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You can construct a total space of a quasicoherent sheaf on an scheme by taking relative spec of the symmetric algebra of the dual sheaf. For locally free sheaves, you get vector bundles, and every vector bundle arises this way. What about sheaves that are not locally free? Are there any other sheaves for which the total space is a useful construction?

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Do you mean Seifert bundles? They are bundles with nonreduced fibers. see Kollar for more info. –  user1290 Oct 29 '09 at 21:17

2 Answers 2

We can back up a step and ask about the relative spec of the symmetric algebra of any coherent sheaf. Then there are lots of nice examples. If $X=\mathrm{Spec}\ k[t]$ and $E$ is the skyscraper sheaf at $0$, then we get $\mathrm{Spec}\ k[t,u]/(tu)$. If $X$ is $\mathrm{Spec}\ k[x,y]$ and $E$ is the ideal sheaf of the origin, then I get the relative spec of $\mathrm{Sym}_X(E)$ is $\mathrm{Spec}\ k[x,y,t,u]/(xu-ty)$. This has a one dimensional fiber over the points of $X$ other then $(0,0)$, and a two dimensional fiber over $(0,0)$. It looks like the fiber over $x \in X$ is the dual vector space to $E_x \otimes_{\mathcal{O}_x} k(x)$, but I am not sure whether this is always true.

If you insist that $E$ be the dual sheaf to some sheaf $F$, then you run into the problem that non-locally-free dual sheaves are hard to come by in dimensions 1 and 2. See my question, to which there are several good answers.

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This sounds like it's related to my old question, here. In my answer, I needed that the sheaf is locally something nice in the etale topology, but if it is locally nice in the Zariski topology, then it will be a fiber bundle, the question is only which open sets are needed for local triviality.

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