From 8a515269cdbdde5eac4d3e1453407d7f41816a52 Mon Sep 17 00:00:00 2001 From: "m.gonzalez" Date: Thu, 26 Sep 2024 17:41:21 +0200 Subject: [PATCH] actualizar Mult3.tex --- Mult3.tex | 92 +++++++++++++++++++++++++++++++++++++++++++++++++------ 1 file changed, 82 insertions(+), 10 deletions(-) diff --git a/Mult3.tex b/Mult3.tex index 08a324c..6fb0838 100644 --- a/Mult3.tex +++ b/Mult3.tex @@ -401,7 +401,7 @@ The singular point is isolated if and only if the discriminant $D_zF$ is reduced The monic Weierstrass polynomial $z^3 -3b_2z + 2b_3$ is non degenerate unless $(c_2M_2)^3 = (c_3M_3)^2$. \end{proposition} -\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 $c_3=(\frac{-b}{2})^3$.} +\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 $c_3=(\frac{-b}{2})^3$.} \section{Jung´s method for power series of multiplicity three } @@ -445,7 +445,7 @@ $N^\Delta_j - 3 N^0_j = N^\Delta_j - 2 N^\infty_j$ is called the \emph{excess of \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 30 blow-ups. The following figure and table give the dual resolution graph of $G$ and its associated numerical data. +$b_3= (y^3+x^4)(x^3+y^4)$. 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. @@ -587,6 +587,9 @@ N^g_i& {\bf12}& {\bf 13}& {\bf 26}& {\bf 39}& {\bf 52}& {\bf 65}& {\bf 78}& {\bf N^\Delta_i& 25& {\bf 26}& {\bf 52}& {\bf 78}& {\bf 104}& {\bf 130}& {\bf 156}& {\bf 52}&{\bf 104}\\ \hline \hline +\pi()& 1& 0& 0& 0& 0& 0& \mathbb{P}^1& 0&0\\ +\hline +\hline \nu_i&2&3&5&7&9&11&13&6&11\\ \chi()&-13&1&-1&0&0&0&-2&1&-1\\ E_i^2& -11& -2& -4& -2& -2& -2& -1&-2&-1\\ @@ -676,7 +679,9 @@ in the case $I= \{i, j\}$. \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} -$$\begin{array}{c|c|c|c} + + + $$\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\\ @@ -688,7 +693,6 @@ i&f_i&\omega_i& \chi()\\ \end{array}$$ - $$\begin{array}{c|c|c} (i,j)&f_{i,j}&\omega_{i,j}\\ \hline @@ -1044,18 +1048,22 @@ Therefore, $F$ is non degenerate with respect to its Newton polyhedron if and on \begin{proposition}\label{formestrpi} The monodromy zeta function $\zeta_F(t)$ and the local topological zeta function $Z(F,s)_0$ of $F$ equal -$$(1 - t^3)^{\alpha} \prod_{I} \zeta_{f_I}(t)^{-\chi(E_I^\circ)}$$ +$$(1 - t^3)^{-\alpha} \prod_{I} \zeta_{f_I}(t)^{\chi(E_I^\circ)}$$ $$\sum_I \chi(E^\circ_I) Z(f_I, \omega_I)_{o_I}$$ 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{proof} It follows from the the formula for $\zeta_f$ in terms of a partial resolution \cite[Theorem 1]{GSLM97} and the Stratum principle for topological zeta functions \cite[Stratum Principle 1.2]{ABCNLM02}. 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. + \begin{proof} It follows from the formulae 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]{ABCNLM02}. 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 $$\zeta_F = \zeta_F_{|_{M_F \setminus A}} \cdot \zeta_F_{|_A} = \zeta_F_{|_{M_F \setminus A}} (1 - t^3) .$$ + 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, +$$\zeta_F= \zeta_{\pi^\ast F} = \zeta_{\pi^\ast F_{|_{\tilde{M} \setminus B}}} \zeta_{\pi^\ast F_{|_B}} = \zeta_{\pi^\ast F_{|_{\tilde{M} \setminus B}}} (1 - t^3)^{a+1}.$$ + Finally, we can express $\zeta_{\pi^\ast F} = \prod_{I} \zeta_{f_I}(t)^{\chi(E_I^\circ)}.$$ + % Each of the $\alpha$ factors of the form $(1-t^3)$ of $\zeta_f$ comes from the intersection of Milnor fiber of $F$ at $o$ and the line thru a center of blow-up of $\pi$. Each of these intersection sets consists of 3 points that are cyclically permuted by the monodromy. \end{proof} -\begin{example} The resolution $\pi$ of Example \ref{extriplepole} consist of $\alpha=15$ blow-ups and the associated stratification has 35 strata. According to Proposition \ref{formestrpi} we have -$$\zeta_F(t) = \frac{(1-t^3)^{15} \zeta^{}_{f_{2}} \zeta^{2}_{f_{4}} \zeta^{2}_{f_{5}} \zeta^{2}_{f_{6}} \zeta^{}_{f_{8}} \zeta^{}_{f_{10}} \zeta^{2}_{f_{12}} \zeta^{2}_{f_{13}} \zeta^{2}_{f_{14}} \zeta^{}_{f_{16}} } -{\zeta^{13}_{f_{1}} \zeta^{}_{f_{3}} \zeta^{4}_{f_{7}} \zeta^{}_{f_{9}} \zeta^{}_{f_{11}} \zeta^{4}_{f_{15}} \zeta^{}_{f_{17}} }= \Phi^3_1 \Phi^{15}_2 \Phi_3 \Phi^2_6 \Phi^2_{13} \Phi^{15}_{26} \Phi^2_{78}$$ +\begin{example} The resolution $\pi$ of Example \ref{extriplepole} consist of $\alpha=17$ blow-ups and the associated stratification has 35 strata. According to Proposition \ref{formestrpi} and using the simmetry of the resolution graph and the fact that the monodromy zeta function coincides for the 16 points of intersection of two different exceptional divisor, we have +$$\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}} } +{(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}$$ where $\Phi_i$ is $i$-th the cyclotomic polymonial, and $Z(F,s)_0$ equals $$-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=$$ $$\frac{-1/26s^3 + 2749/6084s^2 + 1969/4056s + 1/8}{(s+1)(s+1/2)^3}$$ @@ -1304,9 +1312,69 @@ $$Z(F_7, \omega_7)_{o_7}=Z(F_{7,33}, \omega_{7,33})_{o_{7,33}}= Z(F_{15}, \omega \subsection{Third reduction: $\QQ$-resolution of the discriminant} - + + + + + +$$\begin{array}{c|c|c|c|c} +i&f_i&\omega_i& \chi()& \zeta_{f_i}\\ +\hline +1& z^3 + x^4z^2 + x^{13}& (2, 1, 1)& -13& ( \Phi_2^2 \Phi_4 \Phi_6 \Phi_{12} \Phi_{26})^{-1} \\ +2,10& z^3 + x^{10}z + x^{13}& (3, 1, 1)&1& \Phi_1 (\Phi_{39})^{-1}\\ +3,11& z^3 + x^{20}z + x^{26}& (5, 1, 1)& -1& \Phi_1 ({\Phi_6 \Phi_{39} \Phi_{78})^{-1}\\ +7,15& z^3 + \color{red}{x^{52}z} + x^{78}& (13, 1, 1)& -2&( \Phi_1 \Phi_2^2 \textcolor{red}{\Phi_3} \Phi_6^2 \Phi_{13}^2 \Phi_{26}^2 \Phi_{39}^2 \Phi_{78}^2 )^{-1}\\ +8, 16& z^3 + x^{21}z + x^{26}& (6, 1, 1)& 1& \Phi_1 (\Phi_6 \Phi_{39} \Phi_{78})^{-1}\\ + 9,17& z^3 + x^{42}z + x^{52}& (11, 1, 1)& -1& \Phi_1 (\Phi_6 \Phi_{12} \Phi_{39} \Phi_{78} \Phi_{156})^{-1}\\ +\end{array}$$ + +$$\begin{array}{c|c|c|c} +(i,j)&f_{i,j}&\omega_{i,j}\\ +\hline +&&\\ +(1, 7), (1, 15)& z^3 + x^4y^{26}z^2 + x^{13}y^{78}& (2, 13, 1)& \Phi_1 \textcolor{red}{\Phi_3} \\ +(2,3), (10, 11)& z^3 + x^{10}y^{20}z + x^{13}y^{26}& (3, 5, 1)& \Phi_1 \textcolor{red}{\Phi_3} \\ +(3, 9), (11, 17)& z^3 + x^{20}y^{42}z + x^{26}y^{52}& (5, 11, 1)& \Phi_1 \textcolor{red}{\Phi_3} \\ +(3, 4), (11,12)& z^3 + x^{20}y^{28}z + x^{26}y^{39}& (5, 7, 1)& \Phi_1 \textcolor{red}{\Phi_3} \\ +(8, 9), (16,17)& z^3 + x^{21}y^{42}z + x^{26}y^{52}& (6, 11, 1)& \Phi_1 \textcolor{red}{\Phi_3} \\ +(4, 5), (12,13)& z^3 + x^{28}y^{36}z + x^{39}y^{52}& (7, 9, 1)& \Phi_1 \textcolor{red}{\Phi_3} \\ +(5, 6), (13,14)& z^3 + x^{36}y^{44}z + x^{52}y^{65}& (9, 11, 1)& \Phi_1 \textcolor{red}{\Phi_3} \\ +(6, 7), (14,15)& z^3 + x^{44}y^{52}z + x^{65}y^{78}& (11, 13, 1)& \Phi_1 \textcolor{red}{\Phi_3}\\ +(1, i), i=16, \dots, 28& z^3 + x^4z^2 + x^{13}y& (2, 1, 1)& (\Phi_2 \Phi_4 \Phi_6 \Phi_{12})^{-1} \\ +(9, 31), (17, 32)& z^3 + x^{42}yz + x^{52}& (11, 1, 1)&\Phi_1 (\Phi_6 \Phi_{12} \Phi_{39} \Phi_{78} \Phi_{156})^{-1} \\ +(7, 33), (15, 34)& z^3 + x^{52}z + x^{78}y& (13, 1, 1)& (\Phi_1 \Phi_2^2 \textcolor{red}{\Phi_3} \Phi_6^2 \Phi_{13}^2 \Phi_{26}^2 \Phi_{39}^2 \Phi_{78}^2 )^{-1} \\ +(7, 29) (15, 30)& z^3 + x^{26}z^2 + x^{78}y& (13, 1, 1)& (\Phi_2 \Phi_6 \Phi_{13} \Phi_{26} \Phi_{39} \Phi_{78})^{-1} \\ +%(12, 'f')& z^3 + x^{42}yz + x^{52}& (11, 1, 1)\\ +%(16, 'g')& z^3 + x^{52}z + x^{78}y& (13, 1, 1)\\ +%(16, 'd')& z^3 + x^{26}z^2 + x^{78}y& (13, 1, 1) +\end{array}$$ + + +$$\begin{align} +\Phi_1& =& 17 - 13 + 8 - 16 - 2 &=& 18\\ +\Phi_2 &=& 26 + 8 -13 -2 -4 &=& 15\\ +\Phi_3 &=& -17 + 4 + 16 - 2 &= &1\\ +\Phi_4 &=& 13 -13 &=& 0\\ +\Phi_6 &=& 13 +2 + 8 -2 + 2 -13 -2-2-4 &=& 2\\ +\Phi_{12} &=& 13 + 2 -13 -2 &=& 0\\ +\Phi_{13} &= &8 -2 - 4& =& 2\\ +\Phi_{26} &= &13 + 8 -2 -4 & =& 15\\ +\Phi_{39} &= &-2 + 2 - 8 -2 + 2 + 2 + 2 + 4& =& 0 \\ +\Phi_{78} &= &2 + 8 -2 + 2 - 2 - 2 - 4& =& 2\\ +\Phi_{156} &= & 2 - 2& = &0 +\end{align}$$ + + +$$\Phi_1^{18}\Phi_2^{15}\Phi_3\Phi_6^2\Phi_{13}^2\Phi_{26}^{15}\Phi_{78}^2$$ + + + \section{On generic projections} + + + + % 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. @@ -1533,6 +1601,10 @@ and dual resolution graph. \end{example} + + + + \bibliographystyle{amsplain} \bibliography{biblio}