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From Topological Methods to Combinatorial Proofs of Kneser Graphs

From Topological Methods to Combinatorial Proofs of Kneser Graphs. Daphne Der-Fen Liu 劉 德 芬 Department of Mathematics California State University, Los Angeles. Kneser Graphs. Given positive integers n ≥ 2k, the Kneser Graph KG(n, k) has the vertex set of all

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From Topological Methods to Combinatorial Proofs of Kneser Graphs

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

  2. 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 =.

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

  4. 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 Sn ) Lovász Thm Tucker Lemma Borsuk-Ulam Matoušek [2004] Combinatorial proof !

  5. 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 ϵ E(G) then q ≤ |c(u) – c(v)| ≤ p - q. • Circular chromatic number of G is: χc(G) = inf {p/q : G admits a (p,q)-coloring} • Note, a (p, 1)-coloring is a proper coloring.

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

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

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

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

  10. Same conclusion for other symmetric triangulations of a square D2: N X N 3 X 3 These are all symmetric triangulation of D2

  11. Tucker Lemma [1946] • Given an array of N2 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 Nk

  12. Special Triangulations of Sn   D2 S1 D2   D3 S2 D3

  13. Alternating Simplex of a labeling • Let K be a triangulation of Sn. 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 { k1, – k2 , k3 , – k4 ……, (-1) d-1kd}, where 1≤ k1 < k2 < k3 < k4 … < kd  m. • Negative alternating is similar, except the leading label is negative, – k1 , k2 , – k3 ……

  14. Ky Fan Lemma [1956] • Let K be a barycentric derived subdivision of the octahedral subdivision of the n-sphere Sn. 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 Antipodal labeling No complementary edges • Then there exist an odd number of positive alternating n-simplices in K. (Hence, m > n)

  15. Boundary of the First Barycentric Subdivision of Dn 0,1 -1,1 1,1 For each simplex, the vertices can be ordered as V1, V2, … such that: 0,0 -1,0 1,0 If the i-th coordinate of Vi is non-zero, say z, then the i-th coordinate of all Vj, j > i, must be z. 1,-1 -1,-1 0,-1

  16. Special Triangulations of Sn   D2 S1 D2   D3 S2 D3

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

  18. Useful Essence of Fan Lemma Barycentric subdivision of Sn-1 Rn • Fix n. A (nontrivial) signed n-sequenceis: A = (a1 , a2 , a3 , . . . ., an), each ai { 0, +, - }, and at least one ai ≠ 0. • A can be expressed by A = (A+,A-), where A+ = { i : ai = + }, A- = { i : ai = - }. • Opposite of A: - A = (A-, A+). • Example: A = ( -, 0, +, +, +) = ( {3, 4, 5} , {1} ) - A = ( +, 0, -, -, - ) = ( {1}, {3, 4, 5} )

  19. Simplices in the Triangulations of Sn-1 • Let  ndenote all non-trivial signed n-sequences. • A d-set consists of d elements from  n:  = {A1 , A2 , ….. , Ad}, A1 < A2 < ….. < Ad • Let A, B  n, A = (A+, A-),B = (B+, B-). Denote A ≤ B, if A+  B+and A-  B- Denote |A| = | A+ | + | A- | • Note, A+ A- =  and |A|  1. Dimension () = d = |  |.

  20. 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,  (  ) = { k1 , - k2 , k3 , ….. , (-1)d-1 kd }, 0 < k1 < k2 < k3 < ….. < kd ≤ m.

  21. Fan Lemma applied to the 1st barycentric subdivision of octahedral subdivision of Sn-1 • Assume  :  n → { 1, 2, … m } is sign-preservingwithout 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 existssome complementary pair.

  22. 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(’) = | ’ |

  23. 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)

  24. 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 +1 n – 2k +2 .…. n -1 n 2k – 1

  25. 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}.

  26. 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)

  27. Alternative Kneser Coloring LemmaChen [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  {a1, a2, …., an-2k+2}, where • |S| = |T| = k- 1, and • c(S {ai})= c(T {ai})= i, i =1, 2, ..., n-2k+2.  By Alternative Kneser Coloring Lemma, Chen Theorem can be proved easily.

  28. Example (Recall) • 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 +1 n – 2k +2 .…. n -1 n {a1, a2, …., an -2k+1 } 2k – 1 ai = i,  i = 1, 2, …, n – 2k+1 S  T  {an - 2k+2}

  29. Alternative Kneser Coloring Lemma Any proper (n-2k+2)-coloring c for KG(n, k), [n] = S  T  {a1, a2, …., an-2k+2}, where • |S| = |T| = k-1, and • c(S {ai})= c(T {ai})= i,  i =1, 2, ..., n-2k+2. In KG(n, k), the sets S {ai}, T {ai}, i = 1, ..., n–2k+2, Induce a complete bipartite graph Kn- 2k+2, n- 2k+2 minus a perfect matching which used all colors calleda colorful Kn- 2k+2, n- 2k+2

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

  31. 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 }. • Let  be a linear order on subsets of [n] such that if |U| < |W| then U  W . • Define  :  n → { 1,  2, … n } by: where c(U) = max {c(W): W  U and |W| = k}.

  32.  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.

  33. Important Claim Let  be a positive alternating n-set. Then • : A1 < A2 < . . . . < An •  () = {1, - 2, 3, - 4, …, (-1)n-1n } • | Ai | = i, for all i = 1, 2, …, 2k-2 • |A+2k-2 | = |A-2k-2| = k-1 •  a2k-1, a2k, … an  [n] \ (A+2k-2 A-2k-2 ), so that c(A+2k-2  {a2k-1, a2k+1, … a2i+1} ) =2i +1,  i c(A-2k-2  {a2k, a2k+2, … a2i} ) = 2j,  j

  34. 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  .

  35. Proof (continue) • Define ’ : n → { 1,  2, …  n } by: ’(Z) = - (Z); and ’(A) = (A)if A  Z. • ’ is sign-preserving without complementary pairs • By Fan Lemma,  A   (A, ’) is odd. • Since  A  Z (A, ) =  A  Z (A, ’), and (- Z, ) = (Z, ’) = 0, So we have, (Z, )  (- Z, ’)  1 (mod 2)

  36. Then - Z = (T, S). • Let , ’ be the positive alternating n-sets for , ’, containing Z and – Z, respectively. • Let Z = (Z+, Z-) = (S, T). • Apply Claim to both , ’ we get: •  a2k-1, a2k, … an  [n] \ (S  T ),  b2k-1, b2k, … bn  [n] \ (T  S ), so that • c ( S {a2k-1, a2k+1, … a2i -1 } ) = c( T{b2k-1, b2k+1, … b2i-1 })= 2i – 1,  i • c ( S{a2k, a2k+2, … a2i } ) = c ( T{b2k, b2k, … b2i })= 2i,  i

  37. Hence, c (S  { a2k-1 }) = c(T  { b2k-1 }) = 2k – 1. • So, a2k-1 = b2k-1 • By induction, ai = bi and c (S  { ai }) = c(T  { bi }) = i for all i. This completes the proof of Alternative Kneser Coloring Theorem. Fished (3)

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

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