CIMPA 2015 Research School on LPA and Graph C*-algebras Sections - 1 and 2: Direct limits and Complete subgraphs Kulumani M. Rangaswamy University of Colorado Colorado Springs July 5 - 7, 2015 Section 1: DIRECT LIMITS A directed set is a partially ordered set (I; ) with the property that to each i; j 2 I there is a k 2 I such that i k and j k. A direct system of algebras is a system of algebras fAi (i 2 I); ij g where I is a directed set such that for any two algebras Ai ; Aj with i j, there is a homomorphism ij : Ai ! Aj satisfying the following conditions: (i) ii = 1Ai , the identity map of Ai , for all i 2 I; (ii) For all i j k in I, we have the commutative diagram j Ai i ! k i & Aj # jk Ak (1) The direct limit of the direct system of K-algebras (1) above is a K-algebra A that possesses the following universal properties: (a) For each i 2 I there is a homomorphism i : Ai ! A such that the diagrams j Ai i ! i & Aj # j A (2) are commutative for all i; j 2 I; (b) If B is K-algebra and there are homomorphisms following commutative diagrams for all i; j 2 I; j Ai i ! i & Aj # j B 1 (3) i : Ai ! B with the then there exists one and only one homomorphism diagrams i ! i & Ai A # B : A ! B such that all the (4) commute for all i 2 I: The algebra A together with the morphisms i , is uniquely determined by this property, upto an isomorphism. That is, if there is an algebra A0 satisfying property (a) and (b), thern A0 = A Construction: Given aM directed system of K- algebras fAi (i 2 I); ij g, form the direct sum A = Ai . Let N be the ideal generated by the set i2I j fai jg. De…ne, for each i, i : Ai ! A=B by i (ai ) : ai 2 Ai ; i; j 2 I; i i : ai 7 ! ai + N . Then A=N satis…es the commutative diagram (2). Properties: (a) If a = ai1 + + ain 2 N with aik 2 Aik , then ther is an i i1 ; ; in in I such that ii1 (ai1 ) + + iin (ain ) = 0 : Now the generators j j k ai j k, ik (ai ) j ( i (ai )) = i (ai ) of N have this property, namely, for i k k i (ai ) i (ai ) = 0. Hence the element a also has this property. (b) Every a + N is represented by a coset ai + N for some i 2 I: Suppose a = ai1 + + ain with aik 2 Aik . Choose i i1 ; ; in in I. Let ai = i i i i (a )+ + (a ). Then a a = (a (a ))+ +(a i i1 in i1 i1 in in i1 i1 in (ain )) 2 N and so a + N =[ai + N . (c) A=N = i (Ai ) i (d) If every ij is a monomorphism, then all he i are monomorphisms: Suppose i (ai ) = 0 so that ai 2 N . Then by Property (a), there is a j i in I such that ij (ai ) = 0. Since ij is a monomorphism, ai = 0. We claim that the algebra A=N constructed above is isomorphic to the direct limit A . Suppose B is an algebra that satis…es the commutative diagram (3) with the indicated maps. We de…ne a map : A=N ! B as follows: If a = a + N 2 A=N is represented by ai + N , de…ne (a ) = i (ai ). From the commutativity of the diagram (3), this de…nition is independent of the choice of i and hence is a well-de…ned homomorphism from A=N to B. Also makes the diagram commutative since i (ai ) = (a ) = i (ai ). If 0 : A=N ! B is another homomorphism that makes all the diagrams 0 of the form (4) commutative, then ( ) i = 0 for every i 2 I, that is, 0 0 ( ) i (Ai ) = 0 for all i. By Property (c) above, = 0 and thus is unique. Finally, suppose A0 is an algebra with homomorphisms i : Ai ! A0 satisfying the property stated for A and i , then from what we have shown, there will be a unique homomorphism : A=N ! A0 and by our assumption on A0 , 2 0 a homomorphism 0 : A0 ! A=N such that i = i and i = i . We then 0 0 0 obtain i = and = . This yields an isomorphism A A=N . = i i i Notation: A = LimAi . ! Section 2: COMPLETE SUBGRAPHS All the graphs that we consider in these talks are arbitrary connected row…nite graphs E and L will denote the Leavitt path algebra LK (E) of the graph E over a …eld K: First observe that if F is a subgraph of E, then LK (F ) need not be a subalgebra of LK (E). e 2.1. Example: Let E be the graph u v where the bottom edge is e f . Let F be the subgraph u ! v . In Lk (F ), we have , by CK-2 relation, ee = u. But in LK (E), we have ee +f f = u. Thus LK (F ) is not (isomorphic to) a subalgebra of LK (E). To make LK (F ) a subalgebra of LK (E), the CK-2 relations must be preserved. This happens when F is a complete subgraph. 2.2. De…nition: A subgraph F of a graph E is called a complete subgraph if for each regular vertex v in F , we have, sF 1 (v) = sE 1 (v). This means that for every vertex v in F which is not a sink in F , if v is a …nite emitter in F say sF 1 (v) = fe1 ; ; en g, then v must also be a …nite emitter in E and sE 1 (v) = fe1 ; ; en g. 2.3. Proposition: If F is a complete subgraph of E, then there is an algebra monomorphism : LK (F ) ! LK (E) satisfying (v) = v, (e) = e and (e ) = e for all v 2 F 0 and e 2 F 1 . The subalgebra generated by F in LK (E) is isomorphic to LK (F ). Proof : We need only to check the CK-2 relation (4). Since F is a complete 1 1 subgraph, applying toX the equation X for any v 2 F , sF (v) = sE (v) and soX v= ee in LK (F ), we get v = (v) = ( ee ) = ee . e2sF 1 (v) e2sF 1 (v) e2sE 1 (v) 2.4. Exercise: List all the complete subgraphs of the graph E if (a) E is the in…nite clock; (b) E is the Rose graph Rn with n petals. 2.5 Proposition: Let E be a row-…nite graph. Then E is a directed union of …nite complete subgraphs. Proof : Now the …nite subgraphs of E form a directed set under set inclusion whose union is E. So it is enough to show that every non-empty …nite subgraph of E embeds in a …nite complete subgraph of E. Let F be a …nite subgraph of E. De…ne a subgraph F as follows: (F )0 = F 0 [ frE (e) : e 2 E 1 and sE (e) 2 F 0 g and (F )1 = fe 2 E 1 : sE (e) 2 0 F g. It is then clear that for any vertex v which is not a sink in F , we have sF 1 (v) = sE 1 (v). Thus F is a …nite complete subgraph of E containing F . Now 3 the union of a …nitely many complete subgraphs is again a complete subgraph (Verify). Hence E is a directed union of …nite complete subgraphs of E From Propositions 2.3 and 2.5, we obtain the following main result of this section. 2.6 Theorem : Let E be a row-…nite graph. Then the Leavitt path algebra L := LK (E) is a directed union of a direct system of subalgebras Si (i 2 I), where, for each i 2 I, Si = LK (Fi ), the Leavitt path algebra of a …nite complete subgraph Fi . 4 CIMPA 2015 Research School on LPA and Graph C*-algebras Section - 3: Von Neumann Regular rings Kulumani M. Rangaswamy University of Colorado Colorado Springs July 5 - 7, 2015 In this section, we introduce the concept of a von Neumann regular ring and study its properties. We will show that if E is a …nite acyclic graph then the Leavitt path algebra LK (E) is a …nite direct sum of matrix rings over the …eld K and, in particular, von Neumann regular. 3.1 De…nition: A ring R is said to be von Neumann regular if to each element a 2 R there is an element b 2 R such that aba = a. Note that in this case, abab = ab and baba = ba. So both ab and ba are idempotents in R. If = ba, then it is clear that Ra = R and consequently, every principal left ideal of R is a direct summand of R. 3.2 Exercise: Let R be a von Neumann regular ring. (i) Show that every …nitely generated left ideal of R is a direct summand; (ii) For any idempotent in R show that the corner R is von Neumann regular. (iii) Show that for each element a 2 R we can choose an element b such that not only aba = a, but also bab = b. 3.3 Proposition: Let V be a vector space over a …eld K. Then the ring R = EndK (V ) of all linear transformations of V is von Neumann regular. Proof : Let be a linear transformation of the vector space V with Ker( ) = A. Write V = A C, where C is a complementary subspace. If S = Im( ), then it is clear that jC maps C isomorphically onto S. So its inverse ( jC) 1 : S ! C is an isomorphism. Write V = S T . De…ne a linear transformastion : V ! V by setting jS = ( jC) 1 and (T ) = 0. It is then easy to check that = . (Veri…cation: Note that jS = ( jC) 1 means that for any s 2 S, (s) = c if (c) = s. So (c) = (s) = c for all c 2 C. To show = , let x = a + c be any element of V where a 2 A and c 2 C. Then (a) = (c) = (c) = (a). Hence = .) 1 3.4 Corollary: Let K be a …eld. Then for any positive integer n, the ring Mn (K) of all n n matrices with entires from K is von Neumann regular. Proof : Let V be a K-vector space of dimension n. Let B = fv1 ; ; vn g be a …xed basis of V . Then every linear transformation of V can be realized as an n n matrix and so the ring EndK (V ) of all the linear transformations of V is isomorphic to the matrix ring Mn (K). By Proposition 3.4 EndK (V ) is von Neumann regular. So the ring Mn (K) is von Neumann regular. 3.5 Matrix units: In Mn (K), for each 1 i; j n, let eij be the matrix having 1 at (i,j) position and 0 everywhere else. These n2 matrices feij : 1 i; j ng are called the matrix units of Mn (K). They satisfy the condition n M n M eij ekl = eil or 0 according as j = k or not. Moreover, Mn (K) = Keij . i=1 j=1 3.6 De…nition: For any vertex v, de…ne nv to be the number of distinct paths that end at v, (including the trivial path v). Example: 3.7 Proposition: Let E be a …nite acyclic graph. Let v be a sink in E. Then the ideal Iv of LK (E) generated by v is isomorphic to the matrix ring Mnv (K). n X Proof : Let J be the set of all elements of the form ki i i such that i=1 r( X i ) = r i ) = v where ki 2 K and n is some positive integer. Note that ki i i Iv since i = i v 2 Iv and i = v i 2 Iv . Hence J Iv . We wish to show that J is an ideal of LK (E). It is clearly closed under addition. Let s r X X lj j j 2 LK (E). ki i i with r( i ) = r i ) = v. Let b = a 2 J, say a = j=1 i=1 If ab 6= 0 and if ki lj i i j j 6= 0, then i j 6= 0. This means that either 0 or j = i 0 . If i = j 0 , observing that r( 0 ) = v, we have i = j ki lj i i j j = ki lj i ( j 0 ) 2 J. On the other hand , if j = i 0 , then 0 = v as v is a sink and so j = i . In this case, ki lj i i j j = ki lj i j 2 J as r( j ) = r( j ) = r( i ) = r( i ) = v. This shows that J is an ideal. Since J contains v (why ?), J = Iv . Now nv is …nite, because the graph E is …nite, acyclic and row-…nite. Let fp1 ; ; pn g be a listing of the nv paths ending at the sink v so that every n X n n M n X M element of Iv is of the form kij pi pj . Thus Iv = Kpi pj . If we i=1 j=1 i=1 j=1 de…ne ij = pi pj , it is easy to see that ij kl = il or 0 according as j = k or not. Thus the set f ij : i; j = 1; ; ng is a set of matrix units for Iv . Then the map : Iv ! Mn (K) mapping ij 7 ! eij for all i; j = 1; ; n gives rise to an isomorphism of Iv to Mn (K). 3.8 Theorem: Let E be a …nite acyclic graph. Then the Leavitt path algebra L := LK (E) is a direct sum of …nitely many matrix rings of …nite order over K. 2 Proof: Let fv1 ; ; vr g be the set of all sinks in E and, for each i, let Ivi r M be the ideal generated by vi . We wish to show that LK (E) = Ivi . First i=1 observe that, for i 6= j, Ivi Ivj = 0. Beacuse, if 2 Ivi and 2 Ivj , then 6= 0 implies 6= 0 which means that either = 0 or = 0 . But neither of these possibilities can arise since r( ) = vi a sink and r( ) = vj a sink 6= vi . Hence Ivi Ivj = 0, for i 6= j. From Proposition 3.7, it is clear that each ideal Ivi has a multiplicative identity and using this it is clear that, for r r X X M each i, Ivi \ ( Ivj ) = 0. Consequently, I vi = Ivi . i=1 j6=i We next show that LK (E) = r X i=1 sink, say vi for some i, then clearly Then , by CK-2 relations, X = u = ( i=1 Ivi . Let 0 6= 2 LK (E). If r( ) is a 2 Ivi . Suppose r( ) = u is not a sink. ee ) X = e2E 1 ;s(e)=u e( e) . e2E 1 ;s(e)=u If r(e) = vi for someX sink, then e( e) 2 Ivi ; otherwise apply CK-2 relation to replace r(e) by f f . Repeat this process. Now since the graph E f 2E 1 ;s(f )=r(e) is …nite and there are no cycles, every term in the above expression is a term already in some Ivi or we can repeat the process expanding as many times as necessary until we reach the sinks in E. In this way can be written as a sum of the terms of the form ( ) for some such that r( ) = vi This r r X M shows that LK (E) = Ivi = Ivi as desired. i=1 i=1 w1 3.9 Exercise: Describe LK (E) if E is w . - g w2 u1 & % u3 % h & u2 REFERENCE: G. Abrams, G. Aranda Pino and M. Siles Molina, Finite dimensional Leavitt path algebras, J. Pure and Applied Algebra, vol. 209 (2007), 753 - 762. 3 CIMPA 2015 Research School on LPA and Graph C*-algebras Section 4 - Von Neumann Regular Leavitt Path Algebras Kulumani M. Rangaswamy University of Colorado Colorado Springs July 5 - 7, 2015 In this section, we provide a complete characterization of Leavitt path algebras which are von Neumann regular. 4.1 Theorem: Let E be a row-…nite graph. Then the Leavitt path algebra LK (E) is von Neumann regular if and only if E is acyclic. In this case, LK (E) is a directed union of direct sums of matrix rings of …nite order over K. Proof : Assume LK (E) is von Neumann regular. Suppose, by way of contradiction, E contains a cycle c based at a vertex v. Consider the element v c. By hypothesis, there is an element a 2 LK (E) such that (v c)a(v c) = v c ( ) Replacing a by vav, we can assume that va = a = av. Write a as a graded sum n X a= ai where deg(ai ) = i, am 6= 0; an 6= 0 and m n 2 Z. Since deg(v) = 0, i m the equation va = a = av implies vai = ai = ai v for all ai . Substituting for a in the equation ( ), we get (v c) n X ai (v c) = (v c) ( ) i m Equating the lowest degree terms on both sides, we get vam v = am = v. So n X m = deg(am ) = deg(v) = 0. Thus a = ai . i=0 Let deg(c) = s > 0. Since the degree of the terms on the right hand side of the equation ( ) is ks with k = 0 or 1, equating the corresponding graded 1 components on both sides, we conclude (as deg(ai ) = i) that ai = 0 if i is non-zero and not a multiple of s. Rewriting ( ), we then obtain (v c)v(v c) + (v r X c) ats (v c) = (v c): t=1 This simpli…es to c + c2 + (v c) r X ats (v c) = 0: t 1 Comparing the degree s components on both sides, we have c + as = 0 and so as = c. Comparing components of degree 2s on both sides, we get c2 cas as c + a2s = 0. Substituting for as , we obtain a2s = c2 . Proceeding like this, we r X conclude that ats = ct for all t = 1; ; r. Thus a = ci . Clearly a(v c) = i=0 (v c)a and so the equation( ) becomes (v c)2 ( r X ci ) = (v c): i=0 This is impossible, since the highest degree term on the left hand side is cr+2 with degree (r + 2)s where r 0, while the highest degree of the terms on the right hand side is s. This contradiction shows that E cannot contain any cycles. Conversely, suppose the graph E does not contain any cycles. Now by Proposition 2.5, E is a directed union of …nite complete subgraphs Fi where i 2 I, a directed set and, by Theorem 2.6, LK (E) is a directed union of subalgebras Bi = LK (Fi ). Now each Fi is a …nite acyclic graph and so, by Theorem 3.8, Bi is a direct sum of …nitely many matrix rings of …nite order over the …eld K. We then appeal to Corollary 3.4 to conclude that each Bi , (i 2 I) is von Neumann regular. Since each element of LK (E) belongs to some Bi , we then conclude that LK (E) itself is von Neumann regular. 4.2 Remark: The above theorem remains true even if E is not row-…nite. But the proof involves constructing a di¤erent type of directed system of subalgebras of LK (E) and is a bit more involved. See the reference [1] below Leavitt path algebras as algebras with an involution: Recall that an involution on a K-algebra.A is an additive morphism a 7 ! a that satis…es the following properties for all a; b 2 A: (a + b) = a + b ; (ab) = b a and (a ) = a: The Leavitt path algebra LK (E) of a graph E is also endowed with an involution where, for any e 2 E 1 , e denotes the ghost edge, for any path p = e1 en , n X p = en e1 and for any v 2 E 0 , v = v. Thus for any element a = ki i i 2 i=1 2 LK (E), a = n X i=1 ki i i where ki 2 K and where we assume that K is a …eld with identity involution. If the …eld K also possesses an involution k 7 ! k, n n X X then we shall de…ne, for a = ki i i 2 LK (E), a = ki i i . i=1 i=1 In an algebra with involution , an element is said to be a projection if = 2= . We have described Leavitt path algebras L which are von Neumann regular without recognizing and using the fact that L has the structure of an algebra with involution. What happens if we wish to use the existence of an involution in de…ning the von Neumann regularity ? This leads to the de…nition of a *-regular algebra, where projections take over the role of idempotents. 4.3 De…nition: An algebra A is called -regular, if every principal left ideal is generated by a projection, that is, for every a 2 A, there is a projection in A such that Aa = A . This de…nition is left/right symmetric since Aa = A for all a implies that a A = A for all a 2 A. The elementwise de…nition for -regularity is the same as before, namely, to each a 2 A, there is a b 2 A such that aba = a with the additional condition that both ab and ba are projections. 4.4 Notation: In the following theorem, we shall use the following notation. For any vertex v, let nv denote the cardinality of all distinct paths that end at v (including the trivial path v). De…ne = supfnv : v 2 E 0 g if this supremum is …nite. Otherwise, let = !. 4.5 Theorem: Let E be an arbitrary graph and let K be …eld with involution . Then the following conditions are equivalent: (1) LK (E) is -regular; (2) E is acyclic and the …eld K is n- proper for all positive integers n , that is, x1 x1 + + xn xn = 0 implies x1 = = xn = 0 for any n elements x1 ; ; xn in K. For a proof of this theorem see Reference [2] below. 4.6 Example: 1. The …eld R of real numbers with trivial involution x = x satis…es the condition x21 + +x2n = 0 implies x1 = 0; ; xn = 0. Consequently, for any acyclic graph E, LR (E) is -regular. Example 1 2. But for the …eld C of complex numbers with trivial involution, given any acyclic graph E, the Leavitt path algebra LC (E) is not -regular, since in C, 12 + i2 = 0. However, for C with complex conjugation involution , LC (E) is -regular for any acyclic graph E. 4.7 Remark: It is interesting to note that in most of the results on the algebraic structure of a Leavit path algebra LK (E), the undlying …eld K does 3 not seem to play any role at all. But, as is clear fom the preceding theorem, when LK (E) is considered with involution, the nature of the …eld K becomes relevant. Leavitt path algebras as graded algebras Recall that a Leavitt path algebra L := LK (E) over an arbitrary graph E is endowed with a Z-graded algebra structure. 4.8 De…nition: A graded ring R is said to be graded von Neumann regular if for every homogeneous element a 2 R there is a homogeneous element b 2 R such that aba = a. Equivalently, every graded principal one sided ideal is generated by a homogeneous idempotent. As a graded algebra, one can ask when will the von Neumann regular property hold in L:In this connection, we have the following interesting theorem: 4.9 Theorem: (Hazrat) Every Leavitt path algebra is a graded von Neumann regular algebra For a proof see Reference [3] below. REFERENCES: [1] G. Abrams and K.M. Rangaswamy, Regularity conditions for arbotrary Leavitt path algebras, Algebra Representation Theory, vol. 13 (2010), 319 - 334 [2] .G. Aranda Pino, K.M. Rangaswamy and L. Vas, *Regular Leavitt Path Algebras of Arbitrary Graphs, Acta Mathematica Sinica, English Series, vol. 28, (2011), 957 - 968. [3] R. Hazrat, Leavitt pathalgebras are graded von Neumann regular rings, J. Algebra vol. 401 (2014), 220 - 233. 4 CIMPA 2015 Research School on LPA and Graph C*-algebras Section - 5: Condition (K) and Graded Ideals Kulumani M. Rangaswamy University of Colorado, Colorado Springs July 5 - 7, 2015 5.1: De…nition: (a) A graph E is said to satisfy Condition (K) if every vertex in E either does not lie on any closed path or lies on atleast two di¤erent simple closed paths. (b) A graph E is said to satisfy Condition (L) if every closed path contains an exit. 5.2 Condition (K) implies Condition (L), but not conversely. (Verify). The following theorem shows how Condition (K) of the graph E is equivalent several generalized regularity conditions for the Leavitt path algebra LK (E). We begin with some preparatory results. 5.3 Reduction Theorem: ([1]) Given a non-zero element a 2 LK (E), there exist paths ; such that one of the following statements (i), (ii) holds: (i) 0 6= a = kv where k 2 K and v 2 E 0 ; (ii) 0 6= a = f (c; c ) where c is a cycle without exits and 0 6= f (x; x 1 ) 2 K[x; x 1 ]. 5.4 Proposition: Let E be a row-…nite graph. Then an ideal I with I \E 0 = H is a graded ideal of LK (E) if and only if I =< H >, the ideal generated by H. Proof: Suppose I =< H >. A typical element x of I is a …nite sum of elements of the form avb, where a; b 2 LK (E) and v 2 H. Write a; b as graded n l X X sum of homogeneous elements, say a = ai and b = bj , where m; n; k; l 2 Z i=m and so avb = n X l X j=k ai vbj . Here we assume, without loss of generality, that the i=mj=k ai vbj are non-zero for all i; j..Since v 2 I, each element ai vbj 2 I and, since deg(v) = 0, ai vbj is a homogeneous element of degree deg(ai ) + deg(bj ) and n X l X avb = ai vbj is a graded decomposition of avb as a sum of homogeneous i=mj=k 1 elements. In this sum, each ai vbj 2 I as v 2 I. This proves that I is a graded ideal of LK (E). Conversely, suppose I is a graded ideal of LK (E). Let J =< H >. Clearly J I and is a graded ideal by the preceding paragraph. Hence the natural map : LK (E)=J ! LK (E)=I given by (a + J) 7 ! (a + I) is a graded epimorphism Also we have a graded isomorphism : Lk (EnH) ! LK (E)=J. Then : Lk (EnH) ! LK (E)=I is a graded epimorphism and (v) 6= 0 for all v 2 (EnH)0 = E 0 nH. Then, by the graded uniqueness theorem, and hence is injective. This means I = J =< H >. 5.5 Proposition: A graph E satis…es Condition (K) if and only if for every hereditary saturated subset H of E 0 , the graph EnH satis…es Condition (L). Proof: Suppose E satis…es Condition (K). Let H be a hereditary saturated set of vertices in E. Consider EnH. Let c be a cycle based at a vertex v in EnH. Clearly c ias a cycle in E based at v. By Condition (K), v is the base of another closed path g di¤erent from c. Since v 2 = H, all the vertices and all the edges in the path g belong to EnH. Hence g is a closed path in EnH based at v. Since c and g are two distinct closed paths based at v, the cycle c will have an exit in EnH. This proves that EnH satis…es Condition (L). . Conversely, suppose that EnH satis…es Condition (L) for all hereditary saturated subsets H of E 0 . Let v be a vertex based at a simple closed path c. Let H = fw 2 E 0 : w vg. Then H is a hereditary saturated subset of E 0 (Verify). Clearly, c is a closed path in EnH. By hypothesis, c has an exit e in EnH. Now r(e) = w v and so there is a path e2 en from r(e) to v. We can choose the edges ei so that r(ei ) 6= v for all 1 i n. Then = ee2 en is a simple closed path in EnH based at v which is di¤erent from the cycle c in EnH. Clearly, is a simple closed path in E based at v and di¤erent from c. This shows that E satis…es Condition (K). . 5.6 Proposition: If a row-…nite graph E satis…es Condition (K) then every ideal of LK (E) is a graded ideal. Proof : Let I be a non-zero ideal with I \E 0 = H. Let J =< H >. We wish to show that J = I. Suppose, on the contrary, J 6= I. It is known that there is an epimorphism : LK (E) ! LK (EnH) such that (v) = v for all v 2 E 0 nH and such that ker( ) = J. Let I = (I). Now I \ (EnH)0 = ;, because if there is a u 2 I \ (EnH)0 , then 1 (u) = u 2 I \ (E 0 nH) = H \ (E 0 nH) = ;, a contradiction. Thus I is a non-zero ideal of LK (EnH) not containing any vertices. If 0 6= a 2 I, then, by Reduction Theorem 5.3, there exist paths ; in EnH such that a = f (c; c ) 2 I where c is a cycle without exits in EnH. This is a contradiction since EnH satis…es Condition (L) due to the fact that E satis…es Condition (K) (Proposition 5.5). Hence I = J and, by Proposituion 5.4, I is a graded ideal of LK (E). 5.7 The "Hedgehog" graph (see [1]) 2 De…nition: Suppose E a row-…nite graph and H is a non-empty hereditary saturated subset of vertices in E. Let F (H) = fpaths = e1 en : n 1; r(ei ) 2 = H for i = 1; ; n 1 and r(en ) 2 Hg. Let F (H) = f : 2 F (H)g. Then the "hedgehog" graph H E = (H E 0 ;H E 1 ; s0 ; r0 ) is de…ned as follows: (1) H E 0 = H [ F (H). (2) H E 1 = fe 2 E 1 : s(e) 2 Hg [ F (H). (3) For every e 2 E 1 with s(e) 2 H, s0 (e) = s(e) and r0 (e) = r(e). (4) For every =2 F (H), s0 ( ) = and r0 ( ) = r( ). w1 f Example: Let E be the graph ! e1 % ! & w2 . Now H = fw1 g w3 is a hereditary saturated set. F (H) = fe1 ; f e1 g (H E)0 = H [ F (H) = fw1 ; e1 ; f e1 g (H E)1 = fe1 ; (f e1 ) g e1 e1 & w1 The graph H E looks like (f e1 ) % f e1 5.8 Proposition: Let E be a row-…nite graph. If I is a graded ideal of LK (E) with I \ E 0 = H, then I is isomorphic to the leavitt path algebra LK (H E). Proof :De…ne a map : (H E)0 [ (H E)1 [ fe : e 2 (H E)1 g ! I as follows: (v) = v, if v 2 H; ( )= , if 2 F (H); (e) = e and (e ) = e , if e 2 E 1 with s(e) 2 H; ( ) = , ( ) = , if 2 F (H). Observe that for any two di¤erent elements ; 2 F (H), = 0 and so f (u) : u 2 (H E)0 g is a set of pair-wise orthogonal idempotents. Claim: The elements of the set ((H E)0 ) [ ((H E)1 ) [ (fe : e 2 (H E)1 g) satisfy the generating relations (1) - (4) of a Leavitt path algebra. For example, let us verify the CK-2 relations in two cases. (i) Let v 2 H. Then s k X 1 (v) = fe1 ; ; ek g (e) (e ) due to CK-2 condition of LK (E). i=1 3 E 1 and so (v) = v = k X i=1 ee = (ii) Let 2 F (H). Then sH 1E ( ) = f g and so = in LK (H E). On the other hand,by de…nition, ( ) = = ( ) ( ): Hence there is a K-algebra homomorphism : LK (H E) ! I. Now is a graded homomorphism since it prserves the degrees of the vertices and edges in does not vanish on any vertex in H E, is a monomorphism by the H E. As graded uniqueness theorem. To show that is onto, note that every element a 2 I is of the form a = n X ki pi qi with r(pi ) = r(qi ) 2 H. So it is enough if we show that any path p i=1 with r(p) 2 H belongs to im( ). Now H im( ) as (v) = v for all v 2 H. Let p = e1 ek be a path with r(ek ) 2 H. If s(e1 ) 2 H, then s(ei ) 2 H for each i and then p = (e1 ) (ek ) = (e1 ek ) 2 im( ). Suppose s(e1 ) 2 = H. Now r(ek ) 2 H. Let j be the smallest integer with 1 < j k such that r(ej ) 2 H. Then r(ei ) 2 = H for all i = 1; ; j 1 so that = e1 ej 2 F (H): Consequently, p = (e1 ej )ej+1 ek = ej+1 ek = ( ) (ej+1 ) (ek ) 2 im( ). Thus I = im( ) and : LK (H E) ! I is a graded isomorphism. REFERENCES: [1] P. Ara and E. Pardo, Stable rank of Leavitt path algebras, Proc. Amer. Math. Soc., vol. 136 (2008), 2375 - 2386. [2] G. Aranda Pino, D. Martin Barquero, C. Martin Gonzalez and M. Siles Molina, The socle of a Leavitt path algebra, J. Pure and Applied Algebra, vol. 212 (2008), 500 - 509. 4 CIMPA 2015 Research School on LPA and Graph C*-algebras Section - 6: Generalized Regularity Kulumani M. Rangaswamy University of Colorado, Colorado Springs July 5 - 7 , 2015 In this section, we shall show how Condition (K) of the graph E implies various di¤erent generalized regularity conditions on the Leavitt path algebra LK (E). Recall, a ring R is said to be (von Neumann) regular if each a 2 R satis…es a 2 aRa. In the paper [1], this property was also shown to be equivalent to LK (E) being -regular, that is, for each a 2 LK (E) there is a positive integer n such that an 2 an LK (E)an . We wish to consider generalizations of these properties. 6.1 De…nition : A ring R is said to be right (left) weakly regular if each a 2 R satis…es a 2 aRaR (a 2 RaRa). Right (left) weakly regular rings are also known as right (left) fully idempotent rings due to the equivalent condition that I = I 2 for every right (left) ideal I of R. As a common generalization of both the weak regularity and the -regularity, a ring R is called a right (left) weakly -regular ring if for each element a 2 R there is a positive integer n such that an 2 an Ran R (an 2 Ran Ran ). A generalization of the right/left fully idempotent rings are the rings R in which, for every right (left) ideal I, there is a positive integer n such that I n = I n+1 . GOAL: To show how Condition (K) of the graph E is equivalent to LK (E) having all these and other ring theoretical properties. 6.2 Examples. The following examples show that the various generalized regularity conditions for a ring R are not, in general, equivalent to each other. Weak regularity of a ring R trivially implies weak -regularity of R but, in general, weak -regularity of R need not imply weak regularity. For example, consider the ring Z(pn ) of all integers modulo pn , where p is a …xed prime and n is an integer > 1. Z(pn ) is clearly -regular and hence weakly -regular, but it is not weakly regular as is clear by considering the ideal I = pZ(pn ) 6= I 2 . Let T be a simple integral domain which is not a division ring. (For eg., consider the example by J.H. Cozzens of a simple ring T with identity which is a left/right principal ideal domain but not a division ring). This ring T , being 1 simple is clearly both right and left weakly regular, but T is not -regular as it is an integral domain which is not a division ring. Then the ring direct sum T Z(pn ), with n > 1, is weakly -regular, but it is neither weakly regular nor -regular. The ring T Z(pn ) with n > 1, mentioned above, has the property that n I = I n+1 for all right (left) ideals I, but, as already noted, it is neither weakly regular nor -regular. On the other hand, if R is -regular, then every right (left) ideal I of R need not satisfy I n = I n+1 , for some integer n. For example, n consider the ring direct sum S = 1 n=1 Z(p ), where p is a …xed prime. This ring S (which is commutative) is easily seen to be -regular as each element a k P in S lies inside a …nite ring direct sum Z(pn ) for some k, (and, in particular, n=1 n satis…es (aR)k = (aR)k+1 ( (Ra)k = (Ra)k+1 ), but if I = pS = 1 n=1 pZ(p ), k k+1 then I 6= I for all positive integers k. Also it was shown in (R.R.Andruszkiewicz and E.R. Puczylowski, Right fully idempotent rings need not be left fully idempotent, Glasgow Math. J. 37 (1995), 155 - 157). that a right weakly regular ring need not be left weakly regular. We will now see how, for Leavitt path algebras, all these di¤erent properties coincide and are implied by Condition (K) of the graph. 6.3 Theorem: Let E be a row-…nite graph and K be any …eld. Then the following properties are equivalent for the Leavitt path algebra L = LK (E): (i) L is right (left) weakly -regular. (ii) The graph E satis…es Condition (K). (iii) L is right and left weakly regular. (iv) For every right (left) ideal I of L, there is a positive integer n such that I n = I n+1 . (v) For each element a 2 L, there is a positive integer n such that (aL)n = (aL)n+1 ((La)n = (La)n+1 ). Proof : Assume (i). Speci…cally, let L be right weakly -regular. We …rst show that E satis…es Condition (L). Assume, by way of contradiction, that there is a cycle c without exits and based P at a vertex v in E. Since c has no exits, it can be shown that a typical element v ki i i v of vLv simpli…es to a term of the P ti form ki c with ti 2 Z where we have denoted, for notational convenience, c by c 1 (verify this). Then an isomorphism ' : vLv ! K[x; x 1 ] can be de…ned under which v maps to 1, c to x and c to x 1 . Consider the element v c. By hypothesis, there is a positive integer n such that (v c)n 2 (v c)n L(v c)n L. Clearly (v c)n = v(v c)n v is an element of (v c)n vLv(v c)n vLv vLv. Since vLv = K[x; x 1 ] is a commutative ring, we get (v c)n = (v c)2n a for some a 2 vLv. Applying the isomorphism ', we get the equation (1 x)n = (1 x)2n f (x) in K[x; x 1 ] where f (x) is a Laurent polynomial. Since K[x; x 1 ] is an integral domain, canceling (1 x)n on both sides, we get the equation 1 = (1 x)n f (x) = (1 x)g(x) which is impossible by comparing the degrees of terms on both sides. Hence E must satisfy Condition (L). A similar argument works if we assume that L is left weakly -regular. Thus we have shown that, 2 for any graph E, the right/left weakly -regularity of LK (E) implies Condition (L) for E. Now, for every hereditary saturated subset H of E 0 , if I(H) denotes the ideal generated by H, then, being a homomorphic image of L, L=I(H) is left/right weakly -regular. On the other hand, it is known that LK (EnH) = L=I(H). Then the arguments in the preceding paragraph show that the graph EnH satis…es Condition (L). Since this holds for every hereditary saturated set H, we appeal to Proposition 5.5 to conclude that E satis…es Condition (K), thus proving (ii). Assume (ii). Since E is row-…nite, Condition (K) implies that every ideal I of LK (E) is a graded ideal, by Proposition 5.6. Moreover, I is a ring with local units, since I = LK (H E) by Proposition 5.8, where H = I \ E 0 . So given a 2 LK (E); the ideal LK (E)aLK (E), being a graded ideal, contains a local unit u such that a = au = ua. This shows that LK (E) is both left and right weakly regular, thus proving (iii). Now (iii) => (iv) since right/left weak regularity implies I 2 = I for every left//right ideal. It is also clear that (iv) =) (v) and also (iii) =) (i). To complete the proof, we show that (v) =) (ii). But this is almost identical to the proof of (i) =) (ii) where (in order to prove that Condition (L) holds) we replace the equation (v c)n 2 (v c)n R(v c)n R by (v c)n 2 ((v c)R)n+1 (since by supposition ((v c)R)n = ((v c)R)n+1 ) and observe that since again vRv = K[x; x 1 ] is commutative, we have (v c)n 2 ((v c)vRv)n+1 = (v c)n+1 vRv. Proceeding as before, we are lead to a contradiction and to the conclusion that Condition (L) holds. Then follow the arguments in (i) =) (ii) to show that Condition (K) holds. Thus (v) =) (ii). Remark: Theorem 6.3 above actually holds for arbitrary graphs. For a proof of this general result, see the reference [2] below. REFERENCES: [1] G. Abrams and K.M. Rangaswamy, Regularity conditions for arbitrary Leavitt path algebras, Algebra Representation Theory, vol. 13 (2010), 319 334. [2] G. Aranda Pino, D. Martin Barquero, C. Martin Gonzalez and M. Siles Molina, The socle of a Leavitt path algebra, J. Pure and Applied Algebra, vol. 212 (2008), 500 - 509. [3] G. Aranda Pino, K.M. Rangaswamy and M. Siles Molina, Generalized regularity conditions for Leavitt path algebras over arbitrary graphs, Communications in Algebra, vol. 42 (2014), 325 - 331. 3 CIMPA 2015 Research School on LPA and Graph C*-algebras Section - 7: Condition (L) and Zorn Rings Kulumani M. Rangaswamy University of Colorado Colorado Springs July 5 - 7, 2015 In sections 5 and 6, we showed how Condition (K) on the graph E implies a number of algebraic properties for the Leavitt path algebra L := LK (E). In this section, we wish to explore the ring properties implied by Condition (L) on the graph E. 7.1 De…nition: A ring R is said to be a Zorn ring (sometimes also called an I-ring) if given an element a 2 R either a is nilpotent or there is an element b 2 R such that ab is a non-zero idempotent. In the latter case, (bab)a is also a non-zero idempotent. So this de…nition is right/left symmetric. Also, since abab is also an idempotent, we see that b0 = bab can be chosen so that both ab0 and b0 a are idempotents. Zorn rings are among the class of rings with a large supply of idempotents such as von Neumann regular rings. 7.2 If a ring R is von Neumann regular, then it is always a Zorn ring, but the converse is not true as is clear by considering the ring R = Z=pn Z of integers modulo pn where p is a prime and n 2. We state a Lemma which is a special case of the Reduction Theorem 5.3 and will be used in the proof of Theorem 7.4 Lemma 7.3: If E satis…es condition (L), then given any non-zero element a 2 LK (E), there are elements x; y 2 Lk (E) so that xay = v where v is a vertex in E. Theorem 7.4 Let E be an arbitrary graph and K be any …eld. Then the following properties are equivalent: (i) LK (E) is a Zorn ring. (ii) Every non-zero right/left ideal of LK (E) contains a non-zero idempotent. (iii) For every non-zero a 2 LK (E), there is a non-zero b 2 LK (E) such that bab = b. 1 (iv) The graph E satis…es Condition (L). Assume (i). Let I be a non-zero right ideal of LK (E). Since the Jacobson radical J(LK (E)) = 0, I is not a nil ideal. Let a 2 I be an element which is not nilpotent. By hypothesis, there is a b 2 LK (E) such that ab is a non-zero idempotent. Clearly ab 2 I. A similar argument works if I is a left ideal. This proves (ii). Assume (ii). Let 0 6= a 2 LK (E). Since LK (E) is a ring with local units, a 2 aLK (E) and so aLK (E) 6= 0. By hypothesis, aLK (E) contains a non-zero idempotent ax for some x 2 LK (E). Then ax = axaxax. Multiplying by x on the left, we get xax = (xax)a(xax). Clearly xax 6= 0 since xax = 0 implies ax = axax = a(xax) = 0, a contradiction. A similar argument holds for the left ideal LK (E)a. This proves (iii Assume (iii). First observe that, for any idempotent u, the corner uLK (E)u also satis…es condition (iii). To see this, let a = uau 2 uLK (E)u. By hypothesis, there is a b 2 LK (E) so that buaub = b. Multiplying by u on both sides, we get (ubu)uau(ubu) = ubu, thus proving the desired conclusion. We wish to show that the graph E satis…es Condition (L). Suppose, by way of contradiction, there exists a cycle c without exits based at a vertex v in E. An examination of the elements of vLK (E)v shows that every element is a K-linear combination of integral powers of c or c and using this one can de…ne an isomorphism ' : vLK (E)v ! K[x; x 1 ] mapping v to 1, c to x and c to x 1 . We then obtain a contradiction, because while the corner vLK (E)v satis…es condition (iii), K[x; x 1 ] does not, due to the fact that for any a 6= 0 in the integral domain K[x; x 1 ], bab = b for some non-zero b implies that ab = 1, a contradiction. Hence E satis…es Condition (L). This proves (iv). Assume (iv). Let a be a non-zero element of LK (E). By Lemma 7.3, there are elements x; y 2 LK (E) such that xay = v, a vertex in E. Multiplying by v on both sides, we may assume that vx = x and yv = y. Then the element a(yx) is an idempotent since ayxayx = ayvx = ayx. Also ayx 6= 0 since ayx = 0 implies that 0 = xayx = vx = x and this will then imply that v = xay = 0, a contradiction. Hence LK (E) is Zorn ring, thus proving (i). Remark 7.4: The proof of (iv) => (i) shows that if a Leavitt path algebra LK (E) is a Zorn ring, it actually satis…es a stronger condition: For every nonzero element a in LK (E) (even if a is nilpotent), there exists an element b such that ab 6= 0 is an idempotent. This is to be expected since the Jacobson radical J(LK (E)) = 0 and so aLK (E) is not a nil right ideal. Remark 7.5: It is also clear from the proof of Theorem 7.4 that LK (E) is a Zorn ring if and only if every corner eLK (E)e is a Zorn ring, the converse following from the fact that, in a Leavitt path algebra, every element a belongs some corner eLK (E)e where the idempotent e is a sum of suitable …nitely many vertices. 7.6 A homomorphic image of a Zorn ring R need not, in general, be a Zorn ring even if R is a Leavitt path algebra. For example, let R = LK (E) be the 2 Toeplitz algebra (where E is the graph with a single loop c based at a vertex v with an exit e at v with r(e) 6= v). Since E satis…es Condition (L), Theorem 7.3 implies that R is a Zorn ring. But if I is the ideal generated by e, then R=I is not a Zorn ring since R=I = LK (F ) where F = Enfeg is the graph with a single loop c based at a vertex v and obviously does not satisfy Condition (L). The next proposition describes when every homomorphic image of LK (E) is a Zorn ring. Proposition 7.7: Let E be an arbitrary graph and K be any …eld. Then every homomorphic image of the Leavitt path algebra LK (E) is a Zorn ring if and only if LK (E) is a weakly regular ring. Suppose every homomorphic image of LK (E) is a Zorn ring. By Theorem 7.3, E satis…es Condition (L). Now, for any hereditary saturated subset H of E 0 , LK (EnH) = LK (E)=I(H). Hence LK (EnH) is a Zorn ring and this implies, by Theorem 7.3, that the graph EnH satis…es Condition (L). Since this happens for every hereditary saturated set H, we appeal to Proposition 5.5 to conclude that E satis…es Condition (K). Then, by Theorem 6.3, LK (E) is weakly regular. Conversely, suppose LK (E) is weakly regular. Now the weak regularity of LK (E) implies, by Theorem 6.3, that the graph E satis…es Condition (K). Since Condition (K) implies Condition (L), we conclude by Theorem 1 that LK (E) is a Zorn ring. Since a homomorphic image of a weakly regular ring is again weakly regular, we conclude that every homomorphic image of LK (E) is a Zorn ring. Remark 7.8: The above Proposition 7.7 does not hold for arbitrary rings. For instance, let R = Z=pn Z, the ring of integers modulo pn where p is a prime and n is an integer 2. Then R pR :::::: pn 1 R pn R = f0g are all i the ideals of R and, for each i with 1 i n, R=p R is a Zorn ring since each element in it is either nilpotent or a unit. But R is not weakly regular since for the ideal I = pR, I 2 6= I. REFERENCES: J. Levitzky, On the structure of algebraic algebras and related rings, Trans. Amer. Math. Soc. vol. 74 (1953), 384 - 409. K. Nicholson, I-rings, Trans. Amer. Math. Soc. vol 207 (1975), 361 - 373. K.M. Rangaswamy, Leavitt path algebras which are Zorn rings, Contemporary Math., vol. 609 (2014), 277 - 283. 3