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All varieties are assumed over $\mathbb{C}$. Consider a geometrically ruled surface $X$ over a curve $C$, it is known that $X$ can be realized as the projective bundle associated to a rank $2$ vector bundle $E$ over $C$, that is $X=\mathbb{P}(E)$.

But there are two ways to associate a projective space $\mathbb{P}(V)$ to a vector space $V$: $\mathbb{P}_l(V)$ consists of lines in $V$; $\mathbb{P}_h(V)$ consists of hyperplanes in $V$ (all passing through $0$). These two projective spaces are dual to each other but not isomorphic naturally. (This is the problem!)

Hartshorne uses $\mathbb{P}_h(E)$, since he defined it by $\textrm{Proj }S(E)$; while Beauville uses $\mathbb{P}_l(E)$. It is obviously true that $\mathbb{P}_h(E)=\mathbb{P}_l(E^*)$, we denote it by $X$. Let $\pi:X\rightarrow C$ be the projection map.

The difference comes from the definitions of $O_X(1)$ over $X$. Hartshorne defines it using $\textrm{Proj}$ construction, so $O_l(1)$ is a natural quotient of $\pi^*E$; while Beauville defines it using the exact sequence $$0\rightarrow N\rightarrow \pi^\ast E^\ast \rightarrow O_h(1)\rightarrow 0,$$ here $N$ is the natural line bundle over $X=\mathbb{P}_l(E^*)$.

Question: Are these two definitions closely related?

When $E$ is decomposable, I prove that $O_h(1)\cong O_l(1)\otimes \pi^\ast(\wedge^2E^\ast)$, but I cannot prove it in general.

I'm sorry that the description is a little bit long, but I'm really confused by the subtle difference between the two definitions of the projective bundle associate to $E$. Thanks for your help!

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1 Answer 1

up vote 3 down vote accepted

If $E$ is a rank 2 bundle then $E^* \cong E \otimes \Lambda^2E^*$. This gives the required relation for $O(1)$ bundles.

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Dear Sasha, thanks for your answer, but I don't know how to prove $E^\ast\cong E\otimes\wedge^2 E^\ast$? Is this isomorphism natural or not? –  Yuchen Liu Apr 15 '13 at 3:14
2  
It is, you just plug any element of $E$ in the tensor $\wedge^2 E^*$ and get $E^*$. End of the proof. –  IMeasy Apr 16 '13 at 19:41
    
@IMeasy Thanks! –  Yuchen Liu Apr 17 '13 at 3:37

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