Title: From Topological Methods to Combinatorial Proofs of Kneser Graphs
1From Topological Methods to Combinatorial Proofs
of Kneser Graphs
- Daphne Der-Fen Liu ? ? ?
- Department of Mathematics
- California State University, Los Angeles
2Kneser 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 ?.
3Example 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
4Lovász Theorem 1978
- 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 !
5Circular 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.
6Johnson-Holroyd-Stahl Conjecture 1997
- Known result For any graph G,
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
7Chen Theorem 2011Johnson-Holroyd-Stahl
Conjecture 1997
- Proof was by Ky Fan Lemma 1954
Because ??c (G) ? ? (G), so
Lovász Thm
Chen Theorem
8This 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
9Use 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) !
10Same conclusion for other symmetric
triangulations of a square D2
N X N
3 X 3
These are all symmetric triangulation of D2
11Tucker 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.
12Special Triangulations of Sn
?
? D2
S1
D2
?
? D3
S2
D3
13Alternating 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
14Ky 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)
15Boundary 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
16Special Triangulations of Sn
?
? D2
S1
D2
?
? D3
S2
D3
17Example (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
18Useful 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 )
19Simplices 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 ? .
20Antipodal labeling
- ? 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.
21Fan 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.
22Proof 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(?) ?
23Claim
- 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)
24Lovász 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
25Assume ? (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.
26Define ? ?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)
27Alternative 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
- c(S? ai) c(T? ai) i, ?i 1, 2, ..., n-2k2.
?? By Alternative Kneser Coloring Lemma,
Chen Theorem can be proved easily.
28Example (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
29Alternative Kneser Coloring Lemma
- Any proper (n-2k2)-coloring c for KG(n, k),
- n S ? T ? a1, a2, ., an-2k2, where
- 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
30Alternative Kneser Coloring Lemma
- Any proper (n-2k2)-coloring c for KG(n, k), then
- n S ? T ? a1, a2, ., an-2k2, where
- 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
31Proof 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.
33Important 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-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
34Proof (continue)
- 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 ? ?.
-
35Proof (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 ?, ? 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.
c (S ? ai ) c(T ? bi ) i for all
i.
This completes the proof of Alternative Kneser
Coloring Theorem.
Fished (3)
38THANK 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 !!