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
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 in space, which is, however, not always the case.
[1] 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, assuming that both parameters vary continuously in space.
The Borsuk–Ulam theorem has several equivalent statements in terms of odd functions. Recall that
is the
n-sphere and
is the
n-ball:
is a continuous odd function, then there exists an
such that:
.
is a continuous function which is odd on
(the boundary of
), then there exists an
such that:
.
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
is called
odd (aka
antipodal or
antipode-preserving) if for every
:
.
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:
- If the theorem 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 the theorem is correct.
With retractions
Define a retraction as a function
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
must include 0 in its range. However,
so there cannot be a continuous odd function whose range is
.
Conversely, if it 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.
Proofs
1-dimensional case
The 1-dimensional case can easily be proved using the intermediate value theorem (IVT).
Let
be the odd real-valued continuous function on a circle defined by
. Pick an arbitrary
. If
then we are done. Otherwise, without loss of generality,
But
Hence, by the IVT, there is a point
between
and
at which
.
General case
Algebraic topological proof
Assume that
is an odd continuous 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 continuous function
between
real projective spaces, which induces an isomorphism on
fundamental groups. By the
Hurewicz theorem, the induced
ring homomorphism on
cohomology with
coefficients [where <math>\mathbb F_2</math> denotes the [[GF(2)|field with two elements]]],
=H*\left(RPn;F2\right)\leftarrowH*\left(RPn-1;F2\right)=
,
sends
to
. But then we get that
is sent to
, a contradiction.
[3] One can also show the stronger statement that any odd map
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
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
l(v)\in{\pm1,\pm2,\ldots,\pmn}
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 the 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
, so in particular
|g(u)1-g(v)1|=|g(u)1|+|g(v)1|\leq\epsilon
(since
and
have opposite signs) and so
. But since the largest coordinate of
is coordinate #1, this means that
for each
. So
, where
is some constant depending on
and the norm
which you have chosen.
The above is true for every
; since
is compact there must hence be a point
u in which
.
Corollaries
is
homeomorphic to
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
containing the origin (Here, symmetric means that if
x is in the subset then -
x is also in the subset).
is a compact
n-dimensional
Riemannian manifold, and
is continuous, there exists a pair of points
x and
y in
such that
and
x and
y are joined by a geodesic of length
, for any prescribed
.
[6] [7] - Consider the function A which maps a point to its antipodal point:
Note that
The original theorem claims that there is a point
x in which
In general, this is true also for every function
A for which
[8] However, in general this is not true for other functions
A.
[9] See also
References
- Karol. Borsuk. Drei Sätze über die n-dimensionale euklidische Sphäre. Fundamenta Mathematicae. 20. 1933. 177 - 190. Karol Borsuk. https://ghostarchive.org/archive/20221009/http://matwbn.icm.edu.pl/ksiazki/fm/fm20/fm20117.pdf . 2022-10-09 . live. de. 10.4064/fm-20-1-177-190. free.
- Lazar. Lyusternik. Lazar Lyusternik. Lev. Shnirel'man. Lev Schnirelmann. Topological Methods in Variational Problems. Issledowatelskii Institut Matematiki I Mechaniki Pri O. M. G. U.. Moscow. 1930.
- Book: Matoušek, Jiří . Jiří Matoušek (mathematician). Using the Borsuk–Ulam theorem. Using the Borsuk–Ulam Theorem. Springer Verlag. Berlin. 2003. 978-3-540-00362-5. 10.1007/978-3-540-76649-0.
- H.. Steinlein. Borsuk's antipodal theorem and its generalizations and applications: a survey. Méthodes topologiques en analyse non linéaire. Sém. Math. Supér. Montréal, Sém. Sci. OTAN (NATO Adv. Study Inst.) . 95. 1985. 166–235.
- Borsuk-Ulam Implies Brouwer: A Direct Construction . Francis Edward . Su . The American Mathematical Monthly . 104 . 9 . Nov 1997 . 855 - 859 . 10.2307/2975293 . 2975293 . 10.1.1.142.4935 . 2006-04-21 . https://web.archive.org/web/20081013051302/http://www.math.hmc.edu/~su/papers.dir/borsuk.pdf . 2008-10-13 . dead .
External links
Notes and References
- 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 .
- Extensions of the Borsuk–Ulam Theorem . BS . Harvey Mudd College . 2002 . Prescott, Timothy . 10.1.1.124.4120.
- Joseph J. Rotman, An Introduction to Algebraic Topology (1988) Springer-Verlag (See Chapter 12 for a full exposition.)
- 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.
- 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 .
- Hopf . H. . 1944 . Eine Verallgemeinerung bekannter Abbildungs-und Überdeckungssätze . Portugaliae Mathematica.
- Malyutin . A. V. . Shirokov . I. M. . 2023 . Hopf-type theorems for f-neighbors . Sib. Èlektron. Mat. Izv . 20 . 1 . 165–182.
- 10.2307/1969632 . Yang, Chung-Tao . 1954. 60 . 2 . On Theorems of Borsuk-Ulam, Kakutani-Yamabe-Yujobo and Dyson, I . . 262–282. 1969632 .
- Web site: Generalization of Borsuk-Ulam . Math Overflow . 18 May 2015 . Jens Reinhold, Faisal . Sergei Ivanov.