Automorphisms of association schemes whose relations are of valency - PDF document

Automorphisms of association schemes whose relations are of valency one or three Mitsugu Hirasaka Department of Mathematics Pusan National University Joint Work with Jeong Rye Park June 2-5, 2014 Modern Trends in Algebraic Graph Theory at Villanova University, Villanova, Pennsilvania 1

Notations for association schemes X : a finite set ( X, { R i } d i =0 ) : an association scheme { A i } d i =0 : the set of adjacency matrices A 0 = I , A T i = A i ′ , A i A j = � d i =0 p k ij A k , k i = p 0 ii ′ : the valency of R i R i ( x ) = { y ∈ X | ( x, y ) ∈ R i } . X : a finite set ( X, S ) where S is a partition of X × X { σ s | s ∈ S } : the set of adjacency matrices σ 1 = I , σ ∗ s = σ s ∗ , σ s σ t = � u ∈ S a stu σ u , n s = a ss ∗ 1 : the valency of s , xs = { y ∈ X | ( x, y ) ∈ s } . 2

In this talk we assume that S # = S \ { 1 } , N ( S ) = { n s | s ∈ S # } , and we shall write s · t = � u ∈ S a stu u instead of σ s σ t = � u ∈ S a stu σ u , and st = { u ∈ S | a stu > 0 } . r : X × X → S , ( x, y ) �→ r ( x, y ) where ( x, y ) ∈ r ( x, y ). Aut( X, S ) = { ρ ∈ Sym ( X ) | ∀ x, y, r ( x, y ) = r ( x ρ , y ρ ) } . An orbital means an orbit of the induced action of G on X × X where G ≤ Sym ( X ). We say that ( X, S ) is regular if S is the set of orbitals of a regular group. We say that ( X, S ) is Frobenius if S is the set of orbitals of a Frobenius group. We say that ( X, S ) is schurian if S is the set of orbitals of a transitive group. 3

Known Results N ( S ) = { k } (called k -equivalenced). 1. N ( S ) = { 1 } . For example,         1 0 0 0 1 0 0 0 1     0 1 0 0 0 1 1 0 0  ,  ,           0 0 1 1 0 0 0 1 0     Theorem. ([Zieschang, 1996]) N ( S ) = { 1 } ⇐ ⇒ ( X, S ) is regular. 2. N ( S ) = { 2 } . For example,         1 0 0 0 0 0 1 0 0 1 0 0 1 1 0       0 1 0 0 0 1 0 1 0 0 0 0 0 1 1                           0 0 1 0 0 , 0 1 0 1 0 , 1 0 0 0 1             0 0 0 1 0 0 0 1 0 1 1 1 0 0 0                       0 0 0 0 1 1 0 0 1 0 0 1 1 0 0     Theorem. ([Muzychuk, Zieschang, 2008]) N ( S ) = { 2 } ⇒ ( X, S ) is Frobenius. 4

3. N ( S ) = { 3 } . For example, { Cay ( F 7 , { 0 } ) , Cay ( F 7 , { 1 , 2 , 4 } ) , Cay ( F 7 , { 2 , 5 , 6 } ) } Theorem. (see [H, Kyoung-Tark Kim, Jeong Rye Park, 2014]) N ( S ) = { 3 } ⇒ ( X, S ) is Frobenius. 4. N ( S ) = { 4 } . For example, H (2 , 3) : the Hamming scheme Theorem. ([Jeong Rye Park, 2014]) N ( S ) = { 4 } ⇒ Aut( X, S ) is transitive. 5. N ( S ) = { k } . For example, Cyclotomic schemes of valency k Theorem. (see [Muzychuk, Ponomarenko, 2011]) Suppose N ( S ) = { k } , ∀ s ∈ S # , � t ∈ S a s tt ∗ = k − 1. If | S | ≥ 4( k − 1) k 3 then ( X, S ) is Frobenius. 5

6. N ( S ) = { 1 , 2 } (called quasi-thin). For example, The set of orbitals of the action of G on G/H where | H | = 2. Lemma. ([H, Muzychuk, 2002]) Suppose N ( S ) = { 1 , 2 } . For x ∈ X , φ x : X → X, y �→ y ′ where yr ( x, y ) = { y, y ′ } is an element of Aut( X, S ). Remark. (see [Hanaki, Miyamoto], and [M.Klin, M.Muzychuk, C.Pech, A.Woldar, P.-H.Zieschang] ) (1) as16-173 is a non-schurian quasi-thin scheme. (2) as28-176 is a non-schurian quasi-thin scheme. Theorem. ([Muzychuk, Ponomarenko, 2012]) Suppose N ( S ) = { 1 , 2 } . If ( X, S ) is not schurian then ( X, S ) is a Kleinian scheme of index 4 or 7. 6

Our Main Focus N ( S ) = { 1 , 3 } . For example, The set of orbitals of the action of G on G/H where | H | = 3. Remark. ([Hanaki, Miyamoto]) (1) as27-473 is a non-schurian scheme; (2) as27-475 is a non-schurian scheme; (3) as27-477 is a non-schurian scheme. We wish to find a non-trivial automorphism with a fixed point. For the remainder of this talk we assume that ( X, S ) is an association scheme with N ( S ) = { 1 , 3 } . We define τ : S → Z ≥ 0 such that τ ( s ) = max { a ss ∗ u | u ∈ S # } for s ∈ S . (1) For s, t ∈ S , n s n t = � u ∈ S a stu n u ∈ { 1 , 3 , 9 } and lcm( n s , n t ) | a stu n u ; (2) τ ( s ) = 0 ⇐ ⇒ n s = 1 and S = τ − 1 (0) ∪ τ − 1 (1) ∪ τ − 1 (2) ∪ τ − 1 (3); (3) For s ∈ τ − 1 (1) ∪ τ − 1 (2) and t ∈ S there exists u ∈ S such that a tu ∗ s = 1; (4) For s ∈ S , τ ( s ) = τ ( s ∗ ). 7

Please watch a movie on the white board. Fix x ∈ X . We define Γ x to be { ρ ∈ Sym ( X ) | ∀ t ∈ S, xt is an orbit of � ρ � } . For ρ ∈ Γ x we define a binary relation ∼ ρ on S such that s ∼ ρ t if and only if r ( y, z ) = r ( y ρ , z ρ ) for each ( y, z ) ∈ xs × xt . Lemma 1. For ρ ∈ Γ x , ∼ ρ is the trivial equiva- lence relation if and only if ρ ∈ Aut( X, S ). Lemma 2. For s ∈ τ − 1 (1) ∪ τ − 1 (2) there exists ρ ∈ Γ x such that s ∼ ρ t for each t ∈ S and u ∼ ρ v for all u, v ∈ S with n u n v ≤ 3. Theorem. Suppose u ∼ ρ v , u ∼ ρ w , v ≁ ρ w and u ∈ τ − 1 (1) ∪ τ − 1 (2) for v, w ∈ S \ τ − 1 (0). Then we have the following: (1) ( X, S ) is not commutative; (2) For s ∈ uu ∗ \ { 1 } , τ ( s ) = 3. 8

Sketch of the Proof s ∈ S α s ¯ Define � � s ∈ S α s s, � s ∈ S β s s � := � β s n s . Key Lemma. For all a, b, c ∈ S \ τ − 1 (0) with b � = c and ab ∩ ac � = ∅ we have the following: (1) If τ ( a ) = τ ( b ) = 3 then cb ∗ ⊆ τ − 1 (0), (2) If τ ( a ) = 1 and b ∗ b = c ∗ c ⊆ τ − 1 (0) then c · b ∗ = 3 d for some d ∈ a ∗ a with τ ( d ) = 3. (Proof) Note that 0 < � a · b, a · c � = � a ∗ · a, c · b ∗ � = � a ∗ · a − 3 · 1 , c · b ∗ � . (1) τ ( a ∗ ) = τ ( a ) = 3 implies that c = t · b for some t ∈ a ∗ a ∩ τ − 1 (0). Since τ ( b ) = 3, cb ∗ = ( tb ) b ∗ ⊆ τ − 1 (0). (2) By the assumption, � c · b ∗ , c · b ∗ � = � c ∗ · c, b ∗ · b � = 27. This implies that c · b ∗ = 3 d for some d ∈ a ∗ a with d ∈ τ − 1 (3). 9

Please watch a movie on the white board. Suppose that | u ∗ v | = | u ∗ w | = | v ∗ w | = 3, ( u ∗ v ∪ u ∗ w ∪ v ∗ w ) ∩ τ − 1 (0) = ∅ u ∗ v = { a 0 , a 1 , a 2 } , u ∗ w = { b 0 , b 1 , b 2 } , v ∗ w = { c 0 , c 1 , c 2 } . By the assumption on ∼ ρ , we have a ∗ i · b j = c 0 + c 1 + c 2 for all i, j ∈ { 0 , 1 , 2 } ; a i · c j = b 0 + b 1 + b 2 for all i, j ∈ { 0 , 1 , 2 } ; b i · c ∗ j = a 0 + a 1 + a 2 for all i, j ∈ { 0 , 1 , 2 } . s ∈ S α s ¯ Recall � � s ∈ S β s s � := � s ∈ S α s s, � β s n s . Lemma 4. We have � b i · b ∗ i , a j · a ∗ k � = � a ∗ j · b i , a ∗ k · b i � = 9, τ ( a i ) � = 2 and � b ∗ i · b i , a ∗ k · a j � = � b ∗ i · b i − 3 · 1 , a ∗ k · a j � if j � = k . Lemma 5. For s, t ∈ S \ τ − 1 (0), s · s ∗ = t · t ∗ ⇐ ⇒ � t ∗ · s, t ∗ · s � > 9. 10

Lemma 6. We have a 1 · a ∗ 1 = a 2 · a ∗ 2 = a 3 · a ∗ 3 , b 1 · b ∗ 1 = b 2 · b ∗ 2 = b 3 · b ∗ 3 , c 1 · c ∗ 1 = c 2 · c ∗ 2 = c 3 · c ∗ 3 , a ∗ 1 · a 1 = a ∗ 2 · a 2 = a ∗ 3 · a 3 , b ∗ 1 · b 1 = b ∗ 2 · b 2 = b ∗ 3 · b 3 , c ∗ 1 · c 1 = c ∗ 2 · c 2 = c ∗ 3 · c 3 , (Proof for the first case) Recall b ∗ i · a ∗ j = c 0 + c 1 + c 2 for all i, j ∈ { 0 , 1 , 2 } ; τ ( b ∗ i ) � = 2 since � b ∗ i · b i − 3 · 1 , a ∗ k · a j � = 9. i · b i = 3 · 1+ d + d ∗ for some d ∈ If τ ( b ∗ i ) = 1, then b ∗ S \ τ − 1 (0). Since a ∗ k · a j ∈ { d +2 d ∗ , 2 d + d ∗ , 3 d, 3 d ∗ } for all distinct j , k , we have � a ∗ k · a j , a ∗ k · a i � > 9 for each case. Therefore, Lemma 6 follows from Lemma 5. i · b i = 3 · 1 + 3 t + 3 t ∗ for some If τ ( b ∗ j ) = 3, then b ∗ t ∈ τ − 1 (0), and hence a j ∈ { a j · t, a k · t ∗ } , implying j = ( a k · t ) · ( t ∗ · a k ) = a k · a ∗ a j · a ∗ k . 11

Lemma 7. We have ( τ ( a i ) , τ ( b i )) ∈ { (1 , 3) , (3 , 1) } , (Proof) By Lemma 4, 2 / ∈ { τ ( a i ) , τ ( b i ) , τ ( b i ) } . Thus, it suffices to exclude the cases ( τ ( a i ) , τ ( b i )) ∈ { (1 , 1) , (3 , 3) } . Suppose ( τ ( a i ) , τ ( b i )) = (1 , 1), Then a j · a ∗ k ∈ { e + 2 e ∗ , 2 e + e ∗ } where b i · b ∗ i = 3 · 1 + e + e ∗ . This implies that � a i · a ∗ j , a i · a ∗ k � = � a ∗ i · a i , a ∗ k · a j � ≥ 12, which contradicts τ ( a ∗ i ) = τ ( a i ) = 1. Suppose ( τ ( a i ) , τ ( b i )) = (3 , 3), Then t ∈ a j a ∗ k where b i · b ∗ i = 3 · 1 + 3 t + 3 t ∗ . Since τ ( a j ) = 3 and a j = t · a k , a j a ∗ k ⊆ τ − 1 (0). Since u ∗ u ∩ a j a ∗ k � = ∅ , it contradicts τ ( u ) ∈ { 1 , 2 } . Without loss of generality we may assume that ( τ ( a i ) , τ ( b i )) = (1 , 3). By Key Lemma, τ ( d ) = 3 where d ∈ u ∗ u ∩ b 1 b ∗ 2 . Thus, the first second statement of Theorem is proved. Suppose that ( X, S ) is commutative. Then d ∈ u ∗ u ∩ b 1 b ∗ 2 = b ∗ 2 b 1 ⊆ τ − 1 (0) since τ ( b i ) = τ ( b ∗ i ) = 3, a contradiction. This completes the theorem. 12