The Stacks project

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. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0ESB. Beware of the difference between the letter 'O' and the digit '0'.