[PPT] - Coxeter numbers: From fake degree palindromicity to the enumeration PowerPoint Presentation

SLIDE 1

Coxeter numbers: From fake degree palindromicity to the enumeration of reflection factorizations.

Theo Douvropoulos

Paris VII, IRIF

ERC CombiTop
November 21, 2018

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 1 / 23

SLIDE 2

The number of reduced reflection factorizations of c

Theorem (Hurwitz, 1892)

There are nn−2 (minimal length) factorizations t1 · · · tn−1 = (12 · · · n) ∈ Sn where the ti’s are transpositions.

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 2 / 23

SLIDE 3

The number of reduced reflection factorizations of c

Theorem (Hurwitz, 1892)

There are nn−2 (minimal length) factorizations t1 · · · tn−1 = (12 · · · n) ∈ Sn where the ti’s are transpositions. For example, the 31 factorizations (12)(23) = (123) (13)(12) = (123) (23)(13) = (123).

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 2 / 23

SLIDE 4

The number of reduced reflection factorizations of c

Theorem (Hurwitz, 1892)

There are nn−2 (minimal length) factorizations t1 · · · tn−1 = (12 · · · n) ∈ Sn where the ti’s are transpositions. For example, the 31 factorizations (12)(23) = (123) (13)(12) = (123) (23)(13) = (123).

Theorem (Deligne-Arnol’d-Bessis)

For a well-generated, complex reflection group W , with Coxeter number h, there are hnn! |W | (minimal length) reflection factorizations t1 · · · tn = c of the Coxeter element c.

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 2 / 23

SLIDE 5

Arbitrary length reflection factorizations of c

If R denotes the set of reflections of W , we write FactW ,c(N) := #{(t1, · · · , tN) ∈ Rn | t1 · · · tN = c}. Now, consider the exponential generating function: FACSn,c(t) =

N≥0

FactSn,c(N)tN N!.

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 3 / 23

SLIDE 6

Arbitrary length reflection factorizations of c

If R denotes the set of reflections of W , we write FactW ,c(N) := #{(t1, · · · , tN) ∈ Rn | t1 · · · tN = c}. Now, consider the exponential generating function: FACSn,c(t) =

N≥0

FactSn,c(N)tN N!.

Theorem (Jackson, ’88)

If c = (12 · · · n) ∈ Sn, then FACSn,c(t) = et(

n 2)

n!

1 − e−tnn−1.

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 3 / 23

SLIDE 7

Arbitrary length reflection factorizations of c

If R denotes the set of reflections of W , we write FactW ,c(N) := #{(t1, · · · , tN) ∈ Rn | t1 · · · tN = c}. Now, consider the exponential generating function: FACSn,c(t) =

N≥0

FactSn,c(N)tN N!.

Theorem (Jackson, ’88)

If c = (12 · · · n) ∈ Sn, then FACSn,c(t) = et(

n 2)

n!

1 − e−tnn−1.

Notice that

tn−1

(n − 1)!

FACSn,c(t) = 1

n! · (n)n−1 · (n − 1)! = nn−2.

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 3 / 23

SLIDE 8

Arbitrary length reflection factorizations of c

If R denotes the set of reflections of W , we write FactW ,c(N) := #{(t1, · · · , tN) ∈ Rn | t1 · · · tN = c}. Now, consider the exponential generating function: FACW ,c(t) =

N≥0

FactW ,c(N)tN N!.

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 4 / 23

SLIDE 9

Arbitrary length reflection factorizations of c

If R denotes the set of reflections of W , we write FactW ,c(N) := #{(t1, · · · , tN) ∈ Rn | t1 · · · tN = c}. Now, consider the exponential generating function: FACW ,c(t) =

N≥0

FactW ,c(N)tN N!.

Theorem (Chapuy-Stump, ’12)

If W is well-generated, of rank n, and h is the order of the Coxeter element c, then FACW ,c(t) = et|R| |W |

1 − e−thn.

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 4 / 23

SLIDE 10

Arbitrary length reflection factorizations of c

If R denotes the set of reflections of W , we write FactW ,c(N) := #{(t1, · · · , tN) ∈ Rn | t1 · · · tN = c}. Now, consider the exponential generating function: FACW ,c(t) =

N≥0

FactW ,c(N)tN N!.

Theorem (Chapuy-Stump, ’12)

If W is well-generated, of rank n, and h is the order of the Coxeter element c, then FACW ,c(t) = et|R| |W |

1 − e−thn.

Notice that tn n!

FACW ,c(t) =

1 |W | · hn · n! = hnn! |W | .

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 4 / 23

SLIDE 11

A brief history of the W -Hurwitz number hnn! |W |

Some proofs and some mathematical shadows:

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 5 / 23

SLIDE 12

A brief history of the W -Hurwitz number hnn! |W |

Some proofs and some mathematical shadows:

1

Deligne-Tits-Zagier, rediscovered by Reading. Enumerate factorizations t1 · · · tn = c with respect to the c-orbit of tn: Hur(W ) = h 2

s∈S

Hur(Ws)

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 5 / 23

SLIDE 13

A brief history of the W -Hurwitz number hnn! |W |

Some proofs and some mathematical shadows:

1

Deligne-Tits-Zagier, rediscovered by Reading. Enumerate factorizations t1 · · · tn = c with respect to the c-orbit of tn: Hur(W ) = h 2

s∈S

Hur(Ws)

2

Chapoton. Interpretation as the number of maximal chains of NC(W ):

Hur(W ) =

X n

n!

n
i=1

hX + di di

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 5 / 23

SLIDE 14

A brief history of the W -Hurwitz number hnn! |W |

Some proofs and some mathematical shadows:

1

Deligne-Tits-Zagier, rediscovered by Reading. Enumerate factorizations t1 · · · tn = c with respect to the c-orbit of tn: Hur(W ) = h 2

s∈S

Hur(Ws)

2

Chapoton. Interpretation as the number of maximal chains of NC(W ):

Hur(W ) =

X n

n!

n
i=1

hX + di di

3

Lyashko-Looijenga and Bessis. There exist two subgroups G1 ≤ G2 ≤ Bn of the braid group Bn on n strands, with finite indexes ν1 and ν2 such that: ν1 = hnn! |W | ν2 = #{reduced reflection factorizations of c}

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 5 / 23

SLIDE 15

How to count, the Frobenius way

Consider the central element R :=

t∈R t of the group algebra C[W ].

N≥0

#{(t1, · · · , tN) ∈ RN | t1 · · · tN = c} ·tN N!

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 6 / 23

SLIDE 16

How to count, the Frobenius way

Consider the central element R :=

t∈R t of the group algebra C[W ].

N≥0

#{(t1, · · · , tN) ∈ RN | t1 · · · tN = c} ·tN N! =

N≥0
c
RN

·tN N!

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 6 / 23

SLIDE 17

How to count, the Frobenius way

Consider the central element R :=

t∈R t of the group algebra C[W ].

N≥0

#{(t1, · · · , tN) ∈ RN | t1 · · · tN = c} ·tN N! =

N≥0
c
RN

·tN N! =

N≥0
id

RN · c−1 ·tN N!

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 6 / 23

SLIDE 18

How to count, the Frobenius way

Consider the central element R :=

t∈R t of the group algebra C[W ].

N≥0

#{(t1, · · · , tN) ∈ RN | t1 · · · tN = c} ·tN N! =

N≥0
c
RN

·tN N! =

N≥0
id

RN · c−1 ·tN N!

!=! N≥0

1 |W | TrC[W ]

RN · c−1

·tN N!

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 6 / 23

SLIDE 19

How to count, the Frobenius way

Consider the central element R :=

t∈R t of the group algebra C[W ].

N≥0

#{(t1, · · · , tN) ∈ RN | t1 · · · tN = c} ·tN N! =

N≥0
c
RN

·tN N! =

N≥0
id

RN · c−1 ·tN N!

!=! N≥0

1 |W | TrC[W ]

RN · c−1

·tN N! =

N≥0

1 |W | ·

χ∈

W

dim(χ) · χ

RN · c−1

·tN N!

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 6 / 23

SLIDE 20

How to count, the Frobenius way

Consider the central element R :=

t∈R t of the group algebra C[W ].

=

N≥0

1 |W | ·

χ∈

W

dim(χ) · χ

RN · c−1

· tN N!

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 7 / 23

SLIDE 21

How to count, the Frobenius way

Consider the central element R :=

t∈R t of the group algebra C[W ].

=

N≥0

1 |W | ·

χ∈

W

dim(χ) · χ

RN · c−1

· tN N! =

N≥0

1 |W | ·

χ∈

W

χ(1) · χ(R) χ(1) N · χ(c−1)· tN N!

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 7 / 23

SLIDE 22

How to count, the Frobenius way

Consider the central element R :=

t∈R t of the group algebra C[W ].

=

N≥0

1 |W | ·

χ∈

W

dim(χ) · χ

RN · c−1

· tN N! =

N≥0

1 |W | ·

χ∈

W

χ(1) · χ(R) χ(1) N · χ(c−1)· tN N! = 1 |W |

χ∈

W

χ(1) · χ(c−1) · exp

t · χ(R)

χ(1)

Theo Douvropoulos (Paris VII, IRIF)

How to count reflection factorizations November 21, 2018 7 / 23

SLIDE 23

How to count, the Frobenius way

Consider the central element R :=

t∈R t of the group algebra C[W ].

=

N≥0

1 |W | ·

χ∈

W

dim(χ) · χ

RN · c−1

· tN N! =

N≥0

1 |W | ·

χ∈

W

χ(1) · χ(R) χ(1) N · χ(c−1)· tN N! = 1 |W |

χ∈

W

χ(1) · χ(c−1) · exp

t · χ(R)

χ(1)

Remark (Hurwitz 1901)

Exponential generating functions that enumerate factorizations of the form a1 · · · aN = g, where all ai’s belong to a set C closed under conjugation, are finite (weighted) sums of (scaled) exponentials.

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 7 / 23

SLIDE 24

The type-A calculation

FACSn,c(t) = 1 n!

χ∈

Sn

χ(1) · χ(c−1) · exp(t · χ(R) χ(1) )

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 8 / 23

SLIDE 25

The type-A calculation

FACSn,c(t) = 1 n!

χ∈

Sn

χ(1) · χ(c−1) · exp(t · χ(R) χ(1) ) Ingredients to calculate the above sum:

1

c = (12 · · · n), then χ(c−1) = 0 iff χ is a hook ( ).

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 8 / 23

SLIDE 26

The type-A calculation

FACSn,c(t) = 1 n!

χ∈

Sn

χ(1) · χ(c−1) · exp(t · χ(R) χ(1) ) Ingredients to calculate the above sum:

1

c = (12 · · · n), then χ(c−1) = 0 iff χ is a hook ( ).

2

If χk := χ(1k,n−k), we have (using Jucys-Murphy elements on

0 1 2 3 4 5

1
2
3

) χk(1) = n − 1 k

χk(c−1) = (−1)k

χk(R) χk(1) = n 2

− nk

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 8 / 23

SLIDE 27

The type-A calculation

FACSn,c(t) = 1 n!

χ∈

Sn

χ(1) · χ(c−1) · exp(t · χ(R) χ(1) ) Ingredients to calculate the above sum:

1

c = (12 · · · n), then χ(c−1) = 0 iff χ is a hook ( ).

2

If χk := χ(1k,n−k), we have (using Jucys-Murphy elements on

0 1 2 3 4 5

1
2
3

) χk(1) = n − 1 k

χk(c−1) = (−1)k

χk(R) χk(1) = n 2

− nk

Then, FACSn,c(t) = et(

n 2)

n! · n−1

k=0

n−1

k

· (−1)k ·
e−tnk

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 8 / 23

SLIDE 28

The type-A calculation

FACSn,c(t) = 1 n!

χ∈

Sn

χ(1) · χ(c−1) · exp(t · χ(R) χ(1) ) Ingredients to calculate the above sum:

1

c = (12 · · · n), then χ(c−1) = 0 iff χ is a hook ( ).

2

If χk := χ(1k,n−k), we have (using Jucys-Murphy elements on

0 1 2 3 4 5

1
2
3

) χk(1) = n − 1 k

χk(c−1) = (−1)k

χk(R) χk(1) = n 2

− nk

Then, FACSn,c(t) = et(

n 2)

n! · n−1

k=0

n−1

k

· (−1)k ·
e−tnk

= et(

n 2)

n! ·

1 − e−tnn−1.

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 8 / 23

SLIDE 29

Complex reflection groups and regular elements

A finite subgroup G ≤ GLn(V ) is called a complex reflection group if it is generated by pseudo-reflections. There are C-linear maps t that fix a hyperplane (i.e. codim(V t) = 1).

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 9 / 23

SLIDE 30

Complex reflection groups and regular elements

A finite subgroup G ≤ GLn(V ) is called a complex reflection group if it is generated by pseudo-reflections. There are C-linear maps t that fix a hyperplane (i.e. codim(V t) = 1).Shephard and Todd have classified (irreducible) complex reflection groups into:

1

an infinite 3-parameter family G(r, p, n) of monomial groups

2

34 exceptional cases indexed G4 to G37.

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 9 / 23

SLIDE 31

Complex reflection groups and regular elements

A finite subgroup G ≤ GLn(V ) is called a complex reflection group if it is generated by pseudo-reflections. There are C-linear maps t that fix a hyperplane (i.e. codim(V t) = 1).Shephard and Todd have classified (irreducible) complex reflection groups into:

1

an infinite 3-parameter family G(r, p, n) of monomial groups

2

34 exceptional cases indexed G4 to G37.

Definition

An element g ∈ W is called ζ-regular if it has a ζ-eigenvector v that lies in no reflection hyperplane. In particular, a Coxeter element is defined as a e2πi/h-regular element for h = (|R| + |A|)/n.

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 9 / 23

SLIDE 32

You already know this definition of Coxeter elements

Example

1

In Sn, the regular elements are (12 · · · n), (12 · · · n − 1)(n), and their powers. Indeed, (ζn−1, ζn−2, · · · , 1) with ζ = e2πi/n is an eigenvector for (12 · · · n).

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 10 / 23

SLIDE 33

You already know this definition of Coxeter elements

Example

1

In Sn, the regular elements are (12 · · · n), (12 · · · n − 1)(n), and their powers. Indeed, (ζn−1, ζn−2, · · · , 1) with ζ = e2πi/n is an eigenvector for (12 · · · n).

2

For real reflection groups:

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 10 / 23

SLIDE 34

The Chapuy-Stump proof

FACW ,c(t) = 1 |W |

χ∈

W

χ(1) · χ(c−1) · exp(t · χ(R) χ(1) )

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 11 / 23

SLIDE 35

The Chapuy-Stump proof

FACW ,c(t) = 1 |W |

χ∈

W

χ(1) · χ(c−1) · exp(t · χ(R) χ(1) ) Ingredients to calculate the above sum:

1

Well-generated complex reflection groups are classified into two infinite families G(r, 1, n), G(r, r, n) and some exceptional groups among G4 to G37.

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 11 / 23

SLIDE 36

The Chapuy-Stump proof

FACW ,c(t) = 1 |W |

χ∈

W

χ(1) · χ(c−1) · exp(t · χ(R) χ(1) ) Ingredients to calculate the above sum:

1

Well-generated complex reflection groups are classified into two infinite families G(r, 1, n), G(r, r, n) and some exceptional groups among G4 to G37.

2

Characters of the infinite families are essentially indexed by tuples of Young

diagrams. Most of them evaluate to 0 on Coxeter elements.

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 11 / 23

SLIDE 37

The Chapuy-Stump proof

FACW ,c(t) = 1 |W |

χ∈

W

χ(1) · χ(c−1) · exp(t · χ(R) χ(1) ) Ingredients to calculate the above sum:

1

Well-generated complex reflection groups are classified into two infinite families G(r, 1, n), G(r, r, n) and some exceptional groups among G4 to G37.

2

Characters of the infinite families are essentially indexed by tuples of Young

diagrams. Most of them evaluate to 0 on Coxeter elements.

3

All complex reflection groups can be described as permutation groups on a set of roots. GAP can then produce their character tables.

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 11 / 23

SLIDE 38

The Chapuy-Stump proof

FACW ,c(t) = 1 |W |

χ∈

W

χ(1) · χ(c−1) · exp(t · χ(R) χ(1) ) Ingredients to calculate the above sum:

1

Well-generated complex reflection groups are classified into two infinite families G(r, 1, n), G(r, r, n) and some exceptional groups among G4 to G37.

2

Characters of the infinite families are essentially indexed by tuples of Young

diagrams. Most of them evaluate to 0 on Coxeter elements.

3

All complex reflection groups can be described as permutation groups on a set of roots. GAP can then produce their character tables.

Remark

The fact that there is no uniform construction of the irreducible characters Irr(W ) makes it is very difficult to have a uniform proof.

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 11 / 23

SLIDE 39

A uniform argument; the decaf version

Definition

Given a character χ ∈ W , we define the Coxeter number cχ as the normalized trace of

t∈R(1 − t). That is,

cχ := 1 χ(1) ·

|R|χ(1) − χ(R)
= |R| − χ(R)

χ(1) .

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 12 / 23

SLIDE 40

A uniform argument; the decaf version

Definition

Given a character χ ∈ W , we define the Coxeter number cχ as the normalized trace of

t∈R(1 − t). That is,

cχ := 1 χ(1) ·

|R|χ(1) − χ(R)
= |R| − χ(R)

χ(1) . The Frobenius Lemma gives then: FACW ,g(t) = et|R| |W | ·

χ∈

W

χ(1) · χ(g −1) · exp(−t · cχ). (1)

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 12 / 23

SLIDE 41

A uniform argument; the decaf version

Definition

Given a character χ ∈ W , we define the Coxeter number cχ as the normalized trace of

t∈R(1 − t). That is,

cχ := 1 χ(1) ·

|R|χ(1) − χ(R)
= |R| − χ(R)

χ(1) . The Frobenius Lemma gives then: FACW ,g(t) = et|R| |W | ·

χ∈

W

χ(1) · χ(g −1) · exp(−t · cχ). (1)

Lemma

For a cpx reflection group W and a regular element g ∈ W , the total contribution in (1) of those characters χ ∈ W for which cχ is not a multiple of |g| is 0. [Just a whiff of coffee]

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 12 / 23

SLIDE 42

A uniform argument; the decaf version

Definition

Given a character χ ∈ W , we define the Coxeter number cχ as the normalized trace of

t∈R(1 − t). That is,

cχ := 1 χ(1) ·

|R|χ(1) − χ(R)
= |R| − χ(R)

χ(1) . The Frobenius Lemma gives then: FACW ,g(t) = et|R| |W | ·

χ∈

W

χ(1) · χ(g −1) · exp(−t · cχ). (1)

Lemma

For a cpx reflection group W and a regular element g ∈ W , the total contribution in (1) of those characters χ ∈ W for which cχ is not a multiple of |g| is 0. [Just a whiff of coffee]There is a cyclic permutation on the characters, induced by a galois action on the corresponding Hecke characters, that cancels out the contributions in each non-singleton orbit.

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 12 / 23

SLIDE 43

A uniform argument; the decaf version

Remark

We write lR(g) for the reflection length of g, i.e. the smallest number k of (quasi-)reflections ti needed to write g = t1 · · · tk.

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 13 / 23

SLIDE 44

A uniform argument; the decaf version

Remark

We write lR(g) for the reflection length of g, i.e. the smallest number k of (quasi-)reflections ti needed to write g = t1 · · · tk. This forces FACW ,g(t) = 0+0· t 1 +0· t2 2! +· · ·+0· tlR(g)−1 (lR(g) − 1)! +(something)· tlR(g) lR(g)! +· · · .

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 13 / 23

SLIDE 45

A uniform argument; the decaf version

Remark

We write lR(g) for the reflection length of g, i.e. the smallest number k of (quasi-)reflections ti needed to write g = t1 · · · tk. This forces FACW ,g(t) = 0+0· t 1 +0· t2 2! +· · ·+0· tlR(g)−1 (lR(g) − 1)! +(something)· tlR(g) lR(g)! +· · · . FACW ,g(t) = et|R| |W | ·

cχ | |g|

χ(1) · χ(g −1) · exp(−t · cχ)

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 13 / 23

SLIDE 46

A uniform argument; the decaf version

Remark

We write lR(g) for the reflection length of g, i.e. the smallest number k of (quasi-)reflections ti needed to write g = t1 · · · tk. This forces FACW ,g(t) = 0+0· t 1 +0· t2 2! +· · ·+0· tlR(g)−1 (lR(g) − 1)! +(something)· tlR(g) lR(g)! +· · · . FACW ,g(t) = et|R| |W | ·

cχ | |g|

χ(1) · χ(g −1) · exp(−t · cχ) = et|R| |W |

˜

Φ(X)

X=e−t|g|

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 13 / 23

SLIDE 47

A uniform argument; the decaf version

Remark

We write lR(g) for the reflection length of g, i.e. the smallest number k of (quasi-)reflections ti needed to write g = t1 · · · tk. This forces FACW ,g(t) = 0+0· t 1 +0· t2 2! +· · ·+0· tlR(g)−1 (lR(g) − 1)! +(something)· tlR(g) lR(g)! +· · · . FACW ,g(t) = et|R| |W | ·

cχ | |g|

χ(1) · χ(g −1) · exp(−t · cχ) = et|R| |W |

˜

Φ(X)

X=e−t|g|

1

Write ˜ Φ(X) = a(α1 − X)(α2 − X) · · · (αk − X).

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 13 / 23

SLIDE 48

A uniform argument; the decaf version

Remark

We write lR(g) for the reflection length of g, i.e. the smallest number k of (quasi-)reflections ti needed to write g = t1 · · · tk. This forces FACW ,g(t) = 0+0· t 1 +0· t2 2! +· · ·+0· tlR(g)−1 (lR(g) − 1)! +(something)· tlR(g) lR(g)! +· · · . FACW ,g(t) = et|R| |W | ·

cχ | |g|

χ(1) · χ(g −1) · exp(−t · cχ) = et|R| |W |

˜

Φ(X)

X=e−t|g|

1

Write ˜ Φ(X) = a(α1 − X)(α2 − X) · · · (αk − X).

2

Each part αi − X = αi − e−t|g| = αi − 1 + t|g| − · · ·

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 13 / 23

SLIDE 49

A uniform argument; the decaf version

Remark

We write lR(g) for the reflection length of g, i.e. the smallest number k of (quasi-)reflections ti needed to write g = t1 · · · tk. This forces FACW ,g(t) = 0+0· t 1 +0· t2 2! +· · ·+0· tlR(g)−1 (lR(g) − 1)! +(something)· tlR(g) lR(g)! +· · · . FACW ,g(t) = et|R| |W | ·

cχ | |g|

χ(1) · χ(g −1) · exp(−t · cχ) = et|R| |W |

˜

Φ(X)

X=e−t|g|

1

Write ˜ Φ(X) = a(α1 − X)(α2 − X) · · · (αk − X).

2

Each part αi − X = αi − e−t|g| = αi − 1 + t|g| − · · · contributes a factor of αi − 1 or t|g| on the leading term, depending on whether αi = 1 or not.

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 13 / 23

SLIDE 50

A uniform argument; the decaf version

Remark

We write lR(g) for the reflection length of g, i.e. the smallest number k of (quasi-)reflections ti needed to write g = t1 · · · tk. This forces FACW ,g(t) = 0+0· t 1 +0· t2 2! +· · ·+0· tlR(g)−1 (lR(g) − 1)! +(something)· tlR(g) lR(g)! +· · · . FACW ,g(t) = et|R| |W | ·

cχ | |g|

χ(1) · χ(g −1) · exp(−t · cχ) = et|R| |W |

˜

Φ(X)

X=e−t|g|

1

Write ˜ Φ(X) = a(α1 − X)(α2 − X) · · · (αk − X).

2

Each part αi − X = αi − e−t|g| = αi − 1 + t|g| − · · · contributes a factor of αi − 1 or t|g| on the leading term, depending on whether αi = 1 or not.

3

0 ≤ cχ ≤ |R| + |R∗|

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 13 / 23

SLIDE 51

A uniform argument; the decaf version

Remark

We write lR(g) for the reflection length of g, i.e. the smallest number k of (quasi-)reflections ti needed to write g = t1 · · · tk. This forces FACW ,g(t) = 0+0· t 1 +0· t2 2! +· · ·+0· tlR(g)−1 (lR(g) − 1)! +(something)· tlR(g) lR(g)! +· · · . FACW ,g(t) = et|R| |W | ·

cχ | |g|

χ(1) · χ(g −1) · exp(−t · cχ) = et|R| |W |

˜

Φ(X)

X=e−t|g|

1

Write ˜ Φ(X) = a(α1 − X)(α2 − X) · · · (αk − X).

2

Each part αi − X = αi − e−t|g| = αi − 1 + t|g| − · · · contributes a factor of αi − 1 or t|g| on the leading term, depending on whether αi = 1 or not.

3

0 ≤ cχ ≤ |R| + |R∗|

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 13 / 23

SLIDE 52

A uniform argument; the decaf version

Theorem

For a complex reflection group W , and a regular element g ∈ W : FACW ,g(t) = et|R| |W | ·

(1 − X)lR(g) · Φ(X)
X=e−t|g|

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 14 / 23

SLIDE 53

A uniform argument; the decaf version

Theorem

For a complex reflection group W , and a regular element g ∈ W : FACW ,g(t) = et|R| |W | ·

(1 − X)lR(g) · Φ(X)
X=e−t|g|

Here Φ(X) is of degree |R|+|A|

|g|

− lR(g), with Φ(0) = 1, and (1 − X) | Φ(X).

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 14 / 23

SLIDE 54

A uniform argument; the decaf version

Theorem

For a complex reflection group W , and a regular element g ∈ W : FACW ,g(t) = et|R| |W | ·

(1 − X)lR(g) · Φ(X)
X=e−t|g|

Here Φ(X) is of degree |R|+|A|

|g|

− lR(g), with Φ(0) = 1, and (1 − X) | Φ(X). Because deg(Φ(X)) = (|R| + |A|)/|g| − lR(g) is sometimes 0, we have:

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 14 / 23

SLIDE 55

A uniform argument; the decaf version

Theorem

For a complex reflection group W , and a regular element g ∈ W : FACW ,g(t) = et|R| |W | ·

(1 − X)lR(g) · Φ(X)
X=e−t|g|

Here Φ(X) is of degree |R|+|A|

|g|

− lR(g), with Φ(0) = 1, and (1 − X) | Φ(X). Because deg(Φ(X)) = (|R| + |A|)/|g| − lR(g) is sometimes 0, we have:

Corollary

When W is a complex reflection group and g ∈ W a regular element, then

1

If |g| = dn (includes Coxeter elements) FACW ,g(t) = et|R| |W | ·

1 − e−t|g|lR(g)

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 14 / 23

SLIDE 56

A uniform argument; the decaf version

Theorem

For a complex reflection group W , and a regular element g ∈ W : FACW ,g(t) = et|R| |W | ·

(1 − X)lR(g) · Φ(X)
X=e−t|g|

Here Φ(X) is of degree |R|+|A|

|g|

− lR(g), with Φ(0) = 1, and (1 − X) | Φ(X). Because deg(Φ(X)) = (|R| + |A|)/|g| − lR(g) is sometimes 0, we have:

Corollary

When W is a complex reflection group and g ∈ W a regular element, then

1

If |g| = dn (includes Coxeter elements) FACW ,g(t) = et|R| |W | ·

1 − e−t|g|lR(g)

2

Generally, we have that RedFactW (g) = multiple of |g|lR(g)(lR(g))! |W |

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 14 / 23

SLIDE 57

Can anyone guess what is happening?

Example

Below are the polynomials ˜ Φ(X) for W = Sn, n = 4 · · · 6 and all regular classes

1

S4:

1

(1234) : (1 − X)3

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 15 / 23

SLIDE 58

Can anyone guess what is happening?

Example

Below are the polynomials ˜ Φ(X) for W = Sn, n = 4 · · · 6 and all regular classes

1

S4:

1

(1234) : (1 − X)3

2

(13)(24) : (1 − X)2(1 + 2X + 2X 3 + X 4)

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 15 / 23

SLIDE 59

Can anyone guess what is happening?

Example

Below are the polynomials ˜ Φ(X) for W = Sn, n = 4 · · · 6 and all regular classes

1

S4:

1

(1234) : (1 − X)3

2

(13)(24) : (1 − X)2(1 + 2X + 2X 3 + X 4)

3

(123)(4) : (1 − X)2(1 + X)2

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 15 / 23

SLIDE 60

Can anyone guess what is happening?

Example

Below are the polynomials ˜ Φ(X) for W = Sn, n = 4 · · · 6 and all regular classes

1

S4:

1

(1234) : (1 − X)3

2

(13)(24) : (1 − X)2(1 + 2X + 2X 3 + X 4)

3

(123)(4) : (1 − X)2(1 + X)2

2

S5:

1

(12345) : (1 − X)4

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 15 / 23

SLIDE 61

Can anyone guess what is happening?

Example

Below are the polynomials ˜ Φ(X) for W = Sn, n = 4 · · · 6 and all regular classes

1

S4:

1

(1234) : (1 − X)3

2

(13)(24) : (1 − X)2(1 + 2X + 2X 3 + X 4)

3

(123)(4) : (1 − X)2(1 + X)2

2

S5:

1

(12345) : (1 − X)4

2

(1234)(5) : (1 − X)3(1 + 3X + X 2)

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 15 / 23

SLIDE 62

Can anyone guess what is happening?

Example

Below are the polynomials ˜ Φ(X) for W = Sn, n = 4 · · · 6 and all regular classes

1

S4:

1

(1234) : (1 − X)3

2

(13)(24) : (1 − X)2(1 + 2X + 2X 3 + X 4)

3

(123)(4) : (1 − X)2(1 + X)2

2

S5:

1

(12345) : (1 − X)4

2

(1234)(5) : (1 − X)3(1 + 3X + X 2)

3

(13)(24)(5) : (1 − X)2(1 + 2X + 3X 2 + 4X 3 + 10X 4 + 4X 5 + 3X 6 + 2X 7 + X 8)

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 15 / 23

SLIDE 63

Can anyone guess what is happening?

Example

Below are the polynomials ˜ Φ(X) for W = Sn, n = 4 · · · 6 and all regular classes

1

S4:

1

(1234) : (1 − X)3

2

(13)(24) : (1 − X)2(1 + 2X + 2X 3 + X 4)

3

(123)(4) : (1 − X)2(1 + X)2

2

S5:

1

(12345) : (1 − X)4

2

(1234)(5) : (1 − X)3(1 + 3X + X 2)

3

(13)(24)(5) : (1 − X)2(1 + 2X + 3X 2 + 4X 3 + 10X 4 + 4X 5 + 3X 6 + 2X 7 + X 8)

3

S6:

1

(123456) : (1 − X)5

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 15 / 23

SLIDE 64

Can anyone guess what is happening?

Example

Below are the polynomials ˜ Φ(X) for W = Sn, n = 4 · · · 6 and all regular classes

1

S4:

1

(1234) : (1 − X)3

2

(13)(24) : (1 − X)2(1 + 2X + 2X 3 + X 4)

3

(123)(4) : (1 − X)2(1 + X)2

2

S5:

1

(12345) : (1 − X)4

2

(1234)(5) : (1 − X)3(1 + 3X + X 2)

3

(13)(24)(5) : (1 − X)2(1 + 2X + 3X 2 + 4X 3 + 10X 4 + 4X 5 + 3X 6 + 2X 7 + X 8)

3

S6:

1

(123456) : (1 − X)5

2

(135)(246) : (1 − X)4(1 + 4X + 5X 2 + 5X 4 + 4X 5 + X 6).

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 15 / 23

SLIDE 65

Can anyone guess what is happening?

Example

Below are the polynomials ˜ Φ(X) for W = Sn, n = 4 · · · 6 and all regular classes

1

S4:

1

(1234) : (1 − X)3

2

(13)(24) : (1 − X)2(1 + 2X + 2X 3 + X 4)

3

(123)(4) : (1 − X)2(1 + X)2

2

S5:

1

(12345) : (1 − X)4

2

(1234)(5) : (1 − X)3(1 + 3X + X 2)

3

(13)(24)(5) : (1 − X)2(1 + 2X + 3X 2 + 4X 3 + 10X 4 + 4X 5 + 3X 6 + 2X 7 + X 8)

3

S6:

1

(123456) : (1 − X)5

2

(135)(246) : (1 − X)4(1 + 4X + 5X 2 + 5X 4 + 4X 5 + X 6).

3

(14)(25)(36) : (1−X)3(1+3X +6X 2+5X 3+18X 5+24X 6+18X 7+5X 9+6X 10+3X 11+X 12)

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 15 / 23

SLIDE 66

Can anyone guess what is happening?

Example

Below are the polynomials ˜ Φ(X) for W = Sn, n = 4 · · · 6 and all regular classes

1

S4:

1

(1234) : (1 − X)3

2

(13)(24) : (1 − X)2(1 + 2X + 2X 3 + X 4)

3

(123)(4) : (1 − X)2(1 + X)2

2

S5:

1

(12345) : (1 − X)4

2

(1234)(5) : (1 − X)3(1 + 3X + X 2)

3

(13)(24)(5) : (1 − X)2(1 + 2X + 3X 2 + 4X 3 + 10X 4 + 4X 5 + 3X 6 + 2X 7 + X 8)

3

S6:

1

(123456) : (1 − X)5

2

(135)(246) : (1 − X)4(1 + 4X + 5X 2 + 5X 4 + 4X 5 + X 6).

3

(14)(25)(36) : (1−X)3(1+3X +6X 2+5X 3+18X 5+24X 6+18X 7+5X 9+6X 10+3X 11+X 12)

4

(12345)(6) : (1 − X)4(1 + 4X + X 2)

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 15 / 23

SLIDE 67

Can anyone guess what is happening?

Example

Below are the polynomials ˜ Φ(X) for W = Sn, n = 4 · · · 6 and all regular classes

1

S4:

1

(1234) : (1 − X)3

2

(13)(24) : (1 − X)2(1 + 2X + 2X 3 + X 4)

3

(123)(4) : (1 − X)2(1 + X)2

2

S5:

1

(12345) : (1 − X)4

2

(1234)(5) : (1 − X)3(1 + 3X + X 2)

3

(13)(24)(5) : (1 − X)2(1 + 2X + 3X 2 + 4X 3 + 10X 4 + 4X 5 + 3X 6 + 2X 7 + X 8)

3

S6:

1

(123456) : (1 − X)5

2

(135)(246) : (1 − X)4(1 + 4X + 5X 2 + 5X 4 + 4X 5 + X 6).

3

(14)(25)(36) : (1−X)3(1+3X +6X 2+5X 3+18X 5+24X 6+18X 7+5X 9+6X 10+3X 11+X 12)

4

(12345)(6) : (1 − X)4(1 + 4X + X 2)

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 15 / 23

SLIDE 68

The topological braid group B(W )

Theorem (Steinberg)

W acts freely on the complement of the hyperplane arrangement V reg := V \ H. That is, ρ : V reg → W \ V reg is a covering map.

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SLIDE 69

The topological braid group B(W )

Theorem (Steinberg)

W acts freely on the complement of the hyperplane arrangement V reg := V \ H. That is, ρ : V reg → W \ V reg is a covering map. 1 ֒ → π1(V reg) := P(W ) ρ∗ − − − → π1(W \V reg) := B(W ) π − − ։ W → 1

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SLIDE 70

The topological braid group B(W )

Theorem (Steinberg)

W acts freely on the complement of the hyperplane arrangement V reg := V \ H. That is, ρ : V reg → W \ V reg is a covering map. 1 ֒ → π1(V reg) := P(W ) ρ∗ − − − → π1(W \V reg) := B(W ) π − − ։ W → 1

Theorem (Shephard-Todd-Chevalley, GIT)

W is realized as the group of deck transformations of a covering map ρ which is explicitly given via the fundamental invariants fi.

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 16 / 23

SLIDE 71

Hecke algebras for complex reflection groups

Consider a set of parameters u := (uC,j)(C∈A/W , 0≤j≤eC−1) where A := H is the reflection arrangement, C an orbit of hyperplanes, and eC the common order of the pointwise stabilizers WH (H ∈ C).

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 17 / 23

SLIDE 72

Hecke algebras for complex reflection groups

Consider a set of parameters u := (uC,j)(C∈A/W , 0≤j≤eC−1) where A := H is the reflection arrangement, C an orbit of hyperplanes, and eC the common order of the pointwise stabilizers WH (H ∈ C).

Definition

The generic Hecke algebra H(W ) associated to W is the quotient of the group ring Z[u, u−1]B(W ), over the ideal generated by the elements of the form (s − uC,0)(s − uC,1) · · · (s − uC,eC−1), which we call deformed order relations. Here s runs over all possible generators of the monodromy around the stratum C of H.

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 17 / 23

SLIDE 73

Hecke algebras for complex reflection groups

Consider a set of parameters u := (uC,j)(C∈A/W , 0≤j≤eC−1) where A := H is the reflection arrangement, C an orbit of hyperplanes, and eC the common order of the pointwise stabilizers WH (H ∈ C).

Definition

The generic Hecke algebra H(W ) associated to W is the quotient of the group ring Z[u, u−1]B(W ), over the ideal generated by the elements of the form (s − uC,0)(s − uC,1) · · · (s − uC,eC−1), which we call deformed order relations. Here s runs over all possible generators of the monodromy around the stratum C of H.

Theorem (Formerly known as “The BMR-freeness conjecture”)

The generic Hecke algebra is free over Z[u, u−1] of rank |W |.

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 17 / 23

SLIDE 74

Hecke algebras , an example

Example

The generic Hecke algebra of G26 (over the ring Z[x±1

0 , · · · y ±1 2 ]) is:

H(G26) = s, t, u | stst = tsts, su = us, tut = utu,

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 18 / 23

SLIDE 75

Hecke algebras , an example

Example

The generic Hecke algebra of G26 (over the ring Z[x±1

0 , · · · y ±1 2 ]) is:

H(G26) = s, t, u | stst = tsts, su = us, tut = utu, (s − x0)(s − x1) = 0 (t − y0)(t − y1)(t − y2) = 0 (u − y0)(u − y1)(u − y2) = 0

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 18 / 23

SLIDE 76

Hecke algebras , an example

Example

The generic Hecke algebra of G26 (over the ring Z[x±1

0 , · · · y ±1 2 ]) is:

H(G26) = s, t, u | stst = tsts, su = us, tut = utu, (s − x0)(s − x1) = 0 (t − y0)(t − y1)(t − y2) = 0 (u − y0)(u − y1)(u − y2) = 0 After the specializations (x0, x1) = (1, −1), (y0, y1, y2) = (1, ζ3, ζ2

3), we obtain the

following Coxeter-like presentation of G26: G26 = s, t, u | stst = tsts, su = us, tut = utu, s2 = t3 = u3 = 1 .

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 18 / 23

SLIDE 77

Splitting fields for Hecke algebras

Theorem (Malle)

Let K be the field of definition of W . There is a number N such that for parameters v := (vC,j)(C∈A/W , 0≤j≤eC−1), which satisfy v N

C,j = exp(2πi/eC)uC,j

the algebra K(v, v −1)H(W ) is split.

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 19 / 23

SLIDE 78

Splitting fields for Hecke algebras

Theorem (Malle)

Let K be the field of definition of W . There is a number N such that for parameters v := (vC,j)(C∈A/W , 0≤j≤eC−1), which satisfy v N

C,j = exp(2πi/eC)uC,j

the algebra K(v, v −1)H(W ) is split.

Definition

We consider the 1-parameter specialization uC,0 → x and uC,j → exp(2πij/eC). Then, if y is such that y N = x, K(y)Hx(W ) is split.

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 19 / 23

SLIDE 79

Splitting fields for Hecke algebras

Theorem (Malle)

Let K be the field of definition of W . There is a number N such that for parameters v := (vC,j)(C∈A/W , 0≤j≤eC−1), which satisfy v N

C,j = exp(2πi/eC)uC,j

the algebra K(v, v −1)H(W ) is split.

Definition

We consider the 1-parameter specialization uC,0 → x and uC,j → exp(2πij/eC). Then, if y is such that y N = x, K(y)Hx(W ) is split.

Definition (Malle’s Permutation Ψ)

We write Ψ for the permutation of the irreducible modules of Hx(W ) induced by the galois conjugation y → e2πi/N · y ∈ Gal

K(y)/K(x)
.

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 19 / 23

SLIDE 80

Fake degree palindromicity

Let (f1, · · · , fn) be homogeneous generators of the invariant algebra C[V ]W (so they satisfy fi(g −1v) = fi(v) ∀v ∈ V ).

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 20 / 23

SLIDE 81

Fake degree palindromicity

Let (f1, · · · , fn) be homogeneous generators of the invariant algebra C[V ]W (so they satisfy fi(g −1v) = fi(v) ∀v ∈ V ). We define the coinvariant algebra of W as the quotient co(W ) := C[V ]/C[V ]W

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 20 / 23

SLIDE 82

Fake degree palindromicity

Let (f1, · · · , fn) be homogeneous generators of the invariant algebra C[V ]W (so they satisfy fi(g −1v) = fi(v) ∀v ∈ V ). We define the coinvariant algebra of W as the quotient co(W ) := C[V ]/C[V ]W = C[V ]/f1, · · · , fn

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 20 / 23

SLIDE 83

Fake degree palindromicity

Let (f1, · · · , fn) be homogeneous generators of the invariant algebra C[V ]W (so they satisfy fi(g −1v) = fi(v) ∀v ∈ V ). We define the coinvariant algebra of W as the quotient co(W ) := C[V ]/C[V ]W = C[V ]/f1, · · · , fn ∼ = C[W ].

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 20 / 23

SLIDE 84

Fake degree palindromicity

Let (f1, · · · , fn) be homogeneous generators of the invariant algebra C[V ]W (so they satisfy fi(g −1v) = fi(v) ∀v ∈ V ). We define the coinvariant algebra of W as the quotient co(W ) := C[V ]/C[V ]W = C[V ]/f1, · · · , fn ∼ = C[W ].

Definition

The fake degree Pχ(q) := qei(χ) of a character χ ∈ W is a polynomial that records the exponents ei(χ) of χ. These are the degrees of the graded components of co(W ) that contain copies of χ.

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 20 / 23

SLIDE 85

Fake degree palindromicity

Let (f1, · · · , fn) be homogeneous generators of the invariant algebra C[V ]W (so they satisfy fi(g −1v) = fi(v) ∀v ∈ V ). We define the coinvariant algebra of W as the quotient co(W ) := C[V ]/C[V ]W = C[V ]/f1, · · · , fn ∼ = C[W ].

Definition

The fake degree Pχ(q) := qei(χ) of a character χ ∈ W is a polynomial that records the exponents ei(χ) of χ. These are the degrees of the graded components of co(W ) that contain copies of χ.

Theorem (Beynon-Lusztig, Malle, Opdam)

The fake degrees Pχ(q) satisfy the following palindromicity property: Pχ(q) = qcχPΨ(χ∗)(q−1), where cχ are the Coxeter numbers and Ψ is Malle’s permutation on Irr(W ).

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 20 / 23

SLIDE 86

The proof of the technical lemma

FACW ,g(t) = et|R| |W | ·

χ∈

W , cχ||g|

χ(1) · χ(g −1) · exp(−t · cχ).

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 21 / 23

SLIDE 87

The proof of the technical lemma

FACW ,g(t) = et|R| |W | ·

χ∈

W , cχ||g|

χ(1) · χ(g −1) · exp(−t · cχ).

1

There is a special element π ∈ P(W ) = π1(V reg, x0) called full twist, central in the braid group B(W ). It is the geometric circle [0, 1] ∋ t → e2πit · x0.

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 21 / 23

SLIDE 88

The proof of the technical lemma

FACW ,g(t) = et|R| |W | ·

χ∈

W , cχ||g|

χ(1) · χ(g −1) · exp(−t · cχ).

1

There is a special element π ∈ P(W ) = π1(V reg, x0) called full twist, central in the braid group B(W ). It is the geometric circle [0, 1] ∋ t → e2πit · x0.

2

Every ζ-regular element w, with ζ = e2πil/d, lifts to a d-th root of πl.

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 21 / 23

SLIDE 89

The proof of the technical lemma

FACW ,g(t) = et|R| |W | ·

χ∈

W , cχ||g|

χ(1) · χ(g −1) · exp(−t · cχ).

1

There is a special element π ∈ P(W ) = π1(V reg, x0) called full twist, central in the braid group B(W ). It is the geometric circle [0, 1] ∋ t → e2πit · x0.

2

Every ζ-regular element w, with ζ = e2πil/d, lifts to a d-th root of πl. (i.e. there exists w ∈ B(W ) with w d = πl and w → w under B(W ) → W )

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 21 / 23

SLIDE 90

The proof of the technical lemma

FACW ,g(t) = et|R| |W | ·

χ∈

W , cχ||g|

χ(1) · χ(g −1) · exp(−t · cχ).

1

There is a special element π ∈ P(W ) = π1(V reg, x0) called full twist, central in the braid group B(W ). It is the geometric circle [0, 1] ∋ t → e2πit · x0.

2

Every ζ-regular element w, with ζ = e2πil/d, lifts to a d-th root of πl. (i.e. there exists w ∈ B(W ) with w d = πl and w → w under B(W ) → W )

3

[Broue-Michel] The value of a character χx that corresponds to χ ∈ W (after Tits’ deformation theorem) is given on roots of the full twist by: χx(Tw) = χ(w) · x(|R|+|A|−cχ)l/d.

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 21 / 23

SLIDE 91

The proof of the technical lemma

FACW ,g(t) = et|R| |W | ·

χ∈

W , cχ||g|

χ(1) · χ(g −1) · exp(−t · cχ).

1

There is a special element π ∈ P(W ) = π1(V reg, x0) called full twist, central in the braid group B(W ). It is the geometric circle [0, 1] ∋ t → e2πit · x0.

2

Every ζ-regular element w, with ζ = e2πil/d, lifts to a d-th root of πl. (i.e. there exists w ∈ B(W ) with w d = πl and w → w under B(W ) → W )

3

[Broue-Michel] The value of a character χx that corresponds to χ ∈ W (after Tits’ deformation theorem) is given on roots of the full twist by: χx(Tw) = χ(w) · x(|R|+|A|−cχ)l/d.

4

If w is a regular element of order d and χ any character we have: Ψ(χ)(w) = exp

2πi · lcχ

d

· χ(w)

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 21 / 23

SLIDE 92

The proof of the technical lemma

FACW ,g(t) = et|R| |W | ·

χ∈

W , cχ||g|

χ(1) · χ(g −1) · exp(−t · cχ).

1

There is a special element π ∈ P(W ) = π1(V reg, x0) called full twist, central in the braid group B(W ). It is the geometric circle [0, 1] ∋ t → e2πit · x0.

2

Every ζ-regular element w, with ζ = e2πil/d, lifts to a d-th root of πl. (i.e. there exists w ∈ B(W ) with w d = πl and w → w under B(W ) → W )

3

[Broue-Michel] The value of a character χx that corresponds to χ ∈ W (after Tits’ deformation theorem) is given on roots of the full twist by: χx(Tw) = χ(w) · x(|R|+|A|−cχ)l/d.

4

If w is a regular element of order d and χ any character we have: Ψ(χ)(w) = exp

2πi · lcχ

d

· χ(w)

5

If k = d gcd(cχ, d) = 1, we have

k

i=1

Ψk(χ)(w) = 0.

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 21 / 23

SLIDE 93

Weighted enumeration

Definition

Consider a set of variables w := (wC)(C∈A/W ) and the weight function wt : R → {wC | C ∈ A/W }, t → w[V t]

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 22 / 23

SLIDE 94

Weighted enumeration

Definition

Consider a set of variables w := (wC)(C∈A/W ) and the weight function wt : R → {wC | C ∈ A/W }, t → w[V t] and the exponential generating funtion of weighted reflection factorizations: FACW ,g(w, z) :=

(t1,··· ,tN)∈RN

t1···tN=g

wt(t1) · · · wt(tN) · zN N!.

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 22 / 23

SLIDE 95

Weighted enumeration

Definition

Consider a set of variables w := (wC)(C∈A/W ) and the weight function wt : R → {wC | C ∈ A/W }, t → w[V t] and the exponential generating funtion of weighted reflection factorizations: FACW ,g(w, z) :=

(t1,··· ,tN)∈RN

t1···tN=g

wt(t1) · · · wt(tN) · zN N!.

Theorem

For a regular element g ∈ W , the weighted generating function takes the form: FACW ,g(w, z) = ez·wt(R) |W | ·

Φ(X) ·
C∈A/W

(1 − XC)nC

XC=e−zwC|g|.

The exponents nC are equal to the smallest number of reflections from C necessary in any reflection factorization of g.

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 22 / 23

SLIDE 96

Thank you!

Theo Douvropoulos (Paris VII, IRIF) How to count reflection factorizations November 21, 2018 23 / 23