U q ( gl ( m | 1 )) and canonical bases Sean Clark Northeastern - - PowerPoint PPT Presentation

Uq(gl(m|1)) and canonical bases

Sean Clark Northeastern University / Max Planck Institute for Mathematics 2nd US-Mexico Conference on Representation theory, Categorification, and Noncommutative Algebra

Sean Clark Uq(gl(m|1)) and canonical bases June 1, 2016 1 / 23

QUANTUM ENVELOPING gl(m|1) Some history

QUANTUM ALGEBRAS AND CANONICAL BASES

Uq(g): algebra over Q(q) coming from root data of simple Lie algebra g. (∼1990) Lusztig and Kashiwara: miraculous bases for U−

q (g) = Uq(n−)

(CB1) B is a Z[q, q−1]-basis of the integral form AU−

q (g);

(CB2) For any b ∈ B, b = b (· is natural involution q → q−1); (CB3) PBW → B is qZ[q]-unitriangular for any PBW (CB4) B induces a basis on suitable modules Connections to geometry, combinatorics, categorification, . . .

Sean Clark Uq(gl(m|1)) and canonical bases June 1, 2016 2 / 23

QUANTUM ENVELOPING gl(m|1) Some history

QUANTUM SUPERALGEBRAS

Question: what if g is a Lie superalgebra? E.g. gl(m|n): linear maps on vector superspace Cm|n. Super complications:

◮ Different simple roots= Different U− (in general)

Workaround: Work with a standard choice, e.g. α1 = ǫ1 − ǫ2, α2 = ǫ2 − ǫ3, . . ..

◮ Finite-dimensional representations = not semisimple (in general)

Workaround: Work with a nice subcategory, e.g. polynomial V(λ)

⊕

⊂ V⊗t

vec ◮ “Chirality” in parameter q; e.g. Uq(gl(m|n))0 ∼

= Uq(gl(m)) ⊗ Uq−1(gl(n)) Workaround: Restrict rank; e.g. gl(m|1)

Sean Clark Uq(gl(m|1)) and canonical bases June 1, 2016 3 / 23

QUANTUM ENVELOPING gl(m|1) Some history

SOME KNOWN RESULTS

Few examples of CBs known:

◮ easy small rank examples on U−; e.g. gl(1|1), standard gl(2|1), osp(1|2) ◮ osp(1|2n) and anisotropic Kac-Moody super [C-Hill-Wang, ’13] ◮ Partial results for gl(m|n), osp(2|2n) [C-Hill-Wang, ’13; Du-Gu ’14]

Why partial? Problems for gl(m|n):

◮ Only considers standard simple roots case; ◮ In standard case, PBW basis → B = (CB1), (CB2) ◮ If m > 1 and n > 1, this basis is not canonical! (Depends on PBW) ◮ When m = 1 or n = 1, get pseudo-canonical basis (a signed basis)

Sean Clark Uq(gl(m|1)) and canonical bases June 1, 2016 4 / 23

QUANTUM ENVELOPING gl(m|1) Quantum gl(m|1)

ROOT DATA FOR gl(m|1)

P =  m

i=1 Z ǫi

• even

  ⊕ Z ǫm+1

• dd

with (ǫi, ǫi) = (−1)p(ǫi), P∨, ·, · as usual. Φ =

• ǫi − ǫj | 1 ≤ i = j ≤ m + 1
• with simple roots Π = {αi | i ∈ I}

Standard choice: Πstd = {ǫ1 − ǫ2, . . . , ǫm − ǫm+1} Note:

◮ odd roots are isotropic; e.g. (ǫm − ǫm+1, ǫm − ǫm+1) = 1 + −1 = 0; ◮ no odd simple reflection! (Still obvious Sm+1 action: Weyl groupoid) ◮ different choices of simple roots may have different GCMs!

GCMs for m = 3:   2 −1 −1 2 −1 −1  

• standard choice

,   2 −1 −1 1 1  

• Π={ǫ1−ǫ2,ǫ2−ǫ4,ǫ4−ǫ3}

Sean Clark Uq(gl(m|1)) and canonical bases June 1, 2016 5 / 23

QUANTUM ENVELOPING gl(m|1) Quantum gl(m|1)

Uq(gl(m|1))

Fix a choice of Π. We define U = Uq(Π) = Q(q)

• Ei, Fi, qh | i ∈ I, h ∈ P∨

subject to usual relations: e.g. qhEiq−h = qh,αiEi, [Ei, Fj]

super commutator

= EiFj − (−1)p(i)p(j)FjEi = δij qhi − q−hi q − q−1 and both usual and unusual Serre relations: if p(i) = 1, E2

i = F2 i = 0,

[Ei−1, [Ei, [Ei+1, Ei]q]q]q

• super q−commutators

= 0 This has standard structural features (integral form, triangular decomposition, bar-involution, . . . ) NOTE: Different Π yield different U− = Q(q) Fi | i ∈ I in general!

Sean Clark Uq(gl(m|1)) and canonical bases June 1, 2016 6 / 23

QUANTUM ENVELOPING gl(m|1) Quantum gl(m|1)

EXAMPLES FOR m = 2

Let A = 2 −1 −1

• ,

B = 1 1

• ,

both GCMs for gl(2|1). U−(A) has generators F1, F2 subject to the relation F2

1F2 + F2F2 1 = (q + q−1)F1F2F1;

F2

2 = 0

So U−(A) has basis of form Fx

2Fa 1Fy 1 (where a ∈ Z≥0 and x, y ≤ 1)

U−(B) has generators F1, F2 subject to the relations F2

1 = F2 2 = 0.

So U−(B) has basis e.g. F1F2F1F2F1F2. Different GCMs typically yield non-isomorphic half-quantum groups!

Sean Clark Uq(gl(m|1)) and canonical bases June 1, 2016 7 / 23

QUANTUM ENVELOPING gl(m|1) Quantum gl(m|1)

THE GOAL

Goal: Construct (non-signed!) canonical basis for U− = U−

q (Πstd)

Strategy:

• 1. Construct crystal on U− using Benkart-Kang-Kashiwara;
• 2. Globalize using pseudo-canonical basis

⇒ get canonical B satisfying (CB1), (CB2), and (CB4) in many cases

• 3. Prove (CB3) using a braid group action.

Sean Clark Uq(gl(m|1)) and canonical bases June 1, 2016 8 / 23

Results Crystals

U-MODULES AND CRYSTALS

As usual, can consider finite-dimensional weight representations. λ ∈ P+: weights in P which are gl(m)-dominant K(λ): induced module from Uq(gl(m))-rep Vgl(m)(λ) V(λ): simple quotient Crystal basis is a pair (L, B) where

◮ L is a lattice over A ⊂ Q(q) (no poles at 0) ◮ B is a basis of L/qL ↔ nodes on a colored digraph ◮ Ei, Fi operators ˜

ei, ˜ fi on L which, mod q, move along the arrows

Theorem (Benkart-Kang-Kashiwara, Kwon)

Let λ ∈ P+. Then K(λ) has a crystal basis (LK(λ), BK(λ)). If additionally λ is a polynomial, V(λ) admits a crystal basis (L(λ), B(λ)), combinatorially realized by super semistandard Young tableaux.

Sean Clark Uq(gl(m|1)) and canonical bases June 1, 2016 9 / 23

Results Crystals

SUPER SEMISTANDARD TABLEAUX FOR gl(2|1)

Super alphabet: S = {1, 2} ∪

• 3
• Semistandard tableaux in S means a Young diagram colored by S such that:

◮ even (odd) letters are (strictly) increasing along rows ◮ odd (even) letters are (strictly) increasing along columns

“Read” tableaux from right-to-left and top-to-bottom to get element of V⊗t

std

The ˜ fi,˜ ei act via the “tensor product rule” 1 2 3 Vector representation: α1 α2 1 1 2 3 3 ⇒ 2 ⊗ 1 ⊗ 1 ⊗ 3 ⊗ 3 Simple rule for odd root: Always apply ˜ f2 to first 2 or 3 encountered.

Sean Clark Uq(gl(m|1)) and canonical bases June 1, 2016 10 / 23

Results Crystals

CRYSTAL FOR U−

Theorem (C)

U− = U−(Πstd) admits a crystal basis B which is compatible with those on modules. Moreover, these crystals “globalize” to a basis satisfying (CB1), (CB2), and (CB4) for

◮ the half-quantum enveloping algebra U−; ◮ the Kac modules K(λ) for any λ ∈ P+; ◮ the simple modules V(λ) for polynomial λ ∈ P+.

• Crystal for U−

q (gl(2|1))

• Crystal for V(3ǫ1 + ǫ2)

corresponds to corresponds to

1 1 1 2 2 2 3 3

Sean Clark Uq(gl(m|1)) and canonical bases June 1, 2016 11 / 23

Results Crystals

IDEAS IN PROOF

◮ Simplify [BKK] results: for the m ≥ n = 1 case,

◮ no upper crystal part (↔ gl(n) part); ◮ no fake highest weights: ˜

eix = 0 for all i implies x = vλ

◮ not signed basis (odd operators never pass odd-colored boxes).

◮ Construct crystal inductively using (truncated) “grand loop” and [BKK]. ◮ Characterize lattice and (signed) basis with bilinear form. ◮ Construct integral form of lattice in usual way. ◮ Existence of globalizations follows from pseudo-canonical basis

Sean Clark Uq(gl(m|1)) and canonical bases June 1, 2016 12 / 23

Results Braid Group Action

ODD REFLECTIONS

To relate CB and PBW, want a braid group action (following Lusztig, Saito, Tingley). Non-super: automorphism Ti : Uq(g) → Uq(g) “lifting si action on weights” TiUν ⊂ Usi(ν) We want an analogue for super, but there is a Problem: Odd isotropic simple root αi ⇒ no odd reflection si ∈ W! (There is a formal odd “reflection” in a Weyl groupoid.) Consequence: No odd braid automorphism, but what about isomorphisms?

Sean Clark Uq(gl(m|1)) and canonical bases June 1, 2016 13 / 23

Results Braid Group Action

PERSPECTIVE ON BRAIDS: NON-SUPER

Usual definition of braid action: automorphism Ti : Uq(g) → Uq(g) e.g. for Uq(sl(3)) T2(F1) = F2F1 − qF1F2 ∈ U−

q (sl(3))s2·α1

Interpretation: Ti is weight-preserving translation between choices of simples Conjugacy of Borels ↔ presentation is “unique”; e.g.      Fǫ1−ǫ2, Fǫ2−ǫ3, . . . Π = {ǫ1 − ǫ2, ǫ2 − ǫ3} Fǫ1−ǫ3, Fǫ3−ǫ2, . . . Π = {ǫ1 − ǫ3, ǫ3 − ǫ2} F1, F2, . . . Π = {α1, α2} all essentially the same Braid operator is automorphism sending e.g. T2(F1) = F2F1 − qF1F2 ⇒ T2(Fǫ1−ǫ2) = Fǫ3−ǫ2Fǫ1−ǫ3 − qFǫ1−ǫ3Fǫ3−ǫ2

Sean Clark Uq(gl(m|1)) and canonical bases June 1, 2016 14 / 23

Results Braid Group Action

PERSPECTIVE ON BRAIDS: SUPER

Borels are non-conjugate in general → can’t just identify different presentations! Example: for standard Uq(gl(2|1)), F2

1 = 0 yet

“T2(F1)”2 = (F2F1 − qF1F2)2 = 0. Solution: really, T2 : U(s2 · Πstd) → U(Πstd) with T2(Fǫ1−ǫ3) = Fǫ2−ǫ3

F2

Fǫ1−ǫ2

F1

−qFǫ1−ǫ2Fǫ2−ǫ3 where Fǫ1−ǫ3 = F1 ∈ U(s2 · Πstd); note F2

ǫ1−ǫ3 = 0.

Sean Clark Uq(gl(m|1)) and canonical bases June 1, 2016 15 / 23

Results Braid Group Action

BRAID ISOMORPHISMS

Theorem (C)

If Π and Π′ are related via the “reflection” si, then there is an isomorphism Ti : Uq(Π) → Uq(Π′). These maps satisfy braid relations:

Uq(Π) Uq(Πi) Uq(Πj) Uq(Πij)

• Ti

Tj Tj Ti

Uq(Π) Uq(Πi) Uq(Πj) Uq(Πij) Uq(Πji) Uq(Π′)

• Ti

Tj Tj Ti Ti Tj Sean Clark Uq(gl(m|1)) and canonical bases June 1, 2016 16 / 23

Results Braid Group Action

BRAID DEFINITION

Identify generators from each copy by their GCMs X and Y: Ti(EX,j) =      −(−1)pY(i)K−e

Y,iFY,i

if j = i; EY,jEY,i − (−1)pY(i)pY(j)qeYijEY,iEY,j if j = i ± 1; EY,j

• therwise;

Ti(FX,j) =      −(−1)pY(i)EY,iKe

Y,i

if j = i; FY,iFY,j − (−1)pY(i)pY(j)q−eYijFY,jFY,i if j = i ± 1; FY,j

• therwise;

Ti(KX,j) =      (−1)pY(i)K−1

Y,i

if j = i; (−1)pY(i)pY(j)KY,iKY,j if j = i ± 1; KY,j

• therwise;

Sean Clark Uq(gl(m|1)) and canonical bases June 1, 2016 17 / 23

Results Braid Group Action

PBW

w0 = si1 . . . sit: reduced expression for the longest element of Sm+1 → root vectors and a PBW basis: Fβ = Ti1 . . . Tir−1(Fir) Given w0 = sj1 . . . sjt, want same Z[q] lattice of PBWs:

◮ Reduce to braid relations, hence rank 2 computations; ◮ New phenomena: three rank 2 cases:

◮

2 −1 −1 2

• : sl(3)-case, which is known.

◮

2 −1 −1

• : standard gl(2|1)-case, which is easy.

◮

1 1

• : non-standard gl(2|1)-case, which is false, but doesn’t occur.

Theorem (C)

B ≡q PBW(i1,...,it), so (CB3) holds.

Sean Clark Uq(gl(m|1)) and canonical bases June 1, 2016 18 / 23

Results Braid Group Action

NON-STANDARD RANK 2

The non-standard GCM for gl(2|1) is

• 1

1

• .

The PBW bases are given by Fx

1(F2F1 + q−1F1F2)(y)Fz 2,

Fx

2(F1F2 + q−1F2F1)(y)Fz 1.

Clearly not the same Z[q]-span; in fact, also not the same Z[q−1]-span! (Nevertheless, there is a sort of CB: Fx

1(F2F1)yFz 2.)

However, this case doesn’t bother the theorem: e.g. w0 = s3s2s1s3s2s3 = s3s2s1s2s3s2

2

−1 −1 2 −1 −1

• 2

−1 0 −1 1 1

• 1

1 −1 0 −1 2

• −1

−1 2 −1 −1 2

• s3

s2 s1

s1, s2 s1 s3 s2, s3

Way to see this in general: “weight-preserving” ⇒ Fβ1+β2 = Fβ2Fβ1 − qFβ1Fβ2

Sean Clark Uq(gl(m|1)) and canonical bases June 1, 2016 19 / 23

Results Braid Group Action

EXAMPLE: CANONICAL BASIS OF gl(3|1)

Strategy for constructing B from canonical basis for gl(m).:

• 1. Inductively build B for weights with larger αm multiplicity
• 2. Multiplicity ≤ 2m, so done in finite number of steps.

Theorem (C)

For x, y, z ∈ {0, 1} and a, b, c ∈ Z≥0, let u = u(x, y, z, a, b, c) ∈ B be the unique element equal to the PBW vector Fx

3Fy 32Fz 321F(a) 2 F(b) 21 F(c) 1

modulo q. Then u =                      Fx

3 F(a+y) 2

F(b+c+z)

1

Fy

3 F(b+z) 2

Fz

3

if c ≥ a, Fx

3Fy 2F(b+z) 1

Fy

3F(a+b+z) 2

Fz

3F(c) 1

if a > c and y ≤ x,

b

• t=0

(−1)t

• a − c − 1 + t

t

• F(a+1+t)

2

F(b+c+z)

1

F3F(b+z−t)

2

Fz

3

• therwise.

Sean Clark Uq(gl(m|1)) and canonical bases June 1, 2016 20 / 23

Results Braid Group Action

TANTALIZING HINTS OF POSITIVITY

For standard gl(2|1), easy CB: Fx

2F(a) 1 Fy 2 for x, y ∈ {0, 1}, a ∈ Z≥0.

Can directly check lots of positivity, e.g: Fx

2F(a) 1 Fy 2Fw 2 F(b) 1 Fz 2 =

               if y = w = 1;

• a + b

a

• Fx

2F(a+b) 1

Fz

2

if y = w = 0; δz,0

• a + b − 1

b

• Fx

2F(a+b) 1

F2 + δx,0

• a + b − 1

a

• F2F(a+b)

1

Fz

2

• .w.

For gl(3|1), harder to check but no counterexamples so far!

Sean Clark Uq(gl(m|1)) and canonical bases June 1, 2016 21 / 23

Results Braid Group Action

FURTHER QUESTIONS

◮ Categorification

◮ CB interpretation in terms of Khovanov-Sussan? ◮ Categorified modules? ◮ What about non-standard half-quantum enveloping algebras?

◮ CB for other Lie superalgebras

◮ For algebra, plausible for “simply laced with isolated isotropics” ◮ Ex: can get signed CB for hqg associated to the Dynkin

X

iso

. . . X

iso

◮ For modules, unclear: for example, crystals=disconnected in general!

◮ CB for standard gl(2|2) and beyond?

◮ “chirality” is key difficulty; Uq(gl(m|n))0 ∼

= Uq(gl(m)) ⊗ Uq−1(gl(n))

◮ seems (CB3) is not correct condition! Sean Clark Uq(gl(m|1)) and canonical bases June 1, 2016 22 / 23

Results Braid Group Action

THANKS FOR YOUR ATTENTION!

Some references:

◮ S. C., Canonical bases for the quantum enveloping algebra of gl(m|1) and its modules,

arXiv:1605.04266

◮ S. C., D. Hill and W. Wang, Quantum shuffles and quantum supergroups of basic type,

• Quant. Top. (to appear), arXiv:1310.7523.

◮ J. Du and H. Gu, Canonical bases for the quantum supergroups U(glm|n), Math. Zeit.

281, 631-660

◮ G. Benkart, S.-J. Kang and M. Kashiwara, Crystal bases for the quantum superalgebra

Uq(gl(m, n)), Journal of Amer. Math. Soc. 13 (2000), 295–331.

◮ M. Khovanov, How to categorify one-half of quantum gl(1|2), arXiv:1007.3517. ◮ M. Khovanov and J. Sussan, A categorification of the positive half of quantum gl(m|1),

• Trans. AMS (to appear), arXiv:1406.1676.

◮ J.-H. Kwon, Crystal bases of q-deformed Kac modules over the quantum superalgebra

Uq(gl(m|n)), Int. Math. Res. Notices 2 (2014), 512–550.

Sean Clark Uq(gl(m|1)) and canonical bases June 1, 2016 23 / 23

Extra Slides

WHY NOT CANONICAL?

The standard GCM for gl(2|2) is   2 −1 −1 1 1 −2  . Let U− be the half-quantum group associated to this GCM. Dpending on the order, the height 2 root vectors are either: F12 = F2F1 − qF1F2 (if α1 < α2); F21 = F1F2 − qF2F1 (if α2 < α1), F23 = F3F2 − q−1F2F3 (if α2 < α3); F32 = F2F3 − q−1F3F2 (if α3 < α2). With respect to Z[q], bar-invariant basis elements are Bα2<α3 =

• F2F3, F3F2 − (q + q−1)F2F3 = F23 − qF2F3
• Bα3<α2 =
• F3F2, F2F3 − (q + q−1)F3F2 = F32 − qF3F2
• (If we take Z[q−1] lattice, similar problem for α1 + α2)

Sean Clark Uq(gl(m|1)) and canonical bases June 1, 2016 1 / 3

Extra Slides

THE CURIOUS CASE OF “gl(1|1|1)”

U−

q ({ǫ1 − ǫ3, ǫ3 − ǫ2}) ∼

= Q(q) F1, F2 /(F2

1, F2 2) AU− q = Z[q, q−1]Fx i (FiFj)yFz j where x, z ∈ {0, 1}, y ∈ Z≥0

Satisfies (CB1), (CB2), and (CB3) for unnormalized PBW Fk(FℓFk + qFkFℓ)yFz

ℓ

Moreover, has some compatibility with finite-dimensional simple modules: V(aǫ1 + bǫ2 + cǫ3) is spanned by the non-zero Fx

i (FiFj)yFz j vλ, with the linear

dependence [a + c](F1F2)a−b = [−b − c](F2F1)a−b

Sean Clark Uq(gl(m|1)) and canonical bases June 1, 2016 2 / 3

Extra Slides

THE CONJECTURAL gl(1|2|1) CRYSTAL

H ⋆ (⋆ repeats) (⋆ repeats) (⋆ repeats) (⋆ repeats)

Sean Clark Uq(gl(m|1)) and canonical bases June 1, 2016 3 / 3