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Nov 13, 2013 at 14:16 comment added IMeasy @Sasha: very good point. If I don't assume the ampleness from start, anyway apparently the question still has sense.
Nov 13, 2013 at 14:16 comment added IMeasy @FrancescoPolizzi: Yes I agree. I wonder if one has to assume irreducibility of the curve.
Nov 13, 2013 at 14:13 comment added Sasha Note that the line bundle $\sigma^*L$ is not ample, so by blowing up you change the setup.
Nov 13, 2013 at 14:12 comment added Francesco Polizzi @IMeasy: you are pulling back $L$, which is very ample, so the general element of $|L|$ will not pass through the points you are blowing up. Therefore it seems to me that curves in $|\nu^*L|$ with $\delta$ nodes and $\kappa$ cusps in general position should be the pullback of curves in $|L|$ with the same singularities, not passing through the blown-up locus. Am I missing something?
Nov 13, 2013 at 14:09 comment added IMeasy @Francesco: what is the result that you use in order to conclude that they are iso?
Nov 13, 2013 at 14:07 history edited IMeasy CC BY-SA 3.0
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Nov 13, 2013 at 14:06 comment added Francesco Polizzi Outside the exceptional divisor, the pair $(\widetilde{S}, \nu^*L)$ is isomorphic to the pair $(S, L)$, so if I am not mistaken the two Severi varieties should be birationally equivalent. Maybe it is more interesting to compare $V_{\delta, \kappa}$ with the Severi variety $V_{\delta, \kappa}(\widetilde{L})$, where $\widetilde{L}$ is the strict transform of $L$.
Nov 13, 2013 at 14:03 comment added IMeasy Yes, of the dimensions of such varieties. I edit a little the question in order to make it more readable.
Nov 13, 2013 at 13:57 comment added Jason Starr I am not sure this is a well-posed problem. For a blowing up $\nu:\widetilde{S}\to S$, are you asking for a comparison of the Severi variety $V_{\delta,\kappa}(L)$ and the Severi variety $V_{\delta,\kappa}(\nu^*L)$, where $\delta$ is the number of nodes and $\kappa$ is the number of cusps?
Nov 13, 2013 at 13:50 history asked IMeasy CC BY-SA 3.0