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 \mathbb Z_2-equivariant map f:S^n \to \mathbb R^n 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 \mathbb R^2 can be simultaneously  equipartitioned by a single line.

The idea will be as follows: We will construct a space X 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 \mathbb RP^2\setminus \{pt\}, or the open mobius band after identifying equivalent lines. We will basically ignore this identification at first, and compactify to obtain a sphere S^2 equipped with a \mathbb Z_2 (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 f:S^2 \to \mathbb R^2. There will be an obvious action on the areas (flipping above and below the line), so f will in fact be \mathbb Z_2 equivariant. A solution to our problem will be 0 \in \mathbb R^2, since we will consider f(x)-f(-x), which will be the difference between areas above and below the line. The Borsuk Ulam theorem implies that a map f:S^2 \to \mathbb R^2 \setminus\{0\} is impossible given that f 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 \mathbb R^2 would work, and the areas could be Lebesgue measure of the characteristic function on each set, but: a mass is any finite measure of \mathbb R^2 in the Borel \sigma-algebra that is absolutely continuous with respect to Lebesgue measure (hyperplanes have measure 0!)

 

The configuration space: ax+by=c is a line granted that a,b are not simultaneously zero.

The equation can be scaled so that the triple (a,b,c) \in S^2, by dividing by the appropriate factor.

Consider a “half space” determined by a line. Given a point (a,b,c) \in S^2, we define

    \[H^{+}(\mathbf{u}):=\{(x,y) \in \mathbb R^2 \mid ax+by \leq c \}\]

 

In the most natural way, we extend this definition to include the “lines at infinity:”

    \[H^{+}((0,0,1))= \mathbb R^2 \, \, \,\,\,\,, H^{+}((0,0,-1))=\emptyset\]

Hence, Our “test space” of all lines becomes the whole sphere S^2.

Test Map: Consider the function f:S^2 \to \mathbb R^2 given by f(\mathbf{x})=(A_1(H^{+}(\mathbf{x}),A_2H^{+}(\mathbf{x}))**, where A_1,A_2 are area functions for the two blobs.

The first thing to notice here, is that

    \[H^{+}(a,b,c)=\{(x,y) \in \mathbb R^2 \mid ax+by \leq c \},\]

while

    \[H^{+}(-a,-b,-c)=\{(x,y) \in \mathbb R^2 \mid ax+by \geq c \}.\]

Then, a solution to our problem is a root of

    \[g(\mathbf{x})=f(\mathbf{x})-f(-\mathbf{x})=0.\]

The Borsuk-Ulam theorem says this exists!

 

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Proof of 2D Borsuk Ulam: If f:S^2 \to \mathbb R^2 is an odd function such that is nonvanishing, we can define g(x):=f(x)/\|f(x)\|, where g:S^2 \to S^1. However, since this is an odd map, it descends to \mathbb RP^2 \to \mathbb RP^1 \cong S^1, so g_*:\pi_1(\mathbb R P^2) \to \pi_1(S^1) must be trivial, and hence there is a lift to \mathbb R, making the map g_* nullhomotopic, contradicting the fact that g(-x)=-g(x).

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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 \mu_1 and \mu_2. It will suffice to show that for any u_n \to u \in S^2, we also have that f(u_i) \to u. Note that for some x \in \mathbb R^2 that is not on the boundary \partial H^{+}(u) (which has measure zero since it is a line), we will have for sufficiently large n, x \in H^{+}(u_n) \iff x \in H^{+}(u). This in turn means that for the characteristic function \chi_u on H^{+}(U), we have that \chi_{u_n} \to \chi(u) almost everywhere.

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

    \[\mu_i(H^{+}u_n)=\lim_{n \to \infty}\int \chi_{ u_n} d \mu_i\to \int \chi_u d \mu_i=\mu_i(H^+(u)),\]

as desired.