I read that one example of k3 surface is a double cover of $\mathbb{P}^2\mathbb{C}$ ramified over a sextic. My question is why a sextic? i believe that the sextic is isomorphic to the ramification divisor, but why is this? also how can i see that k3 surfaces realized this way are deformation equivalent? thank you
1 Answer
$\begingroup$
$\endgroup$
15
Q1: Hurwitz formula + canonical divisor of $\mathbb P^2$.
Q2: Move the curve in $\mathbb P^2$.
-
17
-
13
-
7$\begingroup$ Seriously, @Garfield? You wouldn't put your real name to your actions, but you make sophisticated judgement calls. This answer teaches rick (and you) to fish instead of giving him (and you) something to eat. The part that's not included in this answer belongs to math.stackexchange.com. Not that I care for those puny points, but I am deeply disappointed in trying to help people who want pre-chewed food instead of learning how to hunt. $\endgroup$ Commented Feb 27, 2013 at 16:37
-
8$\begingroup$ @rick, that comment was addressed to Garfield, not you. This is a perfectly fine question. And yes, your computation is right. That was the point of my comment: I believed that if I tell you to use the Hurwitz formula then you can figure it out yourself. It seems I was right and I think you might appreciate it more this way that you did part of the work. So, please do not take that comment to heart, it wasn't meant for you. Cheers! $\endgroup$ Commented Feb 28, 2013 at 0:22
-
12$\begingroup$ I am sorry if this painted the wrong picture, but there is really no need to get offended. Opinions differ, and to continue your metaphor, I believe that people who come here have already tried to fish for months, are starving for food, and they should receive it. A complete answer on mathoverflow isn't spoonfeeding, it's what the site is made for. The site wants me to downvote when I don't find an answer helpful, and I did. I even left a comment to explain. Please understand that while experienced people might find terseness helpful, less experienced people might not. $\endgroup$– GarfieldCommented Feb 28, 2013 at 6:58