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
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.
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:
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
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:
With retractions[edit]
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.
1-dimensional case[edit]
The 1-dimensional case can easily be proved using the intermediate value theorem (IVT).
Let
be an odd real-valued continuous function on a circle. Pick an arbitrary
. If
then we are done. Otherwise, without loss of generality,
Hence, by the IVT, there is a point
between
and
at which
.
General case[edit]
Algebraic topological proof[edit]
Assume that
is an odd continuous function with
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
denotes the field with two elements],
sends
to
. But then we get that
is sent to
, a contradiction.[2]
One can also show the stronger statement that any odd map
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
be a continuous odd function. Because g is continuous on a compact domain, it is uniformly continuous. Therefore, for every
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 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
(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
is compact there must hence be a point u in which
.
Corollaries[edit]
- No subset of is homeomorphic to
- The ham sandwich theorem: For any compact sets A1, …, An in 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 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: 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 [7] However, in general this is not true for other functions A.[8]
See also[edit]
- ^ a b Prescott, Timothy (2002). Extensions of the Borsuk–Ulam Theorem (BS). Harvey Mudd College. CiteSeerX 10.1.1.124.4120.
- ^ Joseph J. Rotman, An Introduction to Algebraic Topology (1988) Springer-Verlag ISBN 0-387-96678-1 (See Chapter 12 for a full exposition.)
- ^ Freund, Robert M.; Todd, Michael J. (1982). “A constructive proof of Tucker’s combinatorial lemma”. Journal of Combinatorial Theory. Series A. 30 (3): 321–325. doi:10.1016/0097-3165(81)90027-3.
- ^ Simmons, Forest W.; Su, Francis Edward (2003). “Consensus-halving via theorems of Borsuk–Ulam and Tucker”. Mathematical Social Sciences. 45: 15–25. doi:10.1016/s0165-4896(02)00087-2. hdl:10419/94656.
- ^ Nyman, Kathryn L.; Su, Francis Edward (2013), “A Borsuk–Ulam equivalent that directly implies Sperner’s lemma”, The American Mathematical Monthly, 120 (4): 346–354, doi:10.4169/amer.math.monthly.120.04.346, JSTOR 10.4169/amer.math.monthly.120.04.346, MR 3035127
- ^ “Borsuk fixed-point theorem”, Encyclopedia of Mathematics, EMS Press, 2001 [1994]
- ^ Yang, Chung-Tao (1954). “On Theorems of Borsuk-Ulam, Kakutani-Yamabe-Yujobo and Dyson, I”. Annals of Mathematics. 60 (2): 262–282. doi:10.2307/1969632. JSTOR 1969632.
- ^ Jens Reinhold, Faisal; Sergei Ivanov. “Generalization of Borsuk-Ulam”. Math Overflow. Retrieved 18 May 2015.
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