Remark 42.45.1. In the discussion above we have defined the components of the Chern character $ch_ p(\mathcal{E}) \in A^ p(X) \otimes \mathbf{Q}$ of $\mathcal{E}$ even if the rank of $\mathcal{E}$ is not constant. See Remarks 42.38.10 and 42.43.5. Thus the full Chern character of $\mathcal{E}$ is an element of $\prod _{p \geq 0} (A^ p(X) \otimes \mathbf{Q})$. If $X$ is quasi-compact and $\dim (X) < \infty $ (usual dimension), then one can show using Lemma 42.34.6 and the splitting principle that $ch(\mathcal{E}) \in A^*(X) \otimes \mathbf{Q}$.
42.45 The Chern character and tensor products
Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be locally of finite type over $S$. We define the Chern character of a finite locally free $\mathcal{O}_ X$-module to be the formal expression
\[ ch({\mathcal E}) = \sum \nolimits _{i=1}^ r e^{x_ i} \]
if the $x_ i$ are the Chern roots of ${\mathcal E}$. Writing this as a polynomial in the Chern classes we obtain
\begin{align*} ch(\mathcal{E}) & = r(\mathcal{E}) + c_1(\mathcal{E}) + \frac{1}{2}(c_1(\mathcal{E})^2 - 2c_2(\mathcal{E})) + \frac{1}{6}(c_1(\mathcal{E})^3 - 3c_1(\mathcal{E})c_2(\mathcal{E}) + 3c_3(\mathcal{E})) \\ & \quad \quad + \frac{1}{24}(c_1(\mathcal{E})^4 - 4c_1(\mathcal{E})^2c_2(\mathcal{E}) + 4c_1(\mathcal{E})c_3(\mathcal{E}) + 2c_2(\mathcal{E})^2 - 4c_4(\mathcal{E})) + \ldots \\ & = \sum \nolimits _{p = 0, 1, 2, \ldots } \frac{P_ p(\mathcal{E})}{p!} \end{align*}
with $P_ p$ polynomials in the Chern classes as in Example 42.43.6. The degree $p$ component of the above is
\[ ch_ p(\mathcal{E}) = \frac{P_ p(\mathcal{E})}{p!} \in A^ p(X) \otimes \mathbf{Q} \]
What does it mean that the coefficients are rational numbers? Well this simply means that we think of $ch_ p(\mathcal{E})$ as an element of $A^ p(X) \otimes \mathbf{Q}$.
Lemma 42.45.2. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be locally of finite type over $S$. Let $ 0 \to \mathcal{E}_1 \to \mathcal{E} \to \mathcal{E}_2 \to 0 $ be a short exact sequence of finite locally free $\mathcal{O}_ X$-modules. Then we have the equality More precisely, we have $P_ p(\mathcal{E}) = P_ p(\mathcal{E}_1) + P_ p(\mathcal{E}_2)$ in $A^ p(X)$ where $P_ p$ is as in Example 42.43.6.
\[ ch(\mathcal{E}) = ch(\mathcal{E}_1) + ch(\mathcal{E}_2) \]
Proof. It suffices to prove the more precise statement. By Section 42.43 this follows because if $x_{1, i}$, $i = 1, \ldots , r_1$ and $x_{2, i}$, $i = 1, \ldots , r_2$ are the Chern roots of $\mathcal{E}_1$ and $\mathcal{E}_2$, then $x_{1, 1}, \ldots , x_{1, r_1}, x_{2, 1}, \ldots , x_{2, r_2}$ are the Chern roots of $\mathcal{E}$. Hence we get the result from our choice of $P_ p$ in Example 42.43.6. $\square$
Lemma 42.45.3. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be locally of finite type over $S$. Let $\mathcal{E}_1$ and $\mathcal{E}_2$ be finite locally free $\mathcal{O}_ X$-modules. Then we have the equality More precisely, we have in $A^ p(X)$ where $P_ p$ is as in Example 42.43.6.
\[ ch(\mathcal{E}_1 \otimes _{\mathcal{O}_ X} \mathcal{E}_2) = ch(\mathcal{E}_1) ch(\mathcal{E}_2) \]
\[ P_ p(\mathcal{E}_1 \otimes _{\mathcal{O}_ X} \mathcal{E}_2) = \sum \nolimits _{p_1 + p_2 = p} {p \choose p_1} P_{p_1}(\mathcal{E}_1) P_{p_2}(\mathcal{E}_2) \]
Proof. It suffices to prove the more precise statement. By Section 42.43 this follows because if $x_{1, i}$, $i = 1, \ldots , r_1$ and $x_{2, i}$, $i = 1, \ldots , r_2$ are the Chern roots of $\mathcal{E}_1$ and $\mathcal{E}_2$, then $x_{1, i} + x_{2, j}$, $1 \leq i \leq r_1$, $1 \leq j \leq r_2$ are the Chern roots of $\mathcal{E}_1 \otimes \mathcal{E}_2$. Hence we get the result from the binomial formula for $(x_{1, i} + x_{2, j})^ p$ and the shape of our polynomials $P_ p$ in Example 42.43.6. $\square$
Lemma 42.45.4. In Situation 42.7.1 let $X$ be locally of finite type over $S$. Let $\mathcal{E}$ be a finite locally free $\mathcal{O}_ X$-module with dual $\mathcal{E}^\vee $. Then $ch_ i(\mathcal{E}^\vee ) = (-1)^ i ch_ i(\mathcal{E})$ in $A^ i(X) \otimes \mathbf{Q}$.
Proof. Follows from the corresponding result for Chern classes (Lemma 42.43.3). $\square$
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