Skip to content
Permalink
Browse files

Add proof of agreement intersections

  • Loading branch information...
aisejohan committed May 22, 2019
1 parent 8399d45 commit 863eb5ab80a97b89b8a686ecfc569785b0b0d9ab
Showing with 40 additions and 16 deletions.
  1. +40 −16 chow.tex
@@ -9842,13 +9842,19 @@ \section{Gysin maps for diagonals}
\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
Let $X' \to X$ be locally of finite type with $\dim_\delta(X') = n$. Then
$\text{pr}_i^*[X'] = [X \times_Y X']_{n + d}$. We have a cartesian
diagram
$$
\xymatrix{
X' \ar[d] \ar[r] & X \ar[d]^\Delta \\
X \times_Y X' \ar[r] & X \times_Y X
}
$$
The left vertical arrow is a regular immersion of codimension $d$
since it is a section of the smooth morphism $X \times_Y X' \to X'$, see
Divisors, Lemma \ref{divisors-lemma-section-smooth-regular-immersion}.
It follows that $\Delta^! \cap [X \times_Y X']_{n + d} = [X']$ by
Lemma \ref{lemma-gysin-fundamental}.
\end{proof}

@@ -10178,6 +10184,14 @@ \section{Intersection products for smooth varieties}
\alpha \cdot \beta = \Delta^!(\alpha \times \beta)
\in A_*(Y \times_X Z)
$$
In the special case where $X = Y = Z$ we obtain a multiplication
$$
A_*(X) \times A_*(X) \to A_*(X),\quad
(\alpha, \beta) \mapsto \alpha \cdot \beta
$$
which is called the {\it intersection product}. We observe that
this product is clearly symmetric. Associativity follows from
the next lemma (as well as the one following).

\begin{lemma}
\label{lemma-associative}
@@ -10240,11 +10254,6 @@ \section{Intersection products for smooth varieties}
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
@@ -10258,10 +10267,25 @@ \section{Intersection products for smooth varieties}
\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.
Denote $g : X \to \Spec(k)$ the structure morphism.
The map is the composition of the isomorphisms
$$
A^p(X) \to A^{p - d}(X \to \Spec(k)) \to A_{d - p}(X)
$$
The first is the isomorphism $c \mapsto c \circ g^*$ of
Proposition \ref{proposition-compute-bivariant}
and the second is the isomorphism $c \mapsto c \cap [\Spec(k)]$ of
Lemma \ref{lemma-chow-cohomology-towards-point}.
From the proof of Lemma \ref{lemma-chow-cohomology-towards-point}
we see that the inverse to the second arrow sends $\alpha \in A_{d - p}(X)$
to the bivariant class $c_\alpha$ which sends $\beta \in A_*(Y)$
for $Y$ locally of finite type over $k$
to $\alpha \times \beta$ in $A_*(X \times_k Y)$. From the proof of
Proposition \ref{proposition-compute-bivariant} we see the inverse
to the first arrow in turn sends $c_\alpha$ to the bivariant class
which sends $\beta \in A_*(Y)$ for $Y \to X$ locally of finite type
to $\Delta^!(\alpha \times \beta) = \alpha \cdot \beta$.
From this the final result of the lemma follows.
\end{proof}


0 comments on commit 863eb5a

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