Connes v Woronowicz - Cyclic Cohomology v Diff Calculi - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T04:41:59Z http://mathoverflow.net/feeds/question/6271 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/6271/connes-v-woronowicz-cyclic-cohomology-v-diff-calculi Connes v Woronowicz - Cyclic Cohomology v Diff Calculi Abtan Massini 2009-11-20T14:40:46Z 2009-11-26T02:54:59Z <p>Following on from my last two questions <a href="http://mathoverflow.net/questions/5865/classical-calculi-as-universal-quotients" rel="nofollow">link text</a> and <a href="http://mathoverflow.net/questions/6074/kahler-differentials-and-ordinary-differentials" rel="nofollow">link text</a>: 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?</p> http://mathoverflow.net/questions/6271/connes-v-woronowicz-cyclic-cohomology-v-diff-calculi/6689#6689 Answer by Mariano Suárez-Alvarez for Connes v Woronowicz - Cyclic Cohomology v Diff Calculi Mariano Suárez-Alvarez 2009-11-24T14:03:09Z 2009-11-24T14:03:09Z <p>For your specific question (but I will deal exclusively with cyclic <em>homology</em>): 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.</p> http://mathoverflow.net/questions/6271/connes-v-woronowicz-cyclic-cohomology-v-diff-calculi/6859#6859 Answer by Mariano Suárez-Alvarez for Connes v Woronowicz - Cyclic Cohomology v Diff Calculi Mariano Suárez-Alvarez 2009-11-26T00:26:19Z 2009-11-26T02:54:59Z <p>I'll answer here instead of in a comment, because of the character limit...</p> <p>If $A$ is the coordinate algebra of an affine variety which is <em>smooth</em> 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. </p> <p>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.</p> <p>(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.)</p> <p>You should really take a look at Loday's book.</p>