1680 lines
75 KiB
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1680 lines
75 KiB
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%-------------------------------------------------------
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\begin{document}
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\title{On germs of hypersurfaces with multiplicity 3}
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\author[E.~Artal]{Enrique Artal Bartolo}
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\address{Departamento de Matemáticas, IUMA, Universidad de Zaragoza,\\ C. Pedro Cerbuna 12,
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50009, Zaragoza, Spain.
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}
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%\thanks{$^{3}$Partially supported by grant CONACyT.}
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\email{\href{mailto:artal@unizar.es}{artal@unizar.es}}
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\author[M.~Gonz\'alez Villa]{Manuel Gonz\'alez Villa}
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\address{
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Centro de investigaci\'on en
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Matem\'aticas\\
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Apartado Postal
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402,
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C.P. 36000, Guanajuato, GTO, M\'exico.\newline
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\indent Departamento de Matemáticas, IUMA, Universidad de Zaragoza,\\ C. Pedro Cerbuna 12,
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50009, Zaragoza, Spain.
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}
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\email{\href{mailto:manuel.gonzalez@cimat.mx}{manuel.gonzalez@cimat.mx}, \href{mailto:m.gonzalez@unizar.es}{m.gonzalez@unizar.es}}
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\author[E. León-Cardenal]{Edwin León-Cardenal}
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\address{Departamento de Matem\'{a}ticas y Computación \\
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Universidad de La Rioja \\
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26006 Logroño, Spain}
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\email{\href{mailto:edwin.leon@unirioja.es}{edwin.leon@unirioja.es}}
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\maketitle
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Consider a the germ of a isolated singular point of hypersurface in $(\CC^3,0)$ of multiplicity $3$.
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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 $z$, where the multiplicity of $a_j$ is at least $j$.
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The isolated condition implies that $\disc_z F$ are not zero, i.e., the equation is reduced.
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A lot of properties are known when $F$ is non-degenerate with respect to its Newton polyhedron.
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We want to derive properties in the general; let us state how we are going to deal with these singularities.
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% 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 $F$.
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We propose the following general method.
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\medskip
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\begin{paso}
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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 case of a suspension that we have studied elsewhere, see~\cite{AGGL}.
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Actually since the topological type does not change if we add terms of \emph{very} high degree, we can safely assume
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that $b_2b_3\neq 0$. Under this condition, up to a constant, $\disc_zF$ is $D_z:=b_2^3 - b_3^2$.
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\end{paso}
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%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 $b_2b_3 \ne 0$.
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\begin{paso}
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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. {\color{red}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 $f \in \CC[x,y]$.
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\end{paso}
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\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 $z$.
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\medskip
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\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.
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\medskip
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A couple of remarks are in order after these four steps.
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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.
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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.
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The following reductions are possible to perform the Fourth Step.
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\medskip
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\noindent \emph{Reduction A:} It is enough to consider the associated trinomial local forms instead of the quasi-trinomial local forms.
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\medskip
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\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.
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\medskip
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\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.
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\medskip
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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 $0$.
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\medskip
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\noindent \emph{Reduction D:} It is enough to consider a minimal $\QQ$-resolution of the discriminant $D_ZF$.
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\medskip
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After all these reductions, we will prove that ...
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\section{Power series of multiplicity three}
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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.
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\begin{definition} The Tschirnhaus transformation associated to $F$ is the transformation given by $z \mapsto z- \frac{a_1}{3}$.
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\end{definition}
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After the Tschirnhaus transformation, the monic Weierstrass polynomial $F(x,y,z - a_1/3)$ can be written as
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$$ z^3 -3b_2z + 2b_3$$
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with
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$$b_2 = \frac{a_1^2}{9} - \frac{ a_2}{3} \quad {\rm and } \quad b_3= \frac{a_1^3}{27}-\frac{a_1a_2}{6} + \frac{a_3}{2}$$
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and its the discriminant $D_zF$ of $F$ with respect to $F$ is of the form $108(b_2^3 - b_3^2)$.
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Denote the decompositions of $b_2$ and $b_3$ into irreducible components as
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\begin{equation}
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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
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\end{equation}
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where $h_i, k_j$ are irreducible and $h_i, k_j, f, g$ are pairwise coprime and $n_i, m_j \geq 1$.
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\begin{proposition}
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The discriminant $D_zF$ is of the form
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$$ \prod_{i=1}^n h_i^{2} \prod_{j=1}^m k_j^3 \left ( \prod_{i=1}^n h_i^{3n_i-2} f^3 - \prod_{j=1}^m h_i^{2m_j -3} g^2 \right ).$$
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\end{proposition}
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\begin{proposition}
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The singular point is isolated if and only if the discriminant $D_zF$ is reduced.
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\end{proposition}
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\noindent \textcolor{red}{Proof. Ask Enrique.}
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\begin{proposition}\label{prop.non-deg} Let $b_2, b_3 \in \CC[[x, y]]$ with $b_2(0), b_3(0) = 0$. Let us write $b_2= M_2(c_2+A)$ and $b_3 = M_3(c_3 + B)$ with $M_2, M_3 \in \CC[x,y]$ (monic) monomials, $c_2, c_3 \in \CC^\ast$ and $A, B \in \CC[[x,y]]$ such that $A(0), B(0) = 0$.
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The monic Weierstrass polynomial $z^3 -3b_2z + 2b_3$ is non degenerate unless $(c_2M_2)^3 = (c_3M_3)^2$.
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\end{proposition}
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\noindent \textcolor{blue}{Proof. The Newton polyhedron of $z^3 -3b_2z + 2b_3$ can be 2-dimensional, polygonal with two segments or polygonal with a unique segment. In the two first cases, it is non degenerate. The compact faces of the Newton polyhedron of $z^3 -3b_2z + 2b_3$ lie on a segment if and only if $(M_2)^3 = (M_3)^2$. The condition of $z^3 -3b_2z + 2b_3$ being non degenerate is equivalent to $z^3 -3b_2z + 2b_3 \in \CC[z]$ having a double root. Notice a triple root is not possible because the degree 2 monomial is zero. Assume that $z^3 -3b_2z + 2b_3 =(z-a)^2(z-b)$ for some $a, b \in \CC^\ast$. Equivalently, we have $2a+b=0$, and therefor $c_2=(\frac{b}{2})^2$ and $c_3=(\frac{-b}{2})^3$.}
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\section{Jung's method for power series of multiplicity three }
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\subsection{Embbeded resolution of the discriminant and classification of their exceptional divisors}
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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
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$$E_I^\circ = \bigcap_{i \in I} E_j \setminus \bigcup_{j \not \in I} E_j .$$
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\begin{definition}For each irreducible component $E_j$ of of $(\pi \circ G)^{-1} \{G=0\}$ we define the numbers
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$$N^0_j = {\rm ord}_{E_j} \pi^\ast b_2, \quad N^\infty_j = {\rm ord}_{E_j} \pi^\ast b_3, \quad N^\Delta_j = {\rm ord}_{E_j} \pi^\ast D_zF, \quad {\rm and} \quad \nu_j = {\rm ord}_{E_j} \pi^\ast \omega -1$$
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where $\omega = dx \wedge dy$ is the standard volume form of $\CC^2$.
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\end{definition}
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\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.
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\end{proposition}
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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 $(\pi)^{-1} \{G=0\}$
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\begin{itemize}
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\item $3 N^0_j > 2N^\infty_j = N^\Delta_j$ and it is called a \emph{zero} because in this case $\mathcal{M}(E_j)=0$;
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\item $2 N^\infty_j > 3N^0_j = N^\Delta_j$ and it is called a \emph{pole} because in this case $\mathcal{M}(E_j)=\infty$;
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\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 $\mathcal{M}$;
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\item $3 N^0_j = 2N^\infty_j < N^\Delta_j$ and in this case it is mapped constantly to $1$ under $\mathcal{M}$.
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\end{itemize}
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\begin{remark}
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In the third case, $N^\Delta_j$ is divisible by 6. 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.
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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., $\mathcal{M}(E_j)=\mathbb{P}^1$.
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\end{remark}
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\begin{definition}\label{extriplepole} For a divisor of th last type, i.e., a divisor that is mapped constantly to $1$, the difference
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$N^\Delta_j - 3 N^0_j = N^\Delta_j - 2 N^\infty_j$ is called the \emph{excess of the discriminant} and it is denoted by $M^\Delta_j$.
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\end{definition}
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\begin{example} Consider $F=z^3 - 3b_2z + 2b_3$ with
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$b_2= ((y^2 + x^3)^2 + x^5y)((x^2 + y^3)^2 + y^5x)$
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and
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$b_3= (y^3+x^4)(x^3+y^4)$. The minimal resolution $\pi$ of the product $G=b_2 b_3 D_zF$ is obtained after 17 blow-ups. The following figure and table give the dual resolution graph of $G$ and its associated numerical data.
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\begin{figure}[ht]
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\begin{center}
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\begin{tikzpicture}[vertice/.style={draw,circle,fill,minimum size=0.2cm,inner sep=0}]
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\coordinate (U) at (0,0);
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\node[vertice] at (U) {};
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\node[below] at (U) {{\small $E_1$}};
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\coordinate (G1) at (-0.5,0.75);
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\coordinate (G2) at (-0.6,0.75);
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\coordinate (G3) at (-0.4,0.75);
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\coordinate (G4) at (-0.3,0.75);
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\coordinate (G5) at (-0.1,0.75);
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\coordinate (G6) at (-0.2,0.75);
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\coordinate (G7) at (0,0.75);
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\node[above] at (G7) {{\small ${{F^\Delta_i}}_{{}_{i=16}}^{{}{28}}$}};
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\coordinate (G8) at (0.1,0.75);
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\coordinate (G9) at (0.2,0.75);
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\coordinate (G10) at (0.3,0.75);
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\coordinate (G11) at (0.4,0.75);
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\coordinate (G12) at (0.5,0.75);
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\coordinate (G13) at (0.6,0.75);
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\coordinate (A1) at (1,0);
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\coordinate (A2) at (2,0);
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\coordinate (A3) at (3,0);
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\coordinate (A4) at (4,0);
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\coordinate (A5) at (5,0);
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\coordinate (A6) at (6,0);
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\coordinate (A7) at (7,0);
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\coordinate (A8) at (5,1);
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\coordinate (B1) at (-1,0);
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\coordinate (B2) at (-2,0);
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\coordinate (B3) at (-3,0);
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\coordinate (B4) at (-4,0);
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\coordinate (B5) at (-5,0);
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\coordinate (B6) at (-6,0);
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\coordinate (B7) at (-7,0);
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\coordinate (B8) at (-5,1);
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\coordinate (F1) at (1.2,0.75);
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\node[above] at (F1) {{\small $F^{\infty}_{33}$}};
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\coordinate (F2) at (1.95,0.75);
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\node[above] at (F2) {{\small $F^\Delta_{29}$}};
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\coordinate (F3) at (6,0.75);
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\node[above] at (F3) {{\small $F^{0}_{31}$}};
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\coordinate (F4) at (-1.2,0.75);
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\node[above] at (F4) {{\small $F^{\infty}_{34}$}};
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\coordinate (F5) at (-1.95,0.75);
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\node[above] at (F5) {{\small $F^\Delta_{30}$}};
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\coordinate (F6) at (-6,0.75);
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\node[above] at (F6) {{\small $F^{0}_{32}$}};
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\node[vertice] at (A1) {};
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\node[below] at (A1) {{\small $E_7$}};
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\node[vertice] at (B1) {};
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\node[below] at (B1) {{\small $E_{15}$}};
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\node[vertice] at (A2) {};
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\node[below] at (A2) {{\small $E_6$}};
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\node[vertice] at (A3) {};
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\node[below] at (A3) {{\small $E_5$}};
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\node[vertice] at (A4) {};
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\node[below] at (A4) {{\small $E_4$}};
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\node[vertice] at (A5) {};
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\node[below] at (A5) {{\small $E_3$}};
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\node[vertice] at (A6) {};
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\node[below] at (A6) {{\small $E_9$}};
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\node[vertice] at (A7) {};
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\node[below] at (A7) {{\small $E_8$}};
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\node[vertice] at (A8) {};
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\node[above] at (A8) {{\small $E_2$}};
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\node[vertice] at (B2) {};
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\node[below] at (B2) {{\small $E_{14}$}};
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\node[vertice] at (B3) {};
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\node[below] at (B3) {{\small $E_{13}$}};
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\node[vertice] at (B4) {};
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\node[below] at (B4) {{\small $E_{12}$}};
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\node[vertice] at (B5) {};
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\node[below] at (B5) {{\small $E_{11}$}};
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\node[vertice] at (B6) {};
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\node[below] at (B6) {{\small $E_{17}$}};
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\node[vertice] at (B7) {};
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\node[below] at (B7) {{\small $E_{16}$}};
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\node[vertice] at (B8) {};
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\node[above] at (B8) {{\small $E_{10}$}};
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\draw (U)--(A1)--(A2)--(A3)--(A4)--(A5)--(A6)--(A7);
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\draw (A5)--(A8);
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\draw (B5)--(B8);
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\draw (U)--(B1)--(B2)--(B4)--(B4)--(B5)--(B6)--(B7);
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\draw[->] (A1)--(F1);
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\draw[->] (A1)--(F2);
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\draw[->] (A6)--(F3);
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\draw[->] (U)--(G1);
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\draw[->] (U)--(G2);
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\draw[->] (U)--(G3);
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\draw[->] (U)--(G4);
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\draw[->] (U)--(G5);
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\draw[->] (U)--(G6);
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\draw[->] (U)--(G7);
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\draw[->] (U)--(G8);
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\draw[->] (U)--(G9);
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\draw[->] (U)--(G10);
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\draw[->] (U)--(G11);
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\draw[->] (U)--(G12);
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\draw[->] (U)--(G13);
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\draw[->] (B1)--(F4);
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\draw[->] (B1)--(F5);
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\draw[->] (B6)--(F6);
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\end{tikzpicture}
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\end{center}
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\end{figure}
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$$\begin{array}{|c||c|c|c|c|c|c|c|c|c|c|}
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\hline
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&E_1&E_2, E_{10}&E_3, E_{11}&E_4, E_{12}&E_5, E_{13}&E_6, E_{14}&E_7, E_{15} &E_8, E_{16}&E_9, E_{17}\\
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\hline
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\hline
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N^f_i& {\bf 8}& 10& 20& 28& 36& 44& {\bf 52}& 21&42\\
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N^g_i& {\bf12}& {\bf 13}& {\bf 26}& {\bf 39}& {\bf 52}& {\bf 65}& {\bf 78}& {\bf 26}& {\bf 52}\\
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N^\Delta_i& 25& {\bf 26}& {\bf 52}& {\bf 78}& {\bf 104}& {\bf 130}& {\bf 156}& {\bf 52}&{\bf 104}\\
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\hline
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\hline
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\pi()& 1& 0& 0& 0& 0& 0& \mathbb{P}^1& 0&0\\
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\hline
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\hline
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\nu_i&2&3&5&7&9&11&13&6&11\\
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\chi()&-13&1&-1&0&0&0&-2&1&-1\\
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E_i^2& -11& -2& -4& -2& -2& -2& -1&-2&-1\\
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\hline
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\end{array}$$
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The divisor $E_1$ is mapped to the value 1, the divisors $E_7$ and $E_{15}$ are dicriticals, and the rest of exceptionals divisors are zeros. The 13 arrows $F^\Delta_i$ with $i=16, \dots, 29$ attached to the divisor $E_1$ correspond to irreducible components of the strict transforms of $D_zF$, the arrows attached to the divisor $E_7$ (resp. $E_{13}$) correspond to an irreducible component strict transform of $b_3$ and another irreducible component of $D_zF$. Finally the arrow attched to $E_9$ (resp. to $E_{17}$) correspond to the strict transform of $b_2$.
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%$$\pi^{-1} \{\mathcal{M} = 0\} = 2 (F^{0}_1 + F^{0}_2) + 4 (E_2 + E_5 + E_{10} + E_{13}) + 6 (E_4 + E_{12} ) + 8 (E_3 + E_{11}) + 11 (E_8+E_{16}) + 22 (E_9 + E_{17})) , \quad \pi^{-1} \{\mathcal{M} = 1\} = \sum_{i=1}^{15} F^\Delta_i + E_1$$
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$$\pi^{-1} \{\mathcal{M} = 0\} = 2 ( \sum_{31,32}F^{0}_i + \sum_{6,14} E_i ) + 4 \sum_{2,5,10,13}E_i + 6 \sum_{4,12 }E_i + 8 \sum_{3,11}E_i + 11\sum_{8,16} E_i + 22 \sum_{9,17} E_i ,$$
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$$\pi^{-1} \{\mathcal{M} = \infty\} = 3 F^{\infty}_{33} + 3 F^{\infty}_{34}, \quad \pi(E_7) = \pi(E_{15}) = \mathbb{P}^1, \quad \pi^{-1} \{\mathcal{M} = 1\} = \sum_{i=16}^{30} F^\Delta_i + E_1$$
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The stratification associated with the resolution $\pi$ consist of $35$ strata $E^\circ_{i,j}$ of dimension 0 and 11 strata $E_i^\circ$ of dimension 1 and $\chi(E_i^\circ) \ne 0$.
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The former correspond to the elements $(i,j)$ of the set $S$
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\[
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\left\{\!
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\begin{array}{ccccccccc}
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(1,7)& (1,15)& (1,16)& (1,17)& (1,18)& (1,19)& (1,20)& (1,21)& (1,22)\\
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(1,23)& (1,24)& (1, 25)& (1, 26)& (1, 27)& (1, 28)& (2,3)& (3,4)& (3,9)\\
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(4,5)& (5,6)& (6,7)& (7, 29)& (7, 33)& (8, 9)& (9,31)& (10,11)& (11,12)\\
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(11,17)& (12, 13)& (13,14)& (14,15)& (15,30)& (15, 34)& 16,17)& (17,32)
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\end{array}
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\!\right \}
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\]
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%the 13 intersections $E_1 \cap F_i^\Delta$ for $i=16, \dots, 28$ and the 6 intersections
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%$$ E_7 \cap F_{34}^\infty, \quad E_7 \cap F_{30}^\Delta, \quad E_9 \cap F_{32}^0, \quad E_{15} \cap F_{35}^\infty, \quad E_{15} \cap F_{31}^\Delta, \quad E_{17} \cap F_{33}^0 $$
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and the latter correspond with the indexes
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$$1, \quad 2, \quad 3, \quad 7, \quad 8, \quad 9, \quad 10, \quad 11, \quad 15, \quad 16, \quad 19.$$
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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 $b_2$.
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\end{example}
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\subsection{Trinomial forms associated to the strata}
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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 $\tilde{\pi}^\ast F$ can be expressed in terms of the pullbacks of $b_2$ and $b_3$ as
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\begin{equation}
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z^3 -3 \pi^\ast b_2 z + 2 \pi^\ast b_3.
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\end{equation}
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over each non empty strata $E^\circ_I$ of $X$ created by the resolution $\pi$.
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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 $N^0_i$ and $N^\infty_i$ with $i \in I$ as
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\begin{equation}
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(\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
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(\pi^\ast b_3)_{|_I}= x_1^{N^\infty_{i_1}} y_1^{N^\infty_{i_2}} u_1(x_1, y_1)
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\end{equation}
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where $u_1, u_2$ are units in $\CC\{x_1, y_1\}$. Hence, we have expressions
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\begin{equation}\label{quasi-trinomial}
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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),
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\end{equation}
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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 $\CC\{x_1, y_1\}$.
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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 $3N^0_i=2N^\infty_i < N^\Delta_i$.
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|
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\begin{remark}
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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., $\mathcal{M}(E_j)=\mathbb{P}^1$.
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\end{remark}
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\begin{proposition} There exist a change of coordinates of the form
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$$x_1= x_2, \quad y_1= y_2, \quad z = z_1 - x_2^\frac{N^0_i}{2} y_2^\frac{N^0_j}{2} + \cdots$$
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such that (\ref{quasi-trinomial}) becomes
|
|
either
|
|
$$z_1^3 + x_2^{\frac{N^0_i}{2}} y_2^{\frac{N^0_j}{2}} u'_1(x_2,y_2) z_1^2 + x_2^{N_i^\infty + M_i^\Delta} y_2^{N_j^\infty + M_j^\Delta} u'_2(x_2,y_2)$$
|
|
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
|
|
$$\begin{cases}
|
|
z^3 - x^{N^\infty_{i}}, & \mathcal{M}(E_i) \ne 1, \infty \\
|
|
z^3 - x^{N^0_{i}} z + x^{N^\infty_{i}}, & \mathcal{M}(E_i) = \infty \\
|
|
z^3 - x_2^{\frac{N^0_i}{2}}z^2 + x^{N_i^\infty + M_i^\Delta} , & \mathcal{M}(E_i) = 1, \\
|
|
\end{cases}$$
|
|
in the case $I= \{i\}$
|
|
and
|
|
$$\begin{cases}
|
|
z^3 + x^{N^\infty_i}y^{N^\infty_j}, & \mathcal{M}(E_i), \mathcal{M}(E_j) \ne 1, \infty \\
|
|
z^3 + x^{\frac{N^0_i}{2}}y^{\frac{N^0_j}{2}}z^2 + x^{N_i^\infty + M_i^\Delta}y^{N_j^\infty + M_j^\Delta} , & \mathcal{M}(E_i) = \mathcal{M}(E_j) = 1 \\
|
|
z^3 + x^{\frac{N^0_i}{2}}y^{\frac{N^0_j}{2}} z^2 + x^{N_i^\infty + M_i^\Delta}y^{N_j^\infty} , & \mathcal{M}(E_i) = 1, \mathcal{M}(E_j) = \mathbb{P}^1 \\
|
|
z^3 + x^{N^0_{i}}y^{N^0_{j}} z + x^{N^\infty_{i}}y^{N^\infty_{j}}, & {\rm else}\\
|
|
\end{cases}$$
|
|
in the case $I= \{i, j\}$.
|
|
\end{definition}
|
|
|
|
|
|
\begin{example} The following tables show all the trinomial forms and differential forms for the all the 0-dimensional strata and all the 1-dimensional strata with not zero Euler characteristic of Example \ref{extriplepole}
|
|
|
|
|
|
|
|
|
|
$$\begin{array}{c|c|c|c}
|
|
i&f_i&\omega_i& \chi()\\
|
|
\hline
|
|
1& z^3 + x^4z^2 + x^{13}& (2, 1, 1)& -13\\
|
|
2,10& z^3 + x^{10}z + x^{13}& (3, 1, 1)&1\\
|
|
3,11& z^3 + x^{20}z + x^{26}& (5, 1, 1)& -1\\
|
|
7,15& z^3 + \color{red}{x^{52}z} + x^{78}& (13, 1, 1)& -2\\
|
|
8, 16& z^3 + x^{21}z + x^{26}& (6, 1, 1)& 1\\
|
|
9,17& z^3 + x^{42}z + x^{52}& (11, 1, 1)& -1\\
|
|
\end{array}$$
|
|
|
|
|
|
$$\begin{array}{c|c|c}
|
|
(i,j)&f_{i,j}&\omega_{i,j}\\
|
|
\hline
|
|
&&\\
|
|
(1, 7), (1, 15)& z^3 + x^4y^{26}z^2 + x^{13}y^{78}& (2, 13, 1)\\
|
|
(2,3), (10, 11)& z^3 + x^{10}y^{20}z + x^{13}y^{26}& (3, 5, 1)\\
|
|
(3, 9), (11, 17)& z^3 + x^{20}y^{42}z + x^{26}y^{52}& (5, 11, 1)\\
|
|
(3, 4), (11,12)& z^3 + x^{20}y^{28}z + x^{26}y^{39}& (5, 7, 1)\\
|
|
(8, 9), (16,17)& z^3 + x^{21}y^{42}z + x^{26}y^{52}& (6, 11, 1)\\
|
|
(4, 5), (12,13)& z^3 + x^{28}y^{36}z + x^{39}y^{52}& (7, 9, 1)\\
|
|
(5, 6), (13,14)& z^3 + x^{36}y^{44}z + x^{52}y^{65}& (9, 11, 1)\\
|
|
(6, 7), (14,15)& z^3 + x^{44}y^{52}z + x^{65}y^{78}& (11, 13, 1)\\
|
|
(1, i), i=16, \dots, 28& z^3 + x^4z^2 + x^{13}y& (2, 1, 1)\\
|
|
(9, 31), (17, 32)& z^3 + x^{42}yz + x^{52}& (11, 1, 1)\\
|
|
(7, 33), (15, 34)& z^3 + x^{52}z + x^{78}y& (13, 1, 1)\\
|
|
(7, 29) (15, 30)& z^3 + x^{26}z^2 + x^{78}y& (13, 1, 1)\\
|
|
%(12, 'f')& z^3 + x^{42}yz + x^{52}& (11, 1, 1)\\
|
|
%(16, 'g')& z^3 + x^{52}z + x^{78}y& (13, 1, 1)\\
|
|
%(16, 'd')& z^3 + x^{26}z^2 + x^{78}y& (13, 1, 1)
|
|
\end{array}$$
|
|
|
|
|
|
\end{example}
|
|
|
|
|
|
|
|
|
|
\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}[scale=.75]
|
|
\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}[scale=.75]
|
|
\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
|
|
$$z^3 -3 c_1z + 2 c_2.$$
|
|
Therefore, $F$ is non degenerate with respect to its Newton polyhedron if and only if \textcolor{blue}{$(c_1)^3 \ne (c_2)^2$}.
|
|
|
|
|
|
\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]{};
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\filldraw[black] (28/3,1,8/3) circle (2pt) node[anchor=south west]{$M_1z$};
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\end{tikzpicture}
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\end{enumerate}
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\end{enumerate}
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\end{itemize}
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\newpage
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\section{Contibutions to the topological zeta function and the A'Campo monodromy zeta function of the trinomial forms}
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\begin{proposition}\label{formestrpi}
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The monodromy zeta function $\zeta_F(t)$ and the local topological zeta function $Z(F,s)_0$ of $F$ equal
|
|
\begin{align*}
|
|
(1 - t^3)^{-\alpha} \prod_{I} \zeta_{f_I}(t)^{\chi(E_I^\circ)}\\
|
|
\sum_I \chi(E^\circ_I) Z(f_I, \omega_I)_{o_I}
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|
\end{align*}
|
|
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 $E^\circ_I$.
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\end{proposition}
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\begin{proof} It follows from the formul{\ae} for $\zeta_f$ in terms of a resolution \cite{AC}, in terms of a modification \cite[Theorem 1]{GSLM97}, and the Stratum principle for topological zeta functions \cite[Stratum Principle 1.2]{ACNLM02}. Let us discuss the case of $\zeta_F$. The Milnor fiber $M_F$ intersects the $z$-axis in a set $A$ of three points that it is invariant under the action of the Milnor monodromy, i.e., the three points are cyclically permuted. Hence, we have
|
|
\[
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\zeta_F = {\zeta_F}_{|{M_F \setminus A}} \cdot {\zeta_F}_{|A} = {\zeta_F}_{|{M_F \setminus A}} (1 - t^3).
|
|
\]
|
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In order to consider compute the term ${\zeta_F}_{|{M_F \setminus A}} $ we consider the modification $\tilde{X} : =X \times \CC \rightarrow \CC^{3}$. Denote the pull back of the Milnor fibre under $\tilde{\pi}$ by $\tilde{M}$. Assume that $\pi$ is a composition of $\alpha$ point blow-ups. Denote by $B$ the intersection of $\tilde{M}$ with the $\alpha$ vertical planes thru the exceptional divisors of $\pi$. The set $B$ is invariant under the pull back of the Milnor monodromy. Since the exceptional locus of $\pi$ is a tree of $\alpha$ exceptional divisors isomorphic to $\mathbb{P}^1$, its Euler characteristic is $\alpha + 1$. Therefore,
|
|
$$\zeta_F= \zeta_{\pi^\ast F} = \zeta_{\pi^\ast F_{|_{\tilde{M} \setminus B}}} \zeta_{\pi^\ast F_{|_B}} = \zeta_{\pi^\ast F_{|_{\tilde{M} \setminus B}}} (1 - t^3)^{a+1}.$$
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Finally, we can express
|
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\[
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\zeta_{\pi^\ast F} = \prod_{I} \zeta_{f_I}(t)^{\chi(E_I^\circ)}.
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\qedhere
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\]
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% Each of the $\alpha$ factors of the form $(1-t^3)$ of $\zeta_f$ comes from the intersection of Milnor fiber of $F$ at $o$ and the line thru a center of blow-up of $\pi$. Each of these intersection sets consists of 3 points that are cyclically permuted by the monodromy.
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\end{proof}
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\begin{example} The resolution $\pi$ of Example \ref{extriplepole} consist of $\alpha=17$ blow-ups and the associated stratification has 35 strata. According to Proposition \ref{formestrpi} and using the simmetry of the resolution graph and the fact that the monodromy zeta function coincides for the 16 points of intersection of two different exceptional divisor, we have
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$$\zeta_F(t) = \frac{\zeta^{2}_{f_{2}} \zeta^{2}_{f_{8}} \zeta^{16}_{f_{1,7}} \zeta^{13}_{f_{1,16}} \zeta^{2}_{f_{7,29}} \zeta^{2}_{f_{7,33}} \zeta^{2}_{f_{9,31}} }
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{(1-t^3)^{17} \zeta^{13}_{f_{1}} \zeta^{2}_{f_{3}} \zeta^{4}_{f_{7}} \zeta^{2}_{f_{9}} }= \Phi^3_1 \Phi^{15}_2 \Phi_3 \Phi^2_6 \Phi^2_{13} \Phi^{15}_{26} \Phi^2_{78}$$
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where $\Phi_i$ is $i$-th the cyclotomic polymonial, and $Z(F,s)_0$ equals
|
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$$-13Z(f_1,s)_0 + \sum_{i=2,8,10,16}Z(f_i,s)_0 - \sum_{i=3, 9,11,17} Z(f_i,s)_0 - 2 \sum_{i=7,15}Z(f_i,s)_0 + \sum_{(i,j) \in S} Z(f_{i,j},s)_0=$$
|
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$$\frac{-1/26s^3 + 2749/6084s^2 + 1969/4056s + 1/8}{(s+1)(s+1/2)^3}$$
|
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where $S$ is the subset indexing the 0-dimensional strata of Example \ref{extriplepole} and we have omitted the symbols of differential forms associated to each strata for shortness sake.
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\end{example}
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\subsection{Trinomial forms for strata \texorpdfstring{$E^\circ_i$}{Eoi}}
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\subsubsection{First case $f= z^3 - x^{m_3}$ and $\omega = x^{\nu_1-1}dxdz$}
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|
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
|
|
$$ 3p + m_3r = 3m_3$$
|
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The corresponding candidate pole is
|
|
$$ 3m_3 s + 3\nu_1 +m_3$$
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The topological zeta function $Z_{{\rm top}}(f, \omega,s)_0$ equals
|
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%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}
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|
\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
|
|
$$J_{\check \tau_1}= \frac{3}{ \nu_1 (3m_3 s + 3\nu_1 + m_3)}.$$
|
|
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|
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|
\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
|
|
$$J_{\check \sigma}= \frac{(3,m_3)}{(3m_3 s + 3\nu_1 + m_3)}.$$
|
|
|
|
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\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}
|
|
Similarly we have
|
|
$$\zeta_f (t) = \frac{(1-t^3)(1-t^{m_3})}{(1-t^{3m_3})^{(m_3,3)}}$$
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\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
|
|
$$3m_2s + 2 \nu_1 + m_2 \quad {\rm and } \quad m_3s + \nu_1 + m_3-m_2$$
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|
|
|
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}
|
|
Similarly we have
|
|
$$\zeta_f (t) = \frac{(1-t^3)}{(1-t^{3m_2})^{(m_2,2)}}$$
|
|
|
|
|
|
We remark that this case it is equivalent to $f=z(z^2 - x^{m_2})$.
|
|
|
|
|
|
\subsubsection{Third case $f= z^3 - x^{m_1}z^2 - x^{m_3}$ and $\omega = x^{\nu_1-1}dxdz$ such that $m_3 > 3 m_1$}
|
|
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
|
|
\[
|
|
\frac(p(s)){(s+1)(3m_1s + \nu_1 + m_1)(2m_3s + 2\nu_1 + m_3-m_1)}
|
|
\]
|
|
where
|
|
\begin{align*}
|
|
p(s)&:=(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
|
|
\end{align*}
|
|
|
|
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}
|
|
|
|
|
|
Similarly we have
|
|
$$\zeta_f (t) = \frac{(t^{3m_1}-1)(t^{m_3}-1) }{(t-1)(t^3-1)} $$
|
|
$$\zeta_f (t) = \frac{(t^{3m_1}-1)(t^{2m_3}-1) }{(t-1)(t^3-1)(t^{m_3}-1)}$$
|
|
We remark that this case is equivalent to $f=(z-x^{m_1})(z^2 - x^{m_3-m_1})$
|
|
|
|
|
|
\subsection{Trinomial forms for strata \texorpdfstring{$E^\circ_{i,j}$}{Eoi}}
|
|
|
|
\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 $\frac{(2, 0, m_2)}{(m_2,2)}$, $\frac{(0,2,n_2)8}{(n_2,2)}$ 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
|
|
\begin{gather*}
|
|
\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)}
|
|
\end{gather*}
|
|
|
|
\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}
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|
\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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\section{Reductions}
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\subsection{First reduction: Elimination of the 0-dimensional strata due to the strict transforms of \texorpdfstring{$b_2$}{b2} and of \texorpdfstring{$b_3$}{b3}} The following Proposition shows that the o-dimensional strata associated to the strict transforms of $b_2$ and $b_3$ can eliminated. Each of these points can be added to the 1-dimensional strata satisfying that its clousure contains it, and the invariant is computed with the trinomial form of the 1-dimensional strata.
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|
|
|
\begin{proposition}\label{red1}Assume that $E_i$ is an exceptional divisor of the resolution $\pi$ with $\pi(E_i)=0$ (resp. $\pi(E_i)=\infty$) and that $F^0_j$ is an strict transform of $b_2$ (resp. $b_3$) such that $E_i \cap F_j \ne \emptyset$. Then
|
|
$$Z(F_i, \omega_i)_{o_i}=Z(F_{i,j}, \omega_{i,j})_{o_{i,j}} $$
|
|
\end{proposition}
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|
|
|
\begin{example} The proposition \ref{red1} can be applied to the strata
|
|
$$E^\circ_{9,31}=E_9 \cap F^0_{31}, \quad E^\circ_{17,32}=E_{17} \cap F^0_{32}, \quad E^\circ_{7,33}= E_7 \cap F^\infty_{33} \quad \hbox{ and } \quad E^\circ_{15,34}=E_{15} \cap F^\infty_{34}$$
|
|
of Example \ref{extriplepole}.
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|
|
|
For the strata due to the strict transforms $F^0_{31}$ and $F^0_{32}$ of $b_2$, note that $\mathcal{M}(E_9)= \mathcal{M}(E_{17})=0$. The corresponding trinomial forms and differential forms are
|
|
$$f_9=f_{17}=f_{9,31}=f_{17,32}=z^3 - x^{52}, \qquad \omega_9=\omega_{17}=\omega_{9,31}=\omega_{17,32}= x^{11}yzdxdydz,$$
|
|
From the formulas given in the previous section; we have
|
|
\[
|
|
Z(F_p, \omega_p)_{o_p}=Z(F_{p}, \omega_{p})_{o_{p}} = \frac{74s + 85}{11(s + 1)(156s + 85}\quad
|
|
\text{ for } j\in\{9, (9,31), 15, (17,32)\}.
|
|
\]
|
|
For the strata due to the strict transforms $F^\infty_{33}$ and $F^\infty_{34}$ of $b_3$, note that $\mathcal{M}(E_7)= \mathcal{M}(E_{15})=0$. The corresponding trinomial forms and differential forms are
|
|
$$f_7=f_{15}=z^3 - x^{78}, \quad f_{7,33}=f_{15,34}=z^3 - x^{52}z + x^{78}y,$$
|
|
and $$\omega_7=\omega_{15}=\omega_{7,33}=\omega_{15,34}= x^{13}yzdxdydz,$$
|
|
From the formulas given in the previous section; we have
|
|
\[
|
|
Z(F_7, \omega_7)_{o_7}\!=\!Z(F_{7,33}, \omega_{7,33})_{o_{7,33}}\!=\! Z(F_{15}, \omega_{15})_{o_{15}}=Z(F_{15,34}, \omega_{15,34})_{o_{15,34}} = \frac{1}{26s^2 +39 s + 13}.
|
|
\]
|
|
\end{example}
|
|
|
|
\subsection{Second reduction: Recursive elimination of exceptional \texorpdfstring{$(-1)$}{-1}-curve not intersecting the strict transform of the discrimiant}
|
|
|
|
After performing the First reduction, exceptional $(-1)$-curves that do not intersect the strict transform of the discriminant may appear. The next proposition shows that we can eliminante these divisors..
|
|
|
|
\begin{proposition}\label{red2}
|
|
\end{proposition}
|
|
|
|
\begin{example} The Proposition \ref{red2} can be applied twice to eliminate the curves $E_9$ and $E_8$. First, $E_9$ is an exceptional $(-1)$-curve that does not intersect the strict transform of the discriminant, and $E_8$ is an exceptional with $E_8^2=-2$ After blowing down $E_9$, the exceptional curve $E_8$ becomes a $(-1)$-curve. It does not intersect the strict transform of the discriminant.
|
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\end{example}
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|
\subsection{Third reduction: \texorpdfstring{$\QQ$}{Q}-resolution of the discriminant}
|
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|
$$\begin{array}{c|c|c|c|c}
|
|
i&f_i&\omega_i& \chi()& \zeta_{f_i}\\
|
|
\hline
|
|
1& z^3 + x^4z^2 + x^{13}& (2, 1, 1)& -13& ( \Phi_2^2 \Phi_4 \Phi_6 \Phi_{12} \Phi_{26})^{-1} \\
|
|
2,10& z^3 + x^{10}z + x^{13}& (3, 1, 1)&1& \Phi_1 (\Phi_{39})^{-1}\\
|
|
3,11& z^3 + x^{20}z + x^{26}& (5, 1, 1)& -1&
|
|
\Phi_1
|
|
({\Phi_6 \Phi_{39} \Phi_{78}})^{-1}\\
|
|
7,15& z^3 + \color{red}{x^{52}z} + x^{78}& (13, 1, 1)& -2&( \Phi_1 \Phi_2^2 \textcolor{red}{\Phi_3} \Phi_6^2 \Phi_{13}^2 \Phi_{26}^2 \Phi_{39}^2 \Phi_{78}^2 )^{-1}\\
|
|
8, 16& z^3 + x^{21}z + x^{26}& (6, 1, 1)& 1& \Phi_1 (\Phi_6 \Phi_{39} \Phi_{78})^{-1}\\
|
|
9,17& z^3 + x^{42}z + x^{52}& (11, 1, 1)& -1& \Phi_1 (\Phi_6 \Phi_{12} \Phi_{39} \Phi_{78} \Phi_{156})^{-1}\\
|
|
\end{array}$$
|
|
|
|
$$\begin{array}{c|c|c|c}
|
|
(i,j)&f_{i,j}&\omega_{i,j}\\
|
|
\hline
|
|
&&\\
|
|
(1, 7), (1, 15)& z^3 + x^4y^{26}z^2 + x^{13}y^{78}& (2, 13, 1)& \Phi_1 \textcolor{red}{\Phi_3} \\
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|
(2,3), (10, 11)& z^3 + x^{10}y^{20}z + x^{13}y^{26}& (3, 5, 1)& \Phi_1 \textcolor{red}{\Phi_3} \\
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|
(3, 9), (11, 17)& z^3 + x^{20}y^{42}z + x^{26}y^{52}& (5, 11, 1)& \Phi_1 \textcolor{red}{\Phi_3} \\
|
|
(3, 4), (11,12)& z^3 + x^{20}y^{28}z + x^{26}y^{39}& (5, 7, 1)& \Phi_1 \textcolor{red}{\Phi_3} \\
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|
(8, 9), (16,17)& z^3 + x^{21}y^{42}z + x^{26}y^{52}& (6, 11, 1)& \Phi_1 \textcolor{red}{\Phi_3} \\
|
|
(4, 5), (12,13)& z^3 + x^{28}y^{36}z + x^{39}y^{52}& (7, 9, 1)& \Phi_1 \textcolor{red}{\Phi_3} \\
|
|
(5, 6), (13,14)& z^3 + x^{36}y^{44}z + x^{52}y^{65}& (9, 11, 1)& \Phi_1 \textcolor{red}{\Phi_3} \\
|
|
(6, 7), (14,15)& z^3 + x^{44}y^{52}z + x^{65}y^{78}& (11, 13, 1)& \Phi_1 \textcolor{red}{\Phi_3}\\
|
|
(1, i), i=16, \dots, 28& z^3 + x^4z^2 + x^{13}y& (2, 1, 1)& (\Phi_2 \Phi_4 \Phi_6 \Phi_{12})^{-1} \\
|
|
(9, 31), (17, 32)& z^3 + x^{42}yz + x^{52}& (11, 1, 1)&\Phi_1 (\Phi_6 \Phi_{12} \Phi_{39} \Phi_{78} \Phi_{156})^{-1} \\
|
|
(7, 33), (15, 34)& z^3 + x^{52}z + x^{78}y& (13, 1, 1)& (\Phi_1 \Phi_2^2 \textcolor{red}{\Phi_3} \Phi_6^2 \Phi_{13}^2 \Phi_{26}^2 \Phi_{39}^2 \Phi_{78}^2 )^{-1} \\
|
|
(7, 29) (15, 30)& z^3 + x^{26}z^2 + x^{78}y& (13, 1, 1)& (\Phi_2 \Phi_6 \Phi_{13} \Phi_{26} \Phi_{39} \Phi_{78})^{-1} \\
|
|
%(12, 'f')& z^3 + x^{42}yz + x^{52}& (11, 1, 1)\\
|
|
%(16, 'g')& z^3 + x^{52}z + x^{78}y& (13, 1, 1)\\
|
|
%(16, 'd')& z^3 + x^{26}z^2 + x^{78}y& (13, 1, 1)
|
|
\end{array}$$
|
|
|
|
|
|
\begin{align}
|
|
\Phi_1& =& 17 - 13 + 8 - 16 - 2 &=& 18\\
|
|
\Phi_2 &=& 26 + 8 -13 -2 -4 &=& 15\\
|
|
\Phi_3 &=& -17 + 4 + 16 - 2 &= &1\\
|
|
\Phi_4 &=& 13 -13 &=& 0\\
|
|
\Phi_6 &=& 13 +2 + 8 -2 + 2 -13 -2-2-4 &=& 2\\
|
|
\Phi_{12} &=& 13 + 2 -13 -2 &=& 0\\
|
|
\Phi_{13} &= &8 -2 - 4& =& 2\\
|
|
\Phi_{26} &= &13 + 8 -2 -4 & =& 15\\
|
|
\Phi_{39} &= &-2 + 2 - 8 -2 + 2 + 2 + 2 + 4& =& 0 \\
|
|
\Phi_{78} &= &2 + 8 -2 + 2 - 2 - 2 - 4& =& 2\\
|
|
\Phi_{156} &= & 2 - 2& = &0
|
|
\end{align}
|
|
|
|
|
|
$$\Phi_1^{18}\Phi_2^{15}\Phi_3\Phi_6^2\Phi_{13}^2\Phi_{26}^{15}\Phi_{78}^2$$
|
|
|
|
|
|
|
|
\section{On generic projections}
|
|
|
|
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|
|
|
|
|
|
% In general, the $z$-axis is not a generic direction of projection for the embedded topological class of the surface given by the Weierstrass polynomial $F=z^3-2f(x,y)z+3g(x,y)$. Therefore the discriminant of $F$ with respect to $z$ given by $f^2+g^3$ is not generic inside the same equivalence class. Let us describe a situation where it is possible to describe a generic topological deformation $F_t=z^3-2f_t(x,y)z+3g_t(x,y)$ of $F$ whose discriminant $f_t^3+g_t^2$ is generic too.
|
|
|
|
|
|
%\begin{proposition}
|
|
% Assume that the functions $f$ and $g$ admit factorizations of the form $f= x^{2a} f_1$ and $g=x^{3a}g_1$. Moreover, assume that the the multiplicity of the polynomial $F_1(x, y, z)= F(x, y, x^az_1)/x^{3a}$ at the origin is equal or less than $2$. Then, any pair of deformations $f_t$ and $g_t$ of the functions $f$ and $g$ gives a topological deformation $F_t$ of $F$ and a topological deformation $f_t^3+g_t^2$ of $f_t^3+g_t^2$. Moreover, if $f_t^3+g_t^2$ is generic, then $F_t$ is generic.
|
|
%\end{proposition}
|
|
|
|
\begin{example} Consider the Weierstrass polynomial $F=z^3-3f(x,y)z+2g(x,y)$ with $f=y^2 - x^7 + x^5y$ and $g= f^2 + (y + x^2)^6$. It defines a germ of isolated surface singularity with Milnor number $\mu=41$. The discriminant of $F$ is given by
|
|
$$f^3-g^2 = f^3 - f^4 - 2f^2(y+x^2)^6 +(y+x^2)^{12}$$
|
|
and it has the topological type given by three $(2,7)$-cusp with pairwise intersection multiplicity $16$. Its dual resolution graph is pictured (left) bellow.
|
|
|
|
\
|
|
|
|
|
|
\begin{tikzpicture}
|
|
\draw[gray, thick] (6,-2) -- (10,-2);
|
|
\draw[gray, thick] (9,-2) -- (9,0);
|
|
\filldraw[black] (9,-1) circle (2pt) node[anchor=west]{$E_6$};
|
|
\filldraw[black] (9,0) circle (2pt) node[anchor=west]{$E_7$};
|
|
\filldraw[black] (6,-2) circle (2pt) node[anchor=north]{$E_1$};
|
|
\filldraw[black] (7,-2) circle (2pt) node[anchor=north]{$E_2$};
|
|
\filldraw[black] (8,-2) circle (2pt) node[anchor=north]{$E_3$};
|
|
\filldraw[black] (9,-2) circle (2pt) node[anchor=north]{$E_5$};
|
|
\filldraw[black] (10,-2) circle (2pt) node[anchor=north]{$E_4$};
|
|
|
|
\node (Z) at (9, 0) {};
|
|
\node (A) at (9, 1) {$d$};
|
|
\node (B) at (8, 1) {$d$};
|
|
\node (C) at (10, 1) {$d$};
|
|
\draw[->] (Z) edge (A);
|
|
\draw[->] (Z) edge (B);
|
|
\draw[->] (Z) edge (C);
|
|
|
|
|
|
|
|
|
|
\draw[gray, thick] (16,-2) -- (20,-2);
|
|
\draw[gray, thick] (19,-2) -- (19,1);
|
|
\filldraw[black] (19,-1) circle (2pt) node[anchor=west]{$E_6$};
|
|
\filldraw[black] (19,0) circle (2pt) node[anchor=west]{$E_8$};
|
|
\filldraw[black] (19,1) circle (2pt) node[anchor=west]{$E_7$};
|
|
\filldraw[black] (16,-2) circle (2pt) node[anchor=north]{$E_1$};
|
|
\filldraw[black] (17,-2) circle (2pt) node[anchor=north]{$E_2$};
|
|
\filldraw[black] (18,-2) circle (2pt) node[anchor=north]{$E_3$};
|
|
\filldraw[black] (19,-2) circle (2pt) node[anchor=north]{$E_5$};
|
|
\filldraw[black] (20,-2) circle (2pt) node[anchor=north]{$E_4$};
|
|
|
|
\node (W) at (19, -2) {};
|
|
\node (D) at (18, -1) {$d_t$};
|
|
\draw[->] (W) edge (D);
|
|
\node (Y) at (19, 0) {};
|
|
\node (E) at (18, 0) {$d_t$};
|
|
\draw[->] (Y) edge (E);
|
|
\end{tikzpicture}
|
|
|
|
\
|
|
|
|
The polinomial $F_t=z^3 -3f_t(x,y)z - 2 g_t(x,y)$ with
|
|
$$f_t= f + t^2y^2 \qquad \hbox{ and } \qquad g_t = g + \frac{3}{2}tyf + t^3y^3$$ defines a $\mu$-constant topological deformation of $F$ whose discriminant $f_t^3-g_t^2$ has the following embedded topological type. The discriminant of $F_t$ has two branches. One is a $(2,7)$-cusp and the other branch has two Puiseux pairs, namely, $(2,7)$ and $(2,3)$. Their intersection multiplicity is $28$. The dual resolution graph of the discriminant is pictured (right) above.
|
|
|
|
The topological zeta function and the monodromy zeta function of $F$ and $F_t$ are given by
|
|
$$ \frac{-41}{168} \frac{s^2 - 283/123s - 304/123}{(s + 16/21) (s + 19/24) (s + 1) } \quad \hbox{ and } \quad
|
|
\Phi_1 \Phi_2 \Phi_3 \Phi_4 \Phi_6 \Phi_7 \Phi_8 \Phi_{12} \Phi_{21} \Phi_{24}$$
|
|
In particular, they both have Milnor number equal to 41. This can be computed from embedded resolutions of the meromorphic functions $\mathcal{M}$ and $\mathcal{M}_t$ with the following numerical data
|
|
$$\begin{array}{c|ccccccccc}
|
|
& E_1& E_2& E_3& E_4& E_5& E_6& E_7\\
|
|
\hline
|
|
f& 2& 4& 6& 7& 14& 15& 16\\
|
|
g& 4& 8& 12& 12& 24& 24& 24\\
|
|
d& 6&12& 18& 21& 42& 45& 48
|
|
\end{array}
|
|
\qquad \begin{array}{c|ccccccccc}
|
|
& E_1& E_2& E_3& E_4& E_5& E_6& E_7& E_8\\
|
|
\hline
|
|
f_t& 2& 4& 6& 7& 14& 14& 14& 28\\
|
|
g_t& 3& 6& 9& 11& 21& 21& 21& 42\\
|
|
d_t& 6&12& 18& 21& 42& 44& 45& 90
|
|
\end{array}$$
|
|
|
|
\
|
|
|
|
\noindent and dual resolution graphs
|
|
|
|
\
|
|
|
|
\begin{tikzpicture}
|
|
\draw[gray, thick] (6,-2) -- (10,-2);
|
|
\draw[gray, thick] (9,-2) -- (9,0);
|
|
\filldraw[black] (9,-1) circle (2pt) node[anchor=west]{$E_6$};
|
|
\filldraw[black] (9,0) circle (2pt) node[anchor=west]{$E_7$};
|
|
\filldraw[black] (6,-2) circle (2pt) node[anchor=north]{$E_1$};
|
|
\filldraw[black] (7,-2) circle (2pt) node[anchor=north]{$E_2$};
|
|
\filldraw[black] (8,-2) circle (2pt) node[anchor=north]{$E_3$};
|
|
\filldraw[black] (9,-2) circle (2pt) node[anchor=north]{$E_5$};
|
|
\filldraw[black] (10,-2) circle (2pt) node[anchor=north]{$E_4$};
|
|
|
|
\node (Z) at (9, 0) {};
|
|
\node (A) at (9, 1) {$d$};
|
|
\node (B) at (8, 1) {$d$};
|
|
\node (C) at (10, 1) {$d$};
|
|
\node (F) at (8, 0) {$f$};
|
|
|
|
\draw[->] (Z) edge (A);
|
|
\draw[->] (Z) edge (B);
|
|
\draw[->] (Z) edge (C);
|
|
\draw[->] (Z) edge (F);
|
|
|
|
|
|
\node (V) at (8, -2) {};
|
|
\node (G1) at (7.25, -1) {$g$};
|
|
\node (G2) at (7.5, -1) {$g$};
|
|
\node (G3) at (8.25, -1) {$g$};
|
|
\node (G4) at (8.5, -1) {$g$};
|
|
|
|
|
|
\draw[->] (V) edge (G1);
|
|
\draw[->] (V) edge (G2);
|
|
\draw[->] (V) edge (G3);
|
|
\draw[->] (V) edge (G4);
|
|
|
|
|
|
\begin{scope}[xshift=-3cm]
|
|
\draw[gray, thick] (16,-2) -- (20,-2);
|
|
\draw[gray, thick] (19,-2) -- (19,1);
|
|
\filldraw[black] (19,-1) circle (2pt) node[anchor=west]{$E_6$};
|
|
\filldraw[black] (19,0) circle (2pt) node[anchor=west]{$E_8$};
|
|
\filldraw[black] (19,1) circle (2pt) node[anchor=west]{$E_7$};
|
|
\filldraw[black] (16,-2) circle (2pt) node[anchor=north]{$E_1$};
|
|
\filldraw[black] (17,-2) circle (2pt) node[anchor=north]{$E_2$};
|
|
\filldraw[black] (18,-2) circle (2pt) node[anchor=north]{$E_3$};
|
|
\filldraw[black] (19,-2) circle (2pt) node[anchor=north]{$E_5$};
|
|
\filldraw[black] (20,-2) circle (2pt) node[anchor=north]{$E_4$};
|
|
|
|
\node (W) at (19, -2) {};
|
|
\node (D) at (18, -1) {$d_t$};
|
|
\draw[->] (W) edge (D);
|
|
\node (Y) at (19, 0) {};
|
|
\node (E) at (18, 0) {$d_t$};
|
|
\draw[->] (Y) edge (E);
|
|
|
|
|
|
\node (W) at (19, -2) {};
|
|
\node (F) at (17, -1) {$f_t$};
|
|
\node (G) at (20, -1) {$g_t$};
|
|
\draw[->] (W) edge (F);
|
|
\draw[->] (W) edge (G);
|
|
|
|
|
|
\node (V) at (20, -2) {};
|
|
\node (GG) at (21, -2) {$g_t$};
|
|
\draw[->] (V) edge (GG);
|
|
\end{scope}
|
|
\end{tikzpicture}
|
|
|
|
|
|
|
|
|
|
In order to that there check that both $F$ and $F_t$ define the same embedded topological type, notice that
|
|
$$F_t(x,y,z+yt)=F(x,y,t) + 3tyz^2$$
|
|
Moreover, in the left-hand side change $z \mapsto z+ty$ is an isomorphism that does not change the discriminant with respect to $z$, nor the Milnor number. Finally, in the right-hand side we have a $\mu$-constant deformation of isolated hypersurface singularities of the form $f(x) + tg(x)$, that preserves the embedded topological type according to \cite{par:99}.
|
|
|
|
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Finally, it is worth mentioning that the topological type of the discriminant of $F_t$ can be realizable as $f_1^3-g_1^2$ for another pair of functions $(f_1,g_1)$. The corresponding function $F_1= z^3-3f_1z + 2g_1$ does not belong to the embedded topological type equivalence class of $F$ and $F_t$ because their Milnor number and monodromy zeta function do not coincide. The topological and monodromy zeta function of $F_1$ are
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$$ \frac{-5}{21} \frac{s - 16/5}{(s + 16/21)(s + 1)} \qquad \hbox{ and, } \qquad \Phi_1 \Phi_3 \Phi_7^2 \Phi_{21}^2 $$
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In particular, the Milnor number of $F_1$ is $38$.
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This can be computed from embedded resolution of the meromorphic functions $\mathcal{M}_1$ with the following numerical data
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$$\begin{array}{c|ccccccccc}
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& E_1& E_2& E_3& E_4& E_5& E_6& E_7& E_8\\
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\hline
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f_t& 2& 4& 6& 7& 14& 15& 15& 30\\
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g_t& 3& 6& 9& 11& 21& 22& 23& 45\\
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d_t& 6&12& 18& 21& 42& 44& 45& 90
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\end{array}$$
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and dual resolution graph.
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\
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\begin{tikzpicture}
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\draw[gray, thick] (16,-2) -- (20,-2);
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\draw[gray, thick] (19,-2) -- (19,1);
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\filldraw[black] (19,-1) circle (2pt) node[anchor=west]{$E_6$};
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\filldraw[black] (19,0) circle (2pt) node[anchor=west]{$E_8$};
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\filldraw[black] (19,1) circle (2pt) node[anchor=west]{$E_7$};
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\filldraw[black] (16,-2) circle (2pt) node[anchor=north]{$E_1$};
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\filldraw[black] (17,-2) circle (2pt) node[anchor=north]{$E_2$};
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\filldraw[black] (18,-2) circle (2pt) node[anchor=north]{$E_3$};
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\filldraw[black] (19,-2) circle (2pt) node[anchor=north]{$E_5$};
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\filldraw[black] (20,-2) circle (2pt) node[anchor=north]{$E_4$};
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\node (W) at (19, -2) {};
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\node (D) at (18, -1) {$d_t$};
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\draw[->] (W) edge (D);
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\node (Y) at (19, 0) {};
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\node (E) at (18, 0) {$d_t$};
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\draw[->] (Y) edge (E);
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\node (T) at (19, -1) {};
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\node (U) at (19, 1) {};
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\node (FT) at (18, -0.5) {$f_1$};
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\node (GT1) at (18, 1) {$g_1$};
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\draw[->] (T) edge (FT);
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\draw[->] (U) edge (GT1);
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\node (S) at (20, -2) {};
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\node (GT2) at (21, -2) {$g_1$};
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\draw[->] (S) edge (GT2);
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\end{tikzpicture}
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\
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\end{example}
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\bibliographystyle{amsplain}
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\bibliography{biblio}
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\end{document}
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