Lemma 45.14.2. Assume given (D0), (D1), and (D2') satisfying axioms (A1), (A2), (A3), and (A4). There is a unique rule which assigns to every smooth projective $X$ over $k$ a graded ring homomorphism
compatible with pullbacks such that $ch^ H(\alpha ) = \gamma (ch(\alpha ))$ for $\alpha $ in $K_0(\textit{Vect}(X))$.
Proof. Recall that we have an isomorphism
see Chow Homology, Lemma 42.58.1. It is an isomorphism of rings by Chow Homology, Remark 42.56.5. We define $\gamma $ by the formula $\gamma (\alpha ) = ch^ H(\alpha ')$ where $ch^ H$ is as in Lemma 45.14.1 and $\alpha ' \in K_0(\textit{Vect}(X))$ is such that $ch(\alpha ') \cap [X] = \alpha $ in $\mathop{\mathrm{CH}}\nolimits ^*(X) \otimes \mathbf{Q}$.
The construction $\alpha \mapsto \gamma (\alpha )$ is compatible with pullbacks because both $ch^ H$ and taking Chern classes is compatible with pullbacks, see Lemma 45.14.1 and Chow Homology, Remark 42.59.9.
We still have to see that $\gamma $ is graded. Let $\psi ^2 : K_0(\textit{Vect}(X)) \to K_0(\textit{Vect}(X))$ be the second Adams operator, see Chow Homology, Lemma 42.56.1. If $\alpha \in \mathop{\mathrm{CH}}\nolimits ^ i(X)$ and $\alpha ' \in K_0(\textit{Vect}(X)) \otimes \mathbf{Q}$ is the unique element with $ch(\alpha ') \cap [X] = \alpha $, then we have seen in Chow Homology, Section 42.58 that $\psi ^2(\alpha ') = 2^ i \alpha '$. Hence we conclude that $ch^ H(\alpha ') \in H^{2i}(X)(i)$ by Lemma 45.12.5 as desired. $\square$
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