Timeline for How to find the action of an automorphism on the 27 lines on a cubic surface?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Sep 14, 2012 at 16:26 | vote | accept | TonyS | ||
| Sep 14, 2012 at 16:26 | comment | added | TonyS | Now everything fits perfectly. Thanks a lot for your help. | |
| Sep 13, 2012 at 21:33 | comment | added | Jérémy Blanc | 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. | |
| Sep 13, 2012 at 21:32 | history | edited | Jérémy Blanc | CC BY-SA 3.0 |
added 193 characters in body
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| Sep 13, 2012 at 19:24 | comment | added | TonyS | 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. | |
| Sep 13, 2012 at 19:22 | comment | added | TonyS | 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. | |
| Sep 13, 2012 at 18:05 | comment | added | Jérémy Blanc | 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. | |
| Sep 13, 2012 at 17:14 | comment | added | TonyS | 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? | |
| Sep 12, 2012 at 16:19 | history | answered | Jérémy Blanc | CC BY-SA 3.0 |