Is it possible to realize $K3$ surface as a ramified double cover of rational elliptic surface? If so, is there way to see an elliptic fibration structure on $K3$ from such cover? It seems to me one can use the divisor $6H - 2E_{1} - \cdots - 2E_{9}$ and the class $3H - E_{1} - \cdots - E_{9}$ in two fold cover gives the fiber class in $K3$?
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2 Answers
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I slightly disagree with Tony's answer:
If you want to obtain a smooth K3 surface you need to pick a ramification divisor that is smooth and is linear equivalent to twice the anti-canonical class. I.e., you have to pick two smooth fibers.
You can easily see this in terms of Weierstrass equation. Let $S$ be a rational elliptic surface. Since the Picard group of $\mathbb{P}^1$ is trivial you can pick a global minimal Weierstrass model for $S$, i.e., you there exists forms $A(s,t)$ and $B(s,t)$ on $\mathbb{P}^1$ of degree 4, resp., 6. such that $S$ is birational to the solution set of $y^2=x^3+Ax+B$ in some appropriate $\mathbb{P}^2$-bundle. If $S$ is sufficiently general then $S$ is actually isomorphic to its Weierstrass model.
If we pick now two further forms $s(u,v)$ and $t(u,v)$ of degree 2 then $y^2=x^3+A(s(u,v),t(u,v))x+B(s(u,v),t(u,v))$ is the Weierstrass model of an elliptic surface. For a general choice of the forms $s(u,v),t(u,v)$ this defines a smooth K3 surface.
answered Jun 10, 2012 at 9:38
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$\begingroup$ Of course, you are right! I don't know what I was thinking. Somehow I did the calculation in my head without writing and, of course, I got it wrong. $\endgroup$ Commented Jun 10, 2012 at 18:36
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$\begingroup$ Thanks Remke and Tony. Since $6H - 2E_{1} - \cdots -2E_{9} = 2F$, then what I say seems correct. Also, is it true that a pencil of lines in $E(1)$ given by $H$ with one base point pulls back to a pencil of genus two curves with two base points in $K3$? Since the line $H$ intersects the branch curve at $6$ points, it looks to upstairs we have genus two pencil. $\endgroup$ Commented Jun 10, 2012 at 18:59
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Do you want your K3 to be smooth? In that case the answer is no. For the double cover to have a trivial canonical class you will have to chose a branch divisor which is a section in half of the anti-canonical class. But the anti-canonical class of the rational elliptic surface is the class of the fiber and so is never divisible by two.
The closest you can get to such a double cover is when use an Enriques surface as the base surface. Choose a genus one pencil on your Enriques surface. It gives you a genus one fibration whose Jacobian fibration is a rational elliptic surface. However on the Enriques, the fibration has two double fibers. If you take the root double cover that is branched at the two double fibers, it has two curves of nodal singularities. When you normalize this double cover you get a K3 with an elliptic fibration which is a pullback of the genus one fibration on the Enriques.
answered Jun 10, 2012 at 4:55
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$\begingroup$ let say we don't assume K3 is smooth $\endgroup$ Commented Jun 10, 2012 at 6:52
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$\begingroup$ As Remke Kloosterman points out - I got my Hurwitz formula wrong. To get the canonical class of the double cover to be trivial, you need to take a branch divisor that is in the twice the anticanonical linear system of the rational elliptic surface. So you simply need to take two smooth fibers as your branch divisor. This gives a smooth K3. It is the fiber product of the rational elliptic surface and a $\mathbb{P}^{1}$ doubly covering your original $\mathbb{P}^{1}$ with branching at the two points over which the two smooth fibers sit. $\endgroup$ Commented Jun 10, 2012 at 18:43
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$\begingroup$ @alex24: You should accept Remke's answer. It correctly and completely answers your question. $\endgroup$ Commented Jun 10, 2012 at 18:50