From http://michaelnielsen.org/polymath1/index.php?title=The_hot_spots_conjecture and other sources:

The hotspots conjecture can be expressed in simple English as: Suppose a flat piece of metal, represented by a two-dimensional bounded connected domain R, is given a generic initial heat distribution which then flows throughout the metal. Assuming the metal is insulated (i.e. no heat escapes from the piece of metal), then at all sufficiently large times, the hottest and coldest points on the metal both will lie on its boundary... Equivalently, the 2nd eigenfunction of the laplacian has its global max and global min on the boundary of the domain. This conjecture has been proven for some domains and disproven for others. It is conjectured by Rodrigo Banuelos for any bounded simply-connected planar domain. It is false for certain bounded planar domains with 2 holes. The polymath7 project is to prove it for acute triangles. -------------------- I point out this conjecture is "obviously false" in any dimension>=1 for a rotationally-symmetric ball. PROOF: start with generic rotationally symmetric heat distribution. (Hence eternally is rotationally symmetric.) At all future times, temperature is same everywhere on the boundary. Hence conjecture that the boundary includes the hottest point, contradicts conjecture it includes the coldest point. QED. However, I suppose the conjecturers will respond that my initial heat distribution was not generic enough, hence this is not really a disproof. In 1 dimension the 2nd eigenfunction of the interval [-pi, pi] is sin(x/2) which obeys the conjecture (and is odd- not even-symmetric). A proof the hotspots conjecture is false on the surface of a standard Moebius strip would follow if we knew the 2nd eigenfunction was cos(2*pi*y)*sin(2*pi*x) for the unit square 0<x<1, 0<y<1 with wraparound x mod 1 while reversing y. If you have the 2D surface of a 3D sphere, then the conjecture is false in the trivial sense there is no boundary :) If you make an initial heat distribution proportional to the z-coordinate, the min and max will forever be at the z-max and z-min "polar" locations. If you drill a tiny hole on the equator so this surface now does have a boundary, would that be a counterexample now with the same topology as a disc? If so, that would prove that topology and boundedness are not enough -- the actual zero-curvature geometry of the plane is also needed. -- Warren D. Smith http://RangeVoting.org