The Cuntz algebra $O_n$ is the C*-algebra generated by n isometries $S_1$, ..., $S_n$ such that $S_i^* S_j=\delta_{i,j}$ and $\sum_{i=1}^nS_iS_i^*=1$. Cuntz proved that this algebra has the following K-theory: $K_0(O_n)=\mathbb{Z}_{n-1}$, $K_1(O_n)=0$. Does any one know what are the generators of $K_0(O_n)$ ($n\geq 3$)?
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$\begingroup$ I think it should correspond to the idempotent s_1s_1* $\endgroup$– Benjamin SteinbergCommented Feb 20, 2016 at 12:16
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$\begingroup$ @ Benjamin Steinberg: I am not sure. Probably I am wrong, but S_S_1^* is equivalent to 1 (S_1 is an isometry). However, the class of [1] is 0. $\endgroup$– John N.Commented Feb 20, 2016 at 12:19
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$\begingroup$ Yes you are right it is equivalent to 1. But isn't the class of 1 the generator not 0 for n>2? All n of those idempotents are equivalent to 1 so the cuntz relation gives n[1]=[1]. $\endgroup$– Benjamin SteinbergCommented Feb 20, 2016 at 12:25
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$\begingroup$ Ok. So we can choose 1, and say that it is a generator of the K_0 group. Could you give me a reference or expand your comment in an answer that proves that the class of 1 is not zero? Thank you for the help $\endgroup$– John N.Commented Feb 20, 2016 at 13:28
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$\begingroup$ I am not sure where to find a reference for the operator case. You can find this for Leavitt path algebras in the survey of Abrams on the first decade of Leavitt path algebras. I am sure the operator theory proof is essentially the same but as an algebraist I wouldn't know how to write it in a technically correct way. $\endgroup$– Benjamin SteinbergCommented Feb 20, 2016 at 13:36
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1 Answer
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This is answered on the first page of Cuntz's paper "K-Theory for Certain C*-Algebras" (Ann. of Math. 113 (1981), 181-197). If you google "cuntz algebra k-theory" it is the first result.
answered Feb 20, 2016 at 16:38