\begin{info} \section{Results for the cubic $L_4$} We consider the following cubic \begin{equation}\tag{$L_4$}X_0^3+X_1^3+2X_2^3+2X_3^3=0 \end{equation} \subsection{The Galois group} The splitting Galois group for the cubic is given by \[G=\langle c^1,\tau_2^{1}\tau_3^{1}\rangle. \] \subsection{The Picard group}The orbits of the 27 lines are given by \[\begin{split} O_{0}&=\{L(0,0)\},\\ O_{1}&=\{L(1,0),L(2,0)\},\\ O_{2}&=\{L(0,1),L(0,2)\},\\ O_{3}&=\{L(1,1),L(2,2)\},\\ O_{4}&=\{L(2,1),L(1,2)\},\\ O_{5}&=\{M(0,0),M(0,1),M(1,0),M(2,0),M(1,1),M(2,1)\},\\ O_{6}&=\{M(0,2),M(1,2),M(2,2)\},\\ O_{7}&=\{N(0,0),N(0,2),N(1,0),N(2,0),N(1,2),N(2,2)\},\\ O_{8}&=\{N(0,1),N(1,1),N(2,1)\}. \end{split}\] In order to describe the Picard group, we consider the cubic over $\overline {Q}$ as the projective plane blown up in six points so that the $27$ lines are given by \begin{align*} E_1&=L(0,0),&E_2&=L(1,0),&E_3&=L(2,0),\\ E_4&=M(0,0),&E_5&=M(1,0),&E_6&=M(2,0),\\ Q_1&=L(1,1),&Q_2&=L(2,1),&Q_3&=L(0,1),\\ Q_4&=M(2,2),&Q_5&=M(0,2),&Q_6&=M(1,2),\\ R_{1,2}&=L(1,2),&R_{2,3}&=L(2,2),&R_{3,1}&=L(0,2)\\ R_{4,5}&=M(0,1),&R_{5,6}&=M(1,1),&R_{6,4}&=M(2,1)\\ R_{1,4}&=N(2,1),&R_{1,5}&=N(0,1),&R_{1,6}&=N(1,1)\\ R_{2,4}&=N(0,2),&R_{2,5}&=N(1,2),&R_{2,6}&=N(2,2)\\ R_{3,4}&=N(1,0),&R_{3,5}&=N(2,0),&R_{3,6}&=N(0,0). \end{align*} where the $E_i$ are the exceptional lines, the $R_{i,j}$ is the strict lifting of the line $(P_iP_j)$ and $Q_i$ the strict lifting of the conic going through all points except $P_i$. In the basis given by the lifting $\Lambda$ of line not going through any of the six points blown up and by the $6$ exceptional lines $E_i$, the Picard group of the variety over $\mathbf Q$ is given by the matrix \[\begin{pmatrix} 0&0&-3\\ 1&0&0\\ 0&1&0\\ 0&1&0\\ 0&0&1\\ 0&0&1\\ 0&0&1\\ \end{pmatrix} \] The effective cone is generated by \[\begin{split} [O_{1}]&= e_{0},\\ [O_{2}]&= e_{1},\\ [O_{3}]&=-2e_{0}-e_{1}-e_{2},\\ [O_{4}]&=-2e_{1}-e_{2},\\ [O_{5}]&=-2e_{0}-e_{1}-e_{2},\\ [O_{6}]&=-e_{2},\\ [O_{7}]&=-3e_{0}-3e_{1}-2e_{2},\\ [O_{8}]&=-3e_{1}-2e_{2},\\ [O_{9}]&=-3e_{0}-e_{2}.\\ \noalign{and the canonical sheaf is given by}\omega_V^{-1}&=-e_{0}-e_{1}-e_{2}. \end{split}\] The corresponding polyhedron may be described as \[\begin{split} X_1 &> 0,\\ X_2 &> 0,\\ X_1-X_2 &> -1,\\ -X_1-X_2 &> -2,\\ -2X_1+X_2 &> -1. \end{split}\] with $ 1$ as coefficient. Thus the constant $\alpha(V)$ is $ 1$. \subsection{The bad places} \par At the bad place $2$ We use the following reduced form \[X_0^3+X_1^3+2X_2^3+2X_3^3=0 \] of type $(1,a,bp,cp)$. Thus the density at $p$ is $ 3/4$ and the zeta factor at $p$ is $ 9/16$ \par At the bad place $3$ We use the following reduced form \[X_0^3+X_1^3+2X_2^3+2X_3^3=0 .\] And a direct computation gives that the density at $p$ is $ 4/3$ and the zeta factor at $p$ is $ 4/9$ \subsection{Experimental curve} The following graph represents the numerical data \centerline{% {\setlength{\unitlength}{0.240900pt} \begin{picture}(600,900)(0,0) \put(50.000,88.000){\rule[-0.200pt]{120.450pt}{0.400pt}}\put(50.000,88.000){\rule[-0.200pt]{0.400pt}{192.720pt}}\put(50.000,88.000){\rule[-0.200pt]{0.400pt}{4.818pt}} \put(50.000,47.0){\makebox(0,0)[c]{0}} \put(91.692,88.000){\rule[-0.200pt]{0.400pt}{4.818pt}} \put(91.692,47.0){\makebox(0,0)[c]{1}} \put(133.385,88.000){\rule[-0.200pt]{0.400pt}{4.818pt}} \put(133.385,47.0){\makebox(0,0)[c]{2}} \put(175.077,88.000){\rule[-0.200pt]{0.400pt}{4.818pt}} \put(175.077,47.0){\makebox(0,0)[c]{3}} \put(216.769,88.000){\rule[-0.200pt]{0.400pt}{4.818pt}} 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