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Jul 14, 2011 at 12:39 comment added Paul Siegel To be precise one has Thom isomorphisms in both K-theory and De Rham cohomology, but they are not compatible with each other in the sense that $ch \circ Thom_{KT} \neq Thom_{DR} \circ ch$. The Todd class is precisely the "correction term" which makes this equation true.
Jul 11, 2011 at 15:42 comment added Aaron Mazel-Gee Thanks -- your last paragraph is exactly the nail in the coffin of this idea that I was expecting. By the way, I've heard it said that whenever you see the Todd class it just means that you're looking at cohomology when you should be looking at K-theory, and that the Todd class should be thought of as nothing more than a "correction factor".
Jun 30, 2011 at 13:28 history answered Paul Siegel CC BY-SA 3.0