\documentstyle[11pt]{article}
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\title{0.1 Percent of Laura's Dissertation}
\author{Laura Taalman, A.B.D.}
\begin{document}
\pagestyle{empty}
\maketitle
\def\normalbaselines{\baselineskip20pt \lineskip3pt \lineskiplimit3pt}
\def\mapright{\smash{\mathop{\longrightarrow}}}
\def\mapincl{\smash{\mathop{\hookrightarrow}}}
\def\mapup{\Big\uparrow}
\def\mapdown{\Big\downarrow}
\def\mapdowneq{\Big\parallel}
\def\Mapright#1{\smash{\mathop{\longrightarrow}\limits^{#1}}}
\def\Mapincl#1{\smash{\mathop{\hookrightarrow}\limits^{#1}}}
\def\Mapup#1{\Big\uparrow\rlap{$\vcenter{\hbox{$\scriptstyle#1$}}$}}
\def\Mapdown#1{\Big\downarrow\rlap{$\vcenter{\hbox{$\scriptstyle#1$}}$}}
\def\Mapdowneq#1{\Big\parallel\rlap{$\vcenter{\hbox{$\scriptstyle#1$}}$}}
\newcommand{\vsp}{\vspace{1pc}}
\newcommand{\disp}{\displaystyle}
\newcommand{\X}{{\cal X}}
\newcommand{\Y}{{\cal Y}}
\newcommand{\Z}{{\cal Z}}
\def\pitch{\mathbin{\frown \! \! \! \! \mid \, \, \,}}
% had to define because cant get amssymb to work...
% replace by \pitchfork if amssymb is working
% overview?
\section{Smooth Case}
Let $X$ and $Y$ be smooth hypersurfaces in a smooth ambient variety $M$,
where $Z = X \cap Y$ is a smooth global complete intersection and
$X$ meets $Y$ transversely. If $\nu_Z$, $\nu_X$, $\nu_Y$ denote the
normal bundles of $Z$, $X$, and $Y$ in $M$, then the transversality of
$X$ and $Y$ means that $\nu_Z = \nu_X|_Z \oplus \nu_Y|_Z$.
Moreover, the Chern class $c(\nu_Z)$ is given by $1 + Z$, where $Z$
denotes the dual of the fundamental class $[Z]$, {\it i.e.}
$Z \cap [M] = [Z]$ (and similarly for $c(\nu_X)$ and $c(\nu_Y)$). % missing inclusions?
Using these facts we can compute the following formula for the
MacPherson-Chern class $c_*(Z)$ of $Z$ in terms of $X$ and $Y$:
%
\begin{eqnarray*}
c_*(Z) & = & \mbox{Dual }(c(TZ)) \\
& = & c(TZ) \cap [Z] \\
& = & (c(TM|_Z) \cup c(\nu_Z)^{-1}) \cap [Z] \\
& = & c(TM|_Z) \cap (c(\nu_Z)^{-1} \cap [Z]) \\
& = & c(TM|_Z) \cap ((c(\nu_X|_Z) \cup c(\nu_Y|_Z)) \cap [Z]) \\
& = & c(TM|_Z) \cap
(i_X^* (1 + X) \cup i_Y^* (1 + Y)) \cap [Z]) \\
& = & c(TM|_Z) \cap
\left( \frac {[Z]} {i_X^* (1 + X) \, i_Y^* (1 + Y)} \right).
\end{eqnarray*}
%
where $TZ$ is the tangent bundle of $Z$, and $i_X$, $i_Y$ are the inclusions
of $Z$ into $X$, $Y$, respectively.
\section{Guess for General Formula}
Now let $Z = X \cap Y$ be a singular global complete intersection,
where $X = {\cal V}(f)$ and $Y = {\cal V}(g)$ are possibly singular
hypersurfaces intersecting transversely in a smooth ambient variety $M$.
Considering the smooth case, and Aluffi's formula for hypersurfaces,
we might guess that the general formula is of the form:
%
$$c_*(Z) = c(TM|_Z) \cap \pi_* \left( \frac {[\Z] - [E_Z]}
{i_{\X}^*(1 + \X - E_X) \, i_{\Y}^*(1 + \Y - E_Y)} \right).$$
%
\noindent
In the above formula, $\pi \colon Bl_J M \rightarrow M$ is the blowup
along some ideal $J$ (which can be made to factor through the blowup
along the Jacobian ideal
$Jac(f,g)$ consisting of the $2 \times 2$ subdeterminants of the
Jacobian matrix for $f$ and $g$, so that the proper transform of $Z$ in
the blowup along $J$ will be a generalized Nash blowup of $Z$).
The $\Z$, $\X$, $\Y$ and $E_Z$, $E_X$, $E_Y$ then denote the total
transforms and exceptional divisors of $X$, $Y$, and $Z$ under
the blowup, and $i_{\X}$, $i_{\Y}$ denote the inclusions of $\Z$
into $\X$ and $\Y$, respectively.
\section{Special Case where $Y$ is a hyperplane $H$}
If the above guess for the formula in the general case is correct,
we should have the following formula in the case where $Y=H$ is a
hyperplane:
%
$$c_*(Z) = c(TM|_Z) \cap \pi_* \left( \frac {[\Z] - [E_Z]}
{i_{\X}^*(1 + \X - E_X) \, i_{\cal H}^*(1 + {\cal H})} \right). % not cal H, but H?
\eqno({\bf 1})$$
%
\noindent
To prove the above formula we will need the following lemma:
\noindent {\bf Lemma} $\;$
{\it Suppose $X$ is a hypersurface and $H$ a hyperplane
in a smooth ambient variety $M$, meeting transversely not only in the usual
sense, but also in the ``normal direction''. %fix
If $\nu$ is the normal bundle to $X \cap H$ in $X$ (which exists because
of the transversality condition), and $i_{X \cap H}^*$ is the transverse
intersection map $H_*(X) \rightarrow H_{*-2}(X \cap H)$,
Then we have:
%
$$c_*(X \cap H) = (1 + i_H^*H)^{-1} \cap i_{X \cap H}^* c_*(X),$$
%
\noindent
where $i_H \colon X \cap H \rightarrow H$, $i_H^*H = c_1(\nu)$,
and $H$ is the dual of the class $[H]$ in the ambient space $M$.}
\noindent {\bf Proof of Lemma (smooth case)}
A simple calculation; let $i_X$ be the inclusion map for $X \cap H$ in $X$.
Then using the fact that
%
$$TX|_{X \cap H} = \nu \oplus T(X \cap H), \eqno({\bf 2})$$
%
and the definition of the transverse intersection map (in the smooth
case), we have:
%
\begin{eqnarray*}
(1 + i_H^*H)^{-1} \cap i_{X \cap H}^* c_*(X)
& = & (1 + i_H^*H)^{-1} \cap i_X^* (\mbox{Dual }c_*(X)) \cap [X \cap H] \\
& = & (1 + i_H^*H)^{-1} \cap i_X^* c(TX) \cap [X \cap H] \\
& = & (i_X^* c(TX) \cup (1 + c_1(\nu))^{-1}) \cap [X \cap H] \\
& = & (c(TX|_{X \cap H} \cup c(\nu)^{-1}) \cap [X \cap H] \\
& = & c(T(X \cap H)) \cap [X \cap H] \\
& = & c_*(X \cap H) \phantom{XXXXXXXXXXXXXXll} \Box
\end{eqnarray*}
%
\noindent {\bf Proof of Lemma} $\;$
Let $\widetilde{X}$ and $\widetilde{Z}$ be generalized Nash blowups
of $X$ and $Z=X \cap H$, respectively, so that we have a diagram:
%
$$ \matrix{
\widetilde{T}_Z & \mapright & \widetilde{T}_X \cr
\mapdown & & \mapdown \cr
\widetilde{Z} & \mapincl & \widetilde{X} \cr
\Mapdown{\phi} & & \Mapdown{\phi} \cr
Z & \Mapincl{i_X} & X \cr}
$$
%
\noindent where $\widetilde{T}_Z$ and $\widetilde{T}_X$
denote the respective generalized Nash bundles of $Z$ and $X$.
We are abusing notation by writing $\phi$ for the restriction of
$\phi$ to $\widetilde{Z}$ on the left hand side of the
diagram. Note that we are assuming that we have blown up far
enough so that $\phi$ factors through both the Nash blowup of $X$,
and the Nash blowup of $Z$ (when restricted to $\widetilde{Z}$).
We have a statement analagous to (2) in the singular case, namely:
%
$$\widetilde{T}_X|_{Z} = \phi^*\nu \oplus \widetilde{T}_{Z},$$
%
\noindent
which follows from the transversality conditions.
Using this fact, we will show that the Lemma holds when
the MacPherson-Chern classes $c_*(X \cap H)$ and $c_*(X)$
are replaced by
the Mather-Chern classes $c_M(X \cap H)$ and $c_M(X)$.
If $i_{\widetilde{X}}$ is the inclusion of
$\widetilde{Z}$ into $\widetilde{X}$, we have:
%
\begin{eqnarray*}
c_M(X \cap H)
& = & c_M(Z) \\
& = & \phi_*(\mbox{Dual }c(\widetilde{T}_Z)) \\
& = & \phi_*(\mbox{Dual } \left(
\frac {c(\widetilde{T}_X|_Z)} {c(\phi^* \nu)} \right) \\
& = & \phi_*(\mbox{Dual } \left(
\frac {i_{\widetilde{X}}^* c(\widetilde{T}_X)} {\phi^* c(\nu)} \right) \\
& = & \phi_* \left(
\frac {i_{\widetilde{X}}^* c(\widetilde{T}_X)} {\phi^* c(\nu)}
\cap [\widetilde{Z}] \right) \\
& = & \phi_* (\phi^* c(\nu)^{-1} \cap (i_{\widetilde{X}}^* c(\widetilde{T}_X)
\cap [\widetilde{Z}])) \\
& = & c(\nu)^{-1} \cap \phi_* (i_{\widetilde{X}}^* c(\widetilde{T}_X)
\cap [\widetilde{Z}]),
\end{eqnarray*}
%
\noindent
and thus to prove the Lemma (with the MacPherson-Chern classes replaced by
Mather-Chern classes), it suffices to prove that:
%
$$i_{X \cap H}^* c_M(X) =
\phi_* (i_{\widetilde{X}}^* c(\widetilde{T}_X) \cap [\widetilde{Z}]).$$
%
Assuming that $i_{\widetilde{X} \cap \widetilde{H}}^* [\widetilde{X}] =
[\widetilde{X} \pitch \widetilde{H}] = [\widetilde{Z}]$
(here $\widetilde{H}$ is the proper transform of $H$ under the blowup
given by $\phi$), we have:
%
\begin{eqnarray*}
i_{X \cap H}^* c_M(X)
& = & i_{X \cap H}^* \phi_*(\mbox{Dual }c(\widetilde{T}_X)) \\
& = & i_{X \cap H}^* \phi_* (c(\widetilde{T}_X) \cap [\widetilde{X}] \\
& = & \phi_* i_{\widetilde{X} \cap \widetilde{H}}^*
(c(\widetilde{T}_X) \cap [\widetilde{X}] \\
& = & \phi_* (i_{\widetilde{X}}^* c(\widetilde{T}_X) \cap
i_{\widetilde{X} \cap \widetilde{H}}^* [\widetilde{X}] \\
& = & \phi_* (i_{\widetilde{X}}^* c(\widetilde{T}_X) \cap
[\widetilde{Z}],
\end{eqnarray*}
%
\noindent
Now since the Lemma holds for Mather-Chern classes, and the MacPherson-Chern
classes are weighted sums of the Mather-Chern classes, the Lemma as originally
stated has been shown. \hfill $\Box$
We now show that in the special case where $Y=H$ is a hyperplane,
$X$ is a (possibly singular) hypersurface, and $Z=X \cap H$,
we have equation (1). In what follows we assume that the transverse
intersection map satisfies the following ``projection formula'' for
any $\alpha \in H^*(X)$ and $a \in H_*(X)$, where $i_X \colon X \cap H
\rightarrow X$:
%
$$i_{X \cap H}^* (\alpha \cap a) = i_X^* \alpha \cap i_{X \cap H}^* a.$$
%%
Using this, Aluffi's formula, and the Lemma above, we prove formula (1):
%
\begin{eqnarray*}
c_*(Z) & = & c_*(X \cap H) \\
& = & (1 + i_H^*H)^{-1} \cap i_{X \cap H}^* c_*(X) \\
& = & (1 + i_H^*H)^{-1} \cap i_{X \cap H}^*
\left( c(TM|x) \cap \pi_* \left(
\frac {[\X] - [E_X]} {1 + \X - E_X} \right) \right) \\
& = & (1 + i_H^*H)^{-1} \cap \left( i_X^*c(TM|X) \cap
i_{X \cap H}^* \pi_* \left(
\frac {[\X] - [E_X]} {1 + \X - E_X} \right) \right) \\
& = & c(TM|Z) \cap \left( (1 + i_H^*H)^{-1} \cap
i_{X \cap H}^* \pi_* \left(
\frac {[\X] - [E_X]} {1 + \X - E_X} \right) \right) \\
& = & c(TM|Z) \cap \left( (1 + i_H^*H)^{-1} \cap
\pi_* i_{\X \cap {\cal H}}^* \left(
\frac {[\X] - [E_X]} {1 + \X - E_X} \right) \right) \\
& = & c(TM|Z) \cap \left( (1 + i_H^*H)^{-1} \cap \pi_* \left(
\frac {i_{\X \cap {\cal H}}^* ([\X] - [E_X])}
{i_X^* (1 + \X - E_X)} \right) \right) \\
& = & c(TM|Z) \cap \pi_* \left( \pi^* (1 + i_H^*H)^{-1} \cap \left(
\frac {i_{\X \cap {\cal H}}^* ([\X] - [E_X])}
{i_X^* (1 + \X - E_X)} \right) \right) \\
& = & c(TM|Z) \cap \pi_* \left(
\frac {i_{\X \cap {\cal H}}^* ([\X] - [E_X])}
{i_{\X} (1 + \X - E_X) \, \pi^* (1 + i_H^*H)} \right) \\
& = & c(TM|Z) \cap \pi_* \left(
\frac {i_{\X \cap {\cal H}}^*[X] - i_{\X \cap {\cal H}}^*[E_X]}
{i_{\X} (1 + \X - E_X) \, i_H^* \pi^* (1 + H)} \right) \\
& = & c(TM|Z) \cap \pi_* \left(
\frac {[\X \pitch {\cal H}] - [E_X \pitch {\cal H}]}
{i_{\X} (1 + \X - E_X) \, i_H^* (1 + {\cal H})} \right) \\
& = & c(TM|Z) \cap \pi_* \left(
\frac {[\Z] - [E_Z]}
{i_{\X} (1 + \X - E_X) \, i_H^* (1 + {\cal H})} \right).
\end{eqnarray*}
\noindent
We have used the following notations and assumptions:
%
\begin{itemize}
\item $\pi \colon Bl_J M \rightarrow M$ is the blowup along
the Jacobian ideal $J = (Jac(f), f)$ for $X$. Note that if we
wanted to we could have instead used
$J = (Jac(f,g), f, g) \cdot (Jac(f), f)$; this would give
us a blowup that factored through both the
Nash blowup of $Z$, and the blowup that gives us Aluffi's
formula (which holds under further blowups). $\X$, ${\cal H}$
denote the total transforms of $X$ and $H$ under the blowup $Bl_J M$,
and $E_X$ denotes the exceptional divisor over $X$. % actually E_X = E_Y = E_Z = E? or not
\item $i_{X \cap H}^* \pi_* = \pi_* i_{\X \cap {\cal H}}^*$,
where $i_{\X \cap {\cal H}}^*$ is the transverse intersection map
for $\X \cap {\cal H}$.
\item $i_{\X \cap {\cal H}}^*[\X]$ is equal to
$[\X \pitch {\cal H}]$, and this is the same class as $[\Z]$;
similarly for $i_{\X \cap {\cal H}}^*[E_X]$.
\end{itemize}
\section{Special Case where $Y$ is a smooth hypersurface}
When $Y$ is a smooth hypersurface, we can always find local ambient
coordinates so that $Y$ is given by the vanishing of one of the coordinate
functions, {\it i.e.} $Y$ can be locally thought of as a hyperplane.
Thus the arguments given in Section 3 above, which are local in nature,
can be generalized to prove the case when $Y$ is a smooth hypersurface.
\section{The General Case}
It seems that, in the general case, $\pi \colon Bl_J M \rightarrow M$
is likely to be the blowup along the product of the Jacobian ideals for
$X$ and for $Y$. Perhaps it will be possible to, say, blow up the
Jacobian ideal for $Y$, and then apply the special case at the blow-up
level, to arrive at the general formula?
\section{Relaxing Transversality}
Everything here requires that $Y$ be transverse to all the strata of $X$;
thus in the simple case where $X$ is a hypersurface with an isolated singular
point, and thus having a stratum consisting of only that isolated singular
point, $Y$ would have to miss the singular point entirely. Perhaps
this transversality condition can be relaxed somehow, so that although
not true for the intersection of $X$ and $Y$, it would be true on the blow-up
level? Then we might be able apply the above general case (requiring transversality)
to the blow-up level to reach an even more general formula.
\end{document}