Basic Schubert calculus intuition 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.)
 A: (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.
