Bases for Kumjian-Pask algebras over standard k -graphs
aa r X i v : . [ m a t h . K T ] J a n BASES FOR KUMJIAN-PASK ALGEBRAS OVER STANDARD k -GRAPHS RAIMUND PREUSSER Notation
We write N for the set of all natural numbers, including 0, and Z for the set of all integers. For any k ∈ N \ { } and m, n ∈ Z k , m ≤ n means m i ≤ n i for any 1 ≤ i ≤ k , m ∨ n denotes the pointwise maximumand m ∧ n the pointwise minimum. Further we set | m | := m + · · · + m k . We denote the usual basis of Z k by { e i } .In a small category C with object set C , we identify objects v ∈ C with their identity morphisms, andwrite C for the set of morphisms. We write s and r for the domain and codomain maps from C to C .2. k -graphs and Kumjian-Pask algebras Definition 1 ( k -graph) . For a positive integer k , we view the additive monoid N k as a category with oneobject. A k -graph is a countable category Λ = (Λ , Λ , r, s ) together with a functor d : Λ → N k , called the degree map , satisfying the following factorization property : if λ ∈ Λ and d ( λ ) = m + n for some m, n ∈ N k ,then there are unique µ, ν ∈ Λ such that d ( µ ) = m , d ( ν ) = n , and λ = µ ◦ ν .For n ∈ N k , we write Λ n := d − ( n ) (note that it follows from the factorization property that the twodefinitions of Λ coincide), and call the elements λ of Λ n paths of degree n from s ( λ ) to r ( λ ). For v ∈ Λ wewrite v Λ n for the set of paths of degree n with range v . The k -graph Λ is called row-finite if v Λ n is finitefor every v ∈ Λ and n ∈ N k ; Λ has no sources if v Λ n is nonempty for every v ∈ Λ and n ∈ N k . Furtherwe set Λ =0 := { λ ∈ Λ | d ( λ ) = 0 } , and for each λ ∈ Λ =0 we introduce a ghost path λ ∗ (it is assumed thatthe map λ λ ∗ is injective and further that (Λ =0 ) ∗ ∩ Λ = ∅ where (Λ =0 ) ∗ denotes the set of all ghostpaths). We set v ∗ := v for any v ∈ Λ . Definition 2 ( Kumjian-Pask algebra) . Let Λ be a row-finite k -graph without sources and let R be acommutative ring with 1. The (associative, not necessarily unital) R -algebra presented by the generatingset Λ ∪ Λ =0 ∪ (Λ =0 ) ∗ and the relations(KP1) { v ∈ Λ } is a family of mutually orthogonal idempotents,(KP2) for all λ, µ ∈ Λ =0 with r ( µ ) = s ( λ ), we have λµ = λ ◦ µ, µ ∗ λ ∗ = ( λ ◦ µ ) ∗ , r ( λ ) λ = λ = λs ( λ ) , s ( λ ) λ ∗ = λ ∗ = λ ∗ r ( λ ) , (KP3) for all λ, µ ∈ Λ =0 with d ( λ ) = d ( µ ), we have λ ∗ µ = δ λ,µ s ( λ ) , (KP4) for all v ∈ Λ and all n ∈ N k \ { } , we have v = X λ ∈ v Λ n λλ ∗ is called the Kumjian-Pask algebra defined by Λ and is denoted by KP R (Λ). Problem:
Find a basis for KP R (Λ).It follows from the lemma below (which is easy to prove, see [2, Proof of Lemma 3.3]), that the elements λµ ∗ ( λ, µ ∈ Λ) span KP R (Λ). Lemma 3.
Let λ, µ ∈ Λ . Then for each q ≥ d ( λ ) ∨ d ( µ ) , we have λ ∗ µ = X λ ◦ α = µ ◦ β,d ( λ ◦ α )= q αβ ∗ . Standard k -graphs In Definition 4 below we use the following conventions. Let l ∈ N \ { } and s, t ∈ N . We denote by { , . . . , l } s the cartesian product of s copies of { , . . . , l } (we use the convention { , . . . , l } := { () } where() = ∅ is the empty tuple). Further, for any p = ( p s , . . . , p ) ∈ { , . . . , l } s and q = ( q t , . . . , q ) ∈ { , . . . , l } t ,we define p × q := ( p s , . . . , p , q t , . . . , q ) ∈ { , . . . , l } s + t (hence p × q = q if s = 0 and p × q = p if t = 0). Definition 4 ( Standard k -graph) . Let k, l ∈ N \ { } . SetΛ := Z k , Λ := { ( v, w, p ) ∈ Z k × Z k × { , . . . , l } | v − w | | v ≥ w } and define r, s : Λ → Λ by r ( v, w, p ) := v and s ( v, w, p ) := w . Define composition by ( u, v, p ) ◦ ( v, w, q ) =( u, w, p × q ) and define d : Λ → N k by d ( v, w, p ) := v − w . Then the k -graph Λ = (Λ , Λ , r, s, d ) is calledthe standard k -graph of level l .If k, l ∈ N \{ } , Λ is the standard k -graph of level l and λ = ( v, w, p ) ∈ Λ, then p is called the level vectorof λ and is denoted by lv( λ ). In the following we will always use the indexing lv( λ ) = (lv( λ ) | d ( λ ) | , . . . , lv( λ ) )of the components of a level vector lv( λ ). Example 5.
The skeleton (cf. [2, p. 3615]) of the standard 2-graph of level 2 looks as follows: . . . ... . . . (1 , − + + (1 , + + (1 , . . . (0 , − + + H H ✯✤✔ U U ✔✤✯ (0 , + + H H ✯✤✔ U U ✔✤✯ (0 , H H ✯✤✔ U U ✔✤✯ . . . ( − , − , , H H ✯✤✔ U U ✔✤✯ ( − , , , H H ✯✤✔ U U ✔✤✯ ( − , H H ✯✤✔ U U ✔✤✯ . . . ... . . . . Bases for Kumjian-Pask algebras over standard k -graphs In this section R denotes a commutative ring with 1 and Λ the standard k -graph of level l for some k, l ∈ N \ { } . For any v ∈ Λ (= Z k ) and n ∈ N k \ { } set λ v,n := ( v, v − n, (1 , . . . , ∈ v Λ n . If µ ∈ Λ, we sometimes write µ ◦ λ ∼ ,n instead of µ ◦ λ s ( µ ) ,n . Further setˆ A := { ( λ, µ ) | λ, µ ∈ Λ =0 , s ( λ ) = s ( µ ) } ASES FOR KUMJIAN-PASK ALGEBRAS 3 and A := { ( λ, µ ) ∈ ˆ A | there is no n ∈ N k \ { } and λ ′ , µ ′ ∈ Λ such that λ = λ ′ ◦ λ ∼ ,n , µ = µ ′ ◦ λ ∼ ,n } . The proof of the following lemma is straightforward.
Lemma 6.
Let ( λ, µ ) ∈ ˆ A . Then ( λ, µ ) ∈ A iff either lv( λ ) = 1 or lv( µ ) = 1 or d ( λ ) ∧ d ( µ ) = 0 . If ( λ, µ ) , ( λ ′ , µ ′ ) ∈ A , then we write ( λ, µ ) ∼ ( λ ′ , µ ′ ) iff there are m, n ∈ N k \ { } such that λ ◦ λ ∼ ,m = λ ′ ◦ λ ∼ ,n and µ ◦ λ ∼ ,m = µ ′ ◦ λ ∼ ,n . One checks easily that ∼ is an equivalence relation on A . Wedenote the equivalence class of an element ( λ, µ ) ∈ A by [( λ, µ )]. For any equivalence class [( λ, µ )] choose arepresentative ( λ [( λ,µ )] , µ [( λ,µ )] ). We denote by R the subset of A consisting of all the chosen representatives. Lemma 7.
Let ( λ, µ ) ∈ A . Then [( λ, µ )] = { ( λ ′ , µ ′ ) ∈ ˆ A | r ( λ ′ ) = r ( λ ) , r ( µ ′ ) = r ( µ ) , lv( λ ′ ) = lv( λ ) , lv( µ ′ ) = lv( µ ) } . If d ( λ ) ∧ d ( µ ) = 0 , then [( λ, µ )] = { ( λ, µ ) } .Proof. ⊆ : Let ( λ ′ , µ ′ ) ∈ A such that ( λ ′ , µ ′ ) ∼ ( λ, µ ). Then ( λ ′ , µ ′ ) ∈ ˆ A and further there are m, n ∈ N k \ { } suchthat λ ◦ λ ∼ ,m = λ ′ ◦ λ ∼ ,n and µ ◦ λ ∼ ,m = µ ′ ◦ λ ∼ ,n . (1)It follows from (1) that r ( λ ′ ) = r ( λ ) and r ( µ ′ ) = r ( µ ). Assume that | n | > | m | . Then (1) implies thatlv( λ ) , lv( µ ) = 1. Hence, by Lemma 6, d ( λ ) ∧ d ( µ ) = 0. But, by (1), d ( λ ) ∧ d ( µ ) = d ( λ ′ ) + n − m ∧ d ( µ ′ ) + n − m . It follows that n ≤ m (if there were an i ∈ { , . . . , k } such that n i > m i , then min ( d ( λ ′ ) i + n i − m i , d ( µ ′ ) i + n i − m i ) ≥ n i − m i > ). But this contradicts the assumption | n | > | m | .Hence | n | ≤ | m | . By symmetry we also get | m | ≤ | n | and hence | n | = | m | . Now it follows from (1) thatlv( λ ′ ) = lv( λ ) and lv( µ ′ ) = lv( µ ). ⊇ : Let ( λ ′ , µ ′ ) ∈ ˆ A such that r ( λ ′ ) = r ( λ ), r ( µ ′ ) = r ( µ ), lv( λ ′ ) = lv( λ ) and lv( µ ′ ) = lv( µ ). By Lemma 6,lv( λ ) = 1 or lv( µ ) = 1 or d ( λ ) ∧ d ( µ ) = 0.case 1 Assume that lv( λ ) = 1 or lv( µ ) = 1 . Then ( λ ′ , µ ′ ) ∈ A by Lemma 6. Set m := d ( λ ′ ) and n := d ( λ ). One checks easily that λ ◦ λ ∼ ,m = λ ′ ◦ λ ∼ ,n and µ ◦ λ ∼ ,m = µ ′ ◦ λ ∼ ,n . Hence ( λ ′ , µ ′ ) ∼ ( λ, µ ).case 2 Assume that d ( λ ) ∧ d ( µ ) = 0 . Clearly lv( λ ′ ) = lv( λ ), lv( λ ) ∈ { , . . . , l } | r ( λ ) − s ( λ ) | = { , . . . , l } | r ( λ ) |−| s ( λ ) | and lv( λ ′ ) ∈ { , . . . , l } | r ( λ ′ ) − s ( λ ′ ) | = { , . . . , l } | r ( λ ) |−| s ( λ ′ ) | imply that | s ( λ ′ ) | = | s ( λ ) | . (2)Further d ( λ ) ∧ d ( µ ) = 0 ⇔ r ( λ ) − s ( λ ) ∧ r ( µ ) − s ( µ ) |{z} = s ( λ ) = 0 ⇔ r ( λ ) ∧ r ( µ ) = s ( λ ) . (3)It follows that s ( λ ′ ) ≤ r ( λ ′ ) ∧ r ( µ ′ ) = r ( λ ) ∧ r ( µ ) (3) = s ( λ ) . (4)Hence s ( µ ′ ) = s ( λ ′ ) (2) , (4) = s ( λ ) = s ( µ ). Thus ( λ ′ , µ ′ ) = ( λ, µ ).It remains to show that [( λ, µ )] = { ( λ, µ ) } if d ( λ ) ∧ d ( µ ) = 0. Let ( λ ′ , µ ′ ) ∈ A such that ( λ ′ , µ ′ ) ∼ ( λ, µ ).Then, by the inclusion “ ⊆ ” shown above, r ( λ ′ ) = r ( λ ), r ( µ ′ ) = r ( µ ), lv( λ ′ ) = lv( λ ) and lv( µ ′ ) = lv( µ ). Itfollows from case 2 right above that ( λ ′ , µ ′ ) = ( λ, µ ). (cid:3) RAIMUND PREUSSER
If ( λ, µ ) ∈ A such that d ( λ ) ∧ d ( µ ) = 0, it can happen that [( λ, µ )] has more than one element. Forexample let k, l = 2, λ = ((1 , , (1 , , (2)) and λ ′ = ((1 , , (0 , , (2)). Then ( λ, λ ) , ( λ ′ , λ ′ ) ∈ A by Lemma6 and ( λ, λ ) ∼ ( λ ′ , λ ′ ) by Lemma 7. The next lemma shows that the terms λµ ∗ , where ( λ, µ ) ranges overa given equivalence class, agree in KP R (Λ). Lemma 8.
Let ( λ, µ ) , ( λ ′ , µ ′ ) ∈ A such that ( λ, µ ) ∼ ( λ ′ , µ ′ ) . Then λµ ∗ = λ ′ ( µ ′ ) ∗ in KP R (Λ) .Proof. Since ( λ, µ ) ∼ ( λ ′ , µ ′ ), there are m, n ∈ N k \{ } such that λ ◦ λ ∼ ,m = λ ′ ◦ λ ∼ ,n and µ ◦ λ ∼ ,m = µ ′ ◦ λ ∼ ,n .Hence λµ ∗ − X ξ ∈ s ( λ )Λ m ,ξ = λ s ( λ ) ,m ( λ ◦ ξ )( µ ◦ ξ ) ∗ = λs ( λ ) µ ∗ − X ξ ∈ s ( λ )Λ m ,ξ = λ s ( λ ) ,m λξξ ∗ µ ∗ = λ ( s ( λ ) − X ξ ∈ s ( λ )Λ m ,ξ = λ s ( λ ) ,m ξξ ∗ ) µ ∗ = λλ ∼ ,m ( λ ∼ ,m ) ∗ µ ∗ =( λ ◦ λ ∼ ,m )( µ ◦ λ ∼ ,m ) ∗ =( λ ′ ◦ λ ∼ ,n )( µ ′ ◦ λ ∼ ,n ) ∗ = λ ′ λ ∼ ,n ( λ ∼ ,n ) ∗ ( µ ′ ) ∗ = λ ′ ( s ( λ ′ ) − X ζ ∈ s ( λ ′ )Λ n ,ζ = λ s ( λ ′ ) ,n ζζ ∗ )( µ ′ ) ∗ = λ ′ s ( λ ′ )( µ ′ ) ∗ − X ζ ∈ s ( λ ′ )Λ n ,ζ = λ s ( λ ′ ) ,n λ ′ ζζ ∗ ( µ ′ ) ∗ = λ ′ ( µ ′ ) ∗ − X ζ ∈ s ( λ ′ )Λ n ,ζ = λ s ( λ ′ ) ,n ( λ ′ ◦ ζ )( µ ′ ◦ ζ ) ∗ by (KP2) and (KP4). One checks easily that X ξ ∈ s ( λ )Λ m ,ξ = λ s ( λ ) ,m ( λ ◦ ξ )( µ ◦ ξ ) ∗ = X ζ ∈ s ( λ ′ )Λ n ,ζ = λ s ( λ ′ ) ,n ( λ ′ ◦ ζ )( µ ′ ◦ ζ ) ∗ . Thus λµ ∗ = λ ′ ( µ ′ ) ∗ . (cid:3) In the proof of Theorem 14 we will use the definitions and lemmas below.
Definition 9.
Let s, t ∈ N , p = ( p s , . . . , p ) ∈ { , . . . , l } s and q = ( q t , . . . , q ) ∈ { , . . . , l } t . Then we write p ∼ q iff p s − i = q t − i for any i ∈ { , . . . , ( s ∧ t ) − } . Definition 10.
For any v, w ∈ Λ , m, n ∈ N k , p ∈ { , . . . , l } s and q ∈ { , . . . , l } t where s, t ∈ N such that | m | − s = | n | − t ≥ S ( v, w, m, n, p, q ) := { ( α, β ) ∈ v Λ m × w Λ n | lv( α ) = p × r, lv( β ) = q × r for some r ∈ { , . . . , l } | m |− s } . Lemma 11.
Let λ, µ ∈ Λ =0 such that r ( λ ) = r ( µ ) . Set S ( λ, µ ) := { ( α, β ) ∈ Λ × Λ | λ ◦ α = µ ◦ β, d ( λ ◦ α ) = d ( λ ) ∨ d ( µ ) } . ASES FOR KUMJIAN-PASK ALGEBRAS 5
Then S ( λ, µ ) = S ( s ( λ ) , s ( µ ) , ( d ( µ ) − d ( λ )) ∨ , ( d ( λ ) − d ( µ )) ∨ , (lv( µ ) | d ( µ ) |−| d ( λ ) | , . . . , lv( µ ) ) , (lv( λ ) | d ( λ ) |−| d ( µ ) | , . . . , lv( λ ) )) if lv( λ ) ∼ lv( µ ) and S ( λ, µ ) = ∅ otherwise.Proof. Straightforward. (cid:3)
Lemma 12.
Let v, w, m, n, p, q, s, t be as in Definition 10. Further let ˆ n ∈ N k be such that ˆ n ≤ m ∧ n and | ˆ n | ≤ | m | − s . Then X ( α,β ) ∈ S ( v,w,m,n,p,q ) αβ ∗ = X ( α,β ) ∈ S ( v,w,m − ˆ n,n − ˆ n,p,q ) αβ ∗ . Proof.
Follows from (KP4). (cid:3)
Lemma 13.
Let n ∈ N k \ { } , λ, µ ∈ Λ and v ∈ Λ such that s ( λ ) = v = s ( µ ) . Let i , . . . , i | n | ∈ { , . . . , k } such that n = e i + · · · + e i | n | . Then X ξ ∈ v Λ n ,ξ = λ v,n ( λ ◦ ξ )( µ ◦ ξ ) ∗ = X ≤ p ≤| n | , ≤ q ≤ l ( λ ◦ ξ p,q )( µ ◦ ξ p,q ) ∗ where ξ p,q ∈ v Λ p − P j =0 e i | n |− j and lv( ξ p,q ) = (1 , . . . , , q ) for any ≤ p ≤ | n | and ≤ q ≤ l .Proof. We will prove the lemma by induction on m := | n | . m = 1 Obvious. m → m + 1 Suppose that | n | = m + 1. Set n ′ := e i + · · · + e i m . Then n ′ + e i m +1 = n and | n ′ | = m .Further set v ′ := v − e i m +1 . Then X ξ ∈ v Λ n ,ξ = λ v,n ( λ ◦ ξ )( µ ◦ ξ ) ∗ = X ξ ∈ v Λ eim +1 , ξ ∈ v ′ Λ n ′ ,ξ ◦ ξ = λ v,n ( λ ◦ ξ ◦ ξ )( µ ◦ ξ ◦ ξ ) ∗ = X ξ ∈ v Λ eim +1 , ξ ∈ v ′ Λ n ′ ,ξ = λ v,eim +1 , ξ = λ v ′ ,n ′ ( λ ◦ ξ ◦ ξ )( µ ◦ ξ ◦ ξ ) ∗ + X ξ ∈ v Λ eim +1 ,ξ = λ v,eim +1 ( λ ◦ ξ ◦ λ ∼ ,n ′ )( µ ◦ ξ ◦ λ ∼ ,n ′ ) ∗ + X ξ ∈ v ′ Λ n ′ ,ξ = λ v ′ ,n ′ ( λ ◦ λ ∼ ,e im +1 ◦ ξ )( µ ◦ λ ∼ ,e im +1 ◦ ξ ) ∗ RAIMUND PREUSSER
I.A. = ( KP X ξ ∈ v Λ eim +1 , ξ ∈ v ′ Λ n ′ ,ξ = λ v,eim +1 , ξ = λ v ′ ,n ′ ( λ ◦ ξ ◦ ξ )( µ ◦ ξ ◦ ξ ) ∗ + X ξ ∈ v Λ eim +1 ,ξ = λ v,eim +1 (( λ ◦ ξ )( µ ◦ ξ ) ∗ − X ξ ∈ v ′ Λ n ′ ,ξ = λ v ′ ,n ′ ( λ ◦ ξ ◦ ξ )( µ ◦ ξ ◦ ξ ) ∗ )+ X ≤ p ≤| n ′ | , ≤ q ≤ l ( λ ◦ λ ∼ ,e im +1 ◦ ξ ′ p,q )( µ ◦ λ ∼ ,e im +1 ◦ ξ ′ p,q ) ∗ = X ξ ∈ v Λ eim +1 ,ξ = λ v,eim +1 ( λ ◦ ξ )( µ ◦ ξ ) ∗ + X ≤ p ≤| n ′ | , ≤ q ≤ l ( λ ◦ λ ∼ ,e im +1 ◦ ξ ′ p,q )( µ ◦ λ ∼ ,e im +1 ◦ ξ ′ p,q ) ∗ where ξ ′ p,q ∈ v ′ Λ p − P j =0 e i | n ′|− j and lv( ξ ′ p,q ) = (1 , . . . , , q ) for any 1 ≤ p ≤ | n ′ | and 2 ≤ q ≤ l . Clearly v Λ e im +1 \ { λ v,e im +1 } = { ξ ,q | ≤ q ≤ l } . Further λ v,e im +1 ◦ ξ ′ p,q = ξ p +1 ,q for any 1 ≤ p ≤ | n ′ | and 2 ≤ q ≤ l since v ′ − p − P j =0 e i | n ′|− j = v − p − P j = − e i | n ′|− j = v − p P j =0 e i | n |− j . Thus X ξ ∈ v Λ eim +1 ,ξ = λ v,eim +1 ( λ ◦ ξ )( µ ◦ ξ ) ∗ + X ≤ p ≤| n ′ | , ≤ q ≤ l ( λ ◦ λ ∼ ,e im +1 ◦ ξ ′ p,q )( µ ◦ λ ∼ ,e im +1 ◦ ξ ′ p,q ) ∗ = X ≤ q ≤ l ( λ ◦ ξ ,q )( µ ◦ ξ ,q ) ∗ + X ≤ p ≤| n ′ | , ≤ q ≤ l ( λ ◦ ξ p +1 ,q )( µ ◦ ξ p +1 ,q ) ∗ = X ≤ q ≤ l ( λ ◦ ξ ,q )( µ ◦ ξ ,q ) ∗ + X ≤ p ≤| n | , ≤ q ≤ l ( λ ◦ ξ p,q )( µ ◦ ξ p,q ) ∗ = X ≤ p ≤| n | , ≤ q ≤ l ( λ ◦ ξ p,q )( µ ◦ ξ p,q ) ∗ . (cid:3) Theorem 14.
The elements v ( v ∈ Λ ) , λ ( λ ∈ Λ =0 ) , λ ∗ ( λ ∈ Λ =0 ) and λµ ∗ (( λ, µ ) ∈ R ) form a basis for KP R (Λ) .Proof. We want to apply [1, Theorem 15]. Set X := Λ ∪ Λ =0 ∪ (Λ =0 ) ∗ and denote by R h X i the free R -algebra generated by X . Consider the relations(1) for all λ, µ ∈ Λ with s ( λ ) = r ( µ ), λµ = λ ◦ µ, µ ∗ λ ∗ = ( λ ◦ µ ) ∗ , (2) for all λ, µ ∈ Λ, λµ = 0 if s ( λ ) = r ( µ ) , λ ∗ µ = 0 if r ( λ ) = r ( µ ) , λµ ∗ = 0 if s ( λ ) = s ( µ ) , λ ∗ µ ∗ = 0 if r ( λ ) = s ( µ ) , (3) for all λ, µ ∈ Λ =0 such that r ( λ ) = r ( µ ), λ ∗ µ = X ( α,β ) ∈ S ( λ,µ ) αβ ∗ , ASES FOR KUMJIAN-PASK ALGEBRAS 7 (4) for all v ∈ Λ , n ∈ N k \ { } , and λ, µ ∈ Λ such that s ( λ ) = s ( µ ) = v ,( λ ◦ λ v,n )( µ ◦ λ v,n ) ∗ = λµ ∗ − X ξ ∈ v Λ n ,ξ = λ v,n ( λ ◦ ξ )( µ ◦ ξ ) ∗ and(5) for all ( λ, µ ) ∈ A \ R , λµ ∗ = λ [( λ,µ )] ( µ [( λ,µ )] ) ∗ . It follows from Lemma 3 and Lemma 8 that the relations (1)-(5) above generate the same ideal I of R h X i as the relations (KP1)-(KP4) in Definition 2. Denote by S the reduction system for R h X i defined by therelations (1)-(5) (i.e., S is the set of all pairs σ = ( W σ , f σ ) where W σ equals the left hand side of an equationin (1)-(5) and f σ the corresponding right hand side). Denote by h X i the semigroup of all nonempty wordsover X and set h X i := X ∪ { empty word } . For any A = x . . . x n ∈ h X i define • l ( A ) := n (the length of A ), • e ( A ) := P ≤ i ≤ n,x i ∈ Λ =0 i (the entropy of A ), • f ( A ) := P ≤ i ≤ n,x i ∈ Λ =0 | d ( x i ) | (the degree value of A ), • g ( A ) := P ≤ i ≤ n,x i ∈ Λ =0 { j ∈ { , . . . , | d ( x i ) |} | lv( x i ) j = 1 } (the 1 -level value of A ) and • h ( A ) := { i ∈ { , . . . , n − } | x i x i +1 = λµ ∗ for some ( λ, µ ) ∈ A \ R } (the A \ R -value of A ).Define a partial ordering ≤ on h X i by A ≤ B ⇔ (cid:2) A = B (cid:3) ∨ (cid:2) l ( A ) < l ( B ) (cid:3) ∨ (cid:2) l ( A ) = l ( B ) ∧ e ( A ) < e ( B ) (cid:3) ∨ (cid:2) l ( A ) = l ( B ) ∧ e ( A ) = e ( B ) ∧ f ( A ) < f ( B ) (cid:3) ∨ (cid:2) l ( A ) = l ( B ) ∧ e ( A ) = e ( B ) ∧ f ( A ) = f ( B ) ∧ g ( A ) < g ( B ) (cid:3) ∨ (cid:2) l ( A ) = l ( B ) ∧ e ( A ) = e ( B ) ∧ f ( A ) = f ( B ) ∧ g ( A ) = g ( B ) ∧ ∀ C, D ∈ h X i : h ( CAD ) < h ( CBD ) (cid:3) . Then ≤ is a semigroup partial ordering on h X i compatible with S and the descending chain condition issatisfied. It remains to show that all ambiguities of S are resolvable. Below we list all ambiguities.(1) , (1) : λµξ, ξ ∗ µ ∗ λ ∗ ( λ, µ, ξ ∈ Λ , s ( λ ) = r ( µ ) , s ( µ ) = r ( ξ ))(1) , (2) : λµξ, λµζ ∗ etc. ( λ, µ, ξ, ζ ∈ Λ , s ( λ ) = r ( µ ) , s ( µ ) = r ( ξ ) , s ( µ ) = s ( ζ ))(1) , (3) : λ ∗ µξ, ζ ∗ λ ∗ µ ( λ, µ ∈ Λ =0 , ξ, ζ ∈ Λ , r ( λ ) = r ( µ ) , s ( µ ) = r ( ξ ) , r ( ζ ) = s ( λ ))(1) , (4) : ξ ( λ ◦ λ ∼ ,n )( µ ◦ λ ∼ ,n ) ∗ , ( λ ◦ λ ∼ ,n )( µ ◦ λ ∼ ,n ) ∗ ζ ∗ ( n ∈ N k \ { } , λ, µ, ξ, ζ ∈ Λ ,s ( λ ) = s ( µ ) , s ( ξ ) = r ( λ ) , r ( µ ) = s ( ζ ))(1) , (5) : ξλµ ∗ , λµ ∗ ζ ∗ (( λ, µ ) ∈ A \ R , ξ, ζ ∈ Λ , s ( ξ ) = r ( λ ) , r ( µ ) = s ( ζ ))(2) , (2) : λµξ, λµζ ∗ etc. ( λ, µ, ξ ∈ Λ , s ( λ ) = r ( µ ) , s ( µ ) = r ( ξ ) , s ( µ ) = s ( ζ ))(2) , (3) : λ ∗ µξ, λ ∗ µζ ∗ etc. ( λ, µ, ξ, ζ ∈ Λ , r ( λ ) = r ( µ ) , s ( µ ) = r ( ξ ) , s ( µ ) = s ( ζ ))(2) , (4) : ( λ ◦ λ ∼ ,n )( µ ◦ λ ∼ ,n ) ∗ ξ, ( λ ◦ λ ∼ ,n )( µ ◦ λ ∼ ,n ) ∗ ζ ∗ etc. ( n ∈ N k \ { } , λ, µ, ξ, ζ ∈ Λ ,s ( λ ) = s ( µ ) , r ( µ ) = r ( ξ ) , r ( µ ) = s ( ζ )) RAIMUND PREUSSER (2) , (5) : λµ ∗ ξ, λµ ∗ ζ ∗ etc. (( λ, µ ) ∈ A \ R , ξ, ζ ∈ Λ , r ( µ ) = r ( ξ ) , r ( µ ) = s ( ζ ))(3) , (3) : − (3) , (4) : ξ ∗ ( λ ◦ λ ∼ ,n )( µ ◦ λ ∼ ,n ) ∗ , ( λ ◦ λ ∼ ,n )( µ ◦ λ ∼ ,n ) ∗ ζ ( n ∈ N k \ { } , λ, µ ∈ Λ , ξ, ζ ∈ Λ =0 ,s ( λ ) = s ( µ ) , r ( ξ ) = r ( λ ) , r ( µ ) = r ( ζ ))(3) , (5) : ξ ∗ λµ ∗ , λµ ∗ ζ (( λ, µ ) ∈ A \ R , ξ, ζ ∈ Λ =0 , r ( ξ ) = r ( λ ) , r ( µ ) = r ( ζ ))(4) , (4) : λµ ∗ ( λ = λ ◦ λ ∼ ,n = λ ◦ λ ∼ ,n , µ = µ ◦ λ ∼ ,n = µ ◦ λ ∼ ,n , λ , λ , µ , µ ∈ Λ ,s ( λ ) = s ( µ ) , s ( λ ) = s ( µ ) , n , n ∈ N k \ { } , n = n )(4) , (5) : − (5) , (5) : − We will show how to resolve the (3),(4), the (3),(5) and the (4),(4) ambiguities (the other ambiguities arerelatively easy to resolve; for example it follows from Lemma 6 and Lemma 7 that the (1),(5) ambiguitiescan be resolved).(3),(4):Let n ∈ N k \ { } , λ, µ ∈ Λ and ξ, ζ ∈ Λ =0 such that s ( λ ) = s ( µ ), r ( ξ ) = r ( λ ) and r ( µ ) = r ( ζ ). We willshow how to resolve the ambiguity ( λ ◦ λ ∼ ,n )( µ ◦ λ ∼ ,n ) ∗ ζ and leave the ambiguity ξ ∗ ( λ ◦ λ ∼ ,n )( µ ◦ λ ∼ ,n ) ∗ to the reader. Suppose that d ( µ ) = 0. Then ( λ ◦ λ ∼ ,n )( µ ◦ λ ∼ ,n ) ∗ ζ (3) " " ❋❋❋❋❋❋❋❋❋❋❋❋❋❋❋❋❋❋❋ (4) y y rrrrrrrrrrrrrrrrrrrrrrr λµ ∗ ζ − P ξ ∈ s ( λ )Λ n ,ξ = λ s ( λ ) ,n ( λ ◦ ξ )( µ ◦ ξ ) ∗ ζ (3) (cid:15) (cid:15) ( λ ◦ λ ∼ ,n ) P ( α,β ) ∈ S ( µ ◦ λ ∼ ,n ,ζ ) αβ ∗ (1) (cid:15) (cid:15) λ P ( α ′ ,β ′ ) ∈ S ( µ,ζ ) α ′ ( β ′ ) ∗ − P ξ ∈ s ( λ )Λ n ,ξ = λ s ( λ ) ,n ( λ ◦ ξ ) P ( α ′′ ,β ′′ ) ∈ S ( µ ◦ ξ,ζ ) α ′′ ( β ′′ ) ∗ (1) (cid:15) (cid:15) X ( α,β ) ∈ S ( µ ◦ λ ∼ ,n ,ζ ) ( λ ◦ λ ∼ ,n ◦ α ) β ∗ | {z } A := X ( α ′ ,β ′ ) ∈ S ( µ,ζ ) ( λ ◦ α ′ )( β ′ ) ∗ | {z } B := − X ξ ∈ s ( λ )Λ n ,ξ = λ s ( λ ) ,n X ( α ′′ ,β ′′ ) ∈ S ( µ ◦ ξ,ζ ) ( λ ◦ ξ ◦ α ′′ )( β ′′ ) ∗ | {z } C :=ASES FOR KUMJIAN-PASK ALGEBRAS 9 Suppose that lv( µ ) lv( ζ ). Then A = B = C = 0 by Lemma 11. Hence we can assume that lv( µ ) ∼ lv( ζ ).Denote by d ( α ) the degree of an α appearing in A (note that the α ’s appearing in A all have the samedegree by Lemma 11), by d ( β ) the degree of a β appearing in A , by d ( α ′ ) the degree of an α ′ appearingin B and so on. One checks easily that d ( λ ∼ ,n ◦ α ) = d ( α ′ ) + δ = d ( ξ ◦ α ′′ ) and d ( β ) = d ( β ′ ) + δ = d ( β ′′ )where δ = ( d ( µ ◦ λ ∼ ,n ) ∨ d ( ζ )) − ( d ( µ ) ∨ d ( ζ )) ≥
0. It is an easy exercise to show that A = B + C if δ = 0.Hence we can assume that δ = 0.case 1 Assume that | d ( µ ◦ λ ∼ ,n ) | ≤ | d ( ζ ) | . case 1.1 Assume that lv( µ ◦ λ ∼ ,n ) lv( ζ ) . Then, by Lemma 11, A = 0 and C = P ( α ′′ ,β ′′ ) ∈ S ( µ ◦ ξ,ζ ) ( λ ◦ ξ ◦ α ′′ )( β ′′ ) ∗ where ξ ∈ s ( λ )Λ n has the propertythat lv( µ ◦ ξ ) ∼ lv( ζ ). Further, also by Lemma 11, one can write B = P (ˆ α, ˆ β ) ∈ S ˆ α ˆ β ∗ and C = P (˜ α, ˜ β ) ∈ S ˜ α ˜ β ∗ where S = S ( r ( λ ) , s ( ζ ) , d ( λ ) + d ( α ′ ) , d ( β ′ ) , lv( λ ) × (lv( ζ ) | d ( ζ ) |−| d ( µ ) | , . . . , lv( ζ ) ) , ())and S = S ( r ( λ ) , s ( ζ ) , d ( λ ◦ ξ ) + d ( α ′′ ) , d ( β ′′ ) , lv( λ ◦ ξ ) × (lv( ζ ) | d ( ζ ) |−| d ( µ ◦ ξ ) | , . . . , lv( ζ ) ) , ()) . One checks easily that δ ≤ ( d ( λ ◦ ξ ) + d ( α ′′ )) ∧ d ( β ′′ ) and | δ | ≤ | d ( λ ◦ ξ ) + d ( α ′′ ) | − s where s = | d ( λ ◦ ξ ) | + | d ( ζ ) | − | d ( µ ◦ ξ ) | . Hence Lemma 12 shows that C reduces to C ′ := P (˜ α, ˜ β ) ∈ S ˜ α ˜ β ∗ where S = S ( r ( λ ) , s ( ζ ) , d ( λ ◦ ξ ) + d ( α ′′ ) − δ, d ( β ′′ ) − δ, lv( λ ◦ ξ ) × (lv( ζ ) | d ( ζ ) |−| d ( µ ◦ ξ ) | , . . . , lv( ζ ) ) , ()) . Clearly S = S and hence C ′ = B .case 1.2 Assume that lv( µ ◦ λ ∼ ,n ) ∼ lv( ζ ) . Similar to case 1.1.case 2
Assume that | d ( µ ) | < | d ( ζ ) | < | d ( µ ◦ λ ∼ ,n ) | . case 2.1 Assume that lv( µ ◦ λ ∼ ,n ) lv( ζ ) . Then, by Lemma 11, A = 0 and C = P ξ ∈ s ( λ )Λ n , lv( µ ◦ ξ ) ∼ lv( ζ ) P ( α ′′ ,β ′′ ) ∈ S ( µ ◦ ξ,ζ ) ( λ ◦ ξ ◦ α ′′ )( β ′′ ) ∗ . Further, also by Lemma11, one can write B = P (ˆ α, ˆ β ) ∈ S ˆ α ˆ β ∗ and C = P (˜ α, ˜ β ) ∈ S ˜ α ˜ β ∗ where S = S ( r ( λ ) , s ( ζ ) , d ( λ ) + d ( α ′ ) , d ( β ′ ) , lv( λ ) × (lv( ζ ) | d ( ζ ) |−| d ( µ ) | , . . . , lv( ζ ) ) , ())and S = S ( r ( λ ) , s ( ζ ) , d ( λ ◦ ξ ) + d ( α ′′ ) , d ( β ′′ ) , lv( λ ) × (lv( ζ ) | d ( ζ ) |−| d ( µ ) | , . . . , lv( ζ ) ) , ()) . One checks easily that δ ≤ ( d ( λ ◦ ξ ) + d ( α ′′ )) ∧ d ( β ′′ ) and | δ | ≤ | d ( λ ◦ ξ ) + d ( α ′′ ) | − s where s = | d ( λ ) | + | d ( ζ ) | − | d ( µ ) | . Hence Lemma 12 shows that C reduces to C ′ := P (˜ α, ˜ β ) ∈ S ˜ α ˜ β ∗ where S = S ( r ( λ ) , s ( ζ ) , d ( λ ◦ ξ ) + d ( α ′′ ) − δ, d ( β ′′ ) − δ, lv( λ ) × (lv( ζ ) | d ( ζ ) |−| d ( µ ) | , . . . , lv( ζ ) ) , ()) . Clearly S = S and hence C ′ = B .case 2.2 Assume that lv( µ ◦ λ ∼ ,n ) ∼ lv( ζ ) . Similar to case 2.1.case 3
Assume that | d ( ζ ) | ≤ | d ( µ ) | . Since reductions are linear maps, it suffices to show that there is a composition r of reductions such that r ( B ) = B = r ( A + C ). Clearly A + C = P ξ ∈ s ( λ )Λ n P ( α ′′ ,β ′′ ) ∈ S ( µ ◦ ξ,ζ ) ( λ ◦ ξ ◦ α ′′ )( β ′′ ) ∗ . By Lemma 11, one canwrite B = P (ˆ α, ˆ β ) ∈ S ˆ α ˆ β ∗ and A + C = P (˜ α, ˜ β ) ∈ S ˜ α ˜ β ∗ where S = S ( r ( λ ) , s ( ζ ) , d ( λ ) + d ( α ′ ) , d ( β ′ ) , lv( λ ) , (lv( µ ) | d ( µ ) |−| d ( ζ ) | , . . . , lv( µ ) ))and S = S ( r ( λ ) , s ( ζ ) , d ( λ ◦ ξ ) + d ( α ′′ ) , d ( β ′′ ) , lv( λ ) , (lv( µ ) | d ( µ ) |−| d ( ζ ) | , . . . , lv( µ ) )) . One checks easily that δ ≤ ( d ( λ ◦ ξ ) + d ( α ′′ )) ∧ d ( β ′′ ) and | δ | ≤ | d ( λ ◦ ξ ) + d ( α ′′ ) | − s where s = | d ( λ ) | .Hence Lemma 12 shows that A + C reduces to D := P (˜ α, ˜ β ) ∈ S ˜ α ˜ β ∗ where S = S ( r ( λ ) , s ( ζ ) , d ( λ ◦ ξ ) + d ( α ′′ ) − δ, d ( β ′′ ) − δ, lv( λ ) , (lv( µ ) | d ( µ ) |−| d ( ζ ) | , . . . , lv( µ ) )) . Clearly S = S and hence D = B .The case that d ( µ ) = 0 can be treated analogously.(3),(5):Let ( λ, µ ) ∈ A \ R and ξ, ζ ∈ Λ =0 such that r ( ξ ) = r ( λ ) and r ( µ ) = r ( ζ ). We will show how to resolve theambiguity λµ ∗ ζ and leave the ambiguity ξ ∗ λµ ∗ to the reader. Set λ ′ := λ [( λ,µ )] and µ ′ := µ [( λ,µ )] . Clearly λµ ∗ ζ (3) ❍❍❍❍❍❍❍❍❍❍❍❍❍❍❍ (5) y y tttttttttttttttt λ ′ ( µ ′ ) ∗ ζ (3) (cid:15) (cid:15) λ P ( α,β ) ∈ S ( µ,ζ ) αβ ∗ (1) (cid:15) (cid:15) λ ′ P ( α ′ ,β ′ ) ∈ S ( µ ′ ,ζ ) α ′ ( β ′ ) ∗ (1) (cid:15) (cid:15) X ( α,β ) ∈ S ( µ,ζ ) ( λ ◦ α )( β ) ∗ | {z } A := X ( α ′ ,β ′ ) ∈ S ( µ ′ ,ζ ) ( λ ′ ◦ α ′ )( β ′ ) ∗ | {z } B := Denote by d ( α ) the degree of an α appearing in A , by d ( β ) the degree of a β appearing in A , by d ( α ′ ) thedegree of an α ′ appearing in B and by d ( β ′ ) the degree of a β ′ appearing in B . If lv( µ ′ ) L.
7= lv( µ ) lv( ζ ),then A = 0 = B by Lemma 11. Hence we can assume that lv( µ ) ∼ lv( ζ ). By Lemma 11, A = P (ˆ α, ˆ β ) ∈ S ˆ α ˆ β ∗ ASES FOR KUMJIAN-PASK ALGEBRAS 11 and B = P (˜ α, ˜ β ) ∈ S ˜ α ˜ β ∗ where S = S ( r ( λ ) , s ( ζ ) , d ( λ ) + d ( α ) , d ( β ) , lv( λ ) × (lv( ζ ) | d ( ζ ) |−| d ( µ ) | , . . . , lv( ζ ) ) , (lv( µ ) | d ( µ ) |−| d ( ζ ) | , . . . , lv( µ ) ))and S = S ( r ( λ ′ ) , s ( ζ ) , d ( λ ′ ) + d ( α ′ ) , d ( β ′ ) , lv( λ ′ ) × (lv( ζ ) | d ( ζ ) |−| d ( µ ′ ) | , . . . , lv( ζ ) ) , (lv( µ ′ ) | d ( µ ′ ) |−| d ( ζ ) | , . . . , lv( µ ′ ) )) . Set m := ( d ( λ ) + d ( α )) ∧ d ( β ), o := | d ( β ) | − (( | d ( µ ) | − | d ( ζ ) | ) ∨ m ′ := ( d ( λ ′ ) + d ( α ′ )) ∧ d ( β ′ ) and o ′ := | d ( β ′ ) | − (( | d ( µ ′ ) | − | d ( ζ ) | ) ∨ Assume that | m | ≤ o . Clearly | m | ≤ o ⇔ | ( d ( λ ) + d ( α )) ∧ d ( β ) | ≤ | d ( β ) | − (( | d ( µ ) | − | d ( ζ ) | ) ∨ (cid:12)(cid:12) − | d ( β ) |⇔ | ( d ( λ ) + d ( α ) − d ( β )) ∧ | ≤ − (( | d ( µ ) | − | d ( ζ ) | ) ∨ ⇔ | ( d ( λ ) + d ( ζ ) − d ( µ )) ∧ | ≤ − (( | d ( µ ) | − | d ( ζ ) | ) ∨ L. ⇔ | ( d ( λ ′ ) + d ( ζ ) − d ( µ ′ )) ∧ | ≤ − (( | d ( µ ′ ) | − | d ( ζ ) | ) ∨ ... ⇔ | m ′ | ≤ o ′ Lemma 12 shows that A reduces to A ′ := P ( γ,δ ) ∈ S γδ ∗ and B reduces to B ′ := P ( γ ′ ,δ ′ ) ∈ S γ ′ ( δ ′ ) ∗ where S = S ( r ( λ ) , s ( ζ ) , d ( λ ) + d ( α ) − m, d ( β ) − m, lv( λ ) × (lv( ζ ) | d ( ζ ) |−| d ( µ ) | , . . . , lv( ζ ) ) , (lv( µ ) | d ( µ ) |−| d ( ζ ) | , . . . , lv( µ ) ))and S = S ( r ( λ ′ ) , s ( ζ ) , d ( λ ′ ) + d ( α ′ ) − m ′ , d ( β ′ ) − m ′ , lv( λ ′ ) × (lv( ζ ) | d ( ζ ) |−| d ( µ ′ ) | , . . . , lv( ζ ) ) , (lv( µ ′ ) | d ( µ ′ ) |−| d ( ζ ) | , . . . , lv( µ ′ ) )) . Clearly d ( λ ) + d ( α ) − m = d ( λ ) + d ( α ) − (( d ( λ ) + d ( α )) ∧ d ( β ))= d ( λ ) + d ( α ) + ( − ( d ( λ ) + d ( α )) ∨ − d ( β ))= d ( λ ) + ( − d ( λ ) ∨ ( d ( α ) − d ( β )))= d ( λ ) + ( − d ( λ ) ∨ ( d ( ζ ) − d ( µ )))= d ( λ ) − d ( µ ) + (( d ( µ ) − d ( λ )) ∨ d ( ζ )) L. d ( λ ′ ) − d ( µ ′ ) + (( d ( µ ′ ) − d ( λ ′ )) ∨ d ( ζ )) ... = d ( λ ′ ) + d ( α ′ ) − m ′ and analogously d ( β ) − m = d ( β ′ ) − m ′ . Hence S = S and therefore A ′ = B ′ . case 2 Assume that | m | > o . Then | m ′ | > o ′ (see the previous case). Suppose that | d ( ζ ) | ≥ | d ( µ ) | . Then | m | = | ( d ( λ ) + d ( α )) ∧ d ( β ) | ≤ | d ( β ) | = o . Hence | d ( ζ ) | < | d ( µ ) | . Choose n, n ′ ∈ N k such that n ≤ m , n ′ ≤ m ′ , | n | = o and | n ′ | = o ′ . Lemma 12 showsthat A reduces to A ′ := P ( γ,δ ) ∈ S γδ ∗ and B reduces to B ′ := P ( γ ′ ,δ ′ ) ∈ S γ ′ ( δ ′ ) ∗ where S =( r ( λ ) , s ( ζ ) , d ( λ ) + d ( α ) − n, d ( β ) − n, lv( λ ) , (lv( µ ) | d ( µ ) |−| d ( ζ ) | , . . . , lv( µ ) ))and S = S ( r ( λ ′ ) , s ( ζ ) , d ( λ ′ ) + d ( α ′ ) − n ′ , d ( β ′ ) − n ′ , lv( λ ′ ) , (lv( µ ′ ) | d ( µ ′ ) |−| d ( ζ ) | , . . . , lv( µ ′ ) )) . Clearly S = { ( γ, δ ) } and S = { ( γ ′ , δ ′ ) } where γ ∈ r ( λ )Λ d ( λ )+ d ( α ) − n , lv( γ ) = lv( λ ), δ ∈ s ( ζ )Λ d ( β ) − n ,lv( δ ) = (lv( µ ) | d ( µ ) |−| d ( ζ ) | , . . . , lv( µ ) ), γ ′ ∈ r ( λ ′ )Λ d ( λ ′ )+ d ( α ′ ) − n ′ , lv( γ ′ ) = lv( λ ′ ), δ ′ ∈ s ( ζ )Λ d ( β ′ ) − n ′ andlv( δ ′ ) = (lv( µ ′ ) | d ( µ ′ ) |−| d ( ζ ) | , . . . , lv( µ ′ ) ). Clearly d ( γ ) = 0 since d ( λ ) = 0 and d ( γ ′ ) = 0 since d ( λ ′ ) = 0. Fur-ther d ( δ ) , d ( δ ′ ) = 0 since | d ( ζ ) | < | d ( µ ) | = | d ( µ ′ ) | . Hence ( γ, δ ) , ( γ ′ , δ ′ ) ⊆ ˆ A . By Lemma 7, d ( λ ) ∧ d ( µ ) = 0since ( λ, µ ) ∈ A \ R . Hence, by Lemma 7, lv( λ ) = 1 or lv( µ ) = 1. It follows that lv( γ ) = 1 orlv( δ ) = 1. Analogously lv( γ ′ ) = 1 or lv( δ ′ ) = 1. Hence ( γ, δ ) , ( γ ′ , δ ′ ) ⊆ A by Lemma 6. By Lemma7, ( γ, δ ) ∼ ( γ ′ , δ ′ ). Hence, in view of relation (5), A ′ = γδ ∗ and B ′ = γ ′ ( δ ′ ) ∗ can be reduced to the sameelement of R h X i (namely λ [( γ,δ )] ( µ [( γ,δ )] ) ∗ ).(4),(4):Let n , n ∈ N k \ { } and λ , λ , µ , µ ∈ Λ such that n = n , v := s ( λ ) = s ( µ ), v := s ( λ ) = s ( µ ), λ := λ ◦ λ ∼ ,n = λ ◦ λ ∼ ,n and µ := µ ◦ λ ∼ ,n = µ ◦ λ ∼ ,n . Clearly λµ ∗ (4) $ $ ■■■■■■■■■■■■■■■■■■■ (4) z z ✉✉✉✉✉✉✉✉✉✉✉✉✉✉✉✉✉✉✉ λ µ ∗ − X ξ ∈ v Λ n ,ξ = λ v ,n ( λ ◦ ξ )( µ ◦ ξ ) ∗ | {z } A := λ µ ∗ − X ξ ∈ v Λ n ,ξ = λ v ,n ( λ ◦ ξ )( µ ◦ ξ ) ∗ | {z } B := Set m := d ( λ ) ∧ d ( µ ) and o := max { i | lv( λ ) j = 1 = lv( µ ) j ∀ j ∈ { , . . . , i }} . Choose an ˆ n ∈ N k \ { } suchthat ˆ n ≤ m and | ˆ n | = | m | ∧ o . By the factorisation property there are uniquely determined ˆ λ, ˆ µ ∈ Λ suchthat λ = ˆ λ ◦ λ ∼ , ˆ n and µ = ˆ µ ◦ λ ∼ , ˆ n . Set ˆ v := s (ˆ λ ) = s (ˆ µ ). Further choose ˆ i , . . . , ˆ i | ˆ n | ∈ { , . . . , k } such thatˆ n = e ˆ i + · · · + e ˆ i | ˆ n | .case 1 Assume that d (ˆ λ ) , d (ˆ µ ) = 0 . Set C := λ [(ˆ λ, ˆ µ )] ( µ [(ˆ λ, ˆ µ )] ) ∗ − X ≤ p ≤| ˆ n | , ≤ q ≤ l λ [(ˆ λ ◦ ˆ ξ p,q , ˆ µ ◦ ˆ ξ p,q )] ( µ [(ˆ λ ◦ ˆ ξ p,q , ˆ µ ◦ ˆ ξ p,q )] ) ∗ where ˆ ξ p,q ∈ ˆ v Λ p − P j =0 e ˆ i | ˆ n |− j and lv( ˆ ξ p,q ) = (1 , . . . , , q ) for any 1 ≤ p ≤ | ˆ n | and 2 ≤ q ≤ l . Note that(ˆ λ, ˆ µ ) , (ˆ λ ◦ ξ p,q , ˆ µ ◦ ξ p,q ) ∈ A for any 1 ≤ p ≤ | ˆ n | and 2 ≤ q ≤ l by Lemma 6. We will show that A can bereduced to C . It will follow by symmetry that B also can be reduced to C . Clearly | ˆ n | ≥ | n | . ASES FOR KUMJIAN-PASK ALGEBRAS 13 case 1.1
Assume that | ˆ n | = | n | . Choose i , . . . , i | n | ∈ { , . . . , k } such that n = e i + · · · + e i | n | . Lemma 13 shows that A reduces to A ′ := λ µ ∗ − X ≤ p ≤| n | , ≤ q ≤ l ( λ ◦ ξ p,q )( µ ◦ ξ p,q ) ∗ where ξ p,q ∈ v Λ p − P j =0 e i | n |− j and lv( ξ p,q ) = (1 , . . . , , q ) for any 1 ≤ p ≤ | n | and 2 ≤ q ≤ l . Clearly λ ◦ λ ∼ ,n = λ = ˆ λ ◦ λ ∼ , ˆ n , µ ◦ λ ∼ ,n = µ = ˆ µ ◦ λ ∼ , ˆ n and | ˆ n | = | n | imply that lv( λ ) = lv(ˆ λ ) andlv( µ ) = lv(ˆ µ ). It follows that ( λ , µ ) ∈ A and ( λ , µ ) ∼ (ˆ λ, ˆ µ ) (by Lemma 7). Further it follows that forany 1 ≤ p ≤ | n | = | ˆ n | and 2 ≤ q ≤ l , ( λ ◦ ξ p,q , µ ◦ ξ p,q ) ∈ A and ( λ ◦ ξ p,q , µ ◦ ξ p,q ) ∼ (ˆ λ ◦ ˆ ξ p,q , ˆ µ ◦ ˆ ξ p,q )(also by Lemma 7). Hence, in view of relation (5), A ′ reduces to C .case 1.2 Assume that | ˆ n | > | n | . case 1.2.1 Assume that o ≥ | m | . Then clearly ˆ n = m (= d ( λ ) ∧ d ( µ )) since ˆ n ≤ m and | ˆ n | = | m | . Hence n ≤ m = ˆ n . Set ˜ n := ˆ n − n ≥ n = 0 since | ˆ n | > | n | . Clearly λ = ˆ λ ◦ λ ∼ , ˜ n and µ = ˆ µ ◦ λ ∼ , ˜ n . Hence A = λ µ ∗ − X ξ ∈ v Λ n ,ξ = λ v ,n ( λ ◦ ξ )( µ ◦ ξ ) ∗ = (ˆ λ ◦ λ ∼ , ˜ n )(ˆ µ ◦ λ ∼ , ˜ n ) ∗ − X ξ ∈ v Λ n ,ξ = λ v ,n ( λ ◦ ξ )( µ ◦ ξ ) ∗ . In view of relation (4), A reduces to A ′ := ˆ λ ˆ µ ∗ − X ξ ′ ∈ ˆ v Λ ˜ n ,ξ ′ = λ ˆ v, ˜ n (ˆ λ ◦ ξ ′ )(ˆ µ ◦ ξ ′ ) ∗ − X ξ ∈ v Λ n ,ξ = λ v ,n ( λ ◦ ξ )( µ ◦ ξ ) ∗ . Choose i , . . . , i | n | ∈ { , . . . , k } and ˜ i , . . . , ˜ i | ˜ n | ∈ { , . . . , k } such that n = e i + · · · + e i | n | and ˜ n = e ˜ i + · · · + e ˜ i | ˜ n | . Lemma 13 shows that A ′ reduces to A ′′ := ˆ λ ˆ µ ∗ − X ≤ p ≤| ˜ n | , ≤ q ≤ l (ˆ λ ◦ ˜ ξ p,q )(ˆ µ ◦ ˜ ξ p,q ) ∗ − X ≤ p ≤| n | , ≤ q ≤ l ( λ ◦ ξ p,q )( µ ◦ ξ p,q ) ∗ where ˜ ξ p,q ∈ ˆ v Λ p − P j =0 e ˜ i | ˜ n |− j and lv( ˜ ξ p,q ) = (1 , . . . , , q ) for any 1 ≤ p ≤ | ˜ n | and 2 ≤ q ≤ l and ξ p,q ∈ v Λ p − P j =0 e i | n |− j and lv( ξ p,q ) = (1 , . . . , , q ) for any 1 ≤ p ≤ | n | and 2 ≤ q ≤ l . It follows from Lemma 7 thatfor any 1 ≤ p ≤ | ˜ n | and 2 ≤ q ≤ l , (ˆ λ ◦ ˜ ξ p,q , ˆ µ ◦ ˜ ξ p,q ) ∈ A and (ˆ λ ◦ ˜ ξ p,q , ˆ µ ◦ ˜ ξ p,q ) ∼ (ˆ λ ◦ ˆ ξ p,q , ˆ µ ◦ ˆ ξ p,q ). Further,also by Lemma 7, for any 1 ≤ p ≤ | n | and 2 ≤ q ≤ l , ( λ ◦ ξ p,q , µ ◦ ξ p,q ) ∈ A and ( λ ◦ ξ p,q , µ ◦ ξ p,q ) ∼ (ˆ λ ◦ ˆ ξ | ˜ n | + p,q , ˆ µ ◦ ˆ ξ | ˜ n | + p,q ). Hence, in view of relation (5), A ′′ reduces to C .case 1.2.2 Assume that o < | m | . Then | ˆ n | = o . Hence | ˆ n | < | m | = | d ( λ ) ∧ d ( µ ) |⇒ | ˆ n | − | n | < | d ( λ ) ∧ d ( µ ) | − | n |⇒ | ˆ n | − | n | < | d ( λ ) ∧ d ( µ ) | . Therefore one can choose an ˜ n ∈ N k \ { } such that ˜ n ≤ d ( λ ) ∧ d ( µ ) and | ˜ n | = | ˆ n | − | n | >
0. Clearly λ = ˜ λ ◦ λ ∼ , ˜ n and µ = ˜ µ ◦ λ ∼ , ˜ n where ˜ λ ∈ r ( λ )Λ d ( λ ) − n − ˜ n , ˜ µ ∈ r ( µ )Λ d ( µ ) − n − ˜ n , lv(˜ λ ) ∼ lv( λ ) and lv(˜ µ ) ∼ lv( µ ). Hence A = λ µ ∗ − X ξ ∈ v Λ n ,ξ = λ v ,n ( λ ◦ ξ )( µ ◦ ξ ) ∗ = (˜ λ ◦ λ ∼ , ˜ n )(˜ µ ◦ λ ∼ , ˜ n ) ∗ − X ξ ∈ v Λ n ,ξ = λ v ,n ( λ ◦ ξ )( µ ◦ ξ ) ∗ . Set ˜ v := s (˜ λ ) = s (˜ µ ). In view of relation (4), A reduces to A ′ := ˜ λ ˜ µ ∗ − X ξ ′ ∈ ˜ v Λ ˜ n ,ξ ′ = λ ˜ v, ˜ n (˜ λ ◦ ξ ′ )(˜ µ ◦ ξ ′ ) ∗ − X ξ ∈ v Λ n ,ξ = λ v ,n ( λ ◦ ξ )( µ ◦ ξ ) ∗ . Choose i , . . . , i | n | ∈ { , . . . , k } and ˜ i , . . . , ˜ i | ˜ n | ∈ { , . . . , k } such that n = e i + · · · + e i | n | and ˜ n = e ˜ i + · · · + e ˜ i | ˜ n | . Lemma 13 shows that A ′ reduces to A ′′ := ˜ λ ˜ µ ∗ − X ≤ p ≤| ˜ n | , ≤ q ≤ l (˜ λ ◦ ˜ ξ p,q )(˜ µ ◦ ˜ ξ p,q ) ∗ − X ≤ p ≤| n | , ≤ q ≤ l ( λ ◦ ξ p,q )( µ ◦ ξ p,q ) ∗ where ˜ ξ p,q ∈ ˜ v Λ p − P j =0 e ˜ i | ˜ n |− j and lv( ˜ ξ p,q ) = (1 , . . . , , q ) for any 1 ≤ p ≤ | ˜ n | and 2 ≤ q ≤ l and ξ p,q ∈ v Λ p − P j =0 e i | n |− j and lv( ξ p,q ) = (1 , . . . , , q ) for any 1 ≤ p ≤ | n | and 2 ≤ q ≤ l . It follows from Lemma7 that (˜ λ, ˜ µ ) ∈ A and (˜ λ, ˜ µ ) ∼ (ˆ λ, ˆ µ ). Further it follows from Lemma 7 that for any 1 ≤ p ≤ | ˜ n | and2 ≤ q ≤ l , (˜ λ ◦ ˜ ξ p,q , ˜ µ ◦ ˜ ξ p,q ) ∈ A and (˜ λ ◦ ˜ ξ p,q , ˜ µ ◦ ˜ ξ p,q ) ∼ (ˆ λ ◦ ˆ ξ p,q , ˆ µ ◦ ˆ ξ p,q ). Further, also by Lemma 7, forany 1 ≤ p ≤ | n | and 2 ≤ q ≤ l , ( λ ◦ ξ p,q , µ ◦ ξ p,q ) ∈ A and ( λ ◦ ξ p,q , µ ◦ ξ p,q ) ∼ (ˆ λ ◦ ˆ ξ | ˜ n | + p,q , ˆ µ ◦ ˆ ξ | ˜ n | + p,q ).Hence, in view of relation (5), A ′′ reduces to C .case 2 Assume that d (ˆ λ ) = 0 or d (ˆ µ ) = 0 . This case is very similar to case 1 and hence is omitted.Thus all ambiguities are resolvable. It follows from [1, Theorem 15] that KP R (Λ) = R h X i /I is isomorphicto R h X i irr as an R -module where R h X i irr is the submodule of R h X i consisting of all irreducible elements(cf. [1, Definition 11]). Clearly the elements v ( v ∈ Λ ), λ ( λ ∈ Λ =0 ), λ ∗ ( λ ∈ Λ =0 ) and λµ ∗ (( λ, µ ) ∈ R )form a basis for R h X i irr . (cid:3) References [1] R. Hazrat, R. Preusser,
Applications of normal forms for weighted Leavitt path algebras: simple rings and domains ,accepted by Algebr. Represent. Theor.. 6, 14[2] G. A. Pino, J. Clark, A. an Huef, I. Raeburn,
Kumjian-Pask algebras of higher-rank graphs , Trans. Amer. Math. Soc.365