The center and the Lie structure of a Leavitt path algebra

The center and the Lie structure of a Leavitt path algebra Mercedes Siles Molina CIMPA research school Non-commutative algebra From 19 to 28 October 2015 Penonom´e, Cocl´e (Panam´a) Contents Contents i Abstract 1 1 Preliminary results 3 1.1 Leavitt path algebras. ...................... 5 1.2 The grading. ............................ 8 1.3 Cycles. ............................... 8 1.4 Ordering paths. .......................... 10 2 Ideals 11 2.1 Simplicity of a Leavitt path algebra ............... 11 2.2 Hereditary and saturated sets of vertices ............ 12 2.3 The ideal generated by line points. ............... 18 2.4 The ideal generated by vertices in cycles without exits. . . . . 20 2.5 The ideal generated by extreme cycles. ............. 22 3 The center of a Leavitt path algebra 27 3.1 The extended centroid ...................... 28 3.2 The center ............................. 30 4 Simplicity of related Lie algebras 49 References 51 i ii 1 Abstract These notes contain all the explanations and references corresponding to the course I delivered during the first CIMPA research school in Panama in October 2015 (see the complete program of the school here: http://cimpa. up.ac.pa/program.html). The purpose of the course is to explain the essentials on Leavitt path algebras and then to determine the center of a Leavitt path algebra whose associated graph is row finite. We start by the very beginning of the theory of Leavitt path algebras and give the tools needed to understand the structure of the center. Finally, we give a taste of the work that has been done concerning the study of the Lie estructure of a Leavitt path algebra. Panama, October 2015. 2 Chapter 1 Preliminary results We introduce the main object of this course: Leavitt path algebras. Adirected graph is a 4-tuple E= (E0, E1, rE, sE) consisting of two disjoint sets E0,E1and two maps rE, sE:E1→E0. The elements of E0are called the vertices of Eand the elements of E1the edges of Ewhile for e∈E1, rE(e) and sE(e) are called the range and the source of e, respectively. If there is no confusion with respect to the graph we are considering, we simply write r(e) and s(e). 3 4Chapter 1. Preliminary results Given a (directed) graph Eand a field K, the path K-algebra of E, denoted by KE, is defined as the free associative K-algebra generated by the set of paths of Ewith relations: (V) vw =δv,wvfor all v, w ∈E0. (E1) s(e)e=er(e) = efor all e∈E1. If s−1(v) is a finite set for every v∈E0, then the graph is called row-finite. If E0is finite and Eis row-finite, then E1must necessarily be finite as well; in this case we say simply that Eis finite. A vertex which emits no edges is called a sink. A vertex vis called an infinite emitter if s−1(v) is an infinite set, and a regular vertex otherwise. The set of infinite emitters will be denoted by E0 inf while Reg(E) will denote the set of regular vertices. The extended graph of Eis defined as the new graph b E= (E0, E1∪ (E1)∗, rb E, sb E),where (E1)∗={e∗ i|ei∈E1}and the functions rb Eand sb Eare defined as rb E|E1=r, s b E|E1=s, rb E(e∗ i) = s(ei),and sb E(e∗ i) = r(ei). The elements of E1will be called real edges, while for e∈E1we will call e∗ aghost edge. 1.1. Leavitt path algebras. 5 1.1 Leavitt path algebras. The Leavitt path algebra of Ewith coefficients in K, denoted LK(E), is the quotient of the path algebra Kb Eby the ideal of Kb Egenerated by the relations: (CK1) e∗e0=δe,e0r(e) for all e, e0∈E1. (CK2) v=P{e∈E1|s(e)=v}ee∗for every v∈Reg(E). 12 Chapter 2. Ideals that it is a direct limit of matricial algebras; a matricial algebra is a finite direct product of full finite dimensional matrix algebras over the field K). We recall here a characterization of purely infinite rings and the theorem characterizing purely infinite simple Leavitt path algebras. For the definition that follows, see e.g. [10, Definitions 1.2]. Let Rbe a ring. An idempotent ein Ris said to be infinite if there exist orthogonal idempotents f, g ∈Rsuch that e=f+g,g6= 0, and Re ∼ =Rf as left R-modules. In other words, eis infinite if Re is isomorphic to a proper direct summand of itself. In such a situation we say that Re is a directly infinite module. Theorem 2.2. (The Purely Infinite Simplicity Theorem; see, e.g. [2, Theorem 3.1.10.]). Let Ebe an arbitrary graph and Kany field. Then the Leavitt path algebra LK(E)is purely infinite simple if and only if Esatisfies the following conditions: (i) HE={∅, E0}, (ii) Esatisfies Condition (L), and (iii) every vertex in E0connects to a cycle. 2.2 Hereditary and saturated sets of vertices We define a relation ≥on E0by setting v≥wif there exists a path in E from vto w. A subset Hof E0is called hereditary if v≥wand v∈H imply w∈H. A hereditary set is saturated if every regular vertex which feeds into Hand only into His again in H, that is, if s−1(v)6=∅is finite and r(s−1(v)) ⊆Himply v∈H. Denote by HEthe set of hereditary saturated subsets of E0. Hereditary and saturated subsets of vertices play an important role in the theory of Leavitt path algebras. In fact, they are closely related to graded ideals of Leavitt path algebras (as was highlighted for the first time in [11]), and also to general ideals since every ideal Iin a Leavitt path algebra LK(E) contains a graded part: the ideal generated by I∩E0(see [2, Theorem 2.8.6]). Let Xbe a subset of vertices in E0. Denote by I(X) the ideal of LK(E) generated by X. Then, I(X) is a graded ideal. The reason is that it is generated by elements of degree zero. Moreover, if Eis a row-finite graph every graded ideal Jof LK(E) is I(H) for Ha hereditary and saturated subset of E0; concretely, H=J∩E0(see [17, Lemma 2.1 and Remark 2.2]). Although not every hereditary subset has to be saturated, hereditary subsets 2.2. Hereditary and saturated sets of vertices 13 (much more easy to get) give important information about the Leavitt path algebra. Whenever Xis a set of vertices of a graph E, the saturated closure of Xis defined as ∪i∈NΛi(X), where Λ0(X) = Xand by recurrence Λi(X) = Λi−1(X)∪{v∈Reg(E)|r(s−1(v)) ∈Λi−1(X)}. In particular, for a hereditary subset of vertices, say H, this saturated closure is hereditary and saturated and is denoted by H. For Xa subset of vertices in a graph E, the hereditary closure of Xis defined as the minimum hereditary subset of E0containing X. It always exists because is just the intersection of all hereditary subsets of E0which contains X. The hereditary and saturated closure of a set of vertices is defined as the saturated closure of the hereditary closure. Lema 2.3. Let Ebe a graph and Ka field. Let Hbe a hereditary subset of E0. Then I(H) = (n X i=1 kiγiλ∗ i|n≥1, ki∈K×, γi, λi∈Path(E), r(γi) = r(λi)∈H). Moreover, if Hdenotes the saturated closure of H, then I(H) = I(H). Proof. Let Ibe the set in the second part of the identity in the statement. To see that Iis an ideal of LK(E) we show that for every element αβ∗, where r(α) = r(β) = u∈H, and for every a, b ∈LK(E), we have aαuβ∗b∈J. It is enough to prove that γλ∗uµη∗∈Ifor every γ, λ, µ, η ∈Path(E) and u∈H. If γλ∗uµη∗= 0 we are done. Suppose otherwise that γλ∗uµη∗6= 0.Then γλ∗uµη∗=γµ0η∗if µ=λµ0, or γλ∗uµη∗=γ(λ0)∗η∗if λ=µλ0. Note that u=s(µ) and Hhereditary imply r(µ)∈H, therefore, r(µ0) = r(µ)∈H in the first case, and r(λ0) = r(µ)∈Hin the second case, which imply γλ∗uµη∗∈Jin both cases. This shows that Jis an ideal of LK(E); as it contains Hand must be contained in every ideal containing H, it must coincide with I(H). Now we prove I(H) = I(H). Clearly I(H)⊆I(H). The converse can be proved by induction (see [2, Lemma 2.4.1]). Results that will be very useful are the following. Lema 2.4. Let Ebe an arbitrary graph and let H1, H2be non empty hereditary subsets of vertices of E. Then: (i) Λm(Hi)is hereditary for every m∈N. (ii) H1∩H2=H1∩H2. 14 Chapter 2. Ideals Proof. (i). For m= 0 the result is trivial. Suppose the result true for m−1 and let us show it for m. Take u∈Λm(Hi) and let e∈E1be such that s(e) = u. Then r(e)∈r(s−1(u)) ⊆Λm−1(Hi)⊆Λm(Hi). This implies the result. (ii). It is immediate to see H1∩H2⊆H1∩H2. For the converse we will prove: Λm(H1)∩Λm(H2) = Λm(H1∩H2). Note that the first observation implies Λm(H1)∩Λm(H2)⊇Λm(H1∩H2). For the converse containment, use induction. If m= 0 then the result is trivially true. Suppose our assertion is true for m−1 and show it for m. If u∈Λm(H1)∩Λm(H2) then u∈ Λm−1(H1) or r(s−1(u)) ⊆Λm−1(H1). In the first case, and since Λm−1(H1) is hereditary (by (i)) we have also r(s−1(u)) ⊆Λm−1(H1). Analogously we prove r(s−1(u)) ⊆Λm−1(H2). This means r(s−1(u)) ⊆Λm−1(H1)∩Λm−1(H2) = Λm−1(H1∩H2) (by the induction hypothesis) and so u∈Λm(H1∩H2). Lema 2.5. Let Ebe a graph and Ha hereditary subset of E0. Then, for every v∈Hthere exists a finite number of paths α1, . . . , αnsatisfying r(αi)∈H and v=Pn i=1 αiα∗ i. Proof. For vin Hwe get immediately the result. Take vin Λ1(H) not being a sink. Then v=Pf∈s−1(v)ff∗and we have the claim. Suppose the result true for every u∈Λi−1(H) and take v∈Λi(H). Then, for every f∈s−1(v), since r(f)∈r(s−1)(v)⊆Λi−1(H), by the induction hypothesis, there exists a finite number of paths βf 1, . . . , βf min Path(E) such that r(f) = Piβf i(βf i)∗and r(βf i)∈H. Hence v=Pf∈s−1(v)ff∗= Pff(Piβf i(βf i)∗)f∗=Pf,i fβf i(βf i)∗f∗and we have finished. The result that follows was stated for the first time in [15, Proposition 3.1] (although in that paper the statement is slightly different); it has proved to be very useful in many different contexts, for example in order to get the Uniqueness Theorems (see [15, Theorem 3.5]), to prove that every Leavitt path algebra is semisimple, etc. In this notes we also find another context where it can be used. Except otherwise stated, Ewill denote an arbitrary graph and Kan arbitrary field. As usual, we will use the notation K×for K\ {0}. Theorem 2.6. (The Reduction Theorem.) For every nonzero element ain a Leavitt path algebra LK(E), there exist α, β ∈Path(E)such that: (i) 0 6=α∗aβ =kv for some k∈K×and v∈E0, or (ii) 0 6=α∗aβ =p(c, c∗), where cis a cycle without exits in Eand p(c, c∗) denotes the evaluation of a polynomial p(x, x−1)∈K[x, x−1]at c. 2.2. Hereditary and saturated sets of vertices 15 Of special interest will be the following result. Corollary 2.7. Let abe a nonzero homogeneous element in a Leavitt path algebra LK(E). Then, there exist α, β ∈Path(E),k∈K×and v∈E0, such that 06=α∗aβ =kv. Proof. Let α, β ∈Path(E) be such that 0 6=α∗aβ is as in cases (i) or (ii) in Theorem 2.6. In the first case, we have finished. In the second one, use the grading to obtain that in fact p(c, c∗) in (ii) has to be a monomial, that is, α∗aβ =kcmfor some k∈K×and a certain integer m. If m= 0, there is nothing more to do. If m > 0, then(c∗)mα∗aβ =kr(c) and we are done. If m < 0, then α∗aβc−m=kr(c) and the proof is complete. Another useful result which derives from Theorem 2.6 is: Corollary 2.8. Let Hbe a non-empty hereditary subset of a graph E. Then, for every nonzero homogeneous a∈I(H)there exist α, β ∈Path(E)such that α∗aβ =kv for some k∈K×and v∈H. Proof. Given the nonzero element a∈I(H) apply Corollary 2.7 and choose λ, µ ∈Path(E) such that λ∗aµ =kw for some k∈K×and w∈E0. Observe that w∈I(H). Use Lemma 2.3 to write w=Pm i=1 kiλiµ∗ iwith ki∈K×, λi, µi∈Path(E), r(λi) = r(µi)∈H, and suppose λiµ∗ i6=λjµ∗ jfor every i6=j. Then for v=r(µ1), α=λµ1and β=µµ1we have α∗aβ =µ∗ 1λ∗aµµ1= kµ∗ 1wµ1=kµ∗ 1µ1=kr(µ1) = kv, which is nonzero and satisfies v∈H. Proposition 2.9. Let {Hi}i∈Λbe a family of hereditary subsets of a graph Esuch that Hi∩Hj=∅for every i6=j. Then: I∪i∈ΛHi=I(∪i∈ΛHi) = ⊕i∈ΛI(Hi) = ⊕i∈ΛIHi. Proof. The union of any family of hereditary subsets is again hereditary, hence H:= ∪i∈ΛHiis a hereditary subset of E0. By Lemma 2.3 every element ain I(H) can be written as a=Pn l=1 klαlβ∗ l, where kl∈K×, αl, βl∈Path(E) and r(αl) = r(βl)∈H. Separate the vertices appearing as ranges of the αl’s depending on the Hi’s they belong to, and apply again Lemma 2.3. This gives a∈Pi∈ΛI(Hi)⊆I(H) since Hi⊆Hand so Pi∈ΛI(Hi) = I(H). Now we prove that the sum of the I(Hi)’s is direct. If this is not the case, since we are dealing with graded ideals, we may suppose that there exists a homogeneous element 0 6=a∈I(Hj)∩Pj6=i∈ΛI(Hi) for some j∈Λ; by Corollary 2.8 there exist α, β ∈Path(E) and k∈K×such that 0 6= k−1α∗aβ =w∈Hj. Observe that walso belongs to I(∪j6=i∈ΛHi). 16 Chapter 2. Ideals Write w=Pn l=1 klαlβ∗ l, with kl∈K,αl, βl∈Path(E), r(αl) = r(βl)∈ ∪j6=i∈ΛHi, and assume that every summand is non-zero. Then 0 6=r(β1) = β∗ 1β1=β∗ 1wβ1∈ ∪j6=i∈ΛHi. On the other hand, s(α1) = w∈Hjimplies (since Hjis a hereditary set) r(α1)∈Hj; therefore, r(α1) = r(β1)∈Hj∩ (∪j6=i∈ΛHi), a contradiction. To conclude the proof we point out that the first and last identities follow from [17, Lemma 2.1]. A similar relation can be established for the ideal generated by the intersection of a family of hereditary subsets. Lema 2.10. Let {Hi}i∈Λbe a family of hereditary subsets of an arbitrary graph E. Then: (i) I(∩i∈ΛHi) = ∩i∈ΛI(Hi). (ii) If Λis finite, then I(∩i∈ΛHi) = ∩i∈ΛI(Hi). Proof. (i). By the isomorphism given in [2, Theorem 2.4.13] among hereditary and saturated subsets of vertices and a special type of graded ideals, I(∩i∈ΛHi) = ∩i∈ΛI(Hi) = ∩i∈ΛI(Hi). (ii). When Λ is finite, then ∩i∈ΛHi=∩i∈ΛHi(use Lemma 2.4), and consequently I(∩i∈ΛHi) = I(∩i∈ΛHi) = ∩i∈ΛI(Hi) = ∩i∈ΛI(Hi). Before stating the result that will allow us to work with connected graphs, we need to recall that the ideal generated by a hereditary and saturated subset in a Leavitt path algebra is isomorphic to a Leavitt path algebra. Let Ebe a graph. For every non empty hereditary subset Hof E0, define FE(H) = {α=e1. . . en|ei∈E1, s(e1)∈E0\H, r(ei)∈E0\Hfor i < n, r(en)∈H}. Denote by FE(H) another copy of FE(H). For α∈FE(H), we write αto denote a copy of αin FE(H). Then, we define the graph HE= (HE0,HE1, s0, r0) as follows: 1. HE0= (HE)0=H∪FE(H). 2.2. Hereditary and saturated sets of vertices 17 2. HE1= (HE)1={e∈E1|s(e)∈H} ∪ FE(H). 3. For every e∈E1with s(e)∈H,s0(e) = s(e) and r0(e) = r(e). 4. For every α∈FE(H), s0(α) = αand r0(α) = r(α). The following result was first proved in [17, Lemma 5.2] and then in [12, Lemma 1.2] for row-finite graphs, although the result is valid in general (see [2, Theorem 2.4.22]). Lema 2.11. Let Ebe an arbitrary graph and Kany field. For a hereditary subset H⊆E0, the ideal I(H)is isomorphic to the Leavitt path algebra LK(HE). Concretely, there is an isomorphism (which is not graded) ϕ: LK(HE)→I(H)acting as follows: ϕ(v) = vfor every v∈H, ϕ(α) = αα∗for every α∈FE(H), ϕ(e) = eand ϕ(e∗) = e∗for every e∈E1such that s(e)∈H. ϕ(α) = αand ϕ(α∗) = α∗for every α∈FE(H). When we build the Leavitt path algebra of a graph E, we consider paths not only in E, but in the extended graph b E; this means that when we think of a connected graph we have in mind ghost paths too. For this reason we say that a graph Eis connected if b Eis a connected graph in the usual sense, that is, if given any two vertices u, v ∈E0there exist h1, . . . , hm∈E1∪(E1)∗ such that η:= h1. . . hmis a path in Kb E(in particular it is non-zero) such that s(η) = uand r(η) = v. The connected components of a graph Eare the graphs {Ei}i∈Λsuch that Eis the disjoint union E=ti∈ΛEi, where every Eiis connected. Corollary 2.12. Let Ebe a graph and suppose E=ti∈ΛEi, where each Ei is a connected component of E. Then LK(E)∼ =⊕i∈ΛLK(Ei). Proof. By Proposition 2.9,LK(E) = ⊕i∈ΛI(E0 i) and by Lemma 2.11,I(E0 i) is isomorphic to the Leavitt path algebra LK(E0 iE). Since the graph E0 iEis just Ei, the result follows. By means of this corollary, and for our purposes, from now on we will restrict our attention to Leavitt path algebras of connected graphs. Another application of Proposition 2.9 is stated below. Concretely, we will see that essentiality of graded ideals generated by hereditary and saturated subsets in a Leavitt path algebra (which is equivalent to density as one-sided ideals) can be expressed in terms of properties of the underlying graph. 18 Chapter 2. Ideals Proposition 2.13. Let Hbe a hereditary subset of a graph E. Then I(H) is a dense (left/right) ideal if and only if every vertex of E0connects to a vertex in H. Proof. We first remark that since every Leavitt path algebra is left nonsingular (see [26, Proposition 4.1]), the notions of dense left/right ideal and that of essential are equivalent in this context (by [23, (8.7) Proposition]), and by [23, (14.1) Proposition], I(H) is essential as a left/right ideal if and only if it is essential as an ideal. Moreover, as I(H) is a graded ideal, by [24, 2.3.5 Proposition] essentiality and graded essentiality of I(H) are equivalent. Hence, we will show that I(H) is a graded essential ideal if and only if every vertex of E0connects to a vertex in H. Suppose first that I(H) is a graded essential ideal of LK(E). Let v∈E0. If H∩T(v) = ∅, then Lemma 2.10 would imply I(H)∩I(T(v)) = 0, but this cannot happen as I(H) is a graded essential ideal. Hence H∩T(v)6=∅. This implies that vconnects to a vertex in H. Now we prove the converse, i.e., that I(H) is an essential graded ideal. Let Jbe a nonzero graded ideal and pick a nonzero homogeneous element x=uxv ∈J, where u, v ∈E0. Since the Leavitt path algebra LK(E) is an algebra of right quotients of KE, by [25, Proposition 2.2] (which is valid even for non necessarily row-finite graphs) there exists µ∈Path(E) such that 0 6=xµ ∈KE. Denote by wthe range of µ. By the hypothesis wconnects to a vertex in H, hence there exists λ∈Path(E) such that w=s(λ) and r(λ)∈H. If xµλ = 0 then xµ ∈uLK(E)w∩KE would satisfy λ∈Path(E)∩ran(xµ) = ∅, by [20, Lemma 1], a contradiction, hence 06=xµλ ∈I(H)∩Jwhich shows our claim. 2.3 The ideal generated by line points. There is a strong connection among properties of a Leavitt path algebra and properties of the graph. This is one of the advantages of considering these kind of algebras: you can visualize algebraic properties and can build algebras satisfying certain conditions. Here we show some of these connections. 2.3. The ideal generated by line points. 19 Let Ebe a graph. For v∈E0, the tree of v, written T(v), denotes the set {w∈E0|v≥w}. A vertex v∈E0is called a bifurcation vertex (or it is said that there is a bifurcation at v) if s−1 E(v) contains at least two edges of E. A vertex u∈E0is called a line point if there are neither bifurcations nor cycles at any vertex of T(u). Obviously, every sink is a line point. The set of line points of the graph Ewill be denoted by Pl(E). It is always a hereditary subset of E0although it is not necessarily saturated. It is shown in [15,16] that every line point generates a minimal left ideal, hence it is in the socle. It is not difficult to see (apply, for example [15, Propositon 2.6] that: Lema 2.14. Let Ebe an arbitrary graph. Let vbe a line point. Then: vLK(E)v∼ =K. Moreover, Lema 2.15. Let Ebe an arbitrary graph and let vbe a line point. Then I(v)∼ =MΛ(K), where Λdenotes a set whose cardinal coincides with the cardinal of all paths ending at v. 20 Chapter 2. Ideals It is possible to decompose Pl(E) = F i∈Λ Hi, where every Hiis a hereditary subset and I(Hi) = I(vi) for some line point vi∈Hi. Then, using Proposition 2.9 we get I(Pl(E)) = M i∈Λ MΛi(K). Moreover (see [2, Theorem 2.6.14]): Theorem 2.16. Let Ebe an arbitrary graph and Kany field. Decompose Pl(E) = Fi∈ΓHias it has been explained. Then Soc(LK(E)) = I(Pl(E)) ∼ =M i∈Λ MΛi(K), where for every i∈Λ, if viis an arbitrary element of Hithen I(vi)∼ = MΛi(K). 2.4 The ideal generated by vertices in cycles without exits. A similar pattern as for line points can be applied to vertices in cycles without exits. The following result is [16, Lemma 1.5]. Lema 2.17. Let Ebe an arbitrary graph. Let vbe a vertex in E0such that there exists a cycle without exits cbased at v. Then: vLK(E)v=(n X i=−m kici|ki∈K;m, n ∈N)∼ =K[x, x−1], where ∼ =denotes a graded isomorphism of K-algebras, and considering (by abuse of notation) c0=wand c−t= (c∗)t, for any t≥1. Proof. First, it is easy to see that if c=e1. . . enis a cycle without exits based at vand u∈T(v), then s(f) = s(g) = u, for f, g ∈E1, implies f=g. Moreover, if r(h) = r(j) = w∈T(v), with h, j ∈E1, and s(h), s(j)∈T(v) then h=j. We have also that if µ∈E∗and s(µ) = u∈T(v) then there exists k∈N∗,1≤k≤nverifying µ=ekµ0and s(ek) = u. Let x∈vLK(E)vbe given by x=Pp i=1 kiαiβ∗ i+δv, with s(αi) = r(β∗ i) = s(βi) = vand αi, βi∈E∗. Consider A={α∈E∗:s(α) = v}; we prove now that if α∈A,deg(α) = mn +q, m, q ∈Nwith 0 ≤q < n, then α=cme1. . . eq. We proceed by induction on deg(α). If deg(α) = 1 and 2.4. The ideal generated by vertices in cycles without exits. 21 s(α) = s(e1) then α=e1. Suppose now that the result holds for any β∈A with deg(β)≤sn +tand consider any α∈A, with deg(α) = sn +t+ 1. We can write α=α0fwith α0∈A,f∈E1and deg(α0) = sn +t, so by the induction hypothesis α0=cse1. . . et. Since s(f) = r(et) = s(et+1) implies f=et+1, then α=α0f=cse1. . . et+1. We shall show that the elements αiβ∗ iare in the desired form, i.e., cd with d∈Z. Indeed, if deg(αi) = deg(βi) and αiβ∗ i6= 0, we have αiβ∗ i= cpe1. . . eke∗ k. . . e∗ 1c−p=vby (4). On the other hand deg(αi)> deg(βi) and αiβ∗ i6= 0 imply αiβ∗ i=cd+qe1. . . eke∗ k. . . e∗ 1c−q=cd, d ∈N∗. In a similar way, from deg(αi)< deg(βi) and αiβ∗ i6= 0 it follows that αiβ∗ i= cqe1. . . eke∗ k. . . e∗ 1c−q−d=c−d, d ∈N∗. Define ϕ:K[x, x−1]→LK(E) by ϕ(1) = v, ϕ(x) = cand ϕ(x−1) = c∗. It is a straightforward routine to check that ϕis a graded monomorphism with image vLK(E)v, so that vLK(E)vis graded isomorphic to K[x, x−1] as a graded K-algebra. The structure of the ideal generated by any vertex in a cycle without exists is known. See, for example [2, Lemma 2.17.1]. Lema 2.18. For any vertex v∈Pc(E), let Λvdenote the (possibly infinite) set of paths in Ewhich end at v, but which do not contain all the edges of c, where cis the cycle without exits such that s(c) = v. Then: I(c0) = I(v)∼ =MΛv(K[x, x−1]). Proof. It is easy to see that I(c0) = I(v). To prove the isomorphism in the statement, consider the family B:= {µckη∗|µ, η ∈Λv, k ∈Z}, where as usual c0denotes vand ckdenotes (c∗)−kfor k < 0. Then Bis aK-linearly independent set which generates I(v) (see [2, Lemma 2.17.1]). The map ϕ:I(v)→MΛv(K[x, x−1]) given by ϕ(µckη∗) = xkeµ,η where xkeµ,η denotes the element of MΛv(K[x, x−1]) which is xkin the (µ, η) entry, and zero otherwise is a K-algebra isomorphism. A consequence is the structure of the ideal generated by vertices in cycles without exits (see [2, Theorem 2.7.3]). Theorem 2.19. Let Ebe an arbitrary graph and Kany field. Then: I(Pc(E)) ∼ =⊕i∈ΥMΛi(K[x, x−1]), 28 Chapter 3. The center of a Leavitt path algebra As for the general philosophy in Leavitt path algebras, we will see that the center can be computed by looking at the graph. That is, again, algebraic properties can be read from the graph. The key pieces for the description of the center are: the set of line points, the vertices in cycles without exits and the vertices in extreme cycles, jointly with an equivalence relation defined on E0whose set of classes is indexed in a subset of P:= Pl(E)∪Pc(E)∪Pec(E). 3.1 The extended centroid of the Leavitt path algebra of a finite graph The extended centroid of the ideal generated by Pwill coincide with the center of the Martindale symmetric ring of quotients of the Leavitt path algebra, notion that plays an important role. Recall that for an associative algebra A, the center of A, denoted Z(A), is defined by: Z(A) := {x∈A|[x, a] = 0 for every a∈A}, 3.1. The extended centroid 29 where [a, b] := ab−ba and juxtaposition stands for the product in the algebra A. For a semiprime algebra A, the extended centroid of A, denoted by C(A) is defined as: C(A) = Z(Qs(A)) = Z(Ql max(A)) = Z(Qr max(A)), where Qs(A), Ql max(A) and Qr max(A) are the Martindale symmetric ring of quotients of A, the maximal left ring of quotients of Aand the maximal right ring of quotients of A, respectively (see [23, (14.18) Definition]). Lema 3.1. Let Abe a unital simple algebra. Then Z(A) = C(A). Proof. We see first Z(A)⊆C(A). Suppose this containment is not true. Then there exists x∈Z(A) and q∈Qs(A) such that xq −qx 6= 0. Use that Qs(A) is a right ring of quotients of Ato find a∈Asuch that 0 6= (xq −qx)a and qa ∈A. Then 0 6=x(qa)−q(xa) = (qa)x−q(ax) = 0, a contradiction. To prove C(A)⊆Z(A), consider q∈C(A)\ {0}. By [23, (14.22) Corollary], C(A) is a field, hence there exist q−1∈C(A). Use again that Qs(A) is a right ring of quotients of Ato find x∈Asuch that 0 6=q−1x∈A. Since Ais simple and unital Aq−1xA =A, in particular there exists a finite number of elements a1, . . . , am, b1, . . . , bm∈Asuch that 1 = Pm i=1 aiq−1xbi. Multiply this identity by qand use that qis in the center of Qs(A) to get q=Pm i=1 aixbi∈Aas desired. Theorem 3.2. Let Ebe graph such that |E0|<∞and consider I:= Ilce. Then the extended centroid of LK(E)coincides with the extended centroid of I,C(I); moreover, C(LK(E)) = C(I)∼ =(⊕m i=1K)⊕⊕n j=1K[x, x−1]⊕⊕n0 l=1K, where mis the number of sinks, nis the number of cycles without exits and n0is the number of equivalence classes of extreme cycles. If Pl(E),Pc(E)or Pec(E)are empty, then the ideals they generate are zero and the corresponding summands in C(I)do not appear. Proof. As in the proof of Theorem 2.28 we may suppose that our graph is connected. Apply Theorem 2.28 (ii) and [23, (14.14) Theorem] to obtain Qs(LK(E)) = Qs(I). Then by [25, Lemma 1. 3 (i)] C(LK(E)) = C(I) = C(I(Pl(E))) ⊕ C (I(Pc(E))) ⊕ C (I(Pec(E))). 30 Chapter 3. The center of a Leavitt path algebra By Theorem 2.28 (i), Lemma 3.1 and Proposition 2.25,C(I(Pec(E))) = ⊕n0 l=1C(I(˜c0 l)), where ˜cl∈Xec and n0=|Xec|. Use [14, Theorem 4.2] to obtain C(I(˜c0 l)) ∼ =Kfor every l. To finish, use the three pieces of information in the paragraphs before to obtain C(LK(E)) = C(I)=(⊕m i=1K)⊕⊕n j=1K[x, x−1]⊕⊕n0 l=1Kand we get the claim in the statement. 3.2 The center The following remarks and results will be useful to study the center of a Leavitt path algebra. Their proofs are straightforward. Remark 3.3. If Ais an algebra and {Ii}is a set of ideals of Awhose sum is direct, then Z(⊕iIi) = ⊕i(Z(Ii)). Lema 3.4. Let Gbe an abelian group. The center of a G-graded algebra A is G-graded, that is, if x∈Z(A)and x=Pg∈Gxgis the decomposition of x into its homogenous components, then xg∈Z(A)for every g∈G. Proof. Indeed, for every y=Ph∈Gyhin A, 0 = [x, yh]=[Pg∈Gxg, yh] = Pg∈G[xg, yh]; using the grading on Aand that Gis abelian (to be sure that xgyhand yhxgare in the same homogeneous component) we get [xg, yh] = 0 for any g∈G. Hence 0 = Ph∈G[xg, yh]=[xg,Pg∈Gyh]=[xg, y] which means xg∈Z(A). Notation 3.5. The homogeneous component of degree gin the center of a G-graded algebra Awill be denoted by Zg(A). Lema 3.6. Let Ibe an ideal of an algebra A. If for every y∈Z(I)there exist n∈Nand {ai, bi}n i=1 ⊆Z(I)such that y=Pn i=1 aibi, then Z(I) = I∩Z(A). Proof. It is clear that I∩Z(A)⊆Z(I). To show Z(I)⊆Z(A), take y∈Z(I) and x∈A. Write y=Pn i=1 aibi, for ai, bi∈Z(I). Then yx =Pn i=1 aibix= Pn i=1 ai(bix) = Pn i=1(bix)ai=Pn i=1 bi(xai) = Pn i=1(xai)bi=xy. Corollary 3.7. Let Ibe an ideal of a Leavitt path algebra LK(E)such that for every y∈Z(I)there exist a, b ∈Z(I)such that y=ab. Then Z(I) = I∩Z(LK(E)). This happens, in particular, for every ideal of LK(E)generated by vertices in P. 3.2. The center 31 Proof. The first statement follows immediately from Lemma 3.6 For Igenerated by vertices in P, by Remark 2.29,Iis a direct sum of ideals of LK(E) which are isomorphic to Mn(K), Mn(K[x, x−1]) or J, for Jpurely infinite and simple. In this last case, by [20, Theorem 3.6], Z(J) is 0 or isomorphic to K; in the other cases, the centers are zero or isomorphic to K, or to K[x, x−1]. In all of these situations our hypothesis on the ideal is satisfied. Here are some previous papers related to the study of the center and of the derivations of a Leavitt path algebra. 32 Chapter 3. The center of a Leavitt path algebra As for our contribution, we determine the center of some Cohn and graph path algebras, as well as the center of the Leavitt path algebra associated to a row finite graphs in these papers ([20,21]): 3.2. The center 33 The following definition was motivated by our analysis of the centers of different Leavitt path algebras. Here are some examples. 34 Chapter 3. The center of a Leavitt path algebra 3.2. The center 35 Definitions 3.8. Let u, v ∈E0. The element λu,v will denote a path such that s(λu,v) = uand r(λu,v) = v. In E0we define the following relation: given u, v ∈E0we write u∼1vif and only if u=vor: (i) u≤vor v≤uand there are no bifurcations at any vertex in T(u) and T(v). (ii) there exist a cycle c, a vertex w∈c0and λw,u,λw,v ∈Path(E). This relation ∼1is reflexive and symmetric. Consider the transitive closure of ∼1; we shall denote it by ∼. The notation [v] will stand for the class of a vertex v. Example 3.9. The vertices uand vin the following graph are related: •u • 99// ?? •v 36 Chapter 3. The center of a Leavitt path algebra Remark 3.10. The following is an example of a graph which illustrates why do we need to consider the transitive closure of the relation ∼1in Definition 3.8. Note that u∼1v,v∼1wand u6∼1w; however, u∼w. •u • 99// >> •v•ee oo ~~ •w In what follows we are going to describe the zero component of the center of a Leavitt path algebra LK(E) associated to a row-finite graph. Notation 3.11. Let Ebe an arbitrary graph. Consider P=Pl∪Pc∪Pec and define X=P/ ∼. Decompose P=PftP∞, where Pfare those elements vof Psuch that (i) |[v]|<∞, and (ii) |FE([v])|<∞ In the same vein we decompose X=XftX∞, where Xf={[u]∈X|forall v ∈[u],v∈Pf} and X∞={[u]∈X|forsome v ∈[u],v∈P∞}. Finally, we decompose Xf=Xl f∪Xc f∪Xec f, where each of these subsets consists of equivalence classes induced by elements which are in Pf∩Pl, in Pf∩Pcand in Pf∩Pec, respectively. Note that if uand vare vertices in Pf, then u∼vif and only if u, v ∈Pl,u, v ∈Pcor u, v ∈Pec. When we want to emphasize the graph Ewe are considering, we will write Pl(E), X(E), etc. Lema 3.12. Let Ebe an arbitrary graph and u, v ∈P. Then: (i) [u]is a hereditary set. (ii) If u6∼ vthen [u]∩[v] = ∅. Proof. (i). Let w∈[u] and consider w0∈r(s−1(w)). Since w∼uthere exists a finite set {v1. . . vn}of vertices such that w=v1∼1v2∼1· · · ∼1vn=u. Note that w0∼1v2and so w0∈[u]. (ii). By the hypothesis [u]∩[v] = ∅. Use (i) and Lemma 2.4 to get the result. 3.2. The center 37 Definition 3.13. Let Ebe a graph and Kbe any field. An element a∈ LK(E) which can be written as a=X [v]∈Xf k[v]a[v],where k[v] ∈K×and a[v] =X u∈[v] u + X α∈FE([v]) αα∗, will be said to be written in the standard form. We will prove that every element in the zero component of the center of a Leavitt path algebra can be written in the standard form. Lema 3.14. Let Ebe an arbitrary graph and Kbe any field. Consider [v]∈X. For every u∈[v]and α, β ∈FE([v]) we have: (i) If s(α) = s(β), then α∗β6= 0 if and only if α=β. (ii) If s(α)6=s(β)then α∗β= 0. (iii) uα = 0. Proof. (i). If α∗β6= 0 then α=βγ or β=αδ for some γ, δ ∈Path(E). By the definition of FE([v]), necessarily α=β. (ii). This case follows immediately. (iii). For uand αas in the statement, uα 6= 0 implies u=s(α), but this is not possible as α∈FE([v]). Notation 3.15. For a graph Ewe denote by Pethe set of vertices in cycles with exits. Lema 3.16. Let Ebe an arbitrary graph. Then, for every a∈Z0and v∈P∪Pethere exists kv∈Ksuch that if u∈[v]then uau =kvu. Proof. Suppose first u=v. If v∈Pc∪Pethe result follows by [20, Corollary 7]. If v∈Pl, by the proof of [8, Proposition 1.8] vLK(E)v=Kv, hence av =va ∈Kv . Now, suppose u∈[v]. Since the relation ∼is given as the transitive closure of ∼1, we may assume first that there exists a vertex win a cycle, and paths λ, µ satisfying λ=wλu and µ=wµv. By the paragraph before there exists an element k∈Ksuch that waw =kw. Then uau =ua = λ∗λa =λ∗aλ =λ∗(kw)λ=kλ∗λ=ku and analogously we show va =kv. Repeating this argument a finite number of steps we reach our claim. It is easy to see that when uis in the hereditary closure of [v] we also have the result. 44 Chapter 3. The center of a Leavitt path algebra In the second case, 0 6=p(c, c∗)p(c, c∗)∗since p(c, c∗)∈r(c)LK(E)r(c) which is isomorphic to the Laurent polinomial algebra K[x, x−1] (see [13, Proposition 2.3]), hence 0 6=α∗aββ∗a∗α=α∗ββ∗aa∗α, and so aa∗must be nonzero. Theorem 3.24. Let Ebe a row-finite graph. Then, the set Bn=            X m·l(c)=n α∈FE(c0)∪{c0} u∈c0 αcm uα∗|c∈ S             is a basis of Zn(LK(E)) with n∈Z\ {0}. Proof. We will assume n > 0. The case n < 0 follows by using the involution in the Leavitt path algebra. Let a∈Zn(LK(E)). For every u∈E0such that au 6= 0, Lemma 3.21 and Proposition 3.20 imply au =aXαα∗=Xαar(α)α∗=Xαkαcmα αα∗=Xkααcmα αα∗ where α∈FE(c0)∪ {u}, r(α)∈P− cand s(α) = u. Hence a=Xkααcm vα∗,where the cv’s are elements of S, α∈FE(c0)∪ {v}and r(α)∈P− c.(3.3) Now we see that kα=kβfor every α, β ∈FE(c0) appearing in the expresion before. We first note that αcrβ∗is a nonzero element in LK(E) for every r∈N. Since aαβ∗=αβ∗a, using Lemma 3.22 we get kααcm uβ∗=kβαcm uβ∗and so kα=kβ. In what follows we prove that if cuappears in (3.3) then for every v∈c0 u and for every β∈FE([c0 u]) we have that cvand βcvβ∗also appear. Apply (3.3) and Lemmas 3.23 and 3.14 to get 06=aa∗=Xkααcm vα∗Xkααc−m vα∗=Xk2 ααvα∗. By Theorem 3.18,vmust appears in this summand and also every β∈ FE([c0 v]). This shows that the set Bngenerates the n-component of the center. 3.2. The center 45 In what follows we see that the elements of Bnare linearly independent. In fact, we show that elements of the form αcm uα∗are linearly independent. To this end, let kα∈Kand suppose Pkααcmu uα∗= 0, where all the summands are different. Choose a nonzero βcmv vβ∗. Then 0 = β∗Pkααcmu uα∗= (by Lemma 3.22)β∗kββcmv vβ∗=kβcmv vβ∗; this implies kβ= 0 and our claim has been proved. Finally we see that Bn⊆Z(LK(E)). Fix cu∈ S and denote by An= {αcm uα∗|m.l(cu) = n, α ∈FE([c0 u])∪{u}}, i.e., Ancontains all the summands of ain (3.3). We see that for every β∈Path(E), Anβ=βAnand Anβ∗= β∗Anfor all n∈Z; this will prove our statement. So, take αcm uα∗∈An. Then αcm uα∗βis non zero if and only if β=α(by Lemma 3.22); in this case, αcm uα∗β=βcm u∈βAn. Suppose αcm uα∗β= 0; if βαcm uα∗= 0 then we have shown that 0 ∈βAn; otherwise βαcm uα∗6= 0; this implies r(β) = s(α). Now we distinguish two cases: first, α=u. Note that cm uβ= 0 implies β6=cuγ, for any γ∈Path(E) and so βcm uβ∗∈An. Then 0 = ββcm uβ∗∈βAn(because r(β) = u). If αis not a vertex, then r(β)6=u hence 0 = βcm u∈βAn. This shows Anβ⊆βAn. For the converse, note that βαcm uα∗is nonzero if and only if βα is a nonzero element in FE([c0 u]), for w=s(β); in this case βαcm uα∗β∗∈An and so βαcm uα∗=βαcm uα∗β∗β∈Anβ. Now, suppose βαcm uα∗= 0. Then, multiplying on the right hand side by α(cm u)∗and on the left hand side by β∗ we get r(β)α= 0. If cm uβ= 0 we have shown 0 = cm uβ∈Anβ. If cm uβ6= 0, as cuhas no exits, then β=cs uλ, where λis such that custarts by λ. If l(cu) = 1, then β=cs ufor some s∈N; we see that this implies αcm uα∗β= 0. Suppose otherwise αcm uα∗β6= 0. Then α∗β6= 0 and so α≥βor β≥α. The first case is not possible (note that neither c1 u⊆α1nor α=u), hence β≥α, implying s(α) = u. This happens only when α=u, but this is not possible as 0 = βαcm uα∗=cs+m uis a contradiction. Now, suppose l(c)>1. Let ebe the last edge in cu. Then ecm ue∗β, which is an element in Anβ, is zero (because the first edge of βis the first one of cu, which is not e). This proves βAn⊆Anβand therefore Anβ=βAn. Applying the involution we have β∗A−n=A−nβ∗for all n∈Z, as required. Definitions 3.25. Let Ebe an arbitrary graph. Denote by Pb∞the set of all vertices in Esuch that T(v)has infinite bifurcations. Define Hf:= G [v]∈Xf [v] and H∞:=  G [v]∈X∞ [v] ∪Pb∞. 46 Chapter 3. The center of a Leavitt path algebra Proposition 3.26. Let Ebe a row-finite graph and Kbe any field. Then LK(E) = I(Hf)⊕I(H∞). Proof. We will show v∈I(Hf∪H∞) for every v∈E0. If v∈Hf∪H∞ we have finished. Suppose that this is not the case. In particular, this means s−1(v)6=∅. Write v=Pe∈s−1(v)ee∗. If for every ein this sum r(e)∈Hf∪H∞, then we have finished. If for some e,r(e) is a sink, we have finished; if r(e) is a vertex in a cycle, taking into account that every vertex of E0connects to Hf∪H∞then r(e)∼ufor some u∈Hf∪H∞. Note that u /∈Pb∞as otherwise v∈Pb∞, and we are assuming v /∈Hf∪H∞, therefore r(e)∈ ∪w∈X[w]. For those e∈s−1(v) in the remaining cases we apply again Condition (CK2) to u1:= r(e) and write u1=Pf∈s−1(u1)ff∗. Now we proceed in the same way concerning r(f). This process must stop because otherwise there would be infinitely many bifurcations and so v∈Pb∞, a contradiction. Apply Proposition 2.9 to the disjoint hereditary subsets Hf∪H∞to get the result. Theorem 3.27. (Structure Theorem for the center of a row-finite graph). Let Ebe a row-finite graph. Then: Z(LK(E)) ∼ =K|X\Xc f|⊕K[x, x−1]|Xc f| More concretely, Z(LK(E)) ∼ =K|Xl f|⊕K|Xec f|⊕K[x, x−1]|Xc f|. 3.2. The center 47 Proof. By Proposition 3.26,LK(E) = I(Hf)⊕I(H∞). Since Hf:= t[v]∈Xf[v] and H∞:= t[v]∈X∞[v]∪Pb∞, Proposition 2.9 implies LK(E) =  M [v]∈Xf I([v]) MI(H∞), hence Z(LK(E)) =  M [v]∈Xf Z(I([v]) MZ(I(H∞)). By Theorems 3.18 and 3.24 we know Z(LK(E)) ⊆L[v]∈XfZ(I([v])); therefore Z(LK(E)) = M [v]∈Xf Z(I([v]). We claim that Z(I([v])) is isomorphic to Kif v∈Pl∪Pec or isomorphic to K[x, x−1] in case v∈Pc. 48 Chapter 3. The center of a Leavitt path algebra Indeed, take x∈Z(I([v])). By the theorems of the basis of the center we may write x=Pwxw, where [w]∈Xfand xw=kw,0a[w]+X n kw,n       X m·l(cw)=n α∈FE(c0 w)∪{c0 w} u∈c0 w αcm uα∗       , where a[w]is as in Theorem 3.18,kw,0, kw,n ∈Kand kw,n is zero for almost every n∈Zand w∈Pc. Note that xw∈I([w]) and that {I([w])}[w]∈Xfis an independent set (in the sense that its sum is direct). This means that, necessarily, x=xv. If v∈Pl∪Pec, then x=k0 va[v]and we have an isomorphism of K-algebras spaces between Z(I([v])) and K. If v∈Pcthen the following map: a[v]7→ 1 and X m·l(cw)=n α∈FE(c0 w)∪{c0 w} u∈c0 w αcm uα∗7→ xm describes an isomorphism of K-algebras between Z(I([v])) and K[x, x−1]. This finishes the proof. Chapter 4 Simplicity of some Lie algebras related to Leavitt path algebras This last chapter is devoted to explain the advances in the subject concerning the Lie structure of some Lie algebras arising from Leavitt path algebras. For any associative algebra Aits antisymmetrization is the Lie algebra A−defined as the same vector space and product given by [a, b] = ab −ba, for any a, b ∈A. It is well-known that Asimple does not imply, necessarily, A−simple. But, under some mild restrictions, the Lie algebra [A, A]/Z([A, A]) turns out to be simple. Simple Leavitt path algebras are characterized as those Leavitt path algebras whose associated graph is cofinal (there are no more hereditary and saturated sets than the total and the empty set) and satisfies Condition (L), i.e., every cycle has an exit. A natural question is to ask about the simplicity of the Lie algebra [L, L]/Z([L, L]), whenever Lstands for the Leavitt path algebra LK(E) of a graph Eor, more generally, about the simplicity of the Lie algebra [Md(L), Md(L)]/Z([Md(L), Md(L)]), where Md(L) is the algebra of matrices over L, for d∈N. In [6] the authors study the simplicity of such a Lie algebra whenever Lis the algebra Md(LK(n)), for n, d ∈Nand LK(n) the Leavitt algebra of type (1, n), which is known to be isomorphic to the Leavitt path algebra of the n-petal rose. As a remark we recall the reader that it was proved in [1] the following result: 49 50 Chapter 4. Simplicity of related Lie algebras Theorem 4.1. ([1, Theorem 4.14]). Md(LK(n)) ∼ =LK(n)if and only if gcd(d, n −1) = 1. On the other hand, it was also stated ([1, Proposition 2.1]) that if gcd(d, n− 1) >1 then the type of Md(LK(n)) is (1,n−1 gcd(d,n−1) + 1), hence for d, d0∈Nsuch that gcd(d, n −1) 6= gcd(d0, n −1) the algebras Md(LK(n)) and Md0(LK(n)) are not isomorphic. For a field Kand a row-finite graph E, the authors of [7] study the elements of LK(E) which are in the commutator [LK(E), LK(E)], hence obtaining conditions under which this Lie algebra is simple, whenever the algebra LK(E) is a simple associative algebra. Concretely, Lema 4.2. Let α, β be paths. (i) If αis not a closed path, then α, α∗∈[LK(E), LK(E)]. (ii) If α6=βγ and β6=αγ for any closed path γ, then αβ∗is in [LK(E), LK(E)]. Why do they study elements in [LK(E), LK(E)]? The reason is that the center of this Lie algebra is a Lie ideal and that for unital associative algebras (with some mild conditions), this center is nonzero if and only if 1 is an element of it, that is (see [7]): Theorem 4.3. Let Ebe a graph and Ka field. Assume that LK(E)is a simple algebra. (i) If E0is infinite then the Lie algebra [LK(E), LK(E)] is simple; (ii) If E0is finite, then [LK(E), LK(E)] is simple if and only if 1 = Pv∈E0v /∈ [LK(E), LK(E)]. There exist however no non-simple Leavitt path algebras LK(E)such that the Lie algebra [LK(E), LK(E)] simple. For the concrete results characterizing when [LK(E), LK(E)] is simple, for LK(E) a unital purely infinite simple algebra, see [7, Theorem 36]. It will depend on the characteristic of the field and on the order of he unit inside K0(LK(E)). Concretely, the result says: 51 Theorem 4.4. Let Kbe a field, let Ebe a finite graph for which LK(E)is purely infinite simple, and let M=MEdenote the matrix Im−At E(for AE the adjacency matrix of the graph E. (i) Suppose that char(K)=0. Then the Lie K-algebra [LK(E), LK(E)] is simple if and only if 1m+ Im(MZm E)has infinite order in Coker(MZm E); that is, if and only if [1LK(E)]has infinite order in K0(LK(E)). (ii) Suppose that char(K) = p6= 0. Then the Lie K-algebra [LK(E), LK(E)] is simple if and only if 1m+Im(MZm E)is not p-divisible in Coker(MZm E); that is, if and only if [1LK(E)]is not p-divisible in K0(LK(E)). Subsequents results studying the simplicity of the Lie algebra [LK(E), LK(E)], for any field Kand any row-finite graph Eare those contained in [9]. Concretely, Theorem 4.5. Let Ebe a row-finite graph. The Lie algebra [LK(E), LK(E)] is simple if and only if either LK(E)is simple (case covered by Theorem 4.3) or Econtains a simple subgraph Fsuch that every vertex v∈E0\F0is a balloon over Fand Pw∈Xw∈[LK(E), LK(E)], for Xthe set of all ranges of edges in E1connecting vto a vertex in F. 52 Chapter 4. Simplicity of related Lie algebras References [1] G. Abrams, P. N. ´ Ahn, E. Pardo, Isomorphisms between Leavitt algebras and their matrix rings J. Reine Angew. Math. 624 (2008), 103132. [2] Gene Abrams, Pere Ara, Mercedes Siles Molina, Leavitt path algebras. A primer and handbook. Springer. To appear. (See https://www.dropbox.com/s/gqqx735jddrip8f/ AbramsAraSiles_BookC1C2.pdf?dl=0.) [3] Gene Abrams, Gonzalo Aranda Pino, The Leavitt path algebra of a graph, J. Algebra 293 (2) (2005), 319–334. [4] Gene Abrams, Gonzalo Aranda Pino,The Leavitt path algebras of arbitrary graphs,Houston J. Math. 34 (2) (2008), 423–442. [5] Gene Abrams, Gonzalo Aranda Pino, Francesc Perera, Mercedes Siles Molina. Chain conditions for Leavitt path algebras. Forum Math.,22 (2010), 95–114. [6] Gene Abrams, Darren Funk-Neubauer, On the simplicity of Lie algebras associated to Leavitt algebras. Comm. Algebra 39 (11) (2011), 40594069. [7] Gene Abrams, Zakary Mesyan, Simple Lie algebras arising from Leavitt algebras, J. Pure Appl. Algebra 216 (10) (2012), 23022313. [8] Gene Abrams, Kulumani Rangaswamy, Mercedes Siles Molina,The socle series of a Leavitt path algebra,Israel J. Math. 184 (2011), 413–435. [9] Adel Alahmedi, Hamed Alsulami, On the simplicity of the Lie algebra of a Leavitt path algebra, (Preprint). ArXiv:1304.1922v. 53