March2021 DanielTubbenhauer 2 SL andfractals

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SL

₂

and fractals

Or: Modular representation theory in a toy example Daniel Tubbenhauer

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Joint with Lousie Sutton, Paul Wedrich, Jieru Zhu

March 2021

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

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

• ...in the context of the representation theory of Hopf algebras? Fusion rules i.e. tensor products rules.

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

The most amazing things happen if the characteristic of the underlying fieldK=K ofSL2=SL2(K) is finite, and we will see fractals,e.g.

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Spoiler: What will be the take away?

Well, in some sense modular (charp<∞) representation theory so much harder than classical one (char∞a.k.a. char 0)

because secretly we are doing fractal geometry. In my toy exampleSL2we can do everything explicitly.

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

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

• ...in the context of the representation theory of Hopf algebras? Fusion rules i.e. tensor products rules.

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

The most amazing things happen if the characteristic of the underlying fieldK=K ofSL2=SL2(K) is finite, and we will see fractals,e.g.

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300

400

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Spoiler: What will be the take away?

Well, in some sense modular (charp<∞) representation theory so much harder than classical one (char∞a.k.a. char 0)

because secretly we are doing fractal geometry.

In my toy exampleSL2we can do everything explicitly.

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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.

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’s 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

=Q_r

i=0 ai bi

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

i=0ai pi 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

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

a b

=Q_r

i=0 ai bi

i=0ai pi 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

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

a b

=Q_r

i=0 ai bi

i=0ai pi 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

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

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;

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

Lucas∼1878.

a b

=Q_r

i=0 ai bi

i=0ai pi etc.

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Ringel, Donkin∼1991. There is a class of modulesT(v−1) indexed byN. 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) overC.

• They are more well-behaved than simples. ^Analogy

General.

Define them using Weyl and dual Weyl filtrations.

Example ofT(4−1)for characteristic3 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. ExampleT(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).

ExampleT(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.

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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)

...

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

General. Define them using Weyl and dual Weyl filtrations.

Example ofT(4−1)for characteristic3

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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T(v−1) =

∆(v−1)

∆(w−1) ∆(x−1)

∆(y−1)

...

^∆(z⁻¹⁾

∇(v−1)

∇(w−1) ∇(x−1)

∇(y−1)

...

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

Example ofT(4−1)for characteristic3

How many Weyl factors doesT(v−1)have?

w−1

,w≤v}. (Order of vanishing of _w^v⁻₋¹₁ .) 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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T(v−1) =

∆(v−1)

∆(w−1) ∆(x−1)

∆(y−1)

...

^∆(z⁻¹⁾

∇(v−1)

∇(w−1) ∇(x−1)

∇(y−1)

...

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

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

ExampleT(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) =

∆(v−1)

∆(w−1) ∆(x−1)

∆(y−1)

...

^∆(z⁻¹⁾

∇(v−1)

∇(w−1) ∇(x−1)

∇(y−1)

...

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

w−1

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

v−1 w−1

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

ExampleT(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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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? ^Answer

• Find theN_v,w^x ∈N[0] inT(v−1)⊗T(v−1)∼=L

xN_v,w^x T(x−1).

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

• What are the thick⊗-ideals? ^Answer

B For [Tilt] this means finding the ideals.

General.

These facts hold in general, and

tilting modules form the “nicest possible” monoidal subcategory.

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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? ^Answer

• Find theN_v,w^x ∈N[0] inT(v−1)⊗T(v−1)∼=L

xN_v,w^x T(x−1).

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

• What are the thick⊗-ideals? ^Answer

B For [Tilt] this means finding the ideals.

General. These facts hold in general, and

tilting modules form the “nicest possible” monoidal subcategory.

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The morphism. There exists a K-algebra Z_p defined as a (very explicit) quotient of the path algebra of an infinite, fractal-like quiver. LetpMod-Z_p denote the category of finitely-generated, projective (right-)modules forZ_p. There is an equivalence of additive,K-linear categories

F:Tilt−→^∼⁼ pMod-Z_p,

sending indecomposable tilting modules to indecomposable projectives.

v T Δ

012345678910 11 12 13 14 15 161718 19 20 21 22 23 24 252627 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 0123

1 4 0

56 4 7 3

89 7 10

6 11

5 12 10 6 4 13 9 7 3 14 2 15 13 3 1 16 12 4 0

1718 16 19 15 20 14 21 19 15 13 22 18 16 12 23 11 24 22 12 10 25 21 13 9

2627 25 28 24 29 23 30 28 24 22 31 27 25 21 32 20 33 31 21 19 34 30 22 18 35 17 36 34 18 16 37 33 19 15 38 32 20 14 39 37 33 31 21 19 15 13 40 36 34 30 22 18 16 12 41 29 23 11 42 40 30 28 24 22 12 10 43 39 31 27 25 21 13 9 44

8 45 43 9 7 46 42 10 6 47 41 11 5 48 46 42 40 12 10 6 4 49 45 43 39 13 9 7 3 50 38 14 2 51 49 39 37 15 13 3 1 52 48 40 36 16 12 4 0 12345678910 11 12 13 14 15 16 171819 20 21 22 23 24 25 262728 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53

.

Figure:My favorite rainbow: The full subquiver containing the first 53 vertices of the quiver underlyingZ3.

Proof? Time’s up

Example, generation0, i.e. up top. In this case the quiver has no edges.

Continuing this periodically gives a quiver forTiltfor charp=∞. (This is the semisimple case: the quiver has to be boring.)

Example, generation1, i.e. up top².

In this case the quiver is a bunch of type A graphs. The algebra is a zigzag algebra, with arrows acting on the 0th digit.

Continuing this periodically gives a quiver forTilt

for the quantum group at a complex root of unity (due to Andersen∼2014). Example, generation2, i.e. up top³.

In this case every connected component

of the quiver is a bunch of type A graphs glued together in a matrix-grid. Each row and column is a zigzag algebra, with arrows acting on the 0th digit or 1digit,

and there are “squares commute” relations.

Continuing this periodically gives a quiver for projectiveG2T-modules (due to Andersen∼2019).

In general, Zp is basically a bunch of zigzag algebras (there are scalars and a lower-order-error term, but never mind) glued together in a fractal-way, according to the digits ofv = [ar,...,a0]p.

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v T Δ

012345678910 11 12 13 14 15 161718 19 20 21 22 23 24 252627 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 0123

1 4 0

56 4 7 3

89 7 10

6 11

5 12 10 6 4 13 9 7 3 14 2 15 13 3 1 16 12 4 0

1718 16 19 15 20 14 21 19 15 13 22 18 16 12 23 11 24 22 12 10 25 21 13 9

2627 25 28 24 29 23 30 28 24 22 31 27 25 21 32 20 33 31 21 19 34 30 22 18 35 17 36 34 18 16 37 33 19 15 38 32 20 14 39 37 33 31 21 19 15 13 40 36 34 30 22 18 16 12 41 29 23 11 42 40 30 28 24 22 12 10 43 39 31 27 25 21 13 9 44

8 45 43 9 7 46 42 10 6 47 41 11 5 48 46 42 40 12 10 6 4 49 45 43 39 13 9 7 3 50 38 14 2 51 49 39 37 15 13 3 1 52 48 40 36 16 12 4 0 12345678910 11 12 13 14 15 16 171819 20 21 22 23 24 25 262728 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53

.

Proof? Time’s up

Example, generation0, i.e. up top.

In this case the quiver has no edges.

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v T Δ

012345678910 11 12 13 14 15 161718 19 20 21 22 23 24 252627 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 0123

1 4 0

56 4 7 3

89 7 10

6 11

5 12 10 6 4 13 9 7 3 14 2 15 13 3 1 16 12 4 0

1718 16 19 15 20 14 21 19 15 13 22 18 16 12 23 11 24 22 12 10 25 21 13 9

2627 25 28 24 29 23 30 28 24 22 31 27 25 21 32 20 33 31 21 19 34 30 22 18 35 17 36 34 18 16 37 33 19 15 38 32 20 14 39 37 33 31 21 19 15 13 40 36 34 30 22 18 16 12 41 29 23 11 42 40 30 28 24 22 12 10 43 39 31 27 25 21 13 9 44

8 45 43 9 7 46 42 10 6 47 41 11 5 48 46 42 40 12 10 6 4 49 45 43 39 13 9 7 3 50 38 14 2 51 49 39 37 15 13 3 1 52 48 40 36 16 12 4 0 12345678910 11 12 13 14 15 16 171819 20 21 22 23 24 25 262728 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53

.

Proof? Time’s up

for the quantum group at a complex root of unity (due to Andersen∼2014).

Example, generation2, i.e. up top³. In this case every connected component

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v T Δ

012345678910 11 12 13 14 15 161718 19 20 21 22 23 24 252627 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 0123

1 4 0

56 4 7 3

89 7 10

6 11

5 12 10 6 4 13 9 7 3 14 2 15 13 3 1 16 12 4 0

1718 16 19 15 20 14 21 19 15 13 22 18 16 12 23 11 24 22 12 10 25 21 13 9

2627 25 28 24 29 23 30 28 24 22 31 27 25 21 32 20 33 31 21 19 34 30 22 18 35 17 36 34 18 16 37 33 19 15 38 32 20 14 39 37 33 31 21 19 15 13 40 36 34 30 22 18 16 12 41 29 23 11 42 40 30 28 24 22 12 10 43 39 31 27 25 21 13 9 44

8 45 43 9 7 46 42 10 6 47 41 11 5 48 46 42 40 12 10 6 4 49 45 43 39 13 9 7 3 50 38 14 2 51 49 39 37 15 13 3 1 52 48 40 36 16 12 4 0 12345678910 11 12 13 14 15 16 171819 20 21 22 23 24 25 262728 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53

.

Proof? Time’s up

for the quantum group at a complex root of unity (due to Andersen∼2014).

Example, generation2, i.e. up top³. In this case every connected component

of the quiver is a bunch of type A graphs glued together in a matrix-grid.

Each row and column is a zigzag algebra, with arrows acting on the 0th digit or 1digit, and there are “squares commute” relations.

(19)

v T Δ

012345678910 11 12 13 14 15 161718 19 20 21 22 23 24 252627 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 0123

1 4 0

56 4 7 3

89 7 10

6 11

5 12 10 6 4 13 9 7 3 14 2 15 13 3 1 16 12 4 0

1718 16 19 15 20 14 21 19 15 13 22 18 16 12 23 11 24 22 12 10 25 21 13 9

2627 25 28 24 29 23 30 28 24 22 31 27 25 21 32 20 33 31 21 19 34 30 22 18 35 17 36 34 18 16 37 33 19 15 38 32 20 14 39 37 33 31 21 19 15 13 40 36 34 30 22 18 16 12 41 29 23 11 42 40 30 28 24 22 12 10 43 39 31 27 25 21 13 9 44

8 45 43 9 7 46 42 10 6 47 41 11 5 48 46 42 40 12 10 6 4 49 45 43 39 13 9 7 3 50 38 14 2 51 49 39 37 15 13 3 1 52 48 40 36 16 12 4 0 12345678910 11 12 13 14 15 16 171819 20 21 22 23 24 25 262728 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53

.

Proof? Time’s up

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The whole story generalizes to Lusztig’s quantum group over Kwithq∈Kvia:

• We needp, the characteristic ofK, andl, the order ofq².

• Thep-l-adic expansion ofv = [a_r,...,a₀]_p,l isv =Pr

i=0a_ip^(r) withp⁽⁰⁾= 1 andp^(k) =p^k−1l. Here 0≤a₀<l−1 and 0≤a_i <p−1.

B Example. ForK=F7 andq= 2∈F7, we havep= 7 andl= 3.

B Example. 68 = [68]_p,_∞= [66,2]_∞_,3= [1,2,5]_7,7= [3,1,2]_7,3

• Repeat everything I told you for these expansions.

Here is the tilting-Cartan matrix in mixed characteristicp= 5 andl= 2:

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101,

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