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The Chern class yields an isomorphism $K_0(X)\otimes \mathbb Q\cong \bigoplus_{i\ge 0} Chow^i(X)\otimes \mathbb Q$ (for a smooth variety $X$ over a field?), whereas the latter group is isomorphic to $Hom_{Chow\otimes \mathbb Q}(M(X),M(\mathbb P^n))$ for $n\ge \dim X$. Is there a conceptual explanation for the (composite) isomorphism obtained (i.e. of the expression of $K_0$ in terms of cycles on $X\times \mathbb P^n$)?

Certainly, the sides of this isomorphism are related somehow; yet I don't know whether 'standard' arguments can identify them directly without mentioning the sum of Chow groups.

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One very conceptual explanation is that both $X \mapsto CH^*(X)$ and $X \mapsto K_0(X)[\beta^{\pm 1}]$ are oriented cohomology theories corresponding to the additive and multiplicative formal group law respectively. But over a field of characterstic 0, every formal group law is isomorphic to the additive one because we can define its exponential. This is the Chern character: a very fancy exponential. –  YBL Jan 3 '11 at 23:47

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If we consider just the question of constructing the map from $K_0(X)$ to $Hom_{Chow\otimes \mathbb Q}(M(X),M(\mathbb P^n)) $, and leave the question of proving that it is an isomorphism to later, I would say that this follows directly from the expression for the Chern classes of a vector bundle via the splitting principle. I.e., given a vector bundle $E$ of rank $d$ on $X$ look at the $d$-th power of the tautological bundle $\xi$ on $\mathbb P(E) $ as usual. This gives a cycle on $\mathbb P(E) $ which can be written as a linear combination of chern classes times lower powers of $\xi$. We can identify $CH^d(\mathbb P(E))$ with the appropriate $Hom$ of motives by, for example, embedding $\mathbb P(E) $ into $X\times\mathbb P^n$ for some large $n$. (Assuming we don't want to simply appeal to the expression for the motive of a projective bundle.) So it seems to me that if the map is given by sending $E$ to $\xi^d$, viewed as a morphism of motives. I imagine that one can then get properties like Whitney sum in terms of this homomorphism of motives by the classical geometric arguments. (So that the usual expression in terms of individual Chern classes would only appear after the fact.)

I don't see a way to prove that the map is an isomorphism without going through an analog of one of the usual proofs.

Not sure if this is the kind of thing you were looking for.

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I do not know whether this is the best possible answer, but it is interesting; thank you! –  Mikhail Bondarko Jan 4 '11 at 6:02

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