The Borsuk-Ulam theoremA Combinatorial Proof
Shreejit Bandyopadhyay
April 14, 2015
1
Introduction
The Borsuk-Ulam theorem is perhaps among the results in algebraic topology
having the greatest number of applications to problems beyond it, especially in
combinatorics and discrete geometry. Partition theorems like the Ham Sandwich
theorem, several non-embeddibility theorems and even some graph-theoretic results can be solved by a direct or indirect application of this famous result, first
formulated by Ulam and proved independently by Borsuk. Other striking features of this theorem are its several different equivalent versions and numerous
different proofs. While some of these proofs use elementary algebraic topological concepts like the degree of a map or other geometric ideas, some proofs are
highly combinatorial in flavour. In this note, we present one such combinatorial proof which uses an equivalent formulation of the main theorem, usually
referred to as the Tucker’s Lemma.
2
Versions of the Theorem and Their Equivalence
First, we state four different versions of the Borsuk-Ulam theorem and prove
the equivalence of them.
Theorem 1. (Borsuk-Ulam) For every n ≥ 0, the following statements are
equivalent, and true.
(i) For every continuous mapping f : S n → Rn , there exists a point x ∈ S n
with f (x) = f (−x).
(ii) For every antipodal mapping f : S n → Rn , i.e., for every mapping f :
S n → Rn with f continuous and f (−x) = −f (x), there exists a point x ∈ S n
satisfying f (x) = 0.
(iii) There is no antipodal mapping from S n → S n−1 .
1
(iv) There is no continuous mapping f : B n → S n−1 that is antipodal on the
boundary, i.e., satisfies f (−x) = −f (x) for all x ∈ S n−1 = ∂B n .
Proof. (i) =⇒ (ii) is clear.
(ii) =⇒ (i) We apply (i) to the mapping g(x) = f (x) − f (−x) which is clearly
antipodal.
(ii) =⇒ (iii) This holds because any antipodal mapping S n → S n−1 is also
a nowhere zero antipodal mapping S n → Rn .
(iii) =⇒ (ii) Assume that f : S n → Rn is a continuous nowhere zero antipodal
(x)
mapping. Then the antipodal mapping g : S n → S n−1 given by g(x) = ||ff (x)||
contradicts (iii).
(iii) =⇒ (iv) We note that the projection map π(x1 , x2 , ..., xn+1 ) = (x1 , x2 , ..., xn )
is a homeomorphism of the upper hemisphere U of S n with B n . So an antipodal mapping f : S n → S n−1 as in (iii) would yield a mapping g : B n → S n−1
antipodal on ∂B n by g(x) = f (π −1 (x).
(iv) =⇒ (iii) For g : B n → S n−1 as in (iv), we define f (x) = g(π(x)) and
f (−x) = −g(π(x)) for x ∈ U . This specifies f on the whole of S n and it’s
well-defined because of the antipodality of g on the equator of S n . Since f is
thus continuous on both the closed hemispheres, the pasting lemma implies the
continuity of f on S n . This implies (iii)
3
Tucker’s Lemma and its Equivalence to the
Borsuk-Ulam Theorem
We now derive the Borsuk-Ulam theorem from a combinatorial statement called
Tucker’s Lemma. Actually, this lemma is equivalent to the Borsuk-Ulam and we
may thus interpret it as a discrete version of it. We’ll divide the proof into two
parts-first establishing the equivalence of Tucker’s Lemma with version (iv) of
the Borsuk-Ulam theorem in the previous section and then proving the lemma
itself using the language of chains and boundaries.
Let’s consider T to be a finite triangulation of B n . We call such a triangulation
antipodally symmetric on the boundary if the set of simplices of T contained in
S n−1 = ∂B n is an antipodally symmetric triangulation of S n−1 , i.e., if σ ⊂ S n−1
is a simplex of T, then −σ is also a simplex of T.
2
Theorem 2. (Tucker’s Lemma) Let T be a triangulation of B n that is antipodally symmetric on the boundary and let
λ : V(T) −→ {+1, −1, +2, −2, ..., +n, −n}
be a labelling of the vertices of T that satisfies λ(−v) = −λv for every v ∈ ∂B n ,
i.e., λ is antipodal on the boundary. Then there exists a 1-simplex, i.e., an
edge in T which is complementary, that is, has its vertices labelled by opposite
numbers.
We can also reformulate Tucker’s Lemma using simplicial maps into the
boundary of the crosspolytope. For that, we take the abstract simplicial complex ∆n−1 with vertex set V (∆n−1 ) = {+1, −1, +2, −2, ..., +n, −n}, with a
subset F ⊂ V (∆n−1 ) forming a simplex whenever there’s no i ∈ [n] such that
both i ∈ F and −i ∈ F . We can thus interpret ∆n−1 as the boundary complex
of the n-dimensional crosspolytope. In particular, we note that ||∆n−1 || ' S n−1 .
We can reformulate Tucker’s Lemma in terms of abstract simplicial complexes
as follows.
Theorem 3. (Tucker’s Lemma, another version) Let T be a triangulation of
B n antipodally symmetric on the boundary. Then there is no map λ : V (T)−→
V (∆n−1 ) that is a simplicial map of T into ∆n−1 and is simultaneously antipodal
on the boundary.
Theorem 4. Theorem 3 is equivalent to version (iv) of the Borsuk-Ulam theorem.
Proof. Borsuk-Ulam implies Tucker’s lemma: Suppose there is a simplicial map
λ of T into ∆n−1 antipodal on the boundary. Then its canonical affine extension
||λ|| gives a continuous map B n → S n−1 antipodal on the boundary, contradicting Borsuk-Ulam version (iv).
Tucker’s Lemma implies Borsuk-Ulam: Assume that f : B n → S n−1 is a continuous map antipodal on the boundary. We construct T and λ such that Tucker’s
Lemma is contradicted.
We choose T to be any triangulation of B n antipodal on the boundary with simplex diameter at most δ. To choose δ, choose = √1n . Then for every y ∈ S n−1 ,
||y||∞ ≥ , i.e.,P
at least one of the components of y has absolute value at least
n
, as otherwise i=1 yi2 < 1.
A continuous function on a compact set being uniformly continuous, there is
a δ > 0 such that if d(x, x0 ) < δ then ||f (x) − f (x0 )|| < 2. We choose this δ to
be the simplex diameter of our triangulation.
3
Now we define λ : V (T ) = {±1, ±2, ..., ±n} as follows. First let
k(v) = min {i : |f (v)i | ≥ },
and then define
+k(v) f (v)k(v) > 0
λ(v) =
−k(v) f (v)k(v) < 0.
Since f is antipodal on ∂B n , it follows that λ(−v) = −λ(v) for each vertex v
on the boundary. So Tucker’s Lemma applies and yields a complementary edge,
say vv 0 . Let i = λ(v) = λ(v 0 ) > 0. Then f (v)i ≥ and f (v 0 )i ≤ − and hence
||f (x) − f (x0 )|| ≥ 2, contracting our assumption.
This shows that Tucker’s Lemma also implies Borsuk-Ulam and the two are
thus equivalent to each other.
4
Some Facts on Chains and Boundaries
Let K be a simplicial complex. By a k − chain we mean a set Ck of simplices
of K of dimension exactly k with k = 0, 1, ...,dim K. Note that a k − chain
contains simplices of dimension k only and is thus not a simplicial complex. We
denote the empty k − chain by 0. Also, if Ck and Dk are k − chains, then
Ck + Dk is defined to be the symmetric difference of Ck and Dk . So, in particular, Ck + Ck = 0.
If F ∈ K is a k− dimensional simplex, the boundary of F is defined to be the
(k−1)− chain ∂F consisting of the facets of F so that ∂F has k+1 simplices. For
a k− chain Ck = {F1 , F2 , ..., Fm }, the boundary is ∂Ck = ∂F1 + ∂F2 + ... + ∂Fm .
So it contains the (k − 1)-dimensional simplices that occur an odd number of
times as facets of the simplices in Ck .
We note two properties of the boundary operator.
(i) It commutes with chain addition, i.e., ∂(Ck + Dk ) = ∂Ck + ∂Dk . This
follows directly from the definition.
(ii) For any k− chain Ck , ∂∂Ck = 0. To see why this holds, note that it is
sufficient to verify this for Ck consisting of a single k−simplex, which is straightforward.
A simplicial map f of a simplicial complex K into a simplicial complex L induces
a mapping f#k sending k−chains of K into k−chains of L. So if Ck = {F } is a
k−chain consisting of a single simplex, we define f#k (Ck ) as {f (F )} if f (F ) is
a k−dimensional simplex of L and as 0 otherwise. Then we extend it linearly
to arbitrary chains as f#k ({F1 , F2 , ..., Fm }) = f#k ({F1 }) + f#k ({F2 }) + ... +
f#k ({Fm }).
4
It is easy to verify that these chain maps commute with the boundary operator,
satisfying f#k−1 (∂Ck )∂f#k (Ck ) for any k− chain Ck . Note that it is enough to
verify this fact for the case when Ck contains a single simplex.
In the proof of Tucker’s Lemma we now proceed to give we also assume an
additional condition on the triangulation T of B n . For k = 0, 1, 2, ...n − 1, we
define
Hk+ = x ∈ S n−1 : xk+1 ≥ 0, xk+2 = xk+3 = ... = xn = 0 ,
Hk− = x ∈ S n−1 : xk+1 ≤ 0, xk+2 = xk+3 = ... = xn = 0 .
+
−
+
So Hn−1
and Hn−1
are the northern and southern hemispheres of S n−1 , Hn−2
∪
−
+
−
Hn−2 is the (n − 2)-dimensional equator, etc., and finally, H0 and H0 are a
pair of antipodal points. We assume that T respects this structure. For each
i = 0, 1, ...n − 1, there are subcomplexes that triangulate Hi+ and Hi− .
5
Proof of Tucker’s Lemma
We now prove Theorem 3. For this proof, the mapping λ need not go into
∆n−1 ;, it’s enough if it goes into a any antipodally symmetric triangulation L
of S n−1 .
Theorem 5. Let T be a triangulation of B n as described above.and let K be
the antipodally symmetric part of T triangulating S n−1 . Let L be another finite
antipodally symmetric triangulation of S n−1 . Let f :V(K)→V(L) be a simplicial
map from K to L. Then:
(i) Let An−1 be the (n − 1)-chain consisting of all (n − 1)−dimensional simplices of K. Then either the (n − 1)−chain Cn−1 : f#n−1 (An−1 ) is empty or it
contains all the (n − 1)−dimensional simplices of L. In other words, either each
(n − 1)−dimensional simplex of L has an even number of preimages, or each
has an odd number of preimages. In the former case, we say f has even degree
and write deg2 (f ) = 0 and in the latter case, we say that f has odd degree and
write deg2 (f ) = 1.
(ii) If f is any simplicial map of T into L, and if f is the restriction of f
on the boundary, i.e., on V(K), then deg2 (f ) = 0.
(iii) If f is a antipodal simplicial map of K into L, then deg2 (f ) = 1.
5
So a simplicial map λ of T into L that is antipodal on the boundary has even
degree by (ii) and odd degree by (iii). So no such map can exist, and Tucker’s
Lemma is proved.
Proof. (i) If Cn−1 is neither empty nor everything, then there are two (n −
1)−dimensional simplices sharing a facet such that one of them is in Cn−1 and
the other isn’t. Then the common facet is precisely ∂Cn−1 .
But,
∂Cn−1 = ∂f#n−1 (An−1 ) = f#n−2 (∂An−1 ) = 0,
since every (n − 2)−simplex of K is a facet of exactly two simplices of An−1 .
This is a contradiction.
(ii) Let An be the n−chain consisting of all n−simplices of T. Then An−1 = ∂An ,
and at the same time, f #n (An ) = 0 because L has no n−simplices. Thus
Cn−1 = f#n−1 (An−1 ) = ∂f #n (An ) = ∂0 = 0.
(iii) Let A+
k be the k−chain consisting of all k−simplices of K contained in
+
−
the k−dimensional hemisphere Hk+ and similarly for A−
k . Let Ak = Ak + Ak .
For k = 1, 2, ...n − 1, we have
−
∂A+
k = ∂Ak = Ak−1 .
−
If we set Ck+ = f#k (A+
k ), and similarly for Ck and Ck , we get
∂Ck+ = ∂Ck− = Ck−1 .
+
−
We need to show that Cn−1 6= 0. Suppose Cn−1 = Cn−1
+ Cn−1
= 0. Then
+
−
+
−
we get Cn−1 = Cn−1 .. Since An−1 is antipodal to An−1 and f is an antipodal
+
−
map, Cn−1
is antipodal to Cn−1
as well, and since they are also equal, the chain
+
−
+
Dn−1 = Cn−1 = Cn−1 is antipodally symmetric. Thus Cn−2 = ∂Cn−1
= ∂Dn−1
is the boundary of an antipodally symmetric chain.
We now assume inductively for some k > 0 that
Ck = ∂Dk+1
for an antipodally symmetric chain Dk+1 , and infer a similar claim for Ck−1 .
Note that the antipodally symmetric chain Dk+1 can be partitioned into two
antip
antip
chains, Dk+1 = Ek+1 + Ek+1 , such that Ek+1 is antipodal to Ek+1 . This is
done by dividing the simplices of Dk+1 into antipodal pairs and splitting each
antip
antip
pair between Ek+1 and Ek+1 . So we have Ck = Ck+ +Ck− = ∂(Ek+1 +Ek+1 ).
antip
So, Ck+ + ∂Ek+1 = Ck− + ∂Ek+1 .
6
Since the left-hand side is antipodal to the right, Dk = Ck+ + ∂Ek+1 is an antipodally symmetric chain. Applying the boundary operator, we get,
∂Dk = ∂Ck+ + ∂∂Ek+1 = ∂Ck+ = Ck−1 ,
and the induction step is complete.
Proceeding to k = 1, we see that C0 is the boundary of an antipodally symmetric 1-chain. But C0 consists of two antipodal points or 0-simplices, while
the boundary of any antipodally symmetric 1-chain consists of an even number
of antipodal pairs, as can be easily checked. This contradiction finishes the
proof.
We note that our proof of the Borsuk-Ulam theorem can be broken down
into three parts. The most common version of the Borsuk-Ulam, i.e., version
(i) is first proved to be equivalent to its version (iv), then this version (iv)
and Tucker’s Lemma are shown to be equivalent and finally, we give a proof of
Tucker’s lemma itself, using the language of chains and boundaries. This proof
of Tucker’s Lemma is actually a reproduction of Tucker’s original proof, though
other constructive and purely combinatorial proofs of the lemma also exist which
actually give algorithms for finding the complementary edge by tracing a certain
sequence of simplices.
References
[1] Jiri Matousek, Using the Borsuk-Ulam theorem: Lectures on Topological
Methods in Combinatorics and Geometry, Springer.
[2] S.Lefschetz, Introduction to Topology.
[3] J.Munkres, Topology.
[4] Allen Hatcher, Algebraic Topology, Cambridge University Press.
7