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206 J. Cuntzprime numbers p and q, the size of p and q (or even the bigger number among p and q)from the corresponding generators s p and s q . In fact, the C -algebra generated by thes n , n 2 N is the infinite tensor product of one Toeplitz algebra for each prime numberand in this C -algebra there is no way to distinguish the s p for different p.However, the fact that we have added u to the generators allows to retrieve the nfrom s n using the KMS-condition.Definition 4.1. Let .A; t / be a C -algebra equipped with a one-parameter automorphismgroup . t /, a state on A and ˇ 2 .0; 1. We say that satisfies the ˇ-KMSconditionwith respect to . t /, if for each pair x;y of elements in A, there is a holomorphicfunction F x;y , continuous on the boundary, on the strip fz 2 C j Im z 2 Œ0; ˇgsuch thatF x;y .t/ D .x t .y//; F x;y .t C iˇ/ D . t .y/x/for t 2 R.Proposition 4.2. Let 0 be the unique tracial state on F and define on Q N by D 0 ı E. Let . t / denote the one-parameter automorphism group on Q N definedby t .s n / D n it s n and t .u/ D u. Then is a 1-KMS-state for . t /.Proof. According to [2] 5.3.1, it suffices to check that.x i .y// D .yx/for a dense *-subalgebra of analytic vectors for . t /. Here . z / denotes the extensionto complex variables z 2 C of . t / on the set of analytic vectors.However it is immediately clear that this identity holds for x;y linear combinationsof elements of the form as n or sm b, a; b 2 F . Such linear combinations are analyticand dense.Theorem 4.3. There is a unique state on Q N with the following property.There exists a one-parameter automorphism group . t / t2R for which is a1-KMS-state and such that t .u/ D u, t .e n / D e n for all n and t. Moreover wehave• is given by D 0 ı E where 0 is the canonical trace on F ,• the one-parameter group for which is 1-KMS, is unique and is the standardautomorphism group considered above, determined by t .s n / D n it s n and t .u/ D u:Proof. Since t acts as the identity on e n and on u, it is the identity on F D C .u; fe n g/.If is a KMS-state for . t /, it therefore has to be a trace on F . It is well-known (andclear) that there is a unique trace 0 on F . For instance, the relationXu k e n u k D 10k