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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, . . .
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: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)
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: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)
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 =
∑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}
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 = 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!
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 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!
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)-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.
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: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.
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
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
Sean Clark Uq(gl(m|1)) and canonical bases June 1, 2016 11 / 23
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
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.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
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)), 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.
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)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;
Sean Clark Uq(gl(m|1)) and canonical bases June 1, 2016 17 / 23
Results Braid Group Action
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.
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[
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
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 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.
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: 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!
Sean Clark Uq(gl(m|1)) and canonical bases June 1, 2016 21 / 23
Results Braid Group Action
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!
Sean Clark Uq(gl(m|1)) and canonical bases June 1, 2016 22 / 23
Results Braid Group Action
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.
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 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)
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 ({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
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