==== TABLE OF CONTENTS ====
Ambiently-homeomorphic-isolated-hypersurfaces-of-multiplicities-two-and-three//background.tex
Ambiently-homeomorphic-isolated-hypersurfaces-of-multiplicities-two-and-three//certificate.tex
Ambiently-homeomorphic-isolated-hypersurfaces-of-multiplicities-two-and-three//conclusion.tex
Ambiently-homeomorphic-isolated-hypersurfaces-of-multiplicities-two-and-three//global.tex
Ambiently-homeomorphic-isolated-hypersurfaces-of-multiplicities-two-and-three//introduction.tex
Ambiently-homeomorphic-isolated-hypersurfaces-of-multiplicities-two-and-three//local.tex
Ambiently-homeomorphic-isolated-hypersurfaces-of-multiplicities-two-and-three//paper.tex
Ambiently-homeomorphic-isolated-hypersurfaces-of-multiplicities-two-and-three//preamble.tex
Ambiently-homeomorphic-isolated-hypersurfaces-of-multiplicities-two-and-three//proof-route.tex
Ambiently-homeomorphic-isolated-hypersurfaces-of-multiplicities-two-and-three//seeds.tex
Ambiently-homeomorphic-isolated-hypersurfaces-of-multiplicities-two-and-three//references.bib
===========================
==== BEGIN Ambiently-homeomorphic-isolated-hypersurfaces-of-multiplicities-two-and-three//background.tex ====
\section{Classical inputs and conventions}\label{sec:background}
This section records the classical results used below, together with their
hypotheses and the conventions needed to apply them.
\subsection{Milnor lattices, spectra, and integral monodromy}
\begin{theorem}[Milnor]\label{bg:milnor}
Let $s\geq2$, and let $f:(\mathbb C^s,0)\to(\mathbb C,0)$ be a holomorphic germ with an isolated
critical point at $0$, with $f(0)=0$. For sufficiently small $\varepsilon>0$, the link
$K_f=f^{-1}(0)\cap S^{2s-1}_\varepsilon$ is a smooth closed manifold of
dimension $2s-3$, and
\[
f/|f|:S^{2s-1}_\varepsilon\setminus K_f\longrightarrow S^1
\]
is a locally trivial fibration. Each fiber is the interior of its closure
$F_f$, a compact manifold with boundary $K_f$. Both $F_f$ and its interior
are homotopy equivalent to a finite bouquet of $(s-1)$-spheres.
The link is $(s-3)$-connected.
\end{theorem}
See \cite[Theorems~4.8, 5.2, and~6.5, and Lemma~6.1]{Milnor1968}.
The free group $L_f=\widetilde H_{s-1}(F_f;\mathbb Z)$ is the
\emph{Milnor lattice}; for a one-variable power we use the reduced
zeroth homology of its finite Milnor fiber. The unimodular integral Seifert form is
$S_f(x,y)=\operatorname{lk}(x,y^+)$, where the second cycle is pushed
positively to a neighboring page.
We use column vectors and fix the monodromy convention
\begin{equation}\label{eq:bg-monodromy}
M_f=(-1)^sS_f^{-1}S_f^{\mathsf t}.
\end{equation}
This amounts to using the inverse of the monodromy in the convention
$S_f(Mx,y)=(-1)^sS_f(y,x)$; see
\cite[Section~6, Equations~(6.1)--(6.2)]{BalnojanHertling2019}.
Inverting monodromy throughout preserves every comparison made below.
An isomorphism of Seifert forms means integral congruence unless a different
coefficient field or ring is specified.
\begin{theorem}[Thom--Sebastiani--Sakamoto]\label{bg:ts}
Suppose $f$ and $g$ have isolated critical points in $a$ and $b$ variables,
respectively. There is an isomorphism of integral lattices
\[
L_{f+g}\cong L_f\otimes_{\mathbb Z}L_g
\]
under which
\[
M_{f+g}\cong M_f\otimes M_g,
\qquad
S_{f+g}\cong(-1)^{ab}S_f\otimes S_g.
\]
The assertion includes one-variable powers, with reduced zeroth homology.
\end{theorem}
The tensor-product monodromy theorem originates with
Sebastiani and Thom~\cite{SebastianiThom1971}. The integral
Seifert-form formula is \cite[Theorem~2, p.~715]{Sakamoto1974};
the lattice and monodromy formulas are also recorded in
\cite[Section~6, Equations~(6.3)--(6.5)]{BalnojanHertling2019}.
Thus comparisons of sums with the same factor dimensions have the same
overall tensor sign.
A polynomial has \emph{normalized positive weights}
$w_1,\ldots,w_s\in\mathbb Q_{>0}$ if
$f(\lambda^{w_1}z_1,\ldots,\lambda^{w_s}z_s)=\lambda f(z)$, understood
after clearing denominators. We use positive spectral exponents
$\beta\in(0,s)$, shifted by $1$ from the convention with spectrum in
$(-1,s-1)$.
\begin{theorem}[Weighted homogeneous spectrum]\label{bg:spectrum}
If $f$ is weighted homogeneous of normalized degree $1$ and has an isolated
critical point, its spectral polynomial is
\begin{equation}\label{eq:bg-spectrum}
P_f(t)=\sum_{\beta\in\operatorname{Sp}^{+}(f)}t^\beta
=\prod_{j=1}^s\frac{t^{w_j}-t}{1-t^{w_j}}.
\end{equation}
Exponents are counted with multiplicity. Their classes modulo $\mathbb Z$
determine the eigenvalues of $M_f$ by exponentiation, with a fixed choice
of sign in the exponent. The operator $M_f$ has finite order.
\end{theorem}
Here is the algebraic conversion from the usual spectrum formula to
\eqref{eq:bg-spectrum}. Steenbrink's monomial formula
\cite[Section~1, Examples, p.~165]{Steenbrink1989} assigns exponent
$\sum_j(k_j+1)w_j$ to a monomial $z^k$ in a homogeneous basis of
$\mathbb C[z_1,\ldots,z_s]/(\partial f)$.
The derivatives have degrees $1-w_j$ and form a regular sequence:
their common zero set is the origin, so the ideal has height $s$ in the
Cohen--Macaulay polynomial ring
\cite[Tags~00ND and 02JN]{StacksProject}. The local regular-sequence
criterion applies at the homogeneous maximal ideal; positive grading
detects regularity there. Successive multiplication exact
sequences therefore give
\[
\operatorname{Hilb}\bigl(\mathbb C[z]/(\partial f),t\bigr)
=\prod_j\frac{1-t^{1-w_j}}{1-t^{w_j}}.
\]
Multiplication by $t^{\sum_jw_j}$ yields \eqref{eq:bg-spectrum}.
Finally, the weighted coordinate rotation
$z_j\mapsto e^{2\pi i w_j}z_j$ represents geometric monodromy; a common
multiple of the weight denominators kills this rotation. Finite-order
operators in characteristic zero are semisimple.
\begin{theorem}[Real Seifert forms from the spectrum]\label{bg:real}
Let $f$ and $g$ be isolated weighted homogeneous singularities in the same
number of variables. If their multisets of positive spectral exponents
agree modulo $2\mathbb Z$, their real Seifert forms are isomorphic.
\end{theorem}
For clarity, this is a consequence of the spectral-pair classification,
not an assertion about integral forms. N\'emethi's theorem
\cite[Theorem~6.5]{Nemethi1995} and
\cite[Theorem~4.4(b)]{BalnojanHertling2019} say that spectral pairs modulo
$2\mathbb Z$ in the first coordinate, together with the dimension,
determine the real Seifert form. In the present semisimple case the
nilpotent logarithm is zero. Put $m=s-1$. The actual cohomological weight
is $m$ away from eigenvalue $1$ and $m+1$ at eigenvalue $1$; the definition
of spectral pairs subtracts the latter shift, so every second coordinate
is $m$. Thus this coordinate supplies no additional data. Translation
$\beta\mapsto\beta-1$ respects equality modulo $2\mathbb Z$, and the
normalization and passage from cohomology to homology depend only on $s$.
More explicitly, the singularity construction gives a signed Steenbrink
polarized mixed Hodge structure, whose defining replacement of the
nilpotent logarithm by its negative leaves the data unchanged when that
logarithm is zero~\cite[Definition~6.1 and Remark~6.2]{BalnojanHertling2019}.
In that convention the normalized cohomological Seifert form is
$(-1)^{(m+1)(m+2)/2}$ times the dual homological Seifert form
\cite[Equation~(6.14)]{BalnojanHertling2019}. The common sign and duality
preserve isomorphism. This proves the stated consequence, including the
eigenvalue-$1$ part.
\begin{theorem}[Hertling--Mase]\label{bg:orlik}
Suppose $f$ is a sum of chains
\[
x_1^{a_1}x_2+\cdots+x_{r-1}^{a_{r-1}}x_r+x_r^{a_r},
\qquad a_i\geq2,
\]
including chains with $r=1$. Write the characteristic polynomial of
$M_f$ as $\prod_d\Phi_d(t)^{c_d}$, where $\Phi_d$ is cyclotomic.
Then, as an integral module with operator $t=M_f$,
\begin{equation}\label{eq:bg-orlik}
(L_f,M_f)\cong
\bigoplus_{j=1}^{\max c_d}\mathbb Z[t]/\bigl(p_j(t)\bigr),
\qquad p_j(t)=\prod_{c_d\geq j}\Phi_d(t).
\end{equation}
In particular, the eigenvalues with multiplicity determine this integral
monodromy module within the indicated class of singularities.
\end{theorem}
Orlik formulated the integral-monodromy decomposition
\cite{Orlik1972}; the chain calculation of Orlik and Randell
\cite{OrlikRandell1977} is a predecessor of the standard decomposition
proved by Hertling and Mase. The precise result used here is
\cite[Theorem~1.3(a), (c), (d), and Definition~2.6]{HertlingMase2022}.
Reversing the variables puts our chains in that paper's convention.
Each standard block has an integral cyclic generator, hence is exactly
the companion module displayed in \eqref{eq:bg-orlik}. Inverting its
operator preserves cyclicity and its characteristic
polynomial, whose roots are invariant under inversion. Thus \eqref{eq:bg-monodromy} causes no change
to the conclusion. This theorem controls the lattice with monodromy;
it does not identify the integral Seifert form.
\subsection{Arithmetic inputs}
For a number field $F$, write $F_v$ for its completion at a place $v$.
\begin{theorem}[Hasse--Minkowski]\label{bg:hasse}
Two nondegenerate quadratic forms over a number field $F$ are isometric
over $F$ if and only if they are isometric over $F_v$ for every place
$v$, including the archimedean places.
\end{theorem}
We use the number-field form of the theorem; see
\cite[Theorem~25.64(c) and the following discussion]{Milne2022};
see also \cite[Chapter~VI]{OMeara1963}.
\begin{theorem}[Strong approximation]\label{bg:approximation}
Let $G$ be a connected, simply connected, absolutely almost simple
algebraic group over a number field $F$. If
$\prod_{v\mid\infty}G(F_v)$ is noncompact, then $G(F)$ is dense in
$G(\mathbb A_{F,f})$, the group of finite adelic points.
\end{theorem}
See \cite[Theorem~A, p.~1140]{Platonov1969}, with the correction
in~\cite{Platonov1970}, and
\cite[Theorem~2.3, p.~281]{Rapinchuk2014}.
Here the finite adelic group is the restricted product of the $G(F_v)$
over finite places, with respect to integral compact open subgroups.
Consequently the density concerns approximation at finitely many
places together with integrality at all remaining places.
We apply the theorem to special unitary groups of nondegenerate
hermitian forms over a totally imaginary quadratic extension of a
totally real number field. In dimension at least two these groups are
connected, simply connected, and absolutely almost simple; an
indefinite archimedean component makes their archimedean product
noncompact. The construction below verifies these hypotheses for
every factor used.
\subsection{Embedded high-dimensional topology}
\begin{theorem}[Smale and Levine]\label{bg:classification}
A smooth closed manifold of dimension $d\geq5$ that is homotopy
equivalent to $S^d$ is homeomorphic to $S^d$.
Let $q\geq3$, and let $K_1,K_2\subset S^{2q+1}$ be smooth oriented
submanifolds homeomorphic to $S^{2q-1}$. Suppose the knots are
\emph{simple}, meaning
\[
\pi_1(S^{2q+1}\setminus K_j)\cong\mathbb Z,
\qquad
\pi_i(S^{2q+1}\setminus K_j)=0\quad(2\leq i\leq q-1).
\]
If their integral Seifert matrices are congruent, then they are
ambiently isotopic. In particular their embedded sphere pairs are
homeomorphic.
\end{theorem}
The first assertion is the high-dimensional generalized Poincar\'e
theorem \cite[Theorem~A, p.~391]{Smale1961}. The second follows from
\cite[Theorem~3, pp.~185--186]{Levine1970}: integral congruence is a
special case of the matrix equivalence in that theorem. Levine's
knots are required to be homeomorphic to spheres, so exotic smooth
sphere structures are allowed. Related formulations for Milnor
fibered links appear in \cite{Durfee1974},
\cite[pp.~698--699]{Saeki1999}, and
\cite[Section~2]{Neumann2017}.
The proof below checks the spherical and simplicity hypotheses
directly before applying this theorem.
==== END Ambiently-homeomorphic-isolated-hypersurfaces-of-multiplicities-two-and-three//background.tex ====
==== BEGIN Ambiently-homeomorphic-isolated-hypersurfaces-of-multiplicities-two-and-three//certificate.tex ====
\section{The finite arithmetic certificate}
\label{app:certificate}
The following complete Python program verifies the finite calculation in Lemma~\ref{seed:product}. It uses only integer arithmetic and the Python standard library. In the dictionaries representing valuation columns, $(r,p)$ denotes the valuation at the prime $p$ of ratio number $r$. The key $(-1,0)$ represents the appended row of ones. The rank routine performs Gaussian elimination over $\F_{1009}$, storing one normalized pivot for each leading row. Thus its return value is exactly the rank of the augmented matrix. Separately, $1009$ is prime: none of $2,3,5,7,11,13,17,19,23,29,31$ divides it, and $32^2>1009$.
The sieve stores a prime divisor of every integer greater than one. Repeated division therefore computes its prime valuations exactly. The preliminary cutoff and deletion of singly supported columns give the matrix used in the proof; every deleted column receives coefficient zero in the eventual relation. The final assertion is the sole finite calculation needed for existence of the two product identities.
\newpage
\lstinputlisting{../verification/support/supplied_finite_rank.py}
\noindent An explicit alternative relation is supplied in
\nolinkurl{verification/support/explicit-relation.json}. It has $849$ nonzero integer
coefficients, each with at most $140$ decimal digits, and positive
coefficient sum. The standard-library program
\nolinkurl{verification/validation/explicit_relation.py} verifies both product
identities by trial division of all six integers in every row, without
using the matrix or its rank. It also checks every decreasing-remainder
chain and the variable counts in the polynomial construction. This
independent witness is supplementary: the rank computation above and the
adjugate argument already prove Lemma~\ref{seed:product}.
==== END Ambiently-homeomorphic-isolated-hypersurfaces-of-multiplicities-two-and-three//certificate.tex ====
==== BEGIN Ambiently-homeomorphic-isolated-hypersurfaces-of-multiplicities-two-and-three//conclusion.tex ====
\section{From the integral forms to the ambient germs}
\label{topo:section}
We now turn the integral isometry of Proposition~\ref{global:integral}
into an equivalence of hypersurface germs. The two extra prime-power
summands introduced there also ensure that the links are
homotopy spheres. This permits a direct application of the spherical
knot classification in Theorem~\ref{bg:classification}.
\subsection{The final polynomials and their Seifert forms}
For $j\in\{1,2\}$, put
\begin{equation}
\label{topo:polynomials}
f_j=q_j^{[1]}+q_j^{[2]}+q_j^{[3]}+q_j^{[4]}
+z_1^3+\cdots+z_6^3+x^{11}+y^{13}.
\end{equation}
Here $q_j^{[a]}$ denotes a copy of $q_j$ in its own set of variables;
all the displayed summands use disjoint variable sets. Since $q_j$
has $s_0+2$ variables and $s_0$ is even, the common number of variables is
\begin{equation}
\label{topo:dimension}
N=4(s_0+2)+6+2=4s_0+16\equiv0\pmod 8.
\end{equation}
In particular, $N>3$. The weights of the summands, each normalized to
weighted degree one, give positive weights for $f_j$.
Write $(L_j^{\mathrm f},T_j^{\mathrm f},S_j^{\mathrm f})$ for the
Milnor lattice, chosen monodromy operator, and integral Seifert form of $f_j$,
with $T_j^{\mathrm f}=(S_j^{\mathrm f})^{-1}(S_j^{\mathrm f})^{\mathsf t}$.
All monodromy orders under consideration are odd. Choose a common
odd multiple $m_{\mathrm f}$ of them and an integer $r$ with
$2r\equiv1\pmod {m_{\mathrm f}}$.
As $N$ is even, the symmetric form associated with $S_j^{\mathrm f}$ is
\[
B_j^{\mathrm f}=S_j^{\mathrm f}(T_j^{\mathrm f})^r,
\qquad
S_j^{\mathrm f}=B_j^{\mathrm f}(T_j^{\mathrm f})^{-r}.
\]
The integral Thom--Sebastiani formula identifies these operator--form
data with the data $H_j$ of Proposition~\ref{global:integral}, up to the
same overall sign on the two symmetric forms. Indeed, every factor
in~\eqref{topo:polynomials}, grouped as four copies of $q_j$, the sum
of six cubes, and the sum $x^{11}+y^{13}$, has even ambient complex
dimension. Its
symmetrization therefore commutes with taking the indicated tensor
products, and the Thom--Sebastiani signs depend only on these common
dimensions.
Consequently Proposition~\ref{global:integral} supplies an integral
isomorphism $\Psi:L_1^{\mathrm f}\longrightarrow L_2^{\mathrm f}$ with
\[
\Psi T_1^{\mathrm f}=T_2^{\mathrm f}\Psi,
\qquad
\Psi^{\mathsf t}B_2^{\mathrm f}\Psi=B_1^{\mathrm f}.
\]
It also preserves the Seifert forms, because
\begin{equation}
\label{topo:seifert}
\Psi^{\mathsf t}S_2^{\mathrm f}\Psi
=\Psi^{\mathsf t}B_2^{\mathrm f}(T_2^{\mathrm f})^{-r}\Psi
=B_1^{\mathrm f}(T_1^{\mathrm f})^{-r}
=S_1^{\mathrm f}.
\end{equation}
This is a simultaneous integral isometry, which is precisely the
information needed for the topological step.
\begin{lemma}
\label{topo:analytic}
The polynomials $f_1,f_2$ have isolated critical points at the origin,
are reduced both as polynomials and as analytic germs, and satisfy
$\{\ord_0(f_1),\ord_0(f_2)\}=\{2,3\}$.
\end{lemma}
\begin{proof}
The first nonzero-coordinate argument in Lemma~\ref{seed:chain}
shows that each chain has no critical point anywhere in its affine
space except the origin. The gradient equations of a sum in disjoint
variables separate into the gradient equations of its summands.
The pure powers have the same global property. Consequently the only
critical point of each $q_j$, and then of each $f_j$, is the origin.
The orders of $q_1,q_2$ are two and three. Disjoint copies have no
monomials that could cancel one another, and all added terms have
degree at least three. The claimed orders follow. More precisely, the
$4E$ balancing squares in each copy of $p_1$ are the only quadratic
terms. Four copies give Hessian ranks $16E$ for $f_1$ and $0$ for
$f_2$, so both ranks are even.
If an irreducible nonunit analytic germ $a$ occurred with multiplicity
at least two in $f_j$, we could write $f_j=a^2b$. Every first derivative
of $f_j$ would then vanish on $V(a)$. A nonunit hypersurface germ in
$\C^N$ has dimension $N-1>0$, contradicting the isolated critical
point. Thus $f_j$ is reduced as an analytic germ. The same argument
for a repeated nonconstant polynomial factor would force the gradient
to vanish on its positive-dimensional affine zero set. The global
critical-locus statement just proved excludes this as well, so $f_j$
is reduced as a polynomial.
\end{proof}
\subsection{Spherical links and their embedded equivalence}
Let
\[
K_j=V(f_j)\cap S^{2N-1}_\varepsilon,
\qquad
X_j=S^{2N-1}_\varepsilon\setminus K_j,
\qquad n=N-1,
\]
where $\varepsilon>0$ is sufficiently small for both polynomials.
The binding $K_j$ is a smooth closed manifold of dimension $2n-1$.
Let $F_j$ be a compact Milnor page, so the fiber of $X_j\to S^1$ is
$\operatorname{int}F_j$. The inclusion of this interior into $F_j$ is a
homotopy equivalence, and both have the homotopy type of a bouquet of
$n$-spheres. Moreover $K_j$ is $(n-2)$-connected
by Theorem~\ref{bg:milnor} \cite{Milnor1968}.
\begin{lemma}
\label{topo:sphere}
Each $K_j$ is a homotopy sphere, and each embedding
$K_j\subset S^{2n+1}_\varepsilon$ is a simple knot.
\end{lemma}
\begin{proof}
Every eigenvalue of $T_j^{\mathrm f}$ has the form
\[
\xi\zeta_{11}^{a}\zeta_{13}^{b},
\qquad 1\leq a\leq10,\quad1\leq b\leq12,
\]
where the order of $\xi$ divides $3^A\cdot5\cdot7$ for some integer
$A\geq1$. The factors of orders $11$ and $13$ cannot cancel against
each other or against $\xi$. Thus every eigenvalue has order divisible
by both $11$ and $13$.
Since $T_j^{\mathrm f}$ is integral and has finite order, its
characteristic polynomial is a product of cyclotomic polynomials
$\Phi_d$, with every occurring index $d$ divisible by $143$.
For $d>1$, the elementary identity
\[
\Phi_d(1)=
\begin{cases}
p,&d\text{ is a power of the prime }p,\\
1,&d\text{ has at least two distinct prime factors}
\end{cases}
\]
follows by induction from
$d=\prod_{e\mid d,\ e>1}\Phi_e(1)$: the prime-power divisors already
contribute the full prime factorization of $d$ when $d$ is not a
prime power. Hence
\begin{equation}
\label{topo:determinant}
\det(1-T_j^{\mathrm f})=1.
\end{equation}
In particular, $1-T_j^{\mathrm f}$ is an automorphism of the integral
middle homology of $F_j$.
The geometric monodromy acts as $(T_j^{\mathrm f})^{-1}$ in our
convention. The identity
$1-(T_j^{\mathrm f})^{-1}=-(T_j^{\mathrm f})^{-1}(1-T_j^{\mathrm f})$
shows that its difference from the identity is also invertible over
$\Z$. The Wang exact sequence of the Milnor fibration now gives
\[
H_i(X_j;\Z)=
\begin{cases}
\Z,&i=0,1,\\
0,&i\geq2.
\end{cases}
\]
Indeed, the only possible additional groups are, equivalently, the
kernel and cokernel of $1-T_j^{\mathrm f}$ on $H_n(F_j;\Z)$, and both vanish
by~\eqref{topo:determinant}. Alexander duality~\cite[Corollary~3.45]{Hatcher2002} therefore makes $K_j$
an integral homology $(2n-1)$-sphere. Its $(n-2)$-connectivity makes
it simply connected. The Hurewicz and Whitehead Theorems
\cite[Theorem~4.32 and Corollary~4.33]{Hatcher2002} show that
it is a homotopy sphere. Since $2n-1\geq5$, the high-dimensional
Poincar\'e theorem included in Theorem~\ref{bg:classification}
makes it homeomorphic to $S^{2n-1}$.
Finally, the homotopy exact sequence of
$\operatorname{int}F_j\to X_j\to S^1$, together with the
$(n-1)$-connectivity of $\operatorname{int}F_j$, gives
\[
\pi_1(X_j)\cong\Z,
\qquad
\pi_i(X_j)=0\quad(2\leq i\leq n-1).
\]
This is the simplicity condition for a $(2n-1)$-knot.
\end{proof}
Orient the links and their Seifert surfaces by the Milnor fibration,
using the same positive push-off convention on both sides.
By~\eqref{topo:seifert}, their integral Seifert matrices are congruent.
Since $n\geq3$, Levine's classification of simple spherical knots
\cite[Theorem~3]{Levine1970}, in the form of
Theorem~\ref{bg:classification}, yields an isotopy of the embedded links. Smooth isotopy extension
then gives an orientation-preserving diffeomorphism of pairs
\begin{equation}
\label{topo:link-equivalence}
\phi:(S^{2N-1}_\varepsilon,K_1)
\longrightarrow(S^{2N-1}_\varepsilon,K_2).
\end{equation}
Levine's definition allows smooth homotopy spheres that are not
standard smooth spheres. Thus no standard-smooth-sphere assertion
is needed here. Nor is an extra framing hypothesis needed: the
classification is for these oriented embedded knots with their
Seifert forms. The same implication is part of the classification
of high-dimensional simple fibered links
\cite{Durfee1974,Saeki1999}; see also \cite{Neumann2017}.
\subsection{An explicit homeomorphism of ambient germs}
We finish by spelling out the passage from the sphere to an ambient
neighborhood of the origin. For a positive weight vector
$w=(w_1,\ldots,w_N)$, define
\[
a_w(t,x)=(t^{w_1}x_1,\ldots,t^{w_N}x_N),\qquad t>0.
\]
For each nonzero $x$, the function
$t\mapsto\|a_w(t,x)\|^2=\sum_i t^{2w_i}|x_i|^2$
is strictly increasing, with limits zero and infinity at the two
ends of $(0,\infty)$. Each nonzero weighted orbit therefore meets
$S^{2N-1}_\varepsilon$ in exactly one point. Consequently
\begin{equation}
\label{topo:weighted-cone}
A_w:C(S^{2N-1}_\varepsilon)\longrightarrow\overline B^{2N}_\varepsilon,
\qquad [x,t]\longmapsto a_w(t,x),
\quad 0\leq t\leq1,
\end{equation}
is a continuous bijection, where all points $(x,0)$ are identified
and their image is zero. The uniform bound
$\|a_w(t,x)\|\leq\varepsilon t^{\min_i w_i}$ proves continuity
at the cone vertex. The domain is compact and the target Hausdorff, so
$A_w$ is a homeomorphism.
Let $w^{(j)}$ be the weight vector of $f_j$. Weighted homogeneity
gives
\[
f_j(a_{w^{(j)}}(t,x))=t f_j(x).
\]
Thus $A_{w^{(j)}}$ identifies the cone on $K_j$ with
$V(f_j)\cap\overline B^{2N}_\varepsilon$. Extend the map
in~\eqref{topo:link-equivalence} by
\begin{equation}
\label{topo:ambient-map}
h(0)=0,
\qquad
h\bigl(a_{w^{(1)}}(t,x)\bigr)
=a_{w^{(2)}}(t,\phi(x))
\quad(0<t\leq1).
\end{equation}
Equivalently, $h=A_{w^{(2)}}\circ C(\phi)\circ A_{w^{(1)}}^{-1}$,
so $h$ is an ambient ball homeomorphism carrying the first
hypersurface to the second. It preserves the ball boundary;
restricting to the ball interiors gives the required homeomorphism
germ of ambient pairs.
Together with Lemma~\ref{topo:analytic}, this proves
Theorem~\ref{thm:main}: the reduced hypersurface germs in
\eqref{topo:polynomials} are ambiently homeomorphic and have
multiplicities two and three.
\subsection{Consequences for function germs and initial forms}
\label{topo:consequences}
The same pair has unequal Hessian coranks, as the balancing squares
already show. Its real coefficients also allow us to strengthen the
ambient equivalence to topological right equivalence. We then use that
stronger equivalence to compare initial forms after adding one common
square in a new variable.
\begin{corollary}[Negative answer to Arnold's corank problem]
\label{topo:arnold-corank}
In the finite ambient dimension $N>3$ divisible by eight
in~\eqref{topo:dimension}, the polynomials $f_1,f_2$
in~\eqref{topo:polynomials} have different Hessian coranks, although
their reduced complex hypersurface germs are ambiently homeomorphic.
Both polynomials have real coefficients, are weighted homogeneous with
positive rational weights, and have isolated critical points at the origin.
\end{corollary}
\begin{proof}
Theorem~\ref{thm:main} supplies the ambient homeomorphism and the stated
analytic properties. The calculation in the proof of
Lemma~\ref{topo:analytic} gives complex Hessian ranks $16E$ for $f_1$
and $0$ for $f_2$. Since $E>0$ by Lemma~\ref{seed:product}, their
coranks satisfy
\[
\operatorname{corank}_0(f_1)=N-16E<N=\operatorname{corank}_0(f_2).
\]
\end{proof}
\begin{corollary}[Right-topological multiplicity and parity counterexamples]
\label{topo:right-equivalence}
For the same polynomials $f_1,f_2$ and the single finite ambient dimension
$N>3$ divisible by eight in Theorem~\ref{thm:main}, there is a homeomorphism
germ
\[
\varphi:(\C^N,0)\longrightarrow(\C^N,0)
\qquad\text{such that}\qquad
f_1=f_2\circ\varphi.
\]
Nevertheless, their multiplicities are respectively $2$ and $3$. Thus
neither multiplicity nor multiplicity modulo two is invariant under
topological right equivalence, even for reduced real-coefficient weighted
homogeneous holomorphic germs with positive rational weights and isolated
critical points at the origin.
\end{corollary}
\begin{proof}
For holomorphic germs $f,g:(\C^n,0)\to(\C,0)$ with $n\ge2$ and isolated
critical points at the origin, Saeki's topological $V$-equivalence is
precisely an ambient homeomorphism of their zero-set germs
\cite[Definition~3]{Saeki1989}. Saeki's Corollary~2 identifies this with
topological right-left equivalence \cite{Saeki1989}. King's corollary
\cite{King1978}, recalled in \cite[Section~3, p.~28]{Saeki1989}, then
gives right equivalence up to conjugation. In the precise
source-homeomorphism formulation recorded by Sampaio
\cite[proof of Proposition~2.4]{Sampaio2025Milnor}, there is a homeomorphism germ
$\varphi:(\C^n,0)\to(\C^n,0)$ such that
\[
f=g\circ\varphi
\quad\text{or}\quad
f=(c_1\circ g\circ c_n)\circ\varphi,
\]
where $c_k(z_1,\ldots,z_k)=(\overline z_1,\ldots,\overline z_k)$.
Theorem~\ref{thm:main} supplies these hypotheses for $f=f_1$ and $g=f_2$.
Writing $f_2(z)=\sum_\alpha a_\alpha z^\alpha$ with $a_\alpha\in\R$, we have
\[
(c_1\circ f_2\circ c_N)(z)
=\overline{f_2(\overline z)}
=\sum_\alpha\overline{a_\alpha}z^\alpha
=f_2(z).
\]
Consequently both alternatives give $f_1=f_2\circ\varphi$.
The multiplicities are $2$ and $3$ by Theorem~\ref{thm:main}, and these
integers are unequal both in $\Z$ and modulo two.
\end{proof}
The homeomorphism $\varphi$ furnished by this argument need not coincide
with the constructed ambient homeomorphism $h$ in~\eqref{topo:ambient-map}.
The parity conclusion concerns multiplicity; the unequal Hessian coranks
above have the same parity.
We now pass to germs in $N+1$ variables by adding the same square to
each polynomial. The product of $\varphi$ with the identity preserves
right equivalence. Both new germs have order two, but their quadratic
initial forms have different Milnor fibers.
For a polynomial $g$ vanishing at the origin, write $g_*$ for its lowest
nonzero homogeneous component for ordinary total degree, not the weighted
grading used above. We use the global unit fibre $g_*^{-1}(1)$ as the
initial-form Milnor fibre: it has the homotopy type of the local Milnor
fibre of $g_*$ \cite[Proposition~1.6 in arXiv v3]{BudurEtAl2022ContactLoci}.
This convention retains the full polynomial $g_*$, not its square-free
reduction.
\begin{corollary}[Initial-form Milnor-fibre counterexamples]
\label{topo:tangent-milnor}
In the single ambient dimension $N+1$, put
\[
\widetilde f_j(z,t)=f_j(z)+t^2,\qquad j=1,2.
\]
The germs $\widetilde f_1,\widetilde f_2:(\C^{N+1},0)\to(\C,0)$ are
topologically right equivalent and have isolated critical points at the
origin, but
\[
H_0\bigl((\widetilde f_1)_*^{-1}(1);\Z\bigr)\cong\Z,
\qquad
H_0\bigl((\widetilde f_2)_*^{-1}(1);\Z\bigr)\cong\Z^2.
\]
Thus integral homology of the initial-form Milnor fibre is not invariant
under topological right equivalence, already in degree zero. In particular,
the initial-form Milnor-fibre homotopy conjecture of Budur, Fern\'andez de
Bobadilla, L\^e, and Nguyen
\cite[Conjecture~1.7 in arXiv v3]{BudurEtAl2022ContactLoci} fails.
\end{corollary}
\begin{proof}
Apply the quadratic case of Sampaio's stabilization
\cite[proof of Proposition~2.3]{Sampaio2025Milnor}.
For the homeomorphism germ $\varphi$ in
Corollary~\ref{topo:right-equivalence}, the product homeomorphism
$\widetilde\varphi(z,t)=(\varphi(z),t)$ satisfies
\[
\widetilde f_1=\widetilde f_2\circ\widetilde\varphi.
\]
It also carries the corresponding zero-set germs to one another. Moreover,
$\nabla\widetilde f_j(z,t)=(\nabla f_j(z),2t)$, so the critical point
remains isolated on each side.
By the proof of Lemma~\ref{topo:analytic}, the quadratic part of $f_1$ is
\[
Q=\sum_{\alpha=1}^{16E}u_\alpha^2
\]
in the independent balancing-square coordinates, with $E>0$, whereas
$f_2$ has order three. Hence
\[
(\widetilde f_1)_*=Q+t^2,\qquad
(\widetilde f_2)_*=t^2.
\]
Their global unit fibres are respectively
\[
\left\{v\in\C^{16E+1}:
\sum_{\alpha=1}^{16E+1}v_\alpha^2=1\right\}
\times\C^{N-16E},
\qquad
\C^N\times\{-1,1\}.
\]
The quadric factor is connected. Indeed, writing $v=x+iy$, its equation
becomes $|x|^2-|y|^2=1$ and $x\cdot y=0$. The map
$(u,y)\mapsto\sqrt{1+|y|^2}\,u+iy$, with
$u\in S^{16E}$ and $y\perp u$, identifies this quadric with the tangent
bundle of the connected sphere $S^{16E}$. The first unit fibre is therefore
connected, whereas the second has exactly two components. This gives the
displayed zeroth homology groups and rules out homotopy equivalence.
\end{proof}
The isolatedness in Corollary~\ref{topo:tangent-milnor} concerns the
stabilized germs, not their initial forms. In particular, the nonreduced
initial form $t^2$ is retained: replacing it by $t$ would change its unit
fibre. No reducedness or isolatedness hypothesis on the initial forms is
part of the cited homotopy conjecture.
==== END Ambiently-homeomorphic-isolated-hypersurfaces-of-multiplicities-two-and-three//conclusion.tex ====
==== BEGIN Ambiently-homeomorphic-isolated-hypersurfaces-of-multiplicities-two-and-three//global.tex ====
\section{From local isometries to an integral isometry}
\label{global:section}
We now turn the local and rational isometries of
Proposition~\ref{local:rational} into one integral isometry, after two
further operations. A second tensor square removes all eigenspace
determinants from the discrepancy between the local and rational maps.
Two common polynomial summands then put the relevant isometry groups
within the scope of strong approximation.
Write the tensor-square data as
\[
A_j=(M_j,\tau_j,b_j)
=(L_j^{\otimes2},T_j^{\otimes2},B_j^{\otimes2}).
\]
Fix the equivariant rational isometry $R:A_1\longrightarrow A_2$ and
the equivariant local lattice isometries $D_\ell$ supplied by
Proposition~\ref{local:rational}, with $D_\ell=R$ for all but finitely
many primes $\ell$.
\subsection{Cancelling the determinant on each eigenspace}
The ratio $R^{-1}D_\ell$ preserves $b_1$ and commutes with $\tau_1$
over $\Q_\ell$, but need not have determinant one on each eigenspace.
To cancel these
determinants by a second tensor square, we use complex conjugation on
the real-coefficient polynomials $q_j$. Recall that their common
ambient dimension $s$ is even and that $B_j=S_jT_j^r$.
\begin{lemma}[Integral reversers]\label{local:reverser}
There is an integral involution $C_j$ of $L_j$ such that
\[
C_j^{-1}T_jC_j=T_j^{-1},\qquad C_j^{\mathsf t}B_jC_j=B_j.
\]
\end{lemma}
\begin{proof}
Complex conjugation preserves the argument-zero Milnor page and
exchanges its positive and negative normal directions; let $C_j$ be
its action on middle homology. We check the sign directly in the
linking definition of the Seifert form~\cite{Milnor1968}.
The ambient sphere has dimension $2s-1$, and the cycles have dimension
$s-1$. Complex conjugation has degree $(-1)^s=1$ on that sphere, and
interchanging the two linking cycles has sign $(-1)^{s^2}=1$.
If $y^+$ and $y^-$ denote the two push-offs, then
\[
\begin{aligned}
S_j(C_jx,C_jy)
&=\operatorname{lk}(x,y^-)
=\operatorname{lk}(y^-,x)\\
&=\operatorname{lk}(y,x^+)=S_j(y,x).
\end{aligned}
\]
The third equality moves both cycles along the page-normal flow.
Thus $C_j^{\mathsf t}S_jC_j=S_j^{\mathsf t}$. Taking the monodromy
of this congruence gives $C_j^{-1}T_jC_j=T_j^{-1}$, and then
\[
C_j^{\mathsf t}B_jC_j
=S_j^{\mathsf t}T_j^{-r}=S_jT_j^{1-r}=B_j.
\]
The map on oriented cycles already includes conjugation's effect on
page orientation; the calculation uses the orientation of the ambient
sphere for linking.
\end{proof}
Tensoring gives an integral isometry
$C_j^{(2)}=C_j\otimes C_j$ of $b_j$ that reverses $\tau_j$.
On the fourth tensor powers of the original data define
\begin{equation}
\label{global:corrected-maps}
\widehat R
=R\otimes\bigl(C_2^{(2)}R(C_1^{(2)})^{-1}\bigr),
\qquad
\widehat D_\ell
=D_\ell\otimes\bigl(C_2^{(2)}D_\ell(C_1^{(2)})^{-1}\bigr).
\end{equation}
Both second factors intertwine the operators: each conjugating map
reverses its operator, and $R$ and $D_\ell$ intertwine them.
They also preserve the forms. Consequently $\widehat R$ is a rational
equivariant isometry, and $\widehat D_\ell$ is an equivariant isometry
of the fourth-power lattices over $\Z_\ell$.
\begin{lemma}[Eigenspace determinant cancellation]
\label{global:determinants}
For every prime $\ell$, the ratio
$\widehat R^{-1}\widehat D_\ell$ has determinant $1$ on every
eigenspace of $\tau_1\otimes\tau_1$ over $\overline{\Q_\ell}$.
This remains true after tensoring both maps with identity maps on any
common finite-order operator data.
\end{lemma}
\begin{proof}
Put $a=R^{-1}D_\ell$ and $C=C_1^{(2)}$. Multiplying the two maps in
\eqref{global:corrected-maps} gives the exact identity
\begin{equation}
\label{global:ratio}
\widehat R^{-1}\widehat D_\ell=a\otimes CaC^{-1}.
\end{equation}
In particular, no compatibility of $R$ with the reversers is required.
Let $W_\gamma$ be the $\gamma$-eigenspace of $\tau_1$ over
$\overline{\Q_\ell}$, and set
\[
m_\gamma=\dim W_\gamma,
\qquad t_\gamma=\det(a|_{W_\gamma}).
\]
Invariance and nondegeneracy of $b_1$ give a perfect pairing between
$W_\gamma$ and $W_{\gamma^{-1}}$. Since $a$ is an isometry,
\begin{equation}
\label{global:inverse-determinants}
m_{\gamma^{-1}}=m_\gamma,
\qquad t_{\gamma^{-1}}=t_\gamma^{-1}.
\end{equation}
For $\gamma=1$, this says $t_1^2=1$; no choice of its sign is needed.
The reverser carries $W_{\eta^{-1}}$ onto $W_\eta$, so the determinant
of $CaC^{-1}$ on $W_\eta$ is $t_{\eta^{-1}}$.
For each $\lambda$, its entire eigenspace in the tensor square is
\[
\bigoplus_{\gamma\eta=\lambda}W_\gamma\otimes W_\eta,
\]
where the sum ranges over all ordered pairs of occurring eigenvalues.
By the determinant formula for a tensor product, the determinant of
\eqref{global:ratio} on this space is
\begin{align*}
\prod_{\gamma\eta=\lambda}
t_\gamma^{m_\eta}t_{\eta^{-1}}^{m_\gamma}
&=
\frac{\displaystyle\prod_{\gamma\eta=\lambda}t_\gamma^{m_\eta}}
{\displaystyle\prod_{\gamma\eta=\lambda}t_\eta^{m_\gamma}}
=1.
\end{align*}
The last equality exchanges $\gamma$ and $\eta$ in the denominator.
It includes diagonal pairs and all coincidences among product
eigenvalues. Finally, tensoring with an identity gives, on each new
eigenspace, a product of powers of the already unit determinants.
\end{proof}
\subsection{Two common summands}
Use new disjoint variables to form
\begin{equation}
\label{global:common-polynomials}
u=z_1^3+\cdots+z_6^3,
\qquad v=z_7^{11}+z_8^{13}.
\end{equation}
Both have an even number of variables and odd-order monodromy. Denote
their symmetric-form and operator data by $\mathcal U$ and
$\mathcal V$, and put
\begin{equation}
\label{global:final-data}
H_j=(\Lambda_j,\Theta_j,\beta_j)
=A_j^{\otimes2}\otimes\mathcal U\otimes\mathcal V.
\end{equation}
Extend the maps in \eqref{global:corrected-maps} by the identity on
these two common factors, keeping the notation $\widehat R$ and
$\widehat D_\ell$.
The six cubes will make each eigenspace form indefinite. The degrees
$11$ and $13$ exclude real eigenvalues from the final operator, even
though the preceding tensor squares may have acquired eigenvalue $1$.
These two primes also enter the later proof that the links are homotopy
spheres in Lemma~\ref{topo:sphere}.
\begin{lemma}[Spectral properties of the added factors]
\label{global:indefinite}
Every eigenvalue of $\Theta_j$ has odd order divisible by both $11$
and $13$. On every complex eigenspace, the hermitian form
\[
(x,y)\longmapsto\beta_j(x,\overline y)
\]
is nondegenerate and indefinite; the complex dimension of that
eigenspace is at least $21$.
\end{lemma}
\begin{proof}
Before adjoining $\mathcal V$, the operator order divides
$3^a\cdot5\cdot7$ for some $a$. The eigenvalues of $v$ are
\[
\zeta_{11}^b\zeta_{13}^c,
\qquad 1\le b\le10,\quad 1\le c\le12.
\]
Their nontrivial $11$- and $13$-primary components cannot cancel
against an eigenvalue of the earlier tensor factors. Thus every
final order contains both primes and remains odd. In particular,
neither $1$ nor $-1$ is a final eigenvalue.
We calculate the signatures supplied by $u$. Let
\[
A=\begin{pmatrix}1&-1\\0&1\end{pmatrix}.
\]
In the convention of Theorem~\ref{bg:ts}, a one-cube Seifert matrix
is $S_c=-A$~\cite[Corollary~3]{Sakamoto1974}. Its chosen monodromy is
\[
t=-A^{-1}A^{\mathsf t}
=\begin{pmatrix}0&-1\\1&-1\end{pmatrix}.
\]
Here the minus sign is the one-variable convention. The form
\[
J=At^2=\begin{pmatrix}0&1\\-1&0\end{pmatrix}
\]
is alternating and $t$-invariant. Iterating Theorem~\ref{bg:ts}
gives $S_u=(-1)^{15}(-A)^{\otimes6}=-A^{\otimes6}$.
The sixfold monodromy is $t^{\otimes6}$, and
$2\cdot2\equiv1\pmod3$, so its symmetric form is
\[
B_u=S_u(t^{\otimes6})^2=-J^{\otimes6}.
\]
We compute the signatures of $J^{\otimes6}$ below; passing to $B_u$
reverses every sign.
Set $\omega=e^{2\pi i/3}$ and $e_+=(1,-\omega)$,
$e_-=(1,-\omega^2)$. These are eigenvectors of $t$ with
eigenvalues $\omega$ and $\omega^2$, respectively, and
\[
J(e_+,\overline{e_+})=i\sqrt3,
\qquad J(e_-,\overline{e_-})=-i\sqrt3,
\qquad J(e_+,\overline{e_-})=0.
\]
In the resulting orthogonal tensor basis, a vector with $k$ copies
of $e_+$ has eigenvalue $\omega^{-k}$ and length
$27(-1)^{k+1}$; there are $\binom6k$ such vectors. We obtain the
following signatures, up to reversal of every sign:
\[
\begin{array}{c|c|c|c}
\text{eigenvalue}&k&\text{signature}&\text{dimension}\\\hline
1&0,3,6&(20,2)&22\\
\omega&2,5&(6,15)&21\\
\omega^2&1,4&(6,15)&21
\end{array}
\]
Every eigenspace of this common factor is therefore indefinite.
For any real nondegenerate symmetric form $b$ invariant under a
finite-order operator, its complex bilinear extension pairs the
$\gamma$-eigenspace perfectly with the $\gamma^{-1}$-eigenspace.
Consequently $b(x,\overline y)$ restricts to a nondegenerate
hermitian form on each eigenspace. Distinct eigenspaces are
orthogonal for this hermitian form.
Apply this observation to each of the three factors in
\eqref{global:final-data}. A final eigenspace is the orthogonal
direct sum of the tensor products of factor eigenspaces whose
eigenvalues have the prescribed product. Each nonzero summand
contains one of the indefinite spaces in the table. Diagonalizing
the other two hermitian forms shows that its tensor form has both
signs: multiply any nonzero diagonal entry from the other factors
by a positive and a negative entry from the six-cube factor.
Every summand is thus indefinite and has dimension at least $21$.
The same conclusions hold for their orthogonal direct sum.
\end{proof}
\subsection{The special unitary group and strong approximation}
\begin{proposition}[Integral equivariant isometry]
\label{global:integral}
The triples $H_1$ and $H_2$ in \eqref{global:final-data} are
isometric over $\Z$: there is a lattice isomorphism
$U:\Lambda_1\longrightarrow\Lambda_2$ satisfying
\[
U\Theta_1=\Theta_2U,
\qquad U^{\mathsf t}\beta_2U=\beta_1.
\]
\end{proposition}
\begin{proof}
Put $V=\Lambda_1\otimes\Q$. Since $\Theta_1$ has finite order,
its minimal polynomial is a product of distinct cyclotomic
polynomials. Decompose the rational operator space as
\[
V=\bigoplus_{d\in\mathcal D}V_d,
\qquad
E_d=\Q(\zeta_d),\quad F_d=\Q(\zeta_d+\zeta_d^{-1}),
\quad n_d=\dim_{E_d}V_d.
\]
The action of $\Theta_1$ gives $V_d$ its $E_d$-vector space
structure. Distinct $V_d$ are orthogonal for $\beta_1$.
By Lemma~\ref{global:indefinite}, all $d$ are odd and divisible
by $11\cdot13$, so $E_d/F_d$ is quadratic with nontrivial
conjugation. The adjoint of multiplication by $a\in E_d$ is
multiplication by $\overline a$, because the adjoint of $\Theta_1$
is $\Theta_1^{-1}$.
For completeness, the nondegenerate trace pairing of $E_d/\Q$
defines a unique $E_d$-valued form $h_d$ by
\[
\tr_{E_d/\Q}\bigl(a h_d(x,y)\bigr)=\beta_1(ax,y)
\quad(a\in E_d).
\]
This form is linear in its first argument, conjugate-linear in
its second, and satisfies $h_d(y,x)=\overline{h_d(x,y)}$.
These identities follow by testing against every $a$ in the
displayed trace equation. The same equation proves its
nondegeneracy. It also shows that an $E_d$-linear map preserves
$\beta_1$ exactly when it preserves $h_d$.
Let $G$ be the algebraic $\Q$-group of equivariant self-isometries of
$(V,\Theta_1,\beta_1)$ with determinant $1$ on every eigenspace
over an algebraic closure. These conditions are defined over
$\Q$. The preceding description identifies this algebraic group as
\begin{equation}
\label{global:unitary-group}
G=\prod_{d\in\mathcal D}\Res_{F_d/\Q}\SU(h_d).
\end{equation}
Indeed, commuting with $\Theta_1$ is exactly $E_d$-linearity on
each $V_d$. Each embedding of $E_d$ into an algebraic closure
gives one $\Theta_1$-eigenspace, and the determinant on it is
the corresponding image of the $E_d$-determinant. Requiring
all these determinants to be $1$ gives precisely $\SU(h_d)$.
In particular this requirement removes the unitary determinant
tori, and the absence of eigenvalues $\pm1$ leaves no
orthogonal factor.
Every embedding $F_d\hookrightarrow\R$ extends to a conjugate
pair of embeddings of $E_d$ into $\C$. At this real place the trace
identity writes the underlying real form as $2\operatorname{Re}h_d$.
After complexification, its pairing $(x,y)\mapsto\beta_1(x,\overline y)$
on either corresponding operator eigenspace is the associated complex
hermitian form. Lemma~\ref{global:indefinite} therefore makes it
indefinite, and $n_d\ge21$.
The group $\SU(h_d)$ is connected, simply connected, and
absolutely almost simple over $F_d$, since over an algebraic
closure it becomes $\mathrm{SL}_{n_d}$
\cite[Theorem~24.44, Remark~24.46, and Summary~24.66]{Milne2022}. At each real place it
is a noncompact group $\SU(p,q)$ with $p,q>0$.
Theorem~\ref{bg:approximation}, applied to $\SU(h_d)$ over $F_d$
with all archimedean places omitted, now gives density of
$\SU(h_d)(F_d)$ in its finite adelic group
\cite[Theorem~A]{Platonov1969}, with~\cite{Platonov1970}; see also
\cite[Theorem~2.3]{Rapinchuk2014}.
Restriction of scalars and the finite product in
\eqref{global:unitary-group} therefore give
\begin{equation}
\label{global:density}
\overline{G(\Q)}=G(\A_{\Q,f}).
\end{equation}
Here finite adeles mean the restricted product over all finite
places. No isotropy over the number field itself is being
assumed; the required noncompactness is supplied at the real
places just checked.
We spell out how \eqref{global:density} gives an integral map.
For each prime set
\[
a_\ell=\widehat R^{-1}\widehat D_\ell,
\qquad \Lambda_{j,\ell}=\Lambda_j\otimes\Z_\ell.
\]
Lemma~\ref{global:determinants} gives $a_\ell\in G(\Q_\ell)$,
and $a_\ell=1$ for all but finitely many $\ell$. Thus
$a=(a_\ell)$ is a finite adelic point. Define
\[
K_\ell=\{g\in G(\Q_\ell):g\Lambda_{1,\ell}
=\Lambda_{1,\ell}\},
\qquad K=\prod_\ell K_\ell.
\]
In a basis of the actual lattice $\Lambda_1$, the subgroup
$K_\ell$ is $G(\Q_\ell)\cap\GL_{\operatorname{rank}\Lambda_1}(\Z_\ell)$.
It is compact and open in $G(\Q_\ell)$. To identify their product with
an adelic open subgroup, take the schematic closure of $G$ in
$\GL(\Lambda_1)$ over $\Z$. After inverting finitely many primes this
is an integral model of the same affine group, and its
$\Z_\ell$-points are exactly $K_\ell$. Thus the $K_\ell$ agree with
the reference integral subgroups defining the restricted product at
all but finitely many primes. Their product $K$ is consequently compact
and open in $G(\A_{\Q,f})$. This definition uses the full lattice and
requires no integral splitting of its cyclotomic components.
By \eqref{global:density}, the nonempty open coset $aK$ contains
a point $g\in G(\Q)$. For every prime we can write
$g=a_\ell k_\ell$ with $k_\ell\in K_\ell$, and hence
\[
(\widehat Rg)(\Lambda_{1,\ell})
=\widehat R a_\ell(\Lambda_{1,\ell})
=\widehat D_\ell(\Lambda_{1,\ell})
=\Lambda_{2,\ell}.
\]
A full integral lattice satisfies
\[
\Lambda_j
=\{x\in\Lambda_j\otimes\Q:
x\in\Lambda_{j,\ell}\text{ for every prime }\ell\}:
\]
in a lattice basis this is the elementary identity
$\Q\cap\bigcap_\ell\Z_\ell=\Z$.
Applying this characterization to $\widehat Rg$ and its inverse
shows that $U=\widehat Rg$ carries $\Lambda_1$ onto $\Lambda_2$.
It is an equivariant isometry because both factors are such
isometries, which proves the proposition.
\end{proof}
\begin{remark}
\label{global:recover-seifert}
An equivariant isometry of the symmetric data recovers an isometry
of the original Seifert data. Choose one integer $r$ satisfying
$2r\equiv1$ modulo a common odd exponent of the final operators.
If $S_j'=\beta_j\Theta_j^{-r}$, then the isomorphism in
Proposition~\ref{global:integral} satisfies
\[
U^{\mathsf t}S_2'U
=U^{\mathsf t}\beta_2U\Theta_1^{-r}
=\beta_1\Theta_1^{-r}=S_1'.
\]
The tensor-product convention can multiply both final forms by
the same sign, which does not alter this identity.
\end{remark}
==== END Ambiently-homeomorphic-isolated-hypersurfaces-of-multiplicities-two-and-three//global.tex ====
==== BEGIN Ambiently-homeomorphic-isolated-hypersurfaces-of-multiplicities-two-and-three//introduction.tex ====
\section{The multiplicity problem}
\label{sec:introduction}
For a nonzero convergent power series
$f\in\C\{z_1,\ldots,z_N\}$ with $f(0)=0$, the order $\ord_0f$ is
the least total degree of a monomial with nonzero coefficient. When $f$
is reduced, this integer is the multiplicity of the hypersurface germ
$V(f)=\{f=0\}$. Two such germs are \emph{ambiently homeomorphic} if a
homeomorphism between neighborhoods of the origin in $\C^N$ fixes the
origin and carries one hypersurface to the other. Zariski's multiplicity
question asks whether this embedded topological equivalence determines
multiplicity~\cite{Zariski1971}. It asks, in particular, whether topology
can distinguish a hypersurface with a nonzero quadratic term from one
whose first nonzero terms are cubic.
For an irreducible plane curve, the classical relation between embedded
topology and the characteristic data of a branch determines its
multiplicity~\cite{Zariski1932}. Summing the branch multiplicities gives
the same conclusion for a reduced plane curve. In every ambient dimension,
the A'Campo--L\^e theorem shows that a reduced hypersurface germ ambiently
homeomorphic to a smooth one is itself smooth~\cite{ACampo1973,Le1973};
see~\cite[Theorem~2.2]{Sampaio2025Milnor}. Thus multiplicity one cannot
occur in an unequal-multiplicity pair. Eyral surveys the multiplicity
problem and its further positive cases~\cite{Eyral2007}.
In higher dimensions, the Milnor
fibration gives powerful topological invariants of an isolated
singularity~\cite{Milnor1968}. Its integral middle homology carries both
a monodromy operator and a Seifert form. The latter records linking of
cycles in neighboring fibers and, in a suitable high-dimensional range,
classifies the embedded link. Multiplicity, by contrast, is read directly
from the lowest-degree terms of an equation. Our construction separates
these two kinds of information.
A polynomial is \emph{weighted homogeneous with positive rational
weights} if there are positive integers $a_1,\ldots,a_N,d$ such that
$f(t^{a_1}z_1,\ldots,t^{a_N}z_N)=t^d f(z)$; the normalized weights are
$w_i=a_i/d$. We prove the following existence statement.
\begin{theorem}\label{thm:main}
There is a finite integer $N>3$, divisible by eight, and there are reduced
polynomials $f_1,f_2\in\R[z_1,\ldots,z_N]$, weighted homogeneous for
positive rational weights, such that:
\begin{enumerate}[label=\textup{(\roman*)}]
\item each $f_j$ has an isolated critical point at the origin;
\item $\ord_0f_1=2$ and $\ord_0f_2=3$;
\item there is a homeomorphism germ of ambient pairs
\[
(\C^N,V(f_1),0)\ \cong\ (\C^N,V(f_2),0).
\]
\end{enumerate}
\end{theorem}
Theorem~\ref{thm:main} gives a negative answer to the embedded Zariski
multiplicity conjecture. A single finite ambient dimension suffices for
this purpose. The construction determines such a dimension by exact
integer arithmetic; it is not intended to minimize it.
The same examples settle the corresponding questions for topological right
equivalence. Holomorphic function germs $f,g:(\C^N,0)\to(\C,0)$ are
\emph{topologically right equivalent} if $f=g\circ\varphi$ for a
homeomorphism germ $\varphi:(\C^N,0)\to(\C^N,0)$. The right-topological
multiplicity conjecture asks whether reduced germs related in this way have
equal multiplicities; its mod-two version asks only for equal multiplicity
parity \cite[Conjecture~2 and Conjecture~Z mod~2]{Sampaio2025Milnor}.
Corollary~\ref{topo:right-equivalence} gives negative answers to both
questions for the pair in Theorem~\ref{thm:main}. The source homeomorphism
there may differ from the ambient map constructed below and is not claimed
to be holomorphic or bi-Lipschitz.
A further consequence concerns Arnold's corank problem. For a holomorphic
germ $f:(\C^N,0)\to(\C,0)$ with a critical point at the origin, write
\[
\operatorname{corank}_0(f)
=N-\rank_{\C}\operatorname{Hess}_0(f),
\]
where $\operatorname{Hess}_0(f)$ is the Hessian matrix at the origin.
Arnold's question asks whether this number is preserved when a
homeomorphism germ of $(\C^N,0)$ carries $V(f)$ to $V(g)$ for reduced
holomorphic germs $f,g$ with critical points at the origin
\cite[Problem~1]{Sampaio2026}. Corollary~\ref{topo:arnold-corank} gives
a negative answer, even for the isolated weighted homogeneous germs
with real coefficients in Theorem~\ref{thm:main}.
The examples also answer a question about the lowest-degree part of a
function. For a nonzero holomorphic germ $g$ vanishing at the origin,
let $g_*$ denote its lowest nonzero homogeneous component for ordinary
total degree. Its \emph{initial-form Milnor fibre} has the homotopy type
of the global unit fibre $g_*^{-1}(1)$; the full polynomial $g_*$ is
retained, even when it is nonreduced
\cite[Proposition~1.6 in arXiv v3]{BudurEtAl2022ContactLoci}.
Motivated by the relationship between contact loci and monodromy,
Budur, Fern\'andez de Bobadilla, L\^e, and Nguyen conjectured that these
fibres are homotopy equivalent for embedded topologically equivalent
function germs~\cite[Conjecture~1.7 in arXiv v3]{BudurEtAl2022ContactLoci}.
Sampaio studies the explicit right-equivalence formulation
\cite[Conjecture~4]{Sampaio2025Milnor}.
Corollary~\ref{topo:tangent-milnor} disproves both formulations after adding one common
square: the germs $f_j(z)+t^2$ in $\C^{N+1}$ remain topologically right
equivalent and have isolated critical points, but their initial-form Milnor fibres are
connected and have two connected components, respectively. Even their
zeroth integral homology groups are therefore different.
\subsection*{Families, metric equivalence, and the present question}
The question for an arbitrary pair must be distinguished from
multiplicity in a family. L\^e and Ramanujam showed that a one-parameter holomorphic deformation
of isolated hypersurface singularities with constant Milnor number is
topologically constant when the ambient complex dimension differs from
three~\cite{LeRamanujam1976}. In the excluded case $\C^3$, the companion
\cite[Theorem~1.1]{OpenAIMuConstant2026} proves, after shrinking,
topological right-triviality for holomorphic one-parameter families $f_t$ with
$f_t(0)=0$, an isolated critical point at the origin for every $t$,
and constant $\mu(f_t,0)$.
Greuel and O'Shea proved equimultiplicity
for the corresponding deformation problem at an isolated quasihomogeneous
singularity~\cite{Greuel1986,OShea1987}. Fern\'andez de Bobadilla and
Pe\l ka subsequently established equimultiplicity for continuous
families with finite constant Milnor number, including coefficientwise
continuous families of formal power series
\cite[Theorem~1.1 and Corollary~1.3]{BobadillaPelka2022}.
Their Example~8.6 gives topologically equivalent germs in $\C^4$, both
of multiplicity two, that cannot be joined by a continuous
$\mu$-constant family. They obtain the embedded equivalence by adding
two independent squares to plane-curve germs with the same integral
Seifert form, a precedent for the passage from integral forms to
embedded topology used here.
Thus an arbitrary equivalent pair does not
reduce to the family problem. The two different multiplicities in
Theorem~\ref{thm:main} therefore preclude a connecting $\mu$-constant
family.
There are also positive results for pairs under additional geometric
hypotheses. Yau established multiplicity invariance for pairs of isolated
quasihomogeneous surface singularities in $\C^3$
\cite[Theorem~C]{Yau1988}, crediting the detailed proofs jointly with
Xu in the article's added note. Metric and low-dimensional hypotheses
give further positive results. Fernandes, Jelonek, and Sampaio proved that multiplicity two is preserved
by ambient bi-Lipschitz equivalence, and that Hessian corank is preserved
by ambient homeomorphisms in three complex variables
\cite[Corollary~3.5 and Theorem~4.4]{FernandesJelonekSampaio2025}.
The homeomorphism in Theorem~\ref{thm:main} is topological and its
ambient dimension is larger than three. Sampaio's more recent
corank-parity theorem gives an ambient topological invariant in every
dimension~\cite[Theorem~4.1]{Sampaio2026}.
Corollary~\ref{topo:arnold-corank} is consistent with this theorem:
the two Hessian ranks are even, so the unequal coranks have the same parity.
These comparisons describe the precise topological setting of the
counterexample.
\subsection*{The main ideas}
The main difficulty is to preserve different ordinary orders while
making the \emph{integral Seifert forms} coincide. We compare the
spectra, finite multisets of rational exponents whose exponentials give
the monodromy eigenvalues. Equality of spectra alone does not accomplish
the desired integral identification. For the isolated weighted homogeneous
singularities considered here, in a fixed ambient dimension, the spectrum
determines the real Seifert form, through the work of Steenbrink and
N\'emethi
\cite{Steenbrink1977,Steenbrink1989,Nemethi1995,BalnojanHertling2019}.
For sums of chain singularities, the integral-monodromy results of
Hertling and Mase, building on Orlik and Randell, separately determine
the integral module with its monodromy operator
\cite{Orlik1972,OrlikRandell1977,HertlingMase2022}.
The two resulting identifications need not be compatible. The arithmetic
part of the proof produces a single integral map preserving the whole
Seifert form.
\paragraph{1. Seeds from a finite relation.}
We begin with two multiplicative identities among explicitly specified
positive rational numbers. A finite valuation matrix certifies an
integer relation whose coefficient sum is positive. A decreasing
integer-remainder construction completes each required weight to an
isolated chain polynomial. The two collections of chains have different
numbers of variables; balancing the difference by squares on one side
produces multiplicity two there, while common cubes give multiplicity
three on the other. The product identities and a triple-angle identity
show that the spectra agree modulo two. The two classical classification
results just described then supply a real Seifert-form isometry and an
integral monodromy-module isomorphism
(Section~\ref{seed:section}).
\paragraph{2. Compatible local and rational forms.}
After adding common fifth and seventh powers, the monodromy has finite
odd order and no eigenvalue one. In even ambient dimension, this permits
an integral symmetrization of each Seifert form that retains the
monodromy and recovers the original form. We prove that a first tensor
square of these symmetric data has equivariant isometries over every
$\Z_\ell$. At odd primes, square roots and sums of two squares in a
finite operator algebra produce the maps. At two, the operator splits
the lattice into dual pairs and unramified hermitian modules. A
quadratic-form local--global argument then supplies an equivariant
rational isometry (Section~\ref{local:section}).
\paragraph{3. A single integral isometry.}
The rational map and the local maps can still differ by nontrivial
eigenspace determinants. Real coefficients provide integral involutions
reversing monodromy. A second tensor square pairs each map with its
conjugate by these involutions; an exact determinant calculation cancels
the discrepancies on every eigenspace. Six common cubic variables make
every relevant hermitian form indefinite. Two further common pure
powers, of degrees eleven and thirteen, eliminate real eigenvalues.
The determinant-one isometry group is now a product of special unitary
groups. The strong-approximation method of Eichler, Kneser, and Platonov
\cite{Platonov1969,Platonov1970,Rapinchuk2014} gives a rational group
element meeting all prescribed local lattice conditions, and hence an
integral Seifert-form isometry (Section~\ref{global:section}).
\paragraph{4. From forms to ambient germs.}
The Thom--Sebastiani theorem and Sakamoto's integral Seifert-form formula
realize the tensor operations by polynomial sums in disjoint variables
\cite{SebastianiThom1971,Sakamoto1974}. The last two pure powers also
force $\det(1-T)=1$ for the final monodromy $T$. The links are thus
homotopy spheres, and the Milnor fibration verifies that they are simple
knots. Levine's classification converts the integral Seifert congruence
into embedded link equivalence~\cite{Levine1970}. Weighted radial
coordinates extend that equivalence to a homeomorphism of ambient germs
(Section~\ref{topo:section}). Figure~\ref{fig:proof-route} collects
these stages and their distinct outputs.
Section~\ref{topo:section} then applies the results of King and Saeki
to obtain topological right equivalence. Real coefficients remove the
complex-conjugation alternative in that transfer
\cite{King1978,Saeki1989,Sampaio2025Milnor}.
Adding a square in one new variable preserves right equivalence and
gives the different connectedness of the initial-form Milnor fibres.
\input{figures/proof-route}
Section~\ref{sec:background} fixes the conventions and exact classical
inputs. Appendix~\ref{app:certificate} contains the complete integer
program for the finite rank certificate; a separately supplied exact
relation can be verified by trial division alone. All construction,
local algebra, determinant calculations, and the passage from local
lattices to an integral map are proved in the text. The classical
singularity-theoretic, arithmetic, and topological classification
theorems are invoked after checking their hypotheses.
\subsection*{Conventions}
Every sum combining singularities uses disjoint sets of variables.
Superscript $[i]$ marks a fresh copy of a polynomial. An isometry of
lattices with operators is required to intertwine those operators.
The transpose of a matrix is $M^{\mathsf t}$. A lattice is a finite-rank
free abelian group; a bilinear form is unimodular if its determinant is
$\pm1$.
==== END Ambiently-homeomorphic-isolated-hypersurfaces-of-multiplicities-two-and-three//introduction.tex ====
==== BEGIN Ambiently-homeomorphic-isolated-hypersurfaces-of-multiplicities-two-and-three//local.tex ====
\section{Local and rational isometries of tensor squares}
\label{local:section}
Let $q_1,q_2$ be the polynomials of Corollary~\ref{seed:prepared},
in the same even number $s$ of variables. Write $L_j$ for their Milnor
lattices and $S_j$ for their integral Seifert forms. The two inputs from
that corollary are an isomorphism of the integral monodromy modules
and an isomorphism of the real Seifert forms. We now combine these
inputs, first locally and then rationally.
Choose an odd integer $m$ divisible by the orders of both monodromies,
and choose $r\in\Z$ with $2r\equiv1\pmod m$. With $s$ even and the
monodromy convention in Equation~\eqref{eq:bg-monodromy}, put
\begin{equation}\label{local:symmetric-data}
T_j=S_j^{-1}S_j^{\mathsf t},\qquad B_j=S_jT_j^r.
\end{equation}
Throughout, an \emph{equivariant isometry} between data $(L,T,B)$
means an isometry of the forms that intertwines the displayed operators.
Tensor products carry both the tensor product form and the tensor product
operator.
\begin{lemma}[Symmetrization]\label{local:symmetrize}
Each $B_j$ is an integral unimodular symmetric form preserved by $T_j$.
The pair $(B_j,T_j)$ recovers $S_j$ by $S_j=B_jT_j^{-r}$.
The real Seifert isometry is an equivariant real isometry of the
symmetric data. Symmetrization commutes with tensor products of
even-variable factors, up to the common signs in
Theorem~\ref{bg:ts}.
\end{lemma}
\begin{proof}
Suppress the subscript. The defining equality $S^{\mathsf t}=ST$ implies
\[
T^{\mathsf t}ST=S,\qquad
B^{\mathsf t}=(T^r)^{\mathsf t}S^{\mathsf t}
=ST^{1-r}=ST^r=B.
\]
It also gives $T^{\mathsf t}BT=B$. Integrality and unimodularity follow
from those of $S$ and $T$. An isometry of $S$ intertwines
$S^{-1}S^{\mathsf t}$, and therefore preserves $B$ as well.
For tensor products, use the same $r$ modulo a common odd exponent and
the identity
\[
(S_1\otimes S_2)(T_1\otimes T_2)^r
=(S_1T_1^r)\otimes(S_2T_2^r).
\]
The overall Seifert signs supplied by the integral Thom--Sebastiani
theorem depend only on the factor dimensions~\cite{Sakamoto1974};
they are thus the same on the two sides of every comparison here.
\end{proof}
The local comparison will concern the first tensor squares
\[
A_j=(L_j,T_j,B_j)^{\otimes2}.
\]
Our goal is an equivariant isometry between $A_1$ and $A_2$ over every
$\Z_\ell$, together with one over $\Q$. The integral operator-module
isomorphism lets us compare the two forms on a common lattice. At odd
primes a tensor square makes these forms isometric; at $2$ the prepared
operators already give uniqueness before tensoring. The real Seifert
isometry then supplies the archimedean input for rational descent.
\subsection{Odd primes}
We first give the elementary lifting fact needed for a simultaneous
isometry of the form and operator.
\begin{lemma}[Lifting in a finite operator algebra]\label{local:lifting}
Let $\ell$ be an odd prime and let $\mathcal A$ be a commutative
$\Z_\ell$-algebra that is finite as a $\Z_\ell$-module. Write $J$ for
its Jacobson radical. Then $\mathcal A/J$ is a product of finite fields,
and $\mathcal A$ is complete for the $J$-adic topology. For a unit
$u\in\mathcal A$:
\begin{enumerate}
\item if its image in every residue field is a square, then $u$ is a
square in $\mathcal A$;
\item $u=a^2+b^2$ for some $a,b\in\mathcal A$.
\end{enumerate}
\end{lemma}
\begin{proof}
The algebra is complete for the $\ell$-adic topology. For every
$c\in\mathcal A$, the geometric series for $(1-\ell c)^{-1}$ converges,
so $\ell\mathcal A\subseteq J$. The finite algebra
$\mathcal A/\ell\mathcal A$ has nilpotent radical and semisimple quotient
a product of finite fields. Consequently $J^N\subseteq\ell\mathcal A$
for some $N$, proving that the two topologies agree and establishing the
first assertions.
For (1), choose a lift $v$ of a square root in $\mathcal A/J$.
Both $v$ and $2v$ are units. If $e=v^2-u\in J^k$, replace $v$ by
$v-e/(2v)$. The new error is $e^2/(4v^2)\in J^{2k}$.
Iteration converges to a square root of $u$.
In a finite field $k$ of odd cardinality, the sets of squares and of
$\bar u$ minus the squares each have $(|k|+1)/2$ elements. They
intersect, so $\bar u=\bar a^2+\bar b^2$. For (2), make such a
choice in each residue field and lift it to $a_0,b_0\in\mathcal A$.
Then $c=a_0^2+b_0^2$ is a unit and $u/c\in1+J$. By (1), write
$u/c=v^2$. Taking $a=va_0$ and $b=vb_0$ proves the claim.
\end{proof}
\begin{proposition}[Odd-prime tensor squares]\label{local:odd-primes}
Let $L$ be a finite free $\Z_\ell$-module, with $\ell$ odd, and let
$T$ be an automorphism of $L$. If $B_1,B_2$ are unimodular symmetric
forms preserved by $T$, then
\[
(L\otimes L,T\otimes T,B_1\otimes B_1)
\cong
(L\otimes L,T\otimes T,B_2\otimes B_2)
\]
by an equivariant integral isometry.
\end{proposition}
\begin{proof}
Set $P=B_1^{-1}B_2$, so
$B_2(x,y)=B_1(x,Py)$. Unimodularity makes $P$ an integral
automorphism. Symmetry of $B_2$ makes it self-adjoint for $B_1$, and
invariance of both forms gives $PT=TP$.
Write the characteristic polynomial of $P$ modulo $\ell$ as
$\prod_i f_i^{e_i}$, with distinct monic irreducible $f_i$.
Lift this coprime factorization to
$\det(X-P)=\prod_i F_i$ over $\Z_\ell$, where
$F_i\bmod\ell=f_i^{e_i}$. For completeness, the lift is obtained
successively modulo $\ell^n$. For two coprime monic residue factors
$f,g$ of degrees $a,b$, the correction map
\[
(u,v)\longmapsto ug+fv,\qquad \deg u<a,\quad\deg v<b,
\]
is an isomorphism onto the polynomials of degree less than $a+b$:
its kernel is zero by coprimality and the dimensions agree. It
therefore corrects the product by any prescribed error at the next
power of $\ell$, without changing monicity. Completeness gives the
lift, and iteration handles any number of factors. The Chinese remainder
idempotents for the $F_i$, evaluated at $P$, give
\[
L=\bigoplus_i L_i,\qquad F_i(P_i)=0,
\qquad P_i=P|_{L_i}.
\]
These idempotents are polynomials in $P$. They are self-adjoint and
commute with $T$, so the sum is orthogonal for $B_1$ and $B_2$ and each
$L_i$ is $T$-stable. Each restricted form is unimodular.
On $L_i\otimes L_j$, let $\mathcal A_{ij}$ be the algebra generated
by $P_i\otimes1$ and $1\otimes P_j$. Cayley--Hamilton bounds the
degrees in these two generators, so this is a finite commutative
$\Z_\ell$-algebra of endomorphisms. Every element is self-adjoint
for the restricted form $B_1\otimes B_1$ and commutes with
$T\otimes T$. The unit
\[
U_{ij}=P_i\otimes P_j
\]
relates the two restricted tensor forms. Its inverse belongs to
$\mathcal A_{ij}$, since the inverses of the generators are polynomials
in them with integral coefficients.
If $i=j$, consider a residue field of $\mathcal A_{ii}$.
The relations $F_i(P_i)=0$ imply that the two generator images
$x,y$ are roots of $f_i$. They are nonzero because the generators
are units. In a finite field, two roots of the same irreducible
polynomial over $\F_\ell$ are Frobenius conjugates. Hence
$y=x^{\ell^h}$ for some $h\geq0$, and
\[
xy=x^{1+\ell^h}
=\left(x^{(1+\ell^h)/2}\right)^2.
\]
Lemma~\ref{local:lifting} supplies $a\in\mathcal A_{ii}$ with
$a^2=U_{ii}$. Because $a$ is self-adjoint,
\[
(B_1\otimes B_1)(ax,ay)=(B_2\otimes B_2)(x,y)
\qquad(x,y\in L_i\otimes L_i).
\]
For $i<j$, pair $L_i\otimes L_j$ with $L_j\otimes L_i$.
The factor swap identifies their operators and both their forms.
On the resulting two equal blocks, use
Lemma~\ref{local:lifting} to choose $a,b\in\mathcal A_{ij}$ with
$a^2+b^2=U_{ij}$ and set
\[
M=\begin{pmatrix}a&b\\-b&a\end{pmatrix}.
\]
Writing $*$ for adjoint with respect to the first tensor form on
the two blocks gives
\[
M^*M=\begin{pmatrix}U_{ij}&0\\0&U_{ij}\end{pmatrix}.
\]
This is invertible, and $M$ commutes with the tensor operator.
The diagonal-block maps and these paired-block maps combine to an
integral equivariant isometry from the second tensor form to the first.
Its inverse has the direction asserted in the proposition.
\end{proof}
\subsection{The prime two}
Here the odd order and absence of eigenvalue $1$ give a stronger result.
The unramified hermitian classification underlying this step belongs
to the classical local theory developed by Jacobowitz~\cite{Jacobowitz1962}.
We give the needed argument directly, including residue characteristic two.
\begin{proposition}[Dyadic uniqueness]\label{local:dyadic}
Let $L$ be a finite free $\Z_2$-module, and let $T$ satisfy $T^m=1$
for an odd $m$. Suppose $1$ is not an eigenvalue of $T$ on
$L\otimes\Q_2$. Any two $T$-invariant unimodular symmetric forms
on $L$ are equivariantly isometric over $\Z_2$.
\end{proposition}
\begin{proof}
The reduction of $X^m-1$ modulo $2$ is separable, since its derivative
is $X^{m-1}$ there. Lifting its distinct irreducible factors gives
\[
\mathcal O=\Z_2[X]/(X^m-1)=\prod_i\mathcal O_i,
\]
where each $\mathcal O_i$ is the integer ring of an unramified
extension of $\Q_2$. The action $X\mapsto T$ decomposes
$L=\bigoplus_i M_i$. Every $M_i$ is free over $\mathcal O_i$:
it is finite and torsion-free over this discrete valuation ring.
Invariance of a form $B$ says that the adjoint of $X$ is $X^{-1}$.
Let $\sigma$ denote this involution of $\mathcal O$. If it exchanges
the factors indexed by $i$ and $i'$, then $M_i$ and $M_{i'}$ are
isotropic and pair perfectly with each other; they are orthogonal to
all other factors. Such a pairing has a unique equivariant isometry
class for the prescribed modules. Indeed, it identifies $M_{i'}$
with the $\Z_2$-dual of $M_i$, with its $\mathcal O_i$-action
transported by $\sigma$. An isomorphism on $M_i$ extends uniquely by
its inverse dual to an isometry of the pair.
It remains to treat a nonzero $M_i$ whose factor is preserved by
$\sigma$. Write $\mathcal O_i=\mathcal D$ and
$\mathcal D_0=\mathcal D^\sigma$.
The involution on $\mathcal D$ is nontrivial. Otherwise the image
$\zeta$ of $X$ would satisfy $\zeta=\zeta^{-1}$; its odd order would
force $\zeta=1$, contrary to the hypothesis. Thus
$\mathcal D/\mathcal D_0$ is an unramified quadratic extension.
We record explicitly how $B$ becomes an integral hermitian form.
The trace pairing of $\mathcal D$ over $\Z_2$ is perfect: modulo
$2$ it is the nondegenerate trace pairing of a finite separable field
extension, and its determinant is therefore a unit. Hence there is
a unique $\mathcal D$-valued form $h$ such that
\begin{equation}\label{local:trace}
\operatorname{Tr}_{\mathcal D/\Z_2}\bigl(a h(x,y)\bigr)
=B(ax,y)\qquad(a\in\mathcal D).
\end{equation}
The adjoint relation and symmetry of $B$ give
\[
h(ax,by)=a\sigma(b)h(x,y),\qquad
h(y,x)=\sigma(h(x,y)).
\]
Thus $h$ is hermitian, with linear first variable.
It is unimodular: after extending to fraction fields, the $B$-dual
of $M_i$ equals its $h$-dual. Indeed, if $B(x,M_i)\subseteq\Z_2$,
then for every $a\in\mathcal D$ and $y\in M_i$,
\[
B(ax,y)=B(x,\sigma(a)y)\in\Z_2.
\]
Testing all $a$ in~\eqref{local:trace} and using the perfect trace
pairing gives $h(x,M_i)\subseteq\mathcal D$; the converse is immediate.
This argument takes no quotient by $2$.
Every unimodular hermitian form over
$\mathcal D/\mathcal D_0$ is determined by its rank. Here are the
details in residue characteristic two. Let $k/k_0$ be the quadratic
residue extension and let $\bar h$ be the reduced hermitian form.
There is a vector of nonzero length. If all lengths vanished, then
for all $x,y$ and $a\in k$,
\[
0=\bar h(x+ay,x+ay)
=\operatorname{Tr}_{k/k_0}
\bigl(\sigma(a)\bar h(x,y)\bigr).
\]
The finite-field trace pairing would force $\bar h=0$, contradicting
nonsingularity. A lift of a vector of nonzero length has unit length
$u\in\mathcal D_0^\times$. Its line splits off integrally, by the
projection $x\mapsto u^{-1}h(x,v)v$. Its orthogonal complement is
again unimodular. Induction diagonalizes $h$ with all diagonal
entries in $\mathcal D_0^\times$.
Every such unit is a norm from $\mathcal D^\times$. The residue norm
$k^\times\to k_0^\times$ is surjective. After lifting a residue
preimage, improve a norm congruence modulo $2^n$ to one modulo
$2^{n+1}$ using
\[
N_{\mathcal D/\mathcal D_0}(1+2^nz)
\equiv1+2^n\operatorname{Tr}_{\mathcal D/\mathcal D_0}(z)
\pmod{2^{n+1}},\qquad n\geq1.
\]
The trace modulo $2$ is surjective, so the needed $z$ always exists.
The corrections converge. Rescaling each diagonal vector therefore
makes every diagonal entry $1$. The ranks of the $M_i$ depend only
on the operator module, so this proves uniqueness on every fixed
factor and completes the proof.
\end{proof}
\subsection{Rational descent}
Although neither $T_j$ has eigenvalue $1$, its tensor square can:
an eigenvalue and its inverse have product $1$. The rational argument
must therefore include an ordinary quadratic-form component at
eigenvalue $1$, as well as the hermitian components for nonreal
eigenvalues. The odd order still excludes $-1$.
We shall use Hasse--Minkowski in the form stated in
Theorem~\ref{bg:hasse}. The passage from quadratic forms to hermitian
forms needed here has an elementary proof.
\begin{lemma}[Hermitian forms from their lengths]
\label{local:hermitian-lengths}
Let $E/F$ be a quadratic field extension with $\operatorname{char}F\ne2$,
and let $h_1,h_2$ be nonsingular hermitian forms. If the quadratic
forms
\[
Q_{h_j}(x)=h_j(x,x)
\]
on the underlying $F$-spaces are isometric, then $h_1,h_2$ are
$E$-linearly isometric.
\end{lemma}
\begin{proof}
The polar pairing of $Q_h$ is
$\tfrac12\operatorname{Tr}_{E/F}h(x,y)$.
If it annihilates $x$, testing scalar multiples of $y$ and using
the nondegenerate trace pairing shows $h(x,y)=0$ for every $y$.
Thus $Q_h$ is nonsingular. In particular a nonzero hermitian space
contains a vector of nonzero length, and such a vector splits off
an orthogonal hermitian line.
The quadratic cancellation used below is the reflection argument
underlying Witt cancellation~\cite{Witt1937}; we include the needed
form of it. For a quadratic
form $Q$ with polar form $b$ and a vector $z$ of nonzero length, the
reflection
\[
\rho_z(v)=v-\frac{2b(v,z)}{Q(z)}z
\]
is an isometry, as direct expansion shows. If $Q(x)=Q(e)=a\ne0$,
then $Q(x-e)+Q(x+e)=4a$. If $Q(x-e)\ne0$, the reflection in $x-e$
sends $x$ to $e$. Otherwise reflection in $x+e$ sends $x$ to $-e$,
and reflection in $e$ sends $-e$ to $e$. It follows that an isometry
$\langle a\rangle\perp V\cong\langle a\rangle\perp W$ can be
adjusted to match the two displayed vectors, and then restricts to
an isometry $V\cong W$.
Now write $E=F(\sqrt d)$. Choose $x$ of hermitian length $a\ne0$
for $h_1$, and let $y$ be its image under an isometry of the length
quadratic forms. The lines $Ex$ and $Ey$ are hermitian isometric,
and their length quadratic forms are both
\[
aN_{E/F}=\langle a,-ad\rangle.
\]
Split these hermitian lines off. Cancel the two nonzero quadratic
lines $\langle a\rangle$ and $\langle-ad\rangle$ successively by
the preceding reflection argument. The complementary hermitian
forms again have isometric length forms. Induction on their
$E$-dimension proves the lemma.
\end{proof}
\begin{lemma}[Equivariant rational local--global principle]
\label{local:hasse-descent}
Let $(V_j,U_j,\beta_j)$ be rational vector spaces with nonsingular
symmetric forms preserved by operators of finite odd order.
If they are equivariantly isometric over $\R$ and over $\Q_\ell$
for every prime $\ell$, then they are equivariantly isometric over $\Q$.
\end{lemma}
\begin{proof}
Take a common odd exponent for the operators. The cyclotomic
decomposition of each $V_j$ is orthogonal: its idempotents are
unchanged under $U_j\mapsto U_j^{-1}$, which is the adjoint
involution. Each equivariant local isometry respects these
decompositions, and corresponding component dimensions agree.
The eigenvalue-$1$ component carries an ordinary rational symmetric
form. Its local isometries give a rational isometry by
Theorem~\ref{bg:hasse}. Odd order excludes eigenvalue $-1$.
For any remaining cyclotomic component put
\[
E=\Q(\zeta_d),\qquad F=\Q(\zeta_d+\zeta_d^{-1}),
\qquad d>1.
\]
The component is an $E$-vector space, with $U_j$ acting by $\zeta_d$.
Complex conjugation $\sigma$ is a nontrivial involution of $E$.
The same trace construction as~\eqref{local:trace}, now over $\Q$,
gives a unique nonsingular $E/F$-hermitian form $h_j$ satisfying
\[
\operatorname{Tr}_{E/\Q}\bigl(a h_j(x,y)\bigr)
=\beta_j(ax,y)\qquad(a\in E).
\]
An equivariant isometry over a rational completion is linear over
the corresponding algebra $E\otimes_\Q\Q_\ell$, or over
$E\otimes_\Q\R$, and preserves the trace identity. Nondegeneracy
of the extended trace pairing therefore makes it an isometry of
the extended hermitian forms. Decomposing
$F\otimes_\Q\Q_\ell$ into its completion factors supplies an
isometry at every place of $F$ above $\ell$; decomposing
$F\otimes_\Q\R$ supplies every archimedean place. This also
covers finite places where $E/F$ splits, since the trace pairing of
the resulting quadratic \'etale algebra is still nondegenerate.
Consequently the nonsingular quadratic length forms $Q_{h_1}$ and
$Q_{h_2}$ are isometric over every completion of $F$. Hasse--Minkowski
over $F$ gives a global quadratic isometry
\cite[Theorem~25.64(c)]{Milne2022}. Lemma~\ref{local:hermitian-lengths} upgrades its
existence to an $E$-linear hermitian isometry. Such an isometry
preserves the rational symmetric form and intertwines $U_j$.
Combining the cyclotomic components proves the claim.
\end{proof}
\begin{proposition}[Local and rational square isometries]
\label{local:rational}\label{local:conclusion}
Set
\[
A_j=(L_j,T_j,B_j)^{\otimes2}.
\]
There are an equivariant rational isometry
$R:A_1\otimes\Q\longrightarrow A_2\otimes\Q$ and, for every prime
$\ell$, an equivariant isometry
$D_\ell:A_1\otimes\Z_\ell\longrightarrow A_2\otimes\Z_\ell$.
These can be chosen so that $D_\ell=R$ after extension to $\Q_\ell$
for all but finitely many primes.
\end{proposition}
\begin{proof}
Use the integral operator-module isomorphism of
Corollary~\ref{seed:prepared} to transport the two forms to a common
module with a common operator. Proposition~\ref{local:odd-primes}
gives the required square isometry at each odd prime.
Proposition~\ref{local:dyadic} applies at $2$, because the prepared
operators have odd order and no eigenvalue $1$; tensor its isometry
with itself. Lemma~\ref{local:symmetrize} supplies the real isometry,
which also tensors. Lemma~\ref{local:hasse-descent} now gives $R$.
Finally, the matrices of $R$ and $R^{-1}$ in integral
lattice bases have denominators supported at finitely many primes.
At every other prime $R$ itself is an integral lattice isometry,
so choose $D_\ell=R$ there.
\end{proof}
==== END Ambiently-homeomorphic-isolated-hypersurfaces-of-multiplicities-two-and-three//local.tex ====
==== BEGIN Ambiently-homeomorphic-isolated-hypersurfaces-of-multiplicities-two-and-three//paper.tex ====
\documentclass[11pt]{article}
\input{preamble}
\title{Ambiently homeomorphic isolated hypersurfaces\\of multiplicities two and three}
\author{OpenAI}
\date{September 24, 2026}
\begin{document}
\maketitle
\begin{abstract}
We give a negative answer to the embedded Zariski multiplicity conjecture.
We construct two reduced hypersurface germs that are ambiently homeomorphic
but have multiplicities two and three. Both have isolated singularities and
lie in a common complex affine space of dimension divisible by eight. Their
defining function germs are also topologically right equivalent, and the same
examples answer Arnold's corank problem negatively for ambient topological equivalence.
\end{abstract}
\input{sections/introduction}
\input{sections/background}
\input{sections/seeds}
\input{sections/local}
\input{sections/global}
\input{sections/conclusion}
\appendix
\input{sections/certificate}
\setlength{\bibsep}{6pt plus 1pt minus 1pt}
\bibliographystyle{plainnat}
\bibliography{references}
\end{document}
==== END Ambiently-homeomorphic-isolated-hypersurfaces-of-multiplicities-two-and-three//paper.tex ====
==== BEGIN Ambiently-homeomorphic-isolated-hypersurfaces-of-multiplicities-two-and-three//preamble.tex ====
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\newtheorem{remark}[theorem]{Remark}
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\hyphenation{OpenAI}
\hypersetup{pdftitle={Ambiently homeomorphic isolated hypersurfaces of multiplicities two and three},
pdfauthor={OpenAI},
pdfsubject={A construction concerning the embedded Zariski multiplicity problem}}
==== END Ambiently-homeomorphic-isolated-hypersurfaces-of-multiplicities-two-and-three//preamble.tex ====
==== BEGIN Ambiently-homeomorphic-isolated-hypersurfaces-of-multiplicities-two-and-three//proof-route.tex ====
\begin{figure}[htbp]
\centering
\begingroup
\setlength{\tabcolsep}{0pt}
\renewcommand{\arraystretch}{1.08}
\begin{tabular}{c}
\fbox{\parbox{.88\linewidth}{\centering
\textbf{Arithmetic relations and chain polynomials}\\
Equal spectra modulo $2$; ordinary orders $2$ and $3$.}}\\[3pt]
$\Downarrow$\\[3pt]
\fbox{\parbox{.88\linewidth}{\centering
\textbf{Two separate seed identifications}\\
Real Seifert forms \quad and \quad integral monodromy modules.}}\\[3pt]
$\Downarrow\quad$ common fifth and seventh powers;\\
\textit{symmetrize and take one tensor square}\\[3pt]
\fbox{\parbox{.88\linewidth}{\centering
\textbf{Equivariant isometries over $\Q$ and every $\Z_\ell$}\\
A second tensor square cancels their eigenspace determinant defects.}}\\[3pt]
$\Downarrow\quad$ common cubic, eleventh, and thirteenth powers\\[3pt]
\fbox{\parbox{.88\linewidth}{\centering
\textbf{One integral Seifert-form isometry}\\
Indefinite special unitary groups and strong approximation.}}\\[3pt]
$\Downarrow\quad$ simple spherical knots and weighted radial coordinates\\[3pt]
\fbox{\parbox{.88\linewidth}{\centering
\textbf{Ambiently homeomorphic hypersurface germs}\\
Disjoint sums preserve the two different ordinary orders.}}
\end{tabular}
\endgroup
\caption{The proof keeps analytic order separate from the successive
identifications of topological data. The two seed identifications are
independent; the local and global arithmetic makes them compatible.}
\label{fig:proof-route}
\end{figure}
==== END Ambiently-homeomorphic-isolated-hypersurfaces-of-multiplicities-two-and-three//proof-route.tex ====
==== BEGIN Ambiently-homeomorphic-isolated-hypersurfaces-of-multiplicities-two-and-three//seeds.tex ====
\section{Two seeds with different orders}\label{seed:section}
We first construct two sums of chains with the same spectrum modulo~$2$,
but with ordinary orders $2$ and $3$. The arithmetic ingredient is a
finite relation between positive rational numbers.
\begin{lemma}[Finite product relation]\label{seed:product}
Set $d=3^{11}$ and
\[
\mathcal B=\{b\in\Z:0<b<d/2,\ 3\nmid b\},\qquad
A_b=\frac{d-b}{b},\qquad
R_b=\frac{(2d+b)(2d-b)}{(3d-b)(d+b)}.
\]
There are integers $e_b$, $b\in\mathcal B$, such that
\begin{equation}\label{seed:products}
E:=\sum_{b\in\mathcal B}e_b>0,
\qquad \prod_{b\in\mathcal B}A_b^{e_b}
=\prod_{b\in\mathcal B}R_b^{e_b}=1.
\end{equation}
\end{lemma}
\begin{proof}
Form the integer matrix with column $b$ consisting of
$(v_p(A_b))_p$ and $(v_p(R_b))_p$, with separate rows for the two
families of prime valuations. It is enough to find an integer kernel
vector whose coordinate sum is nonzero.
Here is the finite certificate for this assertion. Retain only the
columns for which each of
\[
d-b,\ b,\ 2d+b,\ 2d-b,\ 3d-b,\ d+b
\]
has no prime factor greater than $4450$. Repeatedly delete any column
that is the only column supported on some nonzero row; discard zero
rows. The resulting matrix $M$ and its augmentation $\widetilde M$
by a final row of ones satisfy
\begin{center}
\begin{tabular}{lrr}
& columns & nonzero valuation rows\\\hline
After the prime cutoff & $1118$ & $1100$\\
After all deletions & $896$ & $867$
\end{tabular}
\end{center}
and
\begin{equation}\label{seed:rank-certificate}
\operatorname{rank}_{\F_{1009}}\widetilde M=868.
\end{equation}
Appendix~\ref{app:certificate} gives a complete program, using only
integer arithmetic and elimination modulo the prime $1009$, that
verifies these counts and~\eqref{seed:rank-certificate}. Its sieve
stores a prime divisor of every integer greater than one. Its six
signed factor entries are precisely the two valuation columns above.
In the elimination, a normalized pivot is stored for each leading row,
so the number of stored pivots is the rank. Thus the certificate is a
finite check of the displayed integer matrix, without numerical
approximation.
Since $M$ has $867$ rows,~\eqref{seed:rank-certificate} gives an
$868$-column square submatrix $H$ of $\widetilde M$ whose integer
determinant is nonzero. The vector
\[
z=\operatorname{adj}(H)(0,\ldots,0,1)^{\mathsf t}
\]
is integral, satisfies the valuation equations on these columns, and
has coordinate sum $\det H$. Extend $z$ by zeros to all columns
$b\in\mathcal B$, and change its overall sign if necessary. This
gives $e_b$ with $E=|\det H|>0$. Every omitted valuation row is zero on
the retained columns, so all prime valuations of both products
in~\eqref{seed:products} vanish. A positive rational number with this
property is $1$, proving the Lemma.
\end{proof}
The adjugate prescription also makes the construction entirely finite:
one may select the lexicographically first nonsingular minor, with
columns ordered by $b$ and valuation rows ordered by family and prime.
No bound on the size of its coefficients is needed.
\begin{lemma}[Completing one weight to a chain]\label{seed:chain}
Let $D$ be a positive odd integer and let $c\in\Z$ satisfy
$0<c<D/2$. There is a chain
polynomial
\begin{equation}\label{seed:chain-polynomial}
F_c=x_1^{a_1}x_2+\cdots+x_{\ell-1}^{a_{\ell-1}}x_\ell
+x_\ell^{a_\ell}
\end{equation}
of weighted degree $1$, whose first weight is $c/D$, all of whose
weights are positive and less than $1/2$, and all of whose ordinary
monomial degrees are at least $3$. It has an isolated critical point
at the origin. Removing its first variable and first monomial gives
a chain with exactly the remaining weights, when $\ell>1$.
\end{lemma}
\begin{proof}
Set $c_1=c$ and successively perform the divisions
\[
a_i=\lfloor D/c_i\rfloor,\qquad c_{i+1}=D-a_ic_i,
\]
stopping at the first zero remainder. Each nonzero remainder is a
positive integer strictly smaller than its predecessor, so this
procedure terminates.
Give $x_i$ weight $c_i/D$. Then $a_ic_i+c_{i+1}=D$ for each
nonterminal term, and $a_\ell c_\ell=D$ for the terminal term.
All $a_i\ge2$. Consequently each nonterminal monomial has ordinary
degree $a_i+1\ge3$. The terminal exponent divides the odd integer
$D$ and exceeds $2$, so it too is at least $3$.
To prove isolation, suppose a nonzero point is critical and let $j$
be the first index with $x_j\ne0$. If $j=\ell$, the derivative in
$x_j$ is $a_\ell x_\ell^{a_\ell-1}\ne0$. Otherwise that derivative
forces $x_{j+1}=0$, because the preceding coordinate is zero. The
derivative in $x_{j+1}$ is then $x_j^{a_j}\ne0$: its other term
vanishes since $a_{j+1}\ge2$. Both cases contradict criticality.
The same reasoning applies to the tail, which has the stated weights.
\end{proof}
\begin{proposition}[The seeds]\label{seed:seeds}
There are real-coefficient weighted homogeneous polynomials
$p_1,p_2$, each in an even number $s_0$ of variables, such that:
\begin{enumerate}
\item both are sums, in disjoint variables, of chains with exponents
at least $2$, and both have isolated critical points;
\item $\ord_0p_1=2$ and $\ord_0p_2=3$;
\item their spectral multisets are equal modulo $2$;
\item their real Seifert forms are isomorphic, and, separately, their
integral monodromy modules are isomorphic;
\item both monodromy operators have order dividing $D=3^{12}$.
\end{enumerate}
\end{proposition}
\begin{proof}
Fix the integers in Lemma~\ref{seed:product}, and put $D=3d$.
The four weights below are chosen for the triple-angle identity in the
spectral comparison; their signed counts will give side $2$ an excess
of $4E$ variables, to be balanced by squares on side $1$.
Start with two lists of weights prescribed by the following signed
counts; a positive entry means that many variables on side $2$, and a
negative entry means its absolute value on side $1$:
\begin{equation}\label{seed:weight-table}
\begin{array}{c|rrrr}
w& b/d&b/(3d)&(d-b)/(3d)&(d+b)/(3d)\\\hline
\text{count on side $2$ minus count on side $1$}
&2e_b&-2e_b&2e_b&2e_b.
\end{array}
\end{equation}
All these weights are positive and less than $1/2$. Their numerators
with denominator $D$ are $3b,b,d-b,d+b$, respectively.
Realize each occurrence of a starting weight $c/D$ by its own copy of
the chain in Lemma~\ref{seed:chain}. On the opposite side insert an
independent copy of the tail of that chain, doing nothing there if the
tail is empty. The added tail weights occur equally on both sides.
All variables used for different copies are disjoint. Thus this
operation realizes the signed weight counts exactly by sums of
isolated chains, and all its monomials have degree at least $3$.
The variable excess on side $2$ is $4E$. Add $4E$ independent squares
to side $1$, giving both sides the same number of variables. Every
starting multiplicity in~\eqref{seed:weight-table} is even, as is the
number of copies of each tail, and $4E$ is even. Both variable counts
are therefore even already. Finally add two independent pure cubes
to each side, and call the resulting polynomials $p_1,p_2$. They have
the same even number $s_0$ of variables. Sums in disjoint variables
have critical loci equal to products of the individual critical loci,
so both critical points are isolated. All coefficients can be taken
to be $1$. Squares occur only in $p_1$, and the common cubes occur in
both polynomials, proving the asserted ordinary orders.
We next compare the spectra. In the convention of
Theorem~\ref{bg:spectrum}, write
\begin{equation}\label{seed:spectrum-product}
P_p(t)=\sum_{\alpha\in\operatorname{Sp}^{+}(p)}t^\alpha
=\prod_{w\text{ a weight of }p}\frac{t^w-t}{1-t^w}.
\end{equation}
All exponents lie in $(2D)^{-1}\Z$. Reduction modulo $2$ puts their
multiplicities in the group algebra of the finite cyclic group
$(2D)^{-1}\Z/2\Z$. Its characters are
$\alpha\mapsto e^{\pi i k\alpha}$, for integer $k$. By Fourier
inversion, it suffices to prove that $P_{p_1}$ and $P_{p_2}$ take the
same value at every such character. When an individual quotient
in~\eqref{seed:spectrum-product} is indeterminate, evaluate along
$t^u=\exp((\pi i k+\varepsilon)u)$ and let $\varepsilon\to0$.
\smallskip\noindent
\emph{Case 1: $k=2m$ is even.}
For a weight $w$, its individual factor has value
\begin{equation}\label{seed:even-factor}
\left.\frac{t^w-t}{1-t^w}\right|_k
=\begin{cases}
-1,&mw\notin\Z,\\
(1-w)/w,&mw\in\Z.
\end{cases}
\end{equation}
The second value follows by taking the ratio of the two first
derivatives in $\varepsilon$. All these values are nonzero, so common
weights can be canceled.
The first column of~\eqref{seed:weight-table} has exact reduced
denominator $d$, and the other three have exact reduced denominator
$3d$, since $3\nmid b$. Within either denominator group, therefore,
all factors are exceptional in~\eqref{seed:even-factor}, or none are.
The exceptional signed product for the first group is
$\prod_b A_b^{2e_b}=1$. For the other group it is
\[
\prod_b\left(
\frac{b}{3d-b}\frac{2d+b}{d-b}\frac{2d-b}{d+b}
\right)^{2e_b}
=\prod_b(R_b/A_b)^{2e_b}=1.
\]
The nonexceptional signs disappear because all signed counts are
even. A square has exceptional value $1$ and nonexceptional value
$-1$; its total contribution is $1$, since there are $4E$ squares.
This proves equality in Case~1, including $k=0$.
\smallskip\noindent
\emph{Case 2: $k$ is odd and $3\nmid k$.}
Each factor is
\[
\left.\frac{t^w-t}{1-t^w}\right|_k
=i\cot(\pi k w/2).
\]
All chain weights have a reduced denominator that is a nontrivial
power of $3$. None of their cotangents has a zero or pole in this
case: either would imply that this denominator divides $k$.
Squares also have finite nonzero factors. Hence common weights may
again be canceled.
Put $C(u)=\cot(\pi ku/2)$. The exact signed identity we need is
\begin{equation}\label{seed:cotangent}
C(b/d)=
\frac{C(b/(3d))}{C((d-b)/(3d))C((d+b)/(3d))}.
\end{equation}
Indeed, with $x=\pi kb/(6d)$ the usual triple-angle product gives
\[
\cot x\,\cot(x+k\pi/3)\,\cot(x+2k\pi/3)=-\cot(3x).
\]
The shifts by $k\pi/3$ give the same three factors modulo $\pi$ as
the shifts by $\pi/3$, because $3\nmid k$. Since $k$ is odd,
\[
\cot(x+k\pi/3)=\frac{1}{C((d-b)/(3d))},\qquad
\cot(x+2k\pi/3)=-\frac{1}{C((d+b)/(3d))}.
\]
Substitution proves~\eqref{seed:cotangent}. Its $2e_b$-th power is
exactly the assertion that the signed cotangent product in row $b$
of~\eqref{seed:weight-table} equals $1$.
Each square contributes $C(1/2)=\cot(\pi k/4)\in\{1,-1\}$, whose
$4E$-th power is $1$. Finally the powers of $i$ are the same because
the two complete polynomials have the same number $s_0$ of variables.
This proves equality in Case~2.
\smallskip\noindent
\emph{Case 3: $k$ is odd and $3\mid k$.}
A pure cube has spectrum $t^{1/3}+t^{2/3}$, whose value is $-1+1=0$.
Before the final common cubes were inserted, both sides were already
isolated sums of chains and squares, so their spectra were finite
polynomials. Multiplication by a cube spectrum therefore makes both
character values zero. This argument uses those finite polynomials,
and does not cancel a zero against a pole of an individual weight
quotient. It completes the Fourier comparison.
Weighted homogeneous monodromy has finite order and hence is
semisimple. Theorem~\ref{bg:real} now converts the equality of
spectra modulo $2$ in the common dimension into an isomorphism of
real Seifert forms. The same equality gives equal eigenvalue
multiplicities. All summands are chains, including the pure powers,
so Theorem~\ref{bg:orlik} gives an isomorphism of integral monodromy
modules. These are separate isomorphisms; no compatibility between
them is asserted here.
Finally, every nonsquare weight has denominator dividing $D$.
Each square has spectrum $t^{1/2}$, so the $4E$ squares together
contribute $t^{2E}$. All spectral exponents of either seed therefore
belong to $D^{-1}\Z$. Their monodromy eigenvalues are $D$-th roots of
unity. Semisimplicity then gives $M_{p_j}^D=1$ for $j=1,2$, as claimed.
\end{proof}
\begin{corollary}[Removing eigenvalue one]\label{seed:prepared}
In two new variables on each side, put
\begin{equation}\label{seed:q-definition}
q_j=p_j+x^5+y^7,\qquad j=1,2,
\qquad s=s_0+2.
\end{equation}
These are real-coefficient isolated sums of chains in the same even
number $s$ of variables, with orders $2$ and $3$. Their real Seifert
forms are isomorphic, and their integral monodromy modules are
isomorphic. Both monodromies have odd order dividing
\[
m=3^{12}\cdot5\cdot7,
\]
and neither has eigenvalue $1$.
\end{corollary}
\begin{proof}
The new terms preserve isolation and ordinary orders. Formula
\eqref{seed:spectrum-product} multiplies each seed spectrum by the
same two pure-power spectra, preserving equality modulo $2$.
Theorems~\ref{bg:real} and~\ref{bg:orlik} consequently apply again.
Each new eigenvalue has the form
\[
\alpha\beta\gamma,\qquad
\alpha^D=1,\quad
\beta^5=\gamma^7=1,\quad \beta\ne1,\quad\gamma\ne1.
\]
Because $D$, $5$, and $7$ are pairwise coprime, its order divides
$35D$ and is divisible by both $5$ and $7$. In particular it is not
$1$. Finite-order weighted homogeneous monodromy is semisimple, so
the same bound holds for the order of the operators themselves.
\end{proof}
==== END Ambiently-homeomorphic-isolated-hypersurfaces-of-multiplicities-two-and-three//seeds.tex ====
==== BEGIN Ambiently-homeomorphic-isolated-hypersurfaces-of-multiplicities-two-and-three//references.bib ====
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}
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@misc{FernandesJelonekSampaio2025,
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@misc{Sampaio2025Milnor,
author = {Jos\'e Edson Sampaio},
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@misc{OpenAIMuConstant2026,
author = {{OpenAI}},
title = {{Topological triviality of $\mu$-constant families of surface singularities}},
howpublished = {OpenAI Math Release preprint
\href{https://github.com/openai/math/blob/main/preprints/Topological-triviality-of-mu-constant-families-of-surface-singularities-September-24-2026/main.pdf}{OAI:Topological-triviality-of-mu-constant-families-of-surface-singularities-September-24-2026}},
year = {2026}
}
==== END Ambiently-homeomorphic-isolated-hypersurfaces-of-multiplicities-two-and-three//references.bib ====