I'm trying to compute the self-intersection of a degree 5, 6-nodal, curve in $\mathbb{P}^2$, yet I've only done this computations for smooth curves and I'm not sure if the same procedure works when there are singularities. In other words, I'd like to know if given a degree $d$ singular curve in $\mathbb{P}^2$ then its cohomology class is still $dH$, where $H$ is the class of a line?
self-intersection of plane nodal curves
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algebraic-geometry
intersection-theory
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1The cohomology class does not depend on singularities, so it is still $dH$. – 2017-01-17
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0Great, thanks, @Sasha ! Is this also true in the general case? – 2017-01-17
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1What do you mean? Definitely, if you deform a subvariety, its cohomology class does not change (because on one hand, it changes continuously, and on the other, it stays integral). – 2017-01-17
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0This completely answers my question, thanks again! @Sasha – 2017-01-17
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0What do you mean by "cohomology class" ? – 2017-01-18
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0@GeorgesElencwajg Could you elaborate? I'm thinking about integral cohomology. Maybe you're asking about the topology I'm using? In this case I guess the natural choice is the analytic one. – 2017-01-19
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0Yes, my question is: How do you define the integral cohomology class associated to your curve ? – 2017-01-19
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0@GeorgesElencwajg I see your point. I'm thinking you consider the curve as a topological space itself (as a pinched real sphere for example) and consider the pushforward of its fundamental class along the embedding. For example, for an elliptic curve we'd consider the fundamental class of a torus and do this. However, it could also be that we take the class to be the pushforward of the fundamental class of $\mathbb{P}^1$ along its birational morphism with the curve. This map seems better since it has a degree associated to it. – 2017-01-20