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 Recent trends on Non-commutative algebras From 19 to 30 June 2017 Pune (India) 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 The book [2]. ........................... 10 1.5 Conditions (L) and (K). ..................... 14 1.6 Ordering paths. .......................... 15 2 Ideals 17 2.1 Simplicity of a Leavitt path algebra ............... 17 2.2 Hereditary and saturated sets of vertices ............ 18 2.3 The ideal generated by line points. ............... 25 2.4 The ideal generated by vertices in cycles without exits. . . . . 26 2.5 The ideal generated by extreme cycles. ............. 28 2.6 Decomposable Leavitt path algebras .............. 31 3 The center of a Leavitt path algebra 33 3.1 Introduction to the center .................... 33 3.2 The extended centroid ...................... 34 3.3 The center ............................. 36 3.4 The center of an arbitrary Leavitt path algebra ........ 56 4 Simplicity of related Lie algebras 59 References 61 i ii 1 Abstract These notes contain all the explanations and references corresponding to the course I delivered during the CIMPA research school in Pune (India) in June 2017 (see the complete program of the school here: https://iiserpunecimpa17. wordpress.com/program/). The purpose of the course is to explain the essentials on Leavitt path algebras in order to get the tools needed to determine the center of a Leavitt path algebra. First when the graph is row finite; then, in the general setting (for arbitrary graphs). We start in Chapter 1 by the very beginning of the theory of Leavitt path algebras, explain Conditions (L) and (K) and characterize them in graph terms and also in ring-theoretical terms. In Chapter 2 we characterize the simplicity of Leavitt path algebras and determine the graph pieces involved in every graded ideal. We determine the structure of certain graded ideals, concretely: the ideal generated by line points (which is precisely the socle of the Leavitt path algebra), the ideal generated by vertices in cycles without exists (which is the ideal generated by the primitive non-minimal idempotents of the algebra), and the ideal generated by extreme cycles (which is a direct sum of graded ideals each of which is purely infinite simple). The end of the chapter refers to decomposable Leavitt path algebras. For this part, as well for the structure of the center of an arbitrary graph, we use the Steinberg algebra associated to the groupoid related to a graph. We give not here all the details about goupoids and Steinberg algebras because the course of Professor Lisa O. Clark is devoted to that end. In Chapter 3 we explore the center of a Leavitt path algebra and give its structure. Finally, in chapter 4, we give a taste of the work that has been done concerning the study of the simplicity related to the Lie structure of a Leavitt path algebra. We refer the reader to the book [2], the first one devoted to Leavitt paht algebras. Pune, India, June 2017. 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 1. Preliminary results 1.4. The book [2]. 13 14 Chapter 1. Preliminary results 1.5 Conditions (L) and (K). As it has been said in other courses of this CIMPA research school, Condition (K) is equivalent for the Leavitt path algebra to have every ideal graded, that is, every ideal is generated by the idempotents it contains; concretely, every ideal Iof a Leavitt path algebra LK(E) where Esatisfies Condition (K) is generated by I∩E0. And this is equivalent for the Leavitt path algebra to have the exchange property. As for condition (L), it can be characterized using ring-theoretical properties of the Leavitt path algebra. concretely, using primitive idempotents. Let us recall very briefly this notion. Proposition 1.1. Let ebe an idempotent in a ring R(not necessarily unital). The following conditions are equivalent: (i) eR is an indecomposable right R-module (equivalently, Re is an indecomposable left R-module). (ii) eRe is a ring without nontrivial idempotents. (iii) ehas no decomposition into a+b, where a, b are nonzero orthogonal idempotents in R. Proof. As in [26, Proposition 21.8]. Definition 1.2. Following [26], if an idempotent 0 6=e∈Rsatisfies any of these conditions, we say that eis a primitive idempotent. Examples 1.3. – In Leavitt path algebras, Condition (CK2) says that every vertex not being a sink and being the source of finitely many edges, can be decomposed as the sum of non-trivial orthogonal idempotents. – Every line point is a primitive minimal idempotent. – Every vertex in a cycle without exits is a primitive idempotent which is not minimal. The primitive vertices in Leavitt path algebras are characterized. Proposition 1.4. ([13, Proposition 5.3]) Let Ebe an arbitrary graph and let v∈E0. Then vis a primitive idempotent of LK(E)if and only if its tree T(v)has no bifurcations. Theorem 1.5. ([13, Theorem 5.8]) Let Ebe an arbitrary graph. The following are equivalent: (i) Esatisfies Condition (L). (ii) LK(E)has no non-minimal primitive idempotents. 1.6. Ordering paths. 15 1.6 Ordering paths. Given paths α, β, we say α≤βif β=αα0for some path α0. 16 Chapter 1. Preliminary results Chapter 2 Ideals In this chapter we will speak mainly about graded ideals of Leavitt path algebras. Of course, in an arbitrary Leavitt path algebra not every ideal has to be graded. This is the case, for example, of the Leavitt path algebra associated to the graph Ehaving one vertex, say v, and one edge, say e. It is isomorphic to the Laurent polynomial ring K[x, x−1] and here not every ideal is graded. For example, <1 + x > is not a graded ideal. This, translated to LK(E) means that < v +e > is not a graded ideal. 2.1 Simplicity of a Leavitt path algebra Before stating some results on graded ideals, we will recall here the characterization of simple Leavitt path algebras (those algebras Awhose unique ideals are 0 and A). Recall that for a graph E, the set HEconsists of all hereditary and saturated subsets of vertices of E. Also, we recall that a graph Eis said to satisfy Condition (L) if every cycle has an exit. Theorem 2.1. (The Simplicity Theorem)Let Ebe an arbitrary graph and Kany field. Then the Leavitt path algebra LK(E)is simple if and only if Esatisfies the following conditions: (i) HE={∅, E0}. (ii) Esatisfies Condition (L) The Dicothomy Principle for Leavitt path algebras (see, for example [2, Theorem 3.1.11] states that a simple Leavitt path algebra is locally matricial, or it is a purely infinite simple Leavitt path algebra (locally matricial means 17 18 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 [18, 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 19 (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. 20 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.) (MSM; Aranda Pino; Mart´ın Barquero; Mart´ın Gonz´alez) 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 2.2. Hereditary and saturated sets of vertices 21 (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. Some immediate consequences of this theorem are the Uniqueness Theorems for Leavitt path algebras, that can be proved by taking into account the Reduction Theorem and that the kernel of a (graded) homomorphism is a (graded) ideal. Remark 2.7. The Reduction Theorem is also valid when considering a commutative and unital ring instead of a field. Of special interest will be the following result. Corollary 2.8. 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 2.9. 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 2.8 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 2.10. 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 28 Chapter 2. Ideals A consequence is the structure of the ideal generated by vertices in cycles without exits (see [2, Theorem 2.7.3]). Theorem 2.20. Let Ebe an arbitrary graph and Kany field. Then: I(Pc(E)) ∼ =⊕i∈ΥMΛi(K[x, x−1]), where {ci}i∈Υis the set of different cycles without exits in Eand Λiis the set of paths in Ewhich end at the base viof the cycle ci, but do not contain all the edges of ci. The importance of the ideal I(Pc(E)) relays also in that every ideal non containing vertices is contained in it (see [18] for a proof of this fact for row-finite graphs). 2.5 The ideal generated by extreme cycles. Now we introduce the notion of extreme cycle. Roughly speaking it is a cycle such that every path starting at a vertex of the cycle comes back to it. The ideal generated by extreme cycles will be proved to be a direct sum of purely infinite simple Leavitt path algebras. Definitions 2.21. Let Ebe a graph and ca cycle in E. We say that cis an extreme cycle if chas exits and for every path λstarting at a vertex in c0 there exists µ∈Path(E) such that 0 6=λµ and r(λµ)∈c0. We will denote by Pec(E) the set of vertices which belong to extreme cycles. Definitions 2.22. Let X0 ec be the set of all extreme cycles in a graph E. We define in X0 ec the following relation: given c, d ∈X0 ec, we write c∼dwhenever cand dare connected, that is, T(c0)∩d06=∅, equivalently, T(d0)∩c06=∅. It is not difficult to see that ∼is an equivalence relation. Denote the set of all equivalence classes by Xec =X0 ec/∼. When we want to emphasize the graph we are considering we will write X0 ec(E) and Xec(E) for X0 ec and Xec, respectively. For any c∈X0 ec, let ˜cdenote the class of cand let use ˜c0to represent the set of all vertices which are in the cycles belonging to ˜c. The following will be of use. Remark 2.23. For a graph Eand using the notation described above: (i) For any c∈X0 ec, ˜c0=T(c0), hence ˜c0is a hereditary subset. (ii) Given c, d ∈X0 ec, ˜c6=˜ dif and only if ˜c0∩˜ d0=∅. 2.5. The ideal generated by extreme cycles. 29 (iii) For c, d ∈X0 ec,c∼dif and only if T(c) = T(d). Examples 2.24. Consider the following graphs E≡ • e%%f//• g qq h QQF≡ • e%%f ((• g hh h1 qq h2 QQ Then X0 ec(E) = {g, h},Xec(E) = {˜g},X0 ec(F) = {e, fg, gf, h1, h2}and Xec(F) = {˜e}. Now we will analyze the structure of the ideal generated by Pec(E). Lema 2.25. Let Ebe an arbitrary graph and Kany field. For every cycle csuch that c∈X0 ec, the ideal I(˜c0)is isomorphic to a purely infinite simple Leavitt path algebra. Concretely, to LK(HE), where H= ˜c0. Proof. Use Lemma 2.12 to have I(˜c0) isomorphic to the Leavitt path algebra LK(HE) and let us show that this Leavitt path algebra is purely infinite and simple. For that, we will use the characterization [4, Theorem 4.3], which is valid for arbitrary graphs as can be proved by using the techniques described in [25] or in [13] in order to translate certain characterizations of Leavitt path algebras from the row-finite case to the case of an arbitrary Leavitt path algebra. Every vertex of HEconnects to a cycle: take v∈HE0; if v∈Hthen it connects to c, otherwise there exists a path µ∈Path(E) such that s(µ) = v and r(µ)∈H. Since r(µ) connects to the cycle cin E, the vertex valso connects to cin HE. Every cycle in HEhas an exit: this follows because any cycle in this graph comes from a cycle din Hand, by construction, ˜ d= ˜c; this means that d connects to cand hence it has an exit which is an exit in HE. The only hereditary and saturated subsets of HE0are ∅and HE0: let H0∈ HHEbe non empty and consider v∈H0; if v∈Hthen H⊆H0; since H0is saturated, H0=HE0; if v /∈Hthen there exists f∈HE1such that v=s(f) and r(f)∈H; this implies H⊆H0; now, apply the construction of HEand that H0is saturated to get H0=HE0. Proposition 2.26. Let Ebe any graph and Kany field. Then I(Pec(E)) = ⊕˜c∈Xec I(˜c0)and every I(˜c0)is isomorphic to a Leavitt path algebra which is purely infinite simple. 30 Chapter 2. Ideals Proof. The hereditary set Pec(E) can be decomposed as: Pec(E) = t˜c∈Xec ˜c0. By the Remark 2.23 and the Proposition 2.10,I(Pec(E)) = I(t˜c∈Xec ˜c0) = ⊕˜c∈Xec I(˜c0). Finally, observe that every I(˜c0) is isomorphic to a purely infinite simple Leavitt path algebra by Lemma 2.25. Lema 2.27. For any graph Ethe hereditary sets Pl(E),Pc(E)and Pec(E)are pairwise disjoint. Moreover, the ideal generated by their union is I(Pl(E)) ⊕ I(Pc(E)) ⊕I(Pec(E)). Proof. By the definition of Pl(E), Pc(E) and Pec(E), they are pairwise disjoint. To get the result, apply Proposition 2.10. The following ideal, will be of use in order to establish the center of a Leavitt path algebra, so we name it here. Definition 2.28. For a graph Ewe define Ilce := I(Pl(E)) ⊕I(Pc(E)) ⊕I(Pec(E)). Theorem 2.29. Let Ebe an arbitrary graph and Kany field. Then: (i) Ilce ∼ =(⊕i∈Λ1Mmi(K)) ⊕⊕j∈Λ2Mnj(K[x, x−1])⊕⊕l∈Λ3I(˜c0 nl), where Λ1is the index set of the sinks of E,Λ2is the index set of the cycles without exits in Eand Λ3indexes Xec(E). (ii) If |E0|<∞then Ilce is a dense ideal of LK(E). Proof. For our purposes we may suppose that the graph Eis connected because if E=ti∈ΛEiis the decomposition of Einto its connected components and we show the claims for every connected component, then the result will be true for Eby virtue of Corollary 2.13. (i). We know that I(Pl(E)) is the socle of the Leavitt path algebra LK(E) (see [16, Theorem 5.2] and [8, Theorem 1.10]), moreover I(Pl(E)) ∼ = ⊕i∈Λ1Mmi(K), where Λ1ranges over the sinks of Eand for any i∈Λ1, if si is a sink (finite or not), then miis the cardinal of the set of paths ending at si. The structure of I(Pc(E)) is also known: by [5, Proposition 3.5] (which also works for arbitrary graphs) I(Pc(E)) ∼ =⊕j∈Λ2Mnj(K[x, x−1]), where the cardinal of Λ2is the cardinal of the set of cycles without exits and njis the cardinal of the set of paths ending at a given cycle without exits cjand not containing the edges appearing in the cycle. Finally, the structure of the third summand in Ilce follows by Proposition 2.26. 2.6. Decomposable Leavitt path algebras 31 (ii). Take a vertex v∈E0. Since E0is finite then vconnects to a line point, to a cycle without exists or to an extreme cycle. This means that every vertex of Econnects to the hereditary set H:= Pl(E)∪Pc(E)∪Pec(E). By Proposition 2.14 this means that I(H) is a dense ideal of LK(E) and by Lemma 2.27 it coincides with Ilce. Remark 2.30. Following a similar reasoning as in the proof of Theorem 2.29 it can be shown that any ideal of LK(E) generated by vertices in Pis isomorphic to (⊕i∈Λ1Mmi(K)) ⊕⊕j∈Λ2Mnj(K[x, x−1])⊕⊕l∈Λ3I(˜c0 nl) for some Λ1, Λ2, Λ3. We specialize this result in the case of a prime Leavitt path algebra. Corollary 2.31. Let Ebe a graph with a finite number of vertices such that LK(E)is a prime algebra. Then, we have the following three mutually exclusive cases: (i) There is a unique sink vin Eand every vertex of Econnects to v. In this case I(Pl(E)) = I(T(v)) ∼ =Mm(K), where mis the cardinal of the set of paths ending at v, or (ii) there is a unique cycle without exits cand every vertex of Econnects to it. In this case I(Pc(E)) = I(c0)∼ =Mn(K[x, x−1]), where nis the cardinal of the number of paths ending at cand not containing all the edges of c, or (iii) Xec(E) = {˜c}, where cis an extreme cycle and every vertex of Econnects to c. In this case I(Pec(E)) = I(c0)is a purely infinite simple unital ring. 2.6 Decomposable Leavitt path algebras We have seeing that whenever E0=tHithen LK(E) = ⊕I(Hi). In particular, the Leavitt path algebra LK(E) can be decomposed as a direct sum of ideals. There are other cases in which a Leavitt path algebra can be decomposed. In a paper jointly written with Lisa Orloff Clark, Dolores Mart´ın Barquero and C´andido Mart´ın Gonz´alez (see [21]), we characterize those Leavitt path algebras that are decomposable. We use the Steinberg algebra, but I’m not going to go into the details of these algebras. concretely, we show: 32 Chapter 2. Ideals Example 2.32. Let Ebe the following graph: • w1 • w2 • w3 • u1• u2• u3 • v1• v2• v3· · · E ////// //////  BB  BB  BB Let X={un|n∈N}and Y={wn|n∈N}. Then the only hereditary and saturated sets are X, Y, ∅and E0. Note that LK(E) is decomposable as LK(E) = I(X)⊕I(Y). Definition 2.33. ([17, Definition 4.1]) Let Ebe an arbitrary graph and H a hereditary and saturated set of vertices. A path λ=λ1. . . λ|λ|, where λi∈E1, is said to be H-compatible if r(λ)∈Hand s(λ|λ|)/∈H∪BH. The key point for decomposability of a Leavitt path algebra is precisely compatibility of hereditary and saturated sets. Theorem 2.34. (MSM; Lisa O. Clark; Dolores Mart´ın Barquero; C´andido Mart´ın Gonz´alez) Let Kbe a field and Ean arbitrary graph. Then LK(E)is decomposable if and only if there exist two nontrivial hereditary and saturated subsets H1and H2such that H1∩H2=∅and for every v∈E0\(H1∪H2), there exists at least one but finitely many paths starting at vwhich are either H1-compatible or H2-compatible. Chapter 3 The center of a Leavitt path algebra 3.1 Introduction to the center In this chapter we will completely describe the center of the Leavitt path algebra. We start by the description and the proof of the results concerning row-finite graphs (results belonging to [23,24]) and then we state the results in the general case, which where proved by using the Steinberg algebra associated to the groupoid related to a graph. This work is [22]. Here are some of the reasons because of which we are interested in the center of a Leavitt path algebra. 33 34 Chapter 3. The center of a Leavitt path algebra Relative to the simplicity of the Lie algebra related to an associative algebra, we have: 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). 3.2 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: 3.2. The extended centroid 35 Z(A) := {x∈A|[x, a] = 0 for every a∈A}, 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 [27, (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 [27, (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. (MSM; Corrales Garc´ıa; Mart´ın Barquero, Mart´ın Gonz´alez; Solanilla Hern´andez) 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. 36 Chapter 3. The center of a Leavitt path algebra Proof. As in the proof of Theorem 2.29 we may suppose that our graph is connected. Apply Theorem 2.29 (ii) and [27, (14.14) Theorem] to obtain Qs(LK(E)) = Qs(I). Then by [29, Lemma 1. 3 (i)] C(LK(E)) = C(I) = C(I(Pl(E))) ⊕ C (I(Pc(E))) ⊕ C (I(Pec(E))). By Theorem 2.29 (i), Lemma 3.1 and Proposition 2.26,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. 3.3 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. 3.3. The center 37 Proof. The first statement follows immediately from Lemma 3.6 For Igenerated by vertices in P, by Remark 2.30,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 [23, 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. 44 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. 3.3. The center 45 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 [23, 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. 46 Chapter 3. The center of a Leavitt path algebra Finally, take u∈[v]. We may assume that uis not a sink (this case has been studied yet). By Lemma 3.12 (i) and Lemma 2.5 we may write u=Pn i=1 αiα∗ i, where the αi’s are paths and r(αi) is in the hereditary closure of [v]. Then au =aPn i=1 αiα∗ i=Pn i=1 αiaα∗ i=Pn i=1 αikr(αi)α∗ i= kPn i=1 αiα∗ i=ku. Proposition 3.17. Let Ebe a row-finite graph and consider a nonzero homogeneous element aof degree zero in Z(LK(E)). Then: (i) av = 0 for every v∈P∞. (ii) a=P[v]∈Xfk[v]a[v]where a[v] =Pu∈[v] u + Pα∈FE([v]) αα∗ Proof. (i). Suppose first |[v]|=∞. If av 6= 0, by Lemma 3.16 we have au 6= 0 for every u∈[v], which is an infinite set; this is not possible, so av = 0 and we have finished. Now, assume |[v]|<∞. Since v∈P∞, necessarily |FE([v])|=∞, then we have an infinite collection of paths {αn}n∈N⊆FE([v]) with αm6=αnfor m6=n. Next we prove that as(αn)6= 0 for each n. First, note that r(αn)∈[v]; since av 6= 0 then ar(αn)6= 0 by Lemma 3.16, therefore 06=ar(αn) = aα∗ nαn=α∗ nas(αn)αnwhich implies our claim. Since there is only a finite number of vertices u∈E0such that au 6= 0, the set {s(αn)}n∈Nmust be finite. Moreover, the set [v], which contains all the 3.3. The center 47 ranges of the paths αnis finite. Since |{α0 n}n∈N|<∞(because every vertex in α0 ndoes not annihilate aby Lemma 3.16), v /∈Pfand no vertex in α0 n\r(αn) is an infinite emitter, we may conclude that there is a path αm=γ1. . . γr. . . γt, with γi∈Path(E) such that l(γr)≥1 and s(γr) = r(γr), hence αmcontains a cycle based at s(γr), implying s(γr)∈[v]. This is a contradiction with the fact αm∈FE([v]) and (i) has been proved. (ii). Write a=Pau, with u∈E0and au 6= 0. If u∼v, with v∈P∪Pe, then au =kvuby Lemma 3.16 and kv∈K×. If u6∼ vfor any v∈P∪Pe then, as it is not a sink, by (CK2), we may write u=Ps(e)=uee∗. Then au = aPs(e)=uee∗=Ps(e)=ueae∗. Take ein this summand. If r(e)∈P∪Pethen ar(e) = kr(e)r(e) where kr(e)∈Kand we get a summand as in the statement. Otherwise we apply (CK2) to r(e) and write r(e) = Ps(f)=r(e)ff∗; then aee∗=ae Ps(f)=r(e)ff∗e∗. Every summand aeff∗e∗with r(f)∈P∪Peis kr(ef)eff∗e∗by Lemma 3.16, which is a summand as in the statement. For every nonzero summand not being in this case we apply again (CK2). This process stops because otherwise we would have an infinite path e1e2. . . with ei∈E0such that s(ei)6=s(ej) for every i6=jand aei6= 0 for every i, which is not possible as the number of vertices not annihilating ahas to be finite. Note that the path e1e2. . . enwe arrive at is, by construction, an element in FE([v]). We remark that v∈Pfby (i). Take v∈E0such that [v]∈Xfand av 6= 0; by Lemma 3.16 we have av =kvvfor some kv∈K. Note that kv6= 0. For any u∈[v], Lemma 3.16 shows that au =kvu6= 0. If β∈FE([v]), then by Lemma 3.14 β∗a= kr(β)β∗ββ∗. Since β∗a=aβ∗=ar(β)β∗=kvβ∗, we get kr(β)=kv6= 0. This shows that acan be written as a linear combination of elements of the form (†)X u∈[v] u+X α∈FE([v]) αα∗, where [v]∈Xf. Theorem 3.18. (MSM; Corrales Garc´ıa; Mart´ın Barquero, Mart´ın Gonz´alez; Solanilla Hern´andez) Let Ebe a row-finite graph. For every class [v]∈Xf, denote by a[v]= Pu∈[v]u+Pα∈FE([v]) αα∗. Then B0=a[v]|[v]∈Xf is a basis of the zero component of the center of LK(E). Proof. By Proposition 3.17 every element in the zero component of the center is a linear combination of the elements of B0. Lemmas 3.12 and 3.14 imply that B0is a set of linearly independent elements. 48 Chapter 3. The center of a Leavitt path algebra In what follows we show that every a[v]in B0is in the center of LK(E). To start, consider a vertex win E0. It is not difficult to see wa[v]=a[v]w since a[v]is in ⊕n i=1uiLK(E)uifor a certain finite family of vertices {ui}n i=1. Consider and edge ein Eand denote by A:= [v]∪FE([v]). We claim: (†)r(e)∈Aif and only if s(e)∈Aor e ∈A. Assume r(e)∈A. If s(e)/∈Athen s(e)/∈[v] and since r(e)∈[v] we conclude that e∈FE([v]) ⊆A. Reciprocally, if s(e)∈Athen r(e)∈A because [v] is hereditary (Lemma 3.12 (i)). Finally, if e∈Athen e∈FE([v]) which implies r(e)∈A. Now we see that a[v]commutes with e. Suppose first that r(e)∈A. Then ea[v]=eby Lemma 3.14. On the other hand, if s(e)∈A, then a[v]e=e+ 0 = eby Lemma 3.14; if s(e)/∈A, then a[v]e= 0 + e=eby Lemma 3.14. Now, suppose r(e)/∈A. Then, by (†) we have s(e), e /∈A. For each αwe have eα = 0 or eα 6= 0 and, in this last case, eα ∈Asince s(e)/∈A, by hypothesis, and r(e) = s(α)/∈A. Applying the results explained above, ea[v]=eX α∈FE([v]) αα∗=eX α∈FE([v])α∗e=0 αα∗+eX α∈FE([v]) eαα∗e∗. We claim that the second summand must be zero. Indeed, suppose e2α6= 0; then, s(e)∼wfor a vertex w∈[v]; this implies s(e)∈[v], a contradiction since we assume s(e)/∈A. Therefore ea[v]=X α∈FE([v])α∗e=0 eαα∗(3.1) In what follows we compute a[v]e. a[v]e= X α∈FE([v])α∗e=0 αα∗ e+ X α∈FE([v]) eαα∗e∗ e=X α∈FE([v]) eαα∗e∗e =X α∈FE([v]) eαα∗=X α∈FE([v])α∗e=0 eαα∗+X α∈FE([v])α∗e6=0 eαα∗. (3.2) We claim that the second summand is zero. The reason is again that for αin the second summand, eα must be zero because αstarts by eand 3.3. The center 49 α∈FE([v]). Hence a[v]e=X α∈FE([v])α∗e=0 eαα∗=ea[v] by (3.1). Notation 3.19. Let Ebe a graph. For a cycle c=e1. . . enin Eand ui=s(ei) we will write cuito denote the cycle ei. . . ei−1. Let Sbe the set of all those cycles without exits csuch that there is a finite number of paths ending at cand not containing c. Note that if cis a cycle in Sand u, v ∈c0with u6=v, then cuand cvwill be different elements in S. Proposition 3.20. Let Ebe an arbitrary graph and Kany field. Consider a homogeneous element ain Z(LK(E)) with deg(a)>0. Then au = 0 for all u∈Pl∪Pe. If u∈Pcthen au =kucr u, where ku∈K,cuis a cycle without exits and r∈N. Moreover, if u∈P+ c, then ku= 0. Proof. If u∈Pl, then au =uau ∈uLK(E)u=Ku, by [16, Proposition 4.7 and Remark 4.8], hence there exists λ∈Ksuch that au =λu. Then λmust be zero as the degree of λu is zero and deg(a)>0. Now, take u∈Peand let dbe a cycle such that u=s(d). We will use partially a reasoning that appears in [23, Theorem 6]; we include it here for the sake of completeness. A generator system for uLK(E)uis A∪B, for A={dn(dm)∗|n, m ≥0} and B={dnαβ∗(dm)∗|n, m ≥0, α, β ∈Path(E), s(α) = u=s(β), d 6≤ α, β, α1∪β16⊆ d1}. For n= 0 we understand dn=u. Note that given n, m ≥0 and dnαβ∗(dm)∗∈B, there exists a suitable r∈Nsuch that (dr)∗dnαβ∗(dm)∗dr= 0. This gives us that if we define the map S:LK(E)→LK(E) by S(x) = d∗xd, for every b∈Bthere is an n∈Nsatisfying Sn(b) = 0. Note that au is a fixed point for S. A consequence of this reasoning is that au ∈span(A). Write au =Pnkndn(dm)∗, for m=n−deg(a), where kn∈K. 50 Chapter 3. The center of a Leavitt path algebra Then, for some l∈Nwe have au =Sl(au) = Pnknddeg(a). Since au commutes with every element in uLK(E)u, the same should happen to ddeg(a), but this is not true as it does not commute with d∗, giving au = 0. Next we show the second part of the statement. For u∈Pc,au =uau ∈ uLK(E)u∼ =K[x, x−1] (by [13, Proposition 2.3]). Since deg(a)>0, au =kucr u for some r∈N\{0},cua cycle without exits and ku∈K. To finish, consider u∈P+ cand write au =kucr, for ca cycle without exits and ku∈K. Then, two cases can happen. First, assume that there exists a cycle dsuch that v:= s(d)≥uand take a path αsatisfying α=vαu. Then, as we have proved before, av = 0 and so kucr=au =aα∗α=α∗aα = α∗avα = 0; this implies ku= 0 as required. In the second case there exists infinitely many paths γn, for n∈Nending at v. Since Eis row-finite (and we are assuming that we are not in the first case, so there are no cycles involved), |{s(γn)}| =∞. Then there exists m∈Nsuch that as(γm)=0 and so av =aγ∗ mγm=γ∗ mas(γm)γm= 0 as desired. Lema 3.21. Let Ebe a row-finite graph and Kany field. For any homogeneous element ain Z(LK(E)) with deg(a)>0we have a=Paαα∗, where α∈FE(c0)for some cycle without exits cand r(α)∈P− c. Proof. Let ube in E0such that au 6= 0. Define T1(u) = {v∈T(u)| ∃ e∈E1,such that s(e) = u, r(e) = v}. Let Tn(u) be Tn(u) = {v∈T(u)\∪ i<nTi(u)| ∃ α∈Path(E), l(α) = n, such that s(α) = u, r(α) = v}. Given 0 6=au, write au =aPs−1(u)=eee∗=Ps−1(u)=eeae∗(note that u cannot be a sink and so we may use (CK2)). By Proposition 3.20, the possibly nonzero summands are those eae∗such that r(e)∈T1(u)\(Pl∪Pe∪P+ c). Let S1:= {aee∗such that r(e)∈P− c}. Now, to every element aee∗with r(e)∈T1(u)\(Pl∪Pe∪Pc), apply Condition (CK2). Then aee∗=a(ePs−1(r(e))=fff∗)e∗= (ePs−1(r(e))=ffaf∗)e∗. 3.3. The center 51 Again by Proposition 3.20 the nonzero summands are those efaf∗e∗such that r(f)∈T2(u)\(Pl∪Pe∪P+ c). Let S2:= {aeff∗e∗such that r(f)∈P− c}. This reasoning must stop in a finite number of steps, say m, because otherwise we would have infinitely many edges e1, e2. . . such that ae1. . . en6= 0 for every n. Since for every vertex win a cycle aw = 0, the ranges of these paths are all different. Therefore, there would be infinitely many vertices not annihilating a, a contradiction. Note that, by construction, a=Px∈∪m i=1Six and so acan be written as in the statement, where α∈FE(c0) for some cycle without exits c. Lema 3.22. Let Ebe an arbitrary graph and consider cu, dw∈ S,α∈ FE(c0 u),β∈FE(d0 w)and n, m ∈Z. (i) If w=u, then α∗β6= 0 if and only if α=β. (ii) If w6=uthen cm uα∗βdn w= 0. Proof. (i). If α∗β6= 0 then α=βγ or β=αδ for some γ, δ ∈Path(E). In the first case, since r(β)∈c0 u, by the very definition of FE(c0 u) and taking into account α∈FE(c0 u) we get α=β. In the second case we get analogously the same conclusion. The converse is obvious. (ii). Take w6=uand assume α∗β6= 0; this implies as before α=βγ or β=αδ for some γ, δ ∈Path(E). In the first case, r(β)∈d0 w; since dw has no exits, then γ0⊆d0 w. On the other hand, r(α) = r(γ)∈c0 u∩d0 w, therefore c0 u=d0 wand arguing as in (i) we conclude α=β. This implies cm uα∗βdn w=cm udn w= 0 since w6=u. The other possibility yields the same conclusion. The following result will relate elements in the ncomponent of the center to elements in the 0 component. Lema 3.23. Let Ean arbitrary graph and Kany field. Let abe a nonzero element in Z(LK(E)). Then aa∗6= 0. Proof. By Theorem 2.6, there are α, β ∈Path(E) such that 0 6=α∗aβ =kv for some k∈K×,v∈E0, or 0 6=α∗aβ =p(c, c∗), where p(c, c∗) is a polynomial in a cycle without exits c. In the first case 0 6=k2v= (kv)(kv∗) = α∗aββ∗a∗α=α∗ββ∗aa∗α, since a is in the center of LK(E), and so aa∗must be nonzero. 52 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. (MSM; Corrales Garc´ıa; Mart´ın Barquero, Mart´ın Gonz´alez; Solanilla Hern´andez) 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∗. 3.3. The center 53 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. 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∞. 60 Chapter 4. Simplicity of related Lie algebras Theorem 4.1. ([1, Theorem 4.14]). (Abrams; ´ Ahn; Pardo) 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. (Abrams; Mesyah) 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: 61 Theorem 4.4. (Abrams; Mesyah) 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 AEthe 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. (Alahmedi; Alsulami) Let Ebe a row-finite graph. 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