The Borsuk-Ulam theorem- A Combinatorial Proof

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1 The Borsuk-Ulam theorem- A Combinatorial Proof Shreejit Bandyopadhyay April 14, 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 R n, there exists a point x S n with f(x) = f( x). (ii) For every antipodal mapping f : S n R n, i.e., for every mapping f : S n R n 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

2 (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 R n. (iii) = (ii) Assume that f : S n R n is a continuous nowhere zero antipodal mapping. Then the antipodal mapping g : S n S n 1 given by g(x) = f(x) f(x) contradicts (iii). (iii) = (iv) We note that the projection map π(x 1, x 2,..., x n+1 ) = (x 1, x 2,..., x n ) 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

3 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 ɛ = 1 n. Then for every y S n 1, y ɛ, i.e., at least one of the components of y has absolute value at least ɛ, as otherwise n i=1 y2 i < 1. A continuous function on a compact set being uniformly continuous, there is a δ > 0 such that if d(x, x ) < δ then f(x) f(x ) < 2ɛ. We choose this δ to be the simplex diameter of our triangulation. 3

4 Now we define λ : V (T ) = {±1, ±2,..., ±n} as follows. First let k(v) = min {i : f(v) i ɛ}, 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. Let i = λ(v) = λ(v ) > 0. Then f(v) i ɛ and f(v ) i ɛ and hence f(x) f(x ) 2ɛ, 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 C k 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 C k and D k are k chains, then C k + D k is defined to be the symmetric difference of C k and D k. So, in particular, C k + C k = 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 simplices. For a k chain C k = {F 1, F 2,..., F m }, the boundary is C k = F 1 + F F m. So it contains the (k 1)-dimensional simplices that occur an odd number of times as facets of the simplices in C k. We note two properties of the boundary operator. (i) It commutes with chain addition, i.e., (C k + D k ) = C k + D k. follows directly from the definition. This (ii) For any k chain C k, C k = 0. To see why this holds, note that it is sufficient to verify this for C k 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 C k = {F } is a k chain consisting of a single simplex, we define f #k (C k ) 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 ({F 1, F 2,..., F m }) = f #k ({F 1 }) + f #k ({F 2 }) f #k ({F m }). 4

5 It is easy to verify that these chain maps commute with the boundary operator, satisfying f #k 1 ( C k ) f #k (C k ) for any k chain C k. Note that it is enough to verify this fact for the case when C k 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 H + k = { x S n 1 : x 0, x k+2 = x k+3 =... = x n = 0 }, H k = { x S n 1 : x 0, x k+2 = x k+3 =... = x n = 0 }. So H + n 1 and H n 1 are the northern and southern hemispheres of Sn 1, H + n 2 H n 2 is the (n 2)-dimensional equator, etc., and finally, H+ 0 and H 0 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 H + i and H i. 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 A n 1 be the (n 1)-chain consisting of all (n 1) dimensional simplices of K. Then either the (n 1) chain C n 1 : f #n 1 (A n 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 deg 2 (f) = 0 and in the latter case, we say that f has odd degree and write deg 2 (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 deg 2 (f) = 0. (iii) If f is a antipodal simplicial map of K into L, then deg 2 (f) = 1. 5

6 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 C n 1 is neither empty nor everything, then there are two (n 1) dimensional simplices sharing a facet such that one of them is in C n 1 and the other isn t. Then the common facet is precisely C n 1. But, C n 1 = f #n 1 (A n 1 ) = f #n 2 ( A n 1 ) = 0, since every (n 2) simplex of K is a facet of exactly two simplices of A n 1. This is a contradiction. (ii) Let A n be the n chain consisting of all n simplices of T. Then A n 1 = A n, and at the same time, f #n (A n ) = 0 because L has no n simplices. Thus C n 1 = f #n 1 (A n 1 ) = f #n (A n ) = 0 = 0. (iii) Let A + k be the k chain consisting of all k simplices of K contained in the k dimensional hemisphere H + k and similarly for A k. Let A k = A + k + A k. For k = 1, 2,...n 1, we have A + k = A k = A k 1. If we set C + k = f #k(a + k ), and similarly for C k and C k, we get C + k = C k = C k 1. We need to show that C n 1 0. Suppose C n 1 = C n C n 1 = 0. Then we get C n 1 + = C n 1.. Since A+ n 1 is antipodal to A n 1 and f is an antipodal map, C n 1 + is antipodal to C n 1 as well, and since they are also equal, the chain D n 1 = C n 1 + = C n 1 is antipodally symmetric. Thus C n 2 = C n 1 + = D n 1 is the boundary of an antipodally symmetric chain. We now assume inductively for some k > 0 that C k = D for an antipodally symmetric chain D, and infer a similar claim for C k 1. Note that the antipodally symmetric chain D can be partitioned into two chains, D = E + E antip, such that E antip is antipodal to E. This is done by dividing the simplices of D into antipodal pairs and splitting each pair between E and E antip. So we have C k = C + k +C k = (E +E antip ). So, C + k + E = C k + Eantip. 6

7 Since the left-hand side is antipodal to the right, D k = C + k + E is an antipodally symmetric chain. Applying the boundary operator, we get, D k = C + k + E = C + k = C k 1, and the induction step is complete. Proceeding to k = 1, we see that C 0 is the boundary of an antipodally symmetric 1-chain. But C 0 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