Forks, finitely related clones, and finitely generated varieties - PowerPoint PPT Presentation

Forks, finitely related clones, and finitely generated varieties Erhard Aichinger Department of Algebra Johannes Kepler University Linz, Austria 53 rd SSAOS, Srní, September 2015 Supported by the Austrian Science Fund (FWF) in P24077

Results

Outline In these lectures, we will present the proofs of: Theorem (2009) [AMM14] Every clone with edge operation on a finite set is finitely related. Theorem (2014) [AM14] Every subvariety of a finitely generated variety with edge term is finitely generated.

Classic Clone Theory

Clones Operations O ( A ) := � | f : A k → A } . | k ∈ N { f | Clones A subset C of O ( A ) is a clone on A if � � 1. ∀ k , i ∈ N with i ≤ k : ( x 1 , . . . , x k ) �→ x i ∈ C , 2. ∀ n ∈ N , m ∈ N , f ∈ C [ n ] , g 1 , . . . , g n ∈ C [ m ] : f ( g 1 , . . . , g n ) ∈ C [ m ] . C [ n ] . . . the n -ary functions in C , C [ n ] ⊆ A A n .

Relational Description of Clones Definition I a finite set, ρ ⊆ A I , f : A n → A . f preserves ρ ( f ⊲ ρ ) if ∀ v 1 , . . . , v n ∈ ρ : � f ( v 1 ( i ) , . . . , v n ( i )) | | | i ∈ I � ∈ ρ. In other words: f A I ( v 1 , . . . , v n ) ∈ ρ. Remark ⇒ ρ is a subuniverse of ( A , f ) I . f ⊲ ρ ⇐ Definition (Polymorphisms) Let R be a set of finitary relations on A , ρ ∈ R . Pol ( { ρ } ) { f ∈ O ( A ) | | | f ⊲ ρ } , := � Pol ( R ) := Pol ( { ρ } ) . ρ ∈ R

Relational Description of Clones Theorem Let R be a set of finitary relations on A , and let ρ 1 , ρ 2 ∈ R with ρ 1 � = ∅ , ρ 2 � = ∅ . Then 1. Pol ( R ) is a clone. 2. Pol ( { ρ 1 , ρ 2 } ) = Pol ( { ρ 1 × ρ 2 } ) .

Every clone can be described by relations Theorem (see [PK79]) Let C be a clone on the finite set A . Then there is a set R of finitary relations such that C = Pol ( R ) . Proof: ◮ Observe C [ n ] ⊆ A A n . ◮ Take I n := A n , ρ n := C [ n ] . Then ρ n ⊆ A I n . ◮ Set R := { ρ 1 , ρ 2 , . . . , . . . } . ◮ Prove C ⊆ Pol ( R ) : f ∈ C [ n ] , g 1 , . . . , g n ∈ ρ m implies f ( g 1 , . . . , g n ) ∈ ρ m by the closure properties of clones. ◮ Prove Pol ( R ) ⊆ C : Let f : A n → A in Pol ( R ) . Then f ⊲ ρ n , hence f ( π 1 , . . . , π n ) ∈ ρ n , thus f ∈ C [ n ] .

Finitely related clones vs. DCC Definition A clone C is finitely related if there is a finite set of finitary relations R with C = Pol ( R ) . Theorem [PK79, 4.1.3] Let C be a clone on the finite set A . TFAE: 1. C is not finitely related. 2. There is a strictly decreasing sequence C 1 ⊃ C 2 ⊃ C 3 ⊃ · · · with C = � i ∈ N C i .

Finitely related clones vs. DCC Theorem Let C be a clone on the finite set A . TFAE: 1. C is not finitely related. 2. There is a strictly decreasing sequence C 1 ⊃ C 2 ⊃ C 3 ⊃ · · · with C = � i ∈ N C i . Proof of (1) ⇒ (2): ◮ We know C = Pol ( { ρ 1 , ρ 2 , . . . } ) . ◮ Hence Pol ( { ρ 1 } ) ⊇ Pol ( { ρ 1 , ρ 2 } ) ⊇ Pol ( { ρ 1 , ρ 2 , ρ 3 } ) ⊇ · · · . Proof of (2) ⇒ (1): ◮ Suppose C = Pol ( { ρ } ) , | ρ | = N . [ N ] = C [ N ] . ◮ Then for some n ∈ N , C n ◮ We show ∀ f ∈ C n : f ⊲ ρ on the next slide. ◮ Then C n ⊆ Pol ( { ρ } ) = C ⊆ C n + 1 , a contradiction.

Finitely related clones vs. DCC Proof of (2) ⇒ (1) (continued): ◮ Assumptions: C = Pol ( { ρ } ) , ρ = { b 1 , . . . , b N } , [ N ] = C [ N ] . C n ◮ We want to show: ∀ f ∈ C n : f ⊲ ρ . ◮ To this end, let f ∈ C n , r -ary, and let a 1 , . . . , a r ∈ ρ . ◮ Goal: f ( a 1 , . . . , a r ) ∈ ρ . ◮ We have f ( a 1 , . . . , a r ) = f ( b i ( 1 ) , . . . , b i ( r ) ) with i ( k ) ∈ { 1 , . . . , N } for all k ∈ { 1 , . . . , r } . ◮ Define g ( y 1 , . . . , y N ) := f ( y i ( 1 ) , . . . , y i ( r ) ) for all y ∈ A N . ◮ Then f ( b i ( 1 ) , . . . , b i ( r ) ) = g ( b 1 , . . . , b N ) . ◮ Now g ∈ C [ N ] n , hence g ∈ C [ N ] . Thus g ( b 1 , . . . , b N ) ∈ ρ .

How to establish “finitely related” Theorem Let M be a clone on A . If ( { C | C clone on A , M ⊆ C } , ⊆ ) satisfies the DCC, then every clone containing M is finitely related. Definition ( X , ≤ ) has the DCC : ⇔ there is no ( x i ) i ∈ N with x 1 > x 2 > x 3 > · · · . Theorem ( X , ≤ ) has the DCC ⇔ Every nonempty subset Y of X has a minimal element.

Forks

Groups Let G be a group, n ∈ N . Goal: represent subgroups of G n . The following lemma will motivate the definition of forks and the formulation of the fork lemma. Lemma Let G be a group, n ∈ N , A ≤ B ≤ G n subgroups. Assume 1. A ⊆ B 2. ∀ i ∈ { 1 , . . . , n } , ∀ g ∈ G , ∀ r i + 1 , . . . , r n ∈ G : ( 0 , . . . , 0 , g , r i + 1 , . . . , r n ) ∈ B ⇒ � �� � i − 1 ∃ s i + 1 , . . . , s n ∈ G : ( 0 , . . . , 0 , g , s i + 1 , . . . , s n ) ∈ A . Then A = B .

Mal’cev algebras I A is a Mal’cev algebra ⇔ ∃ d ∈ Clo 3 A ∀ a , b ∈ A : d ( a , a , b ) = d ( b , a , a ) = b . Definition of Forks Let A be an algebra, let m ∈ N , and let F be a subuniverse of A m . For i ∈ { 1 , . . . , m } , we define the relation ϕ i ( F ) on A by | ϕ i ( F ):= { ( a i , b i ) | | ( a 1 , . . . , a m ) ∈ F , ( b 1 , . . . , b m ) ∈ F , ( a 1 , . . . , a i − 1 ) = ( b 1 , . . . , b i − 1 ) } . If ( c , d ) ∈ ϕ i ( F ) , we call ( c , d ) a fork of F at i . If u = ( a 1 , . . . , a i − 1 , c , a i + 1 , . . . , a m ) ∈ F and v = ( a 1 , . . . , a i − 1 , d , b i + 1 , . . . , b m ) ∈ F , then ( u , v ) is a witness of the fork ( c , d ) at i .

Mal’cev algebras II Forks have not been called forks, but are used, e.g., in: [BD06, p.21], [BIM + 10], [Aic00, p.110]

The fork lemma Lemma (cf. [BIM + 10, Cor. 3.9], [Aic10, Lemma 3.1]) Let A be an algebra with Mal’cev term d , and let m ∈ N . Let F , G be subuniverses of A m with F ⊆ G . We assume ∀ i ∈ { 1 , . . . , m } : ϕ i ( G ) ⊆ ϕ i ( F ) . Then F = G . Proof: ◮ For each k ∈ { 1 , . . . , m } , let | F k := { ( f 1 , . . . , f k ) | | ( f 1 , . . . , f m ) ∈ F } { ( g 1 , . . . , g k ) | | | ( g 1 , . . . , g m ) ∈ G } . G k := ◮ We prove ∀ k ∈ { 1 , . . . , m } : G k ⊆ F k . ◮ k = 1: �

The fork lemma ◮ k ≥ 2: Let ( g 1 , . . . , g k ) ∈ G k . ◮ Then ( g 1 , . . . , g k − 1 ) ∈ G k − 1 . ◮ By the induction hypothesis, ( g 1 , . . . , g k − 1 ) ∈ F k − 1 . ◮ Hence ∃ f k : ( g 1 , . . . , g k − 1 , f k ) ∈ F k . ◮ Since ( f k , g k ) ∈ ϕ k ( G ) , we have ( f k , g k ) ∈ ϕ k ( F ) . ◮ Thus ∃ : a 1 , . . . , a k − 1 ∈ A such that ( a 1 , . . . , a k − 1 , f k ) ∈ F k ( a 1 , . . . , a k − 1 , g k ) ∈ F k . ◮ By Mal’cev: ( g 1 , . . . , g k ) ∈ F k .

Limitation of forks Fact Let A be an algebra, α ∈ Aut ( A ) . Then | B = { ( a , a ) | | a ∈ A } C = { ( a , α ( a )) | | | a ∈ A } have the same forks. Fact [BD06, BIM + 10] (Forks + one witness per fork) represent subalgebras if we have a Mal’cev term. Can be modified to edge terms.

Representing Clones By Forks

Representing clones by forks Let C := Clo (( Z 3 , +)) ; A := Z 3 . We represent the binary part C [ 2 ] . C [ 2 ] = { ( x , y ) �→ ax + by | | | a , b ∈ Z 3 } . ◮ Order A : 0 < 1 < 2. ◮ Order A 2 lexicographically: 00 < 01 < 02 < 10 < 11 < 12 < 20 < 21 < 22.

Representing clones by forks ◮ For each x ∈ A 2 , compute F ( C , x ) := { f ( x ) | | | f ∈ C , ∀ z < x : f ( z ) = 0 } . x F ( C , x ) Reason

Representing clones by forks ◮ For each x ∈ A 2 , compute F ( C , x ) := { f ( x ) | | | f ∈ C , ∀ z < x : f ( z ) = 0 } . x F ( C , x ) Reason { 0 } 00

Representing clones by forks ◮ For each x ∈ A 2 , compute F ( C , x ) := { f ( x ) | | | f ∈ C , ∀ z < x : f ( z ) = 0 } . x F ( C , x ) Reason { 0 } 00 01 A f ( x , y ) := y witnesses 1 ∈ F ( C , 01 )

Representing clones by forks ◮ For each x ∈ A 2 , compute F ( C , x ) := { f ( x ) | | | f ∈ C , ∀ z < x : f ( z ) = 0 } . x F ( C , x ) Reason { 0 } 00 01 A f ( x , y ) := y witnesses 1 ∈ F ( C , 01 ) f ( 02 ) = f ( 01 ) + f ( 01 ) 02 0

Representing clones by forks ◮ For each x ∈ A 2 , compute F ( C , x ) := { f ( x ) | | | f ∈ C , ∀ z < x : f ( z ) = 0 } . x F ( C , x ) Reason { 0 } 00 01 A f ( x , y ) := y witnesses 1 ∈ F ( C , 01 ) f ( 02 ) = f ( 01 ) + f ( 01 ) 02 0 10 A f ( x , y ) := x witnesses 1 ∈ F ( C , 10 )

Representing clones by forks ◮ For each x ∈ A 2 , compute F ( C , x ) := { f ( x ) | | | f ∈ C , ∀ z < x : f ( z ) = 0 } . x F ( C , x ) Reason { 0 } 00 01 A f ( x , y ) := y witnesses 1 ∈ F ( C , 01 ) f ( 02 ) = f ( 01 ) + f ( 01 ) 02 0 10 A f ( x , y ) := x witnesses 1 ∈ F ( C , 10 ) 11 0 f ( 11 ) = f ( 01 ) + f ( 10 )

Representing clones by forks ◮ For each x ∈ A 2 , compute F ( C , x ) := { f ( x ) | | | f ∈ C , ∀ z < x : f ( z ) = 0 } . x F ( C , x ) Reason { 0 } 00 01 A f ( x , y ) := y witnesses 1 ∈ F ( C , 01 ) f ( 02 ) = f ( 01 ) + f ( 01 ) 02 0 10 A f ( x , y ) := x witnesses 1 ∈ F ( C , 10 ) 11 0 f ( 11 ) = f ( 01 ) + f ( 10 ) 12 0 20 0 21 0 22 0