On categories of tilting modules

(1)

On categories of tilting modules

Or: Mind your poles Daniel Tubbenhauer

Joint with Paul Wedrich

October 2020

Daniel Tubbenhauer On categories of tilting modules October 2020 1 / 8

(2)

We want to describeTilt(G) as a category, i.e. find an algebraZsuch that there is an equivalence of additive,K-linear categories

F: Tilt(G)−→^∼⁼ pMod-Z_p,

sending indecomposable tilting modules to indecomposable projectives.

Goal. DescribeTilt(G)by generators and relations.

I “do not care about objects”, but the point are morphisms and their relations.

(3)

Folklore, Lucas∼1878. Letq∈K^∗,qchar(K) =p,a=mp+a₀and b=np+b₀(a₀,b₀zeroth digit of the p-adic expansion). Then

a b

q

= m

n a₀

b₀

q

.

Example/Remark.

K=F^p,q= 1(known as characteristicp),

anda= [ar,...,a0]p,b= [br,...,b0]p (thep-adic expansions), then

a b

=_a

b

q=_a_r

b_r

q..._a₀

b₀

q= _b^a^r

r

... ^a_b⁰

0

. Examples fora= 1331 = 11³andb=a−1.

IfK=C,q= 1, thenqchar(K) = 0,a= [1331]0andb= [1330]0

⇒₁₃₃₁

1330

q= ¹³³¹₁₃₃₀

= 1331 does not vanish.

IfK=C,q= exp(2πi/11), thenqchar(K) = 11,a= [121,0]11andb= [120,10]11

⇒1331 1330

q= 121·0 10

qvanishes of order one.

IfK=F¹¹,q= 3, thenqchar(K) = 5,a= [266,1]5,b= [266,0]5and ^a⁻₅¹ =^b₅ = [2,2,2]11

⇒[1331]_q= 1·1·1·[1]q does not vanish.

IfK=F¹¹,q= 1, thenqchar(K) = 11, a= [1,0,0,0]11andb= [0,10,10,10]11

⇒[1331]_q= 1·0·0·[0]qvanishes of order three. I will stick with characteristicp, but

the quantum group is a “zeroth digit only” version of it; the mixed cases is a mixture of the two.

(4)

a b

q

= m

n a₀

b₀

q

.

Philosophy. Only the vanishing order of_v

w

qmatters for this lecture ;-).

Example/Remark.

a b

=_a

b

q=_a_r

b_r

q..._a₀

b₀

q= _b^a^r

r

... ^a_b⁰

0

⇒₁₃₃₁

1330

q= ¹³³¹₁₃₃₀

⇒1331 1330

q= 121·0 10

(5)

a b

q

= m

n a₀

b₀

q

.

w

Corollary. We understand finite-dimensional modules forSL2=SL2(K=K)

• generically;

• for the quantum group overCatq^2`= 1;

• the quantum group overK,char(K) =p andq^2`= 1 (mixed case);

• in prime characteristicchar(K) =p.

Example/Remark.

a b

=_a

b

q=_a_r

b_r

q..._a₀

b₀

q= _b^a^r

r

... ^a_b⁰

0

⇒₁₃₃₁

1330

q= ¹³³¹₁₃₃₀

⇒1331 1330

q= 121·0 10

(6)

a b

q

= m

n a₀

b₀

q

.

w

• generically;

Example/Remark.

a b

=_a

b

q=_a_r

b_r

q..._a₀

b₀

q= ^a_b^r_r ... ^a_b⁰₀

.

Examples fora= 1331 = 11³andb=a−1.

⇒₁₃₃₁

1330

q= ¹³³¹₁₃₃₀

⇒1331 1330

q= 121·0 10

(7)

a b

q

= m

n a₀

b₀

q

.

w

• generically;

Example/Remark.

a b

=_a

b

q=_a_r

b_r

q..._a₀

b₀

q= ^a_b^r_r ... ^a_b⁰₀

⇒₁₃₃₁

1330

q= ¹³³¹₁₃₃₀

⇒1331 1330

q= 121·0 10

IfK=F11,q= 3, thenqchar(K) = 5,a= [266,1]5,b= [266,0]5and ^a−1₅ =^b₅ = [2,2,2]11

IfK=F¹¹,q= 1, thenqchar(K) = 11,a= [1,0,0,0]11andb= [0,10,10,10]11

⇒[1331]_q= 1·0·0·[0]qvanishes of order three.

I will stick with characteristicp, but

(8)

a b

q

= m

n a₀

b₀

q

.

w

• generically;

Example/Remark.

a b

=_a

b

q=_a_r

b_r

q..._a₀

b₀

q= _b^a^r

r

... ^a_b⁰

0

⇒₁₃₃₁

1330

q= ¹³³¹₁₃₃₀

⇒1331 1330

q= 121·0 10

⇒[1331]_q= 1·0·0·[0]qvanishes of order three.

I will stick with characteristicp, but

the quantum group is a “zeroth digit only” version of it;

the mixed cases is a mixture of the two.

(9)

Weyl ∼1923. TheSL₂Weyl modules ∆(v−1).

∆(1−1)

∆(2−1)

∆(3−1)

∆(4−1)

∆(5−1)

∆(6−1)

∆(7−1)

X0Y0

X1Y0 X0Y1

X2Y0 X1Y1 X0Y2

X3Y0 X2Y1 X1Y2 X0Y3

X4Y0 X3Y1 X2Y2 X1Y3 X0Y4

X5Y0 X4Y1 X3Y2 X2Y3 X1Y4 X0Y5

X6Y0 X5Y1 X4Y2 X3Y3 X2Y4 X1Y5 X0Y6

a bc d

7→matrix who’s columns are expansions of (aX+cY)^v⁻ⁱ(bX+dY)ⁱ⁻¹.

The simples

Example∆(7−1) =KX⁶Y⁰⊕ · · · ⊕KX⁰Y⁶.

(^{a b}_{c d}) acts as

The columns are expansions of (aX+cY)⁷⁻ⁱ(bX+dY)ⁱ⁻¹. Binomials!

Example∆(7−1), characteristic0. No common eigensystem⇒∆(7−1) simple.

Example∆(7−1), characteristic2.

(0,0,0,1,0,0,0) is a common eigenvector, so we found a submodule. When is∆(v−1)simple?

∆(v−1) is simple

⇔

v−1 w−1

6= 0 for allw ≤v

⇔(Lucas’s theorem) v = [ar,0,...,0]p.

General. Weyl ∆(λ) and dual Weyl∇(λ)

are easy a.k.a. standard; are parameterized by dominant integral weights;

are highest weight modules; are defined overZ; have the classical Weyl characters;

form a basis of the Grothendieck group unitriangular w.r.t. simples; satisfy (a version of) Schur’s lemma dimKExti(∆(λ),∇(µ)) =δi,0δλ,µ;

are simple generically;

have a root-binomial-criterion to determine whether they are simple (Jantzen’s thesis∼1973).

(10)

∆(1−1)

∆(2−1)

∆(3−1)

∆(4−1)

∆(5−1)

∆(6−1)

∆(7−1)

X0Y0

X1Y0 X0Y1

X2Y0 X1Y1 X0Y2

X3Y0 X2Y1 X1Y2 X0Y3

a bc d

The simples

Example∆(7−1) =KX⁶Y⁰⊕ · · · ⊕KX⁰Y⁶.

(0,0,0,1,0,0,0) is a common eigenvector, so we found a submodule. When is∆(v−1)simple?

⇔

v−1 w−1

6= 0 for allw ≤v

(11)

∆(1−1)

∆(2−1)

∆(3−1)

∆(4−1)

∆(5−1)

∆(6−1)

∆(7−1)

X0Y0

X1Y0 X0Y1

X2Y0 X1Y1 X0Y2

X3Y0 X2Y1 X1Y2 X0Y3

a bc d

The simples

Example∆(7−1) =KX⁶Y⁰⊕ · · · ⊕KX⁰Y⁶.

No common eigensystem⇒∆(7−1) simple.

(0,0,0,1,0,0,0) is a common eigenvector, so we found a submodule.

When is∆(v−1)simple?

⇔

v−1 w−1

6= 0 for allw ≤v

(12)

∆(1−1)

∆(2−1)

∆(3−1)

∆(4−1)

∆(5−1)

∆(6−1)

∆(7−1)

X0Y0

X1Y0 X0Y1

X2Y0 X1Y1 X0Y2

X3Y0 X2Y1 X1Y2 X0Y3

a bc d

The simples

Example∆(7−1) =KX⁶Y⁰⊕ · · · ⊕KX⁰Y⁶.

(0,0,0,1,0,0,0) is a common eigenvector, so we found a submodule.

When is∆(v−1)simple?

⇔

v−1 w−1

6= 0 for allw ≤v

General.

Weyl ∆(λ) and dual Weyl∇(λ) are easy a.k.a. standard;

are parameterized by dominant integral weights;

are highest weight modules;

are defined overZ;

have the classical Weyl characters;

form a basis of the Grothendieck group unitriangular w.r.t. simples;

satisfy (a version of) Schur’s lemma dimKExti(∆(λ),∇(µ)) =δi,0δλ,µ; are simple generically;

(13)

Ringel, Donkin∼1991. The indecomposable SL₂ tilting modulesT(v−1) are the indecomposable summands of ∆(1)^⊗i ∼= (K²)^⊗i

. Tilting modulesT(v−1)

• are those modules with a ∆(w−1)- and a∇(w−1)-filtration;

• are parameterized by dominant integral weights;

• are highest weight modules;

• satisfy reciprocity T(v−1) : ∆(w−1)

= T(v−1) :∇(w−1)

= [∆(w⁰−1) : L(v⁰−1)] = [∇(w⁰−1) :L(v⁰−1)];

• form a basis of the Grothendieck group unitriangular w.r.t. simples;

• satisfy (a version of) Schur’s lemma dim_KHom T(v−1),T(w−1) P =

x<min(v,w) T(v−1) : ∆(x−1)

T(w−1) :∇(x−1)

Why the name?;

• are simple generically;

• have a root-binomial-criterion to determine whether they are simple.

LetTiltbe the category of tilting modules.

Goal. DescribeTilt by generators and relations.

General.

These facts hold in general, and the first bullet point is the general definition.

How many Weyl factors doesT(v−1)have?

# Weyl factors ofT(v−1) is 2^k where k= max{νp v−1

w−1

,w ≤v}. (Order of vanishing of _w^v⁻¹₋₁ .) determined by (Lucas’s theorem)

non-zeronon-leadingdigits ofv = [ar,ar−1,...,a0]p. ExampleT(220540−1)forp= 11?

v = 220540 = [1,4,0,7,7,1]11;

Maximal vanishing forw= 75594 = [0,5,1,8,8,2]11;

v−1 w−1

= (HUGE) = [...,6= 0,0,0,0,0]11.

⇒T(220540−1) has 2⁴Weyl factors.

Which Weyl factors doesT(v−1)have a.k.a. the negative digits game? Weyl factors ofT(v−1) are

∆([ar,±ar−1,...,±a0]p−1) wherev= [ar,...,a0]p.

ExampleT(220540−1)forp= 11? v = 220540 = [1,4,0,7,7,1]11; has Weyl factors [1,±4,0,±7,±7,±1]11; e.g. ∆(218690 = [1,4,0,−7,−7,−1]11−1) appears.

(14)

= T(v−1) :∇(w−1)

= [∆(w⁰−1) : L(v⁰−1)] = [∇(w⁰−1) :L(v⁰−1)];

x<min(v,w) T(v−1) : ∆(x−1)

T(w−1) :∇(x−1)

Why the name?;

General. These facts hold in general, and

the first bullet point is the general definition.

w−1

,w≤v}. (Order of vanishing of _w^v⁻¹₋₁ .) determined by (Lucas’s theorem)

v= 220540 = [1,4,0,7,7,1]11;

v−1 w−1

= (HUGE) = [...,6= 0,0,0,0,0]11.

Which Weyl factors doesT(v−1)have a.k.a. the negative digits game? Weyl factors ofT(v−1) are

ExampleT(220540−1)forp= 11? v = 220540 = [1,4,0,7,7,1]11; has Weyl factors [1,±4,0,±7,±7,±1]11; e.g. ∆(218690 = [1,4,0,−7,−7,−1]11−1) appears.

(15)

= T(v−1) :∇(w−1)

= [∆(w⁰−1) : L(v⁰−1)] = [∇(w⁰−1) :L(v⁰−1)];

x<min(v,w) T(v−1) : ∆(x−1)

T(w−1) :∇(x−1)

Why the name?;

General. These facts hold in general, and

the first bullet point is the general definition.

w−1

,w ≤v}. (Order of vanishing of _w^v⁻¹₋₁ .) determined by (Lucas’s theorem)

v = 220540 = [1,4,0,7,7,1]11;

v−1 w−1

= (HUGE) = [...,6= 0,0,0,0,0]11.

Which Weyl factors doesT(v−1)have a.k.a. the negative digits game?

Weyl factors ofT(v−1) are

ExampleT(220540−1)forp= 11?

v= 220540 = [1,4,0,7,7,1]11; has Weyl factors [1,±4,0,±7,±7,±1]¹¹; e.g. ∆(218690 = [1,4,0,−7,−7,−1]11−1) appears.

(16)

Strategical interlude.

Start.

Define a categoryWeb by generators-relations.

How?

Find a standard basisS ofWeb(splitting off ∆).

How?

Find a tilting basisTof Web(splitting offT).

How?

Find an integral basisI ofWeb.

How?

ProveWeb∼=ZFund (⊗-gen. by ∆(1)).

How?

Write down the base change matrixStoT.

Why?

Make sure that there are no poles.

Original sin

Goal achieved: Tiltvia generators and realtions.

Quiver

reduce modp

General.

This strategy should work in types ABCD.

(I will zoom in on this in a second.)

What remains to be done?

No more sins!

What is the diagrammatic incarnation of the Frobenius (^{a b}_{c d})7→ ^a_c^p^p_d^b^p^p

? The mixed case will be easier but might be a pain to write down.

Up next: the first steps towards higher ranks,

i.e. let us tryUq(sl3) forqa primitive complex 2`th root of unity.

(17)

Start.

How?

ProveWeb∼=ZFund (⊗-gen. by ∆(1)).

How?

Why?

Original sin

Quiver

reduce modp

General.

This strategy should work in types ABCD. (I will zoom in on this in a second.)

What remains to be done?

No more sins!

What is the diagrammatic incarnation of the Frobenius (^{a b}_{c d})7→ _c^a^p^p_d^b^p^p

? The mixed case will be easier but might be a pain to write down.

Up next: the first steps towards higher ranks,

i.e. let us tryUq(sl3) forqa primitive complex 2`th root of unity.

(18)

Start.

Basically the same.

ProveWeb∼=_Z[q±1]Fund (e.g.∆ (1,0)

∈ Fund).

Remains the same.

Same problem as before!

To be done...

specialize q