In mathematics, the Borsuk–Ulam theorem (BUT), states that every continuous function from an n-sphere into Euclidean n-space maps some pair of antipodal points 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 is continuous then there exists an such that: .
The case 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.
The case 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.
- If is a continuous odd function, then there exists an such that: .
- If is a continuous function which is odd on (the boundary of ), then there exists an such that: .
According to Matoušek (2003, p. 25), the first historical mention of the statement of BUT appears in Lyusternik & Shnirel'man (1930). The first proof was given by Karol Borsuk (1933), where the formulation of the problem was attributed to Stanislaw Ulam. Since then, many alternative proofs have been found by various authors, as collected by Steinlein (1985).
The following statements are equivalent to BUT.
With odd functions
A function is called odd (aka antipodal or antipode-preserving) if for every : .
BUT is equivalent to the following statement: A continuous odd function from an n-sphere into Euclidean n-space has a zero. PROOF:
- If BUT is correct, then it is specifically correct for odd functions, and for an odd function, iff . Hence every odd continuous function has a zero.
- For every continuous function , the following function is continuous and odd: . If every odd continuous function has a zero, then has a zero, and therefore, . Hence BUT is correct.
Define a retraction as a function .
BUT is equivalent to the following claim: there is no continuous odd retraction.
PROOF: If BUT is correct, then every continuous odd function from must include 0 in its range. However, so there cannot be a continuous odd function whose range is .
Conversely, if BUT is incorrect, then there is a continuous odd function with no zeroes. Then we can construct another odd function by:
since has no zeroes, is well-defined and continuous. Thus we have a continuous odd retraction.
The 1-dimensional case can easily be proved using the intermediate value theorem (IVT).
Let be an odd real-valued function on a circle. Pick an arbitrary . If then we are done. Otherwise, w.l.o.g. . But . Hence, by the IVT there is a point between and on which .
General case - algebraic topology proof
Assume that is an odd function with (the case is treated above, the case can be handled using basic covering theory). By passing to orbits under the antipodal action, we then get an induced function , which induces an isomorphism on fundamental groups. By the Hurewicz theorem, the induced map on cohomology with coefficients, , sends to . But then we get that is send to , a contradiction.
One can also show the stronger statement that any odd map has odd degree and then deduce BUT from this result.
General case - combinatorial proof
Let be a continuous odd function. Because g is continuous on a compact domain, it is uniformly continuous. Therefore, for every , there is a such that, for every two points of which are within of each other, their images under g are within of each other.
Define a triangulation of with edges of length at most . Label each vertex of the triangulation with a label in the following way:
- The absolute value of the label is the index of the coordinate with the highest absolute value of g: .
- The sign of the label is the sign of g, so that: .
Because g is odd, the labeling is also odd: . Hence, by Tucker's lemma, there are two adjacent vertices with opposite labels. Assume w.l.o.g. that the labels are . By definition of l, this means that in both and , coordinate #1 is the largest coordinate; in this coordinate is positive while in it is negative. By the construction of the triangulation, the distance between and is at most ; this means that both and are bounded by .
The above is true for every ; hence there must be a point u in which .
- No subset of Rn is homeomorphic to Sn.
- The Ham sandwich theorem: For any compact sets A1, ..., An in Rn we can always find a hyperplane dividing each of them into two subsets of equal measure.
Above we showed how to prove BUT from Tucker's lemma. The converse is also true: it is possible to prove Tucker's lemma from BUT. 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 can be reduced to the other result in its column.
|Algebraic topology||Combinatorics||Set covering|
|Brouwer fixed-point theorem||Sperner's lemma||KKM lemma|
|Borsuk–Ulam theorem||Tucker's lemma||Lusternik–Schnirelmann theorem|
1. In the original BUT, 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 Rn containing the origin (Here, symmetric means that if x is in the subset then -x is also in the subset).
2. Consider the function A which maps a point to its antipodal point: A(x)=-x. Note that A(A(x))=x. The original BUT 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.
- Topological combinatorics
- Necklace splitting problem
- Kakutani's theorem (geometry)
- Imre Barany
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- Su, Francis Edward (Nov 1997). "Borsuk-Ulam Implies Brouwer: A Direct Construction" (PDF). The American Mathematical Monthly. 104 (9): 855–859. doi:10.2307/2975293.