# On the analytic structure of double and triple points in the target of
finite holomorphic multi-germs

Juan J. Nuño-Ballesteros, Guillermo Peñafort Sanchis, Cinzia Villa
Departament de Matemàtiques, Universitat de València, Campus de Burjassot,
46100 Burjassot, Spain<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

We study the analytic structure of the double and triple point spaces
$M_{2}(f)$ and $M_{3}(f)$ of finite multi-germs
$f\colon(X,S)\to(\mathbb{C}^{n+1},0)$, based on results of Mond and Pellikaan
for the mono-germ case. We show that these spaces are Cohen-Macaulay, provided
that certain dimensional conditions are satisfied, and give explicit
expressions for their defining ideals in terms of those of their mono-germ
branches.

###### 2010 Mathematics Subject Classification:

32S25, 14J17, 58K20, 14B05, 13C14

##  1\. Introduction

In this text we study double and triple points of finite holomorphic multi-
germs

$f\colon(X,S)\to(\mathbb{C}^{n+1},0),$

where $X$ stands for a Cohen-Macaulay analytic space of dimension $n$. By
saying that $(X,S)$ is a multi-germ, the set $S$ is assumed to be finite, and
by saying that $X$ is an analytic space (in contrast to a complex space), it
is understood that $X$ is reduced. These assumptions and the fact that $f$ is
holomorphic will not be repeated every time these objects appear, but extra
hypothesis will sometimes be added on top of them.

There are several ways to define double, triple and, in general, $k$-multiple
point spaces, and this work focusses on the spaces
$M_{k}(f)\subseteq(\mathbb{C}^{n+1},0)$, defined by means of Fitting ideals of
the pushforward module $f_{*}\mathcal{O}_{(X,S)}$. These spaces were
introduced by Mond and Pellikaan [5] , who studied their analytic structures
in the mono-germ case. They show that, for mono-germs, the double and triple
point spaces $M_{2}(f)$ and $M_{3}(f)$ are determinantal if they have the
right dimension. As it turns out their proofs do not apply in the multi-germ
case, as explained in Example 1.4. The goal of this work is to study the
analytic structure of $M_{2}(f)$ and $M_{3}(f)$ for multi-germs, giving
conditions for them to be Cohen-Macaulay and showing explicit expressions for
their ideals in terms of those of the spaces $M_{k}$ of the mono-germ branches
of $f$.

For a finite multi-germ $f$ as above, the pushforward
$f_{*}\mathcal{O}_{(X,S)}$ is a finitely presented $\mathcal{O}_{n+1}$-module,
and it admits a square presentation matrix $\lambda$ [5, Section 2.2].
Assuming $\lambda$ to be of size $q\times q$, the $k$th Fitting ideal of
$f_{*}\mathcal{O}_{(X,S)}$ is defined as follows:

$F_{k}(f_{*}\mathcal{O}_{(X,S)})=\begin{cases}\langle\text{$q-k$ size minors
of $\lambda$}\rangle,&\text{ if $k<q$}\\\
\mathcal{O}_{n+1},&\text{otherwise}\end{cases}$

Then, the _(target) $k$-multiple point space_ of $f$ is the complex subspace
$M_{k}(f)\subset(\mathbb{C}^{n+1},0)$, defined as

$M_{k}(f)=V(F_{k-1}(f_{*}\mathcal{O}_{(X,S)})).$

This definition is based on the fact that, set-theoretically, $M_{k}(f)$ is
the germ of the set of points $y\in\mathbb{C}^{n+1}$ where
$(f_{*}\mathcal{O}_{X})_{y}$ requires at least a number $k$ of generators [5,
Proposition 1.5]. Geometrically, this means that $M_{k}(f)$ is given by the
points $y\in\mathbb{C}^{n+1}$ with at least $k$ preimages, counting with
multiplicity. Observe that $M_{1}(f)$ is the hypersurface with equation
$\det\lambda$ in $(\mathbb{C}^{n+1},0)$ and coincides, as a set-germ, with the
image $(f(X),0)$.

######  Definition 1.1.

Given $f\colon(X,S)\to(\mathbb{C}^{n+1},0)$, we say that $M_{k}(f)$ is
_dimensionally correct_ if $\operatorname{dim}M_{k}(f)=n-k+1$.

From the fact that $(X,S)$ is assumed to be reduced and $f$ is assumed to be
finite, it follows that $f$ must be locally an embedding on an open dense
subset of $X$. From this it follows that $M_{2}(f)$ is dimensionally correct
if and only if $f$ is generically one-to-one. Moreover, $f$ is generically
one-to-one if and only if $M_{1}(f)$ is reduced [5, Proposition 3.2].

######  Theorem 1.2 (Mond-Pellikaan).

Any mono-germ $f\colon(X,0)\to(\mathbb{C}^{n+1},0)$ satisfies
$\operatorname{dim}M_{2}(f)\geq n-1$. If the equality holds, then $M_{2}(f)$
is Cohen-Macaulay. If furthermore $M_{3}(f)=\emptyset$, then $M_{2}(f)$ is a
complete intersection.

The proof of this result consists on showing that the presentation matrix can
be chosen to be symmetric and in such a way that
$F_{1}(f_{*}\mathcal{O}_{X,0})$ is generated by minors of the submatrix
obtained by removing the first row [5, Theorem 3.4]. This shows $M_{2}(f)$ to
be determinantal whenever it has dimension $n-1$ (observe that in codimension
two the determinantal and Cohen-Macaulay conditions are equivalent, by the
Hilbert-Burch theorem). The complete intersection case also follows from the
fact that we can remove the first row, taking into account that the hypothesis
that $M_{3}(f)=\emptyset$ is equivalent to the fact that the above matrix has
size $2\times 2$.

In the case where $X$ is Gorenstein, the presentation matrix may be chosen in
such a way that $F_{2}(f_{*}\mathcal{O}_{X,0})$ is generated by minors of the
submatrix obtained by removing the first row and first column [5, Theorem
4.3]. This implies what follows:

######  Theorem 1.3 (Mond-Pellikaan).

Any mono-germ $f\colon(X,0)\to(\mathbb{C}^{n+1},0)$, with $(X,0)$ Gorenstein,
satisfies $\operatorname{dim}M_{3}(f)\geq n-2$. If the equality holds, then
$M_{3}(f)$ is Cohen-Macaulay.

Unfortunately, in the case of multi-germs, sometimes no proper submatrix of
the presentation matrix has the right minors to generate
$F_{1}(f_{*}\mathcal{O}_{(X,S)})$ nor $F_{2}(f_{*}\mathcal{O}_{(X,S)})$. This
is illustrated by the following example:

######  Example 1.4.

Consider the multi-germ
$f\colon\bigsqcup_{i=1}^{3}(\mathbb{C}^{2},0)\to(\mathbb{C}^{3},0)$,
parametrizing the planes $\\{X=0\\}$, $\\{Y=0\\}$ and $\\{Z=0\\}$ in
$(\mathbb{C}^{3},0)$. Since the branches are immersive mono-germs, their
presentation matrices are $1\times 1$ matrices whose only entry are the
generators of the ideals of the corresponding planes. The presentation matrix
of $f_{*}\mathcal{O}_{(X,S)}$ is then the diagonal matrix

$\lambda=\begin{bmatrix}X&0&0\\\ 0&Y&0\\\ 0&0&Z\\\ \end{bmatrix}.$

As expected, $M_{2}(f)$ is the union of the coordinate axes of
$(\mathbb{C}^{3},0)$, and $M_{3}(f)$ is the origin, both spaces with reduced
structure. One sees that no proper submatrix of $\lambda$ has the adequate
entries to generate the corresponding equations.

######  Remark 1.5.

There are multiple point spaces $M_{k}(f)$ for finite (global) mappings
$f\colon X\to Y$, defined by the corresponding sheaf theoretical versions
$\mathcal{F}_{k}$ of the Fitting ideals $F_{k}$. The properties that we show
for multi-germs imply the corresponding properties for mappings, because the
stalks of $\mathcal{F}_{k}$ are the ideals $F_{k}$ associated to a multi-germ.
We omit these translations to the global setting, as they are quite obvious.

##  2\. Target multiple-point spaces for multi-germs

To simplify writing, for the remainder of the text we fix some notations
regarding multi-germ branches. Given $f\colon(X,S)\to(\mathbb{C}^{n+1},0)$,
with $S=\\{x^{(1)},\dots,x^{(r)}\\}$, for each $i\in\\{1,\dots,r\\}$, we write

$f^{(i)}=f|_{(X,x^{(i)})},\qquad
F_{j}^{(i)}=F_{j}(f^{(i)}_{*}\mathcal{O}_{X,x^{(i)}})\quad\text{and}\quad
M_{j}^{(i)}=M_{j}(f^{(i)}).$

From the equality
$f_{*}\mathcal{O}_{(X,S)}=\bigoplus_{i=1}^{r}f^{(i)}_{*}\mathcal{O}_{X,x^{(i)}}$,
one observes that the associated Fitting ideals are generated by minors of a
block diagonal matrix, and then

$F_{k}(f_{*}\mathcal{O}_{(X,S)})=\sum_{j_{1}+\dots+j_{r}=k}F_{j_{1}}^{(1)}\cdot\ldots\cdot
F_{j_{r}}^{(r)}.$

In combination with the inclusions $M_{j}^{(i)}\subseteq M_{j-1}^{(i)}$, the
previous equality gives the _set theoretical_ decomposition

(1) $M_{k}(f)=\bigcup_{\begin{subarray}{c}\vskip 3.0pt plus 1.0pt minus
1.0pt1\leq\ell\leq r\\\ \vskip 3.0pt plus 1.0pt minus 1.0pt1\leq
i_{1}<\dots<i_{\ell}\leq r\\\
j_{1}+\dots+j_{\ell}=k\end{subarray}}M_{j_{1}}^{(i_{1})}\cap\ \dots\ \cap
M_{j_{\ell}}^{(i_{\ell})}.$

This decomposition corresponds to the intuitive idea of what multiple points
of multi-germs should be. For example, it states that, as a set, the triple
points of a bigerm are the union of the triple points of the first branch, the
intersection of the double points of the first branch with the image of the
second, the intersection of the image of the first branch with the double
points of the second and, finally, the triple points of the second branch.

It is natural to ask to what extent this decomposition gives information about
$M_{k}(f)$ as a complex space. To be precise, we want to know if the ideal
$F_{k}(f_{*}\mathcal{O}_{(X,S)})$ can be expressed as

$F_{k}(f_{*}\mathcal{O}_{(X,S)})=\bigcap_{\begin{subarray}{c}\vskip 3.0pt plus
1.0pt minus 1.0pt1\leq\ell\leq r\\\ \vskip 3.0pt plus 1.0pt minus 1.0pt1\leq
i_{1}<\dots<i_{\ell}\leq r\\\
j_{1}+\dots+j_{\ell}=k-\ell+1\end{subarray}}F_{j_{1}}^{(i_{1})}+\dots+F_{j_{\ell}}^{(i_{\ell})}.$

We will use the following three basic lemmas. The first follows easily from
localizing at associated primes and the second follows from the interpretation
of $\operatorname{depth}$ in terms of $\operatorname{Ext}$, by looking at the
long exact sequence associated to a short exact sequence (see e.g. [8, Tag
00LX]). For completeness, we include a proof of the third lemma.

######  Lemma 2.1.

Let $R$ be a noetherian local ring, and let $J_{1},\dots,J_{r}$ and $I$ be
ideals in $R$. Assume that for all $i\neq j$ the condition
$J_{i}\not\subseteq\mathfrak{p}$ holds for all
$\mathfrak{p}\in\operatorname{Ass}(J_{j}+I)$. Then,
$\bigcap_{i}(J_{i}+I)=(\bigcap_{i}J_{i})+I$.

######  Lemma 2.2 (Depth Lemma).

Let $R$ be a local Noetherian ring and let $0\to N^{\prime}\to N\to
N^{\prime\prime}\to 0$ be a short exact sequence of nonzero finitely generated
$R$-modules.

  1. (1)

$\operatorname{depth}N\geq\min\\{\operatorname{depth}N^{\prime},\operatorname{depth}N^{\prime\prime}\\}$.

  2. (2)

$\operatorname{depth}N^{\prime\prime}\geq\min\\{\operatorname{depth}N,\operatorname{depth}N^{\prime}-1\\}$.

  3. (3)

$\operatorname{depth}N^{\prime}\geq\min\\{\operatorname{depth}N,\operatorname{depth}N^{\prime\prime}+1\\}$.

######  Lemma 2.3.

Let $I$ and $J$ be ideals in a local noetherian ring $R$, such that
$\operatorname{depth}\frac{R}{I}<\operatorname{depth}R$,
$\operatorname{depth}\frac{R}{J}<\operatorname{depth}R$ and
$\operatorname{depth}\frac{R}{IJ}<\operatorname{depth}R$. Then

$\operatorname{depth}\frac{R}{IJ}\geq\operatorname{depth}\frac{R}{I}+\operatorname{depth}\frac{R}{J}-\operatorname{depth}R+1.$

###### Proof.

The hypotheses imply
$\operatorname{depth}\frac{R}{I}=\operatorname{depth}I-1$,
$\operatorname{depth}\frac{R}{J}=\operatorname{depth}J-1$ and
$\operatorname{depth}\frac{R}{IJ}=\operatorname{depth}IJ-1$. By the third of
these equalities, it suffices to show that
$\operatorname{depth}IJ\geq\operatorname{depth}\frac{R}{I}+\operatorname{depth}\frac{R}{J}-\operatorname{depth}R+2$
and, by the Auslander-Buchsbaum formula, this amounts to showing
$\operatorname{pd}IJ\leq
2\operatorname{depth}R-\operatorname{depth}\frac{R}{I}-\operatorname{depth}\frac{R}{J}-2$.
Again by the Auslander-Buchsbaum formula, we obtain

$\displaystyle\operatorname{pd}IJ=\operatorname{pd}(I\otimes J)$
$\displaystyle\leq\operatorname{pd}I+\operatorname{pd}J=2\operatorname{depth}R-\operatorname{depth}I-\operatorname{depth}J$
$\displaystyle=2\operatorname{depth}R-\operatorname{depth}\frac{R}{I}-\operatorname{depth}\frac{R}{J}-2.\qed$

Now we are ready to prove that, when they are dimensionally correct, double
points of multi-germs are Cohen-Macaulay and to show how to compute them under
favorable circumstances:

######  Theorem 2.4.

If $f\colon(X,S)\to(\mathbb{C}^{n+1},0)$ is finite and generically one-to-one,
then $M_{2}(f)$ is Cohen-Macaulay of dimension $n-1$. If furthermore
$M_{3}(f)$ is dimensionally correct, then the ideal of $M_{2}(f)$ satisfies

$F_{1}(f_{*}\mathcal{O}_{(X,S)})=\bigcap_{1\leq i\leq
r}F_{1}^{(i)}\cap\bigcap_{1\leq i<j\leq
r}\big{(}F_{0}^{(i)}+F_{0}^{(j)}\big{)}.$

###### Proof.

We proceed by induction over $r=|S|$. If $r=1$, then $M_{2}(f)$ is Cohen-
Macaulay by Theorem 1.2 and the expression for the defining ideal reads
$F_{1}(f_{*}\mathcal{O}_{(X,S)})=F_{1}(f_{*}\mathcal{O}_{(X,S)})$, so there is
nothing to prove. Now assume the statement to be true for multi-germs with no
more than $r-1$ branches.

Showing $M_{2}(f)$ to be Cohen-Macaulay amounts to showing

$\operatorname{depth}\frac{\mathcal{O}_{n+1}}{F_{1}(f_{*}\mathcal{O}_{X,S})}=n-1,$

because the generically one-to-one condition implies
$\operatorname{dim}M_{2}(f)=n-1$. To exploit the inductive hypothesis,
consider $S^{\prime}=\\{x^{(1)},\dots,x^{(r-1)}\\}$ and let

$f^{\prime}=f_{(X,S^{\prime})}\quad\text{and}\quad
F^{\prime}_{j}=F_{j}(f^{\prime}_{*}\mathcal{O}_{(X,S^{\prime})}),$

so that
$F_{1}(f_{*}\mathcal{O}_{X,S})=F_{1}^{\prime}F_{0}^{(r)}+F_{0}^{\prime}F_{1}^{(r)}$.

Since $F_{0}^{\prime}$ and $F_{0}^{(r)}$ are principal, it follows from Lemma
2.3 and the induction hypothesis that

$\operatorname{depth}\frac{\mathcal{O}_{n+1}}{F_{1}^{\prime}F_{0}^{(r)}}\geq
n-1\quad\text{and}\quad\operatorname{depth}\frac{\mathcal{O}_{n+1}}{F_{0}^{\prime}F_{1}^{(r)}}\geq
n-1.$

Applying Item (2) of the Depth Lemma to the short exact sequence

$0\to\frac{\mathcal{O}_{n+1}}{F_{1}^{\prime}F_{0}^{(r)}\cap
F_{0}^{\prime}F_{1}^{(r)}}\to\frac{\mathcal{O}_{n+1}}{F_{1}^{\prime}F_{0}^{(r)}}\oplus\frac{\mathcal{O}_{n+1}}{F_{0}^{\prime}F_{1}^{(r)}}\to\frac{\mathcal{O}_{n+1}}{F_{1}^{\prime}F_{0}^{(r)}+F_{0}^{\prime}F_{1}^{(r)}}\to
0$

reduces the problem to showing that

$\operatorname{depth}\frac{\mathcal{O}_{n+1}}{F_{1}^{\prime}F_{0}^{(r)}\cap
F_{0}^{\prime}F_{1}^{(r)}}\geq n.$

From the inclusions $F_{0}^{\prime}\subseteq F_{1}^{\prime}$ and
$F_{0}^{(r)}\subseteq F_{1}^{(r)}$, one obtains

$F^{\prime}_{1}F^{(r)}_{0}\cap F^{\prime}_{0}F^{(r)}_{1}\subseteq
F^{\prime}_{1}\cap F^{(r)}_{0}\cap F^{\prime}_{0}\cap
F^{(r)}_{1}=F^{\prime}_{0}\cap F^{(r)}_{0},$ $F^{\prime}_{1}F^{(r)}_{0}\cap
F^{\prime}_{0}F^{(r)}_{1}\supseteq F^{\prime}_{0}F^{(r)}_{0}.$

Since $F_{0}^{\prime}$ and $F_{0}^{(r)}$ are principal ideals defining the
images of $f^{\prime}$ and $f^{(r)}$, the generically one-to-one condition
forces $F^{\prime}_{0}F^{(r)}_{0}=F^{\prime}_{0}\cap F^{(r)}_{0}$. From the
two inclusions above, now we obtain $F^{\prime}_{1}F^{(r)}_{0}\cap
F^{\prime}_{0}F^{(r)}_{1}=F^{\prime}_{0}F^{(r)}_{0}$. Being the ideal of a
hypersurface in $(\mathbb{C}^{n+1},0)$, its quotient must have depth $n$, as
desired. This finishes the proof that $M_{2}(f)$ is Cohen-Macaulay.

Now assume $M_{3}(f)$ to be dimensionally correct. This forces
$F_{0}^{\prime}$ and $F_{0}^{(r)}$ to be generated by regular elements in
$\mathcal{O}_{n+1}/F_{1}^{(r)}$ and $\mathcal{O}_{n+1}/F^{\prime}_{1}$. This
is true because $M_{2}(f^{\prime})$ and $M_{2}^{(r)}$ are Cohen-Macaulay of
dimension $n-1$, and points in $M_{2}(f^{\prime})\cap M_{1}(f^{(r)})$ or
$M_{1}(f^{\prime})\cap M_{2}(f^{(r)})$ are triple points of $f$, which are
assumed to be have dimension $(n-2)$. Then, we obtain

$\displaystyle F_{1}(f_{*}\mathcal{O}_{(X,S)})$
$\displaystyle=F_{1}^{\prime}F_{0}^{(r)}+F_{0}^{\prime}F_{1}^{(r)}$
$\displaystyle=\big{(}F_{1}^{\prime}\cap
F_{0}^{(r)}\big{)}+\big{(}F_{0}^{\prime}\cap F_{1}^{(r)}\big{)}$
$\displaystyle=F_{1}^{\prime}\cap
F_{1}^{(r)}\cap\big{(}F_{0}^{\prime}+F_{0}^{(r)}\big{)}$
$\displaystyle=\bigcap_{1\leq i\leq r-1}F_{1}^{(i)}\cap\bigcap_{1\leq i<j\leq
r-1}\big{(}F_{0}^{(i)}+F_{0}^{(j)}\big{)}\cap
F_{1}^{(r)}\cap\big{(}F_{0}^{\prime}+F_{0}^{(r)}\big{)},$

where the third equality follows from the inclusions $F_{0}^{\prime}\subseteq
F_{1}^{\prime}$ and $F_{0}^{(r)}\subseteq F_{1}^{(r)}$ and the last equality
follows from the induction hypothesis (observe that $M_{3}(f^{\prime})$ is
dimensionally correct, because every ingredient in decomposition of
$M_{3}(f^{\prime})$ in (1) appears in the decomposition of $M_{3}(f)$ as
well). Comparign this last expression to the formula in the statement shows
that it is enough to prove that

$F_{0}^{\prime}+F_{0}^{(r)}=\bigcap_{1\leq i\leq
r-1}\big{(}F_{0}^{(i)}+F_{0}^{(r)}\big{)}.$

The generically one-to-one condition implies

$F_{0}^{\prime}=\bigcap_{1\leq i\leq r-1}F_{0}^{(i)}$

and the fact that $F_{0}^{(r)}$ is generated by a regular element in
$\mathcal{O}_{n+1}/F_{0}^{\prime}$. The desired equality now follows, taking
into account that the ideals $F_{0}^{(i)}$ are principal. ∎

######  Example 2.5.

Let
$f\colon(\mathbb{C}^{1},0)\sqcup(\mathbb{C}^{1},0)\sqcup(\mathbb{C}^{1},0)\to(\mathbb{C}^{2},0)$
be the map-germ with branches $f^{(1)}(x)=(x,0)$, $f^{(2)}(x)=(0,x)$ and
$f^{(3)}(x)=(x,x)$. Since $f^{(i)}$ is immersive, it follows that
$F_{0}^{(i)}$ is the reduced ideal of the image of $f^{(i)}$ and
$F_{1}^{(i)}=\mathcal{O}_{2}$. Therefore,

$\displaystyle F_{1}(f_{*}(\mathcal{O}_{(X,S)})$
$\displaystyle=F_{1}^{(1)}F_{0}^{(2)}F_{0}^{(3)}+F_{0}^{(1)}F_{1}^{(2)}F_{0}^{(3)}+F_{0}^{(1)}F_{0}^{(2)}F_{1}^{(3)}$
$\displaystyle=\langle X(X-Y),Y(X-Y),XY\rangle.$

At the same time, the right hand side of the equality in Theorem 2.4 gives

$F_{1}^{(1)}\cap F_{1}^{(2)}\cap
F_{1}^{(2)}\cap\big{(}F_{0}^{(1)}+F_{0}^{(2)}\big{)}\cap\big{(}F_{0}^{(1)}+F_{0}^{(3)}\big{)}\cap\big{(}F_{0}^{(2)}+F_{0}^{(3)}\big{)}=\langle
X,Y\rangle.$

The disagreement comes from the fact that the origin is a triple point of $f$
and, in these dimensions, $M_{3}(f)$ must be empty to be dimensionally
correct.

To study triple points, we will use a typical trick in the theory of singular
mappings, consisting on replacing the original mapping by an unfolding having
desired properties. When studying map-germs
$f\colon(\mathbb{C}^{n},S)\to(\mathbb{C}^{n+1},0)$, an unfolding of $f$ is
simply a map germ of the form
$(f_{t},t)\colon(\mathbb{C}^{n},S)\times(\mathbb{C}^{r},0)\to(\mathbb{C}^{n+1}\times\mathbb{C}^{r},0)$,
but when working with map-germs $f\colon(X,S)\to(\mathbb{C}^{n+1},0)$, it is
convenient to extend this notion to allow for the source space $X$ to be
deformed as well. A possible definition is as follows:

Given a multi-germ $f\colon X\to Y$ between complex spaces, an unfolding of
$f$ over $\Delta=(\mathbb{C}^{r},0)$ is a multi-germ $\phi\colon\mathcal{X}\to
Y\times\Delta$ such that, taking the commutative diagram

${\mathcal{X}}$${Y\times\Delta}$${\Delta}$$\scriptstyle{\phi}$

where $Y\times\Delta\to\Delta$ is the usual projection from the product, the
following conditions are satisfied:

  1. (1)

$\mathcal{X}\to\Delta$ is flat family, with central fibre
$\mathcal{X}_{0}=\mathcal{X}\times_{\Delta}\\{0\\}$ isomorphic to $X$.

  2. (2)

The above isomorphism $\mathcal{X}_{0}\to X$ produces a commutative diagram

${\mathcal{X}_{0}}$${Y}$${X}$$\scriptstyle{\phi_{0}}$$\scriptstyle{f}$

In particular, this provides a family of mappings
$\mathcal{X}_{t}\xrightarrow{\phi_{t}}Y$ so that $\phi_{0}$ is equivalent to
$f$.

For our purposes, the crucial property is that unfoldings of finite multi-
germs satisfy

$M_{k}(f)\cong M_{k}(\phi)\times_{\Delta}\\{0\\}.$

This can be found in more general form in [4, Proposition 11.6]. Now we are
ready to state our main result about triple points in the target.

######  Theorem 2.6.

Given $f\colon(X,S)\to(\mathbb{C}^{n+1},0)$, assume $f$ to be generically one-
to-one and that $M_{3}^{(i)}=\emptyset$, for $2\leq i\leq r$. If
$M_{3}^{(1)}\neq\emptyset$, then assume $(X,x^{(1)})$ to be Gorenstein. Assume
also that $f$ admits an unfolding $\phi$ such that $M_{3}(\phi)$ and
$M_{4}(\phi)$ are dimensionally correct. If $M_{3}(f)$ is dimensionally
correct, then it is Cohen-Macaulay. If furthermore $M_{4}(f)$ is dimensionally
correct, then the ideal of $M_{3}(f)$ satisfies

$\displaystyle F_{2}(f_{*}\mathcal{O}_{(X,S)})=$ $\displaystyle
F_{2}^{(1)}\cap\bigcap_{1\leq i<j\leq
r}\big{(}F_{1}^{(i)}+F_{0}^{(j)}\big{)}\cap\big{(}F_{0}^{(i)}+F_{1}^{(j)}\big{)}$
$\displaystyle\cap\bigcap_{1\leq i<j<k\leq
r}\big{(}F_{0}^{(i)}+F_{0}^{(j)}+F_{0}^{(k)}\big{)}.$

###### Proof.

From the isormorphism $M_{k}(f)\cong M_{k}(\phi)\times_{\Delta}\\{0\\}$ and
the fact that $\\{0\\}$ is regularly embedded in $\Delta$, it follows that, if
$M_{3}(\phi)$ is Cohen-Macaulay, then so is $M_{3}(f)$, because the hypothesis
that $M_{3}(f)$ is dimensionally correct implies that it is regularly embedded
in $M_{3}(\phi)$. To keep the notation we are used to, instead of replacing
$f$ by $\phi$ and showing $M_{3}(\phi)$ to be Cohen Macaulay, we may just
assume $M_{3}(f)$ and $M_{4}(f)$ to be dimensionally correct and show
$M_{3}(f)$ to be Cohen-Macaulay.

We proceed by induction on $|S|$ as in the proof of Theorem 2.4. The case of
$r=1$ is precisely Theorem 1.3 and, for $r\geq 2$, we again use the notation
$S^{\prime}=\\{x^{(1)},\dots,x^{(r-1)}\\}$ and let

$f^{\prime}=f_{(X,S^{\prime})}\quad\text{and}\quad
F^{\prime}_{j}=F_{j}(f^{\prime}_{*}\mathcal{O}_{(X,S^{\prime})}).$

The ideal $F_{0}^{(r)}$ is principal and, since $f^{(r)}$ is generically one-
to-one and $M_{3}^{(r)}(f)=\emptyset$, the ideal $F_{1}^{(r)}$ is generated by
a regular sequence, by Theorem 1.2. From the fact that $M_{4}(f)$ is
dimensionally correct, it follows that the generators of $F_{0}^{(r)}$ and
$F_{1}^{(r)}$ may be chosen so that they form regular sequences in
$\mathcal{O}_{n+1}/F_{2}^{\prime}$ and $\mathcal{O}_{n+1}/F_{1}^{\prime}$,
respectively. This implies $F_{2}^{\prime}F_{0}^{(r)}=F_{2}^{\prime}\cap
F_{0}^{(r)}$ and $F_{1}^{\prime}F_{1}^{(r)}=F_{1}^{\prime}\cap F_{1}^{(r)}$.
Then, taking into account that $F_{2}^{(r)}=\mathcal{O}_{n+1}$, and exploiting
the inclusions $F_{0}^{\prime}\subseteq F_{1}^{\prime}\subseteq
F_{2}^{\prime}$ and $F_{0}^{(r)}\subseteq F_{1}^{(r)}$, one obtains

$\displaystyle F_{2}(f_{*}\mathcal{O}_{X,S})$
$\displaystyle=F_{2}^{\prime}F_{0}^{(r)}+F_{1}^{\prime}F_{1}^{(r)}+F_{0}^{\prime}$
$\displaystyle=F_{2}^{\prime}\cap F_{0}^{(r)}+F_{1}^{\prime}\cap
F_{1}^{(r)}+F_{0}^{\prime}$
$\displaystyle=F^{\prime}_{2}\cap\big{(}F^{\prime}_{1}+F^{(r)}_{0}\big{)}\cap\big{(}F^{\prime}_{0}+F^{(r)}_{1}\big{)}.$

The same inclusions also imply that

$\Big{(}F^{\prime}_{2}\cap\big{(}F^{\prime}_{1}+F^{(r)}_{0}\big{)}\Big{)}+F^{\prime}_{0}+F^{(r)}_{1}=\Big{(}F^{\prime}_{2}\cap\big{(}F^{\prime}_{1}+F^{(r)}_{0}\big{)}\Big{)}+F^{(r)}_{1}=F_{1}^{\prime}+F_{1}^{(r)}.$

This gives a short exact sequence

$0\to\frac{\mathcal{O}_{n+1}}{F_{2}(f_{*}\mathcal{O}_{X,S})}\to\frac{\mathcal{O}_{n+1}}{F^{\prime}_{2}\cap\big{(}F^{\prime}_{1}+F^{(r)}_{0}\big{)}}\oplus\frac{\mathcal{O}_{n+1}}{F^{\prime}_{0}+F^{(r)}_{1}}\to\frac{\mathcal{O}_{n+1}}{F^{\prime}_{1}+F^{(r)}_{1}}\to
0.$

The hypotheses that $M_{3}(f)$ and $M_{4}(f)$ are dimensionally correct imply
that

$\operatorname{depth}\frac{\mathcal{O}_{n+1}}{F^{\prime}_{0}+F^{(r)}_{1}}=n-2\quad\text{and}\quad\operatorname{depth}\frac{\mathcal{O}_{n+1}}{F^{\prime}_{1}+F^{(r)}_{1}}=n-3.$

Therefore, by Item (3) of Lemma 3, it suffices to show that

$\operatorname{depth}\frac{\mathcal{O}_{n+1}}{F^{\prime}_{2}\cap\big{(}F^{\prime}_{1}+F^{(r)}_{0}\big{)}}\geq
n-2.$

Since $F_{1}^{\prime}\subseteq F_{2}^{\prime}$, this last ring fits into the
short exact sequence

$0\to\frac{\mathcal{O}_{n+1}}{F^{\prime}_{2}\cap\big{(}F^{\prime}_{1}+F^{(r)}_{0}\big{)}}\to\frac{\mathcal{O}_{n+1}}{F^{\prime}_{2}}\oplus\frac{\mathcal{O}_{n+1}}{F^{\prime}_{1}+F^{(r)}_{0}}\to\frac{\mathcal{O}_{n+1}}{F^{\prime}_{2}+F^{(r)}_{0}}\to
0.$

That the depth is the desired one follows again by Item (3) of Lemma 3, taking
into account that the dimensionality hypotheses imply

$\operatorname{depth}\frac{\mathcal{O}_{n+1}}{F^{\prime}_{2}}=\operatorname{depth}\frac{\mathcal{O}_{n+1}}{F^{\prime}_{1}+F^{(r)}_{0}}=n-2\quad\text{and}\quad\operatorname{depth}\frac{\mathcal{O}_{n+1}}{F^{\prime}_{2}+F^{(r)}_{0}}=n-3.$

This completes the proof of the statement that $M_{3}(f)$ is Cohen-Macaulay.

Now assume $M_{3}(f)$ and $M_{4}(f)$ are dimensionally correct. We have shown
before that, under these hypotheses,

$F_{2}(f_{*}\mathcal{O}_{X,S})=F^{\prime}_{2}\cap\big{(}F^{\prime}_{1}+F^{(r)}_{0}\big{)}\cap\big{(}F^{\prime}_{0}+F^{(r)}_{1}\big{)}.$

From the fact that $f$ is generically one-to-one, it follows that
$F^{\prime}_{0}=\bigcap_{1\leq i\leq r-1}F_{0}^{(i)}$. Applying induction and
Theorem 2.4, we obtain the expression

$\displaystyle F_{2}(f_{*}\mathcal{O}_{(X,S)})=$ $\displaystyle
F_{2}^{(1)}\cap\bigcap_{1\leq i<j\leq
r-1}\big{(}F_{1}^{(i)}+F_{0}^{(j)}\big{)}\cap\big{(}F_{0}^{(i)}+F_{1}^{(j)}\big{)}$
$\displaystyle\cap\bigcap_{1\leq i<j<k\leq
r-1}\big{(}F_{0}^{(i)}+F_{0}^{(j)}+F_{0}^{(k)}\big{)}$
$\displaystyle\cap\bigg{(}\Big{(}\bigcap_{1\leq i\leq
r-1}F_{1}^{(i)}\cap\bigcap_{1\leq i<j\leq
r-1}\big{(}F_{0}^{(i)}+F_{0}^{(j)}\big{)}\Big{)}+F^{(r)}_{0}\bigg{)}$
$\displaystyle\cap\Big{(}\big{(}\bigcap_{1\leq i\leq
r-1}F_{0}^{(i)}\big{)}+F^{(r)}_{1}\Big{)}.$

Comparing this to the desired expression shows that it suffices to prove the
equality

$\Big{(}\bigcap_{1\leq i\leq r-1}F_{1}^{(i)}\cap\bigcap_{1\leq i<j\leq
r-1}\big{(}F_{0}^{(i)}+F_{0}^{(j)}\big{)}\Big{)}+F^{(r)}_{0}=\\\
\bigcap_{1\leq i\leq
r-1}\big{(}F_{1}^{(i)}+F_{0}^{(r)}\big{)}\cap\bigcap_{1\leq i<j\leq
r-1}\big{(}F_{0}^{(i)}+F_{0}^{(j)}+F_{0}^{(r)}\big{)}$

and the equality

$\big{(}\bigcap_{1\leq i\leq r-1}F_{0}^{(i)}\big{)}+F^{(r)}_{1}=\bigcap_{1\leq
i\leq r-1}\big{(}F_{0}^{(i)}+F^{(r)}_{1}\big{)}.$

The conditions that $M_{4}(f)$ and $M_{3}(f)$ are dimensionally correct imply
that, given $i,j,k$, with $i\neq j$, no irreducible component of
$V(F_{0}^{(i)}+F_{0}^{(j)}+F_{0}^{(r)})$ is contained in
$M_{2}^{(k)}=V(F_{1}^{(k)})$, no irreducible component of $M_{2}^{(k)}\cap
M_{1}^{(r)}$ is contained in $V(F_{0}^{(i)}+F_{0}^{(j)})$, no irreducible
$M_{2}^{(i)}\cap M_{1}^{(r)}$ is contained in $M_{2}^{(j)}$ and no irreducible
component of $V(F_{0}^{(k)}+F_{0}^{(i)}+F_{0}^{(r)})$ is contained in
$V(F_{0}^{(k)}+F_{0}^{(j)})$. The first equality then follows from Lemma 2.1.
The same dimensional correctness conditions imply that no irreducible
component of $M_{1}^{(i)}\cap M_{2}^{(r)}$ is contained in $M_{1}^{(j)}$, for
$i\neq j$. Then by Lemma 2.1 the second equality holds as well. ∎

In the case where $X$ is a complete intersection, the hypotheses in Theorem
2.6 can be relaxed substantially:

######  Proposition 2.7.

Given a generically one-to-one multi-germ
$f\colon(X,S)\to(\mathbb{C}^{n+1},0)$, assume $X$ to be a complete
intersection and that $M_{3}^{(i)}=\emptyset$, for $2\leq i\leq r$. If
$M_{3}(f)$ is dimensionally correct, then it is Cohen-Macaulay. If furthermore
$M_{4}(f)$ is dimensionally correct, then $M_{3}(f)$ can be computed with the
same formula as in Theorem 2.6.

###### Proof.

The hypothesis that $X$ is a complete intersection implies that $X$ is
Gorenstein. To see that $f$ admits an unfolding whose spaces $M_{3}$ and
$M_{4}$ are dimensionally correct we use an original idea of Mond and Montaldi
[3]. Take a regular sequence $h_{1},\dots,h_{k}$ defining $(X,S)$ in its
ambient space $(\mathbb{C}^{n+k},S)$ and consider the mapping

$\varphi\colon(\mathbb{C}^{n+k},S)\to(\mathbb{C}^{n+1}\times\mathbb{C}^{k},0),$

given by

$x\mapsto(\bar{f}_{1}(x),\dots,\bar{f}_{n+1}(x),h_{1}(x),\dots,h_{k}(x)),$

where $\bar{f}_{i}$ is any analytic extension in
$\mathcal{O}_{\mathbb{C}^{k+n},S}$ of the $i$th coordinate function $f_{i}$ of
$f$. The mapping $\varphi$ is finite and, from the fact that
$h_{1},\dots,h_{k}$ are a regular sequence defined on the smooth space
$(\mathbb{C}^{n+k},S)$, it follows that $h_{1},\dots,h_{k}$ define a flat
morphism [2, Theorem 23.1], hence $\varphi$ is an unfolding of $f$. Since
$\varphi$ is finite, in particular it is $\mathscr{K}$-finite and, by a
construction due to Mather (see e.g. [4, Theorem 7.2]), it admits a stable
unfolding,

$F\colon(\mathbb{C}^{k+n}\times\mathbb{C}^{r},S\times\\{0\\})\to(\mathbb{C}^{k}\times\mathbb{C}^{n+1}\times\mathbb{C}^{r},0).$

Being an unfolding of $\varphi$, the germ $F$ is an unfolding of $f$ as well.
Therefore, it suffices to show that the multiple point spaces $M_{3}(F)$ and
$M_{4}(F)$ are both dimensionally correct when $F$ is stable. Fix a small
enough finite representative $F\colon U\to V$, where $U,V$ are open
neighbourhoods of $S\times\\{0\\}$ and $0$ in
$\mathbb{C}^{k+n}\times\mathbb{C}^{r}$ and
$\mathbb{C}^{k}\times\mathbb{C}^{n+1}\times\mathbb{C}^{r}$, respectively. The
set $\Sigma^{2}$ of points in $U$ of corank $\geq 2$ is closed analytic and
has codimension $6$ in $U$. In $F(U)\setminus F(\Sigma^{2})$ we consider the
stratification by stable types, whose strata are the the isosingular loci
$\operatorname{Iso}(F,y)$, with $y\in F(U)\setminus F(\Sigma^{2})$. By
construction, $\operatorname{Iso}(F,y)$ is the set of points $y^{\prime}\in
F(U)$ such that the multi-germ of $F$ at $y^{\prime}$ is
$\mathscr{A}$-equivalent to the multi-germ of $F$ at $y^{\prime}$. It follows
that $\operatorname{Iso}(F,y)$ is a submanifold of $V$ of codimension equal to
the $\mathscr{K}_{e}$-codimension of the multi-germ of $F$ at $y$ (see [4,
Theorem 7.4]). Moreover, the $\mathscr{A}$-class of $F$ at $y$ is determined
by the contact algebra $Q_{y}(F)$ and since $y\notin F(\Sigma^{2})$, it must
be of the form

$Q_{y}(F)\cong\frac{\mathbb{C}\\{x\\}}{(x^{a_{1}})}\oplus\dots\oplus\frac{\mathbb{C}\\{x\\}}{(x^{a_{\ell}})},$

for some $a_{1},\dots,a_{\ell}\geq 1$. The $\mathscr{K}_{e}$-codimension is
$(2a_{1}-1)+\dots+(2a_{\ell}-1)$ which is the codimension of the stratum
$\operatorname{Iso}(F,y)$ in $V$. On the other hand, $y\in M_{r}(F)$ if and
only if $\operatorname{dim}_{\mathbb{C}}Q_{y}(F)\geq r$ and we also have
$\operatorname{dim}_{\mathbb{C}}Q_{y}(F)=a_{1}+\dots+a_{\ell}$. In conclusion,
there exists only one stratum of codimension $r$ in $M_{r}(F)$ corresponding
to the transverse $r$-multiple points and all other strata have codimension
$>r$. This shows that $M_{r}(F)$ is dimensionally correct for $r\leq 7$. ∎

######  Example 2.8.

We are going to compute $M_{3}(f)$, for the bi-germ

$f:U^{(1)}\sqcup U^{(2)}\to(\mathbb{C}^{4},0)$

with branches

  * $f^{(1)}\colon U^{(1)}=(\mathbb{C}^{3},0)\to(\mathbb{C}^{4},0),\ (t,x,y)\mapsto(t,x,y^{3}+ty,xy+y^{5})$

  * $f^{(2)}\colon U^{(2)}=(\mathbb{C}^{3},0)\to(\mathbb{C}^{4},0),\ (\tilde{t},\tilde{x},\tilde{y})\mapsto(\tilde{t},\tilde{x},\tilde{y},\tilde{t})$.

The presentation matrices of $f^{(1)}$ and $f^{(2)}$ are

$\lambda^{(1)}=\begin{bmatrix}-Z-TY&Y^{2}&XY-TZ\\\ X+T^{2}&-Z-2TY&Y^{2}\\\
Y&X+T^{2}&-Z-TY\end{bmatrix},\quad\lambda^{(2)}=\begin{bmatrix}-Z+T\end{bmatrix}.$

In order to use the formula in Theorem 2.6, we must check first $M_{4}(f)$ to
be dimensionally correct. The presentation matrix of $f$ is the block diagonal
matrix

$\lambda=\begin{bmatrix}\lambda^{(1)}&0\\\ 0&\lambda^{(2)}\end{bmatrix}.$

We obtain $F_{3}=\langle X,Y,Z,T\rangle$, hence $M_{4}(f)$ has dimension zero,
as desired. Since $\lambda^{(2)}$ is a $1\times 1$ matrix, we see that
$F_{1}^{(2)}=\mathcal{O}_{4}$, hence $M_{3}(f)$ is defined by the ideal

$F_{2}=F_{2}^{(1)}\cap(F_{1}^{(1)}+F_{0}^{(2)}).$

It turns out that this is the primary decomposition of $F_{2}$, expressing it
as the intersection of the two prime ideals

  * $F^{(1)}_{2}=\langle X+T^{2},Z,Y\rangle$,

  * $\begin{aligned} F^{(1)}_{1}+F^{(2)}_{0}=&\langle T-Z,X^{2}+TY+2XT^{2}+2TY^{2}+T^{4},\\\ &XT+XTY+T^{3}+Y^{3}+T^{3}Y,\\\ &T^{2}-XY^{2}+3T^{2}Y+T^{2}Y^{2}\rangle.\end{aligned}$

We see that $M_{3}(f)$ is a curve (i.e., it has dimension one) and it is not
difficult to check that it is reduced. So, it makes sense to compute its
Milnor number $\mu(M_{3}(f))$ in the sense of [1, Definition 1.1.1]. To do it,
we use the Milnor formula $\mu=2\delta-r+1$, where $\delta$ is the delta
invariant and $r$ is the number of branches (see [1, Proposition 1.2.1]).

We split $M_{3}(f)$ as a union of two curves $M_{3}(f)=C\cup D$, where

$C=V(F^{(1)}_{2}),\quad D=V(F^{(1)}_{1}+F^{(2)}_{0}).$

Since $C$ is a regular branch, it has $\delta=0$. The other branch $D$ is
isomorphic to its projection $\tilde{D}$ onto the 3-space $X,Y,T$. To compute
the invariants of $\tilde{D}$, we polar multiplicities, following [7,
Corollary 3.2]:

$\mu=m_{1}-m_{0}+1,$

where $m_{0}$ is the multiplicity and $m_{1}$ is the first polar multiplicity,
defined as the Milnor number of a generic linear function
$p\colon\mathbb{C}^{3}\to\mathbb{C}$, $p(X,Y,T)=b_{1}X+b_{2}Y+b_{3}T$,
$b_{i}\in\mathbb{C}$, restricted to the curve. When the curve is smoothable,
this is equal to the number of critical points of $p$ on the smoothing of the
curve. With the aid of Singular [9], we see that $\tilde{D}$ admits a
determinantal representation as $\tilde{D}=V(I_{2}(M))$, where $I_{2}(M)$ is
the ideal generated by the $2\times 2$ minors of the matrix

$M=\left(\begin{array}[]{cc}Y^{2}&-T-YT\\\ -T-2YT&X+T^{2}\\\
X+T^{2}&Y\end{array}\right).$

A smoothing $\tilde{D}_{s}$ of $\tilde{D}$ can be obtained as
$\tilde{D}_{s}=V(I_{2}(M+sA))$ where $A=(a_{ij})$ is a generic $3\times
2$-matrix with entries $a_{ij}$ in $\mathbb{C}$ and
$s\in\mathbb{C}\setminus\\{0\\}$ is small enough (see [6, Lemma 3.1]). Again
with Singular, we compute the number of critical points of $p$ on
$\tilde{D}_{s}$, which gives $m_{1}=6$, and the multiplicity $m_{0}=3$. This
gives

$\mu(D)=\mu(\tilde{D})=m_{1}-m_{0}+1=4.$

Since $D$ is irreducible, this implies $\delta(D)=2$.

Now $\delta(M_{3}(f))$ follows from a formula due to Hironaka (see [1, Lemma
1.2.2]):

$\delta(M_{3}(f))=\delta(C)+\delta(D)+C\cdot D$

where $C\cdot
D=\operatorname{dim}_{\mathbb{C}}\O_{4}/(F^{(1)}_{2}+(F^{(1)}_{1}+F^{(2)}_{0}))$
is the intersection number of the two branches. Again with Singular we get
$C\cdot D=1$. Thus,

$\delta(M_{3}(f))=0+2+1=3$

and hence

$\mu(M_{3}(f))=2\delta(M_{3}(f))-r+1=5.$

##  3\. Triple and quadruple points in the source

Given a finite multi-germ $f\colon(X,S)\to(\mathbb{C}^{n+1},0)$, the multiple
point spaces $M_{k}(f)$ of the previous section lie on the target of $f$, that
is, on $(\mathbb{C}^{n+1},0)$. There are however other multiple point spaces
that one can define, and the ones this section is devoted are subspace of the
source of $f$, instead of its target. For simplicity, we consider only the
case where the source is smooth. Given a multi-germ

$f\colon(\mathbb{C}^{n},S)\to(\mathbb{C}^{n+1},0),$

the idea is as follows: Start by defining a subspace
$D^{2}(f)\subseteq(\mathbb{C}^{n},S)\times(\mathbb{C}^{n},S)$ which, naively,
should be the closure of the set of pairs $(x,x^{\prime})$ with $x\neq
x^{\prime}$ and $f(x)=f(x^{\prime})$ (the proper definition is given below).
If $f$ is finite, then the projection
$(\mathbb{C}^{n},S)\times(\mathbb{C}^{n},S)\to(\mathbb{C}^{n},S)$ on the first
coordinate restricts to a finite mapping

${D^{2}(f)}$${(\mathbb{C}^{n},S)}$$\scriptstyle{p}$

The _source double point space_ $D(f)=D^{2}_{1}(f)$ is defined as the image of
$p$, with the analytic structure given by the $0$th Fitting ideal, that is,

$D^{2}_{1}(f)=M_{1}(p).$

This makes sense since, given a pair $(x,x^{\prime})$, with $x\neq x^{\prime}$
and $f(x)=f(x^{\prime})$, the point $x\in\mathbb{C}^{n}$ should be considered
a double point of $f$.

The same idea tells us that $M_{2}(p)\subseteq(\mathbb{C}^{n},S)$ should be
regarded as a triple point space, because two pairs $(x,x^{\prime})$ and
$(\tilde{x},\tilde{x}^{\prime})$, with $x^{\prime}\neq x$,
$f(x^{\prime})=f(x)$, $\tilde{x}^{\prime}\neq\tilde{x}$,
$f(\tilde{x}^{\prime})=f(\tilde{x})$,
$(x,x^{\prime})\neq(\tilde{x},\tilde{x}^{\prime})$ and $x=\tilde{x}$, carry
the same information than a tuple $(x,x^{\prime},x^{\prime\prime})$, with
$x\neq x^{\prime}\neq x^{\prime\prime}\neq x$ and
$f(x)=f(x^{\prime})=f(x^{\prime\prime})$, by setting
$x^{\prime\prime}=\tilde{x}^{\prime}$, and then $x$ should be a triple point
of $f$. By the same token, points in $M_{3}(p)$ should be regarded as
quadruple points in the source. Before making this into a formal definition,
we must define the space $D^{2}(f)$.

Consider the ideal $I_{\Delta}$ defining the diagonal
$(\Delta\mathbb{C}^{n+1},0)$ in
$(\mathbb{C}^{n+1},0)\times(\mathbb{C}^{n+1},0)$. The fibered product
$(\mathbb{C}^{n},S)\times_{(\mathbb{C}^{n+1},0)}(\mathbb{C}^{n},S)$ is the
zero locus of the pullback $(f\times f)^{*}I_{\Delta}$. Given
$x^{(i)},x^{(j)}\in S$, we may take coordinates $x$ and $x^{\prime}$ of
$\mathbb{C}^{n}$, centered at $x^{(i)},x^{(j)}$, so that

$((f\times
f)^{*}I_{\Delta})_{(x^{(i)},x^{(j)})}=\big{\langle}f^{(i)}_{1}(x^{\prime})-f^{(j)}_{1}(x),\dots,f^{(i)}_{n+1}(x^{\prime})-f^{(j)}_{n+1}(x)\big{\rangle}.$

Away from the diagonal,
$(\mathbb{C}^{n},S)\times_{(\mathbb{C}^{n+1},0)}(\mathbb{C}^{n},S)$
corresponds with the idea of what a double point should be, but around a point
$(x^{(i)},x^{(i)})$, it contains all points $(x,x)\in\Delta\mathbb{C}^{n}$,
which, a priori, should not be regarded as triple points. To remedy this, we
must add extra conditions on the diagonal. If we think of
$f^{(i)}(x^{\prime})-f^{(i)}(x)$ and $x^{\prime}-x$, the fact that
$f^{(i)}(x^{\prime})-f^{(i)}(x)$ vanishes on
$(\Delta\mathbb{C}^{n},(x^{(i)},x^{(i)}))$ translates, by Hilbert
Nullstellensatz, into the fact that there is an $(n+1)\times n$ matrix
$\alpha$, with entries in $\mathcal{O}_{2n}$, such that

$f^{(i)}(x^{\prime})-f^{(i)}(x)=\alpha^{(i)}(x,x^{\prime})\cdot(x^{\prime}-x).$

The _double point space_ $D^{2}(f)$ is the disjoint union of the germs

$\left(D^{2}(f),{(x^{(i)},x^{(j)})}\right)=V(I^{2}_{ij}),$

where

$I_{ij}=\begin{cases}((f\times f)^{*}I_{\Delta})_{(x^{(i)},x^{(j)})}&\text{ if
$i\neq j$}\\\ ((f\times
f)^{*}I_{\Delta})_{(x^{(i)},x^{(i)})}+\big{\langle}\text{$n\times n$ minors of
$\alpha$},\big{\rangle}&\text{ if $i=j$}\end{cases}$

The space $D^{2}(f)$ has dimension at least $n-1$ and it is Cohen-Macaulay
when it has dimension exactly $n-1$ [4, Section 9.4]. If $f$ has corank one at
$x^{(i)}$, then, after a change of coordinates in source and target, $f^{(i)}$
takes the form of the germ at the origin of a mapping of the form

$(x_{1},\dots,x_{n-1},y)\mapsto(x_{1},\dots,x_{n-1},f_{n}(x,y),f_{n+1}(x,y)).$

Then, $\left(D^{2}(f),{(x^{(i)},x^{(i)})}\right)$ is isomorphic [4, Section
9.5] to germ of

$\\{(x,y,y^{\prime})\in\mathbb{C}^{n}\times\mathbb{C}\mid\frac{f_{n}(x,y^{\prime})-f_{n}(x,y)}{y^{\prime}-y}=\frac{f_{n+1}(x,y^{\prime})-f_{n+1}(x,y)}{y^{\prime}-y}=0\\}.$

From this, it follows that $\left(D^{2}(f),{(x^{(i)},x^{(i)})}\right)$ is a
complete intersection whenever it has dimension $n-1$.

######  Definition 3.1.

For a finite map-germ $f\colon(\mathbb{C}^{n},S)\to(\mathbb{C}^{n+1},0)$, we
define the _source triple point space_ of $f$ as

$D^{2}_{2}(f)=M_{2}(p),$

where $p\colon D^{2}(f)\to(\mathbb{C}^{n},S)$ is as above. Similarly, we
define the _source quadruple point space_ of $f$ as

$D^{2}_{3}(f)=M_{3}(p).$

######  Example 3.2.

We are going to compute the source triple point space $D^{2}_{2}(f)=M_{2}(p)$
and source quadruple space $D^{2}_{3}(f)=M_{3}(p)$, for the map-germ $f\colon
U^{(1)}\sqcup U^{(2)}\to(\mathbb{C}^{4},0)$ of Example 2.8. We also show how
the explicit formula for $F_{1}(f_{*}\mathcal{O}_{(X,S)})$ in Theorem 2.4
helps us compute the Milnor number of $D^{2}_{2}(f)$ and how Theorem 2.6
allows us to compute the number of quadruple points in a stabilization of the
multi-germ in question. Recall that there is a diagram

${D^{2}(f)}$${U^{(1)}\sqcup
U^{(2)}}$${(\mathbb{C}^{4},0)}$$\scriptstyle{p}$$\scriptstyle{f}$

The double space of $f$ decomposes as $D^{2}(f)=X_{1}\sqcup X_{2}\sqcup
X_{3}$, with $X_{1}=D^{2}(f^{(1)})$,
$X_{2}=U^{(1)}\times_{(\mathbb{C}^{4},0)}U^{(2)}$ and
$X_{3}=U^{(2)}\times_{(\mathbb{C}^{4},0)}U^{(1)}$. These spaces are

  * $\begin{aligned} X_{1}=\Big{\\{}(t,x,y,y^{\prime})\mid&\frac{xy^{\prime}+y^{\prime 5}-xy-y^{5}}{y^{\prime}-y}=\frac{y^{\prime 3}+ty^{\prime}-y^{3}-ty}{y^{\prime}-y}=0\Big{\\}},\end{aligned}$

  * $\begin{aligned} X_{2}&=\left\\{(t,x,y,\tilde{t},\tilde{x},\tilde{y})\ |\ \tilde{t}=t,\tilde{x}=x,\tilde{y}=y^{3}+ty,\tilde{t}=xy+y^{5}\right\\}\\\ &\cong\left\\{(t,x,y,\tilde{y})\ |\ y^{3}+ty-\tilde{y}=xy+y^{5}-t=0\right\\},\end{aligned}$

  * $\begin{aligned} X_{3}&=\left\\{(\tilde{t},\tilde{x},\tilde{y},t,x,y)\ |\ \tilde{t}=t,\tilde{x}=x,\tilde{y}=y^{3}+ty,\tilde{t}=xy+y^{5}\right\\}\\\ &\cong\left\\{(\tilde{y},t,x,y)\ |\ y^{3}+ty-\tilde{y}=xy+y^{5}-t=0\right\\}.\end{aligned}$

The projection $p\colon D^{2}(f)\to U^{(1)}\sqcup U^{(2)}$ decomposes as

$p=\pi\sqcup\tilde{\pi}\colon(X_{1}\sqcup X_{2})\sqcup X_{3}\to U^{(1)}\sqcup
U^{(2)},$

where $\tilde{\pi}:X_{3}\to U^{(2)}$ is a mono-germ and $\pi$ is a bi-germ
with two branches $\pi^{(1)}:X_{1}\to U^{(1)}$ and $\pi^{(2)}:X_{2}\to
U^{(1)}$. Therefore, the triple and quadruple points in the source of $f$
decompose as

$D^{2}_{2}(f)=M_{2}(\pi)\sqcup M_{2}(\tilde{\pi})\quad\text{and}\quad
D^{2}_{3}(f)=M_{3}(\pi)\sqcup M_{3}(\tilde{\pi}).$

We start with the computations over $U^{(1)}$, by computing some presentation
matrices of $\pi^{(1)}_{*}\mathcal{O}_{X_{1}}$ and
$\pi^{(2)}_{*}\mathcal{O}_{X_{2}}$, namely

$\xi^{(1)}=\begin{bmatrix}y^{4}+y^{2}t+t^{2}+x&y^{5}+2y^{3}t+yt^{2}\\\
y^{3}+yt&-t^{2}-x\end{bmatrix},\quad\xi^{(2)}=\begin{bmatrix}y^{5}+xy-t\end{bmatrix}.$

Since $M_{4}(p)$ on $U^{(1)}$ is empty and $X_{1}\sqcup X_{2}$ is a complete
intersection, Proposition 2.7 allows us to use the formula in Theorem 2.6 for
$\pi$, which in this case reads

$M_{3}(\pi)=V\big{(}F_{1}^{(1)}+F_{0}^{(2)}\big{)}=V(x,t,y^{3}).$

As $\operatorname{dim}M_{3}(\pi)=0$, we may apply the formula in Theorem 2.4,
which gives

$M_{2}(\pi)=V\Big{(}F_{1}^{(1)}\cap\big{(}F_{0}^{(1)}+F_{0}^{(2)}\big{)}\Big{)},$

with

  * $F_{1}^{(1)}=\langle y,x+t^{2}\rangle\cap\langle x+t^{2},t+y^{2}\rangle$,

  * $\begin{aligned} F_{0}^{(1)}+F_{0}^{(2)}=\langle&t-xy-y^{5},\\\ &x^{2}+2xt^{2}+3yt^{2}-2xy^{2}t+y^{3}t+t^{4}+2y^{2}t^{3}+4y^{4}t^{2}\rangle.\end{aligned}$

Before moving on to the calculations over $U^{(2)}$, we compute
$\mu(M_{2}(\pi))$. We know $M_{2}(\pi)$ decomposes as the union of
$V(F_{1}^{(1)})$ and $F_{0}^{(1)}+F_{0}^{(2)}$ and that, in turn,
$V(F_{1}^{(1)})$ decomposes as the union of two regular curves, which we call
$Z_{1}$ and $Z_{2}$. In turn, the expression of $F_{0}^{(1)}+F_{0}^{(2)}$
allows us to eliminate $t$, turning $V(F_{0}^{(1)}+F_{0}^{(2)})$ into a plane
curve with Milnor number $7$. This, combined with Hironaka’s $\mu=2\delta-r+1$
formula, prevents this curve to be irreducible and, looking at the degree two
part, we conclude that it consists of $2$ regular curves, called $Z_{3}$ and
$Z_{4}$, with $\delta(Z_{3}\cup Z_{4})=4$. Computing the intersection numbers
$Z_{1}\cdot(Z_{2}\cup Z_{3}\cup Z_{4})=2$ and $Z_{2}\cdot(Z_{3}\cup Z_{4})=2$,
we conclude

$\displaystyle\delta(M_{2}(\pi))=$
$\displaystyle\delta(Z_{1})+\delta(Z_{2})+\delta(Z_{3}\cup Z_{4})$
$\displaystyle+Z_{1}\cdot(Z_{2}\cup Z_{3}\cup Z_{4})+Z_{2}\cdot(Z_{3}\cup
Z_{4})$ $\displaystyle=$ $\displaystyle 0+0+4+2+2=8$

Now from Hironaka’s formula we conclude that

$\mu(M_{2}(\pi))=2\cdot 8-4+1=13.$

Now we proceed to the calculations over $U^{(2)}$. A presentation matrix of
$\tilde{\pi}_{*}\mathcal{O}_{X_{3}}$ is

$\tilde{\xi}=\begin{bmatrix}-\tilde{y}&0&t&0&t^{2}\\\ t&-\tilde{y}&-x&t&-xt\\\
0&t&-\tilde{y}&-x&t\\\ 1&0&t&-\tilde{y}&-x\\\ 0&1&0&t&-\tilde{y}\end{bmatrix}$

The source triple space $D^{2}_{2}(f)\cap
U^{(2)}=M_{2}(\tilde{\pi})=V(\tilde{F}_{1})$ is reduced and irreducible, with
defining ideal

  * $\begin{aligned} \tilde{F_{1}}=&\langle t^{2}+3t^{2}\tilde{y}-x\tilde{y}^{2}+t^{2}\tilde{y}^{2},tx+t^{3}+tx\tilde{y}+\tilde{y}^{3}+t^{3}\tilde{y},\\\ &x^{2}+t\tilde{y}+2t^{2}x+2t\tilde{y}^{2}+t^{4}\rangle\end{aligned}$

We observe that $M_{2}(\tilde{\pi})$ coincides with the curve $\tilde{D}$ in
Example 2.8, hence $\mu(M_{2}(\tilde{\pi}))=4$.

The source quadruple space is

$D^{2}_{3}(f)\cap
U^{(2)}=M_{3}(\tilde{\pi})=V(\tilde{F_{2}})=V(x,t,\tilde{y}).$

We may use these calculations to compute the number of quadruple points found
in a stabilization of $f$, namely

$Q(f)=\frac{\operatorname{dim}_{\mathbb{C}}\mathcal{O}_{D^{2}_{3}(\pi)}+\operatorname{dim}_{\mathbb{C}}\mathcal{O}_{D^{2}_{3}(\tilde{\pi})}}{4}=\frac{3+1}{4}=1.$

To understand that this is the number of quadruple points in a stabilization,
one first observes that, if $(f_{t},t)$ is a stabilization of $f=f_{0}$, then
Proposition 2.7 applies to the mapping $D^{2}(f_{t},t)\to(U^{(1)}\sqcup
U^{(2)})\times(\mathbb{C},0)$ and, therefore, the space $D^{2}_{3}(f_{t},t)$
is a reduced Cohen Macaulay curve. The classification of stable multi-germs
$(\mathbb{C}^{3},S)\to(\mathbb{C}^{4},0)$ implies that, for the stable mapping
$f_{\epsilon},\epsilon\neq 0$, the space $D^{2}_{3}(f_{\epsilon})$ consist of
four reduced points for each quadruple point. Now the claim follows by the
principle of conservation of number.

As a finally observation, it seems natural to wonder how the triple and
quadruple point spaces $D^{2}_{2}(f)$ and $D^{2}_{3}(f)$, which are subspaces
of $(X,S)$, compare to the preimages by $f$ of $M_{3}(f)$ and $M_{4}(f)$,
which are subspaces of $(\mathbb{C}^{4},0)$ and were computed in Example 2.8.
A simple computation shows, for the multi-germ $f$ in question, the equalities
$D^{2}_{2}(f)=f^{-1}(M_{3}(f))$ and $D^{2}_{3}(f)=f^{-1}(M_{4}(f))$ hold.

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