Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
Your title is not as descriptive as it could be. –  Qiaochu Yuan Jul 12 '11 at 19:51
add comment

2 Answers

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.

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.