# The Ham Sandwich Theorem Via Borsuk Ulam

I’ve recently been doing some research for my Senior thesis in applications of characteristic classes to plane equipartition problems.These types of problems follow a general procedure known as the Configuration Space-Test Map paradigm. I’ll try to demonstrate how these basic arguments go, using what is probably the most famous theorem in equivariant topology:

The Borsuk Ulam Theorem: Every -equivariant map vanishes.

The theorem can be proven in a more general language that I will try to flesh out in future blog posts, but I will prove the two dimensional case using just basic covering space theory here and apply it to the two-dimensional ham sandwich theorem:

The Ham-Sandwich Theorem: Any two masses in can be simultaneously  equipartitioned by a single line.

The idea will be as follows: We will construct a space that topologizes all possible lines in the plane, and then compactify it in a way consistent with our problem. The bare-boned collection of affine spaces is nothing but , or the open mobius band after identifying equivalent lines. We will basically ignore this identification at first, and compactify to obtain a sphere equipped with a (antipodal) action. From this, we will construct a “test function” that will just measure the areas on either side of this line for each mass, which will be . There will be an obvious action on the areas (flipping above and below the line), so will in fact be equivariant. A solution to our problem will be , since we will consider , which will be the difference between areas above and below the line. The Borsuk Ulam theorem implies that a map is impossible given that is continuous.

First, I’ll show the construction, then prove the Borsuk Ulam theorem, and then do a depressing verification that this map is continuous. The proof will immediately generalize given a higher dimensional Borsuk-Ulam theorem, which can be proven using only basic algebraic results, but it will be more interesting to revisit this from a slightly different point of view.

Question:  what is a mass? We basically define it in the most general terms to make the proof go through. For all intents and purposes, a compact subset of would work, and the areas could be Lebesgue measure of the characteristic function on each set, but: a mass is any finite measure of in the Borel -algebra that is absolutely continuous with respect to Lebesgue measure (hyperplanes have measure !)

The configuration space: is a line granted that are not simultaneously zero.

The equation can be scaled so that the triple , by dividing by the appropriate factor.

Consider a “half space” determined by a line. Given a point , we define In the most natural way, we extend this definition to include the “lines at infinity:” Hence, Our “test space” of all lines becomes the whole sphere .

Test Map: Consider the function given by **, where are area functions for the two blobs.

The first thing to notice here, is that while Then, a solution to our problem is a root of The Borsuk-Ulam theorem says this exists!

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Proof of 2D Borsuk Ulam: If is an odd function such that is nonvanishing, we can define , where . However, since this is an odd map, it descends to , so must be trivial, and hence there is a lift to , making the map nullhomotopic, contradicting the fact that .

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Showing the map is continuous: We need to show that this assignment is continuous. Let’s make these area functions formal by claiming them to be lebesgue measures on compact sets and . It will suffice to show that for any , we also have that . Note that for some that is not on the boundary (which has measure zero since it is a line), we will have for sufficiently large , . This in turn means that for the characteristic function on , we have that almost everywhere.

Hence by the compactness assumption, we can apply the dominated convergence theorem: as desired.