Automorphism group of the configuration of lines on a cubic surface and quadratic transformations

Solution 1:

Let us denote by $(\;\,.\;)$ the intersection pairing on your cubic surface $X$ and by $\alpha$ the divisor class $e_0-e_1-e_2-e_3$, then the quadratic transformation corresponds to

$$s_{\alpha} : \left\{\begin{array}{ccc} \mathrm{Pic\;} X & \longrightarrow & \mathrm{Pic\;} X \\ v & \longmapsto & v+(v.\alpha)\,\alpha \end{array}\right. $$

which is one of the generators of the Weyl group of $E_6$. It's the reflection through the hyperplane orthogonal to $\alpha$ in $\mathrm{Pic}\,X$, since $\alpha.\alpha=-2$.

Thus, you can write down explicitly its matrix in the basis $(h,e_1,\dots,e_6)$, where $h$ is the total transform in $X$ of a general line of $\mathbb P^2$ :

\begin{bmatrix} 2 & 1 & 1 & 1 & & & \\ -1 & 0 & -1 & -1 & & & \\ -1 & -1 & 0 & -1 & & & \\ -1 & -1& -1 & 0 & & & \\ & & & &1 & 0 & 0\\ & & & & 0&1 &0\\ & & & & 0& 0&1\\ \end{bmatrix}

Since $l_{i,j} = e_0-e_i-e_j$ and $\displaystyle c_j = 2h-\sum_{\substack{1\leq i\leq 6\\i\neq j}} e_i$, we could compute $s_{\alpha}(l_{i,j})$ and $s_{\alpha}(c_j)$ in order to obtain the permutation of the 27 lines.

Seeing this matrix can help us to understand why it is the representation of the quadratic transformation. If we denote by $\tau$ this transformation and by $\hat{\tau}$ the corresponding automorphism of your cubic surface $X$, then we can see that :

• $\newcommand{\P}{\mathbb P}$ the pull-backs by $\tau$ of the lines of $\mathbb P^2$ are the conics passing through the points $P_1,P_2,P_3$, thus $\hat{\tau}^*(h) = 2h-e_1-e_2-e_3$ (because $2h-e_1-e_2-e_3$ is the class of the strict transform of a conic passing through $P_1,P_2,P_3$) ;
• if $L_{i,j}$ is the line of $\P^2$ passing through $P_1$ and $P_j$ with $i,j\leq 3$, then $\tau(L_{i,j}) = P_k$ (the third of the points $P_1,P_2,P_3$, which is different from $P_i$ and $P_j$) thus $\hat{\tau}^*(e_k) = l_{i,j} = e_0-e_i-e_j$.

These observations correspond exactly to the matrix above.

Finally, note that if we replace $\alpha$ by $e_i-e_{i+1}, 1\leq i\leq 5$, in the formula above, then we obtain the transposition $(e_i,e_{i+1})$ and this gives the complete set of generators of the Weyl group of $E_6$.