From Topological Methods to Combinatorial Proofs of Kneser Graphs

Daphne Der-Fen Liu 劉 德 芬Department of MathematicsCalifornia 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 53 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 Sn )

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 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.

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 ThmChen 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

-1

-22

-1 -1

X

Same conclusion for other symmetric triangulations of a square D2:

N X N3 X 3

These are all symmetric triangulation of D2

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

D2 D2 S1

D3 D3 S2

Special Triangulations of Sn

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 ……

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

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 Dn

0,01,0

1,1

1,-10,-1

For each simplex, the verticescan be ordered as V1, V2, … such that:

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

-1,0

0,1-1,1

D2 D2 S1

D3 D3 S2

Special Triangulations of Sn

Example (Recall)

1 1 1

-1

-22

-1 -1

X

D2 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 LemmaBarycentric subdivision of Sn-1 Rn

Fix n. A (nontrivial) signed n-sequence is:

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} )

Simplices in the Triangulations of Sn-1

A d-set consists of d elements from n :

= {A1 , A2 , ….. , Ad}, A1 < A2 < ….. < Ad

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,

( ) = { k1 , - k2 , k3 , ….. , (-1)d-1 kd },

0 < k1 < k2 < k3 < ….. < kd ≤ m.

Fan Lemma applied to the 1st barycentric subdivision of octahedral subdivision of Sn-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:

otherwise. 22

;12min if min )(

k-n

k n - (W) (W)Wc

1 2 3 4 5 .… n – 2k +1 n – 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:

A -A and 1,-2k |A| if )c(A- -

-A A and 1,-2k |A| if )c(A

A -A and 2,-2k |A| if |A| -

-A A and 2,-2k |A| if |A|

(A)

where c(U) = max {c(W): W U and |W| = k}.

Define : n → { 1, 2, … (n-1)} by:

A A- and 1,-2k |A| if )c(A -

-A A and 1,-2k |A| if )c(A

A -A and 2,-2k |A| if |A| -

-A A and 2,-2k |A| if |A|

(A)

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 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.

Claim: (KG(n, k)) ≤ n – 2k + 2.

Define a (n–2k+2)-coloring c of KG(n, k) by:

otherwise. 22

;12min if min )(

k-n

k n - (W) (W)Wc

1 2 3 4 5 .… n – 2k +1 n – 2k +2 .…. n -1 n

2k – 1

S T {an - 2k+2}

{a1, a2, …., an -2k+1 }

ai = i, i = 1, 2, …, n – 2k+1

Example (Recall)

Alternative Kneser Coloring LemmaAny 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 called a colorful Kn- 2k+2, n- 2k+2

Alternative Kneser Coloring LemmaAny 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

S {a1}S {a2}

S {an-2k+2}

: :

c (S {ai}) = c (T {ai} ) = i, i = 1, 2, ..., n – 2k+ 2.

T {a1}T {a2}

T {an-2k+2}

: :

1 12 2

n-2k+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:

A -A and 1,-2k |A| if )c(A

-A A and 1,-2k |A| if )c(A

A -A and 2,-2k |A| if |A|

-A A and 2,-2k |A| if |A|

(A)

- -

-

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 ClaimLet be a positive alternating n-set. Then

() = {1, - 2, 3, - 4, …, (-1)n-1n }

: A1 < A2 < . . . . < An

| 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

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.

a2k-1, a2k, … an [n] \ (S T ),

c ( S {a2k-1, a2k+1, … a2i -1 } )

= c( T {b2k-1, b2k+1, … b2i-1 } ) = 2i – 1, i

b2k-1, b2k, … bn [n] \ (T S ), so that

c ( S {a2k, a2k+2, … a2i } )

= c ( T {b2k, b2k, … b2i } ) = 2i, i

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)

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