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Let $S$ be a del Pezzo surface $S$ of degree $4$. There is an exact sequence

$$ 0\to H^0(\mathbb{P}^4,I_S(2)) \to H^0(\mathbb{P}^4,\mathcal{O}(2))\to H^0(S,\mathcal{O}_S(2))\to0$$ where $I_S$ is the ideal sheaf of $S$. Let $K$ be a canonical divisor. We have $$H^0(S,\mathcal{O}(-2K_S))=13.$$

The claim is $$H^0(S,\mathcal{O}(-2K_S))=H^0(S,\mathcal{O}(2))$$ and from this it follows that $S$ is the base locus of a pencil of quadrics. I neither understand the isomorphism nor how the statement about the base locus follows, could anyone elaborate?

This is from Dolgachev:" Classical Algebraic Geometry: a Modern View" (Thm 8.2.6)

Thanks a lot.

PS: a base locus of a pencil of quadrics is the intersection of two quadrics.

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1 Answer 1

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Your Del Pezzo surface $S$ is embedded into $\mathbb P^4$ anticanonically, i.e. $\mathcal O_S(1) = - K_S$. (For this, recall that $-K_{\mathbb P^2} = \mathcal O_{\mathbb P^3}(3)$. Remember also that blowing up adds a copy of the exceptional divisor to the canonical bundle, and that $S$ is obtained by blowing up $\mathbb P^2$ at five points, and is embedded into $\mathbb P^4$ via the linear system of cubics passing through these five points. These remarks taken together will let you verify the claim.) Consequently $\mathcal O_S(-2K_S) = \mathcal O_S(2)$, and the stated isomorphism follows.

Now the dimension of $H^0(\mathbb P^4, \mathcal O(2))$ is $15$, and so, from the exact sequence $$0 \to H^0(\mathbb P^4,\mathcal I_S(2)) \to H^0(\mathbb P^4, \mathcal O(2)) \to H^0(S, \mathcal O_S(2) ) \to 0$$ we find that $H^0(\mathbb P^4, \mathcal I_S(2))$ is $2$-dimensional. This precisely means that there is a two-dimensional space of quadratic forms that vanish on $S$, and hence that $S$ is the base locus of the corresponding pencil of quadrics.

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  • $\begingroup$ great, thanks! I've corrected the typo. $\endgroup$
    – Carsten
    Apr 20, 2012 at 15:47
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    $\begingroup$ @Matt to have that exact sequence, we must have $H^1(\mathbb{P}^4,\mathcal{I}_S(2))=0$. why is this true? $\endgroup$
    – idioteca
    Feb 3, 2014 at 21:56
  • $\begingroup$ @idioteca: Dear idioteca, Good question! It was given in the OP, so I just took it for granted. I didn't think about why it's true. Maybe look in Dolgachev's book, as cited in the OP. Regards, $\endgroup$
    – Matt E
    Feb 4, 2014 at 0:37
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    $\begingroup$ @Matt actually it is not so important that $H^1(\mathbb{P}^4,\mathcal{I}_S(2))=0$. Infact we always have $$h^0(\mathcal{O}(2))\leq h^0(\mathcal{I}_S(2))+h^0(\mathcal{O}_S(2))$$ and this is enough to say that there are at least two quadratic forms linear independent that vanish on $S$. then i think you look at the degree: since the complete intersection of two quadrics has degree 4 and it contains a del pezzo surface that has degree 4, they are the same. $\endgroup$
    – idioteca
    Feb 4, 2014 at 9:48
  • $\begingroup$ @idioteca: Dear idioteca, I agree with what you wrote; thanks for pointing it out. Best wishes, $\endgroup$
    – Matt E
    Feb 4, 2014 at 23:52

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