/sci/ - Science & Math

>>11445470
Thanks for the effort, but I've realized earlier today that trace-class operators are compact, so their operator indices vanish and hence their [math]K[/math]-homology class coincides with that of the identity. There is no way the inclusion can be a quasi-isomorphism in this case.
I believe the trace-class constraint is way too strong. I will need to regularize the trace so it's defined on just the smooth subalgebra [math]\mathcal{R}[/math] instead.