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208 J. Cuntz6 Representations as crossed productsFor each n, define the endomorphism ' n of Q N by ' n .x/ D s n xs n . Since ' n' m D ' nmthis defines an inductive system.Definition 6.1. We define xQ N as the inductive limit of the inductive system .Q N ;' n /.By construction we have a family n of natural inclusions of Q N into the inductivelimit xQ N satisfying the relations nm ' n D m . We denote by 1 k the element k .1/ ofxQ N . We have that 1 k D kl .e l / 1 kl . The union of the subalgebras 1 kxQ N 1 k is densein xQ N and 1 k 1 kl for all k; l. In order to define a multiplier a of xQ N it thereforesuffices to define a1 k and 1 k a for all k.We can extend the isometries s n naturally to unitaries Ns n in the multiplier algebraM. xQ N / of xQ N by requiringNs n 1 k D k .s n /; 1 k Ns n D kn .s n /:Note that this is well defined because, using the identities s n e l D e nl s n and ' l .s n / Ds n e l ,wehave.1 k Ns n /1 l D kn .s n / l .1/ D knl .s n e l e kn / D knl .e nl s n e kn / D k .1/ l .s n / D 1 k .Ns n 1 l /:Proposition 6.2. The elements Ns n , n 2 N, define unitaries in M. xQ N / such that Ns n Ns m DNs nm and such that Ns n Ns m DNs m Ns n.Defining Ns a DNs n Ns m for a D n=m 2 Q C we define unitaries in M. xQ N / such thatNs a Ns b DNs ab for a; b 2 Q C .Proof. In order to show that Ns n is unitary it suffices to show that 1 k Ns n Ns n1 k Ns n Ns n D 1 k for all k, n.D 1 k andWe can also extend the generating unitary u in Q N to a unitary in the multiplieralgebra M. xQ N /. We define the unitary Nu in M. xQ N / by the identityNu 1 k D k .u k /:We can also define fractional powers of Nu by settingNu 1=n 1 kn D kn .u k /:Proposition 6.3. For all a 2 Q Cand b 2 Q we have the identityProof. Check that Ns 1=n Nu DNu 1=n Ns 1=nNs a Nu b DNu ab Ns a :Following Bost–Connes we denote by P C Qthe ax C b-group P C 1 bQ D j a 2 Q 0 aC ;b2 Q :