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Let $X$ be a nonsingular projective algebraic variety over $\mathbb{C}$. Let $L$ be a base point free line bundle. Bertini's theorem asserts that general divisors in the complete linear system $|L|$ are nonsingular.

An example is $X=\mathbb{P}^n$, and $L=\mathcal{O}(m)$, for $m\ge 2$. The singular divisors of $|L|$ always forms a hypersurface (codimension =1).

My question is:

Can singular divisors form an algebraic set of arbitrary codimension in $|L|$ for $(X, L)$?

Thanks.

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2 Answers

up vote 5 down vote accepted

The answer is yes. Here is an example generalising the one of Serge and giving arbitrary codimension.

Consider line bundle $L=O(1)\boxtimes O(1)$ on $\mathbb CP^1\times \mathbb CP^n$. A section of such a bundle has shape $$\sum_{i=0,j=0}^{i=1,j=n}a_{ij}x_iy_j=0,$$ i.e. $|L|=\mathbb C^{2n+2}$. A section is singular iff the matrix $a_{ij}$ has rank $1$. This happen in codimension $n$. If the rank of $a_{ij}$ is two the section is smooth and is a $\mathbb CP^{n-1}$ bundle over $\mathbb P^1$.

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I think \boxtimes is the symbol you want for $O(1)\boxtimes O(1)=pr_1^*O(1)\otimes pr_2^*O(1)$. –  J.C. Ottem Feb 2 '13 at 11:40
    
James, thank you! Of course this is what I meant :) –  Dmitri Feb 2 '13 at 11:59
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I am not sure about "arbitrary", but there exist quite a few examples where the singular divisors do not form a hypersurface. It suffices to consider a smooth variety $X\subset\mathbb P^n$ embedded by a complete linear system and such that the dual variety $X^*\subset (\mathbb P^N)^*$ is not a hypersurface. The simplest example is the Segre variety $\mathbb P^1\times \mathbb P^2\subset\mathbb P^5$ (or any $X=\mathbb P_C(E)$, where $E$ is a vector bundle of rank $>2$ over a cueve $C$, embedded by $\mathcal O_{X|C}(1)$); there are other examples, too.

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@Thunder: You'll find many more examples in the book "Projective duality and homogeneous spaces". –  Olivier Benoist Feb 2 '13 at 11:29
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