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\newcommand{\Suz}{\ZZ[u,u^3v^4,uv]} \newcommand{\Jn}{\langle u-1,u^3v^4-1,uv-1 \rangle^{n+1}} \newcommand{\Gn}{\mathbb{G}_n} \newcommand{\Mn}{\mathbb{M}_n} \newcommand{\cV}{\mathcal{V}} \newcommand{\Gc}{\mathbb{G}_n^0} \newcommand{\Gp}{\mathbb{G}_n^p} \newcommand{\Pnn}{\mathcal{P}_n} \newcommand{\Pnu}{\mathcal{P}_{n-1}} \newcommand{\Cc}{C_{\mathbb{G}_n^{(0)}}} \newcommand{\Cp}{C_{\mathbb{G}_n^{(p)}}} \newcommand{\bfg}{{\bf \left(\frac{f}{g}\right)^s}} \newcommand{\bfga}{{\bf \frac{f^s}{g^{s+\alpha}}}} \newcommand{\bfgt}{{\bf \frac{\tilde{f}^s}{\tilde{g}^{s+\alpha}}}} \newcommand{\bfgOne}{{\bf \frac{f^s}{g^{s+1}}}} \newcommand{\ga}{{\bf \frac{1}{g^{\alpha}}}} \newcommand{\gd}{\operatorname{gl.dim}} \newcommand{\lm}{\operatorname{lm}} \newcommand{\lc}{\operatorname{lc}} \newcommand{\lt}{\operatorname{lt}} \newcommand{\init}{\operatorname{in}} \newcommand{\diff}{\operatorname{diff}} %------------------------------------------------------- \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. \begin{figure}[ht] \begin{center} \begin{tikzpicture}[vertice/.style={draw,circle,fill,minimum size=0.2cm,inner sep=0}] \coordinate (U) at (0,0); \node[vertice] at (U) {}; \node[below] at (U) {{\small $E_1$}}; \coordinate (G1) at (-0.5,0.75); \coordinate (G2) at (-0.6,0.75); \coordinate (G3) at (-0.4,0.75); \coordinate (G4) at (-0.3,0.75); \coordinate (G5) at (-0.1,0.75); \coordinate (G6) at (-0.2,0.75); \coordinate (G7) at (0,0.75); \node[above] at (G7) {{\small ${{F^\Delta_i}}_{{}_{i=1}}^{{}{13}}$}}; \coordinate (G8) at (0.1,0.75); \coordinate (G9) at (0.2,0.75); \coordinate (G10) at (0.3,0.75); \coordinate (G11) at (0.4,0.75); \coordinate (G12) at (0.5,0.75); \coordinate (G13) at (0.6,0.75); \coordinate (A1) at (1,0); \coordinate (A2) at (2,0); \coordinate (A3) at (3,0); \coordinate (A4) at (4,0); \coordinate (A5) at (5,0); \coordinate (A6) at (6,0); \coordinate (A7) at (7,0); \coordinate (A8) at (5,1); \coordinate (B1) at (-1,0); \coordinate (B2) at (-2,0); \coordinate (B3) at (-3,0); \coordinate (B4) at (-4,0); \coordinate (B5) at (-5,0); \coordinate (B6) at (-6,0); \coordinate (B7) at (-7,0); \coordinate (B8) at (-5,1); \coordinate (F1) at (1.2,0.75); \node[above] at (F1) {{\small $F^{\infty}_1$}}; \coordinate (F2) at (1.95,0.75); \node[above] at (F2) {{\small $F^\Delta_{14}$}}; \coordinate (F3) at (6,0.75); \node[above] at (F3) {{\small $F^{0}_1$}}; \coordinate (F4) at (-1.2,0.75); \node[above] at (F4) {{\small $F^{\infty}_2$}}; \coordinate (F5) at (-1.95,0.75); \node[above] at (F5) {{\small $F^\Delta_{15}$}}; \coordinate (F6) at (-6,0.75); \node[above] at (F6) {{\small $F^{0}_2$}}; \node[vertice] at (A1) {}; \node[below] at (A1) {{\small $E_7$}}; \node[vertice] at (B1) {}; \node[below] at (B1) {{\small $E_{15}$}}; \node[vertice] at (A2) {}; \node[below] at (A2) {{\small $E_6$}}; \node[vertice] at (A3) {}; \node[below] at (A3) {{\small $E_5$}}; \node[vertice] at (A4) {}; \node[below] at (A4) {{\small $E_4$}}; \node[vertice] at (A5) {}; \node[below] at (A5) {{\small $E_3$}}; \node[vertice] at (A6) {}; \node[below] at (A6) {{\small $E_9$}}; \node[vertice] at (A7) {}; \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 $3(m_1, n_1) \ne 2(m_3, n_3)$. \bibliographystyle{amsplain} \bibliography{biblio} \end{document}