Abstract:

A line can meet a circle in 2 points, and any pair of points on the circle are the intersection with some line. For a plane curve $X$ of degree 3, i.e. ${f(x,y)=0}$ where $f(x,y)$ is a polynomial of degree 3 in $x$ and $y$, a line usually meets $X$ in 3 points, but since there is only a 2-parameter family of lines in the plane, there is one condition on three points on $X$ for them to be the intersection with a line. What is this condition? We want a condition in terms of intrinsic geometry on $X$?

For a triangle in the plane, i.e. the union of three lines, there is also one condition for 3 points on the triangle to be the intersection with a line. In fact, given $m$ points on each edge of the triangle, there is one condition on the $3m$ points for them to be the intersection of the triangle with some plane curve of degree $m$. What is this condition? Why is this question a lot easier than the first question?

A tetrahedron in 3-space, i.e. the union of 4 planes, meets a line in 4 points, one on each face. The tetrahedron meets a space curve of degree $m$ in $4m$ points, $m$ on each face. There is an answer to this problem once we change it a little bit, but now the answer is really interesting, involving an exotic object called the Milnor K-group.

If we take a surface $X$ of degree $d$ in space, given by one equation of degree $d$ in 3 variables, when does a curve $Csubset X$ defined by algebraic equations turn out to be the intersection with another surface $Y$ of degree $m$? This problem is easier than the tetrahedron problem, but harder than the triangle problem. What is the answer? We will see that the topology of the surface comes in in a natural way.

So far, all of the questions are ones that I will answer in the talk. If we look at the topological issues raised in the preceding paragraph carefully, then we meet up with the Hodge Conjecture, one of the most celebrated and important problems in mathematics. This conjecture has a price on its head--the Clay Foundation has made it one of seven $1,000,000 prize problems. This talk will be an opportunity to find out what the problem is and begin fantasizing about how to spend the money.