Steenbrink/Mult3.tex
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\begin{document}
\title{On germs of hypersurfaces with multiplicity 3}
\author{EAB, MGV, ELC}
\maketitle
Consider a the germ of a multiplicity 3 singular point of hypersurface. An equation for such a germ it is given by a monic Weierstrass polynomial $F = z^3 + a_1z^2 + a_2z + a_3 \in \CC[[{\bf x}]][z]$ of degree 3 with respect to $z$. Assume that $D_z F$ are not zero, and that $F$ is degenerate with respect to its Newton polyhedron. If it is non degenerate, the partial resolution associated with the dual of the Newton polyhedron of $F$ gives plenty information of the geometry and topology of the singular point of the hypersurface defined by $F$.
We propose the following general method.
\medskip
\noindent \emph{First step:} By means of a Tschirnhaus transformation, reduce $F$ to a monic Weierstrass equation $F = z^3 -3 b_2z + 2 b_3$ that does not have terms of degree 2 with respect to $z$. Assume that the coefficients of degree one and zero with respect to $z$ are both non zero, i.e., the product $b_2b_3 \ne 0$. If either $b_2 =0$ and $b_3 \ne 0$ or $b_2 \ne 0$ and $b_3=0$ it reduces to the ase of a suspension that we have studied elsewhere. See \cite{}.
%We propose the following general method to study germs of multiplicity 3 singular points of hypersurfaces. Firstly, consider an equation of the germ given by a monic Weierstrass polynomial $F = z^3 + a_1z^2 + a_2z + a_3 \in \CC[[{\bf x}]][z]$ of degree 3 with respect to $z$. Assume that $D_z F$ is not zero. We will be mainly interested in the case when $F$ is degenerate with respect to its Newton polyhedron. Otherwise, the partial resolution associated with the dual of the Newton polyhedron of $F$ gives plenty information of the geometry and topology of the singular point of the hypersurface defined by $F$. By means of a Tschirnhaus transformation, reduce $F$ to a monic Weierstrass equation $z^3 + b_2z + b_3$ that does not have terms of degree 2 with respect to $z$. Assume that the coefficients of degree one and zero with respect to $z$ are both non zero, i.e., the product $b_2b_3 \ne 0$.
\medskip
\noindent \emph{Second step:} Consider an embedded resolution $\pi: X \rightarrow \CC^2$ of the reduced divisor given by the product $G=b_2b_3D_zF$. Consider the usual stratification of the pair $(X, (\pi \circ G)^{-1} \{G=0\})$ defined in terms of decomposition $\cup_{j \in J} E_j$ of $(\pi \circ G)^{-1} \{G=0\}$ into irreducible components. Pulling back $F$ under the map $\pi \times id$, we obtain for each stratum $E_I^\circ$ a \emph{quasi-trinomial local expression} for $F$. Section \ref{3nomial} clasifies the 5 posible quasi-trinomial local forms having terms of degrees 3, 1, and 0 with respect to $z$. These are the quasi-trinomials forms that appear at this step because we have previously made a Tschirnhaus transformation. Notice that the local expression is quasi-trinomial due to the fact that $\pi$ resolves both $b_2$ and $b_3$. Moreover, the quasi-trinomial local form is quasi-ordinary because $\pi$ also resolves the discriminant $D_zF$. The local quasi-trinomial expression will be non degenerate with respect to its Newton polyhedron unless the monomials $M_1$, $M_2$ and their coefficients $c_1$ and $c_2$ satisfy a precise condition. In principle, this precise condition can be defined on a subset of a stratum of $E_I^\circ$, with $\emptyset \ne I \subset J$. If it is the case, we consider the corresponding refinement of the stratification. Moreover, the degeneracies of the degenerate trinomial local case are either on horizontal faces or of the form $(z-af(x,y))(z-bf(x,y))$ with $a, b \in \CC^\ast$, and $f \in \CC[x,y]$.
\medskip
\noindent \emph{Third step:} Make translations of the type $z \mapsto z + h(x,y)$, where $h(x,y)$ is determined by the local quasi-ordinary expansion, to obtain a new set of coordinates such that the local form is non degenerate. Notice that after the translation terms of degree 2 respect to $z$ appear. In fact, we obtain a new type of quasi-trinomial local form with terms of degree $3$, $2$, and $0$ with respect to $z$.
\medskip
\noindent \emph{Fourth step:} Use the stratification principle with the stratification introduced in the Second Step, the quasi-trinomial local forms and, if needed, the corresponding local differential forms to compute \emph{additive / multiplicative local invariants} of the singular point defined by $f$ at $o$. For instance, we will interested in the A'Campo zeta function and in the topological zeta function.
\medskip
A couple of remarks are in order after these four steps.
First, these steps achieve a resolution of the meromorphic function defined by $\frac{b_2^3}{b_3^2}$ and therefore the irreducible components $E_j$ of $(\pi \circ F)^{-1} \{F=0\}$ can be classified into 5 different types according to the image of $e_j$ under the meromorphic function $\mathcal{M}$ defined by $\frac{b_2^3}{b_3^2}$. There are components $E_j$ that are mapped to $0$ called zeros, components $E_j$ that are mapped to $\infty$ called poles, components $E_j$ that are mapped surjectively onto $\mathbb{P}^1$ called dicritical, components that are mapped to the value 1, and finally components that are mapped to other values different than 0, 1, and $\infty$. The image of a component $E_j$ is determined by the orders of vanishing of $f$, $g$ and $\mathcal{M}= \frac{b_2^3}{b_3^2}$ along the divisor $E_j$ and (in some cases by) their pattern of intersection.
Second, the trinomial part of any quasi-trinomial local form, given by the formal restriction of the quasi-trinomial form to its Newton polyhedron, is determined by the valuations associated with $f$, $g$ and $\mathcal{M}= \frac{b_2^3}{b_3^2}$ too.
The following reductions are possible to perform the Fourth Step.
\medskip
\noindent \emph{Reduction A:} It is enough to consider the associated trinomial local forms instead of the quasi-trinomial local forms.
\medskip
\noindent \emph{Reduction B:} The strata produced by the strict transforms of $b_2$ and $b_3$ can be omitted, i.e., the contribution to the invariant due to a 0-dimensional stratum determined by an irreducible component of the strict transform of $b_2$ (resp. $b_3$) equals the contribution of a single point of the 1-dimensional stratum to whose adherence the 0-dimensional stratum belongs.
\medskip
\noindent \emph{Reduction C:} Bamboos consisting of irreducible components that are mapped to $0$ (resp. to $\infty$ and containing a $(-1)$-curve can be omitted too.
\medskip
The upshot of Reductions A, B, and C is that it is enough to consider the stratification associated to the minimal resolution $\pi' : X' \rightarrow \CC^2$ of $D_zF$ together with the trinomial local forms, and if needed the corresponding differential forms to compute the invariants of the singular point defined by $f$ at $0$.
\medskip
\noindent \emph{Reduction D:} It is enough to consider a minimal $\QQ$-resolution of the discriminant $D_ZF$.
\medskip
After all these reductions, we will prove that ...
\section{Power series of multiplicity three}
Consider a the germ of a multiplicity 3 singular point of hypersurface. An equation for such a germ it is given by a monic Weierstrass polynomial $F = z^3 + a_1z^2 + a_2z + a_3 \in \CC[[{\bf x}]][z]$ of degree 3 with respect to $z$. Assume that $D_z F$ is not zero, and that $F$ is degenerate with respect to its Newton polyhedron.
\begin{definition} The Tschirnhaus transformation associated to $F$ is the transformation given by $z \mapsto z- \frac{a_1}{3}$.
\end{definition}
After the Tschirnhaus transformation, the monic Weierstrass polynomial $F(x,y,z - a_1/3)$ can be written as
$$ z^3 -3b_2z + 2b_3$$
with
$$b_2 = \frac{a_1^2}{9} - \frac{ a_2}{3} \quad {\rm and } \quad b_3= \frac{a_1^3}{27}-\frac{a_1a_2}{6} + \frac{a_3}{2}$$
and its the discriminant $D_zF$ of $F$ with respect to $F$ is of the form $108(b_2^3 - b_3^2)$.
Denote the decompositions of $b_2$ and $b_3$ into irreducible components as
\begin{equation}
b_2 = \prod_{i=1}^n h_i^{n_i} \prod_{j=1}^m k_j f \qquad {\rm and} \qquad b_3 = \prod_{i=1}^n h_i \prod_{j=1}^m k_j^{m_j}g
\end{equation}
where $h_i, k_j$ are irreducible and $h_i, k_j, f, g$ are pairwise coprime and $n_i, m_j \geq 1$.
\begin{proposition}
The discriminant $D_zF$ is of the form
$$ \prod_{i=1}^n h_i^{2} \prod_{j=1}^m k_j^3 \left ( \prod_{i=1}^n h_i^{3n_i-2} f^3 - \prod_{j=1}^m h_i^{2m_j -3} g^2 \right ).$$
\end{proposition}
\begin{proposition}
The singular point is isolated if and only if the disciminant $D_zF$ is reduced.
\end{proposition}
\begin{proposition}\label{prop.non-deg}
The monic Weierstrass polynomial $z^3 -3b_2z + 2b_3$ is non degenerate unless $M_1(c_1+A)$ and $b_3 = M_2(c_2 + B)$ with $M_1^3 = M_2^2$ and \textcolor{red}{$2c_1=3c_2$}
\end{proposition}
\section{Jung´s method for power series of multiplicity three }
Consider an embedded resolution $\pi: X \rightarrow \CC^2$ of the reduced divisor given by the product $G=b_2b_3D_zF$. Consider the usual stratification of the pair $(X, (\pi \circ G)^{-1} \{G=0\})$ defined in terms of decomposition $\cup_{j \in J} E_j$ of $(\pi \circ G)^{-1} \{G=0\}$ into irreducible components, i.e., for each subset $I \subset J$ we define
$$E_I^\circ = \bigcap_{i \in I} E_j \setminus \bigcup_{j \not \in I} E_j .$$
\begin{definition}For each irreducible component $E_j$ of of $(\pi \circ G)^{-1} \{G=0\}$ we define the numbers
$$N^0_j = {\rm ord}_{E_j} \pi^\ast b_2, \quad N^\infty_j = {\rm ord}_{E_j} \pi^\ast b_3, \quad N^\Delta_j = {\rm ord}_{E_j} \pi^\ast D_zF, \quad {\rm and} \quad \nu_j = {\rm ord}_{E_j} \pi^\ast \omega -1$$
where $\omega = dx \wedge dy$ is the standard volume form of $\CC^2$.
\end{definition}
\begin{proposition} Let $\pi': X' \rightarrow \CC^2$ be the minimal resolution of $D_zF$. Then, $\pi'$ is a resolution of the meromorphic function $\mathcal{M} : \CC^2 \rightarrow \mathbb{P}^1$ defined by $\frac{b_2^3}{b_3^2}$ too.
\end{proposition}
There are four possibilities for the behaviour of the numbers $N^0_j, N^\infty_j$ and $N^\Delta_j$ associated with an irreducible component $E_j$ of $(\pi \circ G)^{-1} \{G=0\}$
\begin{itemize}
\item $3 N^0_j > 2N^\infty_j = N^\Delta_j$ and it is called a \emph{zero} because in this case $\mathcal{M}(E_j)=0$;
\item $2 N^\infty_j > 3N^0_j = N^\Delta_j$ and it is called a \emph{pole} because in this case $\mathcal{M}(E_j)=\infty$;
\item $3 N^0_j = 2N^\infty_j = N^\Delta_j$ and in this case either it is called \emph{dicritical} because $\mathcal{M}(E_i)=\mathbb{P}^1$ or it is mapped constantly to a value different than $0$, $1$ or $\infty$ under $\mathcal{M}$;
\item $3 N^0_j = 2N^\infty_j < N^\Delta_j$ and in this case it is mapped constantly to $1$ under $\mathcal{M}$.
\end{itemize}
In order to distinguish between dicritical divisors and divisors that are mapped constantly to a value different than $0$, $1$ or $\infty$ one notices that connected components of dicritical divisors must intersect both to zeros and poles and connected components of divisors mapped constantly to a value different than $0$, $1$ or $\infty$ do not intersect zeros or poles but do intersect dicritical divisors.
\begin{definition} For a divisor of the last type, the difference
$N^\Delta_j - 3 N^0_j = N^\Delta_j - 2 N^\infty_j$ is called the \emph{excess of the discriminant} and it is denoted by $M^\Delta_j$.
\end{definition}
\begin{example} Consider $F=z^3 - 3b_2z + 2b_3$ with
$b_2= ((y^2 + x^3)^2 + x^5y)((x^2 + y^3)^2 + y^5x)$
and
$b_3= (y^3+x^4)(x^3+y^4)$. The minimal resolution $\pi$ of the product $G=b_2 b_3 D_zF$ is obtained after 15 blow-ups. The following figure and table give the dual resolution graph of $G$ and its associated numerical data.
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\node[below] at (A7) {{\small $E_8$}};
\node[vertice] at (A8) {};
\node[above] at (A8) {{\small $E_2$}};
\node[vertice] at (B2) {};
\node[below] at (B2) {{\small $E_{14}$}};
\node[vertice] at (B3) {};
\node[below] at (B3) {{\small $E_{13}$}};
\node[vertice] at (B4) {};
\node[below] at (B4) {{\small $E_{12}$}};
\node[vertice] at (B5) {};
\node[below] at (B5) {{\small $E_{11}$}};
\node[vertice] at (B6) {};
\node[below] at (B6) {{\small $E_{17}$}};
\node[vertice] at (B7) {};
\node[below] at (B7) {{\small $E_{16}$}};
\node[vertice] at (B8) {};
\node[above] at (B8) {{\small $E_{10}$}};
\draw (U)--(A1)--(A2)--(A3)--(A4)--(A5)--(A6)--(A7);
\draw (A5)--(A8);
\draw (B5)--(B8);
\draw (U)--(B1)--(B2)--(B4)--(B4)--(B5)--(B6)--(B7);
\draw[->] (A1)--(F1);
\draw[->] (A1)--(F2);
\draw[->] (A6)--(F3);
\draw[->] (U)--(G1);
\draw[->] (U)--(G2);
\draw[->] (U)--(G3);
\draw[->] (U)--(G4);
\draw[->] (U)--(G5);
\draw[->] (U)--(G6);
\draw[->] (U)--(G7);
\draw[->] (U)--(G8);
\draw[->] (U)--(G9);
\draw[->] (U)--(G10);
\draw[->] (U)--(G11);
\draw[->] (U)--(G12);
\draw[->] (U)--(G13);
\draw[->] (B1)--(F4);
\draw[->] (B1)--(F5);
\draw[->] (B6)--(F6);
\end{tikzpicture}
\end{center}
\end{figure}
$$\begin{array}{|c||c|c|c|c|c|c|c|c|c|c|}
\hline
&E_1&E_2, E_{10}&E_3, E_{11}&E_4, E_{12}&E_5, E_{13}&E_6, E_{14}&E_7, E_{15} &E_8, E_{16}&E_9, E_{17}\\
\hline
\hline
N^f_i& {\bf 8}& 10& 20& 28& 36& 44& {\bf 52}& 21&42\\
N^g_i& {\bf12}& {\bf 13}& {\bf 26}& {\bf 39}& {\bf 52}& {\bf 65}& {\bf 78}& {\bf 26}& {\bf 52}\\
N^\Delta_i& 25& {\bf 26}& {\bf 52}& {\bf 78}& {\bf 104}& {\bf 130}& {\bf 156}& {\bf 52}&{\bf 104}\\
\hline
\hline
\nu_i&2&3&5&7&9&11&13&6&11\\
\chi()&-13&1&-1&2&2&2&-4&1&-1\\
E_i^2& -7& -2& -2& -2& -2& -2& -1&-2&-1\\
\hline
\end{array}$$
The divisor $E_1$ is mapped to the value 1, the divisors $E_7$ and $E_{13}$ are dicriticals, and the rest of exceptionals divisors are poles. The 13 arrows $F^\Delta_i$ with $i=1, \dots, 13$ attached to the divisor $E_1$ correspond to irreducible components of the strict transforms of $D_zF$, the arrows attachef to the divisor $E_7$ (resp. $E_{13}$) correspond to an irreducible component strict transform of $b_3$ and another irreducible component of $D_zF$.
%$$\pi^{-1} \{\mathcal{M} = 0\} = 2 (F^{0}_1 + F^{0}_2) + 4 (E_2 + E_5 + E_{10} + E_{13}) + 6 (E_4 + E_{12} ) + 8 (E_3 + E_{11}) + 11 (E_8+E_{16}) + 22 (E_9 + E_{17})) , \quad \pi^{-1} \{\mathcal{M} = 1\} = \sum_{i=1}^{15} F^\Delta_i + E_1$$
$$\pi^{-1} \{\mathcal{M} = 0\} = 2 ( \sum_{1,2}F^{0}_i + \sum_{6,14} E_i ) + 4 \sum_{2,5,10,13}E_i + 6 \sum_{4,12 }E_i + 8 \sum_{3,11}E_i + 11\sum_{8,16} E_i + 22 \sum_{9,17} E_i ,$$
$$\pi^{-1} \{\mathcal{M} = \infty\} = 3 F^{\infty}_1 + 3 F^{\infty}_2, \quad \pi(E_7) = \pi(E_{15}) = \mathbb{P}^1, \quad \pi^{-1} \{\mathcal{M} = 1\} = \sum_{i=1}^{15} F^\Delta_i + E_1$$
The stratification associated with the resolution $\pi$ consist of $35$ strata $E^\circ_{i,j}$ of dimension 0 and 8 strata $E_i^\circ$ of dimension 1 and $\chi(E_i^\circ) \ne 0$.
The former correspond to the 16 pairs $(i,j)$ listed belllow
$$\begin{array}{cccccccc}
(1,7)& (2,3)& (3,4)& (3,9)& (4,5)& (5,6)& (6,7)& (8, 9) \\
(1,15)& (10,11)& (11,12)& (11,17)& (12, 13)& (13,14)& (14,15)& (16,17)
\end{array}$$
the 13 intersections $E_1 \cap F_i^\Delta$ for $i=1, \dots, 13$ and the 6 intersections
$$ E_7 \cap F_1^\infty, \quad E_7 \cap F_{14}^\Delta, \quad E_9 \cap F_1^0, \quad E_{15} \cap F_2^\infty, \quad E_{15} \cap F_{14}^\Delta, \quad E_{17} \cap F_2^0 $$
and the latter correspond with the indexes
$$1, \quad 2, \quad 3, \quad 7, \quad 8, \quad 10, \quad 11, \quad 15.$$
Notice that the minimal resolution $\pi'$ of the discriminant $D_zF$ is obtained after just 11 blow-ups. The blowups corresponding to divisors $E_8, E_9, E_{14}$ and $E_{15}$ are needed to resolve $b_2$.
\end{example}
Let us consider the birational map $\tilde{\pi} := ( \pi, id ): \tilde{X} : =X \times \CC \rightarrow \CC^{3}$, where $id: \CC \rightarrow \CC$ is the identity map, and the pullback of $F$ under $\pi \times id$. The pullback $\pi^\ast F$ can be expressed in terms of the pullbacks of $b_2$ and $b_3$ as
\begin{equation}
z^3 -3 \pi^\ast b_2 z + 2 \pi^\ast b_3.
\end{equation}
Over each non empty strata $E^\circ_I$ of $X$ created by the resolution $\pi$,
for the pullback of $F$ under the birational map $\pi \times id$.
Since $\pi$ is a resolution of both $b_2$ and $b_3$, over each non empty strata $E^\circ_I$ of $X$ created by the resolution $\pi$, the terms $(\pi^\ast b_3)_{|_I}$ and $(\pi^\ast b_3)_{|_I}$ can be reexpressed in terms of the multiplicities $N^f_i$ and $N^g_i$ with $i \in I$ as
\begin{equation}
(\pi^\ast b_2)_{|_I}= x_1^{N^0_{i_1}} y_1^{N^0_{i_2}} u_1(x_1, y_1) \quad {\rm and } \quad
(\pi^\ast b_3)_{|_I}= x_1^{N^\infty_{i_1}} y_1^{N^\infty_{i_2}} u_1(x_1, y_1)
\end{equation}
where $u_1, u_2$ are units in $\CC\{x_1, y_1\}$. Hence, we have expressions
\begin{equation}\label{quasi-trinomial}
z^3 -3 x_1^{N^0_{i_1}} y_1^{N^0_{i_2}} u_1(x_1, y_1) z + 2 x_1^{N^\infty_{i_1}} y_1^{N^\infty_{i_2}} u_2(x_1,y_1),
\end{equation}
for each non empty strata $E^\circ_I$. Notice that these expressions are quasi-ordinary because $\pi$ is also a resolution of $D_zF$ and therefore $D_z(\pi^\ast F)_{|_I}$ is the product of a monomial in the variables $x_1$ and $y_1$ and a unit in $\CC\{x_1, y_1\}$.
According to Section \ref{3nomial} the expression (\ref{quasi-trinomial}) is non degenerate with respect to its Newton polyhedron except for the case described in Proposition \ref{prop.non-deg}, i.e., the case when the there is a divisor $E_i$ with $i \in I$ such that $\mathcal{M}(E_i)=1$ or equivalently $3N^0_i=2N^\infty_i < N^\Delta_i$.
\begin{remark}
If the strata $E^\circ_{i,j} \ne \emptyset$ and $\mathcal{M}(E_i)=1$ then either $\mathcal{M}(E_j)=1$ or $E_j$ is dicritical, i.e., $\mathcal{M}(E_j)=\mathbb{P}^1$.
\end{remark}
\begin{proposition} There exist a change of coordinates of the form
$$x_1= x_2, \quad y_1= y_2, \quad z = z_1 - x_2^{\frac{N^0_i}{2}}y_2^{\frac{N^0_j}{2}} + \cdots$$
such that (\ref{quasi-trinomial}) becomes
either
$$z_1^3 + x_2^{\frac{N^0_i}{2}}y_2^{\frac{N^0_j}{2}} u'_1(x_2,y_2) z_1^2 + x_2^{N_i^\infty + M_i^\Delta}y_2^{N_i^\infty + M_i^\Delta} u'_2(x_2,y_2)$$
if both $\mathcal{M}(E_i) = \mathcal{M}(E_j)=1$
or
$$z_1^3 + x_2^{\frac{N^0_i}{2}}y_2^{\frac{N^0_j}{2}} u'_1(x_2,y_2) z_1^2 + x_2^{N_i^\infty + M_i^\Delta}y_2^{N_j^\infty } u'_2(x_2,y_2)$$
if $\mathcal{M}(E_i)=1$ and $E_j$ is dicritical
\end{proposition}
\begin{definition}The trinomial form $f_I$ associated with the stratum $E^\circ_I$ is
$$\begin{cases}
z^3 - x^{N^\infty_{i}}, & \mathcal{M}(E_i) \ne 1, \infty \\
z^3 - x^{N^0_{i}} z + x^{N^\infty_{i}}, & \mathcal{M}(E_i) = \infty \\
z^3 - x_2^{\frac{N^0_i}{2}} + x^{N_i^\infty + M_i^\Delta} , & \mathcal{M}(E_i) = 1, \\
\end{cases}$$
in the case $I= \{i\}$
and
$$\begin{cases}
z^3 + x^{N^\infty_i}y^{N^\infty_j}, & \mathcal{M}(E_i), \mathcal{M}(E_j) \ne 1, \infty \\
z^3 + x^{\frac{N^0_i}{2}}y^{\frac{N^0_j}{2}}z^2 + x^{N_i^\infty + M_i^\Delta}y^{N_j^\infty + M_ij^\Delta} , & \mathcal{M}(E_i) = \mathcal{M}(E_j) = 1 \\
z^3 + x^{\frac{N^0_i}{2}}y^{\frac{N^0_j}{2}} z^2 + x^{N_i^\infty + M_i^\Delta}y^{N_j^\infty} , & \mathcal{M}(E_i) = 1, \mathcal{M}(E_j) = \mathbb{P}^1 \\
z^3 + x^{N^0_{i}}y^{N^0_{j}} z + x^{N^\infty_{i}}y^{N^\infty_{j}}, & {\rm else}\\
\end{cases}$$
in the case $I= \{i, j\}$.
\end{definition}
\section{The quasi-trinomial case}\label{3nomial}
In this section we consider the case of
$$F = z^3 + M_1(c_1+A)z + M_2(c_2 + B) \in \CC[[{\bf x}]][z]$$
where $M_1={\bf x}^{m_1}$ and $M_2={\bf x}^{m_2}$ are monomials, $c_1,c_2 \in \CC^\ast$ and $A, B \in ({\bf x})\CC[[{\bf x}]]$
We distinguish the following two cases
\begin{itemize}
\item Assume that the monomials $M_1$ and $M_2$ are not related by the divisibility relation, i.e., neither $M_1 \nmid M_2$ nor $M_2 \nmid M_1$.
In this case, the Newton polyhedra of $F$ contains a unique two dimensional compact face determined by the monomials $z^3$, $M_1z$ and $M_2$. Moreover, $F$ is non degenerate with respect to its Newton polyhedra.
\
\begin{tikzpicture}
\fill [yellow] (1,0) rectangle (3,-2);
\fill [yellow] (-2,0) rectangle (1,2);
\draw[gray, thick] (-2,-2) -- (-2,2);
\draw[gray, thick] (-2,-2) -- (3,-2);
\draw[gray, thick] (-2,0) -- (3,0);
\draw[gray, thick] (1,-2) -- (1,2);
\filldraw[black] (-2,-2) circle (2pt) node[anchor=west]{};
\filldraw[black] (1,0) circle (2pt) node[anchor=south west]{$m_2$};
\filldraw[black] (2,-1) circle (2pt) node[anchor=south west]{$m_1$};
\fill [yellow] (11,0,4)--(11,0,0)--(13,0,0)--(13,0,4)--(11,0,4);
\fill [yellow] (11,0,4)--(11,0,7)--(6,0,7)--(6,0,4)--(11,0,4);
\draw[gray, thick] (6,0,0) -- (6,0,7);
\draw[gray, thick] (6,0,0) -- (6,4,0);
\draw[gray, thick] (6,0,0) -- (13,0,0);
\filldraw[black] (6,0,0) circle (2pt) node[anchor=west]{};
\draw[gray, thick] (6,0,4) -- (13,0,4);
\draw[gray, thick] (11,0,0) -- (11,0,7);
\draw[gray, thick] (6,3,0) -- (11,0,4);
\draw[gray, thick] (9,1,6) -- (11,0,4);
\draw[gray, thick] (6,3,0) -- (9,1,6);
\draw[gray, dashed] (9,0,0) -- (9,0,6);
\draw[gray, dashed] (6,0,6) -- (9,0,6);
\draw[gray, dashed] (9,1,0) -- (9,1,6);
\draw[gray, dashed] (6,1,6) -- (9,1,6);
\draw[gray, dashed] (9,0,6) -- (9,1,6);
\draw[gray, dashed] (6,1,0) -- (6,1,6);
\draw[gray, dashed] (6,1,0) -- (9,1,0);
\draw[gray, dashed] (6,0,6) -- (6,1,6);
\draw[gray, dashed] (9,0,0) -- (9,1,0);
\filldraw[black] (6,3,0) circle (2pt) node[anchor=south west]{$z^3$};
\filldraw[black] (11,0,4) circle (2pt) node[anchor=north west]{$M_2$};
\filldraw[black] (9,0,6) circle (2pt) node[anchor=north]{$M_1$};
\filldraw[black] (9,1,6) circle (2pt) node[anchor=south west]{$M_1z$};
\end{tikzpicture}
\
\item Assume that the monomials $M_1$ and $M_2$ are related by the divisibility relation, i.e., either $M_1 | M_2$ or $M_1 | M_2$. In this case the Newton polyhedron of $F$ is a polygonal path. We distinguish the following cases.
\
\begin{tikzpicture}
\fill [yellow] (6,-2) rectangle (9,0);
\fill [yellow] (9,0) rectangle (11,2);
\draw[gray, thick] (6,-2) -- (6,2);
\draw[gray, thick] (6,-2) -- (11,-2);
\draw[gray, thick] (6,0) -- (11,0);
\draw[gray, thick] (9,-2) -- (9,2);
\filldraw[black] (6,-2) circle (2pt) node[anchor=west]{};
\filldraw[black] (9,0) circle (2pt) node[anchor=north west]{$m_2$};
\filldraw[black] (9.5, 1.3) circle (2pt) node[anchor=north west]{$m_1$};
\end{tikzpicture}
\
\begin{enumerate}
\item If the point $3m_1$ does not belong to the set $2 m_2 + (\RR_{\geq 0})^2$, the Newton polyhedron of $F$ consists of two segments. The first one is determined by the monomials $z^3$ and $M_1z$ and the second is determined by the monomials $M_1z$ and $M_2$. it follows that $F$ is non degenerate with respect to its newton polyhedron.
\
\begin{tikzpicture}
\fill [yellow] (-2,-2) rectangle (1,-2/3);
\fill [yellow] (-2,-2) rectangle (0,0);
\draw[gray, thick] (-2,-2) -- (-2,2);
\draw[gray, thick] (-2,-2) -- (3,-2);
\draw[gray, thick] (-2,0) -- (3,0);
\draw[gray, thick] (1,-2) -- (1,2);
\draw[gray, dashed] (0,-2/3) -- (0,0);
\draw[gray, dashed] (0,-2/3) -- (1,-2/3);
\filldraw[black] (-2,-2) circle (2pt) node[anchor=west]{};
\filldraw[black] (1,0) circle (2pt) node[anchor=north west]{$m_2$};
\filldraw[black] (0, -2/3) circle (2pt) node[anchor=north west]{$\frac{2}{3}m_2$};
\filldraw[black] (-0.5, -0.2) circle (2pt) node[anchor=north east]{$m_1$};
\fill [yellow] (28/3,0,4)--(28/3,0,8/3)--(11,0,8/3)--(11,0,0)--(6,0,0)--(6,0,4)--(28/3,0,4);
\draw[gray, thick] (6,0,0) -- (6,0,7);
\draw[gray, thick] (6,0,0) -- (6,4,0);
\draw[gray, thick] (6,0,0) -- (13,0,0);
\filldraw[black] (6,0,0) circle (2pt) node[anchor=west]{};
\draw[gray, thick] (6,0,4) -- (13,0,4);
\draw[gray, thick] (11,0,0) -- (11,0,7);
\draw[gray, thick] (8,1,3) -- (11,0,4);
\draw[gray, thick] (6,3,0) -- (8,1,3);
\draw[gray, dashed] (8,1,3) -- (8,0,3);
\draw[gray, dashed] (8,0,0) -- (8,0,3);
\draw[gray, dashed] (8,1,3) -- (8,1,0);
\draw[gray, dashed] (8,1,0) -- (8,0,0);
\draw[gray, dashed] (6,1,0) -- (8,1,0);
\draw[gray, dashed] (6,1,0) -- (6,1,3);
\draw[gray, dashed] (6,1,3) -- (8,1,3);
\draw[gray, dashed] (6,1,3) -- (6,0,3);
\draw[gray, dashed] (6,0,3) -- (8,0,3);
\draw[gray, dashed] (28/3,0,8/3) -- (28/3,0,4);
\draw[gray, dashed] (28/3,0,8/3) -- (11,0,8/3);
\filldraw[black] (6,3,0) circle (2pt) node[anchor=south west]{$z^3$};
\filldraw[black] (11,0,4) circle (2pt) node[anchor=north west]{$M_2$};
\filldraw[black] (8,0,3) circle (2pt) node[anchor=north]{$M_1$};
\filldraw[black] (8,1,3) circle (2pt) node[anchor=south west]{$M_1z$};
\end{tikzpicture}
\
\item If the point $3m_1$ belongs to the set $2 m_2 + (\RR_{\geq 0})^2$, the Newton polyhedron of $F$ consists of one segment with vertex corresponding to the monomials $z^3$ and $M_2$. We distinguish to further cases.
\begin{enumerate}
\item If $M_1^3 \ne M_2^2$, the point corresponding to the monomial $M_1z$ is an interior point of the Newton polyhedron of $F$. It follows that $F$ is non degenerate with respect to its Newton polyhedron.
\begin{tikzpicture}
\fill [yellow] (28/3,0,4)--(28/3,0,8/3)--(11,0,8/3)--(11,0,0)--(6,0,0)--(6,0,4)--(28/3,0,4);
\draw[gray, thick] (6,0,0) -- (6,0,7);
\draw[gray, thick] (6,0,0) -- (6,4,0);
\draw[gray, thick] (6,0,0) -- (13,0,0);
\filldraw[black] (6,0,0) circle (2pt) node[anchor=west]{};
\draw[gray, thick] (6,0,4) -- (13,0,4);
\draw[gray, thick] (11,0,0) -- (11,0,7);
\draw[gray, thick] (6,3,0) -- (11,0,4);
\draw[gray, thick] (6,3,0) -- (10.5,1,3) -- (12.75, 0, 4.5);
\draw[gray, thick] (6,3,0) -- (6,3,7);
\draw[gray, thick] (6,3,0) -- (13,3,0);
\draw[gray, dashed] (6,1,3) -- (6,0,3);
\draw[gray, dashed] (10.5,0,0) -- (10.5,0,3);
\draw[gray, dashed] (6,1,3) -- (6,1,0);
\draw[gray, dashed] (10.5,1,0) -- (10.5,0,0);
\draw[gray, dashed] (6,1,0) -- (10.5,1,0);
\draw[gray, dashed] (10.5,1,0) -- (10.5,1,3);
\draw[gray, dashed] (10.5,1,3) -- (6,1,3);
\draw[gray, dashed] (10.5,1,3) -- (10.5,0,3);
\draw[gray, dashed] (10.5,0,3) -- (6,0,3);
\draw[gray, dashed] (28/3,0,8/3) -- (28/3,0,4);
\draw[gray, dashed] (28/3,0,8/3) -- (11,0,8/3);
\filldraw[black] (6,3,0) circle (2pt) node[anchor=south west]{$z^3$};
\filldraw[black] (11,0,4) circle (2pt) node[anchor=north west]{$M_2$};
\filldraw[black] (10.5,0,3) circle (2pt) node[anchor=north]{};
\filldraw[black] (10.5,1,3) circle (2pt) node[anchor=south west]{$M_1z$};
\filldraw[black] (12.75,0,4.5) circle (2pt) node[anchor=north]{};
\end{tikzpicture}
\item If $M_1^3 = M_2^2$, the point corresponding to the monomial $M_1z$ lies in the unique dimensional compact face of the Newton polyhedron of $F$. The face equation is equal to
$$z^3 + c_1z + c_2.$$
Therefore, $F$ is non degenerate with respect to its Newton polyhedron if and only if $3(c_1)^2 \ne 2(c_2)^3$.
\begin{tikzpicture}
\fill [yellow] (28/3,0,4)--(28/3,0,8/3)--(11,0,8/3)--(11,0,0)--(6,0,0)--(6,0,4)--(28/3,0,4);
\draw[gray, thick] (6,0,0) -- (6,0,7);
\draw[gray, thick] (6,0,0) -- (6,4,0);
\draw[gray, thick] (6,0,0) -- (13,0,0);
\filldraw[black] (6,0,0) circle (2pt) node[anchor=west]{};
\draw[gray, thick] (6,0,4) -- (13,0,4);
\draw[gray, thick] (11,0,0) -- (11,0,7);
\draw[gray, thick] (6,3,0) -- (11,0,4);
\draw[gray, thick] (6,3,0) -- (6,3,7);
\draw[gray, thick] (6,3,0) -- (13,3,0);
\draw[gray, dashed] (6,1,8/3) -- (6,0,8/3);
\draw[gray, dashed] (28/3,0,0) -- (28/3,0,8/3);
\draw[gray, dashed] (6,1,8/3) -- (6,1,0);
\draw[gray, dashed] (28/3,1,0) -- (28/3,0,0);
\draw[gray, dashed] (6,1,0) -- (28/3,1,0);
\draw[gray, dashed] (28/3,1,0) -- (28/3,1,8/3);
\draw[gray, dashed] (28/3,1,8/3) -- (6,1,8/3);
\draw[gray, dashed] (28/3,1,8/3) -- (28/3,0,8/3);
\draw[gray, dashed] (28/3,0,8/3) -- (6,0,8/3);
\draw[gray, dashed] (28/3,0,8/3) -- (28/3,0,4);
\draw[gray, dashed] (28/3,0,8/3) -- (11,0,8/3);
\filldraw[black] (6,3,0) circle (2pt) node[anchor=south west]{$z^3$};
\filldraw[black] (11,0,4) circle (2pt) node[anchor=north west]{$M_2$};
\filldraw[black] (28/3,0,8/3) circle (2pt) node[anchor=north]{};
\filldraw[black] (28/3,1,8/3) circle (2pt) node[anchor=south west]{$M_1z$};
\end{tikzpicture}
\end{enumerate}
\end{enumerate}
\end{itemize}
\newpage
\section{Contibutions to the topological zeta function and the A´Campo monodromy zeta function of the trinomial forms}
\begin{proposition}
The monodromy zeta function and the local topological zeta function $Z(F,s)_0$ of $F$ equal
$$(1 - t^2)^{\alpha} \prod_{I} \zeta_{f_I}(t)^{-\chi(E_I^\circ)}$$
$$\sum_I \chi(E^\circ_I) Z(f_I, \omega_I)_{o_I}$$
where $I$ varies over the statification associated to the resolution $\pi$ of $b_2b_3D_zF$, the exponent $\alpha$ is the number of points of blow-ups needed to obtain $\pi$, and $o_I$ is a point in $E^\circ_I$.
\end{proposition}
\begin{example}
\end{example}
\subsection{Trinomial forms for strata $E^\circ_i$}
\subsubsection{First case $f= z^3 - x^{m_3}$ and $\omega = x^{\nu_1-1}dxdz$}
Newton polyhedra has three compact faces: the 0-dimensional faces $\tau_0$ corresponding to the monomial $z^3$ and $\tau_1$ corresponding to the monomial $x^{m_3}$, and the 1-dimensional face $\sigma$ contained in
$$ 3p + m_3r = 3m_3$$
The corresponding candidate pole is
$$ 3m_3 s + 3\nu_1 +m_3$$
The topological zeta function $Z_{{\rm top}}(f, \omega,s)_0$ equals
%a*c^2*s - m*s - 3*a*s - m - 3*a
$$ \frac{3\nu_1 + m_3 + (3\nu_1 + m_3 - \nu_1 (m_3, 3)^2)s}{\nu_1(3m_3 s + 3\nu_1 + m_3)(s+1)}$$
or equivalently $J_{\check \tau_0}-\frac{(3, m_3)J_{\check \sigma}s}{s+1}+J_{\check \tau_1}$ with the terms $J_{\check \tau_0}$, $J_{\check \tau_1}$ and $J_{\check \sigma}$ given as follows.
\begin{enumerate}
\item The dual cone $\check \tau_1$ to the face $\tau_1$ is simplicial and generated by the rays $$\frac{1}{(m_3,3)} (3, m_3), e_2.$$ This cone has multiplicity $\frac{3}{(m_3,3)}$. Its contribution to the local topological zeta function is
$$J_{\check \tau_1}= \frac{3}{ \nu_1 (3m_3 s + 3\nu_1 + m_3)}.$$
\item The dual cone $\check \sigma$ of the face $\sigma$ is the simplical cone generated by the ray $$\frac{1}{(m_3,3)} (3, m_3).$$ Its contribution to the local zeta function is
$$J_{\check \sigma}= \frac{(3,m_3)}{(3m_3 s + 3\nu_1 + m_3)}.$$
\item The dual cone $\check \tau_0$ to the face $\tau_0$ is the cone generated by the rays $$\frac{1}{(m_3,3)} (3, m_3), e_1.$$ This cone has multiplicity $\frac{m_3}{(m_3,3)}$. Its contribution to the the local topological zeta function is
$$J_{\check \tau_0}= \frac{m_3}{(3m_3 s + 3\nu_1 + m_3)}.$$
\end{enumerate}
Simmilarly we have that
$$\zeta_f (t) = \frac{(1-t^3)(1-t^{m_3})}{(1-t^{3m_3})^{(m_3,3)}}$$
\subsubsection{Second case $f= z^3 - x^{m_2}z - x^{m_3}$ and $\omega = x^{\nu_1-1}dxdz$ such that $3m_2 < 2 m_3$.}
Newton polyhedra has five compact faces: the 0-dimensional faces $\tau_0$ corresponding to the monomial $z^3$, $\tau_1$ corresponding to the monomial $x^{m_2}z$, and $\tau_2$ corresponding to the monomial $ x^{m_3}$, and the 1-dimensional faces $\sigma_1$ contained in the hyperplane
$$2p + m_2r = 3m_2$$
and $\sigma_2$ contained in the hyperplane
$$p + (m_3-m_2)r = m_3$$
The candidate poles are
$$3m_2s + 2 \nu_1 + m_2 \quad {\rm and } \quad m_3s + \nu_1 + m_3-m_2$$
The topological zeta function $Z_{{\rm top}}(f, \omega,s)_0$ equals
$$\frac{(2-\nu_1(m_2,2)^2)s + 2(\nu_1 + 1)}{(3m_2s + 2 \nu_1 + m_2)(s+1)}$$
or equivalently $J_{\check \tau_0}-(m_2,2)J_{\check \sigma_1}\frac{s}{s+1}+J_{\check \tau_1} - \frac{J_{\check \sigma_2}s}{s+1}+J_{\check \tau_2}$ with the terms $J_{\check \tau_0}$, $J_{\check \tau_1}$, $J_{\check \tau_2}$, $J_{\check \sigma_1}$ and $J_{\check \sigma_2}$ given as follows. Note that the candidate pole $ m_3s + \nu_1 + m_3-m_2$ is a false candidate pole.
\begin{enumerate}
\item The dual cone $\check \tau_2$ to the face $\tau_2$ is the regular cone generated by the rays
$$(1, m_3 - m_2), e_2$$
give the contribution
$$J_{\check \tau_2}=\frac{1}{(m_3s + \nu_1 + m_3-m_2)}$$
\item The dual cone $\check \sigma_2$ of the face $\sigma_2$ is the regular cone generated by the rays
$$(1, m_3 - m_2)$$
give the contribution
$$J_{\check \sigma_2} = \frac{1}{(m_3s + \nu_1 + m_3-m_2)}$$
\item The dual cone $\check \tau_1$ to the face $\tau_1$ is the simplicial cone generated by the rays
$$(1, m_3 - m_2), \frac{1}{(m_2,2)}(2, m_2)$$
The multiplicity of this cone is $\frac{m_2+2m_3}{(2,m_2)}$. Its contribution to the the local topological zeta function is
$$J_{\check \tau_1} = \frac{3m_3-2m_2}{(m_3s + \nu_1 + m_3-m_2)(3m_2s + 2 \nu_1 + m_2)}$$
\item The dual cone $\check \sigma_1$ of the face $\sigma_1$ is the cone generated by the rays
$$ \frac{1}{(n_2,2)} (2, m_2)$$
It gives the contribution
$$J_{\check \sigma_1} = \frac{(m_2, 2)}{(3m_2s + 2\nu_1 + m_2)}$$
\item The dual cone $\check \tau_0$ of the face $\tau_0$ is the simplicial cone generated by the rays
$$e_1, \frac{1}{(m_2,2)}(2, m_2)$$
The multiplicity of this cone is $\frac{2}{(2,m_2)}$. Its contribution to the the local topological zeta function is
$$J_{\check \tau_0}=\frac{2}{\nu_1(3m_2s + 2\nu_1+m_2)}$$
\end{enumerate}
Simmilarly we have that
$$\zeta_f (t) = \frac{(1-t^3)}{(1-t^{3m_2})^{(m_2,2)}}$$
\subsubsection{Third case $f= z^3 - x^{m_1}z^2 - x^{m_3}$ and $\omega = x^{\nu_1-1}dxdz$ such that $m_3 > m_1$}
Newton polyhedra has five compact faces: the 0-dimensional faces $\tau_0$ corresponding to the monomial $z^3$, $\tau_1$ corresponding to the monomial $x^{m_1}z^2$, and $\tau_2$ corresponding to the monomial $ x^{m_3}$, and the 1-dimensional faces $\sigma_1$ contained in the hyperplanes
$$p + m_1r = 3m_1$$
and $\sigma_2$ contained in the hyperplanes
$$2p + (m_3-m_1)r = 2m_3$$
The candidate poles are
$$3m_1s + \nu_1 + m_1 \quad {\rm and } \quad 2m_3s + 2\nu_1 + m_3-m_1$$
The topological zeta function $Z_{{\rm top}}(f, \omega,s)_0$ equals
$$\frac{ (2m_1m_3 + 6m_1\nu_1 - 2m_3\nu_1 -3m_1\nu_1c^2)s^2 + ( 3m_1m_3 - m_1^2 + 8m_1\nu_1 - \nu_1^2c^2s - m_1\nu_1c^2)s
- m_1^2 + m_1m_3 + m_1\nu_1 + m_3\nu_1 + 2\nu_1^2}{(s+1)(3m_1s + \nu_1 + m_1)(2m_3s + 2\nu_1 + m_3-m_1)}$$
or equivalently $V_0-\frac{As}{s+1}+V_1 - \frac{A_2s}{s+1}+V_2$ with the terms $V_0$, $V_1$ and $A$ given as follows.
\begin{enumerate}
\item The dual cone $\check \tau_2$ to the face $\tau_2$ is the regular cone generated by the rays
$$e_2, \frac{1}{(m_3-m_1,2)}(2, m_3-m_1)$$
give the contribution
$$J_{\check \tau_2}=\frac{2}{(2m_3s + 2\nu_1 + m_3-m_1)}$$
\item The dual cone $\check \sigma_2$ of the face $\sigma_2$ is the regular cone generated by the rays
$$\frac{1}{(m_3-m_1,2)}(2, m_3-m_1)$$
give the contribution
$$J_{\check \sigma_2} = \frac{(m_3-m_1,2)}{(2m_3s + 2\nu_1 + m_3-m_1)}$$
\item The dual cone $\check \tau_1$ to the face $\tau_1$ is the simplicial cone generated by the rays
$$ \frac{1}{(m_3-m_1,2)}(2, m_3-m_1), (1, m_1)$$
The multiplicity of this cone is $\frac{m_2+2m_3}{(2,m_2)}$. Its contribution to the the local topological zeta function is
$$J_{\check \tau_1} = \frac{m_3-3m_1}{(2m_3s + 2\nu_1 + m_3-m_1)(3m_1s + \nu_1 + m_1)}$$
\item The dual cone $\check \sigma_1$ of the face $\sigma_1$ is the cone generated by the ray
$$ (1, m_1)$$
It gives the contribution
$$J_{\check \sigma_1} = \frac{1}{(3m_1s + \nu_1 + m_1)}$$
\item The dual cone $\check \tau_0$ of the face $\tau_0$ is the simplicial cone generated by the ray
$$e_1, (1, m_1)$$
The multiplicity of this cone is $\frac{2}{(2,m_2)}$. Its contribution to the the local topological zeta function is
$$J_{\check \tau_0}=\frac{m_1}{\nu_1(3m_1s + \nu_1+m_1)}$$
\end{enumerate}
\subsection{Trinomial forms for strata $E^\circ_{i,j}$}
\subsubsection{ First case $f= z^3 - x^{m_3}y^{n_3}$ and $\omega = x^{\nu_1-1}y^{\nu_2-1}dxdydz$}
Newton polyhedra has three compact faces: the 0-dimensional faces $\tau_0$ corresponding to the monomial $z^3$ and $\tau_1$ corresponding to the monomial $x^{m_3}y^{n_3}$, and the 1-dimensional face $\sigma$ contained in
$$ 3p + m_3r = 3m_3, \qquad 3q + n_3r = 3n_3$$
The corresponding candidate poles are
$$ 3m_3 s + 3\nu_1 +m_3, \qquad 3n_3 s + 3\nu_2 + n_3$$
The topological zeta function $Z_{{\rm top}}(f, \omega,s)_0$ equals
$$ \frac{((3s+1)m_3n_3 + 3(\nu_1n_3+\nu_2m_3+3\nu_1\nu_2))(s+1) -3\nu_1\nu_2(3, m_3, n_3)^2s}{\nu_1\nu_2(3m_3 s + 3\nu_1 + m_3)(3n_3 s + 3\nu_2 + n_3)(s+1)}$$
or equivalently $V_0-\frac{(3, m_3, n_3)As}{s+1}+V_1 $ with the terms $V_0$, $V_1$ and $A$ given as follows.
\begin{enumerate}
\item The dual cone $\check \tau_1$ to the face $\tau_1$ is simplicial and generated by the rays $$\frac{1}{(m_3,3)} (3, 0, m_3), \frac{1}{(n_3,3)} (0, 3, n_3) , e_3.$$ This cone has multiplicity $\frac{9}{(m_3,3)(n_3,3)}$. Its contribution to the local topological zeta function is
$$V_1= \frac{9}{ (3m_3 s + 3\nu_1 + m_3)(3n_3 s + 3\nu_2+n_3)}.$$
\item The dual cone $\check \sigma$ of the face $\sigma$ is the simplical cone generated by the rays $$\frac{1}{(m_3,3)} (3, 0, m_3), \frac{1}{(n_3,3)} (0, 3, n_3) .$$ This cone has multiplicity $\frac{3 (m_3, n_3, 3)}{(m_3,3)(n_3,3)}$. Its contribution to the local zeta function is
$$A= \frac{3(m_3,n_3,3)}{(3m_3 s + 3\nu_1 + m_3)(3n_3 s + 3\nu_2+n_3)}.$$
\item The dual cone $\check \tau_0$ to the face $\tau_0$ is the cone generated by the rays $$\frac{1}{(m_3,3)} (3, 0, m_3), \frac{1}{(n_3,3)} (0, 3, n_3), e_1, e_2$$ which is not simplicial. Using the expression $\check \tau_0 = \mathbb{R}^3_{\geq 0} - \check \tau_1 - \check \sigma$, its contribution to the the local topological zeta function is
$$V_0=\frac{1}{ \nu_1 \nu_2 (3s+1)} - \frac{9}{(3m_3 s + 3\nu_1 + m_3)(3n_3 s + 3\nu_2 + n_3)(3s+1)}$$
$$ = \frac{m_1m_2(3s+1) + 3(\nu_1m_2+\nu_2m_1)}{\nu_1\nu_2(3m_3 s + 3\nu_1 + m_3)(3n_3 s + 3\nu_2 + n_3)}$$
\end{enumerate}
%Comparing the contributions of the pairs $(f, \omega)$ and the case $(\bar{f}, \bar{\omega})$ with $\bar{f} = z^3 - \bar{x}^{\frac{m_3}{g}} \bar{y}^{\frac{n_3}{g}}$ and $\bar{\omega} = \bar{x}^{\frac{\nu_1}{g}-1} \bar{y}^{\frac{\nu_2}{g}-1} d \bar{x}d \bar{y} dz$
%we get the relations
%$$\bar{V}_0= g V_0 \qquad \bar{V}_1= g V_1 \qquad \bar{A} =
%\begin{cases}
%g A & \hbox{ if either } 3 \mid m \hbox{ and } 3 \mid \frac{m}{g} \hbox{ or } 3 \nmid m \hbox{ and } 3 \nmid \frac{m}{g} \\
%\frac{g}{9} A & \hbox{ if } 3 \mid m \hbox{ and } 3 \nmid \frac{m}{g}
%\end{cases}$$
\subsection{} Second case $f= z^3 - x^{m_2}y^{n_2}z - x^{m_3}y^{n_3}$ and such that $m_2 \leq 2 m_3, 3 n_2 \leq 2 n_3$ and $3(m_2, n_2) \ne 2(m_3, n_3)$.
Newton polyhedra has five compact faces: the 0-dimensional faces $\tau_0$ corresponding to the monomial $z^3$, $\tau_1$ corresponding to the monomial $x^{m_2}y^{n_2}z$, and $\tau_2$ corresponding to the monomial $ x^{m_3}y^{n_3}$, and the 1-dimensional faces $\sigma_1$ contained in the hyperplanes
$$2p + m_2r = 3m_2, \qquad 2q + n_2r = 3n_2$$
and $\sigma_2$ contained in the hyperplanes
$$p + (m_3-m_2)r = m_3, \qquad q + (n_3-n_2)r =n_3$$
The candidate poles are
$$3m_2s + 2 \nu_1 + m_2, \qquad 3n_2 s + 2\nu_2 + n_2$$
and
$$m_3s + \nu_1 + m_3-m_2, \qquad n_3s + \nu_2 + n_3-n_2$$
The topological zeta function is the sum of the following five terms
\begin{enumerate}
\item The dual cone $\check \tau_2$ to the face $\tau_2$ is the regular cone generated by the rays
$$(1, 0, m_3 - m_2), (0, 1 , n_3 - n_2), e_3$$
give the contribution
$$\frac{1}{(m_3s + \nu_1 + m_3-m_2)(n_3s + \nu_2 + n_3-n_2)}$$
\item The dual cone $\check \sigma_2$ of the face $\sigma_2$ is the regular cone generated by the rays
$$(1, 0, m_3 - m_2), (0, 1 , n_3 - n_2)$$
give the contribution
$$\frac{-s}{(m_3s + \nu_1 + m_3-m_2)(n_3s + \nu_2 + n_3-n_2)(s+1)}$$
\item The dual cone $\check \tau_1$ to the face $\tau_1$ is the cone generated by the rays
$$(1, 0, m_3 - m_2), (0, 1 , n_3 - n_2), \frac{1}{(m_2,2)}(2, 0, m_2), \frac{1}{(n_2,2)}(0,2,n_2)$$
which is not simplicial. Denote by $\rho$ the cone generated by $\frac{1}{(m_2,2)}(2, 0, m_2)$, $\frac{1}{(n_2,2)}(0,2,n_2)$ and $e_3$. Using the expression $\check \tau_1 = \rho - \check \tau_2 - \check \sigma_2$, its contribution to the the local topological zeta function is
$$\frac{4}{(m_2,2)(n_2,2)(3m_2s + 2\nu_1 + m_2)(3n_2s + 2\nu_2 + n_2)} - \frac{1}{(m_3s + \nu_1 + m_3-m_2)(n_3s + \nu_2 + n_3-n_2)}$$
\item The dual cone $\check \sigma_1$ of the face $\sigma_1$ is the cone generated by the rays
$$ \frac{1}{(n_2,2)} (2, 0, m_2), \frac{1}{(n_2,2)} (0, 2 , n_2)$$
This cone has multiplicity $2(m_2, n_2, 2)$. It gives the contribution
$$\frac{-2(m_2, n_2, 2)s}{(3m_2s + 2\nu_1 + m_2)(3n_2s + 2\nu_2 + n_2)(s+1)}$$
\item The dual cone $\check \tau_0$ of the face $\tau_0$ is the cone generated by the rays
$$e_1, e_2, \frac{1}{(m_2,2)}(2, 0, m_2), \frac{1}{(n_2,2)}(0,2,n_2)$$
Using the expression $\check \tau_0 = \mathbb{R}^3_{\geq 0} - \rho - \check \sigma_2$, its contribution to the the local topological zeta function is
$$\frac{1}{\nu_1\nu_2(3s+1)} - \frac{2}{(3m_2s + 2\nu_1+m_2)(3n_2s + 2\nu_2+m_2)(3s+1)}$$
\end{enumerate}
\subsection{} Third case $f= z^3 - x^{m_1}y^{n_1}z^2 - x^{m_3}y^{n_3}$ and such that $3m_1 \leq m_3, 3 n_1 \leq n_3$ and $3(m_1, n_1) \ne 2(m_3, n_3)$.
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