Proposition 45.12.4. In the situation above assume $A(X)$ is a $\mathbf{Q}$-algebra for all $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$. Then there is a unique rule which assigns to every $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ a “chern character”

$ch^ A : K_0(\textit{Vect}(X)) \longrightarrow \prod \nolimits _{i \geq 0} A^ i(X)$

with the following properties

1. $ch^ A$ is a ring map for all $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$.

2. If $f : X' \to X$ is a morphism of $\mathcal{C}$, then $f^* \circ ch^ A = ch^ A \circ f^*$.

3. Given $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and $\mathcal{L} \in \mathop{\mathrm{Pic}}\nolimits (X)$ we have $ch^ A([\mathcal{L}]) = \exp (c_1^ A(\mathcal{L}))$.

Proof. Let $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and let $\mathcal{E}$ be a finite locally free $\mathcal{O}_ X$-module. We first show how to define the rank $r^ A(\mathcal{E}) \in A^0(X)$. Namely, let $X = \bigcup X_ r$ be the decomposition into open and closed subschemes such that $\mathcal{E}|_{X_ r}$ has constant rank $r$. Since $X$ is quasi-compact, this decomposition is finite, say $X = X_0 \amalg X_1 \amalg \ldots \amalg X_ n$. Then $A(X) = A(X_0) \times A(X_1) \times \ldots \times A(X_ n)$. Thus we can define $r^ A(\mathcal{E}) = (0, 1, \ldots , n) \in A^0(X)$.

Let $P_ p(c_1, \ldots , c_ p)$ be the polynomials constructed in Chow Homology, Example 42.43.6. Then we can define

$ch^ A(\mathcal{E}) = r^ A(\mathcal{E}) + \sum \nolimits _{i \geq 1} (1/i!) P_ i(c^ A_1(\mathcal{E}), \ldots , c^ A_ i(\mathcal{E})) \in \prod \nolimits _{i \geq 0} A^ i(X)$

where $ci^ A$ are the Chern classes of Proposition 45.12.1. It follows immediately that we have property (2) and (3) of the lemma.

We still have to show the following three statements

1. If $0 \to \mathcal{E}_1 \to \mathcal{E} \to \mathcal{E}_2 \to 0$ is a short exact sequence of finite locally free $\mathcal{O}_ X$-modules on $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$, then $ch^ A(\mathcal{E}) = ch^ A(\mathcal{E}_1) + ch^ A(\mathcal{E}_2)$.

2. If $\mathcal{E}_1$ and $\mathcal{E}_2 \to 0$ are finite locally free $\mathcal{O}_ X$-modules on $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$, then $ch^ A(\mathcal{E}_1 \otimes \mathcal{E}_2) = ch^ A(\mathcal{E}_1) ch^ A(\mathcal{E}_2)$.

Namely, the first will prove that $ch^ A$ factors through $K_0(\textit{Vect}(X))$ and the first and the second will combined show that $ch^ A$ is a ring map.

To prove these statements we can reduce to the case where $\mathcal{E}_1$ and $\mathcal{E}_2$ have constant ranks $r_1$ and $r_2$. In this case the equalities in $A^0(X)$ are immediate. To prove the equalities in higher degrees, by Lemma 45.12.2 we may assume that $\mathcal{E}_1$ and $\mathcal{E}_2$ have filtrations whose graded pieces are invertible modules $\mathcal{L}_{1, j}$, $j = 1, \ldots , r_1$ and $\mathcal{L}_{2, j}$, $j = 1, \ldots , r_2$. Using the multiplicativity of Chern classes we get

$c_ i^ A(\mathcal{E}_1) = s_ i(c_1^ A(\mathcal{L}_{1, 1}), \ldots , c_1^ A(\mathcal{L}_{1, r_1}))$

where $s_ i$ is the $i$th elementary symmetric function as in Chow Homology, Example 42.43.6. Similarly for $c_ i^ A(\mathcal{E}_2)$. In case (1) we get

$c_ i^ A(\mathcal{E}) = s_ i(c_1^ A(\mathcal{L}_{1, 1}), \ldots , c_1^ A(\mathcal{L}_{1, r_1}), c_1^ A(\mathcal{L}_{2, 1}), \ldots , c_1^ A(\mathcal{L}_{2, r_2}))$

and for case (2) we get

$c_ i^ A(\mathcal{E}_1 \otimes \mathcal{E}_2) = s_ i(c_1^ A(\mathcal{L}_{1, 1}) + c_1^ A(\mathcal{L}_{2, 1}), \ldots , c_1^ A(\mathcal{L}_{1, r_1}) + c_1^ A(\mathcal{L}_{2, r_2}))$

By the definition of the polynomials $P_ i$ we see that this means

$P_ i(c^ A_1(\mathcal{E}_1), \ldots , c^ A_ i(\mathcal{E}_1)) = \sum \nolimits _{j = 1, \ldots , r_1} c_1^ A(\mathcal{L}_{1, j})^ i$

and similarly for $\mathcal{E}_2$. In case (1) we have also

$P_ i(c^ A_1(\mathcal{E}), \ldots , c^ A_ i(\mathcal{E})) = \sum \nolimits _{j = 1, \ldots , r_1} c_1^ A(\mathcal{L}_{1, j})^ i + \sum \nolimits _{j = 1, \ldots , r_2} c_1^ A(\mathcal{L}_{2, j})^ i$

In case (2) we get accordingly

$P_ i(c^ A_1(\mathcal{E}_1 \otimes \mathcal{E}_2), \ldots , c^ A_ i(\mathcal{E}_1 \otimes \mathcal{E}_2)) = \sum \nolimits _{j = 1, \ldots , r_1} \sum \nolimits _{j' = 1, \ldots , r_2} (c_1^ A(\mathcal{L}_{1, j}) + c_1^ A(\mathcal{L}_{2, j'}))^ i$

Thus the desired equalities are now consequences of elementary identities between symmetric polynomials.

We omit the proof of uniqueness. $\square$

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