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Intersection products for smooth varieties

TODO: tie this in with the chapter on classical intersection theory
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aisejohan committed May 21, 2019
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  1. +338 −2 chow.tex
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@@ -9765,6 +9765,186 @@ \section{Higher codimension gysin homomorphisms}
with the gysin homomorphism $i^*$ by definition of bivariant classes.
\end{proof}

\begin{remark}[Variant for immersions]
\label{remark-gysin-for-immersion}
Let $(S, \delta)$ be as in Situation \ref{situation-setup}.
Let $X$ be a scheme locally of finite type over $S$.
Let $i : Z \to X$ be an immersion of schemes.
In this situation
\begin{enumerate}
\item the conormal sheaf $\mathcal{C}_{Z/X}$
of $Z$ in $X$ is defined
(Morphisms, Definition \ref{morphisms-definition-conormal-sheaf}),
\item we say a pair consisting of a finite locally free $\mathcal{O}_Z$-module
$\mathcal{N}$ and a surjection $\sigma : \mathcal{N}^\vee \to \mathcal{C}_{Z/X}$
is a virtual normal bundle for the immersion $Z \to X$,
\item choose an open subscheme $U \subset X$ such that $Z \to X$
factors through a closed immersion $Z \to U$ and set
$c(Z \to X, \mathcal{N}) = c(Z \to U, \mathcal{N}) \circ (U \to X)^*$.
\end{enumerate}
The bivariant class $c(Z \to X, \mathcal{N})$ does not depend on the choice
of the open subscheme $U$. All of the lemmas have immediate counterparts
for this slightly more general construction. We omit the details.
\end{remark}







\section{Gysin maps for diagonals}
\label{section-gysin-for-diagonal}

\noindent
Let $(S, \delta)$ be as in Situation \ref{situation-setup}.
Let $f : X \to Y$ be a smooth morphism of schemes locally
of finite type over $S$. Then the diagonal morphism
$$
\Delta : X \longrightarrow X \times_Y X
$$
is a regular immersion (Divisors, Lemma
\ref{divisors-lemma-immersion-smooth-into-smooth-regular-immersion}).
In particular, the conormal sheaf $\mathcal{C}_{X/X \times_Y X}$
is finite locally free. Of course, by
Morphisms, Lemma \ref{morphisms-lemma-differentials-diagonal}
we have $\Omega_{X/Y} \cong \mathcal{C}_{X/X \times_Y X}$
which is another way to see that it is finite locally free
(Morphisms, Lemma \ref{morphisms-lemma-smooth-omega-finite-locally-free}).
Let
$$
\mathcal{T}_{X/Y} = \SheafHom_{\mathcal{O}_X}(\Omega_{X/Y}, \mathcal{O}_X)
$$
be the tangent sheaf of $X$ over $Y$. The identification of the conormal
sheaf of the diagonal with the module of differentials above gives a
surjection
$\sigma : (\mathcal{T}_{X/Y})^\vee \to \mathcal{C}_{X/X \times_Y X}$
(because an isomorphism is also a surjection).
In this way we may think of $\mathcal{T}_{X/Y}$ as a virtual normal sheaf
for the closed immersion $\Delta$, see
Section \ref{section-gysin-higher-codimension}.
The construction in Section \ref{section-gysin-higher-codimension}
gives us a canonical bivariant class
$$
\Delta^! = c(X \to X \times_Y X, \mathcal{T}_{X/Y}) \in A^*(X \to X \times_Y X)
$$
See Remark \ref{remark-gysin-for-immersion}
for how to deal with the fact that $\Delta$ is in\
general only an immersion and not a closed immersion.
Of course, if $X \to Y$ has constant relative dimension $d$, then
$\Delta^! \in A^d(X \to X \times_Y X)$.

\begin{lemma}
\label{lemma-diagonal-identity}
In the situation above we have $\Delta^! \circ \text{pr}_i^* = 1$ in $A^0(X)$.
\end{lemma}

\begin{proof}
After decomposing $X$ into connected components we may and do assume
that $X \to Y$ is smooth of constant relative dimension $d$.
Let $X' \to X$ be locally of finite type with $X'$
integral of $\delta$-dimension $n$. Then
$\text{pr}_i^*[X'] = [X \times_Y X']_{n + d}$.
Observe that $X \times_{\Delta, X \times_Y X} (X \times_Y X')$ is the
graph of $X' \to X \times_Y X'$ as a locally closed subscheme.
Since the dimension is correct we conclude that
$\Delta^! \cap [X \times_Y X']_{n + d} = [X']$ by
Lemma \ref{lemma-gysin-fundamental}.
\end{proof}

\begin{proposition}
\label{proposition-compute-bivariant}
\begin{reference}
\cite[Proposition 17.4.2]{F}
\end{reference}
Let $(S, \delta)$ be as in Situation \ref{situation-setup}.
Let $f : X \to Y$ and $g : Y \to Z$ be morphisms of schemes locally
of finite type over $S$. If $g$ is smooth of relative dimension $d$, then
$A^p(X \to Y) = A^{p - d}(X \to Z)$.
\end{proposition}

\begin{proof}
Denote $g^* \in A^{-d}(Y \to Z)$ the bivariant class given by flat pullback,
see Lemma \ref{lemma-flat-pullback-bivariant}. Then we can send
$c \in A^p(X \to Y)$ to $c \circ g^* \in A^{p - d}(X \to Z)$.

\medskip\noindent
Conversely, let $c' \in A^{p - d}(X \to Z)$. Denote $res(c')$ the restriction
(Remark \ref{remark-restriction-bivariant}) of $c'$ by the morphism $Y \to Z$.
Since the diagram
$$
\xymatrix{
X \times_Z Y \ar[r]_{\text{pr}_2} \ar[d]_{\text{pr}_1} & Y \ar[d]^g \\
X \ar[r]^f & Z
}
$$
is cartesian we find $res(c') \in A^{p - d}(X \times_Z Y \to Y)$.
Similarly, denote $res(\Delta^!)$ the restriction of $\Delta^!$
to $X \times_Z Y$ by the morphism $X \times_Z Y \to Y \times_Z Y$.
Since the diagram
$$
\xymatrix{
X \ar[r] \ar[d] & X \times_Z Y \ar[d] \\
Y \ar[r] & Y \times_Z Y
}
$$
is cartesian we see that $res(\Delta^!) \in A^d(X \to X \times_Z Y)$.
Combining these two restrictions we obtain
$$
res(\Delta^!) \circ res(c') \in A^p(X \to Y)
$$
Thus we have produced maps $A^p(X \to Y) \to A^{p - d}(X \to Z)$
and $A^{p - d}(X \to Z) \to A^p(X \to Y)$. To finish the proof we
will show these maps are mutually inverse.

\medskip\noindent
Let us start with $c \in A^p(X \to Y)$. Consider the diagram
$$
\xymatrix{
X \ar[d] \ar[r] & Y \ar[d] \\
X \times_Z Y \ar[r] \ar[d]^{\text{pr}_1} &
Y \times_Z Y \ar[r]_{p_2} \ar[d]^{p_1} &
Y \ar[d]^g \\
X \ar[r]^f &
Y \ar[r]^g &
Z
}
$$
whose squares are carteisan. The lower two square of this diagram
show that $res(c \circ g^*) = res(c) \cap p_2^*$ where in this formula
$res(c)$ means the restriction of $c$ via $p_1$. Then the upper
square of the diagram combined with Lemma \ref{lemma-gysin-commutes}
shows that $c \circ \Delta^! = res(\Delta^!) \circ res(c)$.
We compute
\begin{align*}
res(\Delta^!) \circ res(c \circ g^*)
& =
res(\Delta^!) \circ res(c) \circ p_2^* \\
& =
c \circ \Delta^! \circ p_2^* \\
& =
c
\end{align*}
The final equality by Lemma \ref{lemma-diagonal-identity}.

\medskip\noindent
Conversely, let us start with $c' \in A^{p - d}(X \to Z)$. Looking
at the lower rectangle of the diagram above we find
$res(c') \circ g^* = \text{pr}_1^* \circ c'$.
We compute
\begin{align*}
res(\Delta^!) \circ res(c') \circ g^*
& =
res(\Delta^!) \circ \text{pr}_1^* \circ c' \\
& =
c
\end{align*}
The final equality holds because the left two squares of
the diagram show that
$\text{id} = res(\Delta^! \circ p_1^*) = res(\Delta^!) \circ \text{pr}_1^*$.
This finishes the proof.
\end{proof}




@@ -9780,6 +9960,8 @@ \section{Exterior product}
Let $k$ be a field. Set $S = \Spec(k)$ and define $\delta : S \to \mathbf{Z}$
by sending the unique point to $0$. Then $(S, \delta)$ is a special case of
our general Situation \ref{situation-setup}, see Example \ref{example-field}.

\medskip\noindent
Consider a cartesian square
$$
\xymatrix{
@@ -9859,8 +10041,7 @@ \section{Exterior product}
\end{lemma}

\begin{proof}
Observe that $A_*(\Spec(k))$ is nonzero only in degree $0$
with generator $[\Spec(k)]$. Hence we get a map
Consider the element $[\Spec(k)] \in A_0(\Spec(k))$. We get a map
$A^p(X \to \Spec(k)) \to A_{-p}(X)$ by sending $c$ to $c \cap [\Spec(k)]$.

\medskip\noindent
@@ -9927,6 +10108,161 @@ \section{Exterior product}
Some details omitted.
\end{proof}

\begin{remark}
\label{remark-commuting-exterior}
The upshot of Lemmas \ref{lemma-chow-cohomology-towards-point}
and \ref{lemma-chow-cohomology-towards-point-commutes} is the following.
Let $k$ be a field. Let $X$ be a scheme locally of finite type over $k$.
Let $\alpha \in A_*(X)$. Let $Y \to Z$ be a morphism of schemes
locally of finite type over $k$. Let $c' \in A^q(Y \to Z)$. Then
$$
\alpha \times (c' \cap \beta) = c' \cap (\alpha \times \beta)
$$
in $A_*(X \times_k Y)$ for any $\beta \in A_*(Z)$. Namely, this
follows by taking $c = c_\alpha \in A^*(X \to \Spec(k))$ the bivariant class
corresponding to $\alpha$, see proof of
Lemma \ref{lemma-chow-cohomology-towards-point}.
\end{remark}

\begin{lemma}
\label{lemma-exterior-product-associative}
Exterior product is associative. More precisely, let $k$ be a field,
let $X, Y, Z$ be schemes locally of finite type over $k$, let
$\alpha \in A_*(X)$, $\beta \in A_*(Y)$, $\gamma \in A_*(Z)$.
Then $(\alpha \times \beta) \times \gamma =
\alpha \times (\beta \times \gamma)$ in $A_*(X \times_k Y \times_k Z)$.
\end{lemma}

\begin{proof}
Omitted. Hint: associativity of fibre product of schemes.
\end{proof}







\section{Intersection products for smooth varieties}
\label{section-intersection-product}

\noindent
Let $k$ be a field. Set $S = \Spec(k)$ and define $\delta : S \to \mathbf{Z}$
by sending the unique point to $0$. Then $(S, \delta)$ is a special case of
our general Situation \ref{situation-setup}, see Example \ref{example-field}.

\medskip\noindent
Let $X$ be a smooth scheme over $k$. The bivariant class $\Delta^!$
of Section \ref{section-gysin-for-diagonal} allows us to define a kind of
intersection product on chow groups of schemes locally of finite type over $X$.
Namely, suppose that $Y \to X$ and $Z \to X$ are morphisms of schemes
which are locally of finite type. Then observe that
$$
Y \times_X Z = (Y \times_k Z) \times_{X \times_k X, \Delta} X
$$
Hence we can consider the following sequence of maps
$$
A_n(Y) \otimes_\mathbf{Z} A_m(Y)
\xrightarrow{\times}
A_{n + m}(Y \times_k Z)
\xrightarrow{\Delta^!}
A_{n + m - *}(Y \times_X Z)
$$
Here the first arrow is the exterior product constructed in
Section \ref{section-exterior-product}. If $X$ is equidimensional
of dimension $d$, then we end up in $A_{n + m - d}(Y \times_X Z)$
and in general we can decompose into the parts lying over the open
and closed subschemes of $X$ where $X$ has a given dimension.
Given $\alpha \in A_*(Y)$ and $\beta \in A_*(Z)$ we will denote
$$
\alpha \cdot \beta = \Delta^!(\alpha \times \beta)
\in A_*(Y \times_X Z)
$$

\begin{lemma}
\label{lemma-associative}
The product defined above is associative. More precisely, let $k$ be a field,
let $X$ be smooth over $k$,
let $Y, Z, W$ be schemes locally of finite type over $X$, let
$\alpha \in A_*(Y)$, $\beta \in A_*(Z)$, $\gamma \in A_*(W)$.
Then $(\alpha \cdot \beta) \cdot \gamma =
\alpha \cdot (\beta \cdot \gamma)$ in $A_*(Y \times_X Z \times_X W)$.
\end{lemma}

\begin{proof}
By Lemma \ref{lemma-exterior-product-associative} we have
$(\alpha \times \beta) \times \gamma =
\alpha \times (\beta \times \gamma)$ in $A_*(Y \times_k Z \times_k W)$.
Consider the closed immersions
$$
\Delta_{12} : X \times_k X \longrightarrow X \times_k X \times_k X,
\quad (x, x') \mapsto (x, x, x')
$$
and
$$
\Delta_{23} : X \times_k X \longrightarrow X \times_k X \times_k X,
\quad (x, x') \mapsto (x, x', x')
$$
Denote $\Delta_{12}^!$ and $\Delta_{23}^!$ the corresponding bivariant
classes; observe that $\Delta_{12}^!$ is the restriction
(Remark \ref{remark-restriction-bivariant}) of $\Delta^!$
to $X \times_k X \times_k X$ by the map $\text{pr}_{12}$ and that
$\Delta_{23}^!$ is the restriction of $\Delta^!$
to $X \times_k X \times_k X$ by the map $\text{pr}_{23}$.
Thus clearly the restriction of $\Delta_{12}^!$ by $\Delta_{23}$
is $\Delta^!$ and the restriction of $\Delta_{23}^!$ by $\Delta_{12}$ is
$\Delta^!$ too. Thus by Lemma \ref{lemma-gysin-commutes} we have
$$
\Delta^! \circ \Delta_{12}^! =
\Delta^! \circ \Delta_{23}^!
$$
Now we can prove the lemma by the following sequence of equalities:
\begin{align*}
(\alpha \cdot \beta) \cdot \gamma
& =
\Delta^!(\Delta^!(\alpha \times \beta) \times \gamma) \\
& =
\Delta^!(\Delta_{12}^!((\alpha \times \beta) \times \gamma)) \\
& =
\Delta^!(\Delta_{23}^!((\alpha \times \beta) \times \gamma)) \\
& =
\Delta^!(\Delta_{23}^!(\alpha \times (\beta \times \gamma)) \\
& =
\Delta^!(\alpha \times \Delta^!(\beta \times \gamma)) \\
& =
\alpha \cdot (\beta \cdot \gamma)
\end{align*}
All equalities are clear from the above except perhaps
for the second and penultimate one. The equation
$\Delta_{23}^!(\alpha \times (\beta \times \gamma)) =
\alpha \times \Delta^!(\beta \times \gamma)$ holds by
Remark \ref{remark-commuting-exterior}. Similarly for the second
equation.
\end{proof}

\noindent
In the special case where $X = Y = Z$ we obtain a multiplication
$A_*(X) \times A_*(X) \to A_*(X)$. There is an alternative description
of this product.

\begin{lemma}
\label{lemma-identify-chow-for-smooth}
Let $k$ be a field. Let $X$ be a smooth scheme over $k$, equidimensional
of dimension $d$. The map
$$
A^p(X) \longrightarrow A_{d - p}(X),\quad
c \longmapsto c \cap [X]_d
$$
is an isomorphism. Via this isomorphism composition of bivariant
classes turns into the intersection product defined above.
\end{lemma}

\begin{proof}
The map is an isomorphism by combining
Lemmas \ref{lemma-chow-cohomology-towards-point} and
Proposition \ref{proposition-compute-bivariant}.
We omit the verification about composition and products.
\end{proof}



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