of some free product C$^{*}$-algebras ]]> Simplicity and the stable rank of some free product C -algebras Kenneth J. Dykema dykema@imada.ou.dk Dykema_Kenneth_J -algebra reduced free product of finite dimensional abelian algebras is found, and it is proved that the stable rank of every such free product is 1. Related results about other reduced free products of C$^{*}$-algebras are proved.

]]> A necessary and sufficient condition for the simplicity of the C--algebra reduced free product of finite dimensional abelian algebras is found, and it is proved that the stable rank of every such free product is 1. Related results about other reduced free products of C--algebras are proved. 1 J. Anderson, B. Blackadar and U. Haagerup, Minimal projections in the reduced group C --algebra of n m , J. Operator Theory 26 (1991), 3-23. 94c:46110 2 D. Avitzour, Free products of C --algebras , Trans. Amer. Math. Soc. 271 (1982), 423-465. 83h:46070 3 L.G. Brown, Stable isomorphism of hereditary subalgebras of C --algebras , Pacific J. Math. 71 (1977), 335-348. 56:12894 4 L.G. Brown, P. Green, M.A. Rieffel, Stable isomorphism and strong Morita equivalence of C --algebras , Pacific J. Math. 71 (1977), 349-363. 57:3866 5 K.J. Dykema, Free products of hyperfinite von Neumann algebras and free dimension , Duke Math. J. 69 (1993), 97-119. 93m:46071 6 , Free products of finite dimensional and other von Neumann algebras with respect to non-tracial states , Fields Institute Communications, (D. Voiculescu, editor), vol. 12, 1997, pp. 41-88. 98c:46131 7 , Faithfulness of free product states , J. Funct. Anal. 154 (1998), 323--329. 98:11 8 , Free Probability Theory and Operator Algebras , Seoul National University GARC lecture notes, in preparation. 9 K.J. Dykema, U. Haagerup, M. Rrdam, The stable rank of some free product C --algebras , Duke Math. J. 90 (1997), 95--121. 98:03 10 K.J. Dykema, M. Rrdam, Purely infinite simple C -algebras arising from free product constructions , Can. J. Math. 50 (1998), 323--341. 11 E. Germain, KK --theory of reduced free product C --algebras , Duke Math. J. 82 (1996), 707-723. 97f:46111 12 E. Germain, KK --theory of the full free product of unital C --algebras , J. reine angew. Math. 485 (1997), 1-10. 98b:46148 13 R.H. Herman, L.N. Vaserstein, The stable range of C --algebras , Invent. Math. 77 (1984), 553-555. 86a:46074 14 W.L. Paschke, N. Salinas, C --algebras associated with free products of groups , Pacific J. Math. 82 (1979), 211-221. 82c:22010 15 R.T. Powers, Simplicity of the reduced C --algebra associated with the free group on two generators , Duke Math. J. 42 (1975), 151-156. 51:10534 16 M.A. Rieffel, Morita equivalence for operator algebras , Proc. Symp. Pure Math. 38 (1982), 285-298. 84k:46045 17 , Dimension and stable rank in the K--theory of C --algebras , Proc. London Math. Soc. (3) 46 (1983), 301-333. 84g:46085 18 M. Rrdam, Advances in the theory of unitary rank and regular approximation , Ann. Math. 128 (1988), 153--172. 90c:46072 19 D. Voiculescu, Symmetries of some reduced free product C --algebras , Operator Algebras and Their Connections with Topology and Ergodic Theory, Lecture Notes in Mathematics, Volume 1132, Springer--Verlag, 1985, pp. 556--588. 87d:46075 20 , Multiplication of certain non--commuting random variables , J. Operator Theory 18 (1987), 223-235. 89b:46076 21 D. Voiculescu, K.J. Dykema, A. Nica, Free Random Variables , CRM Monograph Series vol. 1, American Mathematical Society, 1992. 94c:46133

1.
J. Anderson, B. Blackadar and U. Haagerup, Minimal projections in the reduced group C$^{*}$-algebra of $\mathbf Z_{n}*\mathbf Z_{m}$, J. Operator Theory 26 (1991), 3-23. MR 94c:46110

2.
D. Avitzour, Free products of C$^{*}$-algebras, Trans. Amer. Math. Soc. 271 (1982), 423-465. MR 83h:46070

3.
L.G. Brown, Stable isomorphism of hereditary subalgebras of C$^{*}$-algebras, Pacific J. Math. 71 (1977), 335-348. MR 56:12894

4.
L.G. Brown, P. Green, M.A. Rieffel, Stable isomorphism and strong Morita equivalence of C$^{*}$-algebras, Pacific J. Math. 71 (1977), 349-363. MR 57:3866

5.
K.J. Dykema, Free products of hyperfinite von Neumann algebras and free dimension, Duke Math. J. 69 (1993), 97-119. MR 93m:46071

6.
-, Free products of finite dimensional and other von Neumann algebras with respect to non-tracial states, Fields Institute Communications, (D. Voiculescu, editor), vol. 12, 1997, pp. 41-88. MR 98c:46131

7.
-, Faithfulness of free product states, J. Funct. Anal. 154(1998), 323-329. CMP 98:11

8.
-, Free Probability Theory and Operator Algebras, Seoul National University GARC lecture notes, in preparation.

9.
K.J. Dykema, U. Haagerup, M. Rørdam, The stable rank of some free product C$^{*}$-algebras, Duke Math. J. 90 (1997), 95-121. CMP 98:03

10.
K.J. Dykema, M. Rørdam, Purely infinite simple $C^{*}$-algebras arising from free product constructions, Can. J. Math. 50 (1998), 323-341.

11.
E. Germain, $KK$-theory of reduced free product C$^{*}$-algebras, Duke Math. J. 82 (1996), 707-723. MR 97f:46111

12.
E. Germain, $KK$-theory of the full free product of unital C$^{*}$-algebras, J. reine angew. Math. 485 (1997), 1-10. MR 98b:46148

13.
R.H. Herman, L.N. Vaserstein, The stable range of C$^{*}$-algebras, Invent. Math. 77 (1984), 553-555. MR 86a:46074

14.
W.L. Paschke, N. Salinas, C$^{*}$-algebras associated with free products of groups, Pacific J. Math. 82 (1979), 211-221. MR 82c:22010

15.
R.T. Powers, Simplicity of the reduced C$^{*}$-algebra associated with the free group on two generators, Duke Math. J. 42 (1975), 151-156. MR 51:10534

16.
M.A. Rieffel, Morita equivalence for operator algebras, Proc. Symp. Pure Math. 38 (1982), 285-298. MR 84k:46045

17.
-, Dimension and stable rank in the K-theory of C$^{*}$-algebras, Proc. London Math. Soc. (3) 46 (1983), 301-333. MR 84g:46085

18.
M. Rørdam, Advances in the theory of unitary rank and regular approximation, Ann. Math. 128 (1988), 153-172. MR 90c:46072

19.
D. Voiculescu, Symmetries of some reduced free product C$^{\ast }$-algebras, Operator Algebras and Their Connections with Topology and Ergodic Theory, Lecture Notes in Mathematics, Volume 1132, Springer-Verlag, 1985, pp. 556-588. MR 87d:46075

20.
-, Multiplication of certain non-commuting random variables, J. Operator Theory 18 (1987), 223-235. MR 89b:46076

21.
D. Voiculescu, K.J. Dykema, A. Nica, Free Random Variables, CRM Monograph Series vol. 1, American Mathematical Society, 1992. MR 94c:46133
]]> 1999 American Mathematical Society Transactions of the American Mathematical Society Trans. Amer. Math. Soc. tran 351 01 19990101 January 21, 1997 1 1-40 40 S0002-9947-99-02180-7 10.1090/S0002-9947-99-02180-7 0002-9947 1088-6850 English 46L05 46L35 1991 http://www.ams.org/jourcgi/jour-getitem?pii=S0002-9947-99-02180-7 Introduction The reduced free product of C --algebras with respect to given states was introduced independently by Voiculescu 19 and Avitzour 2. It is the appropriate construction associated to Voiculescu's free probability theory (see 19 , 21 ). The motivating example concerns reduced group C --algebras. For a discrete group G , its reduced group C --algebra is generated by the left regular representation of G on l 2(G) and is denoted C r(G) . Its canonical tracial state (which is the vector state associated to the characteristic function of the identity element of G ) is written G . Then for discrete groups G 1 and G 2 , the reduced free product construction yields equation (C r(G 1), G 1 ) (C r(G 2), G 2 ) (C r(G), G), equation where G G 1 G 2 is the free product of groups. Voiculescu's definition of freeness is an abstraction of some essential facets of the relationship between the copies of C r(G 1) and C r(G 2) embedded in C r(G) , with respect to the trace G . The reduced free product of C --algebras can be described with respect to freeness as follows. Let A 1 and A 2 be unital C --algebras with states 1 and 2 , respectively whose associated GNS representations are faithful. Then the reduced free product of (A 1, 1) and (A 2, 2) is the (unique) unital C --algebra and state with unital embeddings A such that enumerate (1) the GNS representation associated to is faithful on ; (2) A ; (3) A 1 and A 2 are free with respect to ; (4) is generated by A 1A 2 . enumerate We denote this by equation (,) (A 1, 1) (A 2, 2). 1 equation It is further known 19 (or see 2.5.3 21 ) that is a trace if 1 and 2 are traces. Moreover, by 7, is also faithful on if 1 and 2 are faithful. The reduced free product thus provides a multitude of constructions of C --algebras, about which some results are known (see 19 , 2, 11 , 10 , 9). For example, many can be distinguished one from the other using K--theory, (see 11 and 12 ). However, questions abound. Perhaps the most basic question concerns simplicity of reduced free product C --algebras. In 15 , R.T. Powers showed that the reduced group C --algebra of the free group on two generators, C r(F 2) , is simple and has unique tracial state. Paschke and Salinas 14 then proved the same for C r(G) whenever G G 1 G 2 is the free product of groups, where G 1 has at least two elements and G 2 has at least three. Avitzour 2 generalized further and showed that, for the reduced free product (1), is simple if there are unitaries u,vA 1 and wA 2 such that alignat 3 1(u) 1(v) 0 1(u v), 1(u u) 1, 2(w) 0, 2(w w) 2. 2 alignat (Actually, Avitzour required also 1 and 2 to be faithful, but this hypothesis is easily dispensed with.) Avitzour's conditions imply simplicity of many reduced free product C --algebras, but there are plenty of cases where Avitzour's conditions are not satisfied (see 4 of 9), yet intuition (or the analogous result for von Neumann algebras, see 5) suggests the algebra is simple. In this paper we give necessary and sufficient conditions for simplicity of the reduced free product of arbitrary finite dimensional abelian C --algebras. Stated briefly, if equation (,) (A, A) (B, B), 3 equation where A and B are finite dimensional abelian C --algebras satisfying (A)3 and (B)2 and with faithful tracial states A and B , then is simple if and only if whenever p is a minimal projection of A and q is a minimal projection of B , we have equation A(p) B(q) 1. 4 equation The necessity of this condition can be seen from 1. Note that the condition from 5 for the analogous free product of von Neumann algebras to be a factor is (4) but with the strict inequality replaced by . In addition, when the free product algebra from (3) is not simple, our analysis allows one to easily find all ideals of . We also show that for every reduced free product C --algebra as in (3), the stable rank of is 1, regardless of the simplicity of . The topological stable rank was invented by M.A. Rieffel 17 in order to study non--stable'' K--theory and as a sort of dimension for C --algebras. Topological stable rank of C --algebras was in 13 shown to be equal to the Bass stable rank. The first result about the stable rank of reduced free product C --algebras is in 9, where it is proved that the free product with respect to traces of C --algebras A 1 and A 2 has stable rank 1 if the Avitzour conditions (2) are satisfied. Hence the present paper's results regarding stable rank are, for a restricted class of C --algebra reduced free products, a considerable generalization, and they lend support to the plausible conjecture that every reduced free product of C --algebras with respect to faithful, tracial states has stable rank 1. In 1 we state the main results proved in the paper. In 2 concepts and results essential for the sequel are covered, including some dealing with stable rank, full hereditary subalgebras and free products. In 3 we prove simplicity and stable rank 1 for free products of two C --algebras, when one is diffuse in a specific sense. In 4 the results about the free product of finite dimensional abelian algebras are proved, as well as some related results about free products of more general algebras. In 5, results about free products of abelian C --algebras with states that are inductive limits of the algebras considered in 4 are proved. In 6 we consider free products of infinitely many finite dimensional abelian C --algebras. Finally, in 7 we make two conjectures about simplicity of other free product C --algebras. 1. Statement of the main results In this section, we state the main results in more detail. Let equation split (A, A) 1 p 1 2 p 2 n p n , 1ex (B, B) 1 q 1 2 q 2 m q m . split 5 equation This notation means that n , that equation A n times , equation that p k is the projection equation p k 00 k-1 1 00 n-k , equation and that A is the state on A given by A(p k) k . We thus need k0 and 1 n k 1 , and because we want the GNS representation of A to be faithful we will always take k 0 . The same considerations apply to (B, B) in (5). theorem1 Let (A, A) and (B, B) be finite dimensional, abelian C --algebras with faithful states as in (5), with (A)3 and (B)2 . Let equation (,) (A, A) (B, B) equation be the reduced free product of C --algebras. Let align L (i,j) i j 1 , L 0 (i,j) i j 1 . align Then the stable rank of is 1 and equation 0 r 0 (i,j)L i j-1 p iq j , 6 equation where p iq j n (p iq j) n . If L 0 is empty then 0 is a simple C --algebra. Otherwise, for every (i,j)L 0 , although n (p iq j) n 0 , there is a unital --homomorphism (i,j) : 0 such that (i,j) (r 0p i) 1 (i,j) (r 0q j) and equation 00 (i,j)L 0 (i,j) equation is simple, nonunital and with unique tracial state (r 0) -1 00 . theorem1 The notation in (6) means that 0 if L is empty, and otherwise equation 0 L times . equation In addition, for each (i,j)L the corresponding central summand is (p iq j) , and (p iq j) i j-1 . The assertion that n (p iq j) n 0 when i j 1 refers to the strong--operator limit in the GNS representation of with respect to . Finally, r 0 1- (i,j)L p iq j is the projection which is the unit of 000 . Analogous results hold for free products of more than two finite dimensional abelian C --algebras and for free products of direct sums of other abelian C --algebras (see Theorems 4.8, 4.9, 5.3 and 6.1). The following result is used in the proof of Theorem 1 and is also of independent interest. theorem2 Let A and B be unital C --algebras with states A , respectively B , whose GNS representations are faithful. Let equation (,) (A, A) (B, B). equation Suppose B and the centralizer of A has an abelian subalgebra DC(X) containing the unit of A and such that the restriction of A to D is given by an atomless measure on X . Then is simple. If A and B are traces, then has stable rank 1 and is the unique tracial state on . If one of A and B is not a trace, then has no tracial states. theorem2 2. Preliminaries definition1 Let A be a unital C --algebra and a state on A . Let BC(X) be an abelian C --subalgebra of A containing the unit of A . We say that is diffuse on B if B is given by a measure on X having no atoms. It will usually be clear from the context which state we mean, and then we will speak simply of B being a diffuse abelian subalgebra of A . Given a C --algebra with state (A,) , a Haar unitary (with respect to ) is a unitary element uA such that (u n) 0 for every nonzero integer n . definition1 theorem3 Let B be a unital, abelian C --algebra with state . Then is diffuse on B if and only if B contains a Haar unitary (with respect to ). theorem3 Recall that the centralizer of the state is aAxA ;(ax) (xa) . We will often be interested in the situation when the centralizer of contains a Haar unitary. An ideal of a C --algebra always means a closed, two--sided ideal. For unital C --algebras A 1,A 2,,A n , it is an obvious fact that A 1A 2A n has stable rank 1 if and only if for each j , A j has stable rank 1. In addition, we will make use of the following results, due to Rieffel. theorem4 Let n and let A be a C --algebra. Then A has stable rank 1 if and only if AM n() has stable rank 1. theorem4 The following result follows from 4.4 and 4.11 17 together with the fact that in a finite dimensional C --algebra B , the left invertible elements are invertible, hence the connected stable rank of B is one. theorem5 Let A be a C --algebra with an ideal J such that A J is finite dimensional. Then A has stable rank 1 if and only if J has stable rank 1. theorem5 Recall that a hereditary C --subalgebra B of a C --algebra A is said to be full if there is no closed, proper, two--sided ideal of A containing B . theorem6 Let A be a C --algebra with countable approximate identity. Take hA , h0 , and let B be the hereditary subalgebra hAh of A . Suppose that B is full in A . Then itemize (i) A has stable rank 1 if and only if B has stable rank 1. (ii) If B has unique tracial state then A has at most one tracial state. itemize theorem6 proof For (i), note that B has a countable approximate identity for itself because h is strictly positive in B (see p. 327 3). Thus, by 3, A and B are stably isomorphic, i.e. A B , where is the algebra of compact operators on separable Hilbert space. But 3.6 17 states that equation (A) 1(A) 1 equation and similarly for B , hence equation (A) 1(B) 1. equation To see (ii), note that xhahya,x,yA is dense in A . If is a tracial state on A then (xhahy) (h 1 2 ahyxh 1 2 ) and h 1 2 ahyxh 1 2 B , so is determined by B . proof We will say that a positive element hA is full if the hereditary subalgebra hAh is full in A . The following fact is easy to show. theorem7 Let A be a C --algebra and let B be a full hereditary C --subalge-bra of A . Then A is simple if and only if B is simple. theorem7 In fact (see 16 ), it is well--known that the representation theories of a C --algebra A and its full hereditary C --subalgebra B are equivalent. The reduced free product of two two--dimensional C --algebras is the most transparent nontrivial free product one can consider. It is understood completely and described in the proposition below. This description is the starting point for our investigation into reduced free products of more general finite dimensional abelian algebras. theorem8 Let 1 12 and let equation (,) ( p 1- 1-p ) ( q 1- 1-q ). equation If then equation - p(1-q) C( a,b ,M 2()) -1 pq , equation for some 0 a b 1 . Furthermore, in the above picture align p 1( smallmatrix 1 0 0 0 smallmatrix )1, q 0( smallmatrix t t(1-t) t(1-t) 1-t smallmatrix ) 1, align and the faithful trace is given by the indicated weights on the projections p(1-q) and pq , together with an atomless measure whose support is a,b . If 12 then equation f: 0,b M 2()f continuous and f(0) diagonal -1 pq , equation for some 0 b 1 . Furthermore, in the above picture align p ( smallmatrix 1 0 0 0 smallmatrix )1, q ( smallmatrix t t(1-t) t(1-t) 1-t smallmatrix ) 1, align and the faithful trace is given by the indicated weight on the projection pq , together with an atomless measure on 0,b . If 12 then equation f: 0,1 M 2()f continuous and f(0) and f(1) diagonal . equation Furthermore, in the above picture align p ( smallmatrix 1 0 0 0 smallmatrix ), q ( smallmatrix t t(1-t) t(1-t) 1-t smallmatrix ), align and the faithful trace is given by an atomless measure whose support is 0,1 . theorem8 proof Once the traces of p and q are known, the C --algebra and the trace are determined by composed with the functional calculus of pqp . This, in turn, is computed using Voiculescu's multiplicative free convolution 20 . To see this proof in more detail, see 8 or the proof of the analogous result for von Neumann algebras in 1.1 5. proof The following proposition is a variation on Theorem 1.2 of 5 and is proved similarly. theorem9 Let A A 1A 2 be a direct sum of unital C --algebras, write p 10A and let A be a state on A , such that 0 A(p) 1 . Let B be a unital C --algebra with state B and let (,) (A, A) (B, B) . Let 1 be the C --subalgebra of generated by (0A 2) pA together with B . We abbreviate this by writing align (,) ( A 1 p A 2 1- 1-p ) (B, B) ( 1, 1 ) ( p A 2 1- 1-p ) (B, B). align Then pp is generated by p 1p and A 10A , which are free in (pp,1 pp ) . theorem9 The next elementary lemma will come in handy. theorem10 Let B be a unital C --algebra and a state on B whose GNS representation is faithful. If (u) 1 for every unitary uB , then B . theorem10 proof Let the defining embedding BL 2(B,) be denoted b . Let (B) denote the unitary group of B . Whenever u(B) then 1 , but also ,1 1 , so for some . But L 2(B,) u(B) , so L 2(B,) is one--dimensional. This implies B . proof 3. When in the presence of one spread thin Let A and B be unital C --algebras with states A and B , respectively, whose GNS representations are faithful, and let equation (,) (A, A) (B, B). equation In this section we will prove that if the centralizer of A contains a unital, diffuse abelian subalgebra then is simple, and if, furthermore, A and B are traces then has stable rank 1. The diffuse abelian subalgebra is, if you like, one spread thin.'' By 4.1(i) 9 (see Proposition 2.2 above) this condition is equivalent to the centralizer of A containing a Haar unitary u . Denote by 0 the norm dense --subalgebra of that is generated by AB . Then, using the standard notation A and B , every element x of 0 can be written x x 0 x 1 , where x 0A and equation x 1 j 1 Na o (j) b 1 (j) a 1 (j) b 2 (j) a 2 (j) b n(j) (j) a n(j) (j) 7 equation with N , n(j) , a 0 (j) ,a n(j) (j) A , a 1 (j) ,,a n(j)-1 (j) , b 1 (j) ,,b n(j) (j) . Expressed another way, equation x 1( n 1 A n times A). equation We begin with a technical lemma. Let u be a Haar unitary in the centralizer of A and write align u - u -k k , u u kk . align theorem11 With notation as above, suppose that B and the centralizer of A contains a Haar unitary u . Given 0 and x such that (x) 0 , there is a unitary, z such that z xz differs in norm by no more than from a finite linear combination of elements of equation n 1 u - n times u . 8 equation theorem11 proof Since 0 is dense in , we may assume without loss of generality that x 0 . By Lemma 2.9 there is a unitary element vB such that 0 B(v) 1 . Let c 0 B(v) , c 1 1-c 0 2 and y (v-c 01) c 1 , so that 1 and y are orthonormal in L 2(B, B) . Let n,k and let z (u kv) nu k . Write x x 0 x 1 with x 0A and x 1 as in (7). We first concern ourselves with z x 1z . Writing x 1 as in (7), let 0 . Since (u p) p is an orthonormal family in L 2(A, A) , we have equation aA p A(au p) 0 p A(au -p ). 9 equation Using (9), we see that if k is large enough, then for every positive integer p and every j we have A(u -pk a 0 (j) ) and A(a n(j) (j) u pk ) . Since v c 01 c 1y , we have equation a n(j) (j) z 1,, n 0,1 c 1 c n a n(j) (j) u ky 1 u ky n u k, equation where align y - cases y if 1, 1 if 0, cases y cases y if 1, 1 if 0. cases align If not all the j are zero, then a n(j) (j) u ky 1 u ky n u k differs in norm by at most y m (where m is the number of 1jn for which j 1 ) from an element of equation yyy m times y u . equation Since y (1 c 0) c 1 , we obtain that a n(j) (j) z differs in norm by at most c 0 n a n(j) (j) (1 2c 0) n from an element of equation ( m 1 n yyy m times y u ). equation Similarly, z a 0 (j) differs in norm by at most c 0 n a 0 (j) (1 2c 0) n from an element of equation ( m 1 nu - y y y m times y ). equation Therefore z x 1z dffers in norm by no more than -.75pc equation j 1 N(c 0 n a 0 (j) (1 2c 0) n) ; b 1 (j) a 1 (j) b 2 (j) a 2 (j) b n(j) (j) ; (c 0 n a n(j) (j) (1 2c 0) n) equation from a finite linear combination of elements from . Thus, if n is chosen large enough and then k is chosen large enough, then z x 1z can be made arbitrarily close to a finite linear combination of elements from . We now examine z x 0z . Using again v c 01 c 1y , we have equation z x 0z 1,, 2n 0,1 c 1 c 2n u -k y - 1 u -k y - n u -k x 0 u ky n 1 u ky 2n u k. 10 equation We first concentrate on the 2 n 1 -1 terms when either 1 2 n 0 or n 1 n 2 2n 0 . The sum over these terms is equal to equation c 0 n(u -(n 1)k x 0z z x 0u (n 1)k -c 0 nu -(n 1)k x 0u (n 1)k ), equation which has norm no greater than c 0 n x 0 (2 c 0 n) . This can be made arbitrarily small by choosing n large enough (independently of k ). Each of the remaining 2 2n -2 n 1 1 terms of (10) is of the form equation c 0 lc 1 l' u -r pk y u -r 2k y u -r 1k y u -r 0k x 0 u s 0k yu s 1k yu s 2k yu s qk , 11 equation where l',p,q,r j,s j are positive integers and l0 . Clearly equation (u -r 0k x 0u s 0k ) (x 0u (s 0-r 0)k ). 12 equation If r 0 s 0 then (u -r 0k x 0u s 0k ) (x 0) 0 , and hence the term (11) is an element of . Using (9) we see that by choosing k large enough, each quantity (12) can be made arbitrarily small and hence each of the terms (11) can be made arbitrarily close to an element of . Thus if n is chosen large enough and then k is chosen large enough, then z x 0z can be made arbitrarily close to a finite linear combination of elements from . Considering the above analyses for z x 1z and z x 0z at the same time, we can choose n large enough and then k large enough so that z xz is arbitrarily close to a finite linear combination of elements from . proof theorem12 Let A and B be unital C --algebras with states A and B , respectively, whose GNS representations are faithful. Let equation (,) (A, A) (B, B). equation Suppose the centralizer of A has a unital, diffuse abelian subalgebra and B . Then for every x and 0 there are n and unitaries z 1,,z n such that equation (x)1-1n r 1 nz r xz r . 13 equation Consequently, is simple. Moreover, if both A and B are traces then is the unique tracial state on . If one or both of A and B is not a trace then has no tracial states. theorem12 proof To prove the existence of z r such that (13) holds, we may without loss of generality assume that x x and (x) 0 , and we may replace x by a unitary conjugate of itself. Let u be a Haar unitary in the centralizer of A Employing Lemma 3.1, we may assume that x , (see (8)). Now we will find z 1,,z 5u such that equation 15 r 1 5z r xz r 49 50 x. 14 equation These will be found using the technique of 15 . With notation similar to (7), we have equation x j 1 Nu -l j b 1 (j) a 1 (j) b n(j)-1 (j) a n(j)-1 (j) b n(j) (j) u m j , equation for l j,m j . Let K ( j 1 N l j,m j ) 1 . For 1r5 , let z r u rK . Let r be the closed subspace of L 2(,) spanned by all words of the form u -k b 1a 1b na n with k , (r-1)K krK , n 0 , b 1,,b n , a 1,,a n-1 and a nA . Clearly pq implies p q . Since z r xz r is a finite sum of words whose left--most letter lies in u -k (r-1)K krK and whose right--most letter lies in u k(r-1)K krK , we have equation z rxz r( r ) r. equation Denote by E r the projection from L 2(,) onto r . Given a unit vector L 2(,) , there is some 1p5 for which E p 215 . Thus equation 15 r 1 5z rxz r, 45x 15 z p xz p, , equation and since (1-E p)z p xz p(1-E p) 0 we have -1pc equation z p xz p, z p xz pE p, z p xz p(1-E p),E p 2x , E p 2 5 x. equation Hence equation 15 r 1 5z rxz r, (45 2 5 )x 49 50 x. equation This implies (14). To finish the proof of (13), note that the element 15 r 1 5z rxz r obtained above is again in . Hence, by repeating this process as many times as necessary, for any 0 there are n and z 1,,z nu such that 1n r 1 nz r xz r . Now the remaining facts follow by standard arguments of 15 and 2. Indeed, suppose is a nonzero, two--sided, closed ideal of . Let a 0 . Since the GNS representation of associated to is faithful, there must be b such that (b a ab)0 . Then from (13), it follows that (b a ab)1 ; hence and consequently is simple. The property described at (13) implies that any tracial state on must be equal to . If both A and B are traces, then the free product state is also a trace, and is thus the unique tracial state. If one of A and B is not a trace, then neither is a trace; hence has no tracial states. proof theorem13 Let (,) be as in Proposition 3.2. Let x and let 0 . Then there are unitaries z 1,z 2 such that z 1xz 2-x' for some x' , with as in (8). theorem13 proof Let uA be a Haar unitary in the cetralizer of A and let vB a unitary such that 0 B(v) 1 . By Lemma 3.1 there is a unitary z such that equation z xz-((x)1 x'') 2, 15 equation where x'' . Writing v c 01 c 1y as in the proof of Lemma 3.1, we see that (u v) pu p differs in norm by no more that c 0 p from an element of . We similarly see that (u v) px'' . Let p be so large that c 0 p (2 2 (x) ) . Let z 1 (u v) pz and z 2 zu p . Then from (15) we have equation z 1xz 2-((x)(u v) pu p (u v) px''u p) 2, equation and from the above discussion there is x' such that equation (x)(u v) pu p (u v) px''u p-x' 2. equation Hence z 1xz 2-x' . proof The proof of the following proposition uses ideas from 9. theorem14 Let A and B be unital C --algebras with faithful, tracial states A and B , respectively. Let equation (,) (A, A) (B, B). equation Suppose A has a unital, diffuse abelian subalgebra and B . Then has stable rank 1 , i.e. the set of invertible elements of is dense in . theorem14 proof Suppose for contradiction that the set of invertibles in , denoted () , is not dense. Then, by 2.6 18 , there is x such that x 1 and (x,()) 1 . We must have x 2 1 , since x 2 1 would imply that x is unitary. Let 1-x 2 . Let u be a Haar unitary in A and let vB be a unitary such that 0 B(v) 1 . By Lemma 3.3 there are n and unitaries z 1,z 2 such that z 1xz 2-x' 8 for some x' n , where equation n u lb 1a 1b k-1 a k-1 b ku m k,l,m, ,l,m n, ,b j,a j . equation Let p be so large that c 0 p (8(x )) . By writing v c 01 c 1y as in the proof of Lemma 3.1, we see that equation (u nv) pu nx'-x'' 8 equation for some x'' n,p , where -1pc equation n,p u nl b 1a 1b k-1 a k-1 b ku m k,l,m, ,l p, ,m n, ,b j, ,a j . equation For q let E q:AA be equation E q(a) j q 1 q np u j,(u j) ; , equation where we denote the defining embedding AL 2(A, A) by a . Since j ,(u j) ; 0 , we have q E q(a) 0 for every aA . Therefore, there is q , a positive multiple of n , such that x (3) -x'' 8 for some x (3) in n,p,q ' , where equation split n,p,q ' u nl b 1a 1b k-1 a k-1 b ku m k,l,m, ,l p, ,m n, b j, ,a j,E q(a j) 0 . split equation Let X A be a standard orthonormal basis (see 2 9) for (A, A) containing u,u 2,u 3,,u q np and let X B be a standard orthonormal basis for (B, B) . Let Y X A X B be the resulting free product standard orthonormal basis for (,) . (Note that, by definition, Y 1 is the set of all reduced words in X A 1 and X B 1 .) Then there is x (4) Y n,p,q ' such that u qx (3) -x (4) 8 , where Y n,p,q ' is the subset of Y defined by align Y n,p,q ' u q nl b 1a 1b k-1 a k-1 b ku m , l,m,k, ,l p, ,m n,b jX B 1 , a jX A 1,u q 1 ,u q 2 ,,u q np . align Now we see that, since no cancellation occurs when we multiply elements of Y' n,p,q (but only u on u contact''), whenever w 1,,w mY' n,p we have equation w 1w 2w m 2 w 1 2 w 2 2 w m 2. equation Moreover, if w 1w 2w m w 1'w 2'w m' for any m and w j,w' jY n,p,q ' , then w 1' w 1 , w 2' w 2 , , w m' w m . The reason for this is that when we take the reduced word of w 1w 2w m , a letter u k for qkq np appears at every boundary where w j touches w j 1 (1jm-1) , and nowhere else, and writing u k u ru q ln for l,r and rn , we see that u r must have been the last letter in w j and u q ln must have been the first letter in w j 1 , so we can recover the list of letters, w 1,w 2,,w m from their product. Thus we see that (x (4) ) m 2 x (4) 2 m for every m , and (see 3.2 9) K((x (4) ) m) K(x (4) ) . Now we argue as in the proof of 3.8 9 to show that the spectral radius of x (4) , denoted r(x (4) ) , is no greater than x (4) 2 . Indeed, let q be the largest block length of the words in the support of x (4) , so that, in the notation of 2.2 9, equation x (4) j 1 q Y j. equation Then, by 3.5 9, equation m (x (4) ) m (2mk 1) 3 2 K(x (4) ) x (4) 2 m, equation where K(x (4) ) is a constant. Hence equation r(x (4) ) m (x (4) ) m 1 m x (4) 2. equation Therefore (x (4) ,()) x (4) 2 . But x (4) -u k(u nv) pu nz 1xz 2 2 , so align (x,()) (u k(u nv) pu nz 1xz 2,()) x (4) -u k(u nv) pu nz 1xz 2 x (4) 2 2 x (4) -u k(u nv) pu nz 1xz 2 2 u k(u nv) pu nz 1xz 2 2 x 2 1, align contradicting the choice of x . proof Note that Propositions 3.2 and 3.4 combine to prove Theorem 2. theorem15 Let 0 1 , let A be a unital C --algebra with state A whose GNS representation is faithful, and let equation (,) ( p 1- 1-p ) (A, A). equation Suppose the centralizer of A has a unital, diffuse abelian subalgebra. Then the centralizer of pp has a unital, diffuse abelian subalgebra. theorem15 proof Let u be a Haar unitary in the centralizer of A , let q u pu , and let B be the C --algebra generated by 1,p,q . Then p and q are free, so equation (B, B) ( p 1- 1-p ) ( q 1- 1-q ). equation theorem16 1 2 . Then by Proposition 2.7 we have for some 0 b 1 that equation B f: 0,b M 2()f continuous and f(0) diagonal 1-2 (1-p)(1-q) , equation pBpC( 0,b ) and pBp is given by an atomless measure on 0,b . Thus pBp is a diffuse abelian subalgebra of the centralizer of pAp . theorem16 theorem17 1 2 . This is just as in Case I, except now equation B f: 0,1 M 2()f continuous and f(0) and f(1) diagonal . equation theorem17 theorem18 (n) . 1-2 -(n-1) 1-2 -n . We argue by induction on n . The case III 1 reduces to Cases I and II. Let n 1 . Then 1 2 , and by Proposition 2.7 there is some 0 b 1 such that equation B f: 0,b M 2()f continuous and f(0) diagonal 2-1 pq , equation (p-pq)B(p-pq)C( 0,b ), and the restriction of to (p-pq)B(p-pq) is given by an atomless measure on 0,b . Hence it will suffice to find a diffuse abelian subalgebra of the centralizer of (pq)(pq) , because adding it to (p-pq)B(p-pq) will give a diffuse abelian subalgebra of the centralizer of pp . We claim that the family p,q,u 2 is -free in (,) . Indeed, it suffices to show that every reduced word in p-(p)1 , q-(q)1 and nonzero powers of u 2 evaluates to zero under . However, rewriting each q-(q)1 as u (p-(p)1)u , we see that each such word is equal to a word in p-(p)1 and nonzero powers of u . From the freeness of p and u , it follows that this word evaluates to zero under . Hence pq and u 2 are -free. Letting D be the C --algebra generated by pq,u 2 , we have equation (D, D)( 2-1 pq ) (C (), ). equation Since 2-11-2 -(n-1) , by the inductive hypothesis there is a diffuse abelian subalgebra of (pq)D(pq) . As remarked above, this finishes the proof. theorem18 proof theorem19 Let A be a C --algebra with state A whose GNS representation is faithful, and let n , n2 and equation (,) ( 1 p 1 2 p 2 n p n ) (A, A). equation Suppose the centralizer of A has a unital, diffuse abelian subalgebra. Then the centralizer of has a unital, diffuse abelian subalgebra containing p 1,p 2,,p n . theorem19 proof For each j , using Proposition 3.5 and considering the subalgebra of generated by A p j , we see that the centralizer of p jp j has a diffuse abelian subalgebra D j . Then D 1 D 2 D n is the required diffuse abelian subalgebra of the centralizer of . proof 4. Finite dimensional abelian algebras In this section, we examine the reduced free product of (finitely many) finite dimensional abelian C --algebras. The methods used are reminiscent of 5. Some words about notation are in order. The natural notation equation (A, A) 1 p 1 2 p 2 n p n equation for a finite dimensional abelian C --algebra and a faithful state was explained just before Theorem 1. Similarly, the notation equation 0 r 0 k k r k equation was explained after that theorem. Analogously, we will often write expressions like equation (,) (A 0 1 p 1 n p n ) (B, B). 16 equation This will mean that (,) (A, A) (B, B) , where equation A A 0 n times , equation where A 0 is some C --algebra, where equation p k 00 k times 1 00 n-k times equation and where the state A satisfies A(p k) k . In the case of (16) we will always assume that every k 0 , that 1 n k 1 , and that the GNS representation of the restriction of A to A 000 is faithful. Usually, we will also desire that the centralizer of the restriction of A to A 000 have an abelian subalgebra on which it is diffuse (see Definition 2.1) and whose unit is 100 . This is conveniently expressed by writing the centralizer of A 0 has a unital, diffuse abelian subalgebra.'' theorem20 Let equation (,) (A 0 p) ( 1 q 1 2 q 2 ), equation where the centralizer of A 0 has a unital, diffuse abelian subalgebra. Take 1 2 . Then equation cases 0 r 0 if 11, 0 r 0 1-1 pq 1 if 1 1, , 21, 0 r 0 1-1 pq 1 2-1 pq 2 if 2 1, cases 17 equation where the centralizer of 0 has a unital, diffuse abelian subalgebra which contains r 0p and a unital, diffuse abelian subalgebra which contains r 0q 1 , and where r 0p is full in 0 . If A 0 is a trace, then the stable rank of is 1. If 11 and 21 , then 0 is simple. If, in addition, A 0 is a trace, then (r 0) -1 0 is the unique tracial state on 0 and if A 0 is not a trace, then 0 has no tracial states. Whenever i 1 for i 1,2 , there is a --homomorphism i: 0 such that i(r 0p) 1 i(q i) . If 1 1 and 2 1 , then q 1 is full in 0 and 1 is simple. If, in addition, A 0 is a trace, then (r 0) -1 1 is the unique tracial state on 1 , and if A 0 is not a trace, then 1 has no tracial states. If 1 1 and 2 1 , then q 2 is full in 0 and 2 is simple. If, in addition, A 0 is a trace, then (r 0) -1 2 is the unique tracial state on 2 , and if A 0 is not a trace, then 2 has no tracial states. If 1 1 and 2 1 (which implies 1 12 ), then q 1 is full in 2 and q 2 is full in 1 and ( 1)( 2) is simple. If, in addition, A 0 is a trace, then (r 0) -1 1 2 is the unique tracial state on 1 2 and if A 0 is not a trace, then 1 2 has no tracial states. theorem20 proof Let 1 be the C --subalgebra of generated by 1,p,q , so equation ( 1, 1 ) ( 1- 1-p p) ( 1 q 1 2 q 2 ). equation By Proposition 2.8, (1-p)(1-p) is isomorphic to the free product of (1-p) 1(1-p) and A 0 . We use Proposition 2.7 to find 1 . We will also use the fact that is generated by equation (1-p)(1-p)(1-p) 1pp 1p. equation theorem21 1 . Then equation 1 2-1 pq 2 (C( a,b )M 2()) 1-1 pq 1 , equation so (1-p)(1-p)C( a,b ) A 0 is simple by Proposition 3.2. Thus equation 2-1 pq 2 ((C( a,b ) A 0)M 2()) 1-1 pq 1 . equation Letting r 0 1-pq 1-pq 2 , we then have that 0r 0 (C( a,b ) A 0)M 2() is simple. If A 0 is a trace, then (1-p)(1-p) has stable rank 1 by Proposition 3.4. Thus also has stable rank 1. Finally, r 0 1 is clearly in the centralizer of . Hence the centralizer of 0 has a unital, diffuse abelian subalgebra which contains r 0p and another which contains r 0q 1 . theorem21 theorem22 1 12 . Then equation 1 f: 0,b M 2()f continuous and f(0) diagonal -1 pq 1 , 18 equation with p ( smallmatrix 1 0 0 0 smallmatrix )1 and q 1 ( smallmatrix t t(1-t) t(1-t) 1-t smallmatrix )1 . Moreover, pq 1 is minimal and central in and, by Proposition 3.2, (1-p)(1-p)C( 0,b ) A 0 is simple. Consider the central projection r 0 1-pq 1 , and let 0 r 0 . Because r 0 1 is in the centralizer of , the centralizer of 0 has a unital, diffuse abelian subalgebra which contains r 0p and and another which contains r 0q 1 . Let (1) pq 2 :r 0 1 be the --homomorphism defined, in the notation of (18), by equation (1) pq 2 (f) the (1,1)--entry of f(0), equation so that (1) pq 2 (r 0p) 1 (1) pq 2 (q 2) . Clearly r 0 is also central in , and the linear span of equation split r 0p 1p (1-p)(1-p) (1-p)(1-p) 1p p 1(1-p)(1-p) p 1(1-p)(1-p) 1p split equation is dense in 0 . Thus (1) pq 2 extends to a --homomorphism pq 2 : 0 such that equation split (1) pq 2 (1-p)(1-p) (1-p)(1-p) 1p p 1(1-p)(1-p) p 1(1-p)(1-p) 1p split 19 equation spans a dense subset of pq 2 . We now show that pq 2 is simple. Since (1-p)(1-p) is simple, by Proposition 2.6 it will suffice to show that (1-p)(1-p) is full in pq 2 . But clearly 1-p is full in (1) pq 2 , and p 1(1-p) (1) pq 2 . Hence, by the denseness of the span of (19) in pq 2 , there is no proper ideal of pq 2 containing (1-p)(1-p) . Suppose A 0 is a trace. Then (1-p)(1-p) has stable rank 1 by Proposition 3.4. Since 1-p is full in pq 2 , also pq 2 has stable rank 1 by Proposition 2.5(i). Then has stable rank 1 by Proposition 2.4. Finally, we show that q 2 is full in 0 . Suppose is an ideal of 0 containing q 2 . Looking at the ideal of 1 generated by q 2 , we see that contains a nonzero element of pq 2 , hence by simplicity contains all of pq 2 . But q 2 pq 2 and 0 pq 2 is one-dimensional; hence 0 . theorem22 theorem23 1 2 . Then equation 1 1 - q 1(1-p) (C( a,b ) M 2()) 1 -1 q 1p , equation and equation (1-p)(1-p) ( 1- 1- q 1(1-p) C( a,b )) A 0 equation is by Proposition 3.2 simple. Let r 0 1-q 1p and 0 r 0 . Clearly 1-p is full in r 0 1 and thus is full in 0 . Hence 0 is simple. If D is a unital, diffuse abelian subalgebra of the centralizer of A 0 , then D (p-q 1p) 1(p-q 1p) is a unital, diffuse abelian subalgebra of the centralizer of 0 and contains r 0p . By Proposition 3.5 and considering the C --subalgebra of (1-p)(1-p) generated by q 1(1-p) A 0 , we see that there is a unital, diffuse abelian subalgebra D of the centralizer of (q 1(1-p))(q 1(1-p)) . Then equation D q 2 1q 2 (q 1-q 1(1-p)-q 1p) 1(q 1-q 1(1-p)-q 1p) equation is a unital, diffuse abelian subalgebra of the centralizer of 0 and contains r 0q 1 . If A 0 is a trace, then by Proposition 3.4 (1-p)(1-p) has stable rank 1. So by Proposition 2.5(i), 0 has stable rank 1. We know that equation 0 1-1 pq 1 , equation so has stable rank 1. theorem23 theorem24 12 2 . Then equation 1 1- (1-p)q 1 f: a,1 M 2()f continuous and f(1) diagonal , 20 equation with 0 a 1 , p 0( smallmatrix 1 0 0 0 smallmatrix ) and q 1 1( smallmatrix t t(1-t) t(1-t) 1-t smallmatrix ) . Thus equation (1-p)(1-p)( 1- 1- q 1(1-p) C( a,1 )) A 0 equation is by Proposition 3.2 simple. Moreover, if D is a unital, diffuse abelian subalgebra of the centralizer of A 0 , then D p 1p is a unital, diffuse abelian subalgebra of the centralizer of and contains p . By Proposition 3.5 and considering the C --subalgebra of (1-p)(1-p) generated by q 1(1-p) A 0 , we see that there is a unital, diffuse abelian subalgebra D of the centralizer of (q 1(1-p))(q 1(1-p)) . Then equation D q 2 1q 2 (q 1-q 1(1-p)) 1(q 1-q 1(1-p)) equation is a unital, diffuse abelian subalgebra of the centralizer of and contains q 1 . Let (1) pq 1 : 1 be the --homomorphism defined, in the notation of (20), by equation (1) pq 1 (f) the (1,1)--entry of f(1), equation so that (1) pq 1 (p) 1 (1) pq 1 (q 1) . Then the linear span of equation split p 1p (1-p)(1-p) (1-p)(1-p) 1p p 1(1-p)(1-p) p 1(1-p)(1-p) 1p split equation is clearly dense in , so (1) pq 1 extends to a --homomorphism pq 1 : such that equation split (1) pq 1 (1-p)(1-p) (1-p)(1-p) 1p p 1(1-p)(1-p) p 1(1-p)(1-p) 1p split 21 equation spans a dense subset of pq 1 . As in Case II, since 1-p is full in pq 1 (1) and since (1-p)(1-p) is simple, it follows that pq 1 is simple. If A 0 is a trace then by Proposition 3.4, (1-p)(1-p) has stable rank 1. Since 1-p is full in pq 1 , also pq 1 has stable rank 1 by Proposition 2.5(i). Thus by Proposition 2.4, has stable rank 1. Finally, we show that q 1 is full in . Suppose is an ideal of containing q 1 . Looking at the ideal of 1 generated by q 1 , we see that contains a nonzero element of pq 1 , hence by simplicity contains all of pq 1 . But q 1 pq 1 and pq 1 is one-dimensional; hence . theorem24 theorem25 2 . Then equation 1 1- (1-p)q 1 (C( a,b )M 2()) 2- (1-p)q 2 equation and equation (1-p)(1-p)( 1- 1- (1-p)q 1 C( a,b ) 2- 1- (1-p)q 2 ) A 0 equation is by Proposition 3.2 simple. Since 1-p is full in 1 , is follows that is simple. Moreover, if D is a unital, diffuse abelian subalgebra of the centralizer of A 0 , then D p 1p is a unital, diffuse abelian subalgebra of the centralizer of and contains p . By Proposition 3.5 and considering the C --subalgebra of (1-p)(1-p) generated by q 1(1-p),q 2(1-p) A 0 , we see that there is a unital, diffuse abelian subalgebra D 1 and, respectively, D 2 , of the centralizer of (q 1(1-p))(q 1(1-p)) and, respectively, of (q 2(1-p))(q 2(1-p)) . Then equation split D 1 (q 1-q 1(1-p)) 1(q 1-q 1(1-p)) D 2 (q 2-q 2(1-p)) 1(q 2-q 2(1-p)) split equation is a unital, diffuse abelian subalgebra of the centralizer of and contains q 1 . If A 0 is a trace then by Proposition 3.4 (1-p)(1-p) has stable rank 1. Hence by Proposition 2.5(i) also has stable rank 1. theorem25 theorem26 1 12 . Then equation 1 f: 0,1 M 2()f continuous, f(0), ,f(1) diagonal , 22 equation with p ( smallmatrix 1 0 0 0 smallmatrix ) , q 1 ( smallmatrix t t(1-t) t(1-t) 1-t smallmatrix ) . Thus equation (1-p)(1-p)C( 0,1 ) A 0 equation is by Proposition 3.2 simple. Moreover, clearly 1 is in the centralizer of and has unital, diffuse abelian subalgebras containing p and, respectively, q 1 . For i 1,2 let (1) pq i : 1 be the --homomorphism defined, in the notation of (22), by equation (1) pq i (f) cases the (1,1)--entry of f(1) if i 1, the (1,1)--entry of f(0) if i 2, cases equation so that (1) pq i (p) 1 (1) pq i (q i) . Then the linear span of equation split p 1p (1-p)(1-p) (1-p)(1-p) 1p p 1(1-p)(1-p) p 1(1-p)(1-p) 1p split equation is clearly dense in , so (1) pq i extends to a --homomorphism pq i : such that equation split (1) pq 1 (1) pq 2 (1-p)(1-p) (1-p)(1-p) 1p p 1(1-p)(1-p) p 1(1-p)(1-p) 1p split 23 equation spans a dense subset of pq 1 pq 2 . As in Case II, since 1-p is full in pq 1 (1) pq 2 (1) and since (1-p)(1-p) is simple, it follows that pq 1 pq 2 is simple. If A 0 is a trace, then, by Proposition 3.4, (1-p)(1-p) has stable rank 1. Since 1-p is full in pq 1 pq 2 , by Proposition 2.5(i) also pq 1 pq 2 has stable rank 1. Since ( pq 1 pq 2 ) is two--dimensional, it follows from Proposition 2.4 that has stable rank 1. We show that q 1 is full in pq 2 . Suppose is an ideal of pq 2 containing q 1 . Multiplying by elements of 1 , we see that contains a nonzero element of pq 1 , hence by simplicity contains all of pq 1 pq 2 . But q 1 pq 1 and pq 1 is one-dimensional; hence pq 2 . The proof that q 2 is full in pq 1 is the same. We now examine the question of existence and uniqueness of tracial states on the algebras delineated in the statement of the lemma. In all the cases above, it follows from Proposition 3.2 that (1-p)(1-p) has tracial states if and only if A 0 is a trace, and then the free product state gives the unique tracial state on (1-p)(1-p) . Moreover, the element 1-p is full in the simple algebras under consideration, i.e. itemize 1-p is full in 0 in Cases I, III and V, 1-p is full in pq 2 in Case II, 1-p is full in pq 1 in Case IV, 1-p is full in pq 1 pq 2 in Case VI. itemize It then follows from Proposition 2.5(ii) that in each of Cases I--VI, the corresponding algebra has tracial states if and only if A 0 is a trace, and then the restriction of (r 0) -1 to this algebra is its unique tracial state. (In the non--unital Cases II, IV and VI, one easily sees that the above normalization gives a state by looking at the subalgebra r 0 1 .) theorem26 proof theorem27 Let equation (,) (A 0 1 p 1 2 p 2 ) ( 1 q 1 2 q 2 ), equation where the centralizer of A 0 has a unital, diffuse abelian subalgebra. Let equation split L (i,j) i j 1 , L 0 (i,j) i j 1 . split 24 equation Then equation 0 r 0 (i,j)L i j-1 p iq j , equation where the centralizer of 0 has a unital, diffuse abelian subalgebra which contains r 0p 1 and a unital, diffuse abelian subalgebra which contains r 0q 1 . If A 0 is a trace, then the stable rank of is 1. If L 0 is empty, then 0 is simple. If, in addition, A 0 is a trace, then (r 0) -1 0 is the unique tracial state on 0 , and if A 0 is not a trace, then 0 has no tracial states. If L 0 is not empty, then for every (i,j)L 0 there is a --homomorphism (i,j) : 0 such that (i,j) (r 0p i) 1 (i,j) (r 0q j) . Then itemize (i) equation 00 (i,j)L 0 (i,j) equation is simple. If A 0 is a trace then (r 0) -1 00 is the unique tracial state on 00 , and if A 0 is not a trace then 00 has no tracial states. (ii) For each i 1,2 , r 0p i is full in 0 (i',j)L 0 'i (i',j) . (iii) For each j 1,2 , r 0q j is full in 0 (i,j')L 0 'j (i,j') . itemize theorem27 proof We will assume that 1 2 and 1 2 . To prove the lemma in its full generality, we will now be careful to find a unital, diffuse abelian subalgebra of the centralizer of 0 containing r 0p 1,r 0p 2 , not only r 0p 1 , and another containing r 0q 1,r 0q 2 . Let 1 be the C --subalgebra of generated by A 0 (p 1 p 2) together with q 1,q 2 , i.e. equation ( 1, 1 ) (A 0 1 2 p 1 p 2 ) ( 1 q 1 2 q 2 ). 25 equation We find 1 using Lemma 4.1. Then, by Proposition 2.8 equation (p 1 p 2)(p 1 p 2)(p 1 p 2) 1(p 1 p 2) ( 1 1 2 p 1 2 1 2 p 2 ) 26 equation We consider three cases. theorem28 1 2 11 . Then by Lemma 4.1, the centralizer of 1 has a unital, diffuse abelian subalgebra D which contains p 1 p 2 , and p 1 p 2 is full in 1 . Hence by Proposition 3.2 (p 1 p 2)(p 1 p 2) is simple. Also, p 1 p 2 is full in , hence is simple. If A 0 is a trace then, by Proposition 3.4, (p 1 p 2)(p 1 p 2) has stable rank 1. Hence by Proposition 2.5(i) so does . The application of Lemma 4.1 to (25) yields a unital, diffuse abelian subalgebra of the centralizer of which contains q 1,q 2 . Applying Corollary 3.6 to (26) shows that the centralizer of (p 1 p 2)(p 1 p 2) has a unital, diffuse abelian subalgebra D' containing p 1,p 2 . Then (1-p 1-p 2)D(1-p 1-p 2) D' is a unital, diffuse abelian subalgebra of the centralizer of containing p 1,p 2 . theorem28 theorem29 1 2 1 1 and 1 2 21 . Note that this implies equation split L (i,1) i 1 1 , L 0 (i,1) i 1 1 . split equation Applying Lemma 4.1 to (25) shows that equation 1 1,0 r 1,0 1 2 1-1 (p 1 p 2)q 1 , equation where r 1,0 1-(p 1 p 2)q 1 , where the centralizer of 1,0 has a unital, diffuse abelian subalgebra containing r 1,0 (p 1 p 2) and where each of r 1,0 (p 1 p 2) and q 2 is full in 1,0 . Then equation (p 1 p 2) 1(p 1 p 2) (p 1 p 2) 1,0 (p 1 p 2) 1 2 1-1 1 2 (p 1 p 1)q 1 . equation So, from (26) and Lemma 4.1, equation (p 1 p 2)(p 1 p 2) 2,0 r 2,0 (i,1)L i 1-1 1 2 p iq 1 , equation where r 2,0 p 1 p 2- (i,1)L p iq 1. Since is generated by (p 1 p 2)(p 1 p 2) 1 , we have equation 0 r 0 (i,1)L i 1-1 p iq 1 , equation where the linear span of equation split 2,0 2,0 1,0 (1-p 1-p 2) (1-p 1-p 2) 1,0 2,0 (1-p 1-p 2) 1,0 2,0 1,0 (1-p 1-p 2) (1-p 1-p 2) 1,0 (1-p 1-p 2) split 27 equation is dense in 0 . Thus r 0 r 2,0 (1-p 1-p 2) . Now the centralizer of 2,0 has a unital, diffuse abelian subalgebra D which contains r 2,0 p 1,r 2,0 p 2 . Letting D' be a unital, diffuse abelian subalgebra of the centralizer of A 0 , it follows that D D' is a unital, diffuse abelian subalgebra of the centralizer of 0 and contains r 0p 1,r 0p 2 . Now let D be a unital, diffuse abelian subalgebra of the centralizer of 2,0 containing r 2,0 ((p 1 p 2)q 1) , and D' be a unital, diffuse abelian subalgebra of the centralizer of 1,0 containing r 1,0 q 1,q 2 . Then r 2,0 ((p 1 p 2)q 1)D D' is a unital, diffuse abelian subalgebra of the centralizer of 0 and contains q 2,r 0q 1 . Since r 1,0 (p 1 p 2) 2,0 and is full in 1,0 , it follows that 2,0 is full in 0 . If L 0 is empty then 2,0 is simple, hence (by Proposition 2.6) 0 is also simple. Otherwise, if L 0 is nonempty, for every (i,1)L 0 there is a --homomorphism (i,1) (2) : 2,0 such that (i,1) (2) (r 2,0 p i) 1 (i,1) (2) (r 2,0 ((p 1 p 2)q 1)) . Using the denseness of the span of (27) in 0 , we see that (i,1) (2) extends to a --homomorphism (i,1) : 0 such that (i,1) (r 0p i) 1 (i,1) (r 0q 1) and the linear span of align (i,1) (2) 2,0 1,0 (1-p 1-p 2) (1-p 1-p 2) 1,0 2,0 (1-p 1-p 2) 1,0 2,0 1,0 (1-p 1-p 2) (1-p 1-p 2) 1,0 (1-p 1-p 2) align is dense in (i,1) . Let equation 2,00 2,0 (i,1)L 0 (i,1) (2) . equation From the application of Lemma 4.1, 2,00 is simple. Since 2,00 contains r 1,0 (p 1 p 2) , which is full in 1,0 , it follows that 2,00 is full in 00 . Then (by Proposition 2.6), 00 is simple. Let i 1,2 . We now show that r 0p i is full in equation 0 (i',1)L 0 'i (i',1) . 28 equation Suppose is an ideal of the algebra in (28) containing r 0p i . Since r 2,0 r 0 and (by Lemma 4.1) r 2,0 p i is full in equation 2,0 (i',1)L 0 'i (i',1) (2) , 29 equation must contain the algebra in (29). Hence, arguing as above, 1,0 . Thus align 2,0 (i',1)L 0 'i (i',1) (2) 2,0 1,0 (1-p 1-p 2) (1-p 1-p 2) 1,0 2,0 (1-p 1-p 2) 1,0 2,0 1,0 (1-p 1-p 2), align proving that the algebra of (28) is contained in . Similarly, since r 2,0 ((p 1 p 2)q 1) is full in 2,0 , it follows that r 0q 1 is full in 0 . If A 0 is a trace, then from Lemma 4.1 we have that 2 and indeed 2,0 has stable rank 1. The fullness of 2,0 in 0 implies (via Proposition 2.5(i)) that 0 has stable rank 1, hence has stable rank 1. Finally, concerning existence and uniqueness of traces, from Lemma 4.1 we have that 2,00 has a tracial state if and only if A 0 is a trace, and then (r 2,0 ) -1 2,00 is its unique tracial state. The same statement for 00 then follows from fullness of 2,00 in 00 and Proposition 2.5(ii). (One can easily check the normalization.) theorem29 theorem30 1 2 2 1 . Since 212 we must have 12 1 2 . Let n be least such that 1 2 n 1 . Thus n2 . We will proceed by induction on n , proving the case n 2 and the inductive step simultaneously. Applying Lemma 4.1 to (25), we have equation 1 1,0 r 1,0 1 2 1-1 (p 1 p 2)q 1 1 2 2-1 (p 1 p 2)q 2 , equation where r 1,0 1-(p 1 p 2)q 1-(p 1 p 2)q 2 , where the centralizer of 1,0 has a unital, diffuse abelian subalgebra containing r 1,0 (p 1 p 2) and where 1,0 is simple. Thus from (26), equation split (p 1 p 2)(p 1 p 2) ((p 1 p 2) 1,0 (p 1 p 2) 1 2 1-1 1 2 (p 1 p 2)q 1 1 2 2-1 1 2 (p 1 p 2)q 2 ) ( 1 1 2 p 1 2 1 2 p 2 ). split equation Now since n-1 n 1 2 , we have equation 1 2 1-1 1 2 1 2 2-1 1 2 2-1 2 2 n-2 n-1 . equation The inductive hypothesis (or, when n 2,3 , the previously considered Case I or Case II) applies, and we have, with L as in (24), equation (p 1 p 2)(p 1 p 2) 2,0 r 2,0 (i,j)L i j-1 1 2 p iq j , equation where r 2,0 p 1 p 2- (i,j)L p iq j . We obtain that equation 0 r 0 (i,j)L i j-1 p iq j equation and that the span of (27) is dense in 0 . So r 0 r 2,0 (1-p 1-p 2) . Letting D be a unital, diffuse abelian subalgebra of the centralizer of 2,0 containing r 2,0 p 1,r 2,0 p 2 , and D' a unital, diffuse abelian subalgebra of the centralizer of A 0 , we see that D D' is a unital, diffuse abelian subalgebra of the centralizer of and contains r 0p 1, r 0p 2 . Let D be a unital, diffuse abelian subalgebra of the centralizer of 2,0 that contains ((p 1 p 2)q 1)r 2,0 ,((p 1 p 2)q 2)r 2,0 and let D' be a unital, diffuse abelian subalgebra of the centralizer of 1,0 that contains r 1,0 q 1,r 1,0 q 2 . Then equation r 2,0 ((p 1 p 2)q 1)D r 2,0 ((p 1 p 2)q 2)D D' equation is a unital, diffuse abelian subalgebra of the centralizer of 0 and contains r 0q 1,r 0q 2 . Since r 1,0 (p 1 p 2) 2,0 and is full in 1,0 , it follows that 2,0 is full in 0 . If L 0 , defined in (24), is empty, then 2,0 is simple; hence (by Proposition 2.6) 0 is also simple. If L 0 is nonempty, then for every (i,j)L 0 there is a --homomorphism (i,j) (2) : 2,0 such that (i,j) (2) (r 2,0 p i) 1 (i,j) (2) (r 2,0 ((p 1 p 2)q j)) , and this extends to a --homomorphism (i,j) : 0 such that (i,j) (r 0p i) 1 (i,j) (r 0q j) and the linear span of align (i,j) (2) 2,0 1,0 (1-p 1-p 2) (1-p 1-p 2) 1,0 2,0 (1-p 1-p 2) 1,0 2,0 1,0 (1-p 1-p 2) (1-p 1-p 2) 1,0 (1-p 1-p 2) align is dense in (i,j) . Let equation 2,00 2,0 (i,j)L 0 (i,j) (2) . equation From the application of Lemma 4.1, 2,00 is simple. Since 2,00 contains r 1,0 (p 1 p 2) , which is full in 1,0 , it follows that 2,00 is full in 00 . Then (by Proposition 2.6), 00 is simple. Let i 1,2 . We will now show that r 0p i is full in equation 0 (i',j)L 0 'i (i',j) . 30 equation Suppose is an ideal of the algebra in (30) which contains r 0p i . Since r 2,0 r 0 and since r 2,0 p i is full in equation 2,0 (i',j)L 0 'i (i',j) (2) , equation this algebra must be contained in . Then r 1,0 (p 1 p 2) . Since 1,0 is simple, it is then contained in . Hence align 2,0 (i',j)L 0 'i (i',j) (2) 2,0 1,0 (1-p 1-p 2) (1-p 1-p 2) 1,0 2,0 (1-p 1-p 2) 1,0 2,0 1,0 (1-p 1-p 2), align proving that the algebra of (30) is contained in . Let j 1,2 . We now show that r 0q j is full in equation 0 (i,j')L 0 'j (i,j') . 31 equation Suppose is an ideal of the algebra in (31) which contains r 0q j . Since r 2,0 r 0 and since r 2,0 q j is full in equation 2,0 (i,j') L 0 'j (i,j') (2) , equation this algebra must be contained in . Then r 1,0 (p 1 p 2) , which as before shows that the algebra (31) is contained in . theorem30 The required results about the stable rank of and the existence and uniqueness of traces on 00 follow from the inductive hypothesis because (in the simple case) 2,0 is full in 0 or (more generally) 2,00 is full in 00 . proof theorem31 Let n , n3 and let equation (,) ( 1 p 1 n p n ) ( 1 q 1 2 q 2 ). equation Let equation split L (i,j) i j 1 , L 0 (i,j) i j 1 . split 32 equation Then equation 0 r 0 (i,j)L i j-1 p iq j , equation where the centralizer of 0 has a unital, diffuse abelian subalgebra which contains r 0p 1 and a unital, diffuse abelian subalgebra which contains r 0q 1 . Then the stable rank of is 1. If L 0 is empty, then 0 is simple and (r 0) -1 0 is the unique tracial state on 0 . If L 0 is not empty, then for every (i,j)L 0 there is a --homomorphism (i,j) : 0 such that (i,j) (r 0p i) 1 (i,j) (r 0q j) . Then: itemize (i) equation 00 (i,j)L 0 (i,j) equation is simple and (r 0) -1 00 is the unique tracial state on 0 . (ii) For each i 1,2,,n , r 0p i is full in equation 0 (i',j)L 0 'i (i',j) . 33 equation (iii) For each j 1,2 , r 0q j is full in equation 0 (i,j')L 0 'j (i,j') . 34 equation itemize theorem31 proof We proceed by induction on n , proving the initial step n 3 and the inductive step simultaneously. Let 1 be the C --subalgebra of generated by ((p 1 p 2) p 3 p n)(q 1 q 2) . Thus equation ( 1, 1 )( 1 2 p 1 p 2 3 p 3 n p n ) ( 1 q 1 2 q 2 ). equation By the inductive hypothesis when n 3 or by Proposition 2.7 when n 3 , letting equation alignat 3 L (i,j) i j 1 , L 0 (i,j) i j 1 , L (1) (i,j)i3, , i j 1 , L 0 (1) (i,j)i3, , i j 1 , L ' j 1 2 j 1 , L 0' j 1 2 j 1 , L (2) L L (1) , L 0 (2) L 0L 0 (1) , alignat 35 equation we have equation 1 1,0 r 1,0 jL ' 1 2 j-1 (p 1 p 2)q j (i,j)L (1) i j-1 p iq j , equation and there is a unital, diffuse abelian subalgebra of the centralizer of 1,0 containing r 1,0 (p 1 p 2) . By Proposition 2.8, (p 1 p 2)(p 1 p 2) is freely generated by (p 1 p 2) 1(p 1 p 2) and (p 1 p 2) , so equation split (p 1 p 2)(p 1 p 2) ((p 1 p 2) 1,0 (p 1 p 2) jL ' 1 2 j-1 1 2 (p 1 p 2)q j ) ( 1 1 2 p 1 2 1 2 p 2 ). split equation Noting that L ' 2 , we may use Lemma 4.2, Lemma 4.1 or results from 3 to show that equation (p 1 p 2)(p 1 p 2) 2,0 r 2,0 (i,j)L (2) i j-1 p iq j . equation Hence equation 0 r 0 (i,j)L i j-1 p iq j , 36 equation where r 0 r 2,0 r 1,0 (1-p 1-p 2) and the linear span of the set in (27) is dense in 0 . The inductive hypothesis (or Proposition 2.7) yields for every (i,j)L 0 (1) a --ho -mo -morph -ism (i,j) (1) : 1,0 such that (i,j) (1) (r 1,0 p i) 1 (i,j) (1) (r 1,0 q j) . Moreover, for every (i,j)L 0 (2) we have a --homomorphism (i,j) (2) : 2,0 such that (i,j) (2) (r 2,0 p i) 1 (i,j) (2) (r 2,0 q j) . Looking at (27), one easily sees that each of these --homomorphisms can be uniquely extended to a --homomorphism (i,j) : 0 so that (i,j) (r 0p i) 1 (i,j) (r 0q j) . Let align 1,00 1,0 (i,j)L 0 (1) (i,j) (1) , 2,00 2,0 (i,j)L 0 (2) (i,j) (2) . align Since r 1,0 (p 1 p 2) 1,00 2,00 , we see from the denseness of the span of (27) in 0 that the linear span of equation split 2,0 2,00 1,00 (1-p 1-p 2) (1-p 1-p 2) 1,00 2,00 (1-p 1-p 2) 1,00 2,00 1,00 (1-p 1-p 2) (1-p 1-p 2) 1,0 (1-p 1-p 2) split 37 equation is dense in 0 , and hence the linear span of -.75pc equation split 2,00 2,00 1,00 (1-p 1-p 2) (1-p 1-p 2) 1,00 2,00 (1-p 1-p 2) 1,00 2,00 1,00 (1-p 1-p 2) (1-p 1-p 2) 1,00 (1-p 1-p 2) split equation is dense in 00 . Note that r 1,0 (p 1 p 2) is full in 1,00 . Since r 1,0 (p 1 p 2) 2,00 , this implies that 2,00 is full in 00 . Since 2,00 is simple, it follows from Proposition 2.6 that 00 is simple. (This also shows that 0 is simple when L 0 .) Let us now prove part (ii). If i 1,2 then the linear span of -1pc equation split ( 2,0 (i',j)L 0 (2) 'i (i',j) (2) ) 2,00 1,00 (1-p 1-p 2) (1-p 1-p 2) 1,00 2,00 2ex (1-p 1-p 2) 1,00 2,00 1,00 (1-p 1-p 2) (1-p 1-p 2) 1,00 (1-p 1-p 2) split equation is dense in (33). But r 2,0 p i is full in equation 2,0 (i',j)L 0 (2) 'i (i',j) (2) . equation Since this latter algebra contains r 1,0 (p 1 p 2) , which is full in 1,00 , it then follows that r 0p i is full in the algebra (33). Now take i 3,4,,n . For jL 0' let (0,j) (1) : 1,0 be the --homomorphism such that (0,j) (1) (p 1 p 2) 1 (0,j) (1) (q j) . We have that r 1,0 p i is full in equation 1,0 jL 0' (0,j) (1) (i',j)L 0 (1) 'i (i',j) (1) , equation which in turn contains (1-p 1-p 2) 1,0 (p 1 p 2) and equation (1-p 1-p 2)( 1,0 (i',j)L 0 (1) 'i (i',j) (1) )(1-p 1-p 2). equation But (p 1 p 2)( jL 0' (0,j) (1) )(p 1 p 2) meets 2,00 , which is simple. Hence any ideal of the algebra (34) which contains r 0p i must also contain equation split 2,00 2,00 1,00 (1-p 1-p 2) (1-p 1-p 2) 1,00 2,00 (1-p 1-p 2) 1,00 2,00 1,00 (1-p 1-p 2) (1-p 1-p 2)( 1,0 (i',j)L 0 (1) 'i (i',j) (1) )(1-p 1-p 2), split equation which is dense in the algebra (34). This shows that r 0p i is full in (34). We now prove part (iii). We have that r 1,0 q j is full in equation 1,0 j'L 0' 'j (0,j') (1) (i,j')L 0 (1) 'j (i,j') (1) , 38 equation which in turn contains (1-p 1-p 2) 1,0 (p 1 p 2) and equation (1-p 1-p 2)( 1,0 (i,j')L 0 (1) 'j (i,j') (1) )(1-p 1-p 2). equation If i such that (i,j)L 0 (2) then 1 2 j 1 , so (p 1 p 2)q j0 and r 2,0 ((p 1 p 2)q j) is full in equation 2,0 (i,j')L 0 (2) 'j (i,j') (2) . 39 equation Hence any ideal of the algebra (34) that contains r 0q j must contain equation split ( 2,0 (i,j')L 0 (2) 'j (i,j') (2) ) 2,00 1,00 (1-p 1-p 2) (1-p 1-p 2) 1,00 2,00 1.5ex (1-p 1-p 2) 1,00 2,00 1,00 (1-p 1-p 2) 1ex (1-p 1-p 2)( 1,0 (i,j')L 0 (1) 'j (i,j') (1) )(1-p 1-p 2), split equation whose span is dense in (34). On the other hand, if there is no i such that (i,j)L 0 (2) then the algebra (39) is 2,00 , which is simple. By a dimension argument, the algebra (38) meets 2,00 . Hence any ideal of the algebra (34) that contains r 0q j must contain equation split 2,00 2,00 1,00 (1-p 1-p 2) (1-p 1-p 2) 1,00 2,00 (1-p 1-p 2) 1,00 2,00 1,00 (1-p 1-p 2) (1-p 1-p 2)( 1,0 (i,j')L 0 (1) 'j (i,j') (1) )(1-p 1-p 2), split equation whose span is dense in (34). Thus r 0q j is full in (34). The required results about stable rank and uniqueness of the trace follow as in previous lemmas from the fact that 2,00 is full in 00 . proof theorem32 Let n 0 and let equation (,) (A 0 1 p 1 n p n ) ( 1 q 1 2 q 2 ), equation where the centralizer of A 0 has a unital, diffuse abelian subalgebra. Let L and L 0 be as in (32). Then equation 0 r 0 (i,j)L i j-1 p iq j , 40 equation where the centralizer of 0 has a unital, diffuse abelian subalgebra which contains r 0p 1 and a unital, diffuse abelian subalgebra which contains r 0q 1 . If A 0 is a trace then the stable rank of is 1. If L 0 is empty then 0 is simple. If, in addition, A 0 is a trace then (r 0) -1 0 is the unique tracial state on 0 , and if A 0 is not a trace then 0 has no tracial states. If L 0 is not empty, then for every (i,j)L 0 there is a --homomorphism (i,j) : 0 such that (i,j) (r 0p i) 1 (i,j) (r 0q j) . Then: itemize (i) equation 00 (i,j)L 0 (i,j) equation is simple. If A 0 is a trace then (r 0) -1 00 is the unique tracial state on 0 , and if A 0 is not a trace then 00 has no tracial states. (ii) For each i 1,2,,n , r 0p i is full in equation 0 (i',j)L 0 'i (i',j) . 41 equation (iii) For each j 1,2 , r 0q j is full in equation 0 (i,j')L 0 'j (i,j') . 42 equation itemize theorem32 proof The cases n 0,1,2 have been already proved. Let p 0 1- 1 np j and let 1 be the C --subalgebra of generated by (p 0 p 1 p n)(q 1 q 2) , so that equation ( 1, 1 ) ( 0 p 0 1 p 1 n p n ) ( 1 q 1 2 q 2 ). equation Let align L (0) (0,j) 0 j 1 , L 0 (0) (0,j) 0 j 1 . align We use Lemma 4.3 to find that equation 1 1,0 r 1,0 (i,j)L L (0) i j-1 p iq j . equation Then by Proposition 2.8, p 0 1p 0 and A 0 freely generate p 0p 0 . Hence by Proposition 3.2, p 0p 0 is simple. So (40) holds, where r 0 p 0 (1-p 0)r 1,0 and where the linear span of equation split p 0p 0 p 0p 0 1,0 (1-p 0) (1-p 0) 1,0 p 0p 0 (1-p 0) 1,0 p 0p 0 1,0 (1-p 0) (1-p 0) 1,0 (1-p 0) split 43 equation is dense in 0 . The application of Lemma 4.3 gives for each (i,j)L 0L 0 (0) a --ho -mo -morph -ism (i,j) (1) : 1,0 such that (i,j) (1) (r 1,0 p i) 1 (i,j) (1) (r 1,0 q j) , and then r 1,0 p 0 is full in equation 1,0 (i,j)L 0 (i,j) (1) . 44 equation For each (i,j)L 0 , (i,j) (1) extends to a --homomorphism (i,j) : 0 whose kernel is densely spanned by align p 0p 0 p 0p 0 1,0 (1-p 0) (1-p 0) 1,0 p 0p 0 (1-p 0) 1,0 p 0p 0 1,0 (1-p 0) (1-p 0)( (i,j) (1) )(1-p 0). align Since the algebra (44) contains both (1-p 0)( 1,0 (i,j)L 0 (i,j) (1) (1-p 0)) and (1-p 0) 1,0 p 0 , from the denseness of (44) in 0 we see that p 0 is full in 00 . Since p 0p 0 is simple, by Proposition 2.6 this implies that 00 is simple (hence when L 0 is empty, 0 is simple). From the facts, for (i,j)L 0 , that r 1,0 p i is full in equation 1,0 (0,j)L 0 (0) (0,j) (1) (i'.j)L 0 'i (i',j) (1) , equation which by a dimension argument contains a nonzero element of p 0 1,0 p 0 , and that every positive element of p 0p 0 is full in 00 , we obtain that r 0p i is full in (41), proving (ii). Because r 1,0 q j is full in equation 1,0 (0,j')L 0 (0) 'j (0,j) (1) (i.j')L 0 'j (i,j') (1) , equation which meets p 0 1,0 p 0 , we similarly obtain that r 0q j is full in (41), proving (iii). If A 0 is a trace then, by Propositions 3.2 and 3.4, p 0p 0 has unique tracial state and stable rank 1. Since p 0p 0 is full in 00 , by Proposition 2.5 the same hold for 00 (which is just 0 when L 0 is empty), and (r 0) -1 00 is then seen to be the unique tracial state. If A 0 is not a trace, then by Proposition 3.2, p 0p 0 has no tracial state, so neither does 00 have a tracial state. proof The following proposition proves Theorem 1. theorem33 Let equation (,) ( 1 p 1 n p n ) ( 1 q 1 m q m ), equation where n2 and m3 . Then the stable rank of is 1 . Let equation split L (i,j) i j 1 , L 0 (i,j) i j 1 . split equation Then equation 0 r 0 (i,j)L i j-1 p iq j equation where the centralizer of 0 has a unital, diffuse abelian subalgebra containing r 0p 1 . If L 0 is empty then 0 is simple and nonunital, and (r 0) -1 0 is the unique tracial state on 0 . If L 0 is not empty, then for every (i,j)L 0 there is a --homomorphism (i,j) : 0 such that (i,j) (r 0p i) 1 (i,j) (r 0q j) . Then: itemize (i) equation 00 (i,j)L 0 (i,j) equation is simple and nonunital, and (r 0) -1 00 is the unique tracial state on 00 . (ii) For each i 1,2,,n , r 0p i is full in 0 (i',j)L 0 'i (i',j) . itemize theorem33 proof We proceed by induction on (n,m) . If (n,m) 2 then Lemma 4.3 applies to prove the desired results. If (n,m)3 , take nm and let 1 be the C --subalgebra of generated by ((p 1 p 2) p 3 p n)(q 1 q m) . Thus equation ( 1, 1 )( 1 2 p 1 p 2 3 p 3 n p n ) ( 1 q 1 m q m ). equation By the inductive hypothesis, letting L , L 0 , L (1) , L 0 (1) , L ' , L 0' , L (2) and L 0 (2) be as in (35), we have equation 1 1,0 r 1,0 jL ' 1 2 j-1 (p 1 p 2)q j (i,j)L (1) i j-1 p iq j , equation and there is a unital, diffuse abelian subalgebra of the centralizer of 1,0 containing r 1,0 (p 1 p 2) . By Proposition 2.8, (p 1 p 2)(p 1 p 2) is freely generated by (p 1 p 2) 1(p 1 p 2) and (p 1 p 2) , so equation split (p 1 p 2)(p 1 p 2) ((p 1 p 2) 1,0 (p 1 p 2) jL ' 1 2 j-1 1 2 (p 1 p 2)q j ) ( 1 1 2 p 1 2 1 2 p 2 ). split equation Applying Lemma 4.4 yields equation (p 1 p 2)(p 1 p 2) 2,0 r 2,0 (i,j)L (2) i j-1 p iq j . equation Hence equation 0 r 0 (i,j)L i j-1 p iq j , equation where r 0 r 2,0 r 1,0 (1-p 1-p 2) and the linear span of the set in (27) is dense in 0 . Now the proof of this proposition follows verbatim the proof of Lemma 4.3 after equation (36), except we must also show that when L 0 is nonempty then 00 is nonunital. Suppose for contradiction that 00 is unital with identity e . Then er 0 , but ex exe xe for every x 0 . Thus e is in the center of 0 . But by the results of 5, the strong--operator closure of 0 (in the GNS representation associated to 0 ) is a II 1 --factor. This gives a contradiction, because e must be in the center of this von Neumann algebra. proof remark1 By the same proof, one shows that for every nonempty subset F of L 0 , the ideal (i,j)F (i,j) of 0 is nonunital. remark1 The following lemma can be proved from Proposition 4.5 using Proposition 2.8 in the same way that Lemma 4.4 was proved from Lemma 4.3. theorem34 Let equation (,) (A 0 1 p 1 n p n ) ( 1 q 1 m q m ), 46 equation where the centralizer A 0 has a unital, diffuse abelian subalgebra and where n1 , m2 . Or, let equation (,) (A 0 1 p 1 n p n ) (B 0 1 q 1 m q m ), 47 equation where the centralizer of each of A 0 and B 0 has a unital, diffuse abelian subalgebra and where n1 , m1 . Then equation 0 r 0 (i,j)L i j-1 p iq j , equation where the centralizer of 0 has a unital, diffuse abelian subalgebra containing r 0p 1 and another containing r 0q 1 . If L 0 is empty then 0 is simple. If L 0 is not empty, then for every (i,j)L 0 there is a --homomorphism (i,j) : 0 such that (i,j) (r 0p i) 1 (i,j) (r 0q j) . Then: itemize (i) equation 00 (i,j)L 0 (i,j) equation is simple and nonunital, and (r 0) -1 00 is the unique tracial state on 00 . (ii) For each i 1,2,,n , r 0p i is full in 0 (i',j)L 0 'i (i',j) . (iii) For each j 1,2,,m , r 0p j is full in 0 (i,j')L 0 'j (i,j') . itemize Moreover, if A 0 is a trace (and if, in the case of (47), B 0 is a trace), then has stable rank 1 and (r 0) -1 is the unique tracial state on 0 when L 0 is empty or 00 when L 0 is nonempty. Otherwise, 0 , respectively 00 , has no tracial state. theorem34 As promised in 1, a result similar to 4.5 holds for free products of more than two finite dimensional abelian C --algebras. theorem35 Let N , N3 , and for each 1,,N take a finite dimensional abelian C --algebra and faithful tracial state, equation (A , ) ,1 p ,1 ,2 p ,2 ,n() p ,n() , 48 equation where n() , n()2 . Let equation (,) 1 N(A , ) equation and let equation split L (j()) 1 N ; ; 1 N(1- ,j() ) 1 , 1ex L 0 (j()) 1 N ; ; 1 N(1- ,j() ) 1 . split 49 equation Then equation 0 r 0 jL j r j , equation where for j (j()) 1 NL , j 1- 1 N(1- ,j() ) and r j 1 Np ,j() . For each 1N and each 1kn() , 0 has a unital, diffuse abelian subalgebra which contains r 0p ,k . Moreover, has stable rank 1. If L 0 is empty, then 0 is simple and (r 0) -1 0 is the unique tracial state on 0 . If L 0 is nonempty, then for every j (j()) 1 NL 0 there is a --homomorphism j: 0 such that 1N j(p ,j() ) 1 . Then: itemize (i) equation 00 jL 0 j equation is simple and nonunital, and (r 0) -1 00 is the unique tracial state on 00 . (ii) For each 1N and each 1kn() , r 0p ,k is full in equation 0 jL 0 ()k j. equation itemize theorem35 proof The proof is by induction on N , and the case N 3 and the inductive step are proved simultaneously. Let N-1 be the C --subalgebra of generated by 1 N-1 A . Use the inductive hypothesis (or, when N 3 , Proposition 4.5 or Proposition 2.7) to find that equation ( N-1 , N-1 ) 1 N-1 (A , ). equation Then equation (,)( N-1 , N-1 ) (A N, N), equation and one uses Lemma 4.7 to find . proof One can generalize this to the free product of finitely many (say N ) algebras of the form A 0 , either, as Theorem 4.8 was proved, by using Lemma 4.7 and induction on N , or, as Lemma 4.4 was proved, by applying Theorem 4.8 and Lemma 3.2. One obtains the following result. theorem36 Let N , N3 , and for each 1,,N let (A , ) be either a finite dimensional abelian algebra with a faithful state as in (48) or equation (A , ) A ,0 ,1 p ,1 ,2 p ,2 ,n() p ,n() , 50 equation where n() , n()1 , and where the centralizer of A ,0 has a unital, diffuse abelian subalgebra. Let equation (,) 1 N(A , ) equation and let L and L 0 be as in (49). Then equation 0 r 0 jL j r j , equation where for j (j()) 1 NL , j 1- 1 N(1- ,j() ) and r j 1 Np ,j() . If each A ,0 is a trace then has stable rank 1. If L 0 is empty then 0 is simple, and if each A ,0 is a trace then (r 0) -1 0 is the unique tracial state on 0 . If some A ,0 is not a trace then 0 has no tracial states. If L 0 is nonempty, then for every j (j()) 1 NL 0 there is a --homomorphism j: 0 such that 1N j(p ,j() ) 1 . Then equation 00 jL 0 j equation is simple and nonunital, and if each A ,0 is a trace then (r 0) -1 00 is the unique tracial state on 00 . If some A ,0 is not a trace then 00 has no tracial states. theorem36 Restricting the above proposition to the case when each A ,0 is abelian, we obtain the following result about the free product of finitely many abelian C --algebras. theorem37 Let N , N2 , and for each 1N consider the abelian C --algebra and state (C(X ), ) , where is a regular Borel probability measure on X whose support is all of X and having at most finitely many atoms, each of which is an isolated point of X . Let equation (,) 1 N (C(X ), ). equation Let align L x (x ) 1 N x X , , 1 N(1- ( x )) 1 , 2ex L 0 x (x ) 1 N x X , , 1 N(1- ( x )) 1 . align Assume that no X is a one--point space, and also exclude the case when N 2 and X 1 and X 2 are both two--point spaces. Then has stable rank 1, and equation 0 r 0 xL x r x , 51 equation where x 1- 1 N(1- ( x )) . If L 0 is empty, then 0 is simple and has unique tracial state (r 0) -1 0 . If L 0 is nonempty, then there are distinct, surjective --homomorphisms, x: 0 , (xL 0) , such that equation 00 xL 0 x equation is simple and nonunital and has unique tracial state (r 0) -1 00 . theorem37 5. More general abelian C --algebras In this section we will investigate free products, equation (,) 1 N(C(X ), ), equation of abelian C --algebras and states, (C(X ), ) , each of which can be written as an inductive limit of the abelian algebras and states considered in Corollary 4.10. The criterion for simplicity of the free product of such abelian algebras is the same as for finite dimensional abelian algebras, namely, is simple if and only if there are no atoms x X of (1N) such that 1 N(1-( x ))1 . (Compare Corollary 4.10.) However, in the case when there are atoms x satisfying 1 N(1-( x )) 1 , we don't always get a corresponding copy of as a direct summand of . In fact we get such a direct summand, i.e. a minimal and central projection in , like r x in (51), corresponding to these atoms, if and only if each x is an isolated point of X . definition2 Let X be a compact Hausdorff topological space and let be a regular Borel probability measure on X . We say (X,) is an inverse limit of spaces and measures with isolated atoms if X is an inverse limit, X (X n, n) , of compact Hausdorff spaces X n and surjective, continuous maps n:X n 1 X n , (n) , such that, letting n:XX n be the resulting canonical surjective maps and letting n ( n) () be the push--forward measures, each n has at most finitely many atoms and each atom of n is an isolated point of X n , and, moreover, if xX n and if n -1 ( x ) has more than one point, then x is an atom of n . definition2 definition3 In each of the following cases, (X,) is an inverse limit of spaces and measures with isolated atoms. (One can easily cook up more intricate examples as well.) itemize (i) has no atoms; (ii) all the atoms of are isolated points of X ; (iii) X is separable and totally disconnected; (iv) X 0 n 1 1 2n 1 ,1 2n with the relative topology from , and has a unique atom at 0 . itemize definition3 theorem38 Let N , N2 , and for each 1N let (X , ) be a compact Hausdorff space and a regular Borel probability measure, each of which is an inverse limit of spaces and measures with isolated atoms. Assume each X has more than one point and each has support equal to all of X . Exclude the case when N 2 and X 1 and X 2 are both two--point spaces. Let equation (,) 1 N(C(X ), ). equation Let align I x (x ) 1 N 1 NX 1 N(1- ( x )) 1, ; each x is isolated in X , L x (x ) 1 N 1 NX 1 N(1- ( x ))1 I . align Then has stable rank 1, and equation 0 r 0 xI x r x , 52 equation where x 1- 1 N(1- ( x )) and where r x 1 Np x , with p x C(X ) the characteristic function of x . If L is empty, then 0 is simple and has unique tracial state (r 0) -1 0 . If L is nonempty, then for every xL there is a --homomorphism x: 0 such that whenever fC(X ) we have x(f) f(x ) . Moreover, equation 00 xL x equation is simple and nonunital, and has unique tracial state. Finally, for each nonempty subset FL , the ideal xF x of 0 is nonunital. theorem38 proof Let X (X ,n , ,n ) be an inverse limit with properties as in Definition 5.1, and let ,n :X X ,n and ,n ( ,n ) ( ) be the corresponding map and measure. Thus, we may regard C(X ,n ) as a unital C --subalgebra of C(X ) , where the state restricts to ,n , and C(X) n 1 C(X ,n ) . If yX ,n is an atom of ,n and if -1 ,n ( y ) contains no atoms of , then since -1 ,n ( y ) is clopen in X we may change (X ,n , ,n ) by substituting -1 ,n ( y ) for y and changing ,n accordingly. Then we must modify every (X ,n k , ,n k ) too. By doing so, we may assume without loss of generality that whenever n and yX ,n is an atom of ,n , then ,n -1 ( y ) contains an atom of . Let c ( x )xX , ,1N . Then c 1 . If x X appears as one of the coordinates in an element of LI , then 1- ( x ) (N-1)(1-c)1 , so ( x )(N-1)(1-c) . Therefore, LI is finite. Moreover, if has infinitely many atoms then their masses form a sequence tending to zero, so there is 0 such that whenever x X (1N) and (x ) 1 NLI then 1 N(1-( x )) 1 . Let x X be an atom of . If x is an isolated point of X , then for n large enough -1 ,n ( ,n (x ) ) x and hence the characteristic function of x , which we called p x , is in C(X ,n ) . We assume without loss of generality that this holds for every n and for every . If x is not an isolated point of X then ( -1 ,n ( ,n (x ) )) n 1 is a neighborhood base for x and, since is a regular measure whose support is all of X , we have equation ,n ( ,n (x ) ) ,n 1 ( ,n 1 (x ) ) ( x ) equation and n ,n ( ,n (x ) ) ( x ) . Thus, for n large enough we have, for every atom, x X of , equation ,n ( ,n (x ) ) ( x ) N. equation We assume without loss of generality this holds for every n and for every . Then, whenever x ,n X ,n is an atom of ,n and equation 1 N(1- ,n (x ,n ))1, 53 equation there is a unique (x ) 1 NLI such that x ,n ,n (x ) (1N) . Moreover, if (x ) 1 NI then the atoms ,n (x ) and x have the same mass, and if equality holds in (53) then each x is an isolated point of X . Let n be the C --subalgebra of generated by 1 N C(X ,n ) . Then equation n 1 N(C(X ,n ), ,n ), equation and by Corollary 4.10 and the facts discussed above we have equation n ( n,0 r n,0 xL n,x r n,x ) xI x r x , equation where I , x and r x are as in the statement of the proposition and where align L x (x ) 1 NL 1 N(1- ( x )) 1 , n,x 1- 1 N (1- ,n ( ,n (x ) )), r n,x 1 N p x ,n , align where p x ,n C(X ,n ) is the characteristic function of ,n (x ) . Also, letting L 0 LL , for each xL 0 there is a --homomorphism n,x : n,0 sending p x ,n p x to 1 (1N) . We have n n 1 and n 1 n . Let equation n,0 ' n,0 r n,0 xL n,x r n,x . equation Then n,0 ' n 1,0 ' , and (52) holds with 0 n 1 n,0 ' . If L then each n,0 ' n,0 is simple with unique tracial state, and thus their inductive limit 0 is simple with unique tracial state. Suppose L is nonempty. For each x (x ) 1 NL let x,n : n,0 ' be the --homomorphism sending p x ,n 1 (1N) . Then equation n,00 xL x,n equation is simple with unique tracial state. Clearly x,n 1 is an extension of x,n , so taking the inductive limit we get, for each xL , a --homomorphism x: 0 . When restricted to C(X ) , x gives evaluation at x X . Then 00 is the inductive limit of the algebras n,00 , and thus is simple with unique tracial state. We now show that 00 is nonunital. First consider the case when L is nonempty. Then for each x (x ) 1 NL , n,x decreases to the limit 1 N(1- ( x )) but is never equal to this quantity. Suppose for contradiction that e 00 is the identity of 00 . Let n and a n,00 . Then there is m n such that n,x m,x , and thus r n,x -r m,x m,00 is a nonzero projection. Since a n,x , since n,x (r n,x ) 1 and since r n,x is a minimal projection of n,00 , we must have ar n,x 0 , and hence a(r n,x -r m,x ) 0 . But e(r n,x -r m,x ) r n,x -r m,x , so equation 1 r n,x -r m,x (e-a)(r n,x -r m,x ) e-a . equation This contradicts the fact that e n n,00 . If L then L 0 is nonempty, so each n,00 is nonunital. But this implies that their inductive limit, 00 , is nonunital. The same technique shows that each ideal xF x of 0 is nonunital. proof Although the above proposition was stated only for free products of abelian C --algebras, a similar result is easily proved for free products of inductive limits of the algebras of the form A 0 that were considered, for example, in Theorem 4.9. 6. Free products of infinitely many algebras In this section we consider the reduced free product of infinitely many finite dimensional abelian C --algebras. Although such free products of infinitely many algebras can fail to be simple, they never get a copy of as a direct summand, and hence their proper, nontrivial ideals, if any, are always nonunital. Moreover, the center of the free product algebra is always trivial, even when its von Neumann algebra closure (i.e. the strong--operator closure of the GNS representation) has nontrivial projections that are both minimal and central. theorem39 For each let (A , ) be a finite dimensional abelian algebra with faithful state as in (48). Let equation (,) 1 (A , ). equation Then has stable rank 1. Let equation L (j()) 1 ; ; 1 (1- ,j() )1 . equation If L is empty then is simple and is the unique tracial state on . Otherwise, if L is nonempty, then for each j (j()) 1 L there is a --homomorphism j: such that j(p ,j() ) 1 for every . Then equation 00 jL j equation is simple and nonunital. Moreover, letting equation 0 1- jL ( 1 (1-(,j()))), equation 0 -1 00 is the unique tracial state on 00 . Finally, for every nonempty subset FL , the ideal jF j is nonunital. theorem39 proof The proof follows in a straightforward manner by using Theorem 4.8 and taking inductive limits. To show that 00 and each of the other proper, nontrivial ideals of are nonunital, an argument like that in the proof Theorem 5.3 (in the case L nonempty) is used. proof Of course, similar results for free products of infinitely many abelian algebras of the form considered in 5 or infinitely many algebras of the form A 0 considered in Theorem 4.9 are easily obtained. 7. Conjectures This section contains a couple of related open problems which seem likely to have solutions, though I don't yet see how to find them. theorem40 Let C(X) and C(Y) be unital, abelian C --algebras having faithful states given by probability measures X on X and Y on Y . Let equation (,) (C(X), X) (C(Y), Y). equation Then a necessary and sufficient condition for to be simple is that for every xX and yY , X( x ) Y( y ) 1 . theorem40 theorem41 The conditions in Conjecture 7.1 are necesary for simplicity of . theorem41 proof Suppose the conditions of 7.1 are not satisfied. Let CX be the set of atoms of X . If C is finite then let D C . If C is infinite then let denote the closure of C in X and let D be a totally disconnected, separable, compact Hausdorff space equipped with a continuous surjective map :D . (That such a space exists is a well--known result.) For each cC , choose f(c) -1 (D) . Let D be the measure on D given by D( f(c) ) X( c ) and D(Df(C)) 0 . Replace D by the support of D . Let X' be the measure on X that when restricted to XC gives the same measure as X and such that X' (C) 0 , and let X' be the support of X' . Let X a be the compact Hausdorff space that is the disjoint union of X' and D , and let X a be the measure on X a obtained from X' and D . Let X:X aX be the surjective, continous map composed of the inclusion X'X and :DX . Then X a is the push--forward measure, X a ( X) ( X) . Therefore X induces an injective, unital --homomorphism, X:C(X)C(X a) , preserving the states defined by the measures X a and X . Do the same for Y , getting Y:C(Y)C(Y A) . Now Theorem 5.3 applies to the free product equation (',') (C(X a), X a ) (C(Y a), Y a ). equation If the condition of Conjecture 7.1 is not satisfied, then there are xX a and yY a such that X a ( x ) Y a ( y )1 , which by 5.3 implies the existence of a --homomorphism :' that when restricted to C(X a) is evaluation at x . Since ' is faithful and X and Y are trace--preserving inclusions, they induce an inclusion ' . Then is a nonzero --homomorphism, so is not simple. proof We now look at free products of finite dimensional algebras, and we use the notation of 6 for a faithful state on a finite dimensional algebra. Thus, if D j 1 K M n j () , we write equation (D,) j 1 K M n j () j,1 ,, j,n j equation to mean that the restriction of to the j th summand of D is given by (H) , where is the unnormalized trace on M n j () and where H is the diagonal matrix with j,1 ,, j,n j down the diagonal. theorem42 Let equation split (A 1, 1) j 1 K 1 M n j () j,1 ,, j,n j , (A 2, 2) j 1 K 2 M m j () j,1 ,, j,m j split equation be finite dimensional C --algebras with faithful states and let equation (,) (A 1, 1) (A 2, 2) 54 equation be the reduced free product C --algebra. Then necessary and sufficient conditions for to be simple are that if n j 1 for some j then for every 1kK 2 equation 1 1- j,1 i 1 m k 1 k,i , equation and if m j 1 for some j then for every 1kK 1 equation 1 1- j,1 i 1 n k 1 k,i . equation theorem42 If the above conjecture is true, then in particular is simple whenever each n j 1 and each m j 1 . This conjecture is inspired by the results of 5 and 6, which show that necessary and sufficient conditions for the von Neumann algebra free product analogous to (54) to be a factor are that if n j 1 for some j then for every 1kK 2 equation 1 1- j,1 i 1 m k 1 k,i 55 equation and if m j 1 for some j then for every 1kK 1 equation 1 1- j,1 i 1 n k 1 k,i . 56 equation Moreover, using this von Neumann algebra result and an argument similar to the one used in Proposition 7.2, we see that (55) and (56) are necessary conditions for the simplicity of . For some results about certain reduced free product C --algebras with respect to non--faithful states, see 10 . amsplain