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I am trying to understand the stabilization map, which takes a prestable curve (a curve with some marked points, and at worst nodal singularities) and returns a stable curve (a curve with some marked points, at worst nodal singularities, and finite automorphisms). Essentially what the stabilization map does is it contracts all of the unstable components of the curve, that is, those with infinite automorphisms.

There is furthermore supposed to be a map from the moduli of prestable curves to the moduli of stable curves, and therefore we need a nice description of the stabilization map which works well in flat families.

One possible such description is supposed to be: Take the dualizing sheaf omegaC of the prestable curve C; then take L := omegaC(x1 + ... + xn), where the xi are the marked points; then some large power of L will be generated by global sections, and the stabilization of C is the image of C under the corresponding map to projective space.

In particular, if C' is an unstable component of C, then L restricted to C' should be OC', as the image of C' should be a point.

I have several dumb questions:

  1. Why is omegaC an invertible sheaf? Does it follow from, e.g., Hartshorne III.7.11?

  2. Why is some large power of L generated by global sections?

  3. How do we know that this power can be chosen to be constant in families?

  4. How do we see that L restricted to unstable C' is OC'?

My more general question is: Given a singular curve, or at least a prestable curve, what explicit information can we deduce about the dualizing sheaf? For example, is there a way to figure out what the dualizing sheaf looks like when restricted to an irreducible component, or at least a smooth irreducible component?

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I should add that I'm taking curve to mean proper and over an algebraically closed field. –  Kevin H. Lin Oct 11 '09 at 21:51
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1 Answer

The dualizing sheaf looks like the sheaf of 1-forms, with logarithmic singularities at nodes (of the form f(z)dz/z, f regular at 0) such that the residues on each component of a node add to zero. The answer to the second part of number 1 is "yes" but you can also just calculate using the explicit description.

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