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Is there some results for the cyclic homology group $HC_1(A)$, for example, when it is zero, or which case we can compute out it explictly, here $A$ is a commutative algebra over the complex field.

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Your title is not as descriptive as it could be. –  Qiaochu Yuan Jul 12 '11 at 19:51

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There is a formula in the commutative case, $HC_1(A) \cong \Omega^1(A)/(dA)$. Namely there is a Connes exact sequence $HH_0(A) \to HH_1(A) \to HC_1(A) \to 0$. In the case of a commutative algebra over a field $HH_0(A) \cong A$ and $HH_1(A) \cong \Omega^1(A)$. The left hand map is d, giving the above formula. For smooth algebras, you have a similar formula for all of the cyclic homology groups. You can surely find all of this in Loday.

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You should provide more information about your algebra $A$ if you intend a useful, non-generic answer. I doubt you will find a textbook exposition---but for appropriate classes fo algebras, one can surely direct you to papers where computations are carried out.

The one general approach to computing $HC_\bullet$ inthe commutative case is to mimick rational homotopy theorty and construct a model of your algebra $A$, that is, a differential graded algebra $\mathcal A$, and then use the fact that $HC(A)$ and $HC(\mathcal A)$ (this last homology is the homology of graded differential algebras) are isomorphic.

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