# Does the non-commutative Chern class depend on the choice of connection?

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?

-

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.
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