Cyclic cohomology may be defined in several ways: the easiest way to define it is via a subcomplex $C^*_{\lambda}$of Hochschild complex consisting from cyclic cochains. There are also other definitions for example using the cyclic bicomplex or $(b,B)$ (mixed) complex. Relevant definition can be found here. One can define the so called periodicity operator $S:HC^{n}(A) \to HC^{n+2}(A)$ and using this operator one can define the periodic cyclic cohomology as the inductive limit of $HC^{2n}(A)$ or $HC^{2n+1}$ (where the arrows of inductive system are just iterations of $S$).

Why is periodic cyclic cohomology of $A$ isomorphic with the cohomology of the totalization of $(b,B)$-bicomplex?

Several remarks: 1. To define periodic theory one uses the bicomplex which is no longer concentrated in the first quadrant (see again my previous question linked above).
2. I would be happy if I could see the map yielding quasi-isomorphism explicitly.
3. There are still another variants of cyclic theory such as negative theory for instance. If there is a way to prove that the two approaches: using $(b,B)$ bicomplex and ordinary cyclic complex and $S$ operator are equivalent, without invoking the negative theory, I would be happy to see it.


The original reference is Theorem 40, c) of http://www.alainconnes.org/docs/noncommutative_differential_geometry.pdf

A maybe more pedagogical version is Theorem 29 c) of http://www.alainconnes.org/docs/book94bigpdf.pdf

  • $\begingroup$ Thank you for your answer. I've postponed accepting your answer since I was not quite familiar with spectral sequences and I also wanted to go through all details in this theorem. It requires two auxiliary lemmas (Lemma 23 and 25). After reading some texts about spectral sequence I managed to understand a) in Theorem 29. For b) I would like to know why $\varphi$ is cohomologous in $F^qC$ to some $\psi$ in $C^{p-q,q}$ (why does it follow from a)?). For c) I would like to know which map from $H^p(F^qC)$ to $H^p(F^qC)$ Connes had in mind? And also why this is enough to conclude the proof? $\endgroup$ – truebaran Mar 5 '18 at 20:58
  • $\begingroup$ I would be happy if I could see some more details of the argument (which is rather sketchy) $\endgroup$ – truebaran Mar 5 '18 at 20:59

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