Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.)

share|cite|improve this question
What are you familiar with? My intuition behind this is that this corresponds to intersection theory on the Grassmannian, which is projective, and the dimension of the Grassmannian is 4, so as long as the setup ends with a finite number of lines satisfying the given conditions, that will be "the correct number". More concretely, this corresponds to some kind of "moving lemma": If you move a line to intersect another line in a way so that the setup never degenerates, there is a kind of "continuity" that keeps the number the same. But this is unclear to me without Grassmannians. – only Nov 9 '12 at 22:03
So you seem to be saying that in (projective) intersection theory problems, shifting around the setup may change the dimension of the space we're looking at (in this case, lines meeting the four given lines), but should never change the number of irreducible components? This sounds somewhat plausible. (And I hope someone weighs in on this philosophy.) But in this example, it's not clear (to me) that the number of lines that meet a generic set of 4 lines is nonempty. (I assume when you say "a finite number of lines" above, you mean nonzero.) Though I suppose I could perhaps just check that? – Rob Silversmith Nov 9 '12 at 23:53
You can create an intersection theory on Grassmannians. What this means here is that if you pass to rational equivalence, you can define the "intersection" of two rational equivalence classes in a way so that the intersection number is always of the proper dimension. (For example, you can intersect a curve on a surface with itself and get a "finite number of points", or rather, a dimension 0 rational equivalence class). Under rational equivalence on Grassmannians, which are projective, the number of points always stays the same. (continued in next comment) – only Nov 10 '12 at 2:52
So what's really happening is that whenever your intersection is actually a finite set of points, the number will coincide with the formal intersection number. When the intersection is not a finite set of points, the data from the intersection number is harder to use. If you accept the existence of such an intersection theory, then the specific example given proves the intersection number is 2, and proves that whenever you have an intersection consisting of a finite number of points, it will be 2 points. (to be continued) – only Nov 10 '12 at 2:56
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. – only Nov 10 '12 at 2:59
up vote 2 down vote accepted

(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.

share|cite|improve this answer
Look ^^ there. You should see share edit flag buttons. It's possible that you may not have enough reputation, yet, for all of those buttons, but you should have the first two. The edit button is (of course) the one you want. – Cameron Buie Nov 10 '12 at 5:44
I merged your accounts. That was probably the reason you couldn't delete the other post. – robjohn Nov 10 '12 at 9:54

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.