Lemma 42.54.10. Let $(S, \delta )$ be as in Situation 42.7.1. Let $Z \subset Y \subset X$ be closed subschemes of a scheme locally of finite type over $S$. Let $\mathcal{N}$ be a virtual normal sheaf for $Z \subset X$. Let $\mathcal{N}'$ be a virtual normal sheaf for $Z \subset Y$. Let $\mathcal{N}''$ be a virtual normal sheaf for $Y \subset X$. Assume there is a commutative diagram
\[ \xymatrix{ (\mathcal{N}'')^\vee |_ Z \ar[r] \ar[d] & \mathcal{N}^\vee \ar[r] \ar[d] & (\mathcal{N}')^\vee \ar[d] \\ \mathcal{C}_{Y/X}|_ Z \ar[r] & \mathcal{C}_{Z/X} \ar[r] & \mathcal{C}_{Z/Y} } \]
where the sequence at the bottom is from More on Morphisms, Lemma 37.7.12 and the top sequence is a short exact sequence. Then
\[ c(Z \to X, \mathcal{N}) = c(Z \to Y, \mathcal{N}') \circ c(Y \to X, \mathcal{N}'') \]
in $A^*(Z \to X)^\wedge $.
Proof.
Observe that the assumptions remain satisfied after any base change by a morphism $X' \to X$ which is locally of finite type (the short exact sequence of virtual normal sheaves is locally split hence remains exact after any base change). Thus to check the equality of bivariant classes we may use Lemma 42.35.3. Thus we may assume $X$ is an integral scheme and we have to show $c(Z \to X, \mathcal{N}) \cap [X] = c(Z \to Y, \mathcal{N}') \cap c(Y \to X, \mathcal{N}'') \cap [X]$.
If $Y = X$, then we have
\begin{align*} c(Z \to Y, \mathcal{N}') \cap c(Y \to X, \mathcal{N}'') \cap [X] & = c(Z \to Y, \mathcal{N}') \cap c_{top}(\mathcal{N}'') \cap [Y] \\ & = c_{top}(\mathcal{N}''|_ Z) \cap c(Z \to Y, \mathcal{N}') \cap [Y] \\ & = c(Z \to X, \mathcal{N}) \cap [X] \end{align*}
The first equality by Lemma 42.54.3. The second because Chern classes commute with bivariant classes (Lemma 42.38.9). The third equality by Lemma 42.54.3.
Assume $Y \not= X$. By Lemma 42.35.3 it even suffices to prove the result after blowing up $X$ in a nonzero ideal. Let us blowup $X$ in the product of the ideal sheaf of $Y$ and the ideal sheaf of $Z$. This reduces us to the case where both $Y$ and $Z$ are effective Cartier divisors on $X$, see Divisors, Lemmas 31.32.4 and 31.32.12.
Denote $\mathcal{N}'' \to \mathcal{E}$ the surjection of finite locally free $\mathcal{O}_ Z$-modules such that $0 \to \mathcal{E}^\vee \to (\mathcal{N}'')^\vee \to \mathcal{C}_{Y/X} \to 0$ is a short exact sequence. Then $\mathcal{N} \to \mathcal{E}|_ Z$ is a surjection as well. Denote $\mathcal{N}_1$ the finite locally free kernel of this map and observe that $\mathcal{N}^\vee \to \mathcal{C}_{Z/X}$ factors through $\mathcal{N}_1$. By Lemma 42.54.3 we have
\[ c(Y \to X, \mathcal{N}'') = c_{top}(\mathcal{E}) \circ c(Y \to X, \mathcal{C}_{Y/X}^\vee ) \]
and
\[ c(Z \to X, \mathcal{N}) = c_{top}(\mathcal{E}|_ Z) \circ c(Z \to X, \mathcal{N}_1) \]
Since Chern classes of bundles commute with bivariant classes (Lemma 42.38.9) it suffices to prove
\[ c(Z \to X, \mathcal{N}_1) = c(Z \to Y, \mathcal{N}') \circ c(Y \to X, \mathcal{C}_{Y/X}^\vee ) \]
in $A^*(Z \to X)$. This we may assume that $\mathcal{N}'' = \mathcal{C}_{Y/X}$. This reduces us to the case discussed in the next paragraph.
In this paragraph $Z$ and $Y$ are effective Cartier divisors on $X$ integral of dimension $n$, we have $\mathcal{N}'' = \mathcal{C}_{Y/X}$. In this case $c(Y \to X, \mathcal{C}_{Y/X}^\vee ) \cap [X] = [Y]_{n - 1}$ by Lemma 42.54.5. Thus we have to prove that $c(Z \to X, \mathcal{N}) \cap [X] = c(Z \to Y, \mathcal{N}') \cap [Y]_{n - 1}$. Denote $N$ and $N'$ the vector bundles over $Z$ associated to $\mathcal{N}$ and $\mathcal{N}'$. Consider the commutative diagram
\[ \xymatrix{ N' \ar[r]_ i & N \ar[r] & (C_ Y X) \times _ Y Z \\ C_ Z Y \ar[r] \ar[u] & C_ Z X \ar[u] } \]
of cones and vector bundles over $Z$. Observe that $N'$ is a relative effective Cartier divisor in $N$ over $Z$ and that
\[ \xymatrix{ N' \ar[d] \ar[r]_ i & N \ar[d] \\ Z \ar[r]^-o & (C_ Y X) \times _ Y Z } \]
is cartesian where $o$ is the zero section of the line bundle $C_ Y X$ over $Y$. By Lemma 42.54.9 we have $o^*[C_ ZX]_ n = [C_ Z Y]_{n - 1}$ in
\[ \mathop{\mathrm{CH}}\nolimits _{n - 1}(Y \times _{o, C_ Y X} C_ ZX) = \mathop{\mathrm{CH}}\nolimits _{n - 1}(Z \times _{o, (C_ Y X) \times _ Y Z} C_ ZX) \]
By the cartesian property of the square above this implies that
\[ i^*[C_ ZX]_ n = [C_ Z Y]_{n - 1} \]
in $\mathop{\mathrm{CH}}\nolimits _{n - 1}(N')$. Now observe that $\gamma = c(Z \to X, \mathcal{N}) \cap [X]$ and $\gamma ' = c(Z \to Y, \mathcal{N}') \cap [Y]_{n - 1}$ are characterized by $p^*\gamma = [C_ Z X]_ n$ in $\mathop{\mathrm{CH}}\nolimits _ n(N)$ and by $(p')^*\gamma ' = [C_ Z Y]_{n - 1}$ in $\mathop{\mathrm{CH}}\nolimits _{n - 1}(N')$. Hence the proof is finished as $i^* \circ p^* = (p')^*$ by Lemma 42.31.1.
$\square$
Comments (0)
There are also: