80.31 The splitting principle

This section is the analogue of Chow Homology, Section 42.39.

Lemma 80.31.1. In Situation 80.2.1 let $X/B$ be good. Let $\mathcal{E}_ i$ be a finite collection of locally free $\mathcal{O}_ X$-modules of rank $r_ i$. There exists a projective flat morphism $\pi : P \to X$ of relative dimension $d$ such that

1. for any morphism $f : Y \to X$ of good algebraic spaces over $B$ the map $\pi _ Y^* : \mathop{\mathrm{CH}}\nolimits _*(Y) \to \mathop{\mathrm{CH}}\nolimits _{* + d}(Y \times _ X P)$ is injective, and

2. each $\pi ^*\mathcal{E}_ i$ has a filtration whose successive quotients $\mathcal{L}_{i, 1}, \ldots , \mathcal{L}_{i, r_ i}$ are invertible ${\mathcal O}_ P$-modules.

Proof. We prove this by induction on the integer $r = \sum r_ i$. If $r = 0$ we can take $\pi = \text{id}_ X$. If $r_ i = 1$ for all $i$, then we can also take $\pi = \text{id}_ X$. Assume that $r_{i_0} > 1$ for some $i_0$. Let $(\pi : P \to X, \mathcal{O}_ P(1))$ be the projective bundle associated to $\mathcal{E}_{i_0}$. The canonical map $\pi ^*\mathcal{E}_{i_0} \to \mathcal{O}_ P(1)$ is surjective and hence its kernel $\mathcal{E}'_{i_0}$ is finite locally free of rank $r_{i_0} - 1$. Observe that $\pi _ Y^*$ is injective for any morphism $f : Y \to X$ of good algebraic spaces over $B$, see Lemma 80.27.2. Thus it suffices to prove the lemma for $P$ and the locally free sheaves $\pi ^*\mathcal{E}_ i$. However, because we have the subbundle $\mathcal{E}_{i_0} \subset \pi ^*\mathcal{E}_{i_0}$ with invertible quotient, it now suffices to prove the lemma for the collection $\{ \mathcal{E}_ i\} _{i \not= i_0} \cup \{ \mathcal{E}'_{i_0}\}$. This decreases $r$ by $1$ and we win by induction hypothesis. $\square$

Rather than explaining what the splitting principle says, let us use it in the proof of some lemmas.

Lemma 80.31.2. In Situation 80.2.1 let $X/B$ be good. Let $\mathcal{E}$ be a finite locally free $\mathcal{O}_ X$-module with dual $\mathcal{E}^\vee$. Then

$c_ i(\mathcal{E}^\vee ) = (-1)^ i c_ i(\mathcal{E})$

in $A^ i(X)$.

Proof. Choose a morphism $\pi : P \to X$ as in Lemma 80.31.1. By the injectivity of $\pi ^*$ (after any base change) it suffices to prove the relation between the chern classes of $\mathcal{E}$ and $\mathcal{E}^\vee$ after pulling back to $P$. Thus we may assume there exist invertible $\mathcal{O}_ X$-modules ${\mathcal L}_ i$, $i = 1, \ldots , r$ and a filtration

$0 = \mathcal{E}_0 \subset \mathcal{E}_1 \subset \mathcal{E}_2 \subset \ldots \subset \mathcal{E}_ r = \mathcal{E}$

such that $\mathcal{E}_ i/\mathcal{E}_{i - 1} \cong \mathcal{L}_ i$. Then we obtain the dual filtration

$0 = \mathcal{E}_ r^\perp \subset \mathcal{E}_1^\perp \subset \mathcal{E}_2^\perp \subset \ldots \subset \mathcal{E}_0^\perp = \mathcal{E}^\vee$

such that $\mathcal{E}_{i - 1}^\perp /\mathcal{E}_ i^\perp \cong \mathcal{L}_ i^{\otimes -1}$. Set $x_ i = c_1(\mathcal{L}_ i)$. Then $c_1(\mathcal{L}_ i^{\otimes -1}) = - x_ i$ by Lemma 80.18.2. By Lemma 80.30.4 we have

$c(\mathcal{E}) = \prod \nolimits _{i = 1}^ r (1 + x_ i) \quad \text{and}\quad c(\mathcal{E}^\vee ) = \prod \nolimits _{i = 1}^ r (1 - x_ i)$

in $A^*(X)$. The result follows from a formal computation which we omit. $\square$

Lemma 80.31.3. In Situation 80.2.1 let $X/B$ be good. Let $\mathcal{E}$ and $\mathcal{F}$ be a finite locally free $\mathcal{O}_ X$-modules of ranks $r$ and $s$. Then we have

$c_1(\mathcal{E} \otimes \mathcal{F}) = r c_1(\mathcal{F}) + s c_1(\mathcal{E})$
$c_2(\mathcal{E} \otimes \mathcal{F}) = r^2 c_2(\mathcal{F}) + rs c_1(\mathcal{F})c_1(\mathcal{E}) + s^2 c_2(\mathcal{E})$

and so on (see proof).

Proof. Arguing exactly as in the proof of Lemma 80.31.2 we may assume we have invertible $\mathcal{O}_ X$-modules ${\mathcal L}_ i$, $i = 1, \ldots , r$ ${\mathcal N}_ i$, $i = 1, \ldots , s$ filtrations

$0 = \mathcal{E}_0 \subset \mathcal{E}_1 \subset \mathcal{E}_2 \subset \ldots \subset \mathcal{E}_ r = \mathcal{E} \quad \text{and}\quad 0 = \mathcal{F}_0 \subset \mathcal{F}_1 \subset \mathcal{F}_2 \subset \ldots \subset \mathcal{F}_ s = \mathcal{F}$

such that $\mathcal{E}_ i/\mathcal{E}_{i - 1} \cong \mathcal{L}_ i$ and such that $\mathcal{F}_ j/\mathcal{F}_{j - 1} \cong \mathcal{N}_ j$. Ordering pairs $(i, j)$ lexicographically we obtain a filtration

$0 \subset \ldots \subset \mathcal{E}_ i \otimes \mathcal{F}_ j + \mathcal{E}_{i - 1} \otimes \mathcal{F} \subset \ldots \subset \mathcal{E} \otimes \mathcal{F}$

with successive quotients

$\mathcal{L}_1 \otimes \mathcal{N}_1, \mathcal{L}_1 \otimes \mathcal{N}_2, \ldots , \mathcal{L}_1 \otimes \mathcal{N}_ s, \mathcal{L}_2 \otimes \mathcal{N}_1, \ldots , \mathcal{L}_ r \otimes \mathcal{N}_ s$

By Lemma 80.30.4 we have

$c(\mathcal{E}) = \prod (1 + x_ i), \quad c(\mathcal{F}) = \prod (1 + y_ j), \quad \text{and}\quad c(\mathcal{F}) = \prod (1 + x_ i + y_ j),$

in $A^*(X)$. The result follows from a formal computation which we omit. $\square$

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