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In classical geometry the calculation of the Chern classes of a vector bundle using a connection is independent of the choice of connection. Does any such result hold for projective modules in non-commutative geometry?

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up vote 7 down vote accepted


You can see the construction in detail, for example, in Max Karoubi's ‘Homologie cyclique et $K$-theorie’ (Asterisque 149, SMF; you can get this from his web page), where he constructs the Chern classes $K_0(A)\to H(A)$ using connections much à Chern-Weyl (Here $H(A)$ is the non-commutative de Rham theory, or one of the various cyclic homologies of $A$) He also constructs higher Chern classes on the higher algebraic $K$-theory by a similar procedure. The book by Loday on cyclic homology also covers this.

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This leaves me a little confused. I am familiar with noncommutative vector bundles (ie projective modules) endowed with differential calculi and non-commutative connections, and then calculating the Chern Classes from this information. On the other hand, I know what cyclic cohomology is and why it can be viewed as a non-commutative de Rham cohomology. What I don't really understand is how the two link up: the former is more general than the latter? – John McCarthy Nov 18 '09 at 1:01
... and if so, then Karoubi/Loday only prove the result in a special case? – John McCarthy Nov 18 '09 at 1:04
There are maps connecting non commutative de Rham cohomology and cyclic homology (in fact, there is an isomorphism between $H_{\mathrm{dR}}(A)$ and the kernel of the map $B:HC_\bullet(A)\to HH_{\bullet+1}(A)$ appearing in the Connes long exact sequence---here $HH$ is Hochschild homology), and constructions of the Chern character from $K_0(A)$ to both $H_{\mathrm{dR}}(A)$ and $HC(A)$ which are compatible with those maps. A non trivial part of Max's Asterisque is spent in checking lots of such compatibilities. – Mariano Suárez-Alvarez Nov 18 '09 at 1:08
Oh, when I say non commutative de Rham theory I mean something not really involving 'differential calculi' (or rather involving only the universal one): take $\Omega(A)$ to be the kernel of the multiplication map $A\otimes A\to A$, which is an $A$-bimodule, and let $\Omega^\bullet(A)$ be the tensor algebra over $A$ of $\Omega(A)$, which is a differential graded algebra. Now let $\Omega^\bullet(A)_{\mathrm{ab}}=\Omega^\bullet(A)/[\Omega^\bullet(A),\Omega^\bu‌​llet(A)]$ be the 'abelianization', which is a complex, whose cohomology is the non commutative de Rham cohomology I had in mind. – Mariano Suárez-Alvarez Nov 18 '09 at 1:15

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