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mathematics Mathematics is a field of study that discovers and organizes methods, Mathematical theory, theories and theorems that are developed and Mathematical proof, proved for the needs of empirical sciences and mathematics itself. There are many ar ...
, the Borsuk–Ulam theorem states that every
continuous function In mathematics, a continuous function is a function such that a small variation of the argument induces a small variation of the value of the function. This implies there are no abrupt changes in value, known as '' discontinuities''. More preci ...
from an ''n''-sphere into Euclidean ''n''-space maps some pair of
antipodal point In mathematics, two points of a sphere (or n-sphere, including a circle) are called antipodal or diametrically opposite if they are the endpoints of a diameter, a straight line segment between two points on a sphere and passing through its cen ...
s to the same point. Here, two points on a sphere are called antipodal if they are in exactly opposite directions from the sphere's center. Formally: 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 Earth is the third planet from the Sun and the only astronomical object known to Planetary habitability, harbor life. This is enabled by Earth being an ocean world, the only one in the Solar System sustaining liquid surface water. Almost all ...
'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. 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^ (the boundary of B^n), then there exists an x\in B^n such that: g(x)=0.


History

According to , the first historical mention of the statement of the Borsuk–Ulam theorem appears in . The first proof was given by , where the formulation of the problem was attributed to
Stanisław Ulam Stanisław Marcin Ulam ( ; 13 April 1909 – 13 May 1984) was a Polish and American mathematician, nuclear physicist and computer scientist. He participated in the Manhattan Project, originated the History of the Teller–Ulam design, Telle ...
. Since then, many alternative proofs have been found by various authors, as collected by .


Equivalent statements

The following statements are equivalent to the Borsuk–Ulam theorem.


With odd functions

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^. 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). To prove that (1) and (2) are equivalent, we use the following continuous odd maps: * the obvious inclusion i: S^\to \R^n\setminus \ , * and the radial projection map p: \R^n\setminus \ \to S^ given by x \mapsto \frac. The proof now writes itself. ((1) \Longrightarrow (2)) We prove the contrapositive. If there exists a continuous odd function f:S^n\to S^, then i\circ f is a continuous odd function S^n\to \R^n\setminus \. ((1) \Longleftarrow (2)) Again we prove the contrapositive. If there exists a continuous odd function f:S^n\to \R^\setminus\, then p\circ f is a continuous odd function S^n\to S^.


Proofs


1-dimensional case

The 1-dimensional case can easily be proved using the
intermediate value theorem In mathematical analysis, the intermediate value theorem states that if f is a continuous function whose domain contains the interval , then it takes on any given value between f(a) and f(b) at some point within the interval. This has two imp ...
(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.


General case


Algebraic topological proof

Assume that h: S^n \to S^ is an odd continuous function with n > 2 (the case n = 1 is treated above, the case n = 2 can be handled using basic covering theory). By passing to orbits under the antipodal action, we then get an induced continuous function h': \mathbb^n \to \mathbb^ between real projective spaces, which induces an isomorphism on fundamental groups. By the Hurewicz theorem, the induced
ring homomorphism In mathematics, a ring homomorphism is a structure-preserving function between two rings. More explicitly, if ''R'' and ''S'' are rings, then a ring homomorphism is a function that preserves addition, multiplication and multiplicative identity ...
on
cohomology In mathematics, specifically in homology theory and algebraic topology, cohomology is a general term for a sequence of abelian groups, usually one associated with a topological space, often defined from a cochain complex. Cohomology can be viewed ...
with \mathbb F_2 coefficients field with two elements">GF(2).html" ;"title="here \mathbb F_2 denotes the GF(2)">field with two elements : \mathbb F_2[a]/a^ = H^*\left(\mathbb^n; \mathbb_2\right) \leftarrow H^*\left(\mathbb^; \mathbb F_2\right) = \mathbb F_2[b]/b^, sends b to a. But then we get that b^n = 0 is sent to a^n \neq 0, a contradiction.Joseph J. Rotman, ''An Introduction to Algebraic Topology'' (1988) Springer-Verlag ''(See Chapter 12 for a full exposition.)'' One can also show the stronger statement that any odd map S^ \to S^ has odd degree and then deduce the theorem from this result.


Combinatorial proof

The Borsuk–Ulam theorem can be proved from Tucker's lemma. Let g : S^n \to \R^n be a continuous odd function. Because ''g'' is continuous on a
compact Compact as used in politics may refer broadly to a pact or treaty; in more specific cases it may refer to: * Interstate compact, a type of agreement used by U.S. states * Blood compact, an ancient ritual of the Philippines * Compact government, a t ...
domain, it is uniformly continuous. Therefore, for every \epsilon > 0, there is a \delta > 0 such that, for every two points of S_n which are within \delta of each other, their images under ''g'' are within \epsilon of each other. Define a triangulation of S_n with edges of length at most \delta. Label each vertex v of the triangulation with a label l(v)\in in the following way: * The absolute value of the label is the ''index'' of the coordinate with the highest absolute value of ''g'': , l(v), = \arg\max_k (, g(v)_k, ). * The sign of the label is the sign of ''g'' at the above coordinate, so that: l(v) = \sgn (g(v)_) , l(v), . Because ''g'' is odd, the labeling is also odd: l(-v) = -l(v). Hence, by Tucker's lemma, there are two adjacent vertices u, v with opposite labels. Assume w.l.o.g. that the labels are l(u)=1, l(v)=-1. By the definition of ''l'', this means that in both g(u) and g(v), coordinate #1 is the largest coordinate: in g(u) this coordinate is positive while in g(v) it is negative. By the construction of the triangulation, the distance between g(u) and g(v) is at most \epsilon, so in particular , g(u)_1 - g(v)_1, = , g(u)_1, + , g(v)_1, \leq \epsilon (since g(u)_1 and g(v)_1 have opposite signs) and so , g(u)_1, \leq \epsilon. But since the largest coordinate of g(u) is coordinate #1, this means that , g(u)_k, \leq \epsilon for each 1 \leq k \leq n. So , g(u), \leq c_n \epsilon, where c_n is some constant depending on n and the norm , \cdot, which you have chosen. The above is true for every \epsilon > 0; since S_n is compact there must hence be a point ''u'' in which , g(u), =0.


Corollaries

* No subset of \R^n is
homeomorphic In mathematics and more specifically in topology, a homeomorphism ( from Greek roots meaning "similar shape", named by Henri Poincaré), also called topological isomorphism, or bicontinuous function, is a bijective and continuous function betw ...
to S^n * The ham sandwich theorem: For any
compact Compact as used in politics may refer broadly to a pact or treaty; in more specific cases it may refer to: * Interstate compact, a type of agreement used by U.S. states * Blood compact, an ancient ritual of the Philippines * Compact government, a t ...
sets ''A''1, ..., ''An'' in \R^n we can always find a hyperplane dividing each of them into two subsets of equal measure.


Equivalent results

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.


Generalizations

* 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 In differential geometry, a Riemannian manifold is a geometric space on which many geometric notions such as distance, angles, length, volume, and curvature are defined. Euclidean space, the N-sphere, n-sphere, hyperbolic space, and smooth surf ...
, and f: M \rightarrow \mathbb^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''.


See also

*
Topological combinatorics The mathematical discipline of topological combinatorics is the application of topological and algebro-topological methods to solving problems in combinatorics. History The discipline of combinatorial topology used combinatorial concepts in topo ...
*
Necklace splitting problem Necklace splitting is a picturesque name given to several related problems in combinatorics and measure theory. Its name and solutions are due to mathematicians Noga Alon and Douglas B. West. The basic setting involves a necklace with beads of ...
* Ham sandwich theorem * Kakutani's theorem (geometry) * Imre Bárány


Notes


References

* * * * *


External links

*
The Borsuk-Ulam Explorer
An interactive illustration of Borsuk-Ulam Theorem. {{DEFAULTSORT:Borsuk-Ulam Theorem Theorems in algebraic topology Combinatorics Theory of continuous functions Theorems in topology