Borsuk-Ulam theorem

Informally, the Borsuk-Ulam theorem states that, for a "balloon animal" (or any arbitrarily distorted shape) made out of a spherical balloon, and then squashed into a plane (letting the air out somehow), at least one pair of points that were opposite each other on the original sphere will be squashed onto the same point of the plane.

More formally, every continuous map from the sphere to the plane maps some pair of antipodal points to the same point. An analogous statement is true in higher dimensions (see below).

Formally, the theorem states that every continuous function from an n-sphere into n-dimensional Euclidean space must map some pair of antipodal points to the same point. Two points on a sphere are called antipodal if they lie in exactly opposite directions from the center, like the North and South Poles.

The Borsuk-Ulam theorem has several equivalent statements in terms of odd functions. Recall that \(S^n\) is the n-sphere and \(B^n\) is the n-ball:

• If \(g : S^n \to \R^n\) is a continuous odd function, then there exists an \(x\in S^n\) such that: \(g(x)=0\).
• If \(g : B^n \to \R^n\) is a continuous function which is odd on \(S^{n-1}\) (the boundary of \(B^n\)), then there exists an \(x\in B^n\) such that: \(g(x)=0\).

More compactly: if \(f: S^n \to \R^n\) is continuous then there exists an \(x\in S^n\) such that: \(f(-x)=f(x)\).

The case \(n=1\) can be illustrated by saying that there always exist a pair of opposite points on the Earth's equator with the same temperature. The same is true for any circle. This assumes the temperature varies continuously in space, which is, however, not always the case.

The case \(n=2\) is often illustrated by saying that at any moment, there is always a pair of antipodal points on the Earth's surface with equal temperatures and equal barometric pressures, assuming that both parameters vary continuously in space.

According to Matoušek (2003, p. 25), the first historical mention of the statement of the Borsuk-Ulam theorem appears in Lyusternik & Shnirel'man (1930). The first proof was given by Karol Borsuk (1933), where the formulation of the problem was attributed to Stanisław Ulam. Since then, many alternative proofs have been found by various authors, as collected by Steinlein (1985).

A function \(g\) is called odd (aka antipodal or antipode-preserving) if for every \(x\), \(g(-x)=-g(x)\).

The Borsuk-Ulam theorem is equivalent to each of the following statements:

(1) Each continuous odd function \(S^n\to \R^n\) has a zero.

(2) There is no continuous odd function \(S^n \to S^{n-1}\).

Here is a proof that the Borsuk-Ulam theorem is equivalent to (1):

(\(\Longrightarrow\)) If the theorem is correct, then it is specifically correct for odd functions, and for an odd function, \(g(-x)=g(x)\) iff \(g(x)=0\). Hence every odd continuous function has a zero.

(\(\Longleftarrow\)) For every continuous function \(f:S^n\to \R^n\), the following function is continuous and odd: \(g(x)=f(x)-f(-x)\). If every odd continuous function has a zero, then \(g\) has a zero, and therefore, \(f(x)=f(-x)\).

• the obvious inclusion \(i: S^{n-1}\to \R^n\setminus \{0\}\),
• and the radial projection map \(p: \R^n\setminus \{0\} \to S^{n-1}\) given by \(x \mapsto \frac{x}{|x|}\).

Condensed: the full section is in Wikipedia.

The 1-dimensional case can easily be proved using the intermediate value theorem (IVT).

Let \(g\) be the odd real-valued continuous function on a circle defined by \(g(x)=f(x)-f(-x)\). Pick an arbitrary \(x\). If \(g(x)=0\) then we are done. Otherwise, without loss of generality, \(g(x)>0.\) But \(g(-x)<0.\) Hence, by the IVT, there is a point \(y\) at which \(g(y)=0\).

• No subset of \(\R^n\) is homeomorphic to \(S^n\)
• The ham sandwich theorem: For any compact sets A₁, ..., A_n in \(\R^n\) we can always find a hyperplane dividing each of them into two subsets of equal measure.

Above we showed how to prove the Borsuk-Ulam theorem from Tucker's lemma. The converse is also true: it is possible to prove Tucker's lemma from the Borsuk-Ulam theorem. Therefore, these two theorems are equivalent. There are several fixed-point theorems which come in three equivalent variants: an algebraic topology variant, a combinatorial variant and a set-covering variant. Each variant can be proved separately using totally different arguments, but each variant can also be reduced to the other variants in its row. Additionally, each result in the top row can be deduced from the one below it in the same column.

• In the original theorem, the domain of the function f is the unit n-sphere (the boundary of the unit n-ball). In general, it is true also when the domain of f is the boundary of any open bounded symmetric subset of \(\R^n\) containing the origin (Here, symmetric means that if x is in the subset then -x is also in the subset).
• More generally, if \(M\) is a compact n-dimensional Riemannian manifold, and \(f: M \rightarrow \mathbb{R}^n\) is continuous, there exists a pair of points x and y in \(M\) such that \(f(x) = f(y)\) and x and y are joined by a geodesic of length \(\delta\), for any prescribed \(\delta > 0\).
• Consider the function A which maps a point to its antipodal point: \(A(x) = -x.\) Note that \(A(A(x))=x.\) The original theorem claims that there is a point x in which \(f(A(x))=f(x).\) In general, this is true also for every function A for which \(A(A(x))=x.\) However, in general this is not true for other functions A.