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The Stacks project

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 ith 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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