭ ✭✭✭✭✭ U 1 � r ✭ ✭✭✭✭✭ K F � r r � � � � ✭ ✭✭✭✭✭ K 1 r K ¯ r � F { 0 } r Avoiding join: K 1 ∗ ¯ F K 2 = { K ¯ F : K ¯ F ∩ U 2 = K ¯ F ∩ U 1 = K 1 , U 2 + K ¯ F = W } . � U � V/U � � K 1 ⊆ q , K 2 ⊆ k 1 k 2 q K 1 ∗ ¯ F K 2 = ∪{ ( K 1 ∗ ¯ F K 2 ) : K 1 ∈ K 1 , K 2 ∈ K 2 } Covering join: K 1 ∗ F K 2 = { K F : K F ∩ U 1 = K 1 , U 1 + K F = U 2 + K F = W } . � U � V/U � � K 1 ⊆ q , K 2 ⊆ k 1 k 2 q K 1 ∗ F K 2 = ∪{ ( K 1 ∗ F K 2 ) : K 1 ∈ K 1 , K 2 ∈ K 2 } 7

V = V 0 > V 1 > . . . > V v = { 0 } each F i = V i − 1 /V i of dimension 1. v ≥ k + s . Partition by avoiding joins: k � v � � v − s − i � � s + i − 1 � � q ( v − k − s ) i = . k i k − i q q q i =0 Example 1.2. For v = 10 , k = 3 , s = 3 the formula reads as � 10 � � 3 � � 6 � � 4 � � 5 � � 5 � � 4 � � 6 � � 3 � = q 0 + q 4 + q 8 + q 12 . 3 0 3 1 2 2 1 3 0 q q q q q q q q q Partition by covering joins: k = k 1 + k 2 + 1 � i v − k 2 − 1 � v � � � v − i − 1 � � q ( i − k 1 )( k 2 +1) = . k k 1 k 2 q q q i = k 1 Example 1.3. For v = 10 , k 1 = 1 , k 2 = 1 the formula reads as � 10 � � 8 � � 2 � � 7 � � 3 � � 6 � + q 2 + q 4 = + 3 1 1 1 1 1 q q q q q q � 4 � � 5 � � 5 � � 4 � � 6 � � 3 � � 7 � � 2 � q 6 + q 8 + q 10 + q 12 . 1 1 1 1 1 1 1 1 q q q q q q q q 8

� V {B 1 , . . . , B m } ∈ Part ( � q ) is ( N, t ) partitionable: k ∀ i B i = B ( i ) ∪B ( i ) 1 ˙ ∪ . . . ˙ N � V � , ∀B ( i ) |{ B ∈ B ( i ) ∀ T ∈ : T ⊆ B }| = λ ( T, i ) , j j t q independent of j . � V If {B 1 , . . . , B m } ∈ Part ( � q ) is ( N, t ) partition- k able then the designs D j = ∪ m i =1 B ( i ) for j = 1 , . . . , N j form an LS q [ N ]( t, k, v ). 9

Theorem 1.1. Joining Partitions ∗ one of the 3 joins (either U 1 = U 2 or dim ( U 1 /U 2 ) = 1 ): � U 2 � � V/U 1 � {K 1 1 , . . . , K 1 N } ∈ Part ( ) ( N, t ) partition, M ⊆ k 1 k 2 q q ⇒ {K 1 1 ∗ M, . . . , K 1 = N ∗ M ) } is an ( N, t ) partition � U 2 � � V/U 1 � , {K 2 1 , . . . , K 2 M ⊆ N } ∈ Part ( ) ( N, t ) partition k 1 k 2 q q ⇒ { M ∗ K 2 1 , . . . , M ∗ K 2 = N ) } is an ( N, t ) partition � U 2 � {K 1 1 , . . . , K 1 N } ∈ Part ( ) ( N, t 1 ) partition, k 1 q � V/U 1 � {K 2 1 , . . . , K 2 N } ∈ Part ( ) ( N, t 2 ) partition, k 2 q Lat an N × N Latin Square. = ⇒ {∪ Lat ( r,s )= a K 1 r ∗K 2 s : a = 1 , . . . , N } is an ( N, t 1 + t 2 + 1) partition. Proof analog to the classical case. 10

Doubling the point set by Ordinary join: LS 2 [3](2 , 3 , 8) ↔ dual LS 2 [3](2 , 5 , 8) derived LS 2 [3](2 , 5 , 8) = LS 2 [3](1 , 4 , 7) residual LS 2 [3](2 , 3 , 8) = LS 2 [3](1 , 3 , 7) ——————————————– Point extension: LS 2 [3](1 , 4 , 8) Ordinary joins at U < V, dim ( U ) = 8 , dim ( V ) = 16: � V/U � { LS 2 [3](2 , 5 , 8) ∗ U 2 , 0 LS 2 [3](1 , 4 , 8) ∗ U LS 2 [3](0 , 1 , 8), � V/U � LS 2 [3](2 , 3 , 8) ∗ U 2 , 2 � U � 2 ∗ U LS 2 [3](2 , 3 , 8), 2 LS 2 [3](0 , 1 , 8) ∗ U LS 2 [3](1 , 4 , 8) � U � 2 ∗ U LS 2 [3](2 , 5 , 8) } . 0 ———————————————– LS 2 [ 3 ]( 2 , 5 , 16 ) dual to LS 2 [ 3 ]( 2 , 11 , 16 ) 11