Uq gl mj1)) and canonical bases - Northeastern University · U q(gl(mj1)) and canonical bases Sean...

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

Sean ClarkNortheastern 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 7→ 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, . . .



QUANTUM SUPERALGEBRAS

Question: what if g is a Lie superalgebra?

E.g. gl(m|n): linear maps on vector superspace Cm|n.

Super complications:I Different simple roots= Different U− (in general)

Workaround: Work with a standard choice, e.g.α1 = ε1 − ε2, α2 = ε2 − ε3, . . ..

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

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

vec

I “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)



SOME KNOWN RESULTS

Few examples of CBs known:I easy small rank examples on U−; e.g. gl(1|1), standard gl(2|1), osp(1|2)

I osp(1|2n) and anisotropic Kac-Moody super [C-Hill-Wang, ’13]I Partial results for gl(m|n), osp(2|2n) [C-Hill-Wang, ’13; Du-Gu ’14]

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

I Only considers standard simple roots case;

I In standard case, PBW basis→ B = (CB1), (CB2)

I If m > 1 and n > 1, this basis is not canonical! (Depends on PBW)

I When m = 1 or n = 1, get pseudo-canonical basis (a signed basis)


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

ROOT DATA FOR gl(m|1)

P =

∑mi=1 Z εi︸︷︷︸

even

⊕ Z εm+1︸︷︷︸odd

with (εi, εi) = (−1)p(εi), P∨, 〈·, ·〉 as usual.

Φ ={εi − εj | 1 ≤ i 6= j ≤ m + 1

}with simple roots Π = {αi | i ∈ I}

Standard choice: Πstd = {ε1 − ε2, . . . , εm − εm+1}

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

GCMs for m = 3:

2 −1 0−1 2 −10 −1 0

︸︷︷︸

standard choice

,

2 −1 0−1 0 10 1 0

︸︷︷︸

Π={ε1−ε2,ε2−ε4,ε4−ε3}



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 = q〈h,αi〉Ei, [Ei,Fj]︸︷︷︸super commutator

= EiFj − (−1)p(i)p(j)FjEi = δijqhi − q−hi

q− q−1

and both usual and unusual Serre relations: if p(i) = 1,

E2i = 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!



EXAMPLES FOR m = 2

Let A =

[2 −1−1 0

], B =

[0 11 0

], both GCMs for gl(2|1).

U−(A) has generators F1,F2 subject to the relation

F21F2 + F2F2

1 = (q + q−1)F1F2F1; F22 = 0

So U−(A) has basis of form Fx2Fa

1Fy1 (where a ∈ Z≥0 and x, y ≤ 1)

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

F21 = F2

2 = 0.

So U−(B) has basis e.g. F1F2F1F2F1F2.

Different GCMs typically yield non-isomorphic half-quantum groups!



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.


Results Crystals

U-MODULES AND CRYSTALS

As usual, can consider finite-dimensional weight representations.

λ ∈ P+: weights in P which are gl(m)-dominantK(λ): induced module from Uq(gl(m))-rep Vgl(m)(λ)V(λ): simple quotient

Crystal basis is a pair (L,B) whereI L is a lattice over A ⊂ Q(q) (no poles at 0)I B is a basis of L/qL↔ nodes on a colored digraphI 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 apolynomial, V(λ) admits a crystal basis (L(λ),B(λ)), combinatorially realized bysuper semistandard Young tableaux.


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:I even (odd) letters are (strictly) increasing along rowsI odd (even) letters are (strictly) increasing along columns

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

The fi, ei act via the “tensor product rule”

1 2 3

Vector representation:α1 α2

1 1 233

⇒ 2 ⊗ 1 ⊗ 1 ⊗ 3 ⊗ 3

Simple rule for odd root: Always apply f2 to first 2 or 3 encountered.


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

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

•• •• • •• • • •• • • •

Crystal for U−q (gl(2|1))

•• •• • •• • •• ••

Crystal for V(3ε1 + ε2)

corresponds to

corresponds to

1 1 12

2 2 33


Results Crystals

IDEAS IN PROOF

I Simplify [BKK] results: for the m ≥ n = 1 case,I no upper crystal part (↔ gl(n) part);I no fake highest weights: eix = 0 for all i implies x = vλ

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

I Construct crystal inductively using (truncated) “grand loop” and [BKK].

I Characterize lattice and (signed) basis with bilinear form.

I Construct integral form of lattice in usual way.

I Existence of globalizations follows from pseudo-canonical basis


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?



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.Fε1−ε2 ,Fε2−ε3 , . . . Π = {ε1 − ε2, ε2 − ε3}Fε1−ε3 ,Fε3−ε2 , . . . Π = {ε1 − ε3, ε3 − ε2}F1,F2, . . . Π = {α1, α2}

all essentially the same

Braid operator is automorphism sending e.g.

T2(F1) = F2F1 − qF1F2 ⇒ T2(Fε1−ε2) = Fε3−ε2 Fε1−ε3 − qFε1−ε3 Fε3−ε2



PERSPECTIVE ON BRAIDS: SUPER

Borels are non-conjugate in general→ can’t just identify different presentations!

Example: for standard Uq(gl(2|1)), F21 6= 0 yet

“T2(F1)”2 = (F2F1 − qF1F2)2 = 0.

Solution: really, T2 : U(s2 ·Πstd)→ U(Πstd) with

T2(Fε1−ε3) = Fε2−ε3︸︷︷︸F2

Fε1−ε2︸︷︷︸F1

−qFε1−ε2 Fε2−ε3

where Fε1−ε3 = F1 ∈ U(s2 ·Πstd); note F2ε1−ε3

= 0.



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



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)qeYij EY,iEY,j if j = i± 1;

EY,j otherwise;

Ti(FX,j) =

−(−1)pY(i)EY,iKe

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

FY,j otherwise;

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 otherwise;



PBWw0 = 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:I Reduce to braid relations, hence rank 2 computations;I New phenomena: three rank 2 cases:

I

[2 −1−1 2

]: sl(3)-case, which is known.

I

[2 −1−1 0

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

I

[0 11 0

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

Theorem (C)B ≡q PBW(i1,...,it), so (CB3) holds.



NON-STANDARD RANK 2

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

0 11 0

].

The PBW bases are given by

Fx1(F2F1 + q−1F1F2)(y)Fz

2, Fx2(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)yFz2.)

However, this case doesn’t bother the theorem: e.g.

w0 = s3s2s1s3s2s3 = s3s2s1s2s3s2

[ 2 −1 0−1 2 −10 −1 0

] [ 2 −1 0−1 0 10 1 0

] [ 0 1 01 0 −10 −1 2

] [ 0 −1 0−1 2 −10 −1 2

]s3 s2 s1

s1, s2 s1 s3 s2, s3

Way to see this in general: “weight-preserving”⇒ Fβ1+β2 = Fβ2 Fβ1 − qFβ1 Fβ2



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 multiplicity2. 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 uniqueelement equal to the PBW vector Fx

3Fy32Fz

321F(a)2 F(b)

21 F(c)1 modulo q. Then

u =

Fx3 F(a+y)

2 F(b+c+z)1 Fy

3 F(b+z)2 Fz

3 if c ≥ a,

Fx3Fy

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 Fz3 otherwise.



TANTALIZING HINTS OF POSITIVITY

For standard gl(2|1), easy CB: Fx2F(a)

1 Fy2 for x, y ∈ {0, 1}, a ∈ Z≥0.

Can directly check lots of positivity, e.g:

Fx2F(a)

1 Fy2Fw

2 F(b)1 Fz

2 =

0 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 Fz2 o.w.

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



FURTHER QUESTIONS

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

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

Xiso

. . . Xiso

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

I CB for standard gl(2|2) and beyond?I “chirality” is key difficulty; Uq(gl(m|n))0 ∼= Uq(gl(m))⊗Uq−1 (gl(n))I seems (CB3) is not correct condition!



THANKS FOR YOUR ATTENTION!

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

arXiv:1605.04266I S. C., D. Hill and W. Wang, Quantum shuffles and quantum supergroups of basic type,

Quant. Top. (to appear), arXiv:1310.7523.I J. Du and H. Gu, Canonical bases for the quantum supergroups U(glm|n), Math. Zeit.

281, 631-660I 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.I M. Khovanov, How to categorify one-half of quantum gl(1|2), arXiv:1007.3517.I M. Khovanov and J. Sussan, A categorification of the positive half of quantum gl(m|1),

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


Extra Slides

WHY NOT CANONICAL?

The standard GCM for gl(2|2) is

2 −1 0−1 0 10 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)


Extra Slides

THE CURIOUS CASE OF “gl(1|1|1)”

U−q ({ε1 − ε3, ε3 − ε2}) ∼= Q(q) 〈F1,F2〉 /(F21,F

22)

AU−q =⊕

Z[q, q−1]Fxi (FiFj)

yFzj where x, z ∈ {0, 1}, y ∈ Z≥0

Satisfies (CB1), (CB2), and (CB3) for unnormalized PBW Fk(F`Fk + qFkF`)yFz`

Moreover, has some compatibility with finite-dimensional simple modules:V(aε1 + bε2 + cε3) is spanned by the non-zero Fx

i (FiFj)yFz

j vλ, with the lineardependence

[a + c](F1F2)a−b = [−b− c](F2F1)a−b


Extra Slides

THE CONJECTURAL gl(1|2|1) CRYSTAL

H?

(? repeats)(? repeats)

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