Algebraic Braids and Geometric Representation...

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Algebraic Braids and Geometric Representation Theory

Minh-Tâm Quang Trinh

University of Chicago

Plane Curves

The curve C , below, is singular at (0, 0).

y2 = x5.

To picture its solutions over R, graph it. What about over C?

Imagine x = reiθ ∈ C, where r > 0 is very small.

θ runs from 0 to 2π x runs around a circle x5 runs around five times as fast roots y = ±x 5

2 swap five times

∴ The preimage of a small circle around the origin of C is a braidedcircle (on 2 strands) around the singularity of C .

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To the left is a picture of C ⊆ C4, projected into R3. To the right isthe braided circle, projected onto the complex x-axis.

Images from: https://www.math.purdue.edu/~arapura/graph/nodal.html

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Above any small circle in the complex x-axis, a plane curve

yn + a1(x)yn−1 + · · ·+ an−1(x)y + an(x) = 0

gives rise to the annular closure of a braid on n strands.

Equivalently, an element of the group Brn up to conjugation.

Q How is the geometry of the curve reflected by the topology of thebraid closure—equivalently, the algebra of the conjugacy class?

A (Newton–Puiseux) Not all conjugacy classes in Brn arise this way.To arise from a plane algebraic curve is a strong constraint.

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Consider an abstract singularity (C , 0).

For each ` ≥ 0, there is a variety Hilb`(C , 0) whose points classify`-dimensional quotients of the ring of functions OC,0.

The link of (C , 0) is the embedded 1-manifold

λC,0 = S3 ∩ C ↪→ S3,

where S3 is any small 3-sphere centered at 0. Links that arise this wayare called algebraic links.

Conj (Oblomkov–Shende) The series∑`≥0 q`χ(Hilb`(C , 0)) only

depends on λC,0, or more precisely, its homfly polynomial.∗

Thm (Maulik) The Oblomkov–Shende conjecture is true.∗ Above, χ denotes Euler characteristic.

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Ex Let (C , 0) be the singularity of y4 − 2x3y2 − 4x5y + x6 − x7 = 0at the origin. Puiseux expansion shows

OC,0 ' C[[t4, t6 + t7]].

The coefficients of homfly(λC,0) ∈ Z((q))[a±1]:

q−8q−7q−6q−5q−4q−3q−2q−1q0 q1 q2 q3 q4 q5 q6 q7 q8

a16 1 1 1 2 1 3 1 3 1 3 1 2 1 1 1−a18 1 1 2 2 4 3 5 3 5 3 4 2 2 1 1a20 1 1 2 2 3 2 3 2 2 1 1−a22 1 1 1

The a16 row equals q−8(1− q)∑` q`χ(Hilb`(C , 0)).

Rem In full, Oblomkov–Shende–Maulik incorporate a as well as q.

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Maulik’s proof is inductive. Using blowups, he reduces to the caseyn = xm, which Oblomkov–Shende checked combinatorially. Still:

Q What is homfly and why should it appear at all?

A When C takes the form

yn + a1(x)yn−1 + · · ·+ an−1(x)y + an(x) = 0,

both Hilb`(C , 0) and homfly(λC ,0) can be related to the geometricrepresentation theory of GLn. More precisely, its Springer theory.

This viewpoint leads to a new conjecture that is stronger and alsogeneralizes to any reductive algebraic group G.

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Springer Fibers

Let G be a complex reductive group with Lie algebra g.

Let B be the flag variety and let W be the Weyl group.

Each element x ∈ g induces a vector field on B, with fixed-point locus

Bx = {B ∈ B : x ∈ Lie(B)}.

For x generic, we have a simply-transitive action W y Bx .

For x nilpotent, W 6y Bx . Yet Springer showed W y H∗(Bx). Let

Sprx(q) =∑i≥0

qiH2i(Bx) ∈ K0(W )[q],

where K0(W ) is the representation ring.

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Let G = GL4, so that W = S4. Irr(S4) = {1, φ, χ, εφ, ε}.

A certain Z[q]-linear combination of the Sprx(q):

q0 q1 q2 q3 q4 q5 q6 q7 q8 q9 q10 q11 q12 q13 q14 q15 q16

1 1 1 1 2 1 3 1 3 1 3 1 2 1 1 1φ 1 1 2 2 4 3 5 3 5 3 4 2 2 1 1χ 1 2 1 3 1 4 1 3 1 2 1εφ 1 1 2 2 3 2 3 2 2 1 1ε 1 1 1

Compare to the homfly polynomial we saw earlier:

q−8q−7q−6q−5q−4q−3q−2q−1q0 q1 q2 q3 q4 q5 q6 q7 q8

a16 1 1 1 2 1 3 1 3 1 3 1 2 1 1 1−a18 1 1 2 2 4 3 5 3 5 3 4 2 2 1 1a20 1 1 2 2 3 2 3 2 2 1 1−a22 1 1 1

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Let [Brn] be the set of conjugacy classes of Brn. We define a function

ann : [Brn]→ K0(Sn)[[q]].

Thm-Def 1 If β is a positive braid, then

ann(β) = 1|G(Fq)|

∑nilpotent x

|B(β)x(Fq)|Sprx(q),

where B(β)x is a certain variety over Fq generalizing Bx .

Thm 2 If a link λ ⊆ R3 is the closure of a braid β ∈ Brn, then

(q 12 a−1)|β|−n+1homfly(λ) =

∑0≤i≤n−1

(−a2)i(Λi(φ),ann(β))Sn ,

where |β| is the writhe and φ is the standard irrep Sn y Cn−1.

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Sketch of Thm 2 Let R be the vector space of functions B(Fq)→ C.The Hecke algebra of GLn is

Hn = EndGLn(Fq)(R).

Each transposition si = (i, i + 1) ∈ Sn defines a braid βi ∈ Brn and anoperator Ti ∈ Hn. There is a map Brn → H×n that sends βi 7→ Ti .

(Jones–Ocneanu) We can construct homfly via trace functions

trn : Hn → C[q±1](a).

Such traces are C[q±1](a)-linear combinations of irreps of Hn, closelyrelated to irreps of GLn(Fq).

(Kazhdan) We can relate Deligne–Lusztig reps of G(Fq) to Springerreps of its Weyl group W .

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Artin Braids

The function ann can be generalized to any other G.

Let t ⊆ g be a Cartan. As a reflection group, W y t.

Let t◦ ⊆ t be the complement of the hyperplanes of reflection. TheArtin braid group is

BrW = π1(t◦/W ).

In this generality, there is a function ann : [BrW ]→ K0(W )[[q]].

Ex If G = GLn, then W = Sn and t = Cn.

∴ t◦/W = Confn(C), the n-point configuration space of C.

∴ BrW = Brn.

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Let η = SpecC((x)), an infinitesimal loop.

A plane curve f (x, y) = yn +∑

i ai(x)yn−i = 0 gives rise to a map

η → Confn(C),

sending infinitesimal x to the n-tuple of roots y such that f (x, y) = 0.On profinite π1’s, we have:

Z → [Brn]1 7→ [β]

[β] is roughly the braid conjugacy class from the beginning of the talk.

In general, a : η → t◦/W induces a conjugacy class [βa] ∈ [BrW ].

Artin braids that arise this way, we call algebraic braids.

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Affine Springer Fibers

Maps η → t◦/W appear in an “affine” analogue of Springer theory.

The loop group LG is the infinite-dimensional group of maps η → G.The analogue of the flag variety B is a symmetric space

LG y Baff .

Any map a : η → t◦/W lifts to a map η → g.

∴ Vector field on Baff with finite-dimensional fixed-point locus

Baffa ⊆ Baff .

The affine Weyl group W n X∗ acts on H∗(Baffa ).

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To recap, a : η → t◦/W defines:

• A conjugacy class [βa] ∈ [BrW ]. ≈ [βa] ∈ [BrW ]• A scheme Baffa .

Q How is the geometry of Baffa reflected by the topology of a, orequivalently, the algebra of [βa]?

Thm 3 dimBaffa only depends on [βa], not on a.

Sketch of Thm 3 Bezrukavnikov showed

dimBaffa = ordx Da − dim t + dim twa

for some divisor Da → SpecC[[x]] and element wa ∈W .

Show that ordx Da = |βa| and that BrW �W sends [βa] 7→ [wa].

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As for W y H∗(Baffa ):

Conj If a is elliptic, then Hi(Baffa ) admits a W -equivariant filtrationP≤∗ such that

Spraffa (q) =∑i,j≥0

(−1)iqj grPj Hi(Baffa ) ∈ K0(W )[q]

contains ann(βa) as a large summand.

Rem When G is semisimple, ellipticity means twa = 0.

Rem P≤∗ should arise via Ngô’s homeomorphism relating affineSpringer fibers with Hitchin fibers—that is, from the perverse filtrationof the direct image complex of a Hitchin fibration.

Rem There’s a bivariate version in variables q and t.

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For G = SLn, a map a : η → t◦/W defines:

• A plane curve germ Ca → SpecC[[x]].• A conjugacy class [βa] ⊆ [Brn].• A scheme Baffa such that Sn y H∗(Baffa ).

Here, a is elliptic ⇐⇒ Ca is unibranch.

Thm (Maulik–Yun) The filtration P≤∗ is canonical. For elliptic a,∑`≥0

q`χ(Hilb`(Ca, 0)) = 11− q · (1,Spr

affa (q))Sn .

So our conjecture recovers Oblomkov–Shende–Maulik.

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Periodic Braids

The simplest case of Oblomkov–Shende is yn = xm. Here, the braidcloses up into the (m,n)-torus link.

For m = n, the braid is called the full twist. It is central in Brn.

For general W , there is an analogous element

π ∈ BrW .

Def β ∈ BrW is periodic of slope mn iff βn = πm.

It is a fractional twist iff β = γm and π = γn for some γ ∈ BrW .

Conj Periodic ⇐⇒ fractional twist. (True for W = Sn.)

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It turns out fractional twists are related to rational Cherednik algebras.For all c ∈ C, this is an algebra

Aratc ⊇ C[W ].

It is a deformation of Arat0 ' C[W ] nD(t).

Its rep theory closely resembles that of a semisimple Lie algebra.

Each irrep φ ∈ Irr(W ) gives rise to Aratc -modules

Verma ∆c(φ) � simple Lc(φ).

Each Aratc -module M has a graded character [M ]q ∈ K0(W )(q 1

2 ).

Rep theorists try to express the [Lc(φ)]q in terms of the [∆c(φ)]q.

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Thm 4 If β ∈ BrW is a fractional twist of slope mn , then

(q 12 )r−|β|ann(β) =

∑φ∈Irr(W )

Dφ(e2πim/n)[∆m/n(φ)]q

for some Dφ(q) ∈ Q[q]. Contains [Lm/n(1)]q with multiplicity one.

If n � 0 in lowest terms, then (q 12 )r−|β|ann(β) = [Lm/n(1)]q.

Rem Recovers theorem of Varagnolo–Vasserot–Etingof on the valuesof c such that Lc(1) is finite-dimensional.

Rem The Dφ(q) are the so-called generic degrees of the unipotentprincipal series of G(Fq).

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Homogeneity

C× y t◦ induces C× y t◦/W .

a : η → t◦/W is homogeneous of slope mn iff

a(tnx) = tm · a(x)

for all t ∈ C×. In this case, C× y Baffa .

Lem If a is homogeneous of slope mn , then βa ∈ BrW is a fractional

twist of slope mn .

∴ Fractional twists are algebraic braids.

Conj (T–Bapat–Deopurkar–Licata) Nielsen–Thurston classification inBrW , such that algebraic braids are always periodic or reducible.

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Assume G is almost-simple and simply-connected.

For homogeneous elliptic a of slope mn , Oblomkov–Yun exhibit

• A finite group Sa y Baffa .• An action Arat

m/n y grP∗ H∗ε=1(Baffa )Sa ,

where H∗ε=1 is a specialization of C×-equivariant cohomology.

Construction uses comparison to a Hitchin fiber for P(n, 1).

Thm (Oblomkov–Yun) Lm/n(1) ↪→ grP∗ H∗ε=1(Baffa )Sa .

For n the Coxeter number of W , this inclusion is an isomorphism.

Cor (T) Infinitely many cases where Spraffa (q)Sa = ann(βa).

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Ex Take G = SL2, so that W = S2 and Irr(S2) = {1, ε}.

Nilpotent orbits in sl2: Orbits of 0 = ( 0 00 0 ) and e = ( 0 1

0 0 ).

Have a such that Ca = {y2 = x5} and Baffa ' P2 tP1 (P1 × P1).

Spraffa (q) =∑

i,j≥0 (−1)iqj grPj Hi(Baffa ) = q2[L5/2(1)]q

= (1 + q2 + q4) + (q + q3)ε

ann(βa) = |B(βa)0(Fq)||G(Fq)| Spr0(q) + (q2 − 1) |B(βa)e(Fq)|

|G(Fq)| Spre(q)

= (1 + q2) · (1 + qε) + q4 · 1

Above, B(βa)x = {z ∈ (P1)5 : x · z5 6= z1 6= z2 6= · · · 6= z5}.

P = W in the sense of nonabelian Hodge theory?

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Special thanks to Reid Harris and Rachel McEnroe for their feedback onan earlier draft of this talk.

My greatest thanks goes to my mentors, Professor Bao-Châu Ngô andProfessor Victor Ginzburg, for their guidance and support, and to my

parents, for their enormous sacrifices on my behalf.

Thank you for listening.

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