Wow — make one comment on Terry’s blog and you get a ton of traffic.
Terry encouraged me in the comments to experiment with posting math. So I’m going to experiment!
The background: Zeev Dvir this week presented a beautiful and simple proof of the Kakeya conjecture over finite fields. A Kakeya set in F_q^n is a subset containing a line in every direction. Dvir proved that every Kakeya set has at least c_n q^n elements. (Hat tip to The Accidental Mathematician for alerting me to Dvir’s paper!)
In the special case n=2, Dvir’s method shows that a Kakeya set has at least 1/2 q(q+1) elements. In fact (as I learned from Terry’s blog) the best known lower bound is
1/2 q(q+1) + (5/14)q + O(1)
due to Cooper.
Question: Can Dvir’s method be refined to give a better lower bound?
(Note that there are examples of 2-dimensional Kakeya sets, due to Mockenhaupt and Tao, of size 1/2 q(q+1) + (1/2) q + O(1), so Cooper’s bound can’t be improved very much!)
One might naturally start out as follows. The main idea of Dvir’s proof is to show that a Kakeya set can’t be contained in an affine plane curve of degree q-1.
What if S is a Kakeya set contained in an affine plane curve of degree q? That is, what if F in F_q[x,y] is a polynomial of degree q vanishing on S? This places rather strong conditions on F. In each direction m we have a line L_m (say y = mx+b) such that F vanishes on L_m(F_q); since deg F = q, this implies that
F(x,mx + b) = L_m G + c (x^q – x)
for m = 0,1, … q-1. (Of course, there is one more direction, the infinite one, which you can deal with separately.)
Let V be the space of degree-q polynomials vanishing on S. One might like to bound the dimension of V above; because the dimension of the space of degree-q polynomials vanishing on S is at least (1/2)(q+1)(q+2) – |S|, so
|S| <= (1/2)(q+1)(q+2) – dim V.
We note that V is contained in the intersection of q spaces V_0, … V_{q-1}, where
V_m = span of multiples of L_m and x^q-x.
(Note that if not for the x^q – x, we would be looking for polynomials which were multiples of q different linear forms, which indeed makes V very small! This is what happens in Dvir’s case, where the degree is q-1.)
So far, I’m actually not sure whether any of this does more than restate the problem. But one might try to make an argument along the following lines: we can think of V+[x^q-x] as a linear system of degree-q curves in P^2. Now the Kakeya condition on S shows that a whole lot of these curves have one of the lines L_0, … , L_{q-1} as an irreducible component; in particular, the locus R of reducible curves in this linear system contains q+1 hyperplanes.Does this give an upper bound on dim V? One might, for example, observe that if R contains q+1 hyperplanes, then a general pencil of curves in V + [x^q -x] has q+1 reducible fibers; there is a substantial literature about reducible fibers in pencils of hypersurfaces — particularly relevant here seem to be theorems of Lorenzini (1993) and Vistoli (1993). (Beware: these theorems are usually stated in characteristic 0!)
Evening update: I spent a few hours thinking about this and don’t immediately see how to push it through — the hard upper bound on the number of reducible fibers from Lorenzini or Vistoli is q^2 – 1, and it’s not clear to me how you could ensure that some pencil in V + [x^q - x] satisfies the more delicate conditions necessary to get that number down below q. It may be that cutting down to pencils is the wrong thing to do, and one should instead try to show directly that the reducible locus in V + [x^q - x] doesn’t contain q hyperplanes.
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