Algebraic combinatorics applied to finite geometry - SymOmega

Graphs vs Linear algebra Euler’s Theorem on latin squares Finite geometry The bigger picture

Algebraic combinatorics applied tofinite geometry

John Bamberg

Centre for the Mathematics of Symmetry and Computation,The University of Western Australia

December 1, 2011

,

= 2


Graphs vs Linear algebra

Euler’s Theorem on latin squares

Finite geometry

The bigger picture


Graphs

• Vertices

• Edges: pairs of vertices (u, v)

Degree: 3


Graphs

• Vertices

• Edges: pairs of vertices (u, v)

Degree: 3


Good labellings

• Assign a real number to each vertex.

• For each vertex, sum the values of adjacent vertices.

• Goal: Sum at each vertex should be a common multiple of thevalue at the vertex.

?1

?

1?

1

?

1

?

1


Good labellings




?1

?

1?

1

?

1

?

1


Good labellings




?1

?

1?

1

?

1

?

1


Good labellings




?1

?

1?

1

?

1

?

1


Good labellings




−11

−1

1−1

1

−1

1

−1

1


1

1

?

?

??

?

1

1 ?


1

1

-23

-23

-23-23

-23

1

1 -23


Linear algebra

Adjacency relation

Let V be the vertex-set of a graph.u ∼ v

Adjacency operator A on RV

Given f : V → R, we define

Af : V → R

: v 7→∑u∼v

f (u)


Eigenvectors of A

11

1

1

11

1

1

1

1

−11

−1

1

−11

−1

1

−1

1

1

1

- 23

- 23

- 23

- 23

- 23

1

1 - 23

3 1 −2


Algebraic graph theory

• Subject grew in the 1950’s and ‘60’s:

• Graph is regular if 1 is an eigenvector.

• #Edges = 12

∑λ2i

• #Triangles = 16

∑λ3i

• 3 distinct eigenvalues −→ strongly regular

• Smallest eigenvalue −→ independence and chromatic numbers

• Second largest eigenvalue −→ expansion and randomness properties

• Interlacing −→ substructures.





• #Edges = 12

∑λ2i

• #Triangles = 16

∑λ3i









• #Edges = 12

∑λ2i

• #Triangles = 16

∑λ3i









• #Edges = 12

∑λ2i

• #Triangles = 16

∑λ3i









• #Edges = 12

∑λ2i

• #Triangles = 16

∑λ3i






Using the spectral decomposition

Spectral Theorem

The eigenspaces of A form an orthogonal decomposition of RV .

1

1

- 23

- 23

- 23- 2

3

- 23

1

1 - 23

,

−11

−1

1−1

1

−1

1

−1

1

= 0


Intriguing sets

Corollary

Intriguing sets X and Y associated to different eigenvalues satisfy

|X ∩ Y | =|X ||Y ||V |

.

1

1

0

0

00

0

1

1 0

,

10

1

01

0

1

0

1

0

= 2


Intriguing maps

f : V → R is intriguing ⇐⇒ for some α, β ∈ R

Af = α · f + β · 1.

(In fact, α is an eigenvalue of A.)

1

0

1

0

1

0

1

0

1

0

2

1

2

1

2

1

2

1

2

1

f

Af = f + 1


Intriguing maps


Af = α · f + β · 1.


1

0

1

0

1

0

1

0

1

0

2

1

2

1

2

1

2

1

2

1

f

Af = f + 1


Intriguing maps


Af = α · f + β · 1.


1

0

1

0

1

0

1

0

1

0

2

1

2

1

2

1

2

1

2

1

f

Af = f + 1


Intriguing maps


Af = α · f + β · 1.


1

0

1

0

1

0

1

0

1

0

2

1

2

1

2

1

2

1

2

1

f Af = f + 1


Intriguing maps

0

−1

1

0

1

0

1

0

0 0

1

3

−1

1

−1

1

−1

1

1 1

f

Af = −2 · f + 1


Intriguing maps

0

−1

1

0

1

0

1

0

0 0

1

3

−1

1

−1

1

−1

1

1 1

f

Af = −2 · f + 1


Intriguing maps

0

−1

1

0

1

0

1

0

0 0

1

3

−1

1

−1

1

−1

1

1 1

f Af = −2 · f + 1


Af = αf f + βf 1, Ag = αg f + βg1, A1 = k1.

Corollary

Intriguing maps f and g associated to different eigenvalues satisfy

〈f , g〉 =〈f ,1〉〈g ,1〉|V |

.

Proof.Eigenvectors Eigenvalue

(k − αf )f − βf 1 αf

(k − αg )g − βg1 αg

1 k

(k − αf )(k − αg )〈f , g〉−βg (k − αf )〈f ,1〉−βf (k − αg )〈1, g〉+βf βg |V | = 0



Corollary


〈f , g〉 =〈f ,1〉〈g ,1〉|V |

.




1 k




Corollary


〈f , g〉 =〈f ,1〉〈g ,1〉|V |

.




1 k




Cyclic latin squares1 2 3 4 5

2 4 1 5 3

3 5 4 2 1

4 1 5 3 2

5 3 2 1 4

Transversal1 2 3 4 5

2 4 1 5 3

3 5 4 2 1

4 1 5 3 2

5 3 2 1 4



Cyclic latin squares1 2 3 4 5

2 4 1 5 3

3 5 4 2 1

4 1 5 3 2

5 3 2 1 4

Transversal1 2 3 4 5

2 4 1 5 3

3 5 4 2 1

4 1 5 3 2

5 3 2 1 4


Theorem (Euler, 1782)

An n × n cyclic latin square does not have a transversal whenn is even.

The strongly regular graph

Vertices: Cells of the latin squareAdjacency: Same row, same column, or same symbolEigenvalues: 3(n − 1), n − 3, −3.

1 2 3 4 52 4 1 5 33 5 4 2 14 1 5 3 25 3 2 1 4


Theorem (Euler, 1782)

An n × n cyclic latin square does not have a transversal whenn is even.

The strongly regular graph

Vertices: Cells of the latin squareAdjacency: Same row, same column, or same symbolEigenvalues: 3(n − 1), n − 3, −3.

1 2 3 4 52 4 1 5 33 5 4 2 14 1 5 3 25 3 2 1 4


Transversals are intriguing

11 20 30 40 50

20 40 10 51 30

30 50 41 20 10

40 10 50 30 21

50 31 20 10 40

10 23 33 43 53

23 43 13 50 33

33 53 40 23 13

43 13 53 33 20

53 30 23 13 43

1T

A1T = −31T + 31



11 20 30 40 50

20 40 10 51 30

30 50 41 20 10

40 10 50 30 21

50 31 20 10 40

10 23 33 43 53

23 43 13 50 33

33 53 40 23 13

43 13 53 33 20

53 30 23 13 43

1T

A1T = −31T + 31



11 20 30 40 50

20 40 10 51 30

30 50 41 20 10

40 10 50 30 21

50 31 20 10 40

10 23 33 43 53

23 43 13 50 33

33 53 40 23 13

43 13 53 33 20

53 30 23 13 43

1T A1T = −31T + 31


A magical intriguing map µ

11 21 31 41 51

21 40 14 51 34

31 50 42 25 17

41 15 52 35 27

51 34 26 18 46

132 232 332 432 532

232 430 138 532 338

332 530 434 240 144

432 140 534 340 244

532 338 242 146 442

µ

Aµ = 2 · µ+ 30 · 1



11 21 31 41 51

21 40 14 51 34

31 50 42 25 17

41 15 52 35 27

51 34 26 18 46

132 232 332 432 532

232 430 138 532 338

332 530 434 240 144

432 140 534 340 244

532 338 242 146 442

µ

Aµ = 2 · µ+ 30 · 1



11 21 31 41 51

21 40 14 51 34

31 50 42 25 17

41 15 52 35 27

51 34 26 18 46

132 232 332 432 532

232 430 138 532 338

332 530 434 240 144

432 140 534 340 244

532 338 242 146 442

µ Aµ = 2 · µ+ 30 · 1


Two opposing intriguing maps

1T : 〈1T ,1〉 = nµ: 〈µ,1〉 = n2(n + 1)/2

So

〈1T , µ〉 =〈1T ,1〉〈µ,1〉

n2=

n · n2(n + 1)/2

n2=

n(n + 1)

2

But if n is even, then

n(n + 1)

2≡ n

2(mod n).

A contradiction for a CYCLIC latin square!



1T : 〈1T ,1〉 = nµ: 〈µ,1〉 = n2(n + 1)/2

So

〈1T , µ〉 =〈1T ,1〉〈µ,1〉

n2=

n · n2(n + 1)/2

n2=

n(n + 1)

2


n(n + 1)

2≡ n

2(mod n).




1T : 〈1T ,1〉 = nµ: 〈µ,1〉 = n2(n + 1)/2

So

〈1T , µ〉 =〈1T ,1〉〈µ,1〉

n2=

n · n2(n + 1)/2

n2=

n(n + 1)

2


n(n + 1)

2≡ n

2(mod n).



Finitely many points and lines

Generalised quadrangle

Given a point P and ` which are not incident, there is a unique linem on P concurrent with `.

`

P

Order (s, t)

s + 1 points on a line, t + 1 lines through a point


Finitely many points and lines

Generalised quadrangle

Given a point P and ` which are not incident, there is a unique linem on P concurrent with `.

`

P

Order (s, t)

s + 1 points on a line, t + 1 lines through a point


m-coversA set of lines M of a generalised quadrangle is an m-cover ifevery point lies on m elements of M.

Figure: A 2-cover of W(3, 2).


• Known m-covers:W(3, q) Not too manyQ(4, q), q odd Many, m evenQ−(5, q) Many!H(4, q2) Many found (recent), m > 1H(3, q2) Hemisystems, q odd

• Segre (1965):An m-cover of H(3, q2), q odd, has m = q+1

2 (a hemisystem).

• Bruen & Hirschfeld (1978):No m-cover exists of H(3, q2), q even.




2 (a hemisystem).





2 (a hemisystem).



• J. A. Thas (1981):

Hemisystem of H(3, q2) −→ partial quadrangle andstrongly regular graph

• J. A. Thas (1989):An m-cover of a GQ of order (q2, q), q odd, has m = q+1

2 .

• m-covers are intriguing.

Af = (−s + 1) · f + m(s + 1) · 1


• J. A. Thas (1981):



2 .


Af = (−s + 1) · f + m(s + 1) · 1


• J. A. Thas (1981):



2 .


Af = (−s + 1) · f + m(s + 1) · 1


A magical intriguing map (Bamberg, Devillers & Schillewaert)

Suppose we have two disjoint lines ` and m. Then

µ := s · 1{`,m}⊥⊥ + t · 1{`,m}⊥

is intriguing.

`

m


Two opposing intriguing maps 1

In the case that |{`,m}⊥⊥| = t2/s + 1 ...

1m-cover 〈1m-cover,1〉 = m(st + 1)µ 〈µ,1〉 = (s + t)(t + 1)

So

〈1m-cover, µ〉 =〈1m-cover,1〉〈µ,1〉

(t + 1)(st + 1)= m(s + t)

1µ = s · 1{`,m}⊥⊥ + t · 1{`,m}⊥


Two opposing intriguing maps 1

In the case that |{`,m}⊥⊥| = t2/s + 1 ...

1m-cover 〈1m-cover,1〉 = m(st + 1)µ 〈µ,1〉 = (s + t)(t + 1)

So

〈1m-cover, µ〉 =〈1m-cover,1〉〈µ,1〉

(t + 1)(st + 1)= m(s + t)

1µ = s · 1{`,m}⊥⊥ + t · 1{`,m}⊥


Double Counting: Fix ` not in m-cover M. Count concurrent pairs(r , s) ∈M, such that r , s are concurrent with `.

Result: m = t+12 .

Generalisation: by Frederic Vanhove to regular near polygons.



Result: m = t+12 .




Result: m = t+12 .



The bigger picture

Instead of regular graphs:Association schemes and the Bose-Mesner algebra.

More than inner products:Inner/outer distribution, Krein parameters,MacWilliams Transform.

More than latin squares and generalised quadrangles:Projective spaces, polar spaces, partial geometries.


The bigger picture





The bigger picture




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Algebraic combinatorics applied to finite geometry - SymOmega