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I'm thinking about the famous problem from classical algebraic geometry of how many lines in $\mathbb{P}^3$ meet four given general lines. According to some lecture notes on intersection theory that I was reading, Schubert had the intuition that it's general enough to consider the case where the four lines are in fact two pairs of intersecting lines - a highly nongeneric setup. (And then of course, in this case the answer is easily seen to be two.)

My question is, can someone give a basic explanation of this intuition, and perhaps some explanation of which other scenarios admit this sort of logic (i.e. looking at non-generic-but-still-kind-of-generic cases)? I have a vague picture in my head of continuously varying one of the four lines and how a line meeting all four should vary continuously, but I'm not entirely convinced by it yet.

Thanks! (Also thanks to Michael Joyce for pointing me to this sort of problem in a previous answer.)

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    The objects you are intersecting are in this context sections of line bundles. Any two sections are rationally equivalent, so you can pass to rational equivalence. To prove that the intersection is generically finite, I'm fairly sure there is some sheaf which will give you that after invoking the fact that ranks of sheaves are semicontinuous functions. On a sidenote, I realized that I never mentioned that "the number of points in the intersection" is with multiplicity.2012-11-10

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(expanding on the above -- can't figure out how to edit, please delete earlier)

There's some nice discussion of this in Harris's book "3264 and all that", a (legitimately available) draft of which may be found on Google. Section 4.2 works out the Chow ring of $G(1,3)$. In particular, "A Specialization Argument" in Section 4.2.3 seems to be just the example you're after about degenerating to two intersecting lines. I gather that there is a generalization of this argument by Coskun and Vakil that might be of interest; the reference is in 3264.

To elaborate, let $\sigma_1$ be the cycle consisting of lines which meet a given line. What you're trying to do is compute $\sigma_1^4$, which will be the number of lines that meet four given lines. A sensible first step is to compute $\sigma_1^2$. This turns out to be $\sigma_{1,1} + \sigma_2$, where $\sigma_{1,1}$ is the set of lines in a given plane and $\sigma_2$ is the lines through a particular point. Here's the idea of how to get that answer via degeneration. Fix a line $L$ and a family of lines $M_t$ such that $L$ and $M_t$ do not meet, except when $t=0$ they intersection at a point. You're interested in the class of $\sigma_1(L) \cap \sigma_1(M_t)$, and it's sensible to look for the flat limit of this intersection as $t \to 0$. If $\ell$ meets both $L$ and $M_0$, either it is contained in the plane with both of them in it (there's the $\sigma_{1,1}$), or it goes through the point where they intersect (there's the $\sigma_2$).

Degenerating to two pairs of intersecting lines as you suggest lets you write $\sigma_1^4 = (\sigma_{1,1}+\sigma_2) \cdot (\sigma_{1,1} + \sigma_2) = 2$. To see why the above degeneration argument (I use the term loosely!) is actually legitimate, check out the referenced book.

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    I merged your accounts. That was probably the reason you couldn't delete the other post.2012-11-10