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Borsuk–Ulam theorem

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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.

Contents

Formally: if f : S n → R n is continuous then there exists an x ∈ 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.

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.

BUT 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 → R n is a continuous odd function, then there exists an x ∈ S n such that: g ( x ) = 0 .
  • If g : B n → R n is a continuous function which is odd on S n − 1 (the boundary of B n ), then there exists an x ∈ B n such that: g ( x ) = 0 .

    • History

    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).

    Equivalent statements

    The following statements are equivalent to BUT.

    With odd functions

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

    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, g ( − x ) = g ( x ) iff g ( x ) = 0 . Hence every odd continuous function has a zero.
  • For every continuous function f , 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 ) . Hence BUT is correct.

    • With retractions

    Define a retraction as a function h : S n → S n − 1 .

    BUT is equivalent to the following claim: there is no continuous odd retraction.

    PROOF: If BUT is correct, then every continuous odd function from S n must include 0 in its range. However, 0 ∉ S n − 1 so there cannot be a continuous odd function whose range is S n − 1 .

    Conversely, if BUT is incorrect, then there is a continuous odd function g : S n → R n with no zeroes. Then we can construct another odd function h : S n → S n − 1 by:

    h ( x ) = g ( x ) | g ( x ) |

    since g has no zeroes, h is well-defined and continuous. Thus we have a continuous odd retraction.

    1-dimensional case

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

    Let g be an odd real-valued function on a circle. Pick an arbitrary x . If g ( x ) = 0 then we are done. Otherwise, w.l.o.g. g ( x ) > 0 . But g ( − x ) < 0 . Hence, by the IVT there is a point y between x and − x on which g ( y ) = 0 .

    General case - algebraic topology proof

    Assume that h : S n → S n − 1 is an odd 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 function h ′ : R P n → R P n − 1 , which induces an isomorphism on fundamental groups. By the Hurewicz theorem, the induced map on cohomology with F 2 coefficients, F 2 [ a ] / a n + 1 = H ∗ ( R P n ; F 2 ) ← H ∗ ( R P n − 1 ; F 2 ) = F 2 [ b ] / b n , sends b to a . But then we get that b n = 0 is send to a n ≠ 0 , a contradiction.

    One can also show the stronger statement that any odd map S n − 1 → S n − 1 has odd degree and then deduce BUT from this result.

    General case - combinatorial proof

    BUT can be proved from Tucker's lemma.

    Let g : S n → R n be a continuous odd function. Because g is continuous on a compact domain, it is uniformly continuous. Therefore, for every ϵ > 0 , there is a δ > 0 such that, for every two points of S n which are within δ of each other, their images under g are within ϵ of each other.

    Define a triangulation of S n with edges of length at most δ . Label each vertex v of the triangulation with a label l ( v ) ∈ ± 1 , ± 2 , . . . , ± n 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, 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 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 ϵ ; this means that both | g ( u ) | and | g ( v ) | are bounded by ϵ .

    The above is true for every ϵ ; hence there must be a point u in which | g ( u ) | = 0 .

    Corollaries

  • 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.

    • Equivalent results

    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 in the top row can be deduced from the one below it in the same column.

    Generalizations

    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.

    References

    Borsuk–Ulam theorem Wikipedia
    (Text) CC BY-SA


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