Level one conformal blocks - Rank reduction in type A

Chapter 5. Examples

5.1 Rank reduction in type A

5.1.1 Level one conformal blocks

Level one conformal blocks are the simplest non-trivial case, but even here the bundles and

associated divisors we obtain are quite interesting (see for example [19]). Non-trivial level one

conformal blocks correspond to a subset of the vertices of the multiplicative polytope, and tuples of

central elements ofGmultiplying to the identity. We will refer to these vertices of the multiplicative

polytope as thetrivial vertices.

The level one weights

λ

can be identified with integers. For simplicity, we will assume that

V

_sl_r+1_,~_λ,₁

6=

{0}. Now in this case we must have

r+ 1

P

i=1

λ

. Let

P

ni=1

λ

=s(r+ 1). It is a

fact that 2≤s≤n−1. We will start with the simplest walls of the multiplicative polytope, where

k= 1 andd= 0, so that Gr(k, r+ 1)∼=P

. So in this case our subsetsI

can also be identified with

integers, with 1≤I

≤r+ 1. Note thatI

corresponds to a hyperplane of dimension

I

−1 in

P

.

Then

σI

· · ·σI

= [pt] if and only if the codimensions

r−I

+ 1 add up tor, which is equivalent

to

P

Ij

= (n−1)(r+ 1) + 1. Let [a

≤

b] be one if

a

≤

b

and zero otherwise. Then the above

inequalities become

X

j=1

[Ij

≤λj]≤s.

In order to apply our reduction result, we need to find weights and Schubert varieties making the

above inequality an equality. In order for this to be possible, the largests

weights must sum up to

at least (s−1)(r+ 1) + 1, in order for the codimension equation to be satisfied. Therefore we have

the following proposition.

Proposition 5.1.2.

Suppose

λ

, . . . , λn

are level one weights such thatV

_sl

r+1,~λ,1

6={0}. Let

N

be a

subset of{1, . . . , n}such that|N|=s. Then ifP

j=1

λj

=s(r+ 1)and

P

j∈N

λj

≥(s−1)(r+ 1) + 1,

we have

V

_sl_r+1_,~_λ,₁

∼=V

_sl_r_,~_λ0_,₁

.

where

λ

0_j

=λj−1

for

j

∈N

and

λ

0_j

=λj

otherwise.

Note thatP

i=1

λ

=sr, so that one can iterate this reduction to obtain weights such that any

We are also interested in the intersection of the corresponding divisors with so-called F-curves.

Modulo numerical equivalence, classes of the one-dimensional strata of M0

correspond to partitions

of

{1, . . . , n}

into four non-empty subsets

N

, . . . , N

4. The importance of F-curves is that the

numerical equivalence class of the divisors

D

_sl

r+1,~λ,1

is determined by its intersection with the

F-curves. In [17], Fakhruddin gave a formula for the intersection of level-one type A conformal

blocks with F-curves in terms of these partitions. The hypothesis of the above proposition also

implies that the conformal blocks divisorD

_sl_r+1_,~_λ,₁

intersects a number of F-curves trivially:

Proposition 5.1.3.

Suppose

λ

, . . . , λn

are level one weights such that

V

_sl_r+1_,~_λ,₁

6=

{0}, and

P

j=1

λj

=s(r+ 1). Let

N

be a subset of

{1, . . . , n}. Then if

P

j∈N

λj

≥(s−1)(r+ 1),

D

slr+1,~λ,1

intersects trivially all F-curves associated to partitions{1, . . . , n}=t

i=1

Ni

such thatNi

=N

for

some

i.

Proof.

Let

νk=P

j∈Nk

λj

(mod)r+ 1. By the assumed inequality we see that

P

νk

≤r+ 1. But

by Fakhruddin’s formula, for the intersection to be non-zero, we needP

νk= 2(r+ 1). Therefore

all given intersections are trivial.

We would like to understand level-one conformal blocks on walls withk >1. The first observation

is that there is a

duality

among the walls containing a trivial vertex of the multiplicative polytope.

This depends on the action of the center on the multiplicative polytope.

Consider the vertex of the multiplicative polytope given by the level-one weights

λ

, . . . , λn.

This corresponds to a tuple of elements of the center~c= (c

, . . . , cn) such that

ci

is equal toζ

λi

Id,

where

ζ

is a primitive root of unity, and

Q

c

= Id. Then this tuple of central elements acts on

the multiplicative polytope in the obvious way. For a subsetI

of

{1, . . . , r+ 1}, letI−1 be the

set obtained fromI

by subtracting 1 from each element, and replacing any 0 withr+ 1. For any

positive integerλ,I−λis obtained by iterating this process. Then Agnihotri and Woodward proved

the following proposition.

Proposition 5.1.4.

([1] Prop 7.2)

If

λ

, . . . , λ

are level-one weights corresponding to a trivial

vertex of the multiplicative polytope, and if

F

is the facet of the polytope corresponding to

σ

∗ · · · ∗

σ

=q

_[pt]_∈_QH

∗

_(Gr(_{k, r}_{+ 1))}_{, then}_~_c

−1

_·_F

_{is the facet corresponding to}

σI

1−λ1

∗ · · · ∗σI

n−λn

=q

where

d

=

d−P

j=1

[I

≤λ

]−

_r₊₁k

P

nj=1

λ

. In particular, if

F

contains

(λ

, . . . , λ

), then

d

= 0.

The duality also arises from Grassmann duality. For eachσI

class in H

∗

(Gr(k, r+ 1)), there is a

dual class

σ

given in the following way. ToI

= (i

₁

<· · ·< ik) we can associate a partition

λ(I)

defined asλ(I)

₌_r₋_k_{+ 1 +}_j₋_i

. ThenI

is the (r−k+ 1)-subset of{1, . . . r+ 1}corresponding

to the transpose ofλ(I). Then we have the following lemma.

Lemma 5.1.5.

For any

1≤λ≤r, and subsetI

of

{1, . . . , n}, we have(I−λ)

=I

−(r−λ+ 1).

Proof.

Clearly it is sufficient to consider the case

λ= 1. There are then two cases. First, if 1∈I,

thenr+ 1∈/I

. Then

λ(I−1) is the partition obtained fromλby removing the largest part, and

taking the transpose, we get

λ(I

) with 1 subtracted from each part. On the other handI

−r

is

just the set obtained by adding 1 to each element, or in other words subtracting 1 from each part.

Now suppose 1∈/

I. Note that

r+ 1∈I

. Thenλ(I−1) is the partition obtained by adding 1 to

each part ofλ(I), and

λ(I−1)

_{is the partition obtained from}

_λ₍_I

_{) by adding}

_k_{as its highest part.}

The subset

I

−r

is obtained from

I

by replacingr+ 1 with 1 and adding 1 to all other elements,

finishing the proof.

Proposition 5.1.6.

There is a natural bijection between the

d= 0

regular faces of type

k

and type

r−k+ 1containing a trivial vertex of the multiplicative polytope. Given a facet