Multiplicity of singularities is not a bi-Lipschitz invariant

Lev Birbrair111Partially supported by CNPq grant 302655/2014-0., Alexandre
Fernandes222Partially supported by CNPq grant 302764/2014-7., J. Edson
Sampaio, Misha Verbitsky333Partially supported by the Russian Academic
Excellence Project ’5-100’.

Abstract. It was conjectured that multiplicity of a singularity is bi-
Lipschitz invariant. We disprove this conjecture, constructing examples of bi-
Lipschitz equivalent complex algebraic singularities with different values of
multiplicity.

###### Contents

  1. 1 Introduction
  2. 2 Smale-Barden manifolds
  3. 3 Multiplicity of homogeneous singularities
  4. 4 Lipschitz invariance of singularities

##  1 Introduction

The famous multiplicity conjecture, stated by Zariski in 1971 (see [Z2]), is
formulated as follows: if two germs of complex analytic hypersurfaces are
ambient topolological equivalent, then they have the same multiplicity. It was
proved by Zariski [Z1] for germs of plane analytic curves. The results of
Pham-Teisser in [PT] show that this result can be extended in the following
”metric” way: if the two germs of complex analytic curves in n-dimensional
space are bi-Lipschitz equivalent with respect to the outer metric, then the
germs of the space curves have the same multiplicity. Comte in [Co] proves
that the multiplicity of complex analytic germs (not necessarily codimension 1
sets) is invariant under bi-Lipschitz homeomorphism with Lipschitz constants
close enough to 1 (this is a severe assumption). These results motivated the
following question, closely related to the multiplicity conjecture: is the
multiplicity of a germ of analytic singularity a bi-Lipschitz invariant? This
question was stated as a conjecture in [BFS].

The Lipschitz Regularity Theorem in [BFLS] shows that if the multiplicity of a
complex analytic germ is equal to one, then it is a bi-Lipschitz invariant.
Namely, if a germs of an analytic set is bi-Lipschitz equivalent to a smooth
germ, then it is smooth itself. Later, Fernandes and Sampaio in [FS] give a
positive answer to this question for surfaces in 3-dimensional space with
respect to the ambient bi-Lipschitz equivalence. More recently, Neumann and
Pichon ([NP]) showed that the multiplicity is an invariant under bi-Lipschitz
equivalence, for germs of normal surface singularities. Another important
result in [FS] is the following: in order to prove (or disprove) the bi-
Lipschitz invariance of the multiplicity, it is enough to prove it for the
algebraic cones, i.e. for the algebraic sets, defined by homogeneous
polynomials. In [BFS] the authors show that the conjecture has a positive
answer for 1 or 2 dimensional complex analytic sets.

The present paper shows that the multiplicity of complex algebraic sets is not
a bi-Lipschitz invariant for the sets of dimension bigger or equal to three.
Moreover, we show that there exists an infinite family of $3-$dimensional
germs, such that all of them are bi-Lipschitz equivalent, but they have
different multiplicities. The idea of the construction is to consider the
complex cones over different embeddings of
${\mathbb{C}}P^{1}\times{\mathbb{C}}P^{1}$ to complex projective spaces. Using
the topology of Smale-Barden manifolds, we show that all the links of such
singularities are diffeomorphic. That is why the germs of the corresponding
cones are bi-Lipschitz equivalent. From the other hand, the multiplicities of
the cones at the origin may be explicitly calculated in terms of the
embeddings.

##  2 Smale-Barden manifolds

The classification of 5-manifolds is due to S. Smale ([S]) and D. Barden
([B]).

Definition 2.1: A simply connected, compact, oriented 5-manifold is called
Smale-Barden manifold.

The Smale-Barden manifolds are uniquely determined by their second Stiefel-
Whitney class and the linking form.

Theorem 2.2: ([B]) Let $X,X^{\prime}$ be two Smale-Barden manifolds. Assume
that $H^{2}(X)=H^{2}(X^{\prime})$ and this isomorphism is compatible with the
linking form and preserves the second Stiefel-Whitney class. Then $X$ is
diffeomorphic to $X^{\prime}$.

Corollary 2.3: There exists only two Smale-Barden manifolds $M$ with
$H^{2}(M)={\mathbb{Z}}$, the product $S^{2}\times S^{3}$ and the total space
of a non-trivial $S^{3}$-bundle over $S^{2}$ (see [C] for an introduction to
Barden theory, where this manifold is formally introduced).

Proof: Indeed, the linking form on ${\mathbb{Z}}$ vanishes, therefore the
manifold is uniquely determined by the Stiefel-Whitney class $w_{2}(M)$. Hence
we have only two possibilities: $w_{2}(M)=0$ and $w_{2}(M)\neq 0$.

In early 2000-ies, the classification of 5-manifolds attracted interest coming
from algebraic geometry, in the context of Sasakian geometry and geometry of
generalized Seifert manifolds ([K1], [K2]). In the present paper we are
interested in $S^{1}$-bundles over ${\mathbb{C}}P^{1}\times{\mathbb{C}}P^{1}$.

Proposition 2.4:  Let $\pi:\;M{\>\longrightarrow\>}B$ be a simply connected
5-manifold obtained as a total space of an $S^{1}$-bundle $L$ over
$B={\mathbb{C}}P^{1}\times{\mathbb{C}}P^{1}$. Then $H^{2}(M)$ is torsion-free,
and $M$ is diffeomorphic to $S^{2}\times S^{3}$.

Proof. Step 1: Universal coefficients formula gives an exact sequence

$0\to\operatorname{Ext}_{\mathbb{Z}}^{1}(H_{1}(M;{\mathbb{Z}}),{\mathbb{Z}})\to
H^{2}(M;{\mathbb{Z}})\to\operatorname{Hom}_{\mathbb{Z}}(H_{2}(M;{\mathbb{Z}}),{\mathbb{Z}})\to
0.$

This implies that $H^{2}(M;{\mathbb{Z}})$ is torsion-free.

Step 2: Consider the following exact sequence of homotopy groups

$0{\>\longrightarrow\>}\pi_{2}(M){\>\longrightarrow\>}\pi_{2}(B)\stackrel{{\scriptstyle\phi}}{{{\>\longrightarrow\>}}}\pi_{1}(S^{1}){\>\longrightarrow\>}\pi_{1}(M){\>\longrightarrow\>}0$

Since $\pi_{1}(M)=0$, the map $\phi$, representing the first Chern class of
$L$, is surjective. This exact sequence becomes

$0{\>\longrightarrow\>}\pi_{2}(M){\>\longrightarrow\>}{\mathbb{Z}}^{2}{\>\longrightarrow\>}{\mathbb{Z}}{\>\longrightarrow\>}0$

giving $\pi_{2}(M)={\mathbb{Z}}$, and $H^{2}(M)={\mathbb{Z}}$ because
$H^{2}(M)$ is torsion-free.

Step 3: To deduce 2 from the Smale-Barden classification, it remains to show
that $w_{2}(M)=0$. However, $w_{2}(M)=\pi^{*}(\omega_{2}(B))$ ([K2, Lemma
36]), and the latter clearly vanishes, because $w_{2}(S^{2})=0$.

##  3 Multiplicity of homogeneous singularities

Given a projective variety $X\subset{\mathbb{C}}P^{n}$, the projective cone of
$X$ is the union of all 1-dimensional subspaces $l\subset{\mathbb{C}}^{n+1}$
such that $l$, interpreted as a point in ${\mathbb{C}}P^{n}$, belongs to $X$.
The link of $C(X)$ is an intersection of $C(X)$ with a unit sphere
$S^{2n+1}\subset{\mathbb{C}}^{n+1}$.

Let $A$ be complex algebraic set of ${\mathbb{C}}^{n+1}$ and $x\in A$. The
multiplicity of $A$ at $x$, denoted by $mult(A,x)$, is defined to be the
multiplicity of the maximal ideal of the local ring ${{\mathcal{O}}}_{A,x}$.
Given a projective variety $X\subset{\mathbb{C}}P^{n}$, we see that
multiplicity of the projective cone $C(X)$ at the origin
$0\in{\mathbb{C}}^{n+1}$ coincides with degree of $X$ (see [Ch], subsection
11.3).

Next, we shall be interested in the following geometric situation. Let
$X\subset{\mathbb{C}}P^{n}$ be a variety isomorphic to
${\mathbb{C}}P^{1}\times{\mathbb{C}}P^{1}$. Then the Picard group of
$X={\mathbb{C}}P^{1}\times{\mathbb{C}}P^{1}$ is isomorphic to
${\mathbb{Z}}^{2}$, and a line bundle is determined by its bidegree. We shall
denote a line bundle of bidegree $a,b$ by ${\mathcal{O}}(a,b)$.

Proposition 3.1:  Let $X\subset{\mathbb{C}}P^{n}$ be a variety isomorphic to
${\mathbb{C}}P^{1}\times{\mathbb{C}}P^{1}$, and $S\subset C(X)$ be its link.
Assume that and
${\mathcal{O}}(1){\left|{}_{{\phantom{|}\\!\\!}_{X}}\right.}={\mathcal{O}}(a,b)$.
Then $X$ has degree $2ab$. If, in addition, $a$ and $b$ are relatively prime,
the link of $C(X)$ is diffeomorphic to $S^{2}\times S^{3}$.

Proof. Step 1: Since $c_{1}({\mathcal{O}}(a,b))^{2}=2ab$, and degree of a
subvariety $X\subset{\mathbb{C}}P^{n}$ is its intersection with the top power
of ${\mathcal{O}}(1){\left|{}_{{\phantom{|}\\!\\!}_{X}}\right.}$, one has
$\deg X=2ab$.

Step 2: Consider the homotopy exact sequence

$0{\>\longrightarrow\>}\pi_{2}(S){\>\longrightarrow\>}\pi_{2}(X)\stackrel{{\scriptstyle\phi}}{{{\>\longrightarrow\>}}}\pi_{1}(S^{1}){\>\longrightarrow\>}\pi_{1}(S){\>\longrightarrow\>}\pi_{1}(X){\>\longrightarrow\>}0$

for the circle bundle $\pi:\;S{\>\longrightarrow\>}X$. Since the map $\phi$
represents the first Chern class of
${\mathcal{O}}(1){\left|{}_{{\phantom{|}\\!\\!}_{X}}\right.}$, it is obtained
as a quotient of ${\mathbb{Z}}^{2}$ by a subgroup generated by $(a,b)$, and
this map is surjective because $a$ and $b$ are relatively prime. Then
$\pi_{1}(S)=\pi_{1}(X)=0$, and 2 implies that $S$ is diffeomorphic to
$S^{2}\times S^{3}$.

##  4 Lipschitz invariance of singularities

Let $X\subset{\mathbb{C}}^{n}$ be a complex variety. The induced metric from
the Euclidean distance on ${\mathbb{C}}^{n}$ gives a distance on $X$; it is
called the outer metric on $X$.

Definition 4.1: Let $X\subset{\mathbb{C}}^{n}$ and
$X^{\prime}\subset{\mathbb{C}}^{n^{\prime}}$ be complex varieties equipped
with the outer metrics, $x\in X,x^{\prime}\in X^{\prime}$ marked points. We
say that $(X,x)$ is bi-Lipschitz equivalent to $(X^{\prime},x^{\prime})$ if
there exist a neighborhoods $U$ of $x$ in ${\mathbb{C}}^{n}$ and $U^{\prime}$
of $x$ in ${\mathbb{C}}^{n}$, and a bi-Lipschitz homeomorphism of $X\cap U$ to
$X^{\prime}\cap U^{\prime}$ mapping $x$ to $x^{\prime}$.

Definition 4.2: Let $X,X^{\prime}\subset{\mathbb{C}}^{n}$ be complex varieties
equipped with the outer metrics, $x\in X,x^{\prime}\in X^{\prime}$ marked
points. We say that $(X,x)$ is ambient bi-Lipschitz equivalent to
$(X^{\prime},x^{\prime})$ if there exists a bi-Lipschitz equivalence of a
neighbourhood $U$ of $x$ in ${\mathbb{C}}^{n}$ and a neighbourhood
$U^{\prime}$ of $x$ in ${\mathbb{C}}^{n}$ mapping $X\cap U$ to $X^{\prime}\cap
U^{\prime}$ and $x$ to $x^{\prime}$.

Actually, the two definitions above do not coincide. The ambient bi-Lipschitz
equivalence implies bi-Lipschitz equivalence, but the examples presented in
[BG] show that the converse does not hold true in general.

As it was already mentioned in Introduction, it was conjectured in [BFS] that
the multiplicity is a bi-Lipschitz invariant. We prove that this is false.
Here is the main result of this paper.

Theorem 4.3:  For each $n\geq 3$, there exists a family
$\\{Y_{i}\\}_{i\in{\mathbb{Z}}}$ of $n$-dimensional complex algebraic
varieties $Y_{i}\subset{\mathbb{C}}^{n_{i}+1}$ such that:

  * (a)

for each pair $i\neq j$, the germs at the origin of
$Y_{i}\subset{\mathbb{C}}^{n_{i}+1}$ and $Y_{j}\subset{\mathbb{C}}^{n_{j}+1}$
are bi-Lipschitz equivalent, but $(Y_{i},0)$ and $(Y_{j},0)$ have different
multiplicity.

  * (b)

for each pair $i\neq j$, there are $n$-dimensional complex algebraic varieties
$Z_{ij},\widetilde{Z}_{ij}\subset{\mathbb{C}}^{n_{i}+n_{j}+2}$ such that
$(Z_{ij},0)$ and $(\widetilde{Z}_{ij},0)$ are ambient bi-Lipschitz equivalent,
but $mult(Z_{ij},0)=mult(Y_{i},0)$ and
$mult(\widetilde{Z}_{ij},0)=mult(Y_{j},0)$ and, in particular, they have
different multiplicity.

Proof. Let $\\{p_{i}\\}_{i\in{\mathbb{Z}}}$ be the family of odd prime
numbers. For each $i\in{\mathbb{Z}}$, let $L_{i}$ be a very ample bundle on
$X={\mathbb{C}}P^{1}\times{\mathbb{C}}P^{1}$ of bidegree $(2,p_{i})$. Let
$X_{i}$ be projective variety obtained by the embedding of the very ample
bundle $L_{i}$. Consider the link of the singularity $S_{i}:=C(X_{i})\cap
S^{2m_{i}+1}$, where $S^{2m_{i}+1}$ is the unit sphere centered in
$0\in{\mathbb{C}}^{m_{i}+1}$. Then, for each pair $i\neq j$ the links
$S_{i},S_{j}$ are diffeomorphic to $S^{2}\times S^{3}$ (3). In particular,
$S_{i}$ to $S_{j}$ are bi-Lipschitz homeomorphic. Since a bi-Lipschitz map
from $S_{i}$ to $S_{j}$ induces a bi-Lipschitz map of their cones, then the
affine cones $(C(X_{i}),0)$ and $(C(X_{j}),0)$ are bi-Lipschitz equivalent,
but $mult(C(X_{i}),0)=4p_{i}$ and $mult(C(X_{j}),0)=4p_{j}$ (3). Thus, if for
each $i\in{\mathbb{Z}}$ we define $Y_{i}:=C(X_{i})\times{\mathbb{C}}^{n-3}$,
then we have that the family $\\{Y_{i}\\}_{i\in{\mathbb{Z}}}$ satisfies the
item (a), since $mult(Y_{i},0)=mult(C(X_{i}),0)=4p_{i}$, for all
$i\in{\mathbb{Z}}$.

Concerning to the item (b), let $\phi_{ij}:Y_{i}\to Y_{j}$ be a bi-Lipschitz
homeomorphism such that $\phi_{ij}(0)=0$. Let
$\widetilde{\phi_{ij}}:{\mathbb{C}}^{n_{i}+1}\to{\mathbb{C}}^{n_{j}+1}$ (resp.
$\widetilde{\psi_{ij}}:{\mathbb{C}}^{n_{j}+1}\to{\mathbb{C}}^{n_{i}+1}$) be a
Lipschitz extension of $\phi_{ij}$ (resp. $\psi_{ij}=\phi_{ij}^{-1}$) (see
[Ki], [M] and [W]). Let us define
$\varphi,\psi:{\mathbb{C}}^{n_{i}+1}\times{\mathbb{C}}^{n_{j}+1}\to{\mathbb{C}}^{n_{i}+1}\times{\mathbb{C}}^{n_{j}+1}$
as follows:

$\varphi(x,y)=(x-\widetilde{\psi_{ij}}(y+\widetilde{\phi_{ij}}(x)),y+\widetilde{\phi_{ij}}(x))$

and

$\psi(z,w)=(z+\widetilde{\psi_{ij}}(w),w-\widetilde{\phi_{ij}}(z+\widetilde{\psi_{ij}}(w))).$

It easy to verify that $\psi=\varphi^{-1}$ and since $\varphi$ and $\psi$ are
composition of Lipschitz maps, they are also Lipschitz maps. Moreover, if
$Z_{ij}=Y_{i}\times\\{0\\}$ and $\widetilde{Z}_{ij}=\\{0\\}\times Y_{j}$, we
obtain that $\varphi(Z_{ij})=\widetilde{Z}_{ij}$ (see [Sa]). Therefore,
$(Z_{ij},0)$ and $(\widetilde{Z}_{ij},0)$ are bi-Lipschitz equivalent, but
$mult(Z_{ij},0)=mult(Y_{i},0)$ and $mult(\widetilde{Z}_{ij},0)=mult(Y_{j},0)$
and, in particular, they have different multiplicity.

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Misha Verbitsky
Instituto Nacional de Matemática Pura e Aplicada (IMPA)
Estrada Dona Castorina, 110
Jardim Botânico, CEP 22460-320
Rio de Janeiro, RJ - Brasil
also:
Laboratory of Algebraic Geometry,
National Research University Higher School of Economics,
Department of Mathematics, 6 Usacheva street, Moscow, Russia.

Lev Bribrair, Alexandre Fernandes and J. Edson Sampaio
Departamento de Matemática, Universidade Federal do Ceará,
Rua Campus do Pici, s/n, Bloco 914
Pici, CEP 60440-900
Fortaleza-CE, Brazil