The Stacks project

45.6 Projective space bundle formula

Let $k$ be a base field. Let $X$ be a smooth projective scheme over $k$. Let $\mathcal{E}$ be a locally free $\mathcal{O}_ X$-module of rank $r$. Our convention is that the projective bundle associated to $\mathcal{E}$ is the morphism

\[ \xymatrix{ P = \mathbf{P}(\mathcal{E}) = \underline{\text{Proj}}_ X(\text{Sym}^*(\mathcal{E})) \ar[r]^-p & X } \]

over $X$ with $\mathcal{O}_ P(1)$ normalized so that $p_*(\mathcal{O}_ P(1)) = \mathcal{E}$. Recall that

\[ [\Gamma _ p] \in \text{Corr}^0(X, P) \subset \mathop{\mathrm{CH}}\nolimits ^*(X \times P) \otimes \mathbf{Q} \]

See Example 45.3.2. For $i = 0, \ldots , r - 1$ consider the correspondences

\[ c_ i = c_1(\text{pr}_2^*\mathcal{O}_ P(1))^ i \cap [\Gamma _ p] \in \text{Corr}^ i(X, P) \]

We may and do think of $c_ i$ as a morphism $h(X)(-i) \to h(P)$.

Lemma 45.6.1 (Projective space bundle formula). In the situation above, the map

\[ \sum \nolimits _{i = 0, \ldots , r - 1} c_ i : \bigoplus \nolimits _{i = 0, \ldots , r - 1} h(X)(-i) \longrightarrow h(P) \]

is an isomorphism in the category of motives.

Proof. By Lemma 45.5.3 it suffices to show that our map defines an isomorphism on Chow groups of motives after taking the product with any smooth projective scheme $Z$. Observe that $P \times Z \to X \times Z$ is the projective bundle associated to the pullback of $\mathcal{E}$ to $X \times Z$. Hence the statement on Chow groups is true by the projective space bundle formula given in Chow Homology, Lemma 42.36.2. Namely, pushforward of cycles along $[\Gamma _ p]$ is given by pullback of cycles by $p$ according to Lemma 45.3.6 and Chow Homology, Lemma 42.59.5. Hence pushforward along $c_ i$ sends $\alpha $ to $c_1(\mathcal{O}_ P(1))^ i \cap p^*\alpha $. Some details omitted. $\square$

In the situation above, for $j = 0, \ldots , r - 1$ consider the correspondences

\[ c'_ j = c_1(\text{pr}_1^*\mathcal{O}_ P(1))^{r - 1 - j} \cap [\Gamma _ p^ t] \in \text{Corr}^{-j}(P, X) \]

For $i, j \in \{ 0, \ldots , r - 1\} $ we have

\[ c'_ j \circ c_ i = \text{pr}_{13, *}\left( c_1(\text{pr}_2^*\mathcal{O}_ P(1))^{i + r - 1 - j} \cap (\text{pr}_{12}^*[\Gamma _ p] \cdot \text{pr}_{23}^*[\Gamma _ p^ t]) \right) \]

The cycles $\text{pr}_{12}^{-1}\Gamma _ p$ and $\text{pr}_{23}^{-1}\Gamma _ p^ t$ intersect transversally and with intersection equal to the image of $(p, 1, p) : P \to X \times P \times X$. Observe that the fibres of $(p, p) = \text{pr}_{13} \circ (p, 1, p) : P \to X \times X$ have dimension $r - 1$. We immediately conclude $c'_ j \circ c_ i = 0$ for $i + r - 1 - j < r - 1$, in other words when $i < j$. On the other hand, by the projective space bundle formula (Chow Homology, Lemma 42.36.2) the cycle $c_1(\mathcal{O}_ P(1))^{r - 1} \cap [P]$ maps to $[X]$ in $X$. Hence for $i = j$ the pushforward above gives the class of the diagonal and hence we see that

\[ c'_ i \circ c_ i = 1 \in \text{Corr}^0(X, X) \]

for all $i \in \{ 0, \ldots , r - 1\} $. Thus we see that the matrix of the composition

\[ \bigoplus h(X)(-i) \xrightarrow {\bigoplus c_ i} h(P) \xrightarrow {\bigoplus c'_ j} \bigoplus h(X)(-j) \]

is invertible (upper triangular with $1$s on the diagonal). We conclude from the projective space bundle formula (Lemma 45.6.1) that also the composition the other way around is invertible, but it seems a bit harder to prove this directly.

Lemma 45.6.2. Let $p : P \to X$ be as in Lemma 45.6.1. The class $[\Delta _ P]$ of the diagonal of $P$ in $\mathop{\mathrm{CH}}\nolimits ^*(P \times P)$ can be written as

\[ [\Delta _ P] = \left(\sum \nolimits _{i = 0, \ldots , r - 1} {r - 1 \choose i} c_{r - 1 - i}(\text{pr}_1^*\mathcal{S}^\vee ) \cap c_1(\text{pr}_2^*\mathcal{O}_ P(1))^ i\right) \cap (p \times p)^*[\Delta _ X] \]

where $\mathcal{S}$ is the kernel of the canonical surjection $p^*\mathcal{E} \to \mathcal{O}_ P(1)$.

Proof. Observe that $(p \times p)^*[\Delta _ X] = [P \times _ X P]$. Since $\Delta _ P \subset P \times _ X P \subset P \times P$ and since capping with Chern classes commutes with proper pushforward (Chow Homology, Lemma 42.38.4) it suffices to show that the class of $\Delta _ P \subset P \times _ X P$ in $\mathop{\mathrm{CH}}\nolimits ^*(P \times _ X P)$ is equal to

\[ \left(\sum \nolimits _{i = 0, \ldots , r - 1} {r - 1 \choose i} c_{r - 1 - i}(q_1^*\mathcal{S}^\vee ) \cap c_1(q_2^*\mathcal{O}_ P(1))^ i\right) \cap [P \times _ X P] \]

where $q_ i : P \times _ X P \to P$, $i = 1, 2$ are the projections. Set $q = p \circ q_1 = p \circ q_2 : P \times _ X P \to X$. Consider the maps

\[ q_1^*\mathcal{S} \otimes q_2^*\mathcal{O}_ P(-1) \to q^*\mathcal{E} \otimes q^*\mathcal{E}^\vee \to \mathcal{O}_{P \times _ X P} \]

where the final arrow is the pullback by $q$ of the evaluation map $\mathcal{E} \otimes _{\mathcal{O}_ X} \mathcal{E}^\vee \to \mathcal{O}_ X$. The source of the composition is a module locally free of rank $r - 1$ and a local calculation shows that this map vanishes exactly along $\Delta _ P$. By Chow Homology, Lemma 42.44.1 the class $[\Delta _ P]$ is the top Chern class of the dual

\[ q_1^*\mathcal{S}^\vee \otimes q_2^*\mathcal{O}_ P(1) \]

The desired result follows from Chow Homology, Lemma 42.39.1. $\square$

Comments (2)

Comment #6307 by Qingyuan Jiang on

Since this chapter is using the contravariant convention for motives, should the in the formula of Lemma 45.6.1 be or ( as in Remark 45.4.5)? Similarly for the formula about the composition and below.

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