That Andy! Always with the counterexamples! I had been reading about maps from R^n to R^q and talking about (n-q)-dimensional area, but tripped up on 3-2 = 1. Ah, the glories of old age. Anyhow by 2-dimensional area I meant 1-dimensional length. So: Given a continuous g : [0, 1]^3 —> R^2, the problem is: Prove there exists p in R^2 with the *length* of the inverse image g^(-1)(p) of p is >= 1. ( Or more generally, for any continuous map g : [0, 1]^n —> R^q show there exists a point p of R^q such that (the (n-q)-dimensional "area" of g^(-1)(p)) >= 1 ) —Dan -----
Say
f : [0,1]^3 —> R^2
is a continuous map from the unit cube to the plane.
Unsolved problem (I think):
----- Prove there is some point p in R^2 whose inverse image f^(-1)(p) has area >= 1.
Take f(x,y, z) = (x,y). The inverse image of any point is a line segment, which has area 0. Andy
-----
Here "area" means 2-dimensional Hausdorff measure.*
This should properly be called a conjecture. It is related to an apparently difficult — but proven — theorem in geometry, the "waist inequality" of Gromov:
Let S^3 be the unit sphere in 4-space, and let
g : S^3 —> R^2
be any continuous map, where D^3 is the closed unit ball in R^3.
----- Then there exists some point p of R^2 whose inverse image g^(-1)(p) has area at least the area of the equator of S^3. That is, >= 4π^2, since the equator is a unit 2-sphere). -----
—Dan
————— * Hausdorff r-measure is an ingenious way to assign a reasonable meaning to the notion of r-dimensional measure of a compact subset K of Euclidean space R^N, for any positive real number r. When r is an integer it works like this:
Let alph(N) > 0 be the constant for which the volume of an N-dimensional ball D_R of radius R in R^n is equal to
vol(D_R) = alpha(N) R^N.
Now in R^N, let C be any covering of K by open balls of radii <= eps, for some eps > 0.
Define
|C|_(r, eps) = alph(N) * (the sum of the r'th power of each radius of a ball in C).
Finally, define
|C|_r = inf |C|_(r, eps) eps > 0
It's a bit complicated, but it's exactly the right definition.