Timeline for Karoubi versus Kasparov K-theory
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 22, 2013 at 20:10 | comment | added | Johannes Ebert | Yes, that is a convenient picture, but not the one I wish to consider (I have a concrete application in mind). | |
| May 22, 2013 at 14:05 | comment | added | Mkouboi | ... in fact $F$ is self-adjoint! | |
| May 22, 2013 at 14:03 | comment | added | Mkouboi | @Johannes: Sorry, you are right, that was a very stupid mistake. In fact, the most convenient picture of $K^{p,q}(X)$ to use here is the Fredholm one given in Karoubi's paper Algèbres de Clifford et operateurs de Fredholm, CRAS, t. 267, p. 305 (1968) (Section II): $K^{p,q}(X)$ is generated by pairs $(\mathcal{E},F)$ where $\mathcal{E}\longrightarrow X$ is a graded Hilbert bundle equipped with a $Cl^{p,q}$-module structure, and $F\in C(X,Fred_\mathcal{E})$ is a Fredholm section of degree $1$ anticommuting with the generators of $Cl^{p,q}$. Such a pair is but a $KK$-class. | |
| May 21, 2013 at 16:50 | comment | added | Johannes Ebert | This cannot possibly be correct, because there is an important piece of structure missing: there is no $Z/2$-grading on your $H$. This is absolutely essential. Here is the reason: if $X$ is compact and if $(E,\phi)$ is a \emph{finite-dimensional} $Cl^{p,q}-C(X)$-bimodule, then $\mathcal{K}(E)=\mathcal{L}(E)$. Thus EACH graded operator $F$ yields a Kasparov module $(E,\phi,F)$, and they all represent the same element in $KK$. If one wants to represent a KK-class by a finite-dimensional Hilbert module, all the information is contained in the grading. -1; sorry about that. | |
| May 21, 2013 at 16:08 | history | edited | Mkouboi | CC BY-SA 3.0 |
added 12 characters in body; added 1 characters in body
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| May 21, 2013 at 16:02 | history | answered | Mkouboi | CC BY-SA 3.0 |