3
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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, 2012 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, 2012 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, 2012 at 12:18

1 Answer 1

5
$\begingroup$

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

$[4,1,1,1,2,2,2]$

$[-2,-1,0,-1,-1,-1,-1]$

$[-2,-1,-1,0,-1,-1,-1]$

$[-2,0,-1,-1,-1,-1,-1]$

$[-1,0,0,0,-1,-1,0]$

$[-1,0,0,0,0,-1,-1]$

$[-1,0,0,0,-1,0,-1]$

$\endgroup$
6
  • $\begingroup$ 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, 2012 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. $\endgroup$ Sep 13, 2012 at 18:05
  • 1
    $\begingroup$ 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, 2012 at 19:22
  • 1
    $\begingroup$ 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, 2012 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. $\endgroup$ Sep 13, 2012 at 21:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.