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From Topological Methods to Combinatorial Proofs of Kneser Graphs Daphne Der-Fen Liu 劉 德 芬 Department of Mathematics California State University, Los Angeles

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Kneser Graphs Given positive integers n ≥ 2k, the Kneser Graph KG(n, k) has the vertex set of all k-element subsets of [n] = {1, 2, 3, …, n}. Two vertices A and B are adjacent if A and B are disjoint, A∩B = .

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Example: KG(5, 2) is Petersen graph 1 2 1 3 1 5 1 4 3 5 3 4 2 5 2 4 2 3 4 5

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Lovász Theorem [1978] (KG(n, k)) = n – 2k + 2. The proof was by the Borsuk-Ulam theorem, which can be proved by Tucker Lemma [1946] (with extremely fine “triangulations” of S n ) Tucker Lemma Borsuk-Ulam Lovász Thm Matoušek [2004] Combinatorial proof !

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Circular Chromatic Number For positive integers p, q, with p ≥ 2q, a (p, q)-coloring c for a graph G is a mapping c: V(G) → {0, 1, 2, …, p-1} such that if uv E(G) then q ≤ |c(u) – c(v)| ≤ p - q. Circular chromatic number of G is: χ c (G) = inf {p/q : G admits a (p,q)-coloring} Note, a (p, 1)-coloring is a proper coloring.

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Johnson-Holroyd-Stahl Conjecture [1997] c (KG(n, k)) = n – 2k + 2. Zhu Surveys Yes, for even n [Meunier ‘05] and [Simonyi and Tardos ‘06] ? Yes, for sufficiently large n [Hajiabolhassan and Zhu, 2003] Known result: For any graph G, c (G) = (G).

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Chen Theorem [2011] Johnson-Holroyd-Stahl Conjecture [1997] c (KG(n, k)) = n – 2k + 2. Because c (G) = (G), so Lovász Thm Chen Theorem Proof was by Ky Fan Lemma [1954]

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Tucker Lemma Borsuk-Ulam Lovász Thm Matoušek [2004] Fan Lemma Chen Theorem A combinatorial proof modified from Prescott and Su [2005] (2) (1) (3) This talk Chang, L. Zhu [2012]

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Use 1, 2, -1, -2 to label vertices so that it is antipodal on the boundary, Always exists a complementary edge (sum 0) ! and avoid edges with label sum = 0. IMPOSSIBLE !! 1 1 1 -2 2 X

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Same conclusion for other symmetric triangulations of a square D 2 : N X N 3 X 3 These are all symmetric triangulation of D 2

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Tucker Lemma [1946] Given an array of N 2 elements, in N rows and N columns (N > 1), and given four labels 1,-1, 2, -2, let each element of the array be assigned one of the four labels in such a way that each pair of antipodal elements on the boundary of the array is assigned a pair of labels whose sum is zero; then there is at least one pair of adjoining elements of the array that have labels whose sum is zero. Can be extended to N k

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D2D2 D2 D2 S1S1 D3D3 D3 D3 S2S2 Special Triangulations of S n

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Alternating Simplex of a labeling Let K be a triangulation of S n. Let : V(K) { 1, 2, …, m} be a labeling A simplex with d vertices is positive alternating if labels of its vertices can be expressed as { k 1, – k 2, k 3, – k 4 ……, (-1) d-1 k d }, where 1 ≤ k 1 < k 2 < k 3 < k 4 … < k d m. Negative alternating is similar, except the leading label is negative, – k 1, k 2, – k 3 ……

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Ky Fan Lemma [1956] Let K be a barycentric derived subdivision of the octahedral subdivision of the n-sphere S n. Let m be a positive integer. Label each vertex of K with one of 1, 2, ….., m, so that: The labels assigned to any two antipodal vertices of K have sum 0 Any 1-simplex in K have labels sum ≠ 0 Then there exist an odd number of positive alternating n-simplices in K. (Hence, m > n) Antipodal labeling No complementary edges

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Boundary of the First Barycentric Subdivision of D n 0,0 1,0 1,1 1,-1 0,-1 For each simplex, the vertices can be ordered as V 1, V 2, … such that: If the i-th coordinate of V i is non-zero, say z, then the i-th coordinate of all V j, j > i, must be z. -1,-1 -1,0 0,1 -1,1

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D2D2 D2 D2 S1S1 D3D3 D3 D3 S2S2 Special Triangulations of S n

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Example (Recall) 1 1 1 -2 2 X D2 S1D2 S1 By Key Fan Lemma, 1,0 1,1 There is an odd number of positive alternating 1-simplex (an edge).

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Useful Essence of Fan Lemma Barycentric subdivision of S n-1 R n Fix n. A (nontrivial) signed n-sequence is: A = (a 1, a 2, a 3,...., a n ), each a i { 0, +, - }, and at least one a i ≠ 0. A can be expressed by A = (A +, A-), where A + = { i : a i = + }, A - = { i : a i = - }. Opposite of A: - A = (A-, A + ). Example: A = ( -, 0, +, +, +) = ( {3, 4, 5}, {1} ) - A = ( +, 0, -, -, - ) = ( {1}, {3, 4, 5} )

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Simplices in the Triangulations of S n-1 A d-set consists of d elements from n : = {A 1, A 2, ….., A d }, A 1 < A 2 < ….. < A d Let n denote all non-trivial signed n-sequences. Let A, B n, A = (A +, A - ), B = (B +, B - ). Denote A ≤ B, if A + B + and A - B - Denote |A| = | A + | + | A - | Dimension ( ) = d = | |. Note, A + A - = and |A| 1.

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Antipodal labeling Let : n → { 1, 2, … m}. is sign-preserving: (- A) = - (A), A n Complementary pair: A < B n, (A)+ (B) = 0. Positive alternating d-sets: is a d-set, ( ) = { k 1, - k 2, k 3, ….., (-1) d-1 k d }, 0 < k 1 < k 2 < k 3 < ….. < k d ≤ m.

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Fan Lemma applied to the 1 st barycentric subdivision of octahedral subdivision of S n-1 Assume : n → { 1, 2, … m } is sign- preserving without complementary pairs. Then there exist an odd number of positive alternating n-sets. Consequently, m n. Octahedral Tucker Lemma: Assume : n → { 1, 2, … (n-1)} is sign- preserving. Then there exists some complementary pair.

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Proof of Fan Lemma: Construct a graph G Vertices: is a d or (d-1)-sets, max( ) = d: Type I: is an agreeable alternating (d-1)-set, d 2. Type II: is an agreeable almost alternating d-set. Type III: is an alternating d-set. Edges: ’ is and edge if all below are true: (1) < ’ and | | = | ’| - 1 (2) is alternating (positive or negative) (3) ( ’ ) = sign ( ) (4) max( ’) = | ’ |

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Claim All vertices in G have degree 2, except {(+,0, …,0)}, {(-,0, …, 0)} and alternating n-sets which are of degree 1. So, G consists of disjoint paths. The negative of each path is also a path in G. So, there are an even number of paths in G. The two ends of a path are not opposite sets. By symmetry, there are an odd number of positive alternating n-sets. Q.E.D. Finished (1)

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Lovász Theorem (KG(n, k)) = n – 2k + 2. Claim: (KG(n, k)) ≤ n – 2k + 2. Define a (n–2k+2)-coloring c of KG(n, k) by: 1 2 3 4 5.… n – 2k +1n – 2k +2.…. n -1 n 2k – 1

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Assume (KG(n, k)) ≤ n – 2k + 1 Let c be a (n-2k+1)-coloring for KG(n, k) using colors in { 2k-1, 2k, 2k+1, …, n-1 }. Define : n → { 1, 2, … (n-1)} by: where c(U) = max {c(W): W U and |W| = k}.

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Define : n → { 1, 2, … (n-1)} by: where c(U) = max {c(W): W U and |W| = k}. is sign-preserving: (- A) = - (A), A n No complementary pairs. If (A) = - (B) & A < B, then c(A’) = c (B’), for some A’ A +, B’ B -. Impossible as A + B - = , so A’, B’ adjacent. Contradicting Fan Lemma, as (n -1) < n. Lovász Thm Fished (2)

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Alternative Kneser Coloring Lemma Chen [JCTA, 2011], Chang-L-Zhu [ JCTA, 2012] Suppose c is a proper (n-2k+2)-coloring for KG(n, k). Then [n] can be partitioned into: [n] = S T {a 1, a 2, …., a n-2k+2 }, where |S| = |T| = k- 1, and c(S {a i })= c(T {a i })= i, i =1, 2,..., n-2k+2. By Alternative Kneser Coloring Lemma, Chen Theorem can be proved easily.

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Claim: (KG(n, k)) ≤ n – 2k + 2. Define a (n–2k+2)-coloring c of KG(n, k) by: 1 2 3 4 5.… n – 2k +1n – 2k +2.…. n -1 n 2k – 1 S T {a n - 2k+2 } {a 1, a 2, …., a n -2k+1 } a i = i, i = 1, 2, …, n – 2k+1 Example (Recall)

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Alternative Kneser Coloring Lemma Any proper (n-2k+2)-coloring c for KG(n, k), [n] = S T {a 1, a 2, …., a n-2k+2 }, where |S| = |T| = k-1, and c(S {a i })= c(T {a i })= i, i =1, 2,..., n-2k+2. In KG(n, k), the sets S {a i }, T {a i }, i = 1,..., n–2k+2, Induce a complete bipartite graph K n- 2k+2, n- 2k+2 minus a perfect matching which used all colors called a colorful K n- 2k+2, n- 2k+2

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Alternative Kneser Coloring Lemma Any proper (n-2k+2)-coloring c for KG(n, k), then [n] = S T {a 1, a 2, …., a n-2k+2 }, where | S | = | T | = k – 1, and S {a1}S {a1} S {a2}S {a2} S {a n-2k+2 } : : c (S {a i }) = c (T {a i } ) = i, i = 1, 2,..., n – 2k+ 2. T {a1}T {a1} T {a2}T {a2} T {a n-2k+2 } : : 11 2 2 n-2k+2 : : : :

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Proof of Alternative Kneser Coloring Theorem Let c be a proper (n-2k+2)-coloring of KG(n, k) using colors from { 2k-1, 2k-2,…, n }. Define : n → { 1, 2, … n } by: where c(U) = max {c(W): W U and |W| = k}. Let be a linear order on subsets of [n] such that if |U| < |W| then U W.

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is sign-preserving: (- A) = - (A), A n No complementary pairs. As seen before. By Fan Lemma, there exist an odd number of positive alternating n-sets.

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Important Claim Let be a positive alternating n-set. Then ( ) = {1, - 2, 3, - 4, …, (-1) n-1 n } : A 1 < A 2 <.... < A n | A i | = i, for all i = 1, 2, …, 2k-2 |A + 2k-2 | = |A - 2k-2 | = k-1 a 2k-1, a 2k, … a n [n] \ (A + 2k-2 A - 2k-2 ), so that c (A + 2k-2 {a 2k-1, a 2k+1, … a 2i+1 } ) = 2i +1, i c (A - 2k-2 {a 2k, a 2k+2, … a 2i } ) = 2j, j

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Proof (continue) Let = {A n : |A+| = |A-| = k-1}. By Claim every positive alternating n-set contains exactly one element from . For every A , let (A, ) = # positive alternating n-sets containing A. By Fan Lemma, A (A, ) is odd. So (Z, ) 1 (mod 2) for some Z .

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Proof (continue) By Fan Lemma, A (A, ’) is odd. Define ’ : n → { 1, 2, … n } by: ’(Z) = - (Z); and ’(A) = (A) if A Z. ’ is sign-preserving without complementary pairs Since A Z (A, ) = A Z (A, ’), and (- Z, ) = (Z, ’) = 0, So we have, (Z, ) (- Z, ’) 1 (mod 2)

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Let Z = (Z+, Z-) = (S, T). Then - Z = (T, S). Apply Claim to both , ’ we get: Let , ’ be the positive alternating n-sets for, ’, containing Z and – Z, respectively. a 2k-1, a 2k, … a n [n] \ (S T ), c ( S {a 2k-1, a 2k+1, … a 2i -1 } ) = c( T {b 2k-1, b 2k+1, … b 2i-1 } ) = 2i – 1, i b 2k-1, b 2k, … b n [n] \ (T S ), so that c ( S {a 2k, a 2k+2, … a 2i } ) = c ( T {b 2k, b 2k, … b 2i } ) = 2i, i

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Hence, c (S { a 2k-1 }) = c(T { b 2k-1 }) = 2k – 1. So, a 2k-1 = b 2k-1 By induction, a i = b i and c (S { a i }) = c(T { b i }) = i for all i. This completes the proof of Alternative Kneser Coloring Theorem. Fished (3)

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Thanks to Pen-An Chen for great results iii Happy Birthday Professor Chang iii THANK YOU for being a Great Mentor !! THANK YOU ALL !! Thanks to Xuding Zhu for excellent lecture note Thanks to the Conference Committee

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