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Let $C$ be a stable reducible curve. Is there a natural way to define it's canonical model (I guess via the dualizing sheaf)? And does somehow the dualizing sheaf restrict to the (probably twisted) canonical sheaf of the components?

In particular: to what extent the restriction of the dualizing sheaf of the global curve to a component can be described as the dualizing sheaf (possibly twisted by O(p), with $p$ attaching point) of that component?

There are, for instance, some examples that puzzle me. Take a smooth genus 3 curve, then its canonical model is a plane quartic. Then consider the curve given by a central elliptic curve attached to 2 elliptic "tails" (I am not sure this is the correct terminology). The "canonical model" of this should equally be a plane curve. What plane curve??

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I am confused by this question. There is, indeed, a precise definition of the term "canonical model", and it is (usually) defined using the dualizing sheaf. Are you asking for a statement of that definition? Or are you asking for alternative definitions (perhaps more "natural" than the definition you already know)? Also, what do you mean by "twisted"? It is indeed true that the restriction of the dualizing sheaf of the total curve to a component is different from the dualizing sheaf of that component. Is that what you are referring to? –  Jason Starr Apr 26 '13 at 13:53
    
thank you for answering. yes in fact that was my suspect but I didn't find a proper reference. I guess that the canonical model of a stable reducible curve is obtained via the global sections of the dualizing sheaf. On the other hand I was wondering wether the definition of the dualizing sheaf (I added this to the quesiton and edited) ha some sort of "functorial" behaviour. That is: to what extent the restriction of the dualizing sheaf of the global curve to a component can be described as the dualizing sheaf (possibly twisted by $\mathcal{O}(p)$, if I recall properly) of that component –  IMeasy Apr 26 '13 at 14:13
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The canonical sheaf of a stable curve with an elliptic tale is not globally generated. –  Angelo Apr 26 '13 at 15:16
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@IMeasy: As Angelo points out, usually you need to take a sufficiently positive tensor power of the dualizing sheaf before the global sections define a closed immersion into projective space; I believe the third power always works. Mumford proved that for the closed immersion associated to the fifth power, the corresponding point of the Hilbert scheme is stable in the sense of Geometric Invariant Theory (you did not ask, but I thought I would mention it). –  Jason Starr Apr 26 '13 at 15:52
    
thank you to both! –  IMeasy Apr 27 '13 at 9:20
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