Kneser's conjecture, first proved by Lovász in 1978, states that the graph with all k-element subsets of {1, 2, . . . , n} as vertices and with edges connecting disjoint sets has chromatic number n−2k+2. We derive this result from Tucker's combinatorial lemma on labeling the vertices of special triangulations of the octahedral ball. By specializing a proof of Tucker's lemma, we obtain self-contained purely combinatorial proof of Kneser's conjectur

Matoušek, Jiří

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COMBINATORICABolyai Society – Springer-Verlag0209–9683/104/$6.00 c©2004 Ja´nos Bolyai Mathematical SocietyCombinatorica 24 (1) (2004) 163–170A COMBINATORIAL PROOF OF KNESER’S CONJECTURE*JIRˇI´ MATOUSˇEKReceived August 9, 2000Kneser’s conjecture, ﬁrst proved by Lova´sz in 1978, states that the graph with all k-elementsubsets of {1,2, . . . ,n} as vertices and with edges connecting disjoint sets has chromaticnumber n−2k+2. We derive this result from Tucker’s combinatorial lemma on labeling thevertices of special triangulations of the octahedral ball. By specializing a proof of Tucker’slemma, we obtain self-contained purely combinatorial proof of Kneser’s conjecture.1. IntroductionLet([n]k)denote the system of all k-element subsets of the set [n] ={1,2, . . . ,n}. The Kneser graph KG(n,k) has vertex set ([n]k)and edge set{{S,S′} : S,S′ ∈ ([n]k), S ∩ S′ = ∅}. Kneser [7] conjectured in 1955 thatχ(KG(n,k))≥n−2k+2, n≥2k≥2, where χ denotes the chromatic number.This was proved in 1978 by Lova´sz [11], as one of the earliest and most spec-tacular applications of topological methods in combinatorics. Several otherproofs have been published since then, all of them topological; among them,at least those of Ba´ra´ny [2], Dol’nikov [4] (also see e.g. [3]), and Sarkaria [15]can be regarded as substantially diﬀerent from each other and from Lova´sz’original proof. These proofs are also presented in [13].In this paper, we ﬁrst we show how Kneser’s conjecture can be deriveddirectly from Tucker’s lemma, which is a combinatorial statement concern-Mathematics Subject Classiﬁcation (2000): 05C15; 05A05; 55M35* Research supported by Charles University grants No. 158/99 and 159/99 and by ETHZu¨rich.164 JIRˇI´ MATOUSˇEKing labelings of vertices of certain triangulations (somewhat resembling thewell-known Sperner’s lemma). Then we give a proof of Kneser’s conjecturethat avoids any mentioning of topology or triangulations, by reformulat-ing and specializing a known proof of Tucker’s lemma (namely one due toFreund and Todd [6]) for our particular setting. The proof is self-containedand can be read without reference to the part with Tucker’s lemma. Otherknown proofs of Tucker’s lemma, such as the one in [10], seem to lead to sim-ilar considerations when traced down to the “combinatorial core”, althoughmany details can be varied.Currently the topological proofs appear preferable in all respects, exceptpossibly for a short presentation to an audience with no topological back-ground. On the other hand, it is interesting that a reasonable combinatorialproof can be given, and perhaps further investigation might lead to combi-natorial methods of real interest.The proof from Tucker’s lemma is inspired by previous author’s simpli-ﬁed proof [12] of a theorem of Krˇ´ızˇ [8], [9] (a generalization of Kneser’sconjecture). The approach of Krˇ´ızˇ is in turn based on some ideas from Alon,Frankl, and Lova´sz [1], who established a generalized Kneser’s conjecturefor hypergraphs.2. A proof from Tucker’s lemmaTucker [16] in his lecture at the First Canadian Math. Congress gave anelementary proof of the 2-dimensional Borsuk–Ulam theorem using a com-binatorial lemma about labelings of the vertices of a particular triangulationof the square. A slightly diﬀerent version, concerning triangulations of the n-dimensional octahedral sphere, was used in Lefschetz’s textbook [10]. Later,this lemma has been re-proved and generalized in several papers, mainly be-cause it provides an “eﬀective” proof of the Borsuk–Ulam theorem; see, forexample, Freund [5] or Freund and Todd [6]. We use the version of Tucker’slemma given in the latter paper.Let Bn denote octahedral ball in Rn (the unit ball of the 1-norm),let Sn−1 denote its boundary (the octahedral sphere), and let K0 be thenatural triangulation of Bn induced by the coordinate hyperplanes (the n-dimensional simplices are in one-to-one correspondence with the orthants inRn). Call a triangulation K of Bn a special triangulation if it reﬁnes K0 andis antipodally symmetric around the origin.Lemma 2.1 (Tucker’s lemma). Let K be a special triangulation ofBn, and suppose that each vertex v of K is assigned a label λ(u) ∈A COMBINATORIAL PROOF OF KNESER’S CONJECTURE 165{+1,−1,+2,−2, . . . ,+n,−n} in such a way that for the vertices of K lyingin Sn−1 the labeling satisﬁes λ(−u)=−λ(u). Then there exists a 1-simplex(edge) of K which is complementary, i.e. its two vertices are labeled byopposite numbers.Proof of Kneser’s conjecture from Tucker’s lemma. First we deﬁnethe appropriate triangulation K. Let L0 be the subcomplex of K0 consistingof the simplices lying on Sn−1. We note that the nonempty simplices of L0are in one-to-one correspondence with nonzero vectors from V ={−1,0,1}n;see Fig. 1(a) for a 2-dimensional illustration. The inclusion relation on the0(1,−1)(1,1)(−1,1)(−1,−1)(0,−1)(0,1)(1,0)(−1,0)(a) (b)Figure 1. The triangulations K0 and L0 (a), and K (b).simplices of L0 corresponds to the relation  on V , where uv if uivi forall i=1,2, . . . ,n and where 01 and 0−1.Let L′0 be the ﬁrst barycentric subdivision of L0. Thus, the vertices ofL′0 are centers of gravity of the simplices of L0 and the simplices of L′0correspond to chains of simplices of L0 under inclusion. A simplex of L′0can be identiﬁed with a chain in the set V \ {(0, . . . ,0)} under . Finally,we deﬁne the triangulation K: it consists of the simplices of L′0 and of thecones with apex 0 over such simplices. This is a special triangulation of Bnas in Tucker’s lemma.Let n>2k≥2. Suppose that c is a coloring of the vertices of the Knesergraph KG(n,k) with n− 2k + 1 colors (as we will show, it cannot be aproper coloring, so we regard it just as an arbitrary assignments of colorsto the k-tuples in([n]k)). For technical convenience, we suppose that thecolors are numbered 2k,2k+1, . . . ,n. We are going to deﬁne a labeling ofthe vertices of K as in Tucker’s lemma. These vertices include 0 and so they166 JIRˇI´ MATOUSˇEKcan be identiﬁed with the vectors of V ,1 and we want to deﬁne a labelingλ:V →{±1, . . . ,±n}.Let us ﬁx some arbitrary linear ordering ≤ on 2[n] that reﬁnes the partialordering according to size, i.e. such that if |A|< |B| then A<B. Let v∈V bea sign vector. To deﬁne λ(v), we consider the ordered pair (A,B) of disjointsubsets of [n] deﬁned byA = {i ∈ [n] : vi = 1} and B = {i ∈ [n] : vi = −1}.We distinguish two main cases. If |A|+ |B|≤2k−2 (Case I) then we putλ(v) ={ |A|+ |B|+ 1 if A ≥ B−(|A|+ |B|+ 1) if A < B.If |A|+ |B|≥2k−1 (Case II) then at least one of A and B has size at leastk. If, say, |A| ≥ k we deﬁne c(A) as c(A′), where A′ consists of the ﬁrst kelements of A, and for |B|≥k, c(B) is deﬁned similarly. We setλ(v) ={c(A) if A > B−c(B) if B > A.So labels assigned in Case I are in {±1, . . . ,±(2k−1)} while labels assignedin Case II in {±2k, . . . ,±n}.One can check that this is a well-deﬁned mapping from V to{±1,±2, . . . ,±n} and that it has the antipodal property, i.e. for everynonzero v∈V we have λ(−v)=−λ(v). Therefore, by Tucker’s lemma, thereis a complementary edge of K.Suppose that the complementary edge connects vertices v1,v2 ∈ V , soλ(v1) = −λ(v2). Since {v1,v2} is an edge of K, we have v1  v2 (after apossible renaming), and so if (A1,B1) and (A2,B2) are the correspondingset pairs, we have A1⊆A2 and B1⊆B2, with at least one of the inclusionsbeing proper. From this it is seen that neither of the labels λ(v1) and λ(v2)could have been assigned in Case I, since |A1|+ |A2|< |B1|+ |B2|.Suppose, for instance, that λ(v1)>0 (the other case is symmetric). Then,by the deﬁnition in Case II, λ(v1) is the color of a k-tuple contained in A1 and−λ(v2)=λ(v1) is the color of a k-tuple contained in B2. We have B2∩A2=∅and A1⊆A2, and so A1∩B2=∅. We have found two disjoint k-tuples withthe same color and c is not a proper coloring of the Kneser graph KG(n,k).This proves Kneser’s conjecture.1 The coordinate vector of the vertex corresponding to a nonzero v ∈V is v‖v‖1 , where‖v‖1= |v1|+ |v2|+ · · ·+ |vn|.A COMBINATORIAL PROOF OF KNESER’S CONJECTURE 1673. A direct combinatorial proofThe basic objects in the proof are disjoint ordered pairs (A,B), where A,B⊆[n] and A∩B=∅. The forthcoming deﬁnition of a labeling function on disjointordered pairs is identical to that in the preceding section but we repeat itfor reader’s convenience.For contradiction, we suppose that c is a proper coloring of the Knesergraph KG(n,k) by n−2k+1 colors, which for convenience we assume to bethe integers 2k,2k+1, . . . ,n. We let ≤ be a linear ordering of the subsets of [n]such that if |A|< |B| then A<B. For each disjoint ordered pair (A,B), wedeﬁne the label λ(A,B). We distinguish two main cases. If |A|+|B|≤2k−2(Case I) then we putλ(A,B) ={ |A|+ |B|+ 1 if A ≥ B−(|A|+ |B|+ 1) if A < B.If |A|+ |B|≥2k−1 (Case II) thenλ(A,B) ={c(A) if A > B−c(B) if B > A,where c(A) is the color of the ﬁrst k elements of A and c(B) is the colorof the ﬁrst k elements of B. This is a valid deﬁnition since if, for instance,A>B and |A|+ |B|≥2k−1 then |A|≥k. The labels assigned in Case I arein {±1, . . . ,±(2k−1)} while labels assigned in Case II in {±2k, . . . ,±n}.Next, we consider labels of signed sequences. A signed sequence is a se-quence (s1,s2, . . . ,sm), where 0≤m≤ n, s1, . . . ,sm ∈ {±1,±2, . . . ,±n}, and|si| = |sj| for i =j. A signed sequence (s1, . . . ,sm) deﬁnes a sequence of m+1disjoint ordered pairs (A0,B0), (A1,B1),. . . , (Am,Bm), whereAi = {sj : j ∈ [i], sj > 0}, Bi = {−sj : j ∈ [i], sj < 0}.In other words, A0 =B0 = ∅ and (Ai,Bi) is obtained from (Ai−1,Bi−1) byadding si to Ai−1 if si > 0 and by adding −si to Bi−1 if si < 0. The labelsequence associated to (s1, . . . ,sm) is (λ0,λ1, . . . ,λm), where λi = λ(Ai,Bi).By the deﬁnition of λ, each label sequence begins with (1,±2,±3, . . .), upuntil ±(2k−1), and then it has terms ±i with 2k≤ i≤n (but of course itmay have fewer than 2k−1 terms).We claim that if c is a proper coloring of the Kneser graph then the labelseuqence of any signed sequence never contains two complementary labels,i.e. there are no i = j with λi = −λj . This is exactly as in the precedingsection: supposing λi=−λj, we ﬁrst observe that i,j≥2k−1, i.e. the labelsmust have been assigned according to Case II. Supposing, for example, that168 JIRˇI´ MATOUSˇEKi< j and λi> 0, we get that Ai and Bj are two disjoint subsets of [n] withc(Ai) = c(Bj), which means that two disjoint k-tuples received the samecolor under c.We prove that there exists a signed sequence whose label sequence con-tains complementary labels. We proceed again by contradiction. (But theproof actually provides an algorithm for ﬁnding such a signed sequence and,consequently, exhibiting two k-tuples given the same color by c.)Let us call a signed sequence s = (s1, . . . ,sm) a permissible sequence ifsi ∈ {λ0,λ1, . . . ,λm} for all i∈ [m], where (λ0, . . . ,λm) is the label sequenceof s. So, for example, a permissible sequence with t or more terms, wheret≤2k−1, contains +1, one of −2 and +2, . . . , one of −t and +t.Assuming that no label sequence contains complementary labels, we nowdeﬁne one or two neighbor permissible sequences for each permissible se-quence, where the relation of being neighbor is symmetric. The only per-missible sequence with a single neighbor is the empty one; all others haveexactly two neighbors. This, of course, is impossible since there are onlyﬁnitely many permissible sequences, and it provides the desired contradic-tion.The single neighbor of the empty sequence is the (permissible) sequence(+1).Let s = (s1, . . . ,sm) be a nonempty permissible sequence and let (λ0 =1,λ1, . . . ,λm) be its label sequence. There are at least m distinct numbersamong the m+1 labels, namely s1, . . . ,sm (in some arbitrary order). De-pending on the single remaining label, two cases can occur:(i) Two labels coincide: λi=λj, i<j.(ii) There is one extra label λi ∈{s1,s2, . . . ,sm}.In case (i), one of the two neighbors of s is obtained by transposi-tion at (i, i + 1), by which we mean that the neighbor sequence is(s1,s2, . . . ,si−1,si+1,si,si+2, . . . ,sm). (Note that i = 0 is not possible sincethe label λ0 =1 cannot occur at any other position of the label sequence.)Such a transposition preserves all terms of the label sequence except possi-bly for λi, and so the just deﬁned neighbor is a permissible sequence. Thesecond neighbor of s is deﬁned similarly, by transposition at (j,j+1), un-less j =m. For j =m, the second neighbor of s is the contracted sequence(s1, . . . ,sm−1).In case (ii), one of the two neighbors of s is the expanded sequence(s1, . . . ,sm,λi). This sequence is permissible because we assume that thereare no two complementary labels, and therefore |λi| ∈{|s1|, . . . , |sm|}. To de-ﬁne the second neighbor, we must distinguish a few cases. If λi is neither theA COMBINATORIAL PROOF OF KNESER’S CONJECTURE 169ﬁrst nor the last term of the label sequence, i.e. 1≤ i≤m−1, the second neigh-bor is obtained from s by transposition at (i, i+1). If i=m then the secondneighbor arises by contraction, i.e. it is (s1, . . . ,sm−1). Finally if i=0 thenthe second neighbor is obtained by sign change: it is (−s1,−s2, . . . ,−sm).In this way, two distinct neighbors are assigned to each nonempty permis-sible sequence. It remains to check that the neighbor relation is symmetric,which is done by discussing the few possible cases. This concludes the proof.Added in proof. Methods of this paper were further developed and appliedto generelizations of Kneser’s conjecture by Ziegler [17]. Relations of thevarious topological proofs of Kneser’s conjecture were investigated in [14].AcknowledgmentI would like to thank Professor Hans-Jakob Lu¨thi for his insistence that acombinatorial proof of Kneser’s conjecture should be obtainable, in spite ofmy confused objections.References[1] N. Alon, P. Frankl and L. Lova´sz: The chromatic number of Kneser hypergraphs,Trans. Amer. Math. Soc. 298 (1986), 359–370.[2] I. Ba´ra´ny: A short proof of Kneser’s conjecture, J. Combinatorial Theory Ser. A25 (1978), 325–326.[3] V. L. Dol’nikov: Transversals of families of sets in Rn and a connection betweenthe Helly and Borsuk theorems, Russian Acad. Sci. Sb. Math. 79 (1994), 93–107.Translated from Ross. Akad. Nauk Matem. Sbornik, Tom 184, No. 5, 1993.[4] V. L. Dol’nikov: Transversals of families of sets, In Studies in the theory of functionsof several real variables (Russian), pages 30–36, 109. Yaroslav. Gos. Univ., Yaroslavl’,1981.[5] R. M. Freund: Variable dimension complexes II: A uniﬁed approach to some com-binatorial lemmas in topology, Math. Oper. Res., 9 (1984), 498–509.[6] R. M. Freund and M. J. Todd: A constructive proof of Tucker’s combinatoriallemma, J. Combinatorial Theory Ser. A 30 (1981), 321–325.[7] M. Kneser: Aufgabe 360, Jahresbericht der Deutschen Mathematiker-Vereinigung2. Abteilung 58 (1955), pp. 27.[8] I. Krˇ´ızˇ: Equivariant cohomology and lower bounds for chromatic numbers, Trans.Amer. Math. Soc. 333 (1992), 567–577.[9] I. Krˇ´ızˇ: A correction to “Equivariant cohomology and lower bounds for chromaticnumbers” (Trans. Amer. Math. Soc. 333 (1992), 567–577), Trans. Amer. Math. Soc.352(4) (2000), 1951–1952.[10] S. Lefschetz: Introduction to Topology, Princeton, 1949.[11] L. Lova´sz: Kneser’s conjecture, chromatic number and homotopy, J. CombinatorialTheory Ser. A 25 (1978), 319–324.170 J. MATOUSˇEK: A COMBINATORIAL PROOF OF KNESER’S CONJECTURE[12] J. Matousˇek: On the chromatic number of Kneser hypergraphs, Proc. Amer. Math.Soc. 130 (2002), 2509–2514.[13] J. Matousˇek: Using the Borsuk–Ulam theorem, Springer, Berlin 2003.[14] J. Matousˇek and G. M. Ziegler: Topological lower bounds for the chromatic num-ber: A hierarchy, Jahresbericht der Deutschen Mathematiker-Vereinigung, in press,2004.[15] K. S. Sarkaria: A generalized Kneser conjecture, J. Combinatorial Theory Ser. B49 (1990), 236–240.[16] A. W. Tucker: Some topological properties of disk and sphere, In Proc. FirstCanadian Math. Congr., Montreal 1945, pages 285–309, Univ. of Toronto Press, 1946.[17] G. M. Ziegler: Generalized Kneser coloring theorems with combinatorial proofs,Invent. Math. 147 (2002), 671–691.Jiˇr´ı MatousˇekDepartment of Applied MathematicsCharles UniversityMalostranske´ na´m. 25118 00 Praha 1Czech RepublicandInstitut fu¨r InformatikETH ZentrumZu¨richSwitzerlandmatousek@kam.mff.cuni.cz

A Combinatorial Proof of Kneser'sConjecture*

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A Combinatorial Proof of Kneser'sConjecture*

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