I'm following Beauville's book "Complex Algebraic Surfaces".

If $S$ is a K3 surface and $C$ is a smooth not hyperelliptic curve of genus g, then we have a birational morphism $\phi : S\rightarrow\mathbb{P}^g$, whose restriction to $C$ is the canonical morphism $C\rightarrow\mathbb{P}^{g-1}$. Since $S$ is minimal, it is isomorphic to the surface $\phi(S)$. The hyperplane sections of $\phi(S)$ are the curves $H=\phi( C^\prime)$, with $C^\prime\in|C|$.

$H^2=C^2=2g-2$.

We have the exact sequence $$0 \rightarrow I \rightarrow \mathcal{O}_{\mathbb{P^g}}\rightarrow\mathcal{O}_{\phi(S)}\rightarrow 0,$$ where $I$ is the ideal sheaf of $\phi(S)\subset\mathbb{P}^g$.

From this we obtain also the exact sequence $$0\rightarrow I(k)\rightarrow\mathcal{O}_{\mathbb{P}^g}(k)\rightarrow\mathcal{O}_{\phi(S)}(k)\rightarrow 0,$$ where $I(k)$ is the sheaf of polynomials of degree $k$ that vanish on $\phi(S)$.

Using the cohomology sequence we have $$h^0(\mathcal{O}_{\mathbb{P}^g}(k))\leq h^0(I(k))+h^0(\mathcal{O}_{\phi(S)}(k)).$$

If $g=4$, $\phi(S)$ has degree $6$ and we have $$h^0(\mathcal{O}_{\mathbb{P}^4}(2))=15\,\,\,and\,\,\,h^0(\mathcal{O}_{\phi(S)}(2H))=14,$$

so $\phi(S)$ is contained in a quadric of $\mathbb{P}^4$. Repeating the same argument with $k=3$, we obtain that $\phi(S)$ is contained also in a cubic. Beauville then claims that $\phi(S)$ is a complete intersection surface $S_{2,3}\subset\mathbb{P}^4$. And i agree since they both have the same degree and $S_{2,3}$ is a K3 surface, but i think that maybe this is not enough as we see in the next example.

He repeats the same argument with $g=5$ and he finds out that $\phi(S)$ it is contained in 3 quadrics that are linear independent in $\mathbb{P}^5$. This time he can't conclude that $\phi(S)$ is a complete intersection $S_{2,2,2}\subset\mathbb{P}^5$. Infact in this case, if $C$ admits a $g^1_3$ (i.e. an invertible sheaf $L$ of degree 3 such that $h^0(C,L)=2$), then $\phi(S)$ is not a complete intersection $S_{2,2,2}$ (exercise 11 of the same section of the book). But i see no difference with the $g=4$ case, infact $S_{2,2,2}$ is a K3 surface of degree $8$, like $\phi(S)$.

I don't understand why in the case $g=4$ he can conclude all at once that $\phi(S)$ is a complete intersection $S_{2,3}$, since in the second case he can't.

  • I think I found an explanation on Griffiths-Harris "Principle of Algebraic Geometry" pag. 592. – idioteca Feb 6 '14 at 17:50
  • Related: E.g. IV.5.2.2 of Hartshorne. – Andrew Feb 6 '14 at 20:24
up vote 3 down vote accepted

It sounds like you already have a satisfactory answer, but since I tried to understand this too, here is what I've understood.

By considerations of degree, there is a unique quadric $Q$ containing $X = \varphi(S)$. You show that $X$ is contained in a cubic $F$ as well, since $h^0(\mathcal I_X(3))\ge 6$. The reducible cubics that contain $X$ decompose into unions of quadrics and planes, and since $X$ must be contained in $Q$, this space is at most $5$ dimensional (the equation of $Q$ times a linear form). Thus, $X$ is contained in an irreducible cubic $F$, using $h^0(\mathcal I_X(3))\ge 6$. The intersection $Q\cap F$ then is a degree $6$ complete intersection surface containing $X$, which also has degree $6$, so $X = Q\cap F$.

When the genus is $5$, all we get is that $h^0(\mathcal I_X(2))\ge 3$, which implies that $X$ lies on at least $3$ linearly independent quadrics. This is weaker already than the previous case, since uniqueness fails. Also, using the same trick as previously, we only know that $h^0(\mathcal I_X(4))\ge 60$, while the set of reducible quartics in the ideal has dimension at least $63$ (hit the three independent quadratic forms with quadratic terms), so there is no new information from this dimension count. Thus, the same argument does not follow through, and we need more information to proceed. Indeed, as you say, Beauville gives counterexamples when there is a $g^1_3$.

  • That was exactly the answer i found. But then i would like to ask you another thing. Shouldn't it be $h^0(\mathcal{I}_X(3))\geq 6$? And i still don t get why only one quadric. I thought i obtained $h^0(\mathcal{I}_X(2))\geq 1$. – idioteca Feb 7 '14 at 7:13
  • However in G-H it seems to me that there is an "=". They say $h^0(\mathcal{I}_X(3))-1=5$. If there is an equality this means that $h^1(\mathcal{I}_X(k))=0$. – idioteca Feb 7 '14 at 7:23
  • But i can't see if that's true. – idioteca Feb 7 '14 at 7:24
  • Dear @idioteca, the $h^0(\mathcal I(2))\geq 1$ condition implies that $X$ lies on at least one quadric, but since the degree is $6$ it cannot lie on the intersection of two quadrics, since that would mean $X$ is contained in a degree $4$ curve. I got $\ge 5$ for the other twist... – Andrew Feb 8 '14 at 18:00
  • 1
    Dear @idioteca, sorry for the slow reply. You are correct. For some reason I started thinking about curves in $\mathbb P^3$ rather than surfaces in $\mathbb P^4$, which messed up my numbers. In any case, the argument still follows once we fix the numbers. – Andrew Feb 20 '14 at 22:56

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