Assume $C\subset \mathbb{P}^2$ is a smooth cubic curve. Then there is a cyclic triple cover $\pi: S\rightarrow \mathbb{P}^2$ ramified in $C$. Let $\sigma$ be the covering automorphism (sheet interchange automorphism) associated to $\pi$, i.e. $S/<\sigma>\cong\mathbb{P}^2$, here $\sigma^3=id$. If $H:=\pi^{*}l$ denotes the pullback of a line, then the canonical divisor of $S$ is $K_S=-H$ and $K_S^2=3$. So $S$ can also be seen as the blow up of $\mathbb{P}^2$ in 6 points and is therefore a cubic surface with the famous 27 lines on it.

Looking at the triple cover contruction the lines can be found the following way: $C$ has 9 points of inflection. The preimage of tangent line at such a point decomposes as $\pi^{-1}(l)=E\cup\sigma(E)\cup\sigma^2(E)$, where these are 3 (-1)-curves on $S$, so we get 9$\times$3=27 lines on $S$.

If $\pi: S\rightarrow \mathbb{P}^2$ is the triple cover and we pick 6 mutually skew lines $E_1,\cdots,E_6$ in the preimages of inflection lines, then there is a map $\phi: S \rightarrow \mathbb{P}^2$ such that $S$ is the blow up of $\mathbb{P}^2$ in 6 points $P_1,\cdots,P_6$ and the $E_i$ are the exceptional curves. The strict transforms of the lines in $\mathbb{P}^2$ containing two different points $P_i$ and $P_j$, $1\leq i < j \leq 6$ give 15 (-1)-curves $F_{i,j}$ on $S$. Finally there are six strict transforms of the conics $G_i$ in $\mathbb{P}^2$ containing the $P_j$ for $j\neq i$, $1\leq i \leq 6$.

What can we say about the images of the 27 lines under the automorphism $\sigma$? For example if we pick $E_1$ can we say which lines $\sigma(E_1)$ and $\sigma^2(E_1)$ are in terms of the $F_{i,j}$ and $G_j$, e.g something like $\sigma(E_1)=G_1$? Or is there any other description which tells us exactly which 3 lines are in a preimage of an inflection line?

Background: I recently learned about the "Geiser involution": if one has a double cover $Y$ of $\mathbb{P}^2$ ramified in a smooth quartic $Q$, then $Y$ is the blow up of $\mathbb{P}^2$ in 7 points. $Y$ has 56 (-1)-curves, which arise in the preimages of the 28 bitangents to $Q$. The covering automorphism associated to $Y$ is called the "Geiser involution" and one can describe the images of the 56 (-1)-curves under this involution, see for example arxiv.org/pdf/math/0403245.pdf, Remark 3.3. So i was wondering if there is such a description for a cubic surface and its 27 lines.

  • $\begingroup$ I don't know if this remark is of any help, but the triple of lines $E_1$, $\sigma(E_1)$, $\sigma^2(E_1)$ is characterized by the fact that the three lines go through one point (the preimage of the flex). Also, the lines $E_1, \dots E_6$ project to 6 distinct inflection lines $l_1,\dots l_6$. A first step would be to understand whether any subset of the 9 inflection lines can occur as $l_1,\dots l_6$ $\endgroup$ – rita Sep 12 '12 at 8:35
  • $\begingroup$ Note that for generic blowup of 6 points on $P^2$ there are no triple intersections of lines, so this is a blowup of a very special configuration. I would guess that you should take a cubic curve on $P^2$, choose 3 inflection points, blow them up, and then blowup thee points of the intersection of the exceptional divisors with the proper preimage of the cubic curve. $\endgroup$ – Sasha Sep 12 '12 at 9:26
  • $\begingroup$ @Sasha: if you blow up a point and then a point on the corresponding exceptional curve you end up with a -2 curve. So the cubic surface won't be smooth. $\endgroup$ – rita Sep 12 '12 at 12:18

It is true that the surface obtained is a special one (for example, a general cubic surface admits no automorphism). In fact, taking coordinates $X,Y,Z$ on $\mathbb{P}^2$, the cubic curve is given by $F(X,Y,Z)$ for some polynomial of degree $3$ and the equation of the surface is $W^3=F(X,Y,Z)$. The automorphism corresponds to send $(W:X:Y:Z)$ onto $(aW:X:Y:Z)$ where $a$ is a $3$-rd root of unity.

As you said, the orbit of a line $E\subset S$ of the surface consists of three lines $E, \sigma(E), \sigma^2(E)$ with $E+\sigma(E)+\sigma^2(E)=\pi^{-1}(l)$ where $l$ is an inflexion line of $\mathbb{P}^2$. Note that $E+\sigma(E)+\sigma^2(E)$ is equal to the trace of an hyperplane of $\mathbb{P}^3$, in particular each intersects transversally the two others: $E\cdot \sigma(E)=E\cdot \sigma^2(E)=\sigma(E)\cdot \sigma^2(E)=1$.

I will use the same notation as you and write $\phi\colon S\to \mathbb{P}^2$ a birational morphism which is the blow-up of $6$ points. I denote by $E_1,\dots,E_6$ the six curves contracted and by $L\in \mathrm{Pic}(S)$ the pull-back of a general line of $\mathbb{P}^2$. Then $\mathrm{Pic}(S)$ is isomorphic to $\mathbb{Z}^7$ with basis $L,E_1,\dots,E_6$. The $27$ lines correspond to: $E_i$, $i=1,...,6$, $F_{ij}=L-E_i-E_j$ (line through two points) for $i\not=j$ and $G_j=2L-\sum_{i\not= j} E_i$ (conics through $5$ points).

Because $\sigma(E_1)\cdot E_1=\sigma^2(E_1)\cdot E_1=1$, $\sigma(E_1),\sigma^2(E_1)$ are equal to $F_{1i}$ or $G_j$ for $i,j\not=1$. Because $\sigma^2(E_1)\cdot \sigma(E_1)=1$, we have moreover $i=j$. The orbit of $E_1$ is thus $(E_1,F_{1i},G_i)$ for some $i\not=1$.

Doing the same for $E_2,...,E_6$ we find a permutation $\tau$ of $(1,...,6)$ without fix points such that the orbit of $E_i$ is {$E_i, F_{i,\tau(i)},G_{\tau(i)}$} for $i=1,...,6$.

Because the image by $\sigma$ and $\sigma^2$ of the set {$E_1,...,E_6$} is $6$ skew lines, we can find that these two sets contain three of the $G_i$ and three $F_{i,j}$. Up to permutation of the curves $E_i$, we have then $E_1\to F_{12}\to G_2$

$E_2\to F_{23}\to G_3$

$E_3\to F_{13}\to G_1$

$E_4\to G_5\to F_{45}$

$E_5\to G_6\to F_{56}$

$E_6\to G_4\to F_{46}$

The image of the other lines is induced by these maps. We can for example easily write the matrix of $\sigma$ relatively to the basis $L,E_1,...,E_6$. (in particular, it corresponds to a birational map of $\mathbb{P}^2$ of degree $4$).

The matrix is








  • $\begingroup$ Thanks. I tried to compute the matrix of this action like you suggested. I found that $\sigma(L)=3L-\sum E_i$. So i get a 7x7-matrix $A$ with integer entries, by $L\rightarrow 3L-\sum E_i$ and $E_1\rightarrow L-E_1-E_2$ etc. But if i compute $A^2(E_1)=A(A(E_1)))=A(L-E_1-E_2)=L+E_2-E_4-E_5-E_6$ but this is not $G_2$. Also the matrix should satisfy $A^3=id$, but it does not. Where is my error? $\endgroup$ – TonyS Sep 13 '12 at 17:14
  • $\begingroup$ You cannot find $\sigma(L)=3L-\sum E_i$. In fact, $3L-\sum E_i$ is the canonical divisor and is thus fixed by $\sigma$. Here is a way to compute: $F_{12}=L-E_1-E_2$ so $L=F_{12}+E_1+E_2$. Hence $\sigma(L)=\sigma(F_{12})+\sigma(E_1)+\sigma(E_2)=G_2+F_{12}+F_{23}=4L-2E_1-2E_2-2E_3-E_4-E_5-E_6$. That is why I said that $\sigma$ corresponds to a map of degree $4$ on $\mathbb{P}^2$. You can write the matrix and compute its cube. It is the identity. I did it. $\endgroup$ – Jérémy Blanc Sep 13 '12 at 18:05
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    $\begingroup$ Ah, i found my mistake. But i still have a little problem ;-). For example $\sigma(F_{45})=\sigma(L−E_4−E_5)=\sigma(L)−\sigma(E_4)−\sigma(E_5)$ so we have $4L−2(E_1+E_2+E_3)−E_4−E_5−E_6−L+E_4+E_5−L+E_5+E_6$ which gives $\sigma(F_{45})=2L−2(E_1+E_2+E_3)+E_5$ which is not (directly) $G_5$. I think the problem is the computation of $\sigma(L)$, if one uses $F_{45}$ for the computation, one gets $\sigma(L)=4L−E_1−E_2−E_3−2(E_4+E_5+E_6)$. So here one needs to fit in relations between the $E_i$'s. $\endgroup$ – TonyS Sep 13 '12 at 19:22
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    $\begingroup$ By the way, this is the maxtrix i am working with (in Maple code): A := matrix([[4, 1, 1, 1, 1, 1, 1], [-2, -1, 0, -1, 0, 0, 0], [-2, -1, -1, 0, 0, 0, 0], [-2, 0, -1, -1, 0, 0, 0], [-1, 0, 0, 0, -1, 0, -1], [-1, 0, 0, 0, -1, -1, 0], [-1, 0, 0, 0, 0, -1, -1]]). If i compute $A^2$ then the first few columns are okay, but the last ones are wrong. $\endgroup$ – TonyS Sep 13 '12 at 19:24
  • $\begingroup$ You are absolutely right. I made misprints by rewriting the images of $E_4,E_5,E_6$. (I exchanged the image of $\sigma$ with the image of $\sigma^2$). Sorry for this. Now it is correct. I also added the matrix. $\endgroup$ – Jérémy Blanc Sep 13 '12 at 21:33

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