Borsuk–Ulam theorem explained

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

f:Sn\to\Rn

is continuous then there exists an

x\inSn

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.[1]

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

Sn

is the n-sphere and

Bn

is the n-ball:

g:Sn\to\Rn

is a continuous odd function, then there exists an

x\inSn

such that:

g(x)=0

.

g:Bn\to\Rn

is a continuous function which is odd on

Sn-1

(the boundary of

Bn

), then there exists an

x\inBn

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. 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.[2]

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 the following statement: A continuous odd function from an n-sphere into Euclidean n-space has a zero. PROOF:

g(-x)=g(x)

iff

g(x)=0

. Hence every odd continuous function has a zero.

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 the theorem is correct.

With retractions

Define a retraction as a function

h:Sn\toSn-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

Sn

must include 0 in its range. However,

0\notinSn-1

so there cannot be a continuous odd function whose range is

Sn-1

.

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

g:Sn\to\Bbb{R}n

with no zeroes. Then we can construct another odd function

h:Sn\toSn-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.

Proofs

1-dimensional case

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

between

x

and

-x

at which

g(y)=0

.

General case

Algebraic topological proof

Assume that

h:Sn\toSn-1

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':RPn\toRPn-1

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

F2

coefficients [where <math>\mathbb F_2</math> denotes the [[GF(2)|field with two elements]]],
n+1
F
2[a]/a

=H*\left(RPn;F2\right)\leftarrowH*\left(RPn-1;F2\right)=

n
F
2[b]/b

,

sends

b

to

a

. But then we get that

bn=0

is sent to

an0

, a contradiction.[3]

One can also show the stronger statement that any odd map

Sn-1\toSn-1

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

Combinatorial proof

The Borsuk - Ulam theorem can be proved from Tucker's lemma.[2] [4] [5]

Let

g:Sn\to\Rn

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

\epsilon>0

, there is a

\delta>0

such that, for every two points of

Sn

which are within

\delta

of each other, their images under g are within

\epsilon

of each other.

Define a triangulation of

Sn

with edges of length at most

\delta

. Label each vertex

v

of the triangulation with a label

l(v)\in{\pm1,\pm2,\ldots,\pmn}

in the following way:

|l(v)|=\argmaxk(|g(v)k|)

.

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\leqk\leqn

. So

|g(u)|\leqcn\epsilon

, where

cn

is some constant depending on

n

and the norm

||

which you have chosen.

The above is true for every

\epsilon>0

; since

Sn

is compact there must hence be a point u in which

|g(u)|=0

.

Corollaries

\Rn

is homeomorphic to

Sn

\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 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

\Rn

containing the origin (Here, symmetric means that if x is in the subset then -x is also in the subset).

M

is a compact n-dimensional Riemannian manifold, and

f:MRn

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

.[6] [7]

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.

[8] However, in general this is not true for other functions A.[9]

See also

References

External links

Notes and References

  1. Jha . Aditya . Campbell . Douglas . Montelle . Clemency . Wilson . Phillip L. . 2023-07-30 . On the Continuum Fallacy: Is Temperature a Continuous Function? . Foundations of Physics . en . 53 . 4 . 69 . 10.1007/s10701-023-00713-x . 1572-9516. free .
  2. Extensions of the Borsuk–Ulam Theorem . BS . Harvey Mudd College . 2002 . Prescott, Timothy . 10.1.1.124.4120.
  3. Joseph J. Rotman, An Introduction to Algebraic Topology (1988) Springer-Verlag (See Chapter 12 for a full exposition.)
  4. 10.1016/0097-3165(81)90027-3 . 1982. 30 . 3. A constructive proof of Tucker's combinatorial lemma . . Series A . 321–325. Freund, Robert M.. Todd, Michael J.. free.
  5. Simmons, Forest W.. Su, Francis Edward . 10.1016/s0165-4896(02)00087-2 . 2003. 45 . Consensus-halving via theorems of Borsuk - Ulam and Tucker . Mathematical Social Sciences . 15–25. 10419/94656 . free .
  6. Hopf . H. . 1944 . Eine Verallgemeinerung bekannter Abbildungs-und Überdeckungssätze . Portugaliae Mathematica.
  7. Malyutin . A. V. . Shirokov . I. M. . 2023 . Hopf-type theorems for f-neighbors . Sib. Èlektron. Mat. Izv . 20 . 1 . 165–182.
  8. 10.2307/1969632 . Yang, Chung-Tao . 1954. 60 . 2 . On Theorems of Borsuk-Ulam, Kakutani-Yamabe-Yujobo and Dyson, I . . 262–282. 1969632 .
  9. Web site: Generalization of Borsuk-Ulam . Math Overflow . 18 May 2015 . Jens Reinhold, Faisal . Sergei Ivanov.