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Title: From Topological Methods to Combinatorial Proofs of Kneser Graphs


1
From Topological Methods to Combinatorial Proofs
of Kneser Graphs
  • Daphne Der-Fen Liu ? ? ?
  • 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, AnB ?.

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 ?
    E(G) then q c(u) c(v) p - q.
  • Circular chromatic number of G is
  • ?c(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,

Zhu Surveys
??c (G) ? ? (G).
?
  • ?c (KG(n, k)) n 2k 2.
  • Yes, for sufficiently large n
  • Hajiabolhassan and Zhu, 2003
  • Yes, for even n
  • Meunier 05 and Simonyi and Tardos 06

7
Chen Theorem 2011Johnson-Holroyd-Stahl
Conjecture 1997
  • ?c (KG(n, k)) n 2k 2.
  • Proof was by Ky Fan Lemma 1954

Because ??c (G) ? ? (G), so
Lovász Thm
Chen Theorem
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 gt 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 lt k2 lt k3 lt k4 lt 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 gt 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 gt i, must
be z.
1,-1
0,-1
-1,-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-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 )
19
Simplices in the Triangulations of Sn-1
  • 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-
  • Note, A ? A- ? and A ? 1.
  • A d-set ? consists of d elements from ? n
  • ? A1 , A2 , .. , Ad, A1 lt A2 lt .. lt Ad

Dimension (?) d ? .
20
Antipodal labeling
  • Let ? ? n ? ?1, ?2, ?m.
  • ? is sign-preserving ?(- A) - ? (A), ? A ??n
  • Complementary pair A lt B ?? n, ?(A)?(B) 0.
  • Positive alternating d-sets ? is a d-set,
  • ? ( ? ) k1 , - k2 , k3 , .. , (-1)d-1 kd
    ,
  • 0 lt k1 lt k2 lt k3 lt .. lt kd m.

21
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.

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) ? lt ? 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 (n2k2)-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-2k1)-coloring for KG(n, k) using
    colors in 2k-1, 2k, 2k1, , 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 lt 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) lt n.
Lovász Thm
Fished (2)
27
Alternative Kneser Coloring LemmaChen JCTA,
2011, Chang-L-Zhu JCTA, 2012
  • Suppose c is a proper (n-2k2)-coloring for
  • KG(n, k). Then n can be partitioned into
  • n S ? T ? a1, a2, ., an-2k2, where
  • S T k- 1, and
  • c(S? ai) c(T? ai) i, ?i 1, 2, ..., n-2k2.

?? By Alternative Kneser Coloring Lemma,
Chen Theorem can be proved easily.
28
Example (Recall)
  • Claim ? (KG(n, k)) n 2k 2.
  • Define a (n2k2)-coloring c of KG(n, k) by

1 2 3 4 5 . n 2k 1
n 2k 2 .. n -1 n
a1, a2, ., an -2k1
2k 1
ai i, ? i 1, 2, , n 2k1
S ? T ? an - 2k2
29
Alternative Kneser Coloring Lemma
  • Any proper (n-2k2)-coloring c for KG(n, k),
  • n S ? T ? a1, a2, ., an-2k2, where
  • S T k-1, and
  • c(S? ai) c(T? ai) i, ? i 1, 2, ..., n-2k2.

In KG(n, k), the sets S? ai, T? ai, i 1,
..., n2k2, Induce a complete bipartite graph
Kn- 2k2, n- 2k2 minus a perfect matching which
used all colors
called a colorful Kn- 2k2, n- 2k2
30
Alternative Kneser Coloring Lemma
  • Any proper (n-2k2)-coloring c for KG(n, k), then
  • n S ? T ? a1, a2, ., an-2k2, where
  • S T k 1, and
  • c (S ? ai) c (T? ai ) i, ? i 1, 2, ...,
    n 2k 2.

T? a1
S? a1
1
1
2
2
T? a2
S? a2








T? an-2k2
S? an-2k2
n-2k2
n-2k2
31
Proof of Alternative Kneser Coloring Theorem
  • Let c be a proper (n-2k2)-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 lt 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 lt A2 lt . . . . lt An
  • ? (?) 1, - 2, 3, - 4, , (-1)n-1n
  • Ai i, for all i 1, 2, , 2k-2
  • A2k-2 A-2k-2 k-1
  • ? a2k-1, a2k, an ? n \ (A2k-2 ?A-2k-2 ), so
    that

c (A2k-2 ? a2k-1, a2k1, a2i1 ) 2i 1,
? i
c (A-2k-2 ? a2k, a2k2, 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
  • Let Z (Z, Z-) (S, T).
  • Then - Z (T, S).
  • Let ?, ? be the positive alternating n-sets for
    ?, ?, containing Z and Z, respectively.
  • 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, a2k1, a2i -1 )
  • c( T ? b2k-1, b2k1, b2i-1 ) 2i 1,
    ? i
  • c ( S ? a2k, a2k2, 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 Happy Birthday Professor Chang iii
THANK YOU for being a Great Mentor !!
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