Weighted Leavitt path algebras that are isomorphic to unweighted Leavitt path algebras
Raimund Preusser
[email protected]
Abstract.
Let K be a field. We characterise the row-finite weighted graphs (E,w) such that the weighted Leavitt path algebra LK(E,w) is isomorphic to an unweighted Leavitt path algebra. Moreover, we prove that if LK(E,w) is locally finite, or Noetherian, or Artinian, or von Neumann regular, or has finite Gelfand-Kirillov dimension, then LK(E,w) is isomorphic to an unweighted Leavitt path algebra.
Key words and phrases:
Leavitt path algebra, weighted Leavitt path algebra
2000 Mathematics Subject Classification:
16S10, 16W10, 16W50, 16D70
1. Introduction
In a series of papers [8, 9, 10, 11] William Leavitt studied algebras that are now denoted by LK(n,n+k) and have been coined Leavitt algebras. Let X=(xij) and Y=(yji) be (n+k)×n and n×(n+k) matrices consisting of symbols xij and yji, respectively. Then for a field K, LK(n,n+k) is the unital K-algebra generated by all xij and yji subject to the relations XY=In+k and YX=In. The algebra LK(n,n+k) can be described as the K-algebra A with a universal left A-module isomorphism An→An+k, cf. [4, second paragraph on p. 35].
(Unweighted) Leavitt path algebras are algebras associated to directed graphs. They were introduced by G. Abrams and G. Aranda Pino in 2005 [1] and independently by P. Ara, M. Moreno and E. Pardo in 2007 [3]. For the directed graph
[TABLE]
with one vertex and k+1 loops one recovers the Leavitt algebra LK(1,k+1). The definition and the development of the theory were inspired on the one hand by Leavitt’s construction of LK(1,k+1) and on the other hand by the Cuntz algebras On [5] and the Cuntz-Krieger algebras in C∗-algebra theory [16]. The Cuntz algebras and later Cuntz-Krieger type C∗-algebras revolutionised C∗-theory, leading ultimately to the astounding
Kirchberg-Phillips classification theorem [12]. The Leavitt path algebras have created the same type of stir in the algebraic community.
There have been several attempts to introduce a generalisation of the Leavitt path algebras which would cover the algebras LK(n,n+k), n≥2 as well. In 2013, R. Hazrat [6] introduced weighted Leavitt path algebras. These are algebras associated to weighted graphs. For the weighted graph
[TABLE]
with one vertex and n+k loops of weight n one recovers the Leavitt algebra LK(n,n+k). If the weights of all the edges are 1, then the weighted Leavitt path algebras reduce to the unweighted Leavitt path algebras.
Which are the new examples in the class of weighted Leavitt path algebras? In [7] it was shown that any simple or graded simple weighted Leavitt path algebra is isomorphic to an unweighted Leavitt path algebra. In [14] and [15] it was shown that any finite-dimensional or Noetherian weighted Leavitt path algebra is isomorphic to an unweighted Leavitt path algebra. Furthermore, graph-theoretic criterions that are sufficient and necessary for LK(E,w) being finite-dimensional/Noetherian were found (see [15, Theorems 25 and 52]). On the other hand, it was shown in [13, Corollary 16], that the class of weighted Leavitt path algebras contains infinitely many domains which are neither isomorphic to an unweighted Leavitt path algebra nor to a Leavitt algebra LK(n,n+k).
As examples consider the weighted graphs
[TABLE]
where a number above or below an edge indicates the weight of that edge. In [14, Example 40] it was shown that LK(E,w)≅LK(F) where F is the directed graph
[TABLE]
In [13, Example 21] it was shown that LK(E′,w′)≅LK(F′) where F′ is the directed graph
[TABLE]
But it remained unclear if LK(E′′,w′′) is isomorphic to an unweighted Leavitt path algebra. It will follow from the results of this paper that LK(E′′,w′′) cannot be isomorphic to an unweighted Leavitt path algebra.
In this paper we obtain a graph-theoretic criterion that is sufficient and necessary for LK(E,w) being isomorphic to an unweighted Leavitt path algebra (Condition (LPA), cf. Definition 15). Moreover, we prove that if LK(E,w) is Artinian, or von Neumann regular, or has finite Gelfand-Kirillov dimension, then LK(E,w) is isomorphic to an unweighted Leavitt path algebra.
The rest of the paper is organised as follows.
In Section 2 we recall some standard notation which is used throughout the paper.
In Section 3 we recall the definitions of the unweighted and weighted Leavitt path algebras.
In Section 4 we introduce Condition (LPA).
In Section 5 we prove that if (E,w) is a row-finite weighted graph that satisfies Condition (LPA), then LK(E,w) is isomorphic to an unweighted Leavitt path algebra.
In Section 6 we prove that if (E,w) is a row-finite weighted graph that does not satisfy Condition (LPA), then LK(E,w) is not isomorphic to an unweighted Leavitt path algebra. Moreover, we prove that if LK(E,w) is Artinian, or von Neumann regular, or has finite Gelfand-Kirillov dimension, then LK(E,w) is isomorphic to an unweighted Leavitt path algebra. We also prove again that if LK(E,w) is locally finite or Noetherian, then LK(E,w) is isomorphic to an unweighted Leavitt path algebra (that has already been shown in [15], but the paper was never published in a journal).
In Section 7 we summarise the main results of this paper.
2. Notation
Throughout the paper K denotes a field. By a K-algebra we mean an associative (but not necessarily commutative or unital) K-algebra. By an ideal we mean a two-sided ideal. N denotes the set of positive integers, N0 the set of nonnegative integers, Z the set of integers and R+ the set of positive real numbers.
3. Unweighted and weighted Leavitt path algebras
Definition 1**.**
A (directed) graph is a quadruple E=(E0,E1,s,r) where E0 and E1 are sets and s,r:E1→E0 maps. The elements of E0 are called vertices and the elements of E1 edges. If e is an edge, then s(e) is called its source and r(e) its range.
Remark 2**.**
- (a)
Let E be a graph, v∈E0 a vertex and e∈E1 an edge. Then we say that v emits e if s(e)=v and v receives e if r(e)=v.
2. (b)
In this article all graphs are assumed to be row-finite. Recall that a graph E=(E0,E1,s,r) is called row-finite if s−1(v) is a finite set for any vertex v.
Definition 3**.**
Let E be a graph. The K-algebra LK(E) presented by the generating set {v,e,e∗∣v∈E0,e∈E1} and the relations
- (i)
uv=δuvu(u,v∈E0),
2. (ii)
s(e)e=e=er(e), r(e)e∗=e∗=e∗s(e)(e∈E1),
3. (iii)
e∗f=δefr(e)(v∈E0,e,f∈s−1(v)) and
4. (iv)
e∈s−1(v)∑ee∗=v(v∈E0,s−1(v)=∅)
is called the (unweighted) Leavitt path algebra of E.
Remark 4**.**
Let E be a graph and A a K-algebra that contains a set X={αv,βe,γe∣v∈E0,e∈E1} such that
- (i)
the αv’s are pairwise orthogonal idempotents,
2. (ii)
αs(e)βe=βe=βeαr(e), αr(e)γe=γe=γeαs(e)(e∈E1),
3. (iii)
γeβf=δefαr(e)(v∈E0,e,f∈s−1(v)) and
4. (iv)
e∈s−1(v)∑βeγe=αv(v∈E0,s−1(v)=∅).
We call X an E-family in A. By the relations defining LK(E), there exists a unique K-algebra homomorphism ϕ:LK(E)→A such that ϕ(v)=αv, ϕ(e)=βe and ϕ(e∗)=γe for all v∈E0 and e∈E1. We will refer to this as the Universal Property of LK(E).
Definition 5**.**
A weighted graph is a pair (E,w) where E is a graph and w:E1→N is a map. If e∈E1, then w(e) is called the weight of e. For a vertex v∈E0 we set w(v):=max{w(e)∣e∈s−1(v)} with the convention max∅=0.
Definition 6**.**
Let (E,w) be a weighted graph. The K-algebra LK(E,w) presented by the generating set {v,ei,ei∗∣v∈E0,e∈E1,1≤i≤w(e)} and the relations
- (i)
uv=δuvu(u,v∈E0),
2. (ii)
s(e)ei=ei=eir(e), r(e)ei∗=ei∗=ei∗s(e)(e∈E1,1≤i≤w(e)),
3. (iii)
1≤i≤w(v)∑ei∗fi=δefr(e)(v∈E0,e,f∈s−1(v)) and
4. (iv)
e∈s−1(v)∑eiej∗=δijv(v∈E0,1≤i,j≤w(v))
is called the weighted Leavitt path algebra of (E,w). In relations (iii) and (iv) we set ei and ei∗ zero whenever i>w(e).
Example 7**.**
If (E,w) is a weighted graph such that w(e)=1 for all e∈E1, then LK(E,w) is isomorphic to the unweighted Leavitt path algebra LK(E).
Example 8**.**
Let n≥1 and k≥0. Let (E,w) be the weighted graph
[TABLE]
with one vertex v and n+k edges e(1),…,e(n+k) each of which has weight n. Then LK(E,w) is isomorphic to the Leavitt algebra LK(n,n+k), for details see [6, Example 5.5] or [7, Example 4].
Remark 9**.**
Let (E,w) be a weighted graph and A a K-algebra that contains a set X={αv,βe,i,γe,i∣v∈E,e∈E1,1≤i≤w(e)} such that
- (i)
the αv’s are pairwise orthogonal idempotents,
2. (ii)
αs(e)βe,i=βe,i=βe,iαr(e), αr(e)γe,i=γe,i=γe,iαs(e)(e∈E1,1≤i≤w(e)),
3. (iii)
1≤i≤w(v)∑γe,iβf,i=δefαr(e)(v∈E0,e,f∈s−1(v)) and
4. (iv)
e∈s−1(v)∑βe,iγe,j=δijαv(v∈E0,1≤i,j≤w(v)).
In relations (iii) and (iv) we set βe,i and γe,i zero whenever i>w(e). We call X an (E,w)-family in A. By the relations defining LK(E,w), there exists a unique K-algebra homomorphism ϕ:LK(E,w)→A such that ϕ(v)=αv, ϕ(ei)=βe,i and ϕ(ei∗)=γe,i for all v∈E0, e∈E1 and 1≤i≤w(e). We will refer to this as the Universal Property of LK(E,w).
Remark 10**.**
Let (E,w) be a weighted graph. Then LK(E,w) has the properties (a)-(d) below, for details see [6, Proposition 5.7].
- (a)
If E0 is a finite set, then LK(E,w) is a unital ring (with v∈E0∑v as multiplicative identity).
2. (b)
LK(E,w) has a set of local units, namely the set of all finite sums of distinct elements of E0. Recall that an associative ring R is said to have a set of local units X in case X is a set of idempotents in R having the property that for each finite subset S⊆R there exists an x∈X such that xsx=s for any s∈S.
3. (c)
There is an involution ∗ on LK(E,w) mapping k↦k, v↦v, ei↦ei∗ and ei∗↦ei for any k∈K, v∈E0, e∈E1 and 1≤i≤w(e).
4. (d)
Set n:=sup{w(e)∣e∈E1}. One can define a Zn-grading on LK(E,w) by setting deg(v):=0, deg(ei):=ϵi and deg(ei∗):=−ϵi for any v∈E0, e∈E1 and 1≤i≤w(e). Here ϵi denotes the element of Zn whose i-th component is 1 and whose other components are [math].
4. The Condition (LPA)
We start with a couple of definitions.
Definition 11**.**
Let E be a graph. A path is a nonempty word p=x1…xn over the alphabet E0∪E1 such that either xi∈E1 (i=1,…,n) and r(xi)=s(xi+1) (i=1,…,n−1) or n=1 and x1∈E0. By definition, the length ∣p∣ of p is n in the first case and [math] in the latter case. We set s(p):=s(x1) and r(p):=r(xn) (here we use the convention s(v)=v=r(v) for any v∈E0).
Definition 12**.**
Let E be a graph and v∈E0. A closed path (based at v) is a path p such that ∣p∣>0 and s(p)=r(p)=v. A cycle (based at v) is a closed path p=x1…xn based at v such that s(xi)=s(xj) for any i=j.
Definition 13**.**
Let E be a graph. If u,v∈E0 and there is a path p in E such that s(p)=u and r(p)=v, then we write u≥v. If u∈E0, then T(u):={v∈E0∣u≥v} is called the tree of u. If X⊆E0, we define T(X):=v∈X⋃T(v). Two edges e,f∈E1 are called in line if e=f or r(e)≥s(f) or r(f)≥s(e)
Definition 14**.**
Let (E,w) be a weighted graph. An edge e∈E1 is called unweighted if w(e)=1 and weighted if w(e)>1. The subset of E1 consisting of all unweighted edges is denoted by Euw1 and the subset consisting of all weighted edges by Ew1.
Now we can introduce Condition (LPA).
Definition 15**.**
We say that a weighted graph (E,w) satisfies Condition (LPA) if the following holds true:
- (LPA1)
Any vertex v∈E0 emits at most one weighted edge.
2. (LPA2)
Any vertex v∈T(r(Ew1)) emits at most one edge.
3. (LPA3)
If two weighted edges e,f∈Ew1 are not in line, then T(r(e))∩T(r(f))=∅.
4. (LPA4)
If e∈Ew1 and c is a cycle based at some vertex v∈T(r(e)), then e belongs to c.
Each of the Conditions (LPA1)-(LPA4) in Definition 15 above “forbids” a certain constellation in the weighted graph (E,w). The pictures below illustrate these forbidden constellations. Symbols above or below edges indicate the weight. A dotted arrow stands for a path.
- (LPA1)
[TABLE]
2. (LPA2)
[TABLE]
3. (LPA3)
[TABLE]
4. (LPA4)
[TABLE]
Remark 16**.**
Conditions (LPA1), (LPA2) and (LPA3) already appeared in [14] and [15]. They were independently found by N. T. Phuc. Condition (LPA4) is new, this condition is slightly weaker then Condition (iv) in [15, Definition 19].
5. Presence of Condition (LPA)
Lemma 17**.**
Let (E,w) be a weighted graph that satisfies Condition (LPA). If e and f are distinct edges such that s(e),s(f)∈T(r(Ew1)), then r(e)=r(f).
Proof.
Let e,f∈E1 such that s(e),s(f)∈T(r(Ew1)) and r(e)=r(f). We will show that e=f. Since s(e),s(f)∈T(r(Ew1)), there are g,h∈Ew1 such that s(e)∈T(r(g)) and s(f)∈T(r(h)). It follows that r(e)=r(f)∈T(r(g))∩T(r(h)). Since (E,w) satisfies Condition (LPA3), g and h are in line. It follows that s(e),s(f)∈T(r(g)) or s(e),s(f)∈T(r(h)). W.l.o.g. assume that s(e),s(f)∈T(r(g)).
- Case 1
Assume that there is a cycle c based at some vertex v∈T(r(g)). Since (E,w) satisfies (LPA4), g belongs to c. Write c=α(1)…α(n) where α(1),…,α(n)∈E1. Set xi:=s(α(i)) (1≤i≤n). Then, in view of (LPA2), we have T(r(g))={x1,…,xn}. Moreover, each xi emits precisely one edge, namely α(i). Since s(e),s(f)∈T(r(g)), we get that s(e)=xi and s(f)=xj for some 1≤i,j≤n. Hence e=α(i) and f=α(j). Since r(e)=r(f), it follows that i=j and hence e=f.
2. Case 2
Assume that no cycle is based at a vertex in T(r(g)). Since s(e),s(f)∈T(r(g)), there are paths p and q such that s(p)=r(g)=s(q), r(p)=s(e) and r(q)=s(f). Clearly pe and qf are paths starting at r(g) and ending at r(e)=r(f). It follows from (LPA2) and the assumption that no cycle is based at a vertex in T(r(g)), that pe=qf. Hence e=f.
∎
Recall that if E is a graph, then a vertex v that does not emit any edges is called a sink.
Lemma 18**.**
Let (E,w) be a weighted graph that satisfies Condition (LPA). Then there is a weighted graph (E~,w~) such that the ranges of the weighted edges in (E~,w~) are sinks, no vertex in (E~,w~) emits or receives two distinct weighted edges, and LK(E~,w~)≅LK(E,w).
Proof.
Set Z:=T(r(Ew1)). Define a weighted graph (E~,w~) by E~0=E0, E~1=E~Z1⊔E~Zc1 where
[TABLE]
s~(e(i))=r(e), r~(e(i))=s(e) and w~(e(i))=1 for any e(i)∈E~Z1 and s~(e)=s(e), r~(e)=r(e) and w~(e)=w(e) for any e∈E~Zc1. We have divided the rest of the proof into three parts. In Part I we show that the ranges of the weighted edges in (E~,w~) are sinks, in Part II we show that no vertex in (E~,w~) emits or receives two distinct weighted edges, and in Part III we show that LK(E~,w~)≅LK(E,w).
Part I Let e~∈E~w1. We will show that r~(e~) is a sink in (E~,w~). Clearly e~∈E~Zc1 since all the edges in E~Z1 have weight one in (E~,w~). Hence there is an e∈E1,s(e)∈Z such that e~=e. Clearly w(e)=w~(e)=w~(e~)>1. Now suppose that there is an f~∈E~1 such that s~(f~)=r~(e~).
- Case 1
Assume that f~∈E~Z1. Then there is an f∈E1,s(f)∈Z and an i∈{1,…,w(f)} such that f~=f(i) (note that e=f, since s(e)∈Z). It follows that r(e)=r~(e)=r~(e~)=s~(f~)=s~(f(i))=r(f). Since s(f)∈Z=T(r(Ew1)), there is a g∈Ew1 such that s(f)∈T(r(g)). It follows that r(f)∈T(r(e))∩T(r(g)). Since (E,w) satisfies Condition (LPA3), we get that e and g are in line and hence e=g or r(e)≥s(g) or r(g)≥s(e).
- Case 1.1
Assume that e=g. Since s(f)∈T(r(g))=T(r(e)), there is a path p such that s(p)=r(e) and r(p)=s(f). Since r(f)=r(e), we have a closed path pf based at r(e). That implies the existence of a cycle c based at r(e). Since (E,w) satisfies (LPA4), e belongs to c and therefore s(e)∈T(r(e)). Now we get the contradiction s(e)∈T(r(e))⊆T(r(Ew1))=Z.
2. Case 1.2
Assume that r(e)≥s(g). Then there is a path p such that s(p)=r(e) and r(p)=s(g). Since s(f)∈T(r(g)), there is a path q such that s(q)=r(g) and r(q)=s(f). Since r(f)=r(e), we have a closed path pgqf based at r(e). Now we can proceed as in Case 1.1 to get a contradiction.
3. Case 1.3
Assume that r(g)≥s(e). Then we get the contradiction s(e)∈T(r(g))⊆T(r(Ew1))=Z.
2. Case 2
Assume that f~∈E~Zc1. Then there is an f∈E1,s(f)∈Z such that f~=f. It follows that r(e)=r~(e)=r~(e~)=s~(f~)=s~(f)=s(f). Hence we get the contradiction s(f)=r(e)∈T(r(Ew1))⊆Z.
Thus the ranges of the weighted edges in (E~,w~) are sinks.
Part II Assume that there are distinct e~,f~∈E~w1 such that s~(e~)=s~(f~). Clearly e~,f~∈E~Zc1 since all the edges in E~Z1 have weight one in (E~,w~). Hence there are distinct e,f∈E1,s(e),s(f)∈Z such that e~=e and f~=f. It follows that s(e)=s~(e)=s~(e~)=s~(f~)=s~(f)=s(f) which contradicts the assumption that (E,w) satisfies Condition (LPA1) (note that w(e)=w~(e~)>1 and w(f)=w~(f~)>1). Thus no vertex emits two distinct weighted edges in (E~,w~).
Now assume that there are distinct e~,f~∈E~w1 such that r~(e~)=r~(f~). Clearly e~,f~∈E~Zc1 since all the edges in E~Z1 have weight one in (E~,w~). Hence there are distinct e,f∈E1,s(e),s(f)∈Z such that e~=e and f~=f. It follows that r(e)=r~(e)=r~(e~)=r~(f~)=r~(f)=r(f). Since (E,w) satisfies Condition (LPA3), we get that e and f are in line. Since e and f are distinct, it follows that r(e)≥s(f) or r(f)≥s(e). But in the first case we get the contradiction s(f)∈Z and in the second case the contradiction s(e)∈Z. Thus no vertex receives two distinct weighted edges in (E~,w~).
Part III It remains to show that LK(E~,w~)≅LK(E,w). Set X:={v,ei,ei∗∣v∈E0,e∈E1,1≤i≤w(e)} and X~:={v~,e~i,e~i∗∣v~∈E~0,e~∈E~1,1≤i≤w~(e~)}. Let K⟨X⟩ and K⟨X~⟩ be the free K-algebras generated by X and X~, respectively. Then the bijection X→X~ mapping
[TABLE]
induces an isomorphism ϕ:K⟨X⟩→K⟨X~⟩. Let I and I~ be the ideals of K⟨X⟩ and K⟨X~⟩ generated by the relations (i)-(iv) in Definition 6, respectively
(hence LK(E,w)≅K⟨X⟩/I and LK(E~,w~)≅K⟨X~⟩/I~). In order to show that LK(E,w)≅LK(E~,w~) it suffices to show that ϕ(I)=I~. Set
[TABLE]
[TABLE]
and for any v∈E0
[TABLE]
and
[TABLE]
Then I is generated by A(i), A(ii), the Av(iii)’s and the Av(iv)’s. Analogously define subsets B(i),B(ii),Bv(iii) (v∈E~0),Bv(iv) (v∈E~0) of K⟨X~⟩. Then I~ is generated by B(i), B(ii), the Bv(iii)’s and the Bv(iv)’s. Clearly ϕ(A(i))=B(i) and ϕ(A(ii))=B(ii). One checks easily that ϕ(Av(iii))=Bv(iii) and ϕ(Av(iv))=Bv(iv) if v∈Z.
Let now v∈Z be not a sink in (E,w) (if v∈Z is a sink in (E,w), then Av(iii)=Av(iv)=∅). Then we have s−1(v)={e} for some e∈E1 since (E,w) satisfies Condition (LPA2). Set vˉ:=r(e). Clearly
[TABLE]
and
[TABLE]
It follows from Lemma 17 that s~−1(vˉ)={e(1),…,e(w(e))}. Hence
[TABLE]
and
[TABLE]
Clearly ϕ(Av(iii))=Bvˉ(iv) and ϕ(Av(iv))=Bvˉ(iii). It follows from Lemma 17 that the map ˉ :v↦vˉ defines a bijection between the elements of Z that are not a sink in (E,w) and the elements of Z that are not a sink in (E~,w~). Hence ϕ(I)=I~ and thus LK(E~,w~)≅LK(E,w).
∎
Example 19**.**
Consider the weighted graph
[TABLE]
One checks easily that (E,w) satisfies Condition (LPA) (note that T(r(Ew1))={t,u,x,y,z}). Let (E~,w~) be defined as in the proof of Lemma 18. Then (E~,w~) is the weighted graph
[TABLE]
There is only one weighted edge in (E~,w~), namely f, and its range is a sink. The proof of Lemma 18 shows that LK(E,w)≅LK(E~,w~).
Lemma 20**.**
Let (E,w) be a weighted graph such that the ranges of the weighted edges are sinks and no vertex emits or receives two distinct weighted edges. Then there is a graph E~ such that LK(E,w)≅LK(E~).
Proof.
If v∈r(Ew1), then there is a unique edge gv∈Ew1 such that r(gv)=v (since no vertex in (E,w) receives two distinct weighted edges). Define a graph E~ by
[TABLE]
(note that if e∈E1, then s(e)∈E0∖r(Ew1) since the elements of r(Ew1) are sinks).
We have divided the rest of the proof into three parts. In Part I we define a homomorphism ϕ:LK(E,w)→LK(E~), in Part II we define a homomorphism ϕ~:LK(E~)→LK(E,w), and in Part III we show that ϕ and ϕ~ are inverse to each other.
Part I Set
[TABLE]
In order to show that X:={αv,βe,i,γei∣v∈E0,e∈E1,1≤i≤w(e)} is an (E,w)-family in LK(E~), one has to show that the relations (i)-(iv) in Remark 9 are satisfied. We leave (i) and (ii) to the reader and show (iii) and (iv).
First we check (iii). Let v∈E0 and e,f∈s−1(v). We have to show that 1≤i≤w(v)∑γe,iβf,i=δefαr(e).
- Case 1
Assume that e,f∈Euw1.
- Case 1.1
Assume that r(e),r(f)∈r(Ew1). Then
[TABLE]
2. Case 1.2
Assume that r(e)∈r(Ew1) and r(f)∈r(Ew1). Then
[TABLE]
3. Case 1.3
Assume that r(e)∈r(Ew1) and r(f)∈r(Ew1). Then
[TABLE]
4. Case 1.4
Assume that r(e),r(f)∈r(Ew1). Then
[TABLE]
2. Case 2
Assume that e∈Euw1 and f∈Ew1.
- Case 2.1
Assume that r(e)∈r(Ew1). Then
[TABLE]
2. Case 2.2
Assume that r(e)∈r(Ew1). Then
[TABLE]
3. Case 3
Assume that e∈Ew1 and f∈Euw1. This case is similar to Case 2 and therefore is ommitted.
4. Case 4
Assume that e,f∈Ew1. Since no vertex emits two distinct weighted edges in (E,w), it follows that e=f and w(v)=w(e). Clearly
[TABLE]
(note that e(j) is the only edge emitted by r(e)(j) in E~).
Thus (iii) holds.
Next we check (iv). Let v∈E0 and 1≤i,j≤w(v). Note that the existence of i,j with the property 1≤i,j≤w(v) implies that w(v)≥1, i.e. that v is not a sink in (E,w). It follows that v∈E0∖r(Ew1). We have to show that e∈s−1(v)∑βe,iγe,j=δijαv.
- Case (a)
Assume that i=j=1. Clearly
[TABLE]
Since r~(e(j))=r(e)(j) for any e∈Euw1,r(e)∈r(Ew1), we have e(j)(e(k))∗=0 in LK(E~) whenever j=k. Hence
[TABLE]
One checks easily that
[TABLE]
Hence T2=v=δ11αv.
2. Case (b)
Assume that i=1 and j>1. Then w(v)≥j>1 and hence v emits precisely one weighted edge f. Since γe,j=0 whenever j≥w(e), we have
[TABLE]
(note that r~(f(1))=r(f)(1)=r(f)(j)=s~(f(j))).
3. Case (c)
Assume that i>1 and j=1. Then v emits precisely one weighted edge f. Clearly
[TABLE]
(note that s~(f(i))=r(f)(i)=r(f)(1)=r~(f(1))).
4. Case (d)
Assume that i,j>1. Then v emits precisely one weighted edge f. Clearly
[TABLE]
Thus (iv) holds too and hence X is an (E,w)-family in LK(E~). By the Universal Property of LK(E,w) there is a unique K-algebra homomorphism ϕ:LK(E,w)→LK(E~) such that ϕ(v)=αv, ϕ(ei)=βe,i and ϕ(ei∗)=γe,i for all v∈E0, e∈E1 and 1≤i≤w(e).
Part II
Set
[TABLE]
In order to show that X~:={α~v~,β~e~,γ~e~∣v~∈E~0,e~∈E~1} is an E~-family in LK(E,w), one has to show that the relations (i)-(iv) in Remark 4 are satisfied. We leave (i) and (ii) to the reader and show (iii) and (iv).
First we check (iii). Let v~∈E~0 and e~,f~∈s~−1(v~). We have to show that γ~e~β~f~=δe~f~α~r~(e~).
- Case 1
Assume that v~∈M. Then e~,f~∈A∪B∪C since s~−1(D)⊆N.
- Case 1.1
Assume that e~,f~∈A. Then there are e,f∈Euw1,r(e),r(f)∈r(Ew1) such that e~=e and f~=f. Clearly
γ~e~β~f~=e1∗f1=δefr(e)=δe~f~α~r~(e~).
2. Case 1.2
Assume that e~∈A and f~∈B. Then there is an e∈Euw1,r(e)∈r(Ew1) such that e~=e. Moreover, there is an f∈Euw1,r(f)∈r(Ew1) and an 1≤i≤w(gr(f)) such that f~=f(i). Clearly e=f and e~=f~. Hence
γ~e~β~f~=e1∗f1(gir(f))∗gir(f)=δef(gir(f))∗gir(f)=0=δe~f~α~r~(e~).
3. Case 1.3
Assume that e~∈A and f~∈C. Then there is an e∈Euw1,r(e)∈r(Ew1) such that e~=e. Moreover, there is an f∈Ew1 such that f~=f(1). Clearly e=f and e~=f~. Hence γ~e~β~f~=e1∗f1=δefr(e)=0=δe~f~α~r~(e~).
4. Case 1.4
Assume that e~∈B and f~∈A. Then there is an e∈Euw1,r(e)∈r(Ew1) and an 1≤i≤w(gr(e)) such that e~=e(i). Moreover, there there is an f∈Euw1,r(f)∈r(Ew1) such that f~=f. Clearly e=f and e~=f~. Hence
γ~e~β~f~=(gir(e))∗gir(e)e1∗f1=δef(gir(e))∗gir(e)=0=δe~f~α~r~(e~).
5. Case 1.5
Assume that e~,f~∈B. Then there are e,f∈Euw1,r(e),r(f)∈r(Ew1) and 1≤i≤w(gr(e)),1≤j≤w(gr(f)) such that e~=e(i) and f~=f(j). Clearly γ~e~β~f~=(gir(e))∗gir(e)e1∗f1(gjr(f))∗gjr(f)=δef(gir(e))∗gir(e)(gjr(f))∗gjr(f)=δefδij(gir(e))∗gir(e)=δe~f~α~r~(e~).
6. Case 1.6
Assume that e~∈B and f~∈C. Then there is an e∈Euw1,r(e)∈r(Ew1) and a 1≤i≤w(gr(e)) such that e~=e(i). Moreover, there is an f∈Ew1 such that f~=f(1). Clearly e=f and e~=f~. Hence γ~e~β~f~=(gir(e))∗gir(e)e1∗f1=δef(gir(e))∗gir(e)=0=δe~f~α~r~(e~).
7. Case 1.7
Assume that e~∈C and f~∈A. Then there is an e∈Ew1 such that e~=e(1). Moreover, there is an f∈Euw1,r(f)∈r(Ew1) such that f~=f. Clearly e=f and e~=f~. Hence
γ~e~β~f~=e1∗f1=δefr(e)=0=δe~f~α~r~(e~).
8. Case 1.8
Assume that e~∈C and f~∈B. Then there is an e∈Ew1 such that e~=e(1). Moreover, there is an f∈Euw1,r(f)∈r(Ew1) and an 1≤i≤w(gr(f)) such that f~=f(i). Clearly e=f and e~=f~. Hence
γ~e~β~f~=e1∗f1(gir(f))∗gir(f)=δef(gir(f))∗gir(f)=0=δe~f~α~r~(e~).
9. Case 1.9
Assume that e~,f~∈C. Then there are e,f∈Ew1 such that e~=e(1) and f~=f(1). Since s(e)=s~(e(1))=s~(e~)=s~(f~)=s~(f(1))=s(f), we have e=f (because no vertex in (E,w) emits two distinct weighted edges). It follows that e~=f~. Clearly γ~e~β~f~=e1∗e1=δe~f~α~r~(e~).
2. Case 2
Assume that v~∈N. Then v~=v(i) for some v∈r(Ew1) and 1≤i≤w(gv). One checks easily that s~−1(v~)=∅ if i=1 and s~−1(v~)={(gv)(i)} if i>1. It follows that i>1 and e~=f~=(gv)(i). Hence γ~e~β~f~=giv(giv)∗=s(gv)=δe~f~α~r~(e~).
Thus (iii) holds.
Next we check (iv). Let v~∈E~0 such that s~−1(v)=∅. We have to show that e~∈s~−1(v~)∑β~e~γ~e~=α~v~.
- Case (a)
Assume that v~∈M. Then v~=v for some v∈E0∖r(Ew1). One checks easily that
[TABLE]
Hence
[TABLE]
2. Case (b)
Assume that v~∈N. Then v~=v(i) for some v∈r(Ew1) and 1≤i≤w(gv). As mentioned above we have s~−1(v~)=∅ if i=1 and s~−1(v~)={(gv)(i)} if i>1. Since by assumption s~−1(v~)=∅, it follows that i>1 and e~∈s~−1(v~)∑β~e~γ~e~=β~(gv)(i)γ~(gv)(i)=(giv)∗giv=α~v~.
Thus (iv) holds too and hence X~ is an E~-family in LK(E,w). By the Universal Property of LK(E~) there is a unique K-algebra homomorphism ϕ~:LK(E~)→LK(E,w) such that ϕ~(v~)=α~v~, ϕ~(e~)=β~e~ and ϕ~(e~∗)=γ~e~ for all v~∈E~0 and e~∈E~1.
Part III First we show that ϕ~∘ϕ=idLK(E,w). Clearly it suffices to show that ϕ~∘ϕ fixes all elements of {v,ei,ei∗∣v∈E0,e∈E1,1≤i≤w(e)} since these elements generate LK(E,w) as a K-algebra. One checks easily that ϕ~∘ϕ fixes all elements v,ei,ei∗ where v∈E0 and e∈Ew1 or e∈Euw1,r(e)∈r(Ew1). Let now e∈Euw1,r(e)∈r(Ew1). Then
[TABLE]
Similarly one can show that ϕ(ϕ(e1∗))=e1∗ in this case. Hence ϕ~∘ϕ=idLK(E,w).
Now we show that ϕ∘ϕ~=idLK(E~). Clearly it suffices to show that ϕ∘ϕ~ fixes all elements of {v~,e~,e~∗∣v~∈E~0,e~∈E~1} since these elements generate LK(E~) as a K-algebra. One checks easily that ϕ∘ϕ~ fixes all elements v~,e~,e~∗ where v~∈E~0 and e~∈E~1∖B. Let now e~∈B. Then e~=e(i) for some e∈Euw1,r(e)∈r(Ew1) and 1≤i≤w(gr(e)). Clearly
[TABLE]
But ((gr(e))(1))∗(gr(e))(1)=r~((gr(e))(1))=r(gr(e))(1)=r(e)(1) in LK(E~). Since r~(e(j))=r(e)(j), it follows that j=1∑w(gr(e))e(j)((gr(e))(1))∗(gr(e))(1)=e(1)=e~ if i=1. Now assume that i>1. One checks easily that s~−1(r(e)(i))={(gr(e))(i)}. Hence (gr(e))(i)((gr(e))(i))∗=r(e)(i) in LK(E~). Since r~(e(j))=r(e)(j), it follows that j=1∑w(gr(e))e(j)(gr(e))(i)((gr(e))(i))∗=e(i)=e~. Hence we have shown that ϕ(ϕ~(e~))=e~ if e~∈B. Similarly one can show that ϕ(ϕ~(e~∗))=e~∗ in this case. Hence ϕ∘ϕ~=idLK(E~) and thus LK(E,w)≅LK(E~).
∎
Example 21**.**
Consider the weighted graph
[TABLE]
Let E~ be defined as in the proof of Lemma 20. Then E~ is the graph
[TABLE]
The proof of Lemma 20 shows that LK(E,w)≅LK(E~).
Lemma 18 and 20 directly imply the theorem below.
Theorem 22**.**
Let (E,w) be a weighted graph that satisfies Condition (LPA). Then the weighted Leavitt path algebra LK(E,w) is isomorphic to an unweighted Leavitt path algebra.
Example 23**.**
Consider the weighted graph
[TABLE]
which satisfies Condition (LPA), and the graph
[TABLE]
By Examples 19 and 21 we have LK(E,w)≅LK(E~).
6. Abscence of Condition (LPA)
Throughout this subsection (E,w) denotes a weighted graph. We start by recalling the basis result of [7]. Set X:={v,ei,ei∗∣v∈E0,e∈E1,1≤i≤w(e)}, let ⟨X⟩ the set of all nonempty words over X and set ⟨X⟩:=⟨X⟩∪{empty word}. Together with juxtaposition of words ⟨X⟩ becomes a semigroup and ⟨X⟩ a monoid. If A,B∈⟨X⟩, then B is called a subword of A if there are C,D∈⟨X⟩ such that A=CBD and a suffix of A if there is a C∈⟨X⟩ such that A=CB.
Definition 24**.**
Let p=x1…xn∈⟨X⟩. Then p is called a d-path if either x1,…,xn∈X∖E0 and r(xi)=s(xi+1) (1≤i≤n−1) or x1∈E0 and n=1. Here we use the convention s(v):=v, r(v):=v, s(ei):=s(e), r(ei):=r(e), s(ei∗):=r(e) and r(ei∗):=s(e) for any v∈E0, e∈E1 and 1≤i≤w(e).
Remark 25**.**
Let E^ be the directed graph associated to (E,w) and E^d the double graph of E^ (see [14, Definitions 2 and 8]). The d-paths are precisely the paths in the double graph E^d.
Fix for any v∈E0 such that s−1(v)=∅ an edge ev∈s−1(v) such that w(ev)=w(v). The ev’s are called special edges.
Definition 26**.**
The words eiv(ejv)∗ (v∈E0,1≤i,j≤w(v)) and e1∗f1 (e,f∈E1)
in ⟨X⟩ are called forbidden. A normal d-path or nod-path is a d-path p such that none of its subwords is forbidden.
Let K⟨X⟩ the free K-algebra generated by X (i.e. the K-vector space with basis ⟨X⟩ which becomes a K-algebra by linearly extending the juxtaposition of words). Then LK(E,w) is the quotient of K⟨X⟩ by the ideal generated by the relations (i)-(iv) in Definition 6. Let K⟨X⟩nod be the linear subspace of K⟨X⟩ spanned by the nod-paths.
Theorem 27** (Hazrat, Preusser, 2017).**
The canonical map K⟨X⟩nod→LK(E,w) is an isomorphism of K-vector spaces. In particular the images of the nod-paths under this map form a linear basis for LK(E,w).
Proof.
See [7, Theorem 16] and its proof.
∎
The following lemma will be used in the proofs of Theorems 29,30,31,32,33,34 and 37.
Key Lemma 28**.**
Suppose that (E,w) does not satisfy Condition (LPA). Then there is a nod-path whose first letter is e2 and whose last letter is e2∗ for some e∈Ew1.
Proof.
[14, Proof of Lemma 35] shows that if one of the Conditions (LPA1), (LPA2) and (LPA3) is not satisfied, then then there is a nod-path whose first letter is e2 and whose last letter is e2∗ for some e∈Ew1. Assume now that (E,w) does not satisfy Condition (LPA4). Then there is an e∈Ew1, a path p and a cycle c such that s(p)=r(e), r(p)=s(c) and e does not belong to c. Write c=f(1)…f(m) where f(1),…,f(m)∈E1. If p=r(e), then e2f1(1)…f1(m)e2∗ is a nod-path (since f(m)=e). Now assume that p=g(1)…g(n) where g(1),…,g(n)∈E1. Clearly we assume that no letter of p is a letter of c. One checks easily that e2g1(1)…g1(n)f1(1)…f1(m)(g1(n))∗…(g1(1))∗e2∗ is a nod-path (note that f(m)=g(n)).
∎
Theorem 29**.**
Suppose that (E,w) does not satisfy Condition (LPA). Then LK(E,w) is neither simple nor graded simple.
Proof.
By Lemma 28, there is a nod-path p whose first letter is e2 and whose last letter is e2∗ for some e∈Ew1. One checks easily that the ideal I generated by p equals the linear span of all nod-paths that contain p as a subword (note that e2 is not the second letter of a forbidden word and e2∗ not the first letter of a forbidden word). It follows that I is a proper ideal of LK(E,w) (it is not the zero ideal since it contains the basis element p and it is not equal to LK(E,w) since it does not contain any vertex). Since I is generated by a homogeneous element, it is a graded ideal.
∎
Recall that a group graded K-algebra A=g∈G⨁Ag is called locally finite if dimKAg<∞ for every g∈G.
Theorem 30**.**
Suppose that (E,w) does not satisfy Condition (LPA). Then LK(E,w) is not locally finite.
Proof.
By Lemma 28, there is a nod-path p=x1…xn such that x1=e2 and xn=e2∗ for some e∈Ew1. Set p∗:=xn∗…x1∗ (where (fi∗)∗=fi for any f∈E1 and 1≤i≤w(f)). One checks easily that for any n∈N, (pp∗)n is a nod-path that lies in the homogeneous [math]-component LK(E,w)0. It follows from Theorem 27 that dimK(LK(E,w)0)=∞.
∎
Theorem 31**.**
Suppose that (E,w) does not satisfy Condition (LPA). Then LK(E,w) is not Noetherian.
Proof.
By Lemma 28, there is a nod-path p whose first letter is e2 and whose last letter is e2∗ for some e∈Ew1. Let q be the nod-path one gets by replacing the first letter of p by e1. For any n∈N let In be the left ideal generated by the nod-paths p,pq,…,pqn. One checks easily that In equals the linear span of all nod-paths o such that one of the words p,pq,…,pqn is a suffix of o. It follows that
In⊊In+1 (clearly none of the words p,pq,…,pqn is a suffix of pqn+1 since p and q have the same length but are distinct; hence pqn+1∈In).
∎
Theorem 32**.**
Suppose that (E,w) does not satisfy Condition (LPA). Then LK(E,w) is not Artinian.
Proof.
By Lemma 28, there is a nod-path p whose first letter is e2 and whose last letter is e2∗ for some e∈Ew1. For any n∈N let In be the left ideal generated by pn. One checks easily that In equals the linear span of all nod-paths o such that pn is a suffix of o. Hence In⊋In+1 (clearly pn+1 is not a suffix of pn and hence pn∈In+1).
∎
Theorem 33**.**
Suppose that (E,w) does not satisfy Condition (LPA). Then LK(E,w) is not von Neumann regular.
Proof.
By Lemma 28, there is a nod-path p whose first letter is e2 and whose last letter is e2∗ for some e∈Ew1. One checks easily that for any x∈LK(E,w), pxp is a linear combination of nod-paths of length ≥2∣p∣. Hence the equation pxp=p has no solution x∈LK(E,w).
∎
We recall some general facts on the growth of algebras. Let A={0} be a finitely generated K-algebra. Let V be a finite-dimensional generating subspace of A, i.e. a finite-dimensional subspace of A that generates A as a K-algebra. For n≥1 let Vn denote the linear span of the set {v1…vk∣k≤n,v1,…,vk∈V}. Then
[TABLE]
Given functions f,g:N→R+, we write f≼g if there is a c∈N such that f(n)≤cg(cn) for all n. If f≼g and g≼f, then the functions f,g are called asymptotically equivalent and we write f∼g. If W is another finite-dimensional generating subspace of A, then dV∼dW. The Gelfand-Kirillov dimension or GK dimension of A is defined as
[TABLE]
The definition of the GK dimension does not depend on the choice of the finite-dimensional generating subspace V. If dV≼nm for some m∈N, then A is said to have polynomial growth and we have GKdimA≤m. If dV∼an for some real number a>1, then A is said to have exponential growth and we have GKdimA=∞. If A does not happen to be finitely generated over K, then the GK dimension of A is defined as
[TABLE]
For the algebra A={0} we set GKdimA:=0.
Theorem 34**.**
Suppose that (E,w) does not satisfy Condition (LPA). Then GKdim(LK(E,w))=∞.
Proof.
Suppose first that (E,w) is finite (in our setting that means that E0 is a finite set). By Lemma 28, there is a nod-path p in (E,w) whose first letter is e2 and whose last letter is e2∗ for some e∈Ew1. Let q be the nod-path one gets by replacing the first letter of p by e1. Let n∈N. Consider the nod-paths
[TABLE]
where k,i1,…,ik∈N satisfy
[TABLE]
Clearly different solutions (k,i1,…,ik) and (k′,i1′,…,ik′′) of inequality (2) correspond to different nod-paths in (1) since ∣p∣=∣q∣. Let V denote the finite-dimensional subspace of LK(E,w) spanned by {v,fi,fi∗∣v∈E0,f∈E1,1≤i≤w(f)}. By Theorem 27 the nod-paths in (1) are linearly independent in Vn. The number of solutions of (2) is ∼2n and hence LK(E,w) has exponential growth.
Now suppose that (E,w) is not finite. One checks easily that there is a finite complete weighted subgraph (E~,w~) of (E,w) that does not satisfy Condition (LPA) (see [6, p. 884 and Proof of Lemma 5.19]). By the previous paragraph LK(E~,w~) has exponential growth. Clearly the inclusion (E~,w~)↪(E,w) induces an algebra monomorphism LK(E~,w~)→LK(E,w) since one can choose the special edges such that distinct nod-paths are mapped to distinct nod-paths. Hence LK(E,w) has a finitely generated subalgebra with exponential growth. It follows from the definition of the GK dimension that GKdimLK(E,w)=∞.
∎
The main result of this section is Theorem 37. In order to prove it we need two lemmas.
Lemma 35**.**
Let p be a nod-path starting with e2 and ending with e2∗ for some e∈Ew1. Then the ideal I of LK(E,w) generated by p contains no nonzero idempotent.
Proof.
For a nod-path q=x1…xn define m(q) as the largest nonnegative integer m such that there are indices i1,…,im∈{1,…,n} such that ij+∣p∣−1<ij+1 (1≤j≤m−1), im+∣p∣−1≤n and xij…xij+∣p∣−1=p (1≤j≤m). Hence m(q) is maximal with the property that q contains m(q) not overlapping copies of p.
Now let a∈I∖{0}. By Theorem 27 we can write a=r=1∑tkrqr where k1,…,kt∈K∖{0} and q1,…,qt are pairwise distinct nod-paths. Clearly m(qr)≥1 for any 1≤r≤t, since I consists of all linear combinations of nod-paths containing p as a subword. It easy to show, using the fact that e2 is not the second letter of a forbidden word and e2∗ not the first letter of a forbidden word, that for any 1≤r,s≤t the product qrqs is a linear combination of nod-paths o such that m(o)≥m(qr)+m(qs) (cf. [7, Proof of Proposition 40]). It follows that a2=r,s=1∑tkrksqrqs is a linear combination of nod-paths o such that m(o)≥2m(qrmin)>m(qrmin) where 1≤rmin≤t is chosen such that m(qrmin) is minimal. Hence a2 is a linear combination of nod-paths none of which equals qrmin. Thus a2 cannot be equal to a.
∎
If Λ is an infinite set and S is a unital ring, then we denote by MΛ(S) the K-algebra consisting of all square matrices M, with rows and columns indexed by Λ, with entries from S, for which there are at most finitely many nonzero entries in M (cf. [2, Notation 2.6.3]).
Lemma 36**.**
Let Λ be an infinite set and S a left Noetherian, unital ring. Let I1⊆I2⊆… be an ascending chain of left ideals of MΛ(S). Suppose there is a finite subset Λfin of Λ such that σλμ=0 for any n∈N, σ∈In, λ∈Λ and μ∈Λ∖Λfin. Then the chain I1⊆I2⊆… eventually stabilises.
Proof.
Write Λfin={λ1,…,λm}. Fix a τ∈Λ. For any n∈N, let Nn be the left S-submodule of Sm consisting of all row vectors (στλ1,…,στλm) where σ varies over all matrices in In. Then In equals the set of all matrices σ∈MΛ(S) such that σλμ=0 for any λ∈Λ,μ∈Λ∖Λfin and (σλλ1,…,σλλm)∈Nn for any λ∈Λ. Since S is a left Noetherian ring, Sm is a Noetherian module. It follows that the chain N1⊆N2⊆… eventually stabilises and thus the chain I1⊆I2⊆… eventually stabilises.
∎
Theorem 37**.**
Suppose that (E,w) does not satisfy Condition (LPA). Then LK(E,w) is not isomorphic to an unweighted Leavitt path algebra.
Proof.
Assume there is a graph F and an isomorphism ϕ:LK(E,w)→LK(F). By Lemma 28, there is a nod-path p whose first letter is e2 and whose last letter is e2∗ for some e∈Ew1. Let q be the nod-path one gets by replacing the last letter of p by e1∗. By Lemma 35, the ideal I of LK(E,w) generated by p contains no nonzero idempotent. Similarly, for any n∈N, the ideal In of LK(E,w) generated by qpn contains no nonzero idempotent. It follows from [2, Proposition 2.7.9], that ϕ(I),ϕ(In)⊆I(Pc(F)) (n∈N) where I(Pc(F)) is the ideal of LK(F) generated by all vertices in F0 which belong to a cycle without an exit. It follows that ϕ(p),ϕ(qpn)∈I(Pc(F)) (n∈N). By [2, Theorem 2.7.3] we have
[TABLE]
as a K-algebra. The sets Γ and Λi (i∈Γ) in (3) might be infinite if F is not finite.
It follows from the previous paragraph that there is a subalgebra A of LK(E,w) such that p,qpn∈A (n∈N) and A≅i∈Γ⨁MΛi(K[x,x−1]). For any n∈N let Jn be the left ideal of A generated by qp2,…,qpn+1. Then Jn is contained in the linear span of all nod-paths o such that one of the words qp2,…,qpn+1 is a suffix of o. It follows that
Jn⊊Jn+1 (clearly none of the words qp2,…,qpn+1 is a suffix of qpn+2 since p and q have the same length but are distinct). If the sets Γ and Λi (i∈Γ) are finite, then we already have a contradiction since it is well-known that i∈Γ⨁MΛi(K[x,x−1]) is Noetherian in this case. Hence the next two paragraphs are only needed if one of the sets Γ and Λi (i∈Γ) is infinite.
If a∈A, then we identify a with its image in i∈Γ⨁MΛi(K[x,x−1]) and write ai for the i-th component of a. Set Γfin:={i∈Γ∣pi=0}. Then Γfin is a finite subset of Γ. Clearly (qpn)i=0 for any i∈Γ∖Γfin and n≥2 (since (qpn)i=(qpn−1p)i=(qpn−1)ipi for any n≥2). Hence we can reduce to the case that Γ is finite.
For any n∈N and i∈Γ, let Jn,i be the left ideal of MΛi(K[x,x−1]) generated by (qp2)i,…,(qpn+1)i. Then J_{n}=\limits{}_{i\in\Gamma}J_{n,i} since each MΛi(K[x,x−1]) has local units. Now fix an i∈Γ. Let Λifin be the finite subset of Λi consisting of all λ∈Λi such that the λ-th column of pi has a nonzero entry. Then clearly σλμ=0 for any n∈N, σ∈Jn,i, λ∈Λi and μ∈Λi∖Λifin (since any element of Jn,i is a left multiple of pi). Hence, by Lemma 36, the chain J1,i⊆J2,i⊆… eventually stabilises. Since this holds for any i∈Γ, we get the contradiction that the chain J1⊆J2⊆… eventually stabilises.
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7. Summary
Theorem 38**.**
Let (E,w) be a row-finite weighted graph and K a field. Then LK(E,w) is isomorphic to an unweighted Leavitt path algebra iff (E,w) satisfies Condition (LPA) (see Definition 15).
Proof.
Follows from the Theorems 22 and 37.
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Theorem 39**.**
Let (E,w) be a row-finite weighted graph and K a field. If LK(E,w) is simple, or graded simple, or locally finite, or Noetherian, or Artinian, or von Neumann regular, or has finite GK dimension, then LK(E,w) is isomorphic to an unweighted Leavitt path algebra.
Proof.
Follows from the Theorems 29,30,31,32,33,34 and 22.
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