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This is part of Proposition $VIII.13$ in Beauville's "Complex algebraic surfaces": Let $S$ be a $K3$ surface and $C \subset S$ a smooth curve of genus $g=2$. Then the morphism $\phi$ defined by the linear system $|C|$ is of degree 2, whose branch locus is a sextic of $\mathcal{P}^2$.

The proof is not detailed. He just says that from the fact that $C^2 = 2$ it follows that the degree of $\phi$ is 2. Why is that? Moreover, $C$ is a double cover of a line $l = \phi(C)$ branched in $n$ points of $l \cap \Delta$, where $\Delta$ is the branch locus. Then he just says that $g=2$ implies $n=6$. Could you help me figure out what is implicit in these conclusions? Thanks.

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  • $\begingroup$ $C^2=2$ implies $\deg(\phi)=2$ is from the adjunction formula $2=C^2=(\phi^* H)^2=(\deg \phi)H^2=\deg(\phi)$, and the fact that a hyperelliptic curve of genus $g$ mapping to $\mathbb{P}^1$ has $2g+2$ branch points is from Riemann Hurwitz $\endgroup$ – DCT Mar 12 '16 at 4:01
  • $\begingroup$ @Dtseng how are you using the adjunction formula? I know it just as $\mathcal{O}_S(K_S + C)|_C \cong \mathcal{O}_C(K_C)$. $\endgroup$ – baltazar Mar 12 '16 at 13:27
  • $\begingroup$ Sorry, I meant the projection formula for intersection products. It's exercise 20.1.J in Vakil's notes math.stanford.edu/~vakil/216blog/FOAGjun1113public.pdf $\endgroup$ – DCT Mar 12 '16 at 19:12
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Let me expand slightly Dtseng comment. I will actually try to emphasize the geometric picture, rather than the blind application of formulae. The morphism $\phi\colon S\to \Bbb{P}^2$ is constructed via the sections of the line bundle $\mathcal{O_S}(C)$, which has $3$ independent sections, by Riemann-Roch. But this, I guess, you already know :)

With respect to this morphism, lines in $\Bbb{P}^2$ correspond to divisors in $|C|$. Now consider a point $p\in\Bbb{P}^2$ and two lines $\ell_1,\ell_2$ intersecting at $p$. Suppose $\ell_1$,$\ell_2$ correspond to $C_1,C_2\in|C|$. Where do $C_1$ and $C_2$ intersect? Well, $C_1\cdot C_2 = C^2=2$. This means that they intersect at two distinct points $x,y$ of $S$, at least for a generic choice of $p$, i.e. outside some closed Zariski subset $\Delta\subset\Bbb{P}^2$. Now before going on to determine what $\Delta$ actually is (the sextic curve), notice that we just proved $\phi$ to be generically $2$-to-$1$, i.e. a morphism of degree $2$.

Now let $\ell$ be the line corresponding to $C$ itself. By restricting $\phi$ we get a morphism of degree $d=2$ $$f\colon C\to\ell$$

Riemann-Hurwitz: $2g(C)-2=d(2g(\ell)-2)+r$, where $r$ is the degree of the ramification, i.e. the number of points of $\ell\cap\Delta$, but counted with multiplicity of course, so that more precisely $$r = \ell\cdot\Delta \quad (= \deg\Delta)$$ Of course plugging all in the Riemann-Hurwitz formula we just get $r=6$, so $\Delta$ is a sextic. The fact that it is smooth is because $S$ is smooth (in general, the singularities of a double covering $S\to S'$ between two surfaces are precisely along the singular points of the ramification locus). I hope this helps a bit.

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