Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

up vote 4 down vote accepted

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.

share|improve this answer
    
great, thanks! I've corrected the typo. –  Carsten Apr 20 '12 at 15:47
    
@Matt to have that exact sequence, we must have $H^1(\mathbb{P}^4,\mathcal{I}_S(2))=0$. why is this true? –  idioteca Feb 3 at 21:56
    
@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, –  Matt E Feb 4 at 0:37
    
@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. –  idioteca Feb 4 at 9:48
    
@idioteca: Dear idioteca, I agree with what you wrote; thanks for pointing it out. Best wishes, –  Matt E Feb 4 at 23:52

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.