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Assume that you are given a smooth projective algebraic variety $X$ and a divisor $D$ with normal crossings on it. Put $U=X-D$. Assume now that $L$ is a torsion line bundle(=invertible sheaf) on $U$, i.e. there exists an integer $N$ such that $L^{\otimes N}=\mathcal{O}_U$. I have the following question:

How to extend $L$ to a line bundle $L_X$ on $X$ such that $L_X^N=\mathcal{O}_X(D)$?

Will the direct image with compact support $j_! L$ with respect to the inclusion $j: U \hookrightarrow X$ be such an extension?

I have the feeling that this should be pretty obvious what I cannot see how.

Thanks for your help

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Are you allowing multiplicities for $D$? Otherwise, it is certainly false, take $X=\mathbb{P}^1$, $D=point$, and $N>1$. –  Donu Arapura Jun 29 '13 at 12:31
    
Yes, $D$ need not be reduced. How do you it in that case? –  isaac90 Jun 29 '13 at 13:34
    
I guess the more precise statement will be $\mathcal{O}_X(\sum a_i D_i)$ for some $a_i \geq 1$ and $D_i$ the irreducible components of $D$. –  isaac90 Jun 29 '13 at 13:47

1 Answer 1

The line bundle $L$ gives you a degree $N$ Galois cover $V \to U$. Let $Y$ be an equivariant compactification of $V$. Then there is a rational morphism $Y \to X$. By an additional equivariant blowup $Y' \to Y$ you can get a regular morphism $f:Y' \to X$ which is invariant under the Galois action. Then take $f_*O_{Y'}$ (this is a rank $N$ vector bundle on $X$) and decompose it with respect to the Galois action. You will get a sum of powers of a line bundle $L_X$, which extends $L$.

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Hi Sasha. Thanks for the answer. Could you explain more in detail why there is a rational morphism $Y \to X$ and how do you do the "additional equivariant blowup $Y' \to Y$"? And why the resulting $L_X$ will satisfy $L_X^N=\mathcal{O}_X$? This will be very helpful for me. –  isaac90 Jun 29 '13 at 11:24
    
1. Because $Y$ has an open subset $V$ and there is a map $V \to U \to X$. 2. The usual construction (see Hartshorne) of the resolution of a rational morphism will be equivariant in this case. 3. It will not. But $(L_X)_{|U} = L$, hence $(L_X^N)_{|U} = O_U$, hence $L_X^N = O_X(D)$ (where $D$ is the divisor set-theoretically supported on $X\setminus U$. –  Sasha Jun 29 '13 at 14:27
    
Thanks again Sasha! A last question: why is any line bundle restricting to $\mathcal{O}_U$ in $U$ of the form $\mathcal{O}_X(D)$? What happens with multiplicities? –  isaac90 Jun 29 '13 at 17:21
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If $D$ is a simple divisor you can find in Hartshorne an exact sequence ${\mathbb Z}D \to {\mathrm{Pic}}\, X \to {\mathrm{Pic}}\, U \to 1$ which shows the fact you need. If $D$ has several components you can iterate this argument. –  Sasha Jun 29 '13 at 18:02

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