-1
$\begingroup$

During linear algebra class, I was explaining that given 2 equations + 2 unknowns where we expect there to be a unique solution but sometimes there can be 0 solutions or a line's worth.

At some point I started counting configurations of 3 lines in the plane. We expect three different configurations with three intersection point for the three pairs of lines. I wonder which configurations are the next order down in degeneracy.

My gut is that setting a line parallel to another and having all three intersect at a point should count the same, since they can be both deformed by a single line.

alt text http://www.freeimagehosting.net/newuploads/ok7lz.gif

Theory of simultaneous equations is intersection theory in the plane. Can this hierarchy of degenerate configurations be interpreted geometrically with Schubert calculus?

Improve this question
$\endgroup$
2
  • 4
    $\begingroup$ My feeling is that this should be simpler in the projective plane than in the Euclidean plane — arrangements of lines in the Euclidean plane as you are trying to classify are basically arrangements of lines in the projective plane where you have one more line and you have fixed it at infinity, so the two theories aren't really different, but when you look at it projectively you don't have to make a distinction between triple intersections and parallel lines, they become the same thing. $\endgroup$
    – David Eppstein
    Commented Jan 26, 2012 at 5:15
  • $\begingroup$ David's comment, and Jack's answer are really what you are looking for. However, since you brought up Schubert Calculus specifically, let me point you at this paper:arxiv.org/pdf/math/0608784v1.pdf by Ronga, which talks about a more classical perspective on the Schubert calculus. In fact, in this paper, you can see a picture similar to yours about configuration of lines. I think this purely expository paper gives a very beautiful picture of the incidence geometry. $\endgroup$
    – B. Bischof
    Commented Jan 26, 2012 at 14:10

1 Answer 1

Reset to default
6
$\begingroup$

Thinking of a configuration of 3 lines as a cubic plane curve, we can represent such a configuration by a degree 3 homogeneous polynomial which factors completely into linear factors, modulo scalars. This naturally embeds such configurations into the projective space of cubic forms $\mathbb{P} H^0(\mathcal{O_{\mathbb P^2}}(3))$, a 9-dimensional projective space.

The general configuration of lines, consisting of three lines meeting at three distinct points in the (finite) plane forms a 6-dimensional (quasiprojective) subvariety of this projective space, as it is the image of the finite map $\mathbb{P}^{2*}\times \mathbb{P}^{2*}\times \mathbb{P}^{2*} \to \mathbb{P}^9$ given by unioning three lines in the plane.

When three lines all meet at the same point, the configuration is determined by selecting the point of intersection and the slopes of the 3 lines. This is therefore a 5-dimensional locus.

If two of the lines become parallel, the configuration is essentially determined by choosing the slope of the parallel lines (equivalently, the point they meet along the line at infinity in projective space) along with say the $y$-intercepts of the parallel lines and the position of the third, nonparallel line. This gives a 5-dimensional space as well.

Other degeneracy loci can be analyzed similarly, essentially by determining the number of degrees of freedom in a configuration.

Schubert calculus and intersection theory don't seem particularly relevant, as these degeneracy loci aren't Schubert cycles and aren't naturally described as intersections.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .