Fusion rules for SL

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Fusion rules for SL

₂

Or: A toy example of modular representation theory Daniel Tubbenhauer

T(v−1)⊗T(1)∼=T(v)⊕T(v−2)

⇐⇒

v−1 = v +







−v−1 v ·

v−1

v−2 −explicit scalar· v−1

v−1

| {z }

nilpotent correction term







Joint with Lousie Sutton, Paul Wedrich, Jieru Zhu

July 2021

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Question. What can we say about finite-dimensional modules ofSL₂...

• ...in the context of representations of classical groups? The modules and their structure.

• ...in the context of representations of Hopf algebras? Object fusion rulesi.e.

tensor products rules.

• ...in the context of categories? Morphisms of representations and their structure.

If the characteristic of the underlying fieldK=KofSL₂=SL₂(K) is finite we will see inverse fractals ,e.g.

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Spoiler. What will be the take away?

In some sense modular (charp<∞) representation theory

is much harder than the classical one (char∞a.k.a. char 0 a.k.a. generically) because secretly we are doing fractal geometry.

In my toy exampleSL2everything is explicit.

(3)

Question. What can we say about finite-dimensional modules ofSL₂...

• ...in the context of representations of classical groups? The modules and their structure.

• ...in the context of representations of Hopf algebras? Object fusion rulesi.e.

tensor products rules.

• ...in the context of categories? Morphisms of representations and their structure.

If the characteristic of the underlying fieldK=KofSL₂=SL₂(K) is finite we will see inverse fractals ,e.g.

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400

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Spoiler. What will be the take away?

In some sense modular (charp<∞) representation theory

is much harder than the classical one (char∞a.k.a. char 0 a.k.a. generically) because secretly we are doing fractal geometry.

In my toy exampleSL2everything is explicit.

(4)

Weyl ∼1923. TheSL₂(dual)Weyl modules ∆(v−1).

∆(1−1)

∆(2−1)

∆(3−1)

∆(4−1)

∆(5−1)

∆(6−1)

∆(7−1)

X0Y0

X1Y0 X0Y1

X2Y0 X1Y1 X0Y2

X3Y0 X2Y1 X1Y2 X0Y3

X4Y0 X3Y1 X2Y2 X1Y3 X0Y4

X5Y0 X4Y1 X3Y2 X2Y3 X1Y4 X0Y5

X6Y0 X5Y1 X4Y2 X3Y3 X2Y4 X1Y5 X0Y6

a bc d

7→matrix who’s rows are expansions of (aX +cY)^v−i(bX+dY)ⁱ⁻¹.

Pascals triangle modulop= 5 picks out the simples, e.g. an unbroken east-west line is a Weyl module which is simple.

Example∆(7−1) =KX⁶Y⁰⊕ · · · ⊕KX⁰Y⁶.

(^{a b}_{c d}) acts as

The rows are expansions of (aX+cY)⁷⁻ⁱ(bX +dY)ⁱ⁻¹. Binomials!

Example∆(7−1), characteristic0. No common eigensystem⇒∆(7−1) simple.

Example∆(7−1), characteristic2.

(0,0,0,1,0,0,0) is a common eigenvector, so we found a submodule. When is∆(v−1)simple?

∆(v−1) is simple

⇔

v−1 w−1

6= 0 for allw ≤v

⇔(Lucas’ theorem) v = [ar,0,...,0]p.

General. Weyl ∆(λ) and dual Weyl∇(λ)

are easy a.k.a. standard; are parameterized by dominant integral weights;

are highest weight modules; are defined overZ; have the classical Weyl characters;

form a basis of the Grothendieck group unitriangular w.r.t. simples; satisfy (a version of) Schur’s lemma dimKExti(∆(λ),∆(µ)) = ∆i,0∆λ,µ;

are simple generically;

have a root-binomial-criterion to determine whether they are simple (Jantzen’s thesis∼1973).

Lucas∼1878.

“Binomials mod p are the product of binomials of the p-adic digits”:

a b

=Qr i=0

a_i b_i

modp, wherea= [ar,...,a0]p=Pr

i=0aipⁱ etc.

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Weyl ∼1923. TheSL₂simplesL(v−1) in ∆(v−1) forp= 5.

∆(1−1) L(1−1)

∆(2−1) L(2−1)

∆(3−1) L(3−1)

∆(4−1) L(4−1)

∆(5−1) L(5−1)

∆(6−1) L(6−1)

∆(7−1) L(7−1)

X0Y0

X1Y0 X0Y1

X2Y0 X1Y1 X0Y2

X3Y0 X2Y1 X1Y2 X0Y3

∆(7−1) has(its head)L(7−1) andL(3−1) as factors.

Example∆(7−1) =KX⁶Y⁰⊕ · · · ⊕KX⁰Y⁶.

⇔

v−1 w−1

6= 0 for allw ≤v

Lucas∼1878.

a b

=Qr i=0

a_i b_i

i=0aipⁱ etc.

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Weyl ∼1923. TheSL₂simplesL(v−1) in ∆(v−1) forp= 5.

∆(1−1) L(1−1)

∆(2−1) L(2−1)

∆(3−1) L(3−1)

∆(4−1) L(4−1)

∆(5−1) L(5−1)

∆(6−1) L(6−1)

∆(7−1) L(7−1)

X0Y0

X1Y0 X0Y1

X2Y0 X1Y1 X0Y2

X3Y0 X2Y1 X1Y2 X0Y3

∆(7−1) has(its head)Pascals triangle moduloL(7−1) andL(3p−= 5 picks out the simples,1) as factors.

e.g. an unbroken east-west line is a Weyl module which is simple.

Example∆(7−1) =KX⁶Y⁰⊕ · · · ⊕KX⁰Y⁶.

⇔

v−1 w−1

6= 0 for allw ≤v

Lucas∼1878.

a b

=Qr i=0

a_i b_i

i=0aipⁱ etc.

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∆(1−1)

∆(2−1)

∆(3−1)

∆(4−1)

∆(5−1)

∆(6−1)

∆(7−1)

X0Y0

X1Y0 X0Y1

X2Y0 X1Y1 X0Y2

X3Y0 X2Y1 X1Y2 X0Y3

a bc d

Example∆(7−1) =KX⁶Y⁰⊕ · · · ⊕KX⁰Y⁶.

⇔

v−1 w−1

6= 0 for allw ≤v

Lucas∼1878.

a b

=Qr i=0

a_i b_i

i=0aipⁱ etc.

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∆(1−1)

∆(2−1)

∆(3−1)

∆(4−1)

∆(5−1)

∆(6−1)

∆(7−1)

X0Y0

X1Y0 X0Y1

X2Y0 X1Y1 X0Y2

X3Y0 X2Y1 X1Y2 X0Y3

a bc d

Example∆(7−1) =KX⁶Y⁰⊕ · · · ⊕KX⁰Y⁶.

⇔

v−1 w−1

6= 0 for allw ≤v

Lucas∼1878.

a b

=Qr i=0

a_i b_i

i=0aipⁱ etc.

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∆(1−1)

∆(2−1)

∆(3−1)

∆(4−1)

∆(5−1)

∆(6−1)

∆(7−1)

X0Y0

X1Y0 X0Y1

X2Y0 X1Y1 X0Y2

X3Y0 X2Y1 X1Y2 X0Y3

a bc d

Example∆(7−1) =KX⁶Y⁰⊕ · · · ⊕KX⁰Y⁶.

No common eigensystem⇒∆(7−1) simple.

(0,0,0,1,0,0,0) is a common eigenvector, so we found a submodule.

When is∆(v−1)simple?

⇔

v−1 w−1

6= 0 for allw ≤v

Lucas∼1878.

a b

=Qr i=0

a_i b_i

i=0aipⁱ etc.

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∆(1−1)

∆(2−1)

∆(3−1)

∆(4−1)

∆(5−1)

∆(6−1)

∆(7−1)

X0Y0

X1Y0 X0Y1

X2Y0 X1Y1 X0Y2

X3Y0 X2Y1 X1Y2 X0Y3

a bc d

Example∆(7−1) =KX⁶Y⁰⊕ · · · ⊕KX⁰Y⁶.

(0,0,0,1,0,0,0) is a common eigenvector, so we found a submodule.

When is∆(v−1)simple?

⇔

v−1 w−1

6= 0 for allw ≤v

General.

Weyl ∆(λ) and dual Weyl∇(λ) are easy a.k.a. standard;

are parameterized by dominant integral weights;

are highest weight modules;

are defined overZ;

have the classical Weyl characters;

form a basis of the Grothendieck group unitriangular w.r.t. simples;

satisfy (a version of) Schur’s lemma dimKExti(∆(λ),∆(µ)) = ∆i,0∆λ,µ; are simple generically;

Lucas∼1878.

a b

=Qr i=0

a_i b_i

i=0aipⁱ etc.

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Ringel, Donkin∼1991. There is a class of indecomposables T(v−1) indexed by N. They are a bit tricky to define, but:

• They have ∆- and∇filtrations, which look the same if you tilt your head:

T(v−1) =

∆(v−1)

∆(w−1) ∆(x−1)

∆(y−1)

...

^∆(z⁻¹⁾

∇(v−1)

∇(w−1) ∇(x−1)

∇(y−1)

...

^∇(z⁻¹⁾ “tilting symmetry”

• Play the role of projective modules.

• T(v−1)∼=L(v−1)∼= ∆(v−1)∼=∇(v−1) generically.

• They are a bit better behaved than simples.

General.

Define them using Weyl and dual Weyl filtrations.

Example. T(4−1)in characteristic 3. How many Weyl factors doesT(v−1)have?

# Weyl factors ofT(v−1) is 2^k where k= max{νp v−1

w−1

,w ≤v}. (Order of vanishing of _w−1^v⁻¹ .) determined by (Lucas’s theorem)

non-zeronon-leadingdigits ofv = [ar,ar−1,...,a0]p. Example. T(220540−1)forp= 11?

v = 220540 = [1,4,0,7,7,1]11;

Maximal vanishing forw= 75594 = [0,5,1,8,8,2]11;

v−1 w−1

= (HUGE) = [...,6= 0,0,0,0,0]11.

⇒T(220540−1) has 2⁴Weyl factors.

Which Weyl factors doesT(v−1)have a.k.a. the negative digits game? Weyl factors ofT(v−1) are

∆([ar,±ar−1,...,±a0]p−1) wherev = [ar,...,a0]p (appearing exactly once).

Example. T(220540−1)forp= 11? v = 220540 = [1,4,0,7,7,1]11; has Weyl factors [1,±4,0,±7,±7,±1]11; e.g. ∆(218690 = [1,4,0,−7,−7,−1]11−1) appears. The tilting-Cartan matrix a.k.a. T(v−1) : ∆(w−1)

.

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This is characteristic 3. T(v−1) : ∆(w−1)

vs.

∆(v−1) :L(w−1)

– flawed reciprocity.

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This is characteristic 3.

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T(v−1) =

∆(v−1)

∆(w−1) ∆(x−1)

∆(y−1)

...

^∆(z⁻¹⁾

∇(v−1)

∇(w−1) ∇(x−1)

∇(y−1)

...

General. Define them using Weyl and dual Weyl filtrations.

Example. T(4−1)in characteristic 3.

How many Weyl factors doesT(v−1)have?

w−1

v = 220540 = [1,4,0,7,7,1]11;

v−1 w−1

= (HUGE) = [...,6= 0,0,0,0,0]11.

.

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vs.

∆(v−1) :L(w−1)

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T(v−1) =

∆(v−1)

∆(w−1) ∆(x−1)

∆(y−1)

...

^∆(z⁻¹⁾

∇(v−1)

∇(w−1) ∇(x−1)

∇(y−1)

...

Example. T(4−1)in characteristic 3.

How many Weyl factors doesT(v−1)have?

w−1

,w≤v}. (Order of vanishing of _w−1^v⁻¹ .) determined by (Lucas’s theorem)

v= 220540 = [1,4,0,7,7,1]11;

v−1 w−1

= (HUGE) = [...,6= 0,0,0,0,0]11.

.

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vs.

∆(v−1) :L(w−1)

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T(v−1) =

∆(v−1)

∆(w−1) ∆(x−1)

∆(y−1)

...

^∆(z⁻¹⁾

∇(v−1)

∇(w−1) ∇(x−1)

∇(y−1)

...

w−1

v = 220540 = [1,4,0,7,7,1]11;

v−1 w−1

= (HUGE) = [...,6= 0,0,0,0,0]11.

Which Weyl factors doesT(v−1)have a.k.a. the negative digits game?

Weyl factors ofT(v−1) are

Example. T(220540−1)forp= 11?

v= 220540 = [1,4,0,7,7,1]11; has Weyl factors [1,±4,0,±7,±7,±1]11; e.g. ∆(218690 = [1,4,0,−7,−7,−1]11−1) appears.

The tilting-Cartan matrix a.k.a. T(v−1) : ∆(w−1) .

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vs.

∆(v−1) :L(w−1)

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T(v−1) =

∆(v−1)

∆(w−1) ∆(x−1)

∆(y−1)

...

^∆(z⁻¹⁾

∇(v−1)

∇(w−1) ∇(x−1)

∇(y−1)

...

w−1

v = 220540 = [1,4,0,7,7,1]11;

v−1 w−1

= (HUGE) = [...,6= 0,0,0,0,0]11.

Example. T(220540−1)forp= 11? v = 220540 = [1,4,0,7,7,1]11; has Weyl factors [1,±4,0,±7,±7,±1]11; e.g. ∆(218690 = [1,4,0,−7,−7,−1]11−1) appears.

The tilting-Cartan matrix a.k.a. T(v−1) : ∆(w−1) .

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T(v−1) : ∆(w−1) vs.

∆(v−1) :L(w−1)

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T(v−1) =

∆(v−1)

∆(w−1) ∆(x−1)

∆(y−1)

...

^∆(z⁻¹⁾

∇(v−1)

∇(w−1) ∇(x−1)

∇(y−1)

...

w−1

v = 220540 = [1,4,0,7,7,1]11;

v−1 w−1

= (HUGE) = [...,6= 0,0,0,0,0]11.

.

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T(v−1) : ∆(w−1) vs.

∆(v−1) :L(w−1)

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Tilting modules form a braided monoidal category Tilt.

Simple⊗simple6=simple, Weyl⊗Weyl6=Weyl, but tilting⊗tilting=tilting . The Grothendieck algebra [Tilt] ofTiltis a commutative algebra with basis [T(v−1)]. So what I would like to answer on the object level,i.e. for [Tilt]:

• What are the fusion rules? I start here – fusion forT(1)

• Find theN_v^x_,_w ∈N0in T(v−1)⊗T(w−1)∼=L

xN_v^x_,_wT(x−1).

B For [Tilt] this means finding the structure constants.

This appears to be tricky and I do not have an answer

• What are the thick⊗-ideals?

B For [Tilt] this means finding the ideals. This is discussed second

General.

These facts hold in general, and

tilting modules form the “nicest possible” monoidal subcategory.

Thick⊗-ideal = generated by identities on objects.

⊗-ideal = generated by any sets of morphism.

Prime power Verlinde categories. The idealJp^k ⊂ Tilt/Jp^k+1 is the cell of projectives.

The abelianizationsVer_pk ofTilt/Jp^k+1 are called Verlinde categories. The Cartan matrix ofVer_pk is ap^k−p^k⁻¹-square matrix with entries given by the common Weyl factors ofT(v−1) andT(w−1).

Example (Cartan matrix ofVer₃4).

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Fusion graphs.

The fusion graph Γv = Γ_T(v−1)ofT(v−1) is:

• Vertices of Γ_v arew ∈N, and identified withT(w−1).

• k edges w −→^k x ifT(x−1) appearsk times in T(v−1)⊗T(w−1).

• T(v−1) is a⊗-generator if Γ_v is strongly connected.

• This works for any reasonable monoidal category, with vertices being indecomposable objects and edges count multiplicities in⊗-products.

Baby example. Assume that we have two indecomposable objects1andX, with X^⊗2=1⊕X. Then:

Γ₁= 1 X

not a⊗-generator , ΓX= 1 X a⊗-generator

General. These facts hold in general, and

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Fusion graphs forT(1): char3 vs. generic.

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T(1)’s fusion graph via a Bratteli-type diagram

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Formulas, for friends of formulas Letv = [aj,...,a0]p. We have

T(v−1)⊗T(1)∼=T(v)⊕ Mtl

i=0

T(v−2pⁱ)^⊕xⁱ,x_i =







0 ifa_i= 0 or i=j anda_j = 1, 2 ifa_i= 1,

1 ifa_i>1.

tl=tail length=length of [...,6=p−1,p−1,p−1,...,p−1]p

Proof strategy.

• Feed the problem into a machine;

• let it do a lot of calculations;

• guess the formula;

• prove the formula using character computations. Easy

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