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Following on from my last two questions link text and link text: Is it correct (and useful) to say that the relationship between Connes' cyclic cohomology approach to de Rham cohomology and Woronowicz's differential calculi approach, is a noncommutative generalisation of the difference between ordinary differentials and Kahler differentials respectively?

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I suppose I should add here that by cyclic cohomology I mean periodic cohomology, since this is the actual object that generalises de Rham cohomology – Abtan Massini Nov 20 '09 at 19:51
This is too hand-wavy to have a sensible answer, in my opinion. Maybe you could make more precise what exactly you have in mind? – Mariano Suárez-Alvarez Nov 23 '09 at 22:39
Ok, let me ask a more specific question: Is the cyclic cohomology of the coordinate ring $A$ of an affine variety equal to its noncommutative de Rham cohomology (noncommutative Kahler cohomology) $\Omega_u/[\Omega_u,\Omega_u]$, where $\Omega_u$ is the universal calculus of $A$? – Abtan Massini Nov 24 '09 at 14:00
up vote 3 down vote accepted

I'll answer here instead of in a comment, because of the character limit...

If $A$ is the coordinate algebra of an affine variety which is smooth and the base field $k$ contains $\mathbb{Q}$, then $$HC\_n(A) \cong \Omega^n\_{A/k} / d\Omega^{n-1}\_{A/k} \oplus H\_{\mathrm{dR}}^{n-2}(A) \oplus H\_{\mathrm{dR}}^{n-4}(A) \oplus \cdots$$ for all $n\geq0$. Here $\Omega^n\_{A/k}$ is the $n$-th exterior power of the $A$-module $\Omega^1\_{A/k}$ of Kähler differentials of $A$ over $k$, and $H\_{\mathrm{dR}}^\bullet(A)$ denotes the cohomology of the complex $$A\to \Omega^1\_{A/k} \to \Omega^2\_{A/k} \to \Omega^3\_{A/k} \to\cdots $$ whose differential is the exterior differential.

The summand $\Omega^n\_{A/k} / d\Omega^{n-1}\_{A/k}$ appearing in $HC\_\bullet(A)$ is slightly ugly. If we consider instead periodic cyclic homology, we get instead $$HC^{\mathrm{per}}\_n(A) \cong \bigoplus\_{i\in\mathbb{Z}}H\_{\mathrm{dR}}^{n+2i}(A),$$ which is manifestly nicer.

(If $k$ is not of characteristic zero you only have a spectral sequences going from de Rham cohomology to the cyclic homology). If, on the other hand, $A$ is not smooth then André-Quillen cohomology intervenes, and everythng is rather more complicated.)

You should really take a look at Loday's book.

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For your specific question (but I will deal exclusively with cyclic homology): the answer is mostly no, as measured by Hochschild homology. There is a short exact sequence $$0\to \overline{\mathit{H}}\mathit{DR}_n(A)\to \overline{HC}_n(A)\to \overline{HH}_n(A)\to0$$ where $\mathit{HDR}$ means non-commutative de Rham homology, $\mathit{HC}$ cyclic homology, $\mathit{HH}$ is Hochschild homology, and the bars mean «reduced». You will find the details in Loday's book, section 2.6, Chap. 2.

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For every algebra $A$, does there exist a differential complex whose (co)homology is equal to the cyclic (co)homology of $A$? – John McCarthy Nov 24 '09 at 14:56
@John: I am probably not understanding your question... The cyclic homology of an algebra is defined to be the homology of a specific complex (same thing with cyclic cohomology) – Mariano Suárez-Alvarez Nov 24 '09 at 16:46
I suppose what I'm asking (and what I think Abtan is asking) is how cohomology groups arising from differential calculi are related to cyclic homology. They both generalise (more or less) the same classical objects, and should (I feel) have some connection. – John McCarthy Nov 24 '09 at 19:41
The short exact sequence I wrote above is precisely the relationship between the cohomology groups arising from the universal differential calculus and cyclic homology. – Mariano Suárez-Alvarez Nov 24 '09 at 21:36
Let me ask one last question then: In the classical case, what geometric object does the cyclic cohomology of the coordinate ring of an affine variety correspond to? For an explicit answer to this question I am willing to award the bounty. – Abtan Massini Nov 25 '09 at 19:10

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