Hedetniemi's conjecture for Kneser hypergraphs

Hedetniemi’s conjecture for Kneser hypergraphs

Fr ´ed ´eric Meunier

March 26th, 2015

Joint work with Hossein Hajiabolhassan

CERMICS, Optimisation et Syst`emes

Product of graphs and Hedetniemi’s conjecture

LetG,Hbe two graphs.

Categorical productG×H:

V(G×H) =V(G)×V(H)

E(G×H) ={(u,v)(u⁰,v⁰) : uu⁰∈E(G),vv⁰∈E(H)}

Hedetniemi’s conjecture (1966)

χ(G × H) = min(χ(G), χ(H))

Some known cases

Hedetniemi’s conjecture has been proved in (not so many) special cases.

Theorem (El-Zahar-Sauer 1985)

If min(χ(G), χ(H)) ≤ 4, Hedetniemi’s conjecture is true.

Theorem (Duffus-Sands-Woodrow 1985)

Let G and H be connected graphs with min(χ(G), χ(H)) > n. If both G and H contain a clique of size n, then χ(G × H) > n.

In particular, it implies that Hedetniemi’s conjecture is true for

such graphs when min(χ(G), χ(H)) = n + 1.

Product of hypergraphs and Zhu’s conjecture

u1

u2

u3

{u1, u2, u3} ∈E(G) {v1, v2, v3, v4} ∈E(H)

v1

v2

v3

v4

{(u1, v1),(u1, v2),(u2, v1),(u3, v3),(u3, v4)} ∈E(G × H)

LetG,Hbe two hypergraphs.Categorical productG × H:

V(G × H) =V(G)×V(H)

E(G × H) ={e⊆V(G × H) : πG(e)∈E(G), πH(e)∈E(H)}

Proper coloringof a hypergraph: no monochromatic edges.

Zhu’s conjecture (1992)

χ( G × H ) = min(χ( G ), χ( H ))

Product of graphs vs of hypergraphs, and coloring

Product of two graphs seen as hypergraphs: will have edges of cardinality 3 and 4.

G

G

G G

H

H

H H

G H

G H

G H

But the chromatic number does not depend of the definition of the product.

Zhu’s conjecture is a true generalization of Hedetniemi’s

conjecture.

Kneser graphs

n, k two integers s.t. n ≥ 2k .

KG(5, 2) =

{1,2}

{3,4}

{1,5}

{2,3} {4,5}

{2,4} {2,5} {3,5}

{1,3}

{1,4}

Kneser graph KG(n, k):

V (KG(n, k )) =

E (KG(n, k ))) = n

AB : A, B ∈

, A ∩ B = ∅ o

Kneser hypergraphs and main result

n, k , r three integers s.t. n ≥ rk . Kneser hypergraph KG

(n, k ):

V (KG

(n, k )) =

E(KG

(n, k))) = n

{ A

, . . . , A

} : A

∈

, A

∩ A

= ∅ for i 6 = j o

Theorem (Hajiabolhassan-M.)

χ KG

(n, k ) × KG

(n

, k

)

= min χ KG

(n, k )

, χ KG

(n

, k

)

Known to be true forr=2 (Hell 1980).

Hedetniemi’s conjecture for Kneser graphs

Proof of the general case explained for:

Theorem (Hell, 1980)

χ (KG(n, k ) × KG(n

, k

)) = min (χ (KG(n, k)) , χ (KG(n

, k

))).

Proof uses a combinatorial approach proposed by Matouˇsek (2003) for proving

χ(KG(n, k )) = n − 2k + 2 (Lov ´asz’ theorem 1979).

Approach extended by Ziegler for computing χ(KG

(n, k )).

Tucker’s lemma

Theorem

Suppose there exists a map

λ : { +, − , 0 }

\ { 0, . . . , 0 } −→ {− 1, +1, − 2, +2, . . . , − `, +` }

s.t.

• λ( − y ) = − λ(y ) for all y

• λ(y ) + λ(z ) 6 = 0 if y z . Then ` ≥ m.

“yz” means “yi 6=0⇒yi=zi”.

−1 +1

−2 +2

Proof of Hedetniemi’s conjecture for Kneser graphs

Assume

•

•

With the help of coloring, build a map

λ: {+,−,0}ⁿ⁺ⁿ⁰\ {0, . . . ,0} −→ {−1,+1, . . . ,−(t+n⁰+2k−2),t+n⁰+2k−2}

(x,x⁰) 7−→ s(x,x⁰)

| {z } sign

v(x,x⁰)

| {z } absolute value satisfying condition of Tucker’s lemma

• λ(−x,−x⁰) =−λ(x,x⁰)

• λ(x,x⁰) +λ(y,y⁰)6=0 ifxyandx⁰y⁰.

Second point satisfied because ofcoloring condition: no two adjacent vertices get the same color.

Thus,t+n⁰+2k−2≥n+n⁰, i.e.

t≥n−2k+2

(

x

z }| { 0, +, +, − , 0, . . . , 0, − , 0,

x⁰

z }| {

− , +, 0, 0, +, . . . , 0, +, +)

x⁺={i: xi= +} x⁰⁺={i: x_i⁰= +}

x⁻={i: x_i=−} x⁰⁻={i: x_i⁰=−}

?If|x⁺| ≥k,|x⁻| ≥k,|x⁰⁺| ≥k⁰, and|x⁰⁻| ≥k⁰: existence of

◦ (A⁺,A⁰⁺)s.t.A⁺⊆x⁺andA⁰⁺⊆x⁰⁺.

◦ (A⁻,A⁰⁻)s.t.A⁻⊆x⁻andA⁰⁻⊆x⁰⁻.

(A⁺,A⁰⁺)and(A⁻,A⁰⁻)are bothverticesof KG(n,k)×KG(n⁰,k⁰)and are colored.

s(x,x⁰) = corresponding sign v(x,x⁰) = minimal such color.

?If not:most difficult part

Back to Kneser hypergraphs

n, k , r three integers s.t. n ≥ rk.

Kneser hypergraph KG

(n, k ):

V (KG

(n, k )) =

E (KG

(n, k ))) = n

{ A

, . . . , A

} : A

∈

, A

∩ A

= ∅ for i 6 = j o

Categorical product G × H : V( G × H ) = V ( G ) × V ( H )

E( G × H ) = { e ⊆ V ( G × H ) : π

(e) ∈ E( G ), π

(e) ∈ E ( H ) }

Proof

Theorem

χ KG

(n, k ) × KG

(n

, k

)

= min(χ(KG

(n, k)), χ(KG

(n

, k

))).

Proof via two steps:

1. Proposition

Theorem is true for r prime.

Proved via Ziegler’s extension of Matouˇsek approach, used to compute

χ(KG^r(n,k)) =

n−r(k−1) r−1

(Alon-Frankl-Lov ´asz theorem)

2. Proposition

If theorem true for r

and r

, then it is true for r

r

.

Case r prime

Assume

•

•

•

Zr=rth roots of unity

With the help of coloring, build a map

λ: (Zr∪ {0})ⁿ⁺ⁿ⁰\ {0, . . . ,0} −→ Zr×[t+n⁰+rk−2]

(x,x⁰) 7−→ (s(x,x⁰)

| {z } sign

, v(x,x⁰)

| {z } absolute value

)

satisfying condition of a “Zr-Tucker” lemma

• λ(ωx, ωx⁰) =ωλ(x,x⁰)for allω∈Zr

• condition on{λ(x¹,x⁰¹), . . . , λ(x^{`</sup>,x<sup>0`})}when(x¹,x⁰¹) · · · (x^{`</sup>,x<sup>0`}) Second point satisfied bycoloring condition: noradjacent vertices get the same color.

Thus,(r−1)(t−1) +n⁰+rk−1≥n+n⁰, i.e.

t≥n−r(k−1) r−1

Case r nonprime

Assume

• r = r

r

• KG

(n, k ) × KG

(n

, k

) colored with t colors.

Theorem is true for KG

(A, k ) × KG

(A

, k

), where A ⊆ [n] and A

⊆ [n

] are of carefully chosen size.

Theorem is true for KG

(n, | A | ) × KG

(n

, | A

| ): if t too small,

contradiction.

A general lower bound

General Kneser hypergraphKG^r(H)defined for a hypergraphH= (V(H),E(H))and integerr≥2 by

V(KG^r(H)) = E(H)

E(KG^r(H)) = {{e₁, . . . ,er}:e1, . . . ,er∈E(H),ei∩ej=∅for alli,jwithi6=j}.

cdr(H)=minimum number of vertices to be removed s.t. induced subhypergraph r-colorable

Theorem (Hajiabolhassan-M.) Let H

, . . . , H

be hypergraphs. Then

χ KG

( H

) × · · · × KG

( H

)

≥ 1

r − 1 min

`=1,...,t

cd

( H

).

New graphs for Hedetniemi’s conjecture

U

, . . . , U

pairwise disjoint, `

, . . . , `

integers

B =

A ∈ [n]

k

: | A ∩ U

| ≤ `

bases of the truncation of a partition matroid Proposition (Hajiabolhassan-M.)

Graphs KG

( B ) satisfy Hedetniemi’s conjecture.

Holds also if B is the set of spanning trees of some “dense”

graph.

Open question

Categorical productof two hypergraphsG × H:

V(G × H) =V(G)×V(H)

E(G × H) =

e⊆V(G × H) : π_G(e)∈E(G), π_H(e)∈E(H)

Let r 6 = r

. Do we have

χ

KG

(n, k ) × KG

(n

, k

)

= min

χ KG

(n, k )

, χ(KG

n

, k

)

?

Thank you

Proof, cont’d

Building ofλ(x,x⁰) =s(x,x⁰)v(x,x⁰):

?If|x⁺| ≥k,|x⁻| ≥k,|x⁰⁺| ≥k⁰, and|x⁰⁺| ≥k⁰: existence of

◦ (A⁺,A⁰⁺)s.t.A⁺⊆x⁺andA⁰⁺⊆x⁰⁺.

◦ (A⁻,A⁰⁻)s.t.A⁻⊆x⁻andA⁰⁻⊆x⁰⁻.

(A⁺,A⁰⁺)and(A⁻,A⁰⁻)are bothverticesof KG(n,k)×KG(n⁰,k⁰)and are colored.

s(x,x⁰) = corresponding sign v(x,x⁰) = minimal such color.

?If not:

g(x) =







2k−1 if|x⁺| ≥kand|x⁻| ≤k−1 2k−1 if|x⁺| ≤k−1 and|x⁻| ≥k

|x⁺|+|x⁻| otherwise.

v(x,x⁰) =g(x) +g(x⁰) s(x,x⁰) =(most difficult part)

Case r nonprime

Assume

•

•

•

Supposefor a contradictionthatt<^n−r(k−1)_r−1 . A⊆[n]with|A|=t(r1−1) +r1(k−1) +1 A⁰⊆[n⁰]with|A⁰|=t(r1−1) +r1(k⁰−1) +1

⇓

Monochromatic edge{(X⁽¹⁾,X⁰⁽¹⁾), . . . ,(X^(r1),X^0(r1))}in KG^r1(A,k)×KG^r1(A⁰,k⁰).

⇒Coloring of(A,A⁰).

⇓

Monochromatic edge{(A⁽¹⁾,A⁰⁽¹⁾), . . . ,(A^(r2),A^0(r2))}in KG^r2(n,|A|)×KG^r2(n⁰,|A⁰|).

⇓ Contradiction