A tale of dihedral groups, SL(2)q

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A tale of dihedral groups, SL(2)

_q

, and beyond

Or: Who colored my Dynkin diagrams?

Daniel Tubbenhauer

t • 1 s

· · · ·

Joint work with Marco Mackaay, Volodymyr Mazorchuk, Vanessa Miemietz and Xiaoting Zhang

February 2019

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LetA(Γ) be the adjacency matrix of a finite, connected, loopless graphΓ. Let U_e+1(X) be the Chebyshev polynomial .

Classification problem (CP).Classify allΓsuch that Ue+1(A(Γ)) = 0.

fore= 2

fore= 4 Smith∼1969. The graphs solutions to (CP) are precisely

ADE graphs fore+ 2 being(at most)the Coxeter number.

Type Am: • • • · · · • • • fore=m−1

Type Dm: • • · · · • •

•

for e= 2m−4

Type E6:

• • • • •

•

fore= 10 Type E7:

• • • • • •

•

fore= 16 Type E8:

• • • • • • •

•

fore= 28

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A3= 1 3 2

• • • A(A3) =



0 0 1

0 0 1

1 1 0



 S_A3={2 cos(^π₄),0,2 cos(^3π₄)}

U3(X) = (X−2 cos(^π₄))X(X−2 cos(^3π₄))

fore= 2

fore= 4

∼

Type Am: • • • · · · • • • fore=m−1

Type Dm: • • · · · • •

•

for e= 2m−4

Type E6:

• • • • •

•

fore= 10 Type E7:

• • • • • •

•

fore= 16 Type E8:

• • • • • • •

•

fore= 28

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A3= 1 3 2

• • • A(A3) =



0 0 1

0 0 1

1 1 0



 S_A3={2 cos(^π₄),0,2 cos(^3π₄)}

D4= 1 4 2

3

• •

•

A(D4) =





0 0 0 1

1 1 1 0



 S_D4={2 cos(^π₆),0²,2 cos(^5π₆)}

U3(X) = (X−2 cos(^π₄))X(X−2 cos(^3π₄))

U5(X) = (X−2 cos(^π₆))(X−2 cos(^2π₆))X(X−2 cos(^4π₆))(X−2 cos(^5π₆))

fore= 2

fore= 4 Smith∼1969. The graphs solutions to (CP) are precisely

Type Am: • • • · · · • • • fore=m−1

Type Dm: • • · · · • •

•

for e= 2m−4

Type E6:

• • • • •

•

fore= 10 Type E7:

• • • • • •

•

fore= 16 Type E8:

• • • • • • •

•

fore= 28

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A3= 1 3 2

• • • A(A3) =



0 0 1

0 0 1

1 1 0



 S_A3={2 cos(^π₄),0,2 cos(^3π₄)}

D4= 1 4 2

3

• •

•

A(D4) =





0 0 0 1

1 1 1 0



 S_D4={2 cos(^π₆),0²,2 cos(^5π₆)}

U3(X) = (X−2 cos(^π₄))X(X−2 cos(^3π₄))

U5(X) = (X−2 cos(^π₆))(X−2 cos(^2π₆))X(X−2 cos(^4π₆))(X−2 cos(^5π₆)) fore= 2

fore= 4

∼

Type Am: • • • · · · • • • fore=m−1

Type Dm: • • · · · • •

•

for e= 2m−4

Type E6:

• • • • •

•

fore= 10 Type E7:

• • • • • •

•

fore= 16 Type E8:

• • • • • • •

•

fore= 28

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A3= 1 3 2

• • • A(A3) =



0 0 1

0 0 1

1 1 0



 S_A3={2 cos(^π₄),0,2 cos(^3π₄)}

D4= 1 4 2

3

• •

•

A(D4) =





0 0 0 1

1 1 1 0



 S_D4={2 cos(^π₆),0²,2 cos(^5π₆)}

U3(X) = (X−2 cos(^π₄))X(X−2 cos(^3π₄))

U5(X) = (X−2 cos(^π₆))(X−2 cos(^2π₆))X(X−2 cos(^4π₆))(X−2 cos(^5π₆))

fore= 2

fore= 4

Smith∼1969. The graphs solutions to (CP) are precisely ADE graphs fore+ 2 being(at most)the Coxeter number.

Type Am: • • • · · · • • • fore=m−1

Type Dm: • • · · · • •

•

fore= 2m−4

Type E6:

• • • • •

•

for e= 10 Type E7:

• • • • • •

•

for e= 16 Type E8:

• • • • • • •

•

for e= 28

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Dihedral representation theory The classical representation theory TheN0-representation theory DihedralN0-representation theory

2 Non-semisimple fusion rings The asymptotic limit Cell modules

The dihedral example

3 Beyond

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The dihedral groups are of Coxeter type I2(e+ 2):

We+2=hs,t|s²=t²= 1, se+2 =. . .| {z }sts

e+2

=w0=. . .| {z }tst

e+2

=te+2i, e.g. : W₄=hs,t|s²=t²= 1, tsts=w₀=ststi

Example. These are the symmetry groups of regulare+ 2-gons,e.g. fore= 2 the Coxeter complex is:

• ^H

H

H F F

F F

1

I will sneak in the Hecke case, later on.

I will explain in a few minutes what cells are. For the moment: Never mind!

Lowest cell. Biggest cell.

s-cell. t-cell.

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We+2=hs,t|s²=t²= 1, se+2 =. . .| {z }sts

e+2

=w0=. . .| {z }tst

e+2

• ^H

H

H F F

F F

1 t s

what cells are. For the moment: Never mind!

s-cell. t-cell.

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We+2=hs,t|s²=t²= 1, se+2 =. . .| {z }sts

e+2

=w0=. . .| {z }tst

e+2

• ^H

H

H F F

F F

1 t s

ts

st

s-cell. t-cell.

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We+2=hs,t|s²=t²= 1, se+2 =. . .| {z }sts

e+2

=w0=. . .| {z }tst

e+2

• ^H

H

H F F

F F

1 t s

ts

st tst sts

what cells are. For the moment: Never mind!

s-cell. t-cell.

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We+2=hs,t|s²=t²= 1, se+2 =. . .| {z }sts

e+2

=w0=. . .| {z }tst

e+2

• ^H

H

H F F

F F

1 t s

ts

st tst sts

w0

s-cell. t-cell.

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We+2=hs,t|s²=t²= 1, se+2 =. . .| {z }sts

e+2

=w0=. . .| {z }tst

e+2

• ^H

H

H F F

F F

1 t s

ts

st tst sts

w0

I will explain in a few minutes what cells are.

For the moment: Never mind!

Lowest cell.

s-cell. t-cell.

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We+2=hs,t|s²=t²= 1, se+2 =. . .| {z }sts

e+2

=w0=. . .| {z }tst

e+2

• ^H

H

H F F

F F

1 t s

ts

st tst sts

w0

Lowest cell.

Biggest cell.

s-cell. t-cell.

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We+2=hs,t|s²=t²= 1, se+2 =. . .| {z }sts

e+2

=w0=. . .| {z }tst

e+2

• ^H

H

H F F

F F

1 t s

ts

st tst sts

w0

Lowest cell.

Biggest cell.

s-cell.

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We+2=hs,t|s²=t²= 1, se+2 =. . .| {z }sts

e+2

=w0=. . .| {z }tst

e+2

• ^H

H

H F F

F F

1 t s

ts

st tst sts

w0

Lowest cell.

Biggest cell.

s-cell.

t-cell.

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Dihedral representation theory on one slide.

One-dimensional modules. Mλs,λt, λs, λt∈C, θs7→λs, θt7→λt.

e≡0 mod 2 e6≡0 mod 2

M_0,0,M_2,0, M_0,2,M_2,2 M_0,0,M_2,2

Two-dimensional modules. Mz,z∈C, θs7→(²_{0 0}^z), θt7→(^{0 0}_z₂).

M_z,z ∈V^±_e−{0} M_z,z ∈V^±_e V_e =roots(U_e+1(X)) andV^±_e theZ/2Z-orbits under z7→ −z.

The Bott–Samelson (BS) generatorsθs=s+ 1, θt=t+ 1.

There is also a Kazhdan–Lusztig (KL) bases. Explicit formulas do not matter today.

Proposition (Lusztig?).

The list of one- and two-dimensionalWe+2-modules is a complete, irredundant list of simple modules.

I learned this construction from Mackaay in 2017.Example.

M0,0is the sign representation andM2,2is the trivial representation. In casee is odd, Ue+1(X) has a constant term, soM2,0,M0,2are not representations.

Example.

Mz forz being a root of the Chebyshev polynomial is a representation because the braid relation in terms of BS generators

involves the coefficients of the Chebyshev polynomial.

These representations are indexed byZ/2Z-orbits of the Chebyshev roots:

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M_0,0,M_2,0, M_0,2,M_2,2 M_0,0,M_2,2

There is also a Kazhdan–Lusztig (KL) bases. Explicit formulas do not matter today.

Proposition (Lusztig?).

I learned this construction from Mackaay in 2017.

Example.

involves the coefficients of the Chebyshev polynomial. Example.

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M_0,0,M_2,0, M_0,2,M_2,2 M_0,0,M_2,2

There is also a Kazhdan–Lusztig (KL) bases. Explicit formulas do not matter today. Proposition (Lusztig?).

I learned this construction from Mackaay in 2017.

Example.

M0,0is the sign representation andM2,2is the trivial representation.

In casee is odd, Ue+1(X) has a constant term, soM2,0,M0,2are not representations.

Example.

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M_0,0,M_2,0, M_0,2,M_2,2 M_0,0,M_2,2

Example.

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M_0,0,M_2,0, M_0,2,M_2,2 M_0,0,M_2,2

Example.

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An algebraPwith afixedbasisB^P is called a (multi)N0-algebra if xy∈N0B^P (x,y∈B^P).

AP-moduleMwith afixedbasisB^Mis called aN0-module if xm∈N0B^M (x∈B^P,m∈B^M).

These areN0-equivalent if there is aN0-valued change of basis matrix.

Example. N0-algebras andN0-modules arise naturally as the decategorification of 2-categories and 2-modules, andN0-equivalence comes from 2-equivalence.

Example.

Group algebras of finite groups with basis given by group elements areN⁰-algebras. The regular module is aN⁰-module.

Example.

Fusion rings are with basis given by classes of simples elements areN⁰-algebras. Key example: K0(Rep(G)) (easyN⁰-representation theory).

Key example: K0(Rep^ss_q(Uq(g)) =Gq) (intricateN⁰-representation theory). Example.

Hecke algebras of (finite) Coxeter groups with their KL basis areN⁰-algebras.

TheirN⁰-representation theory is mostly widely open.

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Example.

Group algebras of finite groups with basis given by group elements areN⁰-algebras.

The regular module is aN⁰-module.

Fusion rings are with basis given by classes of simples elements areN⁰-algebras. Key example: K0(Rep(G)) (easyN⁰-representation theory).

Key example: K0(Rep^ss_q(Uq(g)) =Gq) (intricateN⁰-representation theory). Example.

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Example.

Fusion rings are with basis given by classes of simples elements areN⁰-algebras.

Key example: K0(Rep(G)) (easyN⁰-representation theory).

Key example: K0(Rep^ss_q(Uq(g)) =Gq) (intricateN⁰-representation theory).

Example.

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Example.

Fusion rings are with basis given by classes of simples elements areN⁰-algebras.

Key example: K0(Rep(G)) (easyN⁰-representation theory).

Key example: K0(Rep^ss_q(Uq(g)) =Gq) (intricateN⁰-representation theory).

Example.

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Clifford, Munn, Ponizovski˘ı∼1942++, Kazhdan–Lusztig ∼1979. x≤Lyify appears inzxwith non-zero coefficient forz∈B^P. x∼Lyifx≤Lyandy≤Lx.

∼Lpartitions Pinto left cells L. Similarly for right R, two-sided cells LR or N0-modules.

AN0-moduleMis transitive if all basis elements belong to the same∼L

equivalence class. AnapexofMis a maximal two-sided cell not killing it.

Fact. Each transitiveN0-module has a unique apex.

Hence, one can study them cell-wise.

Example. TransitiveN0-modules arise naturally as the decategorification of simple 2-modules.

Philosophy.

Imagine a graph whose vertices are thex’s or them’s. v1→v2ifv1appears inzv2.

cells = connected components transitive = one connected component

“The atoms ofN⁰-representation theory”.

Question (N⁰-representation theory).Classify them! Example.

Group algebras with the group element basis have only one cell,G itself. TransitiveN⁰-modules areC[G/H] forH⊂G subgroup/conjugacy. The apex isG.

Example.

Fusion rings in general have only one cell since each basis element [Vi] has a dual [V_i^∗] such that [Vi][Vi^∗] contains 1 as a summand.

Cell theory is useless for them! Example (Lusztig ≤2003).

Hecke algebras for the dihedral group with KL basis have the following cells:

1

s ts sts tsts ststs

t st tst stst tstst w0

We will see the transitiveN⁰-modules in a second.

Left cells. Right cells.

Two-sided cells. Morally.

The further away anN⁰-algebra is from being semisimple, the more useful and interesting is its cell structure.

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Philosophy.

Imagine a graph whose vertices are thex’s or them’s.

v1→v2ifv1appears inzv2.

x1

x2

x3

x4 m1

m2

m3

m4

Question (N⁰-representation theory).Classify them!

Example.

1

s ts sts tsts ststs

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Philosophy.

x1

x2

x3

x4 m1

m2

m3

m4

Example.

1

s ts sts tsts ststs

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Philosophy.

x1

x2

x3

x4 m1

m2

m3

m4

Example.

1

s ts sts tsts ststs

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Philosophy.

Example.

Group algebras with the group element basis have only one cell,G itself.

TransitiveN⁰-modules areC[G/H] forH⊂G subgroup/conjugacy. The apex isG.

Example.

1

s ts sts tsts ststs

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Example.

TransitiveN⁰-modules areC[G/H] forH⊂G subgroup/conjugacy. The apex isG. Example.

Fusion rings in general have only one cell since each basis element [Vi] has a dual [Vi^∗] such that [Vi][Vi^∗] contains 1 as a summand.

Cell theory is useless for them!

Example (Lusztig ≤2003).

1

s ts sts tsts ststs

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Philosophy.

Example.

1

s ts sts tsts ststs w0

Left cells.

Right cells. Two-sided cells.

Morally.

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Example.

1

s ts sts tsts ststs

Left cells.

Right cells.

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Philosophy.

Example.

1

s ts sts tsts ststs w0

Morally.

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Philosophy.

Example.

1

s ts sts tsts ststs

Two-sided cells.

Morally.

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N0-modules via graphs.

Construct aW_∞-moduleMassociated to a bipartite graphΓ:

M=Ch1,2,3,4,5i

1 3 2 4 5

H F H

F

θ M =

2 0 1 0 0 0 2 1 1 1 0 0 0 0 0











, θ M =

0 0 0 0 0 0 0 0 0 0 1 1 2 0 0













The adjacency matrixA(Γ) ofΓis

A(Γ) =

0 0 1 0 0 0 0 1 1 1 1 1 0 0 0 0 1 0 0 0 0 1 0 0 0













These areWe+2-modules for somee

only ifA(Γ) is killed by the Chebyshev polynomial Ue+1(X). Morally speaking: These are constructed as the simples but with integral matrices having the Chebyshev-roots as eigenvalues. It is not hard to see that the Chebyshev–braid-like relation can not hold otherwise.

Hence, by Smith’s (CP) and Lusztig: We get a representation ofWe+2

ifΓis a ADE Dynkin diagram fore+ 2 being the Coxeter number. That these areN⁰-modules follows from categorification.

‘Smaller solutions’ are neverN⁰-modules. Classification.

Complete , irredundant ^list of transitiveN⁰-modules ofWe+2:

apex 1 cell s – t cell w0 cell

N⁰-reps. M0,0 MADE+bicolering fore+ 2 = Cox. num. M2,2

I learned this from/with Kildetoft–Mackaay–Mazorchuk–Zimmermann∼2016.

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M=Ch1,2,3,4,5i

1 3 2 4 5

θs

action

H F H

F

θs Ms=

2 0 1 0 0 0 2 1 1 1 0 0 0 0 0 0 0 0 0 0













, θt Mt=

0 0 0 0 0 0 0 0 0 0 1 1 2 0 0 0 1 0 2 0













A(Γ) =

0 0 1 0 0 0 0 1 1 1 1 1 0 0 0 0 1 0 0 0 0 1 0 0 0













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M=Ch1,2,3,4,5i

1 3 2 4 5

θs

action

H F H

F

θ M =

2 0 1 0 0 0 2 1 1 1 0 0 0 0 0











, θ M =

0 0 0 0 0 0 0 0 0 0 1 1 2 0 0













A(Γ) =

0 0 1 0 0 0 0 1 1 1 1 1 0 0 0 0 1 0 0 0 0 1 0 0 0













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M=Ch1,2,3,4,5i

1 3 2 4 5

θs

action

H F H

F

θs Ms=

2 0 1 0 0 0 2 1 1 1 0 0 0 0 0 0 0 0 0 0













, θt Mt=

0 0 0 0 0 0 0 0 0 0 1 1 2 0 0 0 1 0 2 0













A(Γ) =

0 0 1 0 0 0 0 1 1 1 1 1 0 0 0 0 1 0 0 0 0 1 0 0 0













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M=Ch1,2,3,4,5i

1 3 2 4 5

θs

action

H F H

F

θ M =

2 0 1 0 0 0 2 1 1 1 0 0 0 0 0











, θ M =

0 0 0 0 0 0 0 0 0 0 1 1 2 0 0













A(Γ) =

0 0 1 0 0 0 0 1 1 1 1 1 0 0 0 0 1 0 0 0 0 1 0 0 0











