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Maximal Dominant Weights for Affine Lie Algebra Representations Suzanne Crifo Advisor: Kailash Misra Department of Mathematics North Carolina State University June 5, 2018 Crifo (NCSU) Maximal Dominant Weights June 5, 2018 1 / 15 Outline

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Maximal Dominant Weights for Affine Lie Algebra Representations

Suzanne Crifo Advisor: Kailash Misra

Department of Mathematics North Carolina State University

June 5, 2018

Crifo (NCSU) Maximal Dominant Weights June 5, 2018 1 / 15

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Outline

1

Definitions

2

Integrable Highest Weight Modules

3

B(1)

n , V (kΛ0)

n = 5, k = 3 Results for arbitrary n and k

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Definitions

g = B(1)

n

Consider B(1)

n

with Cartan matrix and Dynkin diagram: A =            2 −1 . . . 2 −1 . . . −1 −1 2 −1 . . . −1 2 −1 . . . . . . ... . . . . . . −1 2 −1 . . . −2 2            1 2 2 . . . 2 2 1 ⇒

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Definitions

Definitions

Let h = span{h0, h1, . . . , hn, d} be the Cartan subalgebra Π = {α0, α1, . . . , αn} ⊂ h∗ is the set of simple roots. Note that αj(hi) = aij. Let δ = α0 + α1 + 2α2 + · · · + 2αn be the null root c = h0 + h1 + 2h2 + · · · + 2hn−1 + hn is the canonical central element The set {Λ0, Λ1, . . . , Λn} is the set of fundamental weights such that Λj(hi) = δij and Λj(d) = 0.

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Integrable Highest Weight Modules

Integrable Highest Weight Modules

A weight λ is dominant integral if λ(hi) ∈ Z≥0 for i ∈ I. Let P+ be the set of dominant integral weights.

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Integrable Highest Weight Modules

Integrable Highest Weight Modules

A weight λ is dominant integral if λ(hi) ∈ Z≥0 for i ∈ I. Let P+ be the set of dominant integral weights. For Λ ∈ P+ there exists an unique (up to isomorphism) irreducible, integrable highest weight module V (Λ) generated by a highest weight vector vΛ.

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Integrable Highest Weight Modules

V (Λ)

Let P(Λ) be the set of all weights of V (Λ). If λ ∈ P(Λ) then λ = Λ − n

i=0 biαi where bi ∈ Z≥0.

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Integrable Highest Weight Modules

V (Λ)

Let P(Λ) be the set of all weights of V (Λ). If λ ∈ P(Λ) then λ = Λ − n

i=0 biαi where bi ∈ Z≥0.

Definition A weight λ ∈ P(Λ) is called maximal if λ + δ ∈ P(Λ). Denote the set of all maximal weights as max(Λ).

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Integrable Highest Weight Modules

V (Λ)

Let P(Λ) be the set of all weights of V (Λ). If λ ∈ P(Λ) then λ = Λ − n

i=0 biαi where bi ∈ Z≥0.

Definition A weight λ ∈ P(Λ) is called maximal if λ + δ ∈ P(Λ). Denote the set of all maximal weights as max(Λ). Then P(Λ) =

  • λ∈max(Λ)

{λ − nδ|n ∈ Z≥0}. Any λ ∈ max(Λ) is W -conjugate to some µ ∈ max(Λ) ∩ P+, which is known to be a finite set. However, only partial results for an explicit description of this set are known.

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Integrable Highest Weight Modules

Find max(Λ) ∩ P+

θ = δ − a0α0 λ = λ − λ(c)Λ0 − (λ|Λ0)δ kCaf ∩ (Λ + Q) = {λ ∈ ˚ h∗

R|λ(hi) ≥ 0 for all i ∈ ˚

I, (λ|θ) ≤ k} Proposition (Kac) The map λ → λ defines a bijection from max(Λ) ∩ P+ onto kCaf ∩ (Λ + Q). In particular, the set of dominant maximal weights of V (Λ) is finite. We want to determine the maximal dominant weights for the integrable B(1)

n -module V (kΛ0).

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B(1)

n

, V (kΛ0) n = 5, k = 3

Example: B(1)

5 , V (3Λ0)

λ ∈ max(3Λ0) ∩ P+ = ⇒ λ = 3Λ0 − 5

i=0 biαi

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B(1)

n

, V (kΛ0) n = 5, k = 3

Example: B(1)

5 , V (3Λ0)

λ ∈ max(3Λ0) ∩ P+ = ⇒ λ = 3Λ0 − 5

i=0 biαi

Then λ = 3Λ0 −

5

  • i=0

biαi − (3Λ0 −

5

  • i=0

biαi)(c)Λ0 − (3Λ0 −

5

  • i=0

biαi | Λ0)δ

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B(1)

n

, V (kΛ0) n = 5, k = 3

Example: B(1)

5 , V (3Λ0)

λ ∈ max(3Λ0) ∩ P+ = ⇒ λ = 3Λ0 − 5

i=0 biαi

Then λ = 3Λ0 −

5

  • i=0

biαi − (3Λ0 −

5

  • i=0

biαi)(c)Λ0 − (3Λ0 −

5

  • i=0

biαi | Λ0)δ = −

5

  • i=0

biαi + b0δ

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B(1)

n

, V (kΛ0) n = 5, k = 3

Example: B(1)

5 , V (3Λ0)

λ ∈ max(3Λ0) ∩ P+ = ⇒ λ = 3Λ0 − 5

i=0 biαi

Then λ = 3Λ0 −

5

  • i=0

biαi − (3Λ0 −

5

  • i=0

biαi)(c)Λ0 − (3Λ0 −

5

  • i=0

biαi | Λ0)δ = −

5

  • i=0

biαi + b0δ = −

5

  • i=0

biαi + b0(α0 + α1 + 2α2 + 2α3 + 2α4 + 2α5)

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B(1)

n

, V (kΛ0) n = 5, k = 3

Example: B(1)

5 , V (3Λ0)

λ ∈ max(3Λ0) ∩ P+ = ⇒ λ = 3Λ0 − 5

i=0 biαi

Then λ = 3Λ0 −

5

  • i=0

biαi − (3Λ0 −

5

  • i=0

biαi)(c)Λ0 − (3Λ0 −

5

  • i=0

biαi | Λ0)δ = −

5

  • i=0

biαi + b0δ = −

5

  • i=0

biαi + b0(α0 + α1 + 2α2 + 2α3 + 2α4 + 2α5) = (b0 − b1)α1 + (2b0 − b2)α2 + (2b0 − b3)α3 + (2b0 − b4)α4 + (2b0 − b5)α5

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B(1)

n

, V (kΛ0) n = 5, k = 3

Example: B(1)

5 , V (3Λ0)

Let x1 = b0 − b1, xi = 2b0 − bi for i = 2, . . . , 5.

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B(1)

n

, V (kΛ0) n = 5, k = 3

Example: B(1)

5 , V (3Λ0)

Let x1 = b0 − b1, xi = 2b0 − bi for i = 2, . . . , 5. (5

i=1 xiαi)(h1) =

2x1 − x2 ≥ 0 (0, 0, 0, 0, 0) (5

i=1 xiαi)(h2) =

−x1 + 2x2 − x3 ≥ 0 (1, 1, 1, 1, 1) (5

i=1 xiαi)(h3) =

−x2 + 2x3 − x4 ≥ 0 (1, 2, 2, 2, 2) (5

i=1 xiαi)(h4) =

−x3 + 2x4 − x5 ≥ 0 = ⇒ (1, 2, 3, 3, 3) (5

i=1 xiαi)(h5) =

−2x4 + 2x5 ≥ 0 (1, 2, 3, 4, 4) (5

i=1 xiαi | θ) =

x2 ≤ 3 (1, 2, 3, 4, 5) (2, 2, 2, 2, 2) (2, 3, 3, 3, 3) (2, 3, 4, 4, 4) (2, 3, 4, 5, 5) (2, 3, 4, 5, 6) (3, 3, 3, 3, 3)

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B(1)

n

, V (kΛ0) n = 5, k = 3

What are the weights?

Recall λ = 3Λ0 − 5

i=0 biαi. Given an element of kCaf ∩ (Λ + Q),

(x1, x2, x3, x4, x5), we need to find the corresponding λ.

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B(1)

n

, V (kΛ0) n = 5, k = 3

What are the weights?

Recall λ = 3Λ0 − 5

i=0 biαi. Given an element of kCaf ∩ (Λ + Q),

(x1, x2, x3, x4, x5), we need to find the corresponding λ. Example (x = (2, 3, 4, 5, 5)) b0 − b1 = 2 = ⇒ b0 ≥ 2 2b0 − b2 = 3 = ⇒ b0 ≥ 3

2

2b0 − b3 = 4 = ⇒ b0 ≥ 2 = ⇒ b0 ≥ 3 2b0 − b4 = 5 = ⇒ b0 ≥ 5

2

2b0 − b5 = 5 = ⇒ b0 ≥ 5

2

Assume b0 = 3 + r for r ∈ Z>0. Then λ + δ =3Λ0 − (3 + r − 1)α0 − (1 + r − 1)α1 − (3 + 2r − 2)α2 − (2 + 2r − 2)α3 − (1 + 2r − 2)α4 − (1 + 2r − 2)α5 Therefore, b0 = 3 and λ = 3Λ0 − 3α0 − α1 − 3α2 − 2α3 − α4 − α5

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B(1)

n

, V (kΛ0) n = 5, k = 3

max(3Λ0) ∩ P+

x2 x vector Element of max(3Λ0) ∩ P+ (0, 0, 0, 0, 0) 3Λ0 1 (1, 1, 1, 1, 1) 3Λ0 − α0 − α2 − α3 − α4 − α5 2 (1, 2, 2, 2, 2) 3Λ0 − α0 2 (1, 2, 3, 3, 3) 3Λ0 − 2α0 − α1 − 2α2 − α3 − α4 − α5 2 (1, 2, 3, 4, 4) 3Λ0 − 2α0 − α1 − 2α2 − α3 2 (1, 2, 3, 4, 5) 3Λ0 − 3α0 − 2α1 − 4α2 − 3α3 − 2α4 − α5 2 (2, 2, 2, 2, 2) 3Λ0 − 2α0 − 2α2 − 2α3 − 2α4 − 2α5 3 (2, 3, 3, 3, 3) 3Λ0 − 2α0 − α2 − α3 − α4 − α5 3 (2, 3, 4, 4, 4) 3Λ0 − 2α0 − α2 3 (2, 3, 4, 5, 5) 3Λ0 − 3α0 − α1 − 3α2 − 2α3 − α4 − α5 3 (2, 3, 4, 5, 6) 3Λ0 − 3α0 − α1 − 3α2 − 2α3 − α4 3 (3, 3, 3, 3, 3) 3Λ0 − 3α0 − 3α2 − 3α3 − 3α4 − 3α5

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B(1)

n

, V (kΛ0) Results for arbitrary n and k

Elements of kCaf ∩ (Λ + Q)

Theorem Given ˚ A the Cartan matrix for type Bn of finite type, the set of solutions to ˚ Ax ≥ 0 x2 ≤ k is {x = 0} ∪ {x = (x1, x2, . . . , xn) ∈ Zn

≥0|1 ≤ x2 ≤ k, x1 =

x2

2

  • + l1, 0 ≤

l1 ≤ x2

2

  • , xi = x2 + i

j=3 lj, 0 ≤ l3 ≤

x2

2

  • − l1, 0 ≤ ln ≤ ln−1 ≤ ln−2 ≤

· · · ≤ l4 ≤ l3 for all 2 ≤ i ≤ n}.

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B(1)

n

, V (kΛ0) Results for arbitrary n and k

x2 x vector Element of max(4Λ0) ∩ P+ 4 (2, 4, 4, 4, 4) 4Λ0 − 2α0 4 (2, 4, 5, 5, 5) 4Λ0 − 3α0 − α1 − 2α2 − α3 − α4 − α5 4 (2, 4, 5, 6, 6) 4Λ0 − 3α0 − α1 − 2α2 − α3 4 (2, 4, 5, 6, 7) 4Λ0 − 4α0 − 2α1 − 4α2 − 3α3 − 2α4 − α5 4 (2, 4, 6, 6, 6) 4Λ0 − 3α0 − α1 − 2α2 4 (2, 4, 6, 7, 7) 4Λ0 − 4α0 − 2α1 − 4α2 − 2α3 − α4 − α5 4 (2, 4, 6, 7, 8) 4Λ0 − 4α0 − 2α1 − 4α2 − 2α3 − α4 4 (2, 4, 6, 8, 8) 4Λ0 − 4α0 − 2α1 − 4α2 − 2α3 4 (2, 4, 6, 8, 9) 4Λ0 − 5α0 − 3α1 − 6α2 − 4α3 − 2α4 − α5 4 (2, 4, 6, 8, 10) 4Λ0 − 5α0 − 3α1 − 6α2 − 4α3 − 2α4 4 (3, 4, 4, 4, 4) 4Λ0 − 3α0 − 2α2 − 2α3 − 2α4 − 2α5 4 (3, 4, 5, 5, 5) 4Λ0 − 3α0 − 2α2 − α3 − α4 − α5 4 (3, 4, 5, 6, 6) 4Λ0 − 3α0 − 2α2 − α3 4 (3, 4, 5, 6, 7) 4Λ0 − 4α0 − α1 − 4α2 − 3α3 − 2α4 − α5 4 (4, 4, 4, 4, 4) 4Λ0 − 4α2 − 4α3 − 4α4 − 4α5

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B(1)

n

, V (kΛ0) Results for arbitrary n and k

max kΛ0 ∩ P+

Theorem Let n ≥ 3, Λ = kΛ0, k ≥ 2. Then max(Λ) ∩ P+ = {Λ} ∪ {Λ − lα0 − (l − ( x2

2

  • + l1))α1 − (2l − x2)α2 −

n

i=3(2l − (x2 + i j=3 lj))αi

  • where

1 ≤ x2 ≤ k l = max{x1, xn

2

  • }

0 ≤ l1 ≤ x2

2

  • 0 ≤ l3 ≤ ⌊ x2

2 ⌋ − l1

0 ≤ ln ≤ ln−1 ≤ · · · ≤ l4 ≤ l3 l2 = x2 − x1 for n = 3, l2 = 0 else

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B(1)

n

, V (kΛ0) Results for arbitrary n and k

Thank you!

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B(1)

n

, V (kΛ0) Results for arbitrary n and k

Type C (1)

n

1 2 . . . 2 1 ⇐ ⇒ Theorem Let n ≥ 2, Λ = kΛ0, k ≥ 2, m ∈ Z≥0. Then max(Λ)∩P+ = {Λ}∪{Λ−lα0−(2l−x1)α1− n−1

i=2 (2l−(x1+i j=2 lj))αi

  • where

1 ≤ x1 ≤ k 0 ≤ l2 ≤ x1 0 ≤ ln−1 ≤ ln−2 ≤ · · · ≤ l3 ≤ l2 and ln−1 = x1 for n = 2 xn−1

2

  • ≤ l ≤
  • xn−1+ln−1

2

  • Crifo (NCSU)

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B(1)

n

, V (kΛ0) Results for arbitrary n and k

Type D(1)

n

1 2 2 . . . 2 1 1 1

Theorem Let n ≥ 4, Λ = kΛ0, k ≥ 2, m ∈ Z≥0. Then max(Λ) ∩ P+ = {Λ} ∪ {Λ − lα0 − (l − ( x2

2

  • + l1))α1 − (2l − x2)α2 −

n−2

i=3 (2l − (x2 + i j=3 lj))αi

  • − (l − xn−1)αn−1 − (l − xn)αn where

2 ≤ x2 ≤ k l = max{x1, xn−1, xn} 0 ≤ l1 ≤ x2

2

  • 0 ≤ l3 ≤ ⌊ x2

2 ⌋ − l1

0 ≤ ln−2 ≤ ln−3 ≤ · · · ≤ l4 ≤ l3, l2 = x2 − x1 for n = 4, l2 = 0 else xn−1 + xn ≤ xn−2 + ln−2 min{xn−1, xn} ≥ xn−2

2

  • Crifo (NCSU)

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B(1)

n

, V (kΛ0) Results for arbitrary n and k

Type G (1)

2

1 2 3 ⇛ Theorem Let Λ = kΛ0, k ≥ 2. Then max(Λ) ∩ P+ = {Λ} ∪ {Λ − lα0 − (2l − x1)α1 − (3l − x2)α2 where 1 ≤ x1 ≤ k 3x1

2

  • ≤ x2 ≤ 2x1

l = x2

3

  • Crifo (NCSU)

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B(1)

n

, V (kΛ0) Results for arbitrary n and k

Type F (1)

4

1 2 3 4 2 ⇒ Theorem Let Λ = kΛ0, k ≥ 2. Then max(Λ) ∩ P+ = {Λ} ∪ {Λ − lα0 − (2l − x1)α1 − (3l − (x1 + l2))α2 − (4l − (x1 + l2 + l3))α3 − (2l − x4)α4 where 1 ≤ x1 ≤ k x1

2

  • ≤ l2 ≤ x1
  • x1+l2

3

  • ≤ l3 ≤ l2
  • x1+l2+l3

2

  • ≤ x4 ≤ 2l3

l = x4

2

  • Crifo (NCSU)

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B(1)

n

, V (kΛ0) Results for arbitrary n and k

Type A(2)

2n

2 2 . . . 2 1 ⇐ ⇐ Theorem Let n ≥ 2, Λ = kΛ0, k ≥ 2, m ∈ Z≥0. Then max(Λ)∩P+ = {Λ}∪{Λ−lα0−(l −x1)α1− n−1

i=2 (l −(x1+i j=2 lj))αi

  • where

1 ≤ x1 ≤ k

2

  • 0 ≤ l2 ≤ x1 for n > 2

0 ≤ ln−1 ≤ ln−2 ≤ · · · ≤ l3 ≤ l2 and ln−1 = x1 for n = 2 xn−1

2

  • ≤ l ≤
  • xn−1+ln−1

2

  • Crifo (NCSU)

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B(1)

n

, V (kΛ0) Results for arbitrary n and k

Type A(2)

2n−1

1 2 2 . . . 2 1 1 ⇐

Theorem Let n ≥ 3, Λ = kΛ0, k ≥ 2, m ∈ Z≥0. Then max(Λ) ∩ P+ = {Λ} ∪ {Λ − lα0 − (l − ( x2

2

  • + l1))α1 − (2l − x2)α2 −

n−1

i=3 (2l − (x2 + i j=3 lj))αi

  • − (l − xn)αn where

2 ≤ x2 ≤ k l = max{x1, xn} 0 ≤ l1 ≤ x2

2

  • 0 ≤ l3 ≤ ⌊ x2

2 ⌋ − l1 for n > 3

0 ≤ ln−1 ≤ ln−2 ≤ · · · ≤ l4 ≤ l3, l2 = x2 − x1 for n = 3, l2 = 0 else xn−1

2

  • ≤ xn ≤
  • xn−1+ln−1

2

  • Crifo (NCSU)

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B(1)

n

, V (kΛ0) Results for arbitrary n and k

Type D(2)

n+1

1 1 . . . 1 1 ⇒ ⇐ Theorem Let n ≥ 2, Λ = kΛ0, k ≥ 2, m ∈ Z≥0. Then max(Λ)∩P+ = {Λ}∪{Λ−lα0 −(l −x1)α1 − n

i=2(l −(x1 +i j=2 lj))αi

  • where

1 ≤ x1 ≤ k

2

  • l = xn

0 ≤ l2 ≤ x1 0 ≤ ln ≤ ln−1 ≤ · · · ≤ l4 ≤ l3

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B(1)

n

, V (kΛ0) Results for arbitrary n and k

Type E (2)

6

1 2 3 2 1 ⇐ Theorem Let Λ = kΛ0, k ≥ 2. Then max(Λ) ∩ P+ = {Λ} ∪ {Λ − lα0 − (2l − x1)α1 − (3l − x2)α2 − (2l − x3)α3 where 2 ≤ x1 ≤ k 3 x1

2

  • ≤ x2 ≤ 2x1

2

3x2

  • ≤ x3 ≤ x2 −

x1

2

  • x3

2

  • ≤ l ≤ 2x3 − x2

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B(1)

n

, V (kΛ0) Results for arbitrary n and k

Type D(3)

4

1 2 1 ⇚      2x1 − 3x2 ≥ 0 −x1 + 2x2 ≥ 0 x1 ≤ k

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B(1)

n

, V (kΛ0) Results for arbitrary n and k

Elements of kCaf ∩ (Λ + Q)

Theorem The set of solutions to 2x1 − 3x2 ≥ 0 −x1 + 2x2 ≥ 0 x1 ≤ k is {x = 0} ∪ {x = (x1, x2)|2 ≤ x1 ≤ k, x1

2

  • ≤ x2 ≤

2x1

3

  • }.

Proof. Note that since A is of finite type, x ≥ 0 and so x1 ≥ 0 and x1 ≤ k by definition. Therefore, x1 = 0, which implies x = 0, or x1 ≥ 1. However, by (Ax)1 ≥ 0 and (Ax)2 ≥ 0, x1

2 ≤ x2 ≤ 2 3x1 and since x2 must be an integer, x1 = 1.

Now, fix x1 such that 2 ≤ x1 ≤ k. Then by (Ax)1 ≥ 0, x2 ≤ 2

3x1 and by

(Ax)2 ≥ 0, x2 ≥ 1

  • 2x1. Since x2 must be an integer, we have

x1

2

  • ≤ x2 ≤

2x1

3

  • .

This describes all possible solutions.

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B(1)

n

, V (kΛ0) Results for arbitrary n and k

max kΛ0 ∩ P+

Theorem Let Λ = kΛ0, k ≥ 2. Then max(Λ) ∩ P+ = {Λ} ∪ {Λ − lα0 − (2l − x1)α1 where 2 ≤ x1 ≤ k x1

2

  • ≤ l ≤

2x1

3

  • Crifo (NCSU)

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