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Suppose I have a smooth projective surface $S\subset P^n$ embedded by a very ample linear system $|L|$. Consider now the generalized Severi varieties that parametrize curves on $S$ belonging to $|L|$, with a given number of singularities of prescribed type. E.g.: one nodes, two nodes, one cusp, one node and one cusp, one tacnode, one triple point and one node... etc.

My question is: is the dimension of such varieties invariant when the surface undergoes blow-up and blow-down transformations? For example: let $\sigma:\tilde{S}\to S$ be a blow up and $V(L)$ a certain Severi variety associated to $L$, are the dimensions of $V(L)$ and $V(\sigma^*L)$ the same?

Well-posing the question in the case of blow-downs seems a little trickier, but ideas are welcome.

I imagine that possibly one needs some hypothesis of irreducibilty on the curves.

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    $\begingroup$ 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? $\endgroup$
    – Jason Starr
    Commented Nov 13, 2013 at 13:57
  • $\begingroup$ Yes, of the dimensions of such varieties. I edit a little the question in order to make it more readable. $\endgroup$
    – IMeasy
    Commented Nov 13, 2013 at 14:03
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    $\begingroup$ 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 $L$. $\endgroup$
    – Francesco Polizzi
    Commented Nov 13, 2013 at 14:06
  • $\begingroup$ @Francesco: what is the result that you use in order to conclude that they are iso? $\endgroup$
    – IMeasy
    Commented Nov 13, 2013 at 14:09
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    $\begingroup$ Note that the line bundle $\sigma^*L$ is not ample, so by blowing up you change the setup. $\endgroup$
    – Sasha
    Commented Nov 13, 2013 at 14:13

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