# 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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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 k-element subsets of [n] = {1, 2, 3, …, n}. Two vertices A and B are adjacent if A and B are disjoint, A∩B = .

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

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 !

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.

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).

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]

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]

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

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

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

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

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 ……

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

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

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

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).

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} )

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.

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.

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.

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(  ’) = |  ’ |

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)

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

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}.

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)

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.

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)

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

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 : : : :

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.

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.

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

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  .

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)

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

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)

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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