Borsuk–Ulam theorem – Wikipedia

Theorem in topology

In mathematics, the Borsuk–Ulam theorem 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

${displaystyle f:S^{n}to mathbb {R} ^{n}}$

is continuous then there exists an

${displaystyle xin S^{n}}$

such that:

${displaystyle f(-x)=f(x)}$

.

The case

${displaystyle n=1}$

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.

The case

${displaystyle n=2}$

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

${displaystyle S^{n}}$

is the n-sphere and

${displaystyle B^{n}}$

is the n-ball:

History[edit]

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 Stanislaw Ulam. Since then, many alternative proofs have been found by various authors, as collected by Steinlein (1985).

Equivalent statements[edit]

The following statements are equivalent to the Borsuk–Ulam theorem.^[1]

With odd functions[edit]

A function

${displaystyle g}$

is called odd (aka antipodal or antipode-preserving) if for every

${displaystyle x}$

:

${displaystyle g(-x)=-g(x)}$

.

The Borsuk–Ulam theorem is equivalent to the following statement: A continuous odd function from an n-sphere into Euclidean n-space has a zero. PROOF:

With retractions[edit]

Define a retraction as a function

${displaystyle h:S^{n}to S^{n-1}.}$

The Borsuk–Ulam theorem is equivalent to the following claim: there is no continuous odd retraction.

Proof: If the theorem is correct, then every continuous odd function from

${displaystyle S^{n}}$

must include 0 in its range. However,

${displaystyle 0notin S^{n-1}}$

so there cannot be a continuous odd function whose range is

${displaystyle S^{n-1}}$

.

Conversely, if it is incorrect, then there is a continuous odd function

${displaystyle g:S^{n}to {mathbb {R}}^{n}}$

with no zeroes. Then we can construct another odd function

${displaystyle h:S^{n}to S^{n-1}}$

by:

${displaystyle h(x)={frac {g(x)}{|g(x)|}}}$

since

${displaystyle g}$

has no zeroes,

${displaystyle h}$

is well-defined and continuous. Thus we have a continuous odd retraction.

1-dimensional case[edit]

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

Let

${displaystyle g}$

be an odd real-valued continuous function on a circle. Pick an arbitrary

${displaystyle x}$

. If

${displaystyle g(x)=0}$

then we are done. Otherwise, without loss of generality,

${displaystyle g(x)>0.}$

${displaystyle g(-x)<0.}$

Hence, by the IVT, there is a point

${displaystyle y}$

between

${displaystyle x}$

and

${displaystyle -x}$

at which

${displaystyle g(y)=0}$

.

General case[edit]

Algebraic topological proof[edit]

Assume that

${displaystyle h:S^{n}to S^{n-1}}$

is an odd continuous function with

${displaystyle n>2}$

${displaystyle n=1}$

is treated above, the case

${displaystyle n=2}$

can be handled using basic covering theory). By passing to orbits under the antipodal action, we then get an induced continuous function

${displaystyle h’:mathbb {RP} ^{n}to mathbb {RP} ^{n-1}}$

between real projective spaces, which induces an isomorphism on fundamental groups. By the Hurewicz theorem, the induced ring homomorphism on cohomology with

${displaystyle mathbb {F} _{2}}$

coefficients [where

${displaystyle mathbb {F} _{2}}$

denotes the field with two elements],

${displaystyle mathbb {F} _{2}[a]/a^{n+1}=H^{*}left(mathbb {RP} ^{n};mathbb {F} _{2}right)leftarrow H^{*}left(mathbb {RP} ^{n-1};mathbb {F} _{2}right)=mathbb {F} _{2}[b]/b^{n},}$

sends

${displaystyle b}$

to

${displaystyle a}$

. But then we get that

${displaystyle b^{n}=0}$

is sent to

${displaystyle a^{n}neq 0}$

, a contradiction.^[2]

One can also show the stronger statement that any odd map

${displaystyle S^{n-1}to S^{n-1}}$

has odd degree and then deduce the theorem from this result.

Combinatorial proof[edit]

The Borsuk–Ulam theorem can be proved from Tucker’s lemma.^[1][3][4]

Let

${displaystyle g:S^{n}to mathbb {R} ^{n}}$

be a continuous odd function. Because g is continuous on a compact domain, it is uniformly continuous. Therefore, for every

${displaystyle epsilon >0}$

${displaystyle delta >0}$

${displaystyle S_{n}}$

which are within

${displaystyle delta }$

of each other, their images under g are within

${displaystyle epsilon }$

of each other.

Define a triangulation of

${displaystyle S_{n}}$

with edges of length at most

${displaystyle delta }$

. Label each vertex

${displaystyle v}$

of the triangulation with a label

${displaystyle l(v)in {pm 1,pm 2,ldots ,pm n}}$

in the following way:

• The absolute value of the label is the index of the coordinate with the highest absolute value of g:
  ${displaystyle |l(v)|=arg max _{k}(|g(v)_{k}|)}$

  .

• The sign of the label is the sign of g, so that:
  ${displaystyle l(v)=operatorname {sgn}(g(v))|l(v)|}$

  .

Because g is odd, the labeling is also odd:

${displaystyle l(-v)=-l(v)}$

. Hence, by Tucker’s lemma, there are two adjacent vertices

${displaystyle u,v}$

with opposite labels. Assume w.l.o.g. that the labels are

${displaystyle l(u)=1,l(v)=-1}$

. By the definition of l, this means that in both

${displaystyle g(u)}$

and

${displaystyle g(v)}$

, coordinate #1 is the largest coordinate: in

${displaystyle g(u)}$

this coordinate is positive while in

${displaystyle g(v)}$

it is negative. By the construction of the triangulation, the distance between

${displaystyle g(u)}$

and

${displaystyle g(v)}$

is at most

${displaystyle epsilon }$

, so in particular

${displaystyle |g(u)_{1}-g(v)_{1}|=|g(u)_{1}|+|g(v)_{1}|leq epsilon }$

(since

${displaystyle g(u)_{1}}$

and

${displaystyle g(v)_{1}}$

have opposite signs) and so

${displaystyle |g(u)_{1}|leq epsilon }$

. But since the largest coordinate of

${displaystyle g(u)}$

is coordinate #1, this means that

${displaystyle |g(u)_{k}|leq epsilon }$

for each

${displaystyle 1leq kleq n}$

. So

${displaystyle |g(u)|leq c_{n}epsilon }$

, where

${displaystyle c_{n}}$

is some constant depending on

${displaystyle n}$

and the norm

${displaystyle |cdot |}$

which you have chosen.

The above is true for every

${displaystyle epsilon >0}$

${displaystyle S_{n}}$

is compact there must hence be a point u in which

${displaystyle |g(u)|=0}$

.

Corollaries[edit]

• No subset of
  ${displaystyle mathbb {R} ^{n}}$

  is homeomorphic to

  ${displaystyle S^{n}}$

  

• The ham sandwich theorem: For any compact sets A₁, …, A_n in
  ${displaystyle mathbb {R} ^{n}}$

  we can always find a hyperplane dividing each of them into two subsets of equal measure.

Equivalent results[edit]

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.^[5]

Generalizations[edit]

• 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
  ${displaystyle mathbb {R} ^{n}}$

  containing the origin (Here, symmetric means that if x is in the subset then –x is also in the subset).^[6]

• Consider the function A which maps a point to its antipodal point:
  ${displaystyle A(x)=-x.}$

  Note that

  ${displaystyle A(A(x))=x.}$

  The original theorem claims that there is a point x in which

  ${displaystyle f(A(x))=f(x).}$

  In general, this is true also for every function A for which

  ${displaystyle A(A(x))=x.}$

  ^[7] However, in general this is not true for other functions A.^[8]

See also[edit]

References[edit]

External links[edit]