Skip to content
Browse files

Blow up formula

  • Loading branch information
aisejohan committed Nov 27, 2019
1 parent 0f3b2fd commit 4f25a75a3b67d90253802049bc7e8f4fc01e3fd7
Showing with 60 additions and 0 deletions.
  1. +60 −0 chow.tex
@@ -12668,6 +12668,66 @@ \section{Gysin maps for local complete intersection morphisms}
$(f')^! = (i')^! \circ (g')^*$, and $(g')^* = res(g^*)$ we conclude.
\begin{lemma}[Blow up formula]
Let $(S, \delta)$ be as in Situation \ref{situation-setup}.
Let $i : Z \to X$ be a regular closed immersion of schemes
locally of finite type over $S$. Let $b : X' \to X$ be the
blowing up with center $Z$. Picture
E \ar[r]_j \ar[d]_\pi & X' \ar[d]^b \\
Z \ar[r]^i & X
Assume that the gysin map exists for $b$. Then we have
res(b^!) = c_{top}(\mathcal{F}^\vee) \circ \pi^*
in $A^*(E \to Z)$ where $\mathcal{F}$ is the kernel of the canonical map
$\pi^*\mathcal{C}_{Z/X} \to \mathcal{C}_{E/X'}$.
Observe that the morphism $b$ is a local complete intersection morphism
by More on Algebra, Lemma \ref{more-algebra-lemma-blowup-regular-sequence}
and hence the statement makes sense. Since $Z \to X$ is a regular
immersion (and hence a fortiori quasi-regular) we see that $\mathcal{C}_{Z/X}$
is finite locally free and the map
$\text{Sym}^*(\mathcal{C}_{Z/X}) \to \mathcal{C}_{Z/X, *}$
is an isomorphism, see
Divisors, Lemma \ref{divisors-lemma-quasi-regular-immersion}.
Since $E = \text{Proj}(\mathcal{C}_{Z/X, *})$ we conclude
that $E = \mathbf{P}(\mathcal{C}_{Z/X})$
is a projective space bundle over $Z$.
Thus $E \to Z$ is smooth and certainly a local complete intersection
morphism. Thus Lemma \ref{lemma-compare-gysin-base-change}
applies and we see that
res(b^!) = c_{top}(\mathcal{C}^\vee) \circ \pi^!
with $\mathcal{C}$ as in the statement there.
Of course $\pi^* = \pi^!$ by Lemma \ref{lemma-lci-gysin-flat}.
It remains to show that $\mathcal{F}$ is equal to
the kernel $\mathcal{C}$ of the map
$H^{-1}(j^*\NL_{X'/X}) \to H^{-1}(\NL_{E/Z})$.
Since $E \to Z$ is smooth we have $H^{-1}(\NL_{E/Z}) = 0$, see
More on Morphisms, Lemma \ref{more-morphisms-lemma-NL-smooth}.
Hence it suffices to show that $\mathcal{F}$ can be identified
with $H^{-1}(j^*\NL_{X'/X})$. By More on Morphisms, Lemmas
\ref{more-morphisms-lemma-get-triangle-NL} and
\ref{more-morphisms-lemma-NL-immersion} we have an exact sequence
0 \to H^{-1}(j^*\NL_{X'/X}) \to H^{-1}(\NL_{E/X}) \to
\mathcal{C}_{E/X'} \to \ldots
By the same lemmas applied to $E \to Z \to X$ we obtain an isomorphism
$\pi^*\mathcal{C}_{Z/X} = H^{-1}(\pi^*\NL_{Z/X}) \to H^{-1}(\NL_{E/X})$.
Thus we conclude.
Let $(S, \delta)$ be as in Situation \ref{situation-setup}.

0 comments on commit 4f25a75

Please sign in to comment.
You can’t perform that action at this time.