THE ROLE OF INVOLUTION IN GRAPH ALGEBRAS

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1 THE ROLE OF INVOLUTION IN GRAPH ALGEBRAS LIA VAŠ Abstract. Both Leavitt path and graph C -algebras are equipped with involution. After a brief introduction to involutive rings, we study the impact of the presence of involution on some algebraic properties of these two classes of algebras. Whenever possible, we shall point out the similarities and differences between Leavitt path and graph C -algebras. We shall also present a class of open conjectures related to the presence of the involution in these algebras. Introduction and Overview Let us start by an overview of your previous lectures, motivation for our study of involution and an outline of the following lectures. Question 1. What have we learned from previous lecture? What is a Leavitt path algebra? A graph C -algebra? Answer. E= directed graph (E 0, E 1, s, r), K= field (or commutative ring). The Leavitt path algebra L K (E) is a free K-algebra with basis consisting of vertices, edges, and ghost edges {e e E 1 } with r(e ) = s(e) and s(e ) = r(e) of E such that (V) vv = v and vw = 0 if v w, (E1) e = s(e)e = er(e), (E2) e = r(e)e = e s(e), (CK1) e e = r(e), and e f = 0 if e f for all e, f E 1, (CK2) v = ee for all e E 1 with v = s(e) and all v E 0 with 0 < s 1 (v) < (v regular) for all vertices v and w and all edges e. Recall that a universal C -algebra is presented in terms of generators and relations such that the generators are some bounded operators on a Hilbert space, and the relations imply a uniform bound on the norm of each generator. The C -algebra C (E) is universal C -algebra generated by {p v v E 0 } and {s e e E 1 } subject to (V) p 2 v = p v = p v and p v p w = 0 for v 0 (p v are orthogonal projections), (PI) s e s es e = s e (s e are partial isometries) (CK1) Same as (CK1) above. (CK2) Same as (CK2) above. (CK3) s e s e p s(e) (needed for infinite emitters). Date: July Mathematics Subject Classification. 16W99,16W10, 16S99. Key words and phrases. -ring, Leavitt path algebra, graph C -algebra, -regular, directly finite, Isomorphism Conjecture. 1

2 2 LIA VAŠ Convince yourself that these relations imply (E1). Indeed, er(e) = ee e = e and s(e)e = ee e + ff e = ee e + 0 = e if s(e) is regular. If s(e) is singular, note that relation s e s e p s(e) implies that ee = s(e)ee. Thus s(e)e = s(e)ee e = ee e = e. (E2) follows from this by starring the relations in (E1). Question 2. Why are these algebras relevant? Answer. Graph C -algebras are examples of many different classes of C -algebras having a more convenient and manageable representation than some other algebras in these classes. As a consequence, some invariants (e.g. K-theory) can be computed easier than for many non-graph C -algebras. Then Leavitt path algebras emerged as algebraic simplification of graph C -algebras so that we have the following. Operator Theory world Algebra world Graph C -algebras Leavitt Path Algebras Von Neumann algebras and their algebraic analogue, Baer -rings are in similar relation: one inhabits operator theory world and the other is its algebraic counterpart. So, the following quote from the introduction to Berberian s 1972 book [8], holds for graph C -algebras as well. Von Neumann algebras are blessed with an excess of structure algebraic, geometric, topological so much, that one can easily obscure, through proof by overkill, what makes a particular theorem work. The algebraic counterpart is needed because of the following. If all the functional analysis is stripped away... what remains should (be) completely accessible through algebraic avenues. With this in mind, Leavitt path algebras can also be seen as an algebraic avenue paving the road to understanding its more complex operator theory sibling. Question 3. What is the goal of these lectures? Answer. To study one of the umbrella concepts common in both worlds the involution. Both Leavitt path algebras and C - algebras have operation defined on their elements. So, we are going to look at rings with involution first and then study the presence of involution on Leavitt path algebras and graph C -algebras. The consideration of this particular umbrella concept should help each of you find your own umbrella.

3 THE ROLE OF INVOLUTION IN GRAPH ALGEBRAS 3 Question 4. What are some open questions or directions of research? Answer. Characterization of ring theoretic properties by graph theory, description of ideals, K-theory, and various generalizations are just some of the directions of research. We will be particularly interested in a group of conjectures known as the Isomorphism Conjectures because they are tied to the involution. Outline. Below is an outline of the following five lectures. (1) The driving force. After introducing some basic concepts of rings with involution, we present some motivating examples for study of -rings. In particular, we look at examples related to Leavitt path and graph C -algebras. (2) We exhibit and study some ring-theoretic properties of involutive rings. (3) We explore the presence of involution on Leavitt path and graph C -algebras. In particular, we characterize when a Leavitt path algebra is positive definite and when it is -regular by properties of the underlying graph and the underlying field. (4) We characterize when a Leavitt path algebra is directly finite. (5) We consider some open conjectures related to involution referred to as the Isomorphism Conjectures. Lecture 1. The Driving Force. Involution. Let R be an (associative) ring with or without the identity. Definition 1. R is a -ring (or ring with involution) if there is an operation : R R which is (1) additive (x+y) = x +y (2) anti-multiplicative (xy) = y x (3) involutive (x ) = x for all x, y R. Let us point out some direct corollaries. Claim 2. If R is a -ring, then the following hold. (1) 0 = 0. (2) ( x) = x for any x R. (3) If the ring has the identity 1, then 1 = 1. Proof. (1) This follows from additivity since for any x we have that x + 0 = (x + 0) = x which implies that 0 = 0. (2) This follows from (1) and additivity since x + ( x) = 0 x + ( x) = 0 ( x) = x. (3) This follows from anti-multiplicativity since for any x, x 1 = (1x) = x and so 1 = 1.

4 4 LIA VAŠ The maps of -rings of interest are not just ring homomorphisms, but ring homomorphisms f which agree with, i.e. such that f(x ) = f(x) for every element x. Such maps are called -homomorphisms. Definition 3. If R is also an algebra over a field (or a commutative ring) K with involution, then R is an -algebra if (ax) = a x for a K, x R. Let us try to think of some examples of involutive rings and algebras. Here are some of interest. Example 4. (1) Any commutative ring is a -ring for being the identity map. In particular Z, all fields, polynomial rings over fields, etc are -rings. (2) One of the fields has a special role for C -algebras the field of complex numbers C. It has an involution of special importance the complex-conjugate involution a + ib = a ib. (3) My favorite non-commutative ring is the ring of matrices. If R is a ring, M n (R) denotes the ring of n n matrices with entries from R. Some special cases of interest are the following. (a) The transpose given by (a ij ) t = (a ji ) defines an involution on M n (R). If R is commutative, this makes M n (R) a -algebra with respect to the identity involution on R. (b) If R = C with complex-conjugate involution, the map on M n (C) given by transposing and complex-conjugating each entry : (a ij ) (a ji ) makes M n (C) a -algebra over C. This is the only involution considered in C -algebra world. This example generalizes to the following. (c) Most important matrix example. If R = K any field and k k is an involution on K, then the map : (a ij ) (a ji ) makes M n (K) a -algebra over K. It turns out that this is not just a random example of involution on M n (K) but the involution which determines all the other i.e. if is any other involution, then A and A are similar matrices for any matrix A. To see this, note that the algebra M n (K) is simple and its center is isomorphic to K. So, by the Skolem Noether Theorem, every automorphism on it is inner. This implies the next claim. Proposition 5. If is any involution on M n (K) making it a -algebra for involution on K, then there is an invertible matrix U such that. A = UA U 1. Proof. The composition of involutions and is an automorphism of M n (K). By Skolem Noether Theorem, there is an invertible matrix V such that the map is the same as A V 1 AV. Thus A = (V 1 AV ) = V A (V 1 ) for any matrix A. Take U = V (which is also invertible with inverse (V 1 ) ) and obtain that A = UA U 1. Let us also point out a preferred K-basis of M n (K) the matrix units e ij, i, j = 1,..., n. Here e ij is the matrix with all entries zero except ij-th which is 1. The basis elements multiply as follows e ij e kl = δ jk e il where δ jk = 0 if j k and δjj = 1.

5 THE ROLE OF INVOLUTION IN GRAPH ALGEBRAS 5 (d) The claim analogous to Proposition 5, holds for the transpose and any involution making M n (K) a -algebra for the identity involution on K. Note also that no such relation exits for the transpose t and complex-conjugate transpose involution on M n (C). (4) Most relevant example for us. Let K be a field with involution and E be a directed graph. The operation, defined on the elements of K and the edges of E, extends to paths by (e 1... e n ) = e n... e 1 and vertices by v = v. It extends to the entire algebra L K (E) since every element of L K (E) can be written as apq where the sum is finite, a K and p, q are paths. So, we can define of such element as follows. ( ) apq = aqp. Let us explore some specific Leavitt path algebras in the next several examples. (5) If E is v e w, then L K (E) is -isomorphic to M 2 (K) by v e 11 w e 22 e e 12 e. e 21 You can show that this map induces a -isomorphism by using the universal property of Leavitt path algebras. It states that if R is a K-algebra which contains a set {a v, b e, c e v E 0, e E 1 } such that a v, b e, c e satisfy axioms (V), (E1), (E2), (CK1), and (CK2) (such set is called a Leavitt E-family) then there is a unique K-algebra homomorphism f : L K (E) R such that f(v) = a v, f(e) = b e, and f(e ) = c e for all v E 0 and e E 1. It is direct to show that if a v = a v and b e = c e, then this homomorphism is a -homomorphism. Thus, to show the above map induces a -isomorphism L K (E) = M 2 (K), it is sufficient to check that the matrix units are a Leavitt E-family. Similarly, if E is v 1 e 1 v 2 e 2 v n 1 e n 1 vn Then L K (E) is -isomorphic to M n (K) via v i e ii e i e i i+1. Moreover, if K = C, L C (E) and C (E) are both -isomorphic to M n (C). (6) The algebra of Laurent polynomials over a field K is K[x, x 1 ] = { n i= m a ix i m, n nonnegative integers, a i K}. If is an involution on K, the following map is an involution on this algebra. ( n ) n a i x i = a i x i i= m If E is v e, then L K (E) is -isomorphic to K[x, x 1 ] by v 1 e x (thus e x 1 ). (7) Let us turn to an important example of a C -algebras. In fact, this is the universal example of a commutative C -algebra. Let C(X) be the algebra of continuous functions X C on a compact Hausdorff space X. Addition and multiplication are by coordinates. The involution is f (x) = f(x) where is the complex-conjugate involution. This is an universal example of a commutative C -algebra since every abelian C -algebra is -isomorphic to C(X) for some X. For example, if E = v e, C (E) is the universal C -algebra generated by single unitary operator, i.e. one generator u with relations uu = u u = 1. In this case, C (E) is -isomorphic to the algebra of continuous maps on the unit sphere C(S 1 ). The algebra i= m

6 6 LIA VAŠ L C (E) embeds in C(S 1 ) by mapping the vertex 1 to the map t 1 and the edge e to the map t e it (thus e is t e it ). The algebra C(S 1 ) is also the closure of C[x, x 1 ] in the norm topology. e (8) Combining examples (5) and (6), let us consider E to be the graph v w. Then v e 11 L K (E) = M 2 (K[x, x 1 e e ]) by 12 x and f e 21 x C (E) is M 2 (C(S1 )). w e 22 e 11 v e 11 x ef e Equivalently, one can also do 12 f e 12 x e e 21 f e 21 x fef e 22 w e 22 x fe e 11 v (9) If E is v e w f, then L K (E) = M 2 (K[x, x 1 e ]) are -isomorphic by 12 e e 22 w xe 11 f. (10) Analogously, if E is any finite graph in which no cycle has an exit (i.e. an edge leading out of any cycle), then every path leads either to a sink or a cycle. In this case, we have that L K (E) is isomorphic to a direct sum of matrices over K or K[x, x 1 ]. Say that there are n sinks v 1,..., v n and there are k i paths p ij, j = 1,..., k i, i = 1,..., n, ending in v i and m cycles c 1,..., c m and l i paths q ij, j = 1,..., l i, i = 1,..., m, ending in a fixed arbitrary vertex of c i, such that q ij does not contain c i. Then L K (E) n m = M ki (K) M li (K[x, x 1 ]) i=1 i=1 by p ijp il e jl M ki (K) j, l = 1,..., k i, i = 1,..., n, q ij c k i q il x k e jl M li (K[x, x 1 ]) j, l = 1,..., l i, i = 1,..., m, k = 0, 1,.... Practice Problem 1. Identify Leavitt path algebras of the following three graphs as matrix algebras over K and K[x, x 1 ] or their direct sums. f (11) Not all Leavitt path algebras have a matricial representation. One of the main examples are the (classical) Leavitt algebras. Consider E to be the graph e f. We shall refer to this graph as the rose with two petals. Recall that the Leavitt algebra L(1, 2) is the quotient of the free K-algebra on four generators x 1, x 2 and y 1, y 2 subject to relations y i x j = δ ij 1 and x i y i = 1. L K (E) is -isomorphic to L(1, 2) by e x 1 e y 1 f x 2 f The graph C y 2. -algebra of this graph is the Cuntz algebra O 2. This is the universal C -algebra generated by two generators s 1 and s 2 subject to relations s i s j = δ ij 1 for all i, j = 1, 2 and s 1 s 1 + s 2 s 2 = 1.

7 THE ROLE OF INVOLUTION IN GRAPH ALGEBRAS 7 e 2 The graph e 1 v w f 2 also has its Leavitt path algebra -isomorphic to f 1 L(1, 2) and its C -algebra to O 2. The following induces a -isomorphism between the v ee e 1 ee w ff Leavitt path algebras of the 2-petal rose and of this graph. e 2 ef f 1 fe f 2 ff e 3 e 2 (12) If E is a rose with n petals e 1, L K (E) is the Leavitt algebra L(1, n) and C (E) e n is the Cuntz algebra O n, the universal C -algebra generated by n generators s 1,..., s n subject to relations s i s j = δ ij 1 for all i, j and i s is i = 1. Combining this with previous examples, L K (E) is -isomorphic to M m (L(1, n)) for the graph E below. v 1 f 1 v 2 e 3 f 2 v 3 v m 1 f m 1 vm e 1 e 2 (13) If E is a finite graph in which every path leads either to (1) a sink, (2) a cycle without an exit or (3) a rose with two or more petals, then E is called a polycephaly graph. The algebra L K (E) is isomorphic to a direct sum of the matrix algebras over (1) K, (2) K[x, x 1 ] and (3) L(1, n). (14) Toeplitz algebra is another example of a Leavitt path algebra without a matricial representation. Consider the algebra K[x, y yx = 1], i.e. the free K-algebra over K with generators x, y subject to relation yx = 1. This algebra is called the Toeplitz algebra over K. If K is an involutive field, we define the involution on this algebra by x = y y = x. If E is the graph v v 1 xy w xy map e xxy f x xxy e w f e n, then L K (E) is -isomorphic to K[x, y yx = 1] by the. In this case we have that e + f x e + f y. v 1 xx w xx Equivalently, we can define e xxx and obtain a -isomorphism. The algebra C (E) is the universal C -algebra with one generator x and one relation x x = 1. f x xxx (15) We turn now to one of the most important examples of C -algebras which, in a way, is an origin of all other involutive rings. Let B(H) be the algebra of bounded operators on a Hilbert space H. Recall that a Hilbert space is a complex vectors space with an inner product (a C-valued map on H H which is linear, adjoint-symmetric, positive, and faithful) which is complete in the topology given by the norm x = x, x 1/2. For a continuous operator T : H H, a finite value of sup x H T (x) x defines the norm

8 8 LIA VAŠ T. This norm makes B(H) into a C -algebra. The addition of this algebra is by coordinates, the multiplication is the operator composition, and the involution is given by the unique map such that T (x), y = x, T (y) for any operator T and any x, y H. If H is finitely dimensional of dimension n, then B(H) = M n (C) and the involution is given by the complex-conjugate transpose from example (3)(b). We show how our non-matricial algebras examples, Cuntz C -algebras and Toeplitz C -algebra can be represented by operators on a Hilbert space. Let ω denote the set of non-negative integers and l 2 (ω) denote the Hilbert space of all sequences of complex numbers a i, i ω such that i a i 2 is finite. Both the Cuntz algebras and the Toeplitz algebra can be represented as operators on this space. For the Cuntz algebra O 2 one can take the generators s 1 and s 2 to be the following operators: s 1 : (a 0, a 1,...) (0, a 0, 0, a 1,...) and s 2 : (a 0, a 1,...) (a 0, 0, a 1, 0...). Then s 1 : (a 0, a 1,...) (a 1, a 3, a 5,...) and s 2 : (a 0, a 1,...) (a 0, a 2, a 4,...) so that s i s i = 1, i = 1, 2 and s 1 s 1+s 2 s 2 = 1. When considering the graph e 1 v e 2 f 1 w f 2,. v s 1 s 1 e 1 s 1 s 1 we have that the map with f 2 s 2 s 2 w s 2 s 2 e 2 s 1 s 2 f 1 s 2 s 1 induces a -isomorphism between C (E) and the universal C -algebra with generators s 1, s 2 and relations s i s i = 1, i = 1, 2 and s 1 s 1 + s 2 s 2 = 1. For Toeplitz algebra, one can take the generator x to be the unilateral shift x : (a 0, a 1,...) (0, a 0, a 1,...). Then x is the operator (a 0, a 1,...) (a 1, a 2,...) so that x x = 1. If π 0 denotes the operator (a 0, a 1,...) (a 0, 0, 0,...), then xx = 1 π 0. So in the graph representation v π 0 as in example (14), we have the correspondences w 1 π 0 e x(1 π 0 ) f xπ 0 (16) Let us conclude our example list with an example of a non-row-finite graph (i.e. there are infinite emitters) which Leavitt path algebra is still somewhat related to matrices. If E is the graph v w with infinitely countably many edges e 1, e 2,... from v to w, then L K (E) is -isomorphic to M (K) K1 where M (K) stands for matrices with infinitely countably many rows and columns but with finitely many elements which are nonzero, and 1 stands for the diagonal matrix with infinitely countably many rows and columns with 1 in all the diagonal entries. The following map induces a -isomorphism. w e 11 e i e i+1 1 Note that in this representation e i e i e i+1 i+1. v 1 e 11

9 THE ROLE OF INVOLUTION IN GRAPH ALGEBRAS 9 Positivity, involutions with some nice properties. Having involution, one can define when an element of a -ring is positive by generalizing the next example. Example 6. With complex-conjugate involution, for any z j = a j + ib j, j = 1,..., n n n z j z j = a 2 j + b 2 j 0. j=1 j=1 Definition 7. An element of a -ring R is positive if it is a finite sum of elements of the form xx for x R. The notation x > 0 usually denotes positive complex numbers. We abuse this notation slightly and denote the fact that x is positive element by x 0. If x is positive and nonzero, we write x > 0. One may argue that we should refer to positive elements as nonnegative instead. Although this may be a valid point, we continue to use the terminology which is well established in operator theory and keep referring to such elements as positive. Definition 8. An involution on R is (1) proper if xx = 0 x = 0 for any x R. (2) n-proper if n x i x i = 0 x i = 0 for each i = 1,..., n i=1 for all x 1,..., x n R. (3) positive definite if it is n-proper for any n. Example 9. (1) The identity involution on R is positive definite. (2) The identity involution on C is proper but not 2-proper. (3) The complex-conjugate involution on C is positive definite. (4) Involution in any C -algebra is proper. This is because the following. xx = 0 0 = xx = x 2 x = 0 x = 0. (5) Since L C (E) embeds in C (E) with proper involution for any E, the involution in L C (E) is proper as well by the previous example. Questions for the next lecture. (1) Is the involution of a Leavitt path algebra L K (E) proper for any K? (2) When is the involution of a Leavitt path algebra positive definite? We will answer these questions next time. You will be able to make most of the arguments needed for the answers on your own, if you go over the following practice problems. Practice Problem 2. Let R be a -ring. Show that the -transpose involution ((a ij ) = (a ji)) of M n (R) is proper iff the involution of R is n-proper. Deduce that every involution of algebras M n (R), n = 1, 2,... is proper iff the involution of R is positive definite. Practice Problem 3. Assuming that the involution on L K (E) is proper for every E, show that the involution on L K (E) is positive definite for every E. You can assume the following: adding a line of length n 1 ending at every vertex of E produces a graph M n E such that M n (L K (E)) = L K (M n E). Practice Problem 4. Recall that a ring is regular if for every a, there is b with a = aba. Show that R is regular if and only if every principal right (left) ideal is generated by an idempotent.

10 10 LIA VAŠ Lectures 2 and 3. Some -Ring Theoretic Properties of Leavitt Path Algebras Let us start with the solutions to the first three practice problems. Practice Problem 1 Solution. Identify Leavitt path algebras of the following three graphs as matrix algebras over K and K[x, x 1 ] or their direct sums. v1 f e c v2 e c v f e 1 v 1 f 1 f 2 e 2 v 2 g e 3 (1) The first graph is a no-exit graph with one sink and one cycle. There are two paths, v 1 and e, ending in v 1 and two paths, v 2 and f, ending in v 2. Thus, L K (E) is -isomorphic to M 2 (K) M 2 (K[x, x 1 ]). (2) The second graph is a no-exit graph with no sinks and a single cycle. There are three paths v, e and f, ending in v so L K (E) is -isomorphic to M 3 (K[x, x 1 ]). (3) The third graph is a no-exit graph with two sinks v 1 and v 2 and one cycle cd. There are four paths v 1, f 2, f 1 f 2, and e 1 f 2 ending in v 1, and five paths, v 2, g, e 2 g, e 1 e 2 g, and f 1 e 2 g ending in v 2. Choosing w for a base of the cycle cd, there are six paths w, e 3, e 2 e 3, e 1 e 2 e 3, f 1 e 2 e 3, and d ending in w. Choosing w instead of w for a base of cd would lead to the same number. Thus, L K (E) is -isomorphic to M 4 (K) M 5 (K) M 6 (K[x, x 1 ]). Practice Problem 2 Solution. Let R be a -ring. Show that the -transpose involution ((a ij ) = (a ji)) of M n (R) is proper iff the involution of R is n-proper. Deduce that every involution of algebras M n (R), n = 1, 2,... is proper iff the involution of R is positive definite. Proof. Assume that the involution is n-proper in R. Suppose A A = 0 for some matrix A = (a ij ) M n (R). Then the diagonal entries of the product A A are zero and so n j=1 a ija ij = 0 for every i = 1,..., n. Since is n-proper, a ij = 0 for all i, j. Hence A = 0. Conversely, suppose n i=1 a ia i = 0 for a i R. Consider A to be the matrix in M n (R) that has the elements a 1,..., a n in its first row and zeros in the rest of its rows. Then AA = 0. Since M n (R) is proper, A = 0. So a i = 0 for every i. Practice Problem 3 Solution. If the involution on L K (E) is proper for every E, then it is positive definite for every E. Proof. By assumptions, of L K (M n E) is proper. Thus, of M n (L K (E)) is proper and so of L K (E) is n-proper by Practice Problem 2. Since this holds for every n, we have that of L K (E) is positive definite. This gives us one of the implications in the following theorem. Theorem 10. ([6, Proposition 2.4]) Let K a field with involution. The following conditions are equivalent. w c d w

11 THE ROLE OF INVOLUTION IN GRAPH ALGEBRAS 11 (1) The involution on K is positive definite. (2) The involution on L K (E) is positive definite for every graph E. (3) The involution on L K (E) is positive definite for some graph E. Thus, if E is an arbitrary graph, of L K (E) is positive definite if and only if of K is positive definite. Proof. Showing that (1) implies that of L K (E) is proper for every E is long and we won t be going over that. But, assuming that, we have that of L K (E) is positive definite by Practice Problem 3. So, we have (2). (2) implies (3) is a tautology. (3) (1). Suppose that n i=1 k ik i = 0 for k i K. Let E be a graph such that the involution on L K (E) is positive definite. Let v E 0. Since v = v and v 2 = v, 0 = ( n i=1 k ik i )v = n i=1 (k iv)(k i v) and therefore k i v = 0 for all i by hypothesis. But E 0 is a set of linearly independent elements in L K (E), so that k i = 0 for all i, as needed. Idempotents. Idempotent elements of any ring and the following equivalence relation between them are rather important. e a f iff er = fr You should convince yourself that a is indeed an equivalence. The following list of equivalent conditions may come useful. Practice Problem 5. Part 1. If e and f are idempotents of R, show that the following are equivalent (1) e a f (recall that we defined this to mean that er = fr). (2) e = xy and f = yx for some x, y R. (3) e = xy and f = yx for some x, y R with ex = x, ye = y, xf = x and fy = y. (4) Re = Rf. Part 2 Relating the equivalence relations with cancellation properties. Show that for a unital ring R, idempotent e R and a R-module P, e a 1 e = 1 if and only if R P = R P = 0. Definition 11. If e a f for idempotents e, f of a ring R, then e and f are said to be algebraically equivalent. The idempotent matrices of M n (R) and relation a on them have special significance because of the correspondence from the following claim. This correspondence is relevant in order to better understand the Grothendieck group K 0 (R). Claim 12. Every idempotent matrix E M n (R) determines a direct summand of R n. Conversely, if P is a direct summand of R n then there is an idempotent matrix E M n (R) such that E(R n ) is isomorphic to P. As a consequence, the set of isomorphism classes of finitely generated projective modules is in bijective correspondence with the set of algebraic equivalence classes on the matrix algebras M n (R), n = 1, 2,.... Proof. If E M n (R) is an idempotent, then E(R n ) is a direct summand of R n since R n = E(R n ) (1 E)(R n ). Conversely, if Q is the complement of a direct summand P in R n, f an isomorphism R n P Q, π P the projection P Q P and i P the inclusion P P Q, then the map E = f 1 i P π P f

12 12 LIA VAŠ maps R n into R n. So E is an element of M n (R) and its image f 1 (P 0) is isomorphic to P. The matrix E is an idempotent since E 2 = f 1 i P π P ff 1 i P π P f = f 1 i P π P i P π P f = f 1 i P 1 P π P f = f 1 i P π P f = E. To prove the second part of the claim, note that if P and T are isomorphic direct summands of R n and h is the isomorphism P = T, then the complement Q of P can be used as the complement of T as well. If h 1 Q denotes the map P Q T Q given by (p, q) (h(p), q), then it is direct to show that i T π T (h 1 Q ) = (h 1 Q )i P π P. Thus, if f is an isomorphism R n P Q and g an isomorphism R n T Q, we have that the idempotent matrices E = f 1 i P π P f and F = g 1 i T π T g are conjugated by the invertible matrix U = g 1 (h 1 Q )f (i.e. F = UEU 1 ). Thus E a F since E = (EU 1 )U and F = U(EU 1 ). Conversely, if E a F in M n (R) and if X, Y M n (R) are such that XY = E, Y X = F and X = EX = XF, Y = Y E = F Y (such ( elements exists ) by Practice Problem 5), then 1 E X it is direct to check that the matrix U = has square equal to the identity Y 1 F ( ) ( ) ( ) E in M 2n (R) and so U 1 = U. Moreover U U =. Since V = F 1 0 ( ) ( ) ( ) ( ) 0 0 F 0 E 0 F 0 conjugates and, we have that E 0 = and F 0 = 0 F are conjugated and so (E 0)(R 2n ) = (F 0)(R 2n ). But then we have that. E(R n ) = (E 0)(R 2n ) = (F 0)(R 2n ) = F (R n ). The next step of our study of idempotents involves right ideals and this requires us to make a short digression into rings with local units. Definition 13. A ring R is said to have local units if for every finite set x 1,..., x n R there is an idempotent u such that x i u = ux i = x i for all i = 1,..., n. Even though some algebras of our interest may not be unital, they all have local units. Let us recall the related facts in the next example. Example 14. If the number of vertices of the graph E is finite and v 1,..., v n are all the vertices, then L K (E) is unital algebra and 1 = v v n. If the number of vertices of the graph E is not finite, then L K (E) is locally unital algebra. Indeed, if x 1,..., x n L K (E) and v 1,..., v m are the sources of the paths p ij, q ij in the representation of x i and j a ijp ij q ij, i = 1,..., n, then u = v v m is an idempotent such that x i u = ux i = x i for all i = 1,..., n. Locally unital rings retain many properties of unital rings. For example, in a non-unital ring R an element x may not be in the right ideal xr. However, this does not happen in a locally unital ring since x = xu = ux for some idempotent u and so x = xu xr. In the following, let R stand for a ring with local units. The idempotents e and f of a ring R can be compared as follows. e f iff e = fe = ef (iff er fr and Re Rf). You should convince yourself that is indeed an order. The equivalence of conditions e = fe and er fr follows from the following useful observation. Claim 15. Useful Observation. If e is an idempotent of a ring R, then x er iff x = ex. Proof. x er x = ey for some y. But then ex = eey = ey = x. So, x = ex. The converse clearly holds.

13 THE ROLE OF INVOLUTION IN GRAPH ALGEBRAS 13 The condition e = fe is equivalent with e fr by the previous claim. This last condition is equivalent with er fr since R has local units. Projections. Although the conditions er = fr and Re = Rf together imply that e = f, it is not the case that the single condition er = fr implies e = f as it may be desirable. Luckily, there is a fix if the involution is present. Namely, in -ring, you can work with projections instead of the idempotents and then the downside does not happen. Definition 16. An element p of a -ring is a projection iff it is (1) idempotent p 2 = p and (2) self-adjoint p = p. The order is nicer for projections than for idempotents since we have the following. Claim 17. If R is a -ring, the following conditions are equivalent. 1. p q; 2. p = qp; 3. pr qr; 4. Rp Rq. As a consequence, p = q iff pr = qr. Proof. 1. trivially implies 2. and 2. trivially implies 3. Starring 3. produces 4. Lastly, 4. implies p = pq and starring this relation produces p = qp so 1. holds. The equivalence relation a can also be strengthened in case of projections as follows. Definition 18. Define a relation on the set of projections of a -ring R by p q iff x x = p and xx = q for some x R. In this case p and q are said to be -equivalent and the relation p q is said to be implemented by x. Claim 19. If p q is implemented by x, then x can be chosen so that x = xp and qx = x. Proof. If x x = p and xx = q, let y = xp = xx x = qx and note that y y = p and yy = q. Element x from the previous claim is called a partial isometry. 1 Claim 20. (1) An element x of a -ring is a partial isometry iff x = xx x. (2) If x is a partial isometry, then xx and x x are -equivalent projections with x implementing the -equivalence. Proof. (1) ( ) x = xp = xx x. ( ) If x = xx x, put p = x x and q = xx and check that p and q are projections with x = xp and qx = x. (2) If x is a partial isometry, then (xx )(xx ) = (xx x)x = xx and (xx ) = (x ) x = xx so xx is a projection. Similarly, x x is also a projection. The rest trivially follows. Example 21. The vertices of a Leavitt path algebra L K (E) are projections and the paths are partial isometries. Indeed, the vertices are idempotents by axiom (V) and they are selfadjoint by definition of. If x is a path, then xx x = xr(x) = x so paths are partial isometries. Claim 22. For every path x of L K (E), r(x) xx and the equivalence is implemented by x. Proof. Since r(x) = x x, the claim follows directly from part (2) of Claim Recall that you have encountered both terms partial isometry and projection in definition of a graph C -algebra.

14 14 LIA VAŠ This claim can be useful for understanding the monoid of the isomorphism classes of finitely generated projectives V ((L K (E)) which paves the way to K 0 (L K (E)). If E is row-finite, [5, Theorem 3.5] states that the monoid V ((L K (E)) is isomorphic to the monoid M E generated by the elements a v, v E 0 regular, subject to relations a v = e s 1 (v) a r(e). With the knowledge we accumulated and, in particular, Claim 22, we can understand the arguments for having a monoid homomorphism M E V (L K (E)) given by a v [v] for all vertices v which emit edges where [v] is the a -equivalence class of v. Without going into the details of the rest of the proof, let us go over just this following argument. Proposition 23. If v is a regular vertex, then v and e s 1 (v) r(e) determine the same isomorphism class in V ((L K (E)). Before the proof, we just need one preparatory claim. Let diag(a 1,..., a n ) for a 1..., a n in a ring R denote the diagonal matrix in M n (R) with a 1..., a n on the diagonal. Claim 24. If p 1,..., p n are projections in a -ring R with p i p i = 0 for i j (i.e. the projections are orthogonal), then diag( p i, 0,..., 0) diag(p 1, p 2,..., p n ) in M n (R). Proof. Convince yourself that the matrix with p 1, p 2,... p n in the first row and zeros in all other rows implements the -equivalence. Let us prove the proposition now. Proof. Denote the edges that v emits by e i, i = 1,..., n. We have that diag(v, 0,..., 0) = diag( e i e i, 0,..., 0) (by CK2 axiom) diag(e 1 e 1, e 2 e 2,..., e n e n) (by Claim 24) diag(r(e 1 ), r(e 2 ),..., r(e n )) (by Claim 22) diag( r(e i ), 0,..., 0). (by Claim 24) Since the relation implies a, we have that diag(v, 0,..., 0) a diag( r(e i ), 0,..., 0) and finitely generated projective modules determined by these two matrices are isomorphic by the proof of 12. -regular rings. Several ring-theoretic properties feature the term idempotent in their definitions. For example, Baer, Rickart, and clean rings. We have seen that projections are better behaved than idempotents. Thus, if the word idempotent is replaced by projection, in definitions of such properties for a -ring, one obtains a better behaved property. After recalling Practice Problem 4 from the last lecture, we illustrate this phenomenon by comparing definitions of regular and -regular rings. Practice Problem 4 Solution. A ring R is regular if and only if every principal right (left) ideal is generated by an idempotent. Proof. If R is regular, a R and a = aba, then e = ab is an idempotent with ar = er. Indeed, br R implies abr ar and the converse follows since a = ea er and so ar er. Conversely, if ar = er, then a = ea and e = ar for some r R. Then a = ea = ara. Definition 25. A ring R is regular if and only if every principal right (left) ideal is generated by an idempotent. A -ring R is -regular if and only if every principal right (left) ideal is generated by an idempotent a projection.

15 THE ROLE OF INVOLUTION IN GRAPH ALGEBRAS 15 The following practice problems relate regular and -regular properties. These practice problems will help us figure out when a Leavitt path algebra is -regular. Practice Problem 6. Show that if a -ring R is -regular that then it is regular and is proper. Practice Problem 7. Show the converse of the previous claim. Note that it is sufficient to show that for every idempotent e, there is a projection p such that er = pr. The following three claims may be helpful. (1) If is proper, then ann r (x) = ann r (x x) (recall that ann r (X) = {y R xy = 0 for all x X}). (2) If e is an idempotent, then ann r (e) = (1 e)r. (3) If e is an idempotent in a (locally) unital ring R, then Re = ann l (ann r (e)). (4) If e is an idempotent and R is regular (locally) unital -ring with proper, then Re = Re e. Using the above four claims, try to show that for every idempotent e, there is a projection p such that Re = Rp. Staring the previous relation will give you er = pr. Now, let us recall the Abrams Rangaswami characterization of regular Leavitt path algebras. Theorem 26. ([2, Theorem 1]) L K (E) is regular if and only if E is acyclic. This theorem also holds if regular is replaced by unit-regular in the case when E is finite. We present now the following characterization of the -regular Leavitt path algebras. Let µ(v) denotes the cardinality of the set of all the paths ending in v. Theorem 27. ([6, Theorem 3.3]) Let E be an arbitrary graph, K be a field with involution and let σ = sup{µ(v) : v E 0 } in case the supremum is finite or σ = ω otherwise. The following conditions are equivalent. (1) L K (E) is -regular (equivalently, regular and is proper). (2) E is acyclic and of K is n-proper for every finite n σ. The equivalence stated in condition (1) follows from Practice Problems 6 and 7. The theorem follows from Abrams Rangaswami characterization together with the fact that the involution of K is n-proper for every n σ if and only if the involution in L K (E) is proper (note a certain analogy with Theorem 10). This result is relevant since it differs from known characterizations of various other algebraic properties of a Leavitt path algebra L K (E) all of which are given in terms of graph-theoretic properties of E alone. As opposed to such characterizations, this one involves the underlying field as well. We can now prove the following corollary of Theorem 27. Corollary 28. ([6, Corollary 3.4]) The following conditions are equivalent. (1) The involution on K is positive definite. (2) L K (E) is -regular for every acyclic graph E. Proof. Condition (1) implies that of L K (E) is positive definite for any graph by Theorem 10. If E is acyclic, then L K (E) is -regular by Theorem 27 so (2) holds. Conversely, if (2) holds, let L n denote a line of length n 1. M n (K) = L K (L n ) is -regular, its involution is proper, thus of K is n-proper. Since this holds for any n, we have that of K is positive definite by Practice Problem 2.

16 16 LIA VAŠ Lectures 4 and 5. Finiteness and Direct Finiteness. Isomorphism Conjecture. Practice Problem 5 Solution. Part 1. If e and f are idempotents of R, show that the following are equivalent (1) e a f (recall that we defined this to mean that er = fr). (2) e = xy and f = yx for some x, y R. (3) e = xy and f = yx for some x, y R with ex = x, ye = y, xf = x and fy = y. (4) Re = Rf. Part 2. Show that for a unital ring R, idempotent e R and a R-module P, e a 1 e = 1 if and only if R P = R P = 0. Proof. Part 1. (1) (2) Note that a right module isomorphism er = fr is determined by its value at e. If φ is one such isomorphism and φ(e) = y = fy with φ 1 (f) = x = ex, then φ(er) = φ(e)r = yr and φ 1 (fr) = xr. Thus, yx = φ(x) = φ(φ 1 (f)) = f and xy = φ 1 (y) = φ 1 (φ(e)) = e. (2) (3) If x and y are as in (2), then it is direct to check that x = exf and y = fye are as in (3). In particular, x y = exffye = exfye = exyxye = eeeee = e. (3) (1) Define module homomorphisms φ : er fr by φ(er) = yr and ψ : fr er by ψ(fr) = xr and check that they are mutually inverse. (3) (4) Define module homomorphisms φ : Re Rf by φ(re) = rx and ψ : Rf Re by ψ(rf) = ry and check that they are mutually inverse. (4) (2) If φ : Re = Rf is a left module isomorphism and φ(e) = x = xf with φ 1 (f) = y = ye, then φ(re) = rφ(e) = rx and φ 1 (rf) = ry. Thus, yx = φ(y) = φ(φ 1 (f)) = f and xy = φ 1 (x) = φ 1 (φ(e)) = e. Part 2. Let us now show the second part of the problem. ( ) Assume that R P = R for some R-module P 0. From relation R P = R it follows that P is projective since it is a direct summand of the free module R. Let f denote the isomorphism R R P and let π be the projection of R P onto R and i injection R R 0 R P. Consider the compositions x = πf and y = f 1 i which are in Hom R (R, R) = R. Then we have that xy = πff 1 i = πi = 1 R and yx = f 1 iπf. This is an idempotent since yxyx = f 1 iπff 1 iπf = f 1 iπiπf = f 1 i1 R πf = f 1 iπf = yx. If we denote yx by e. Then we have that xy = 1 and yx = e so e a 1. However, e = yx 1 since for any r R with f(r) = (s, p) we have that (1 yx)(r) = r f 1 iπf(r) = r f 1 (s, 0) = r f 1 ((s, p) (0, p)) = r r + f 1 (0, p) = f 1 (0, p). Taking 0 p P, and r = f 1 (1, p) we obtain that (1 yx)(r) 0 and so e = yx 1. ( ) Conversely, let e R be an idempotent with xy = e and yx = 1 and we can choose x such that ex = x and y such that ye = y. Thus er = 1R = R and the relations xy = e and ex = x imply that xr = er. The relation yx = 1 implies that yr = R. Thus xr = R and so R = er + (1 e)r = xr (1 xy)r = R (1 xy)r (1 xy)r = 0. Since R is unital, this implies xy = 1 so e = 1. Practice Problem 6 Solution. Show that if a -ring R is -regular that then it is regular and is proper. Proof. If R is -regular then it is also regular because every projection is an idempotent. Now assume that a a = 0 for some a R. Then ar = pr for some projection p so a = pa (remember the Useful Observation ) and p = ay for some y R. So, a = a p = a ay = 0. Hence a = 0. Thus is proper.

17 THE ROLE OF INVOLUTION IN GRAPH ALGEBRAS 17 Practice Problem 7 Solution. Show the converse of the previous claim. Let us show the four claims first. (1) If is proper, then ann r (x) = ann r (x x). Proof. ann r (x) ann r (x x) is clear since xy = 0 implies x xy = 0. For the converse, assume that x xy = 0. Left multiply by x and get (xy) xy = y (x xy) = 0 so that xy = 0 for any y R. (2) If e is an idempotent, then ann r (e) = (1 e)r. Proof. If ex = 0 then x = x ex = (1 e)x (1 e)r. Conversely, if x (1 e)r, then x = (1 e)x (remember the Useful Observation ) so x = x ex ex = 0 ex = 0 x ann r (e). (3) If e is an idempotent in a (locally) unital ring R, then Re = ann l (ann r (e)). Proof. If x Re, then x = xe. Let y ann r (e). By the second claim y (1 e)r, and so y = (1 e)y. Thus we have that xy = xe(1 e)y = 0. This implies that x is a left annihilator of any y ann r (e). Hence x ann l (ann r (e)). To show the converse, let x ann l (ann r (e)) = ann l ((1 e)r), then x(1 e)y = 0 for all y R, that is xy = xey for all y R. If you assumed that R is unital, this holds for y = 1 and so x = xe x Re If you assumed that R is a ring with local units, there is an idempotent u R such that xu = x and eu = e. Hence x = xu = xeu = xe Re. (4) If e is an idempotent and R is regular (locally) unital -ring with proper, then Re = Re e. Proof. By the regularity of R, Re e = Rf for some idempotent f R. Thus ann r (e e) = ann r (f) and so ann r (e) = ann r (e e) = ann r (f) by the first claim (which uses the assumption that R is proper). So ann l (ann r (e)) = ann l (ann r (f)). By the third claim, this implies that Re = Rf and so Re = Rf = Re e. Proof. Now let us show the main claim. Assume that R is regular and is proper. Since every principal right ideal is generated by an idempotent, it is enough to show that for an arbitrary idempotent e in R, Re = Rp for some projection p R. By the fourth claim Re = Re e and so e = ae e for some a R. Let p = ae. We claim that p is a projection with Re = Rp. To see this, note that e = pe and so pp = pea = ea = p. Starring the relation p = pp we obtain that p = (pp ) = pp = p. Thus p = p. This also implies that p 2 = pp = p = p. So, p is a projection. The relation p = ae implies that Rp Re. The relation e = pe implies that e = e p and this gives us Re Rp. Thus Re = Rp. Starring this we have er = pr. So, for arbitrary idempotent e, we have found projection p with er = pr. Regularity and this fact imply that for any x R there is a projection p with xr = pr. Finiteness and direct finiteness. Let us recall the two equivalence relations we introduced. For idempotents: e a f iff e = xy and f = yx for some x and y. For projections: p q iff p = xx and q = x x for some x. If R is unital, the case when f = 1 or q = 1 is of special interest. Definition 29. A (unital) ring R is said to be directly finite if xy = 1 yx = 1 for all x, y R. If R is also a -ring, it is said to be finite if xx = 1 x x = 1 for all x R.

18 18 LIA VAŠ When we say that a -ring is finite, this will always mean that the ring is finite in this sense, not that it has finitely many elements. An operator algebra is a -rings with infinitely many elements since it contain a copy of C in it. Such algebra being finite means that the identity is not equivalent to any of its proper subprojections (as the following claim will illustrate). For von Neumann algebras, this concept correlates exactly to the existence of C-valued dimension function which takes finite values on all the algebra elements. Claim 30. (1) A ring is directly finite iff e a 1 e = 1 for any idempotent e. (2) A -ring is finite iff p 1 p = 1 for any projection p. The claim relates direct finiteness to a property of a and finiteness to a property of. It follows directly from the definitions. By Practice Problem 5, the condition e a 1 e = 1 for idempotent e is equivalent with R P = R P = 0 for any module P. This condition is related to the following cancellation property of the unit-regular rings. P T = Q T P = Q for all finitely generated projective modules P, Q, T. So, the following is a direct corollary of your Practice Problem 5. Corollary 31. Any unit-regular ring is directly finite. We present some motivating examples. Example 32. (1) M n (K) is directly finite (since it is unit-regular). If you are doubting this, recall that M n (K) can be represented as a Leavitt path algebra of a line of length n 1 which is an acyclic graph. Thus M n (K) is unit-regular by Abrams Rangaswami characterization. (2) Any commutative ring is trivially directly finite. Thus, K[x, x 1 ] is directly finite. Using Abrams Rangaswami characterization and the fact that this algebra is isomorphic to the Leavitt path algebra of v e, this ring is not unit-regular. So, K[x, x 1 ] is an example of directly finite ring which is not unit-regular. (3) Let E be the graph e f. Then e e = 1 and ee + ff = 1 so if ee were 1, that would lead us to a contradiction that ff = 0 f = ff f = 0. So, L K (E) is not finite and thus not directly finite as well. (4) The Toeplitz algebra is a universal example of a non-directly finite algebra. Indeed recall that it can be given as the free K-algebra over K with generators x, y subject to relation yx = 1. The freeness implies that xy 1 and so this algebra is not directly finite. (5) Let V be a countably infinite dimensional vector space over a field K. Consider the algebra R of all endomorphisms of V. Convince yourself that this algebra can also be represented as the algebra of matrices with countably many rows and columns and finitely many nonzero entries in each row. [ Consider ] the matrix unit e 11 and let P = e 11 R. v Then every element of P has the form where v is a 1 matrix with finitely 0 [ ] [ ] v v many nonzero entries and 0 is 0 of R. The map (A, ) is an isomorphism 0 A R P = R. Since P 0, R is not directly finite. So, some Leavitt path algebras are directly finite while some are not. Our next agenda is the following.

19 THE ROLE OF INVOLUTION IN GRAPH ALGEBRAS 19 Agenda. Characterize exactly when a Leavitt path algebra is directly finite. A necessary condition is relatively easy to get but before we even go there we have a problem. Problem. What is 1 if the graph is not finite? To overcome this obstacle, we have to adapt finiteness and direct finiteness to non-unital rings with local units. Recall that a ring R has local units if for every finite set x 1,..., x n R there is an idempotent u such that x i u = ux i = x i for all i = 1,..., n. Definition 33. A ring with local units R is said to be directly finite if for every x, y R and an idempotent element u R such that xu = ux = x and yu = uy = y, we have that xy = u implies yx = u. A -ring with local units R is said to be finite if for every x R and an idempotent u R such that xu = ux = x, we have that xx = u implies x x = u. Condition xx = u implies that u is a projection (selfadjoint idempotent) since u = (xx ) = xx = u. Thus, x u = ux = x as well. Next on the agenda is to show that these definitions agree with definitions in the unital case, i.e. that a unital ring R is directly finite in the unital sense iff it is directly finite in the locally-unital sense. This is taken case of by the following proposition. Proposition 34. Let R be a unital ring. R is directly finite ring iff xy = e implies yx = e for any idempotent e and x, y ere. If R is a -ring, R is finite iff xx = p implies x x = p for any projection p and x, y prp. Proof. ( ) Let xy = e for x, y ere. The condition x, y ere implies that xe = ex = e and ye = ey = e and so x(1 e) = (1 e)x = 0 and y(1 e) = (1 e)y = 0. Then (1 e + x)(1 e + y) = 1 e + xy = 1 e + e = 1. The direct finiteness implies that (1 e + y)(1 e + x) = 1 e + yx = 1. Thus e + yx = 0 yx = e. ( ) Take e = 1. The analogous claim for finiteness is proven similarly. The algebras in Example 32 indicate that the presence of cycles with exits obstructs direct finiteness. So, our eyes are on the class of no-exit graphs (recall that we encountered them in the first lecture). Definition 35. A graph E is said to be no-exit if s 1 (v) has just one element for every vertex v of every cycle. If this fails for some cycle, we say that that cycle has an exit. The next proposition shows that a necessary condition for L K (E) to be (directly) finite is that E is a no-exit graph. Proposition 36. ([7, Proposition 3.1], [12, Proposition 4.3]) If E is a graph with a cycle p which has an exit, then L K (E) is not (directly) finite. Proof. Let p be a cycle with an exit e. Let v be the source of e and w the range of e. Case 1. v w. Take x = p + w, and projection u = v + w. Then ux = ux = x and

20 20 LIA VAŠ x x = (p + w)(p + w) = v + w = u. However, xx = (p + w)(p + w) = pp + w. If xx is u, then pp = v and so e = e v = e pp = 0 which is a contradiction. So, pp v and xx u. Thus L K (E) is not finite and so it is not directly finite as well. Case 2. v = w. Take x = p, u = v. Then x x = p p = v and xx = pp v similarly as in Case 1. So we reach the same conclusion. p p v e w Our goal is to prove that the converse holds as well, in particular, we want to prove the following. Theorem 37. ([12, Theorem 4.12]) The following conditions are equivalent. (1) L K (E) is directly finite. (2) L K (E) is finite. (3) E is no-exit. Note that (1) trivially implies (2) and we have shown that (2) implies (3) in Proposition 36. Thus, we need to show that (3) implies (1). If E is finite, this is (relatively) manageable given the matricial representation of a no-exit graph from the first lecture. Proposition 38. ([1, Theorem 3.7] and [7, Theorem 3.3]) If E is a finite graph, the following are equivalent. (1) E is a no-exit graph. (2) L K (E) is -isomorphic to n i=1 M n i (K) m i=1 M m i (K[x, x 1 ]), for some positive integers m, n n i, i = 1,..., n, and m i, i = 1,..., m. (3) L K (E) is directly finite. v Proof. The idea of the proof of (1) (2) has been sketched during the first lecture. By studying the properties of matrix algebras, one can obtain (2) (3). The implication (3) (1) is in Proposition 36. Idea of the proof of remaining implication of Theorem 37 for any graph. (1) Start with x, y in L K (E) for some E no-exit. (2) Consider a local unit u for x and y such that xy = u. We want to show that yx = u. (3) Consider a finite subgraph F determined by the paths appearing in x, y, u. (4) F is a finite no-exit graph and so L K (F ) is directly finite. Thus yx = u. Done. Problem. L K (F ) may not be a subalgebra of L K (E). So yx = u in L K (F ) does not mean that yx = u in L K (E) as well. Hence step (4) is problematic. It turns out that the cure can be found in a concept you may have encountered during Pere Ara s lectures a Cohn-Leavitt algebra. Recall that (CK2) axiom of Leavitt path algebra holds for the regular vertices (i.e. vertices v which emit nonzero and finitely many edges). Definition 39. The Cohn-Leavitt algebra CL K (E, S) of a graph E and a subset S of the set of all regular vertices is obtained in exactly the same way as the Leavitt path algebra except that the (CK2) axiom holds exactly for those vertices which are in S. In the case when S is the set of all regular vertices, we obtain the Leavitt path algebra and we still denote it by L K (E). In the case when S is the empty set, CL K (E, ) is call the Cohn path algebra and it is denoted by C K (E). e