It's well known that the automorphism group of the configuration of 27 lines on a smooth cubic surface in $\mathbb{P}^3$ (over a field containing all 27 lines) is isomorphic to the Coxeter group of type $ E_6 $.

Labelling the vertices of $ E_6 $ as $ x_1, x_2, x_3, x_4, x_5, y$ (so that $y$ is the unique leaf neighbouring the degree 3 node $x_3$) and labelling the 27 lines coming from looking at blowing up $ \mathbb{P}^2 $ at 6 points $ P_i $ in general position (and then making the anticanonical embedding) as $ e_i $ (exceptional lines of the blowup), $ l_{i,j} $ (proper transforms of lines through pairs of blown-up points) and $ c_j $ (proper transforms of conics going through all but one point), it's claimed in Hartshorne (V, Exercise 4.11) that one possible isomorphism between the automorphism group of the 27 lines and the Coxeter group of type $ E_6 $ is given by:
$x_1 \mapsto (e_1 \leftrightarrow e_2)$
$ \qquad \quad \vdots $
$x_5 \mapsto (e_5 \leftrightarrow e_6)$
and that $y$ is associated with the quadratic transformation of $ \mathbb{P}^2 $ based at the points $P_1, P_2, P_3$. (This is the birational self-map of $ \mathbb{P}^2 $ obtained by blowing up $P_1,P_2,P_3$, and then blowing down the proper transforms of the lines between them).

My question: how exactly does this birational transformation of $ \mathbb{P}^2$ induce an automorphism of the 27 lines, and how does one go about writing it down explicitly in terms of the $e_i$, $l_{i,j}$, $c_j$?


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 $E_6$.


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