2010 summer school

7/31/2019 2010 Summer School

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ORBIT CLASSIFICATION FOR THE REPRESENTATIONS

ASSOCIATED TO GRADED LIE ALGEBRAS

Jerzy Weyman

July 14-18, 2010

Abstract. These are the lectures given by the author during the summer school ontensor decompositions in Norway in June 2010.

Content

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21. Root systems and simple Lie algebras2. Gradings on simple Lie algebras related to simple roots3. Kacs classification of representations with finitely many orbits. Examples involving tensors andexterior powers

4. Vinberg method of classifying orbits, statements of the results5. Vinberg method in practice. Working out of concrete examples6. Defining ideals and syzygies7. ApplicationsReferences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

Introduction

The irreducible representations of the reductive groups with finitely many orbitswere classified by Kac in [K80]. They were divided into classes II, III and IV. All ofthese classes are related to gradings of the root systems, and to the corresponding groups.

The representations of type II are parametrized by a pair ( Xn, k) where Xn isa Dynkin diagram with a distinguished node x Xn. This data defines a grading

g = si=sgi

The author was partially supported by NSF grant DMS-0600229

Typeset by AMS-TEX

1


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2 JERZY WEYMAN

of a simple algebra g of type Xn such that the Cartan subalgebra h is contained ing0 and the root space g is contained in gi where i is the coefficient of the simpleroot corresponding to the node x in the expression for as a linear combination ofsimple roots. The representation corresponding to (Xn, x) is the g1 with the actionof the group G0 C where G0 is the adjoint group corresponding to g0 and C isthe copy ofC that occurs in maximal torus ofG (the adjoint group correspondingto g) but not in maximal torus of G0.

The orbit closures for the representations of type II were described in two waysby Vinberg in [V75], [V87]. The first description states that the orbits are theirreducible components of the intersections of the nilpotent orbits in g with thegraded piece g1. In the second paper Vinberg gave a more precise description interms of some graded subalgebras of the graded algebra g = gi In these noteswe give the introduction to these methods. We use mainly the examples of tripletensor products and third exterior powers.

The notes are organized as follows. In section 1 we discuss root systems and sim-ple Lie algebras. Section 2 introduces representations related to gradings on simpleLie algebras related to a choice of a simple root. Section 3 we state Kacs theorem

and interpret special cases of triple tensor products and third exterior powers. Insection 4 we expose the Vinberg method from [V87] that classfies orbits in repre-sentations with finitely many orbits. In section 5 we work out some examples. Inthe following sections we give information about singularities and defining ideals ofthe orbit closures. We briefly describe the geometric method of calculating syzygiesand give examples of its use. We give some comments on how one should be ableto calculate resolutions of all the coordinate rings of orbit closures.

1. Simple Lie algebras and root systems

We recall the Cartan classification of simple Lie algebra and corresponding rootsystems.

For a simple Lie algebra g we have the root decomposition

g = h g

where is the associated root system.The simple Lie algebra g has also a symmetric bilinear non-degenerate form

(, ) : g g C

given by the formula (X, Y) = T r(ad(X)ad(Y)).

Definition 1.1. LetV be an Euclidean space overR with a scalar product (, ). Aroot system is by definition a finite collection of vectors in V such that

a) If then the only multiples of in are ,


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GEOMETRY OF ORBIT CLOSURES 3

b) The values of

, :=2(, )

(, )

are inZ for , in .

c) If and is a reflection in the hyperplane orthogonal to then() = .d) spans V.

If 1 and 2 are two root systems in the Euclidean spaces V1, V2, then 1 2is a root system in the orthogonal direct sum V1 V2. A root system is irreducibleif it cannot be decomposed into a direct sum of two root systems.

In every root systems one can choose a basis = {1, . . . , l} of simple rootswhich have the following property. The root system decomposes

= +

where positive roots can be written as nonnegative linear combinations of vectorsin the basis, and negative roots, as non-positive linear combinations.

Example 1.2. The root system if type An.

V = {(a1, . . . , an+1) Rn+1 | a1 + . . . + an+1 = 0}

= {(i j) | 1 i < j n + 1}.

= {1 2, 2 3, . . . , n n+1}.

Corresponding Lie algebra is sln+1.

Example 1.3. The root system if type Bn.

V = Rn

= {(i j) | 1 i < j n, i, 1 i n}.

= {1 2, 2 3, . . . , n1 n, n}.

Corresponding Lie algebra is so2n+1.

Example 1.4. The root system if type Cn.

V = Rn

= {(i j) | 1 i < j n, 2i, 1 i n}.

= {1 2, 2 3, . . . , n1 n, 2n}.

Corresponding Lie algebra is sp2n.


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4 JERZY WEYMAN

Example 1.5. The root system if type Dn.

V = Rn

= {(i j) | 1 i < j n}.

= {1 2, 2 3, . . . , n1 n, n1 + n}.

Corresponding Lie algebra is so2n.

Example 1.6. The root system if type E6.

V = Rn

= {(i j) | 1 i < j 5,

1

2(1 2 3 4 5 6 7 + 8), with even number of minus signs}.

= {

1

2 (1 2 . . . 7 + 8), 1 2, 2 3, 3 4, 4 5, 4 + 5}.

To every root system we associate its Dynkin diagram which is a graph whosenodes are the simple roots and the nodes i and j are joined by i, jj , iedges.

Finally we mention that the basic representations of the Lie algebra correspond-ing to (or the corresponding simply connected group) are the fundamentalrepresentations. They correspond to fundamental weights i defined via

i(j) = i,j

for all simple roots j , 1 j .The fundamental representations V(i) are basic in that all others can be ob-

tained from them as summands of tensor products. For classical groups most ofthem have very concrete descriptions.

Example 1.7.

g = sln+1, V(i) =iC

n+1.

Example 1.8.

g = so2n+1, V(i) =iC2n+1, 1 i n 1, V(n) = spin(2n + 1).


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Example 1.9.

g = sp2n, V(i) =iC2n/

i2C2n, 1 i n.

Example 1.10.g = so2n, V(i) =

iC2n, 1 i n 2,

V(n1) = spin+(2n), V(n) = spin(2n).

We will not need the description of spinor and half-spinor representations here.To a representation V of a reductive group G (like a triple tensor product or an

exterior power) we can associate its diagram by choosing a black node (correspond-ing to a representation) and the white nodes corresponding to a Dynkin diagramof the group G. Let the Dynkin diagram of G have several connected componentsi. Then the representation V is the tensor product iV(

(i)) where

(i) =j

(I)j

(i)j .

The diagram of V is obtained from by joining the black node with the vertex

(I)j with

(I)j edges. The main point is that (forgetting a few exceptions that will

be listed) a representaton V has finitely many G-orbits if and only if the diagramof V is a Dynkin diagram.

Exercises for section 1.1. LetF be an orthogonal space of dimension2n+1 with a non-degenerate quadratic

form (, ). Choose a hyperbolic basis {e1, . . . , en, e0, en, . . . , e1}. The orthogonalLie algebra so2n+1 consists of endomorphisms of F such that

((u), v) + (u, (v)) = 0 u, v F.

Find the root decomposition ofso(2n + 1) and identify it with the set2

F.

2. Let F be an symplectic space of dimension 2n with a non-degenerate skew-symmetric form(, ). Choose a hyperbolic basis {e1, . . . , en, en, . . . , e1}. The sym-plectic Lie algebrasp2n consists of endomorphisms of F such that

((u), v) + (u, (v)) = 0 u, v F.

Find the root decomposition ofsp(2n) and identify it with the setS2F.3. Let F be an orthogonal space of dimension 2n with a non-degenerate quadratic

form (, ). Choose a hyperbolic basis {e1, . . . , en, en, . . . , e1}. The orthogonal Liealgebra so2n consists of endomorphisms of F such that

((u), v) + (u, (v)) = 0 u, v F.

Find the root decomposition ofso(2n) and identify it with the set2

F.


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6 JERZY WEYMAN

Hint. The weight of ei is i, the weight of ei is i. The Cartan subalgebra h isthe subspace of vectors of weight zero ing.

2. The representations of type II and groups

Let Xn be a Dynkin diagram and let g be the corresponding simple Lie algebra.Let us distinguish a node x Xn. Let be a corresponding simple root in the rootsystem corresponding to Xn. The choice of determines a Z- grading on byletting the degree of a root be equal to the coefficient of when we write as alinear combination of simple roots. On the level of Lie algebras this corresponds toa Z-grading

g = iZ gi.

We define the group G0 := (G, G) C where (G, G) is a connected semisimplegroup with the Dynkin diagram Xn \ x. A representation of type II is the represen-tation of G0 on g1.

By construction the representations of type II correspond to the Dynkin diagramsXn with distinguished nodes.Denoting by l the Levi factor g0 we have

l = l z(l)

where l denotes the Lie algebra associated to Xn with the omitted node x.Let us look at some interesting examples of these gradings. We look at all possible

grading of Lie algebra of type E6.Notice that in each case the bracket

gi gj gi+j

has to be G0-equivariant, so it i determined up to a non-zero scalar. So in each casewe get a model of Lie algebra of type E6 in terms of classical Lie algebras and theirrepresentations. In each case we exhibit only gi for i 0 because we always have

gi = gi .

In each case we exhibit the restriction of the Killing form of g to g1.

k = 1, 6. The representation in question is V = V(4, D5) is a half-spinor rep-resentation for the group G = Spin(10). Here 4 = (

12 ,

12 ,

12 ,

12 ,

12). dim(X) = 16.

The weights ofX are vectors in 5 dimensional space, with coordinates equal to 12 ,with even number of negative coordinates.


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We label the weight vectors in g1 by [I] where I is the subset of {1, 2, 3, 4, 5} ofeven cardinality where the sign of the component is negative.

The graded Lie algebra of type E6 is

g(E6) = g1 g0 g1

with g0 = Cso(10).The invariant scalar product (, ) on g restricted to g1 is given by the formula

([I], [J]) = 2 1

2#({I\ J} {J \ I}).

Thus possible scalar products are only 2, 1, 0. So the possible root systems wecan get are A1 A1 and A1. Indeed, the triple product A1 A1 A1 is not possiblebecause there are no three subsets I , J, K with cardinalities of three symmetricdifferences being 4. This means we get

k = 2. V =3

F, F = C6, G = GL(F).

The graded Lie algebra of type E6 isg(E6) = g2 g1 g0 g1 g2

with g0 = Csl(6), g1 =3

C6, g2 =

6C6.

The weights ofg1 are i+j+k for 1 i < j < k 6. We label the correspondingweight vector by [I] where I is a cardinality 3 subset of {1, 2, 3, 4, 5, 6}.

The invariant scalar product on g restricted to g1 is

([I], [J]) = 1

where = #(I J).

k = 3, 5. V = E2

F, E =C2

, F =C5

, G = SL(E) SL(F) C


g(E6) = g2 g1 g0 g1 g2

with g0 = Csl(2) sl(5), g1 = C2 2

C5, g2 =

2C2 4

C5.

Let {e1, e2} be a basis ofE, {f1, . . . , f 5} be a basis of F. We denote the tensorea fi fj by [a; ij]. The invariant scalar product on g restricted to g1 is

([a; ij], [b; kl]) = 1

where = #({a} {b}) + #({i, j} {k, l}).


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8 JERZY WEYMAN

k = 4. V = EFH, E = C2, F = H = C3, G = SL(E)SL(F)SL(H)C.


g(E6) = g3 g2 g1 g0 g1 g2 g3

with g0 = Csl(2) sl(3) sl(3), g1 = C2 C3 C3, g2 =

2C2 2

C3 2

C3,

g3 = S2,1C2 3

C3 3

C3.

Let {e1, e2} be a basis of E, and {f1, f2, f3}, {h1, h2, h3} bases of F, H respec-tively. We label ea fi hu by [a; i; u].

The invariant scalar product on g restricted to g1 is

([a; i; u], [b;j; v]) = 1

where = #({a} {b}) + #({i} {j}) + #({u} {v}).It is wortwhile exhibiting explicitly the center z(l). Denote the 1, . . . , n the

simple roots, and let x = i0. Define the elements xi h by setting

j(xi) = i,j.

Then

1.1. Proposition.a) We have z(l) = Cxi0 .b) For a representationV ofg we get the weights of the restrictionV|l by applying

the induced epimorphism h h(l) whose kernel is xi0 , and then expressingi0 by means of other i by modifying the wieght by a multiple of xi0 .

c) Consider the irreducible representation V of g with the highest weight =i mii. Let us describe the restriction Vl. We want to label every irreducible

representation in this restriction by the labeling of the Dynkin diagram Xn withthe marked node i0. This is done as follows. We use b) to determine the weightsof (V)l and identify the highest weights. This will give us the markings on all

the nodes except i0. To determine the marking at the node i0 we need to describethe action of z(l). We have

(xi0) =i

mii(xi0) =i

mi(

mijj)xi0 =i

mimii0

,

where we writei =

j

mijj .

The numbers mij can be read from the Bourbaki tables.

Vinberg in [V75] gave a description of the G0-orbits in the representations oftype II in terms of conjugacy classes of nilpotent elements in g. Let e g1 be a


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nilpotent element in g. Consider the irreducible components of the intersection ofthe conjugacy class of e in g

C(e) g1 = C1(e) . . . Cn(e)(e).

The sets Ci(e) are clearly G0-stable. Vinbergs result shows that these are preciselythe G0-orbits in g1.

1.2 Theorem. The G0-orbits of the action ofG0 ong1 are the components Ci(e),for all choices of the conjugacy classes C(e) and all i, 1 i n(e).

Theorem 1.2. makes the connection between the orbits in g1 and the nilpotentorbits in g. This means we will be needing the classification of nilpotent orbits insimple Lie algebras. This was obtained by Bala and Carter in the papers [BC76a],[BC76b]. A good account of this theory is the book [CM93]. Here we recall thatthe nilpotent orbit of an element e in a simple Lie algebra g is characterized by thesmallest Levi subalgebra l containing e. One must be careful because sometimes lis equal to g. If the element e is a principal element in l, then this orbit is denoted

by the Dynkin diagram of l (but there might be different ways in which the rootsystem R(l) sits as a subroot system of R(g).

There are, however the nonprincipal nilpotent orbits that are not contained ina smaller reductive Lie algebra l. These are called the distinguished nilpotent orbitsand are described in sections 8.2, 8.3, 8.4 of [CM93]. They are characterized bytheir associated parabolic subgroups (as their Dynkin characteristics are eve, seesection 8 in [CM93]). Let us remark that for Lie algebras of classical types, for typeAn the only distinguished nilpotent orbits are the principal ones, and for typesBn, Cn, Dn these are orbits corresponding to the partitions with different parts.For exceptional Lie algebras the distinguished orbits can be read off the tables insection 8.4. of [CM93].

This theorem is not easy to use because it is not very explicit. In section 4 wedescribe more precise method from second Vinberg paper [V87].

Exercises for section 2. 1. Identify the groups G0 and the representations g1 forthe gradings ofAn, Bn, Cn, Dn, E7 and E8 related to simple roots. 2. For (E7, 2)and E8, 2) find the models of corresponding Lie algebras.

3. Kacs classification of representations with finitely many orbits

The representations of reductive groups with finitely many orbits were classifiedby Kac in [K80] and [DK85]. The result is


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10 JERZY WEYMAN

Theorem 3.1. LetH be a reductive group operating on an irreducible representa-tion with finitely many orbits. Then the pair (H, V) is on the following list

a) H = G0, V = g1 where g = iZg1 is a grading on a simple Lie algebra g relatedto a simple root k,

b) H = SL(2) SL(3) SL(n) C operating onC2 C3 Cn, n 6,c) H = SL(2) Spin(7), V = C2 V(3).

In fact Kac classifies a more general class of representations, so-called visiblerepresentations.

The nullcone N(V) of a representation V of G is the set of v V for whichf(v) = 0 for every G-invariant function f Sym(V).

Definition 3.2. The nullcone N(V) of a representationV ofG is the set ofv Vfor which f(v) = 0 for every G-invariant function f Sym(V). A representationV of G is visible if there are finitely many G-orbits in the null-cone N(V) of V.

A very important class of visible representations is related to finite order auto-morphisms of the root systems classified by Kac.

Let Xn be an extended Dynkin graph corresponding to the Dynkin graph Xn.

Let xk be a node in Xn. Let ak be the coefficient of the isotropic root of the graphXn at the vertex xk. Then there is a grading on g by Z/akZ where g0 is a subalgebra

corresponding to Xn \ k and g1 is given by the same recipy as for the cases inprevious section.

Alternative description os that the Kac-Moody Lie algebra g has a grading

g = iZgi

and the components gi are periodic with period ak.The visible representations g1 have analogous properties to the adjoint represen-

tations. Thy have a Cartan subspace h which is the slice representation. The Weylgroup W := N(h)/Z(h) acts on h as a reflection group generated by psedoreflec-tions. Vinberg proves the analogue of Chevalley Theorem. The natural map

Sym(g1)G Sym(h)W

is an isomorphism. Both rings are polynomial rings. Vinberg proves only existenceofh and identifies W using Sheppard-Todd table and the knowledge of degrees ofinvariants. It would be very interesting to find the direct and explicit proofs of theseresults.

Example 3.1. Let us consider the case (E6, 4). The order of is 3. We have

g0 = sl3 sl3 sl3, g1 = C3 C3 C3, g2 =

2C3

2C3

2C3.


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Exercises for section 3.1. Determine the graded components gk for the case (E7, 4). Answer:

g0 = sl2 sl4 sl4, g1 = C2 C4 C4,

g2 =2C2

2C4

2C4, g3 = S2,1C

2 3C4

3C4.

2. Do the same for the case (E7, 5).

4. The Vinberg method for classifying orbits.

All Lie algebras g we will consider will be Lie algebras of some algebraic groupG.

Let (Xn, k) be one of the representations on our list. It defines the grading

g = iZgi

where gi is the span of the roots which, written as a combination of simple roots,

have k with coefficient i. The component g0 contains in addition a Cartan subal-gebra. G0 denotes the connected component of the subgroup of G corresponding tog0.

In the sequel Z(x) denotes the centalizer of an element x G. Z0(x) is Z(x)G0.The gothic letters, z, z0 denote corresponding Lie algebras. Similarly, N(x) denotesthe normalizer of an element x G. N0(x) is N(x) G0.

We denote R(g) the set of roots of a reductive Lie algebra g, and (g) denotesthe set of simple roots.

In order to state Vinberg Theorem we need some definitions.We will be dealing with the graded Lie subalgebras s = iZsi.

Definition 4.1. A graded Lie subalgebra s of g is regular if it is normalized by amaximal torus in g0. A reductive graded Lie algebra s of g is complete if it is nota proper graded Lie subalgebra of any regular reductive Z-graded Lie algebra of thesame rank.

Definition 4.2. A Z-graded Lie algebra g is locally flat if one of the followingequivalent conditions is satisfied, for e a point in general position ing1:

(1) The subgroup Z0(e) is finite,(2) z0(e) = 0,(3) dimg0 = dimg1.

Let g = iZgi be a graded reductive Lie algebra. Let us fix a nonzero nilpotentelement e ga. Let us choose some maximal trous H in N0(e). Its Lie algebra h isthe accompanying torus of the element e. We denote the character of the torusH defined by the condition

[u, e] = (u)e


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12 JERZY WEYMAN

for u h. Consider the graded Lie subalgebra g(h, ) ofg defined as follows

g(h, ) = iZ g(h, )i

where g(h, )i = {x gia | [u, x] = i(u)u H }.

Definition 4.3. The support s of the nilpotent element e ga is the commutantofg(h, ) considered as aZ graded Lie algebra.

Clearly e s1.We are ready to state the main theorem of [V87].

Theorem 4.4. (Vinberg). The supports of nilpotent elements of the space giare exactly the complete regular locally flat semisimple Z-graded subalgebras of thealgebra g. The nilpotent element e can be recovered from the support subalgebra sas the generic element in s1.

It follows from the theorem that the nilpotent e is defined uniquely (up to con-jugation by an element of G0) by its support. This means it is enough to classifythe regular semisimple Z-graded subalgebras s ofg.

Let us choose the maximal torus tofg0. The Z-graded subalgebra s is standardif it is normalized by t, i.e. if for all i Z we have

[t,si] si.

Vinberg also proves that every Z graded subalgebra s is conjugated to a standardsubalgebra by an element ofG0. Moreover, he shows that if two standard Z-gradedsubalgebras are conjugated by an element of G0, then they are conjugated by anelement of N0(t).

This allows to give a combinatorial method for classifying regular semisimpleZ

-graded subalgebras ofg.Let s be a standard semisimple Z-graded subalgebra of g. The subalgebra sdefines the degree map deg : R(s) Z. For a standard Z-graded subalgebra s wealso get the map

f : R(s) R(g).

The map f has to be additive, i.e. it satisfies

f( + ) = f() + f() , R(s),

f() = f() R(s).

Moreover we have


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Proposition 4.5. The map f satisfies the following properties:a)

(f(), f())

(f(), f())=

(, )

(, ), R(s).

b) f() f() / R(g) , (s),c) deg f() = deg , (s).

Conversely, every map satisfying a), b), c) defines a standard regularZ-gradedsubalgebra s ofg.

Remark 4.6. The subalgebras corresponding to the map f is complete if and onlyif there exists an element w in the Weyl group W ofg such that wf((s)) (g)(see [V87], p.25) .

Proposition 4.5 means that in order to classify the nilpotent elements e g1we need to classify the possible maps f corresponding to its support, i.e. the cor-responding complete regular Z-graded subalgebra s. Since we are interested in thenilpotents e g1, we need to classify the maps f for which degf() {0, 1} forevery (s).

One should also mention the connection with the Bala-Carter classification. TheBala-Carter characteristic of a nilpotent element is given by the type of the algebras. So to find the components of a given nilpotent orbit with Bala-Carter character-istic s it is enough to see in how many ways the Lie algebra s can be embedded asa Z-graded Lie algebra into g.

One can determine the orbits explicitly using the following strategy.

Proposition 4.7. The following procedure allows to find the components of theintersection of a nilpotent orbit ing withg1.1. Establish the restriction of the invariant scalar product(, ) on g to g1.2. Fix a nilpotent element e g with Bala-Carter characteristic s.

3. Find in how many ways s embeds into g as a standardZ

-graded Lie subalgebraby exhibiting corresponding map f as in Proposition 4.5.4. For a characteristic s such thate is a principal nilpotent ins the map f sends all

simple roots ofs to weight vectors ing1. Thus it is enough to classify the subsetsof weights vectors in g1 for which the pattern of scalar products is the same asthe one for simple roots of s.

5. For non-principal orbits one needs to make a more detailed analysis, but stillone reduces to finding sets of weight vectors ing1 with certain patterns of scalarproducts.

Exercises for section 4. Classiify the subalgebras s of type A3+2A1 andA2+2A1in bigwedge3C8 (the case (E8, 2). Prove that we do not have any cases of type A5and D4.


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14 JERZY WEYMAN

5. Vinberg method: examples.

Example 5.1. Consider the diagram Xn = An. Let x be the note correspondingto m := m m+1. We have g = sln+1, G0 = SL(m,C) SL(n m + 1,C) C

.The Z grading is

g = g1 g0 g1

where g0 = C sl(m) sl(n m + 1), g1 = HomC(Cn+1m,Cm).

The weight vector ei fj corresponds to the root i m+j. Let us denote it bythe label [i, j]. The invariant scalar product (, ) on g restricted to g1 is

([i1, j1], [i2, j2]) =

where = #([{i1} {i2}) + #([{j1} {j2}). This means the scalar products cantake only values 2, 1, 0. So the only root systems whose simple roots can be embeddedinto g1 are (A1)

r for 1 r min(m, n). The corresponding set of weights has tobe orthogonal, so (up to permutation of indices which means conjugation) we haveone r-tuple {[1, 1], . . . , [r, r]}. The representative is the tensor e1f1+ . . . + er fr.The corresponding orbit is the set of tensors of rankr.

Consider a matrix of rankr ing1. After changing the bases inCm and inCn+1m

we can assume that the corresponding nilpotent e(r) =r

j=1 Ej,m+j. The intersec-

tion of the orbit of e(r) with g1 is a determinantal variety of m (n + 1 m)matrices of rank r.

Lets go to triple tensor product cases.

Example 5.2. Consider the case (E6, 4).

g(E6) = g3 g2 g1 g0 g1 g2 g3

withg0 = Csl(2) sl(3) sl(3), g1 = C2 C3 C3, g2 =

2C2 2

C3 2

C3,

g3 = S2,1C2 3

C3 3

C3.


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number s representative0 zero 01 A1 [1;1;1]

2 2A1 [1; 1; 1] + [2;2; 1]3 2A1 [1; 1; 1] + [2;1; 2]4 2A1 [1; 1; 1] + [1;2; 2]5 3A1 [1; 1; 1] + [1;2; 2] + [2;1; 2]6 3A1 [1; 1; 1] + [1;2; 2] + [1;3; 3]7 A2 [1; 1; 1] + [2;2; 2]8 A2 + A1 [1; 1; 1] + [2;2; 2] + [1;3; 2]9 A2 + A1 [1; 1; 1] + [2;2; 2] + [1;2; 3]

10 A2 + 2A1 [1; 1; 1] + [2;2; 2] + [1;3; 2] + [2;3; 1]11 A2 + 2A1 [1; 1; 1] + [2;2; 2] + [1;2; 3] + [2;1; 3]12 A2 + 2A1 [1; 1; 1] + [2;2; 2] + [1;2; 3] + [1;3; 2]13 2A2 [1; 1; 1] + [2;2; 2] + [1;2; 3] + [2;3; 1]14 A3 [1; 1; 1] + [2;2; 2] + [1;3; 3]15 2A2 + A1 [1; 1; 1] + [2; 2; 2] + [1; 2; 3] + [2; 3;1] + [1; 3;2]16 A3 + A1 [1; 1; 1] + [2;2; 2] + [1;3; 3] + [2;1; 3]17 D4(a1) [1;1;1] + [2;1;1] + [1;2;2] [2; 2;2] + [1;3; 3]

Example 5.3. Consider he case (E7, 4). The graded Lie algebra of type E7 is

g(E7) = g4 g3 g2 g1 g0 g1 g2 g3 g4

withg0 = Csl(2) sl(3) sl(4), g1 = C2 C3 C4, g2 =

2C2 2

C3 2

C4,

g3 = S2,1C2 3

C3 3

C4, g4 = S2,2C

2 S2,1,1C3 4

C4.

Let {e1, e2} be a basis of E, and {f1, f2, f3}, {h1, h2, h3, h4} bases of F, H re-spectively. We label ea fi hu by [a; i; u].

The invariant scalar product ong restricted to g1 is

([a; i; u], [b;j; v]) = 1

where = #({a} {b}) + #({i} {j}) + #({u} {v}).There are six H-nondegenerate orbits. They can be described by observing that

the castling transform establishes a bijection between H-nondegenerate orbits andH-nondegenerate orbits for the 2 3 2 matrices corresponding to representationE F H. The six orbits in the representation E F H are: generic, hy-perdeterminant hypersurface and four F-degenerate orbits, coming from 2 2 2matrices: generic, hyperdeterminant and two determinantal varieties. Combiningthis knowledge with the case (E6, 4) we get 23 orbits in our representation.


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16 JERZY WEYMAN


2 2A1 [1; 1; 1] + [2;2; 1]3 2A1 [1; 1; 1] + [2;1; 2]4 2A1 [1; 1; 1] + [1;2; 2]5 3A1 [1; 1; 1] + [1;2; 2] + [2;1; 2]6 3A1 [1; 1; 1] + [1;2; 2] + [1;3; 3]7 A2 [1; 1; 1] + [2;2; 2]8 A2 + A1 [1; 1; 1] + [2;2; 2] + [1;3; 2]9 A2 + A1 [1; 1; 1] + [2;2; 2] + [1;2; 3]

10 A2 + 2A1 [1; 1; 1] + [2;2; 2] + [1;3; 2] + [2;3; 1]11 A2 + 2A1 [1; 1; 1] + [2;2; 2] + [1;2; 3] + [2;1; 3]12 A2 + 2A1 [1; 1; 1] + [2;2; 2] + [1;2; 3] + [1;3; 2]13 2A2 [1; 1; 1] + [2;2; 2] + [1;2; 3] + [2;3; 1]14 A3 [1; 1; 1] + [2;2; 2] + [1;3; 3]

15 2A2 + A1 [1; 1; 1] + [2; 2; 2] + [1; 2; 3] + [2; 3;1] + [1; 3;2]16 A3 + A1 [1; 1; 1] + [2;2; 2] + [1;3; 3] + [2;1; 3]17 D4(a1) [1; 1 ; 1] + [2; 1 ; 1] + [1; 2 ; 2] [2; 2;2] + [1;3; 3]18 2A2 [1; 1; 1] + [2;2; 2] + [2;1; 3] + [1;2; 4]19 A3 + A1 [1; 1; 1] + [2;2; 2] + [1;3; 3] + [1;2; 4]20 A3 + A2 [1; 1; 1] + [2; 2; 2] + [1; 3; 3] + [2; 1;3] + [1; 2;4]21 A4 [1; 1; 1] + [2;2; 2] + [1;3; 3] + [2;1; 4]22 A4 + A1 [1; 1; 1] + [2; 2; 2] + [1; 3; 3] + [2; 1;4] + [1; 2;4]23 A4 + A2 [1; 1; 1] + [2; 2; 2] + [1; 3; 3] + [2; 1;4] + [1; 2;4] + [2; 3;1]

Example 5.4. Consider the case (E8, 4).The graded Lie algebra of type E8 is

g(E8) = g6 g5 g4 g3 g2 g1 g0 g1 g2 g3 g4 g5 g6

withg0 = Csl(2) sl(3) sl(5), g1 = C2 C3 C5, g2 =2

C2 2

C3 2

C5,

g3 = S2,1C2 3

C3 3

C5, g4 = S2,2C

2 S2,1,1C3 4

C5, g5 = S3,2C

2

S2,2,1C3 5

C5, g6 = S3,3C

2 S2,2,2C3 S2,14C

5.Let {e1, e2} be a basis of E, and {f1, f2, f3}, {h1, h2, h3, h4, h5} bases of F, H

respectively. We label ea fi hu by [a; i; u].The invariant scalar product ong restricted to g1 is

([a; i; u], [b;j; v]) = 1

where = #({a} {b}) + #({i} {j}) + #({u} {v}).


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2 2A1 [1; 1; 1] + [2;2; 1]3 2A1 [1; 1; 1] + [2;1; 2]4 2A1 [1; 1; 1] + [1;2; 2]5 3A1 [1; 1; 1] + [1;2; 2] + [2;1; 2]6 3A1 [1; 1; 1] + [1;2; 2] + [1;3; 3]7 A2 [1; 1; 1] + [2;2; 2]8 A2 + A1 [1; 1; 1] + [2;2; 2] + [1;3; 2]9 A2 + A1 [1; 1; 1] + [2;2; 2] + [1;2; 3]

10 A2 + 2A1 [1; 1; 1] + [2;2; 2] + [1;3; 2] + [2;3; 1]11 A2 + 2A1 [1; 1; 1] + [2;2; 2] + [1;2; 3] + [2;1; 3]12 A2 + 2A1 [1; 1; 1] + [2;2; 2] + [1;2; 3] + [1;3; 2]13 2A2 [1; 1; 1] + [2;2; 2] + [1;2; 3] + [2;3; 1]14 A3 [1; 1; 1] + [2;2; 2] + [1;3; 3]15 2A2 + A1 [1; 1; 1] + [2; 2; 2] + [1; 2; 3] + [2; 3;1] + [1; 3;2]16 A3 + A1 [1; 1; 1] + [2;2; 2] + [1;3; 3] + [2;1; 3]17 D4(a1) [1; 1 ; 1] + [2; 1 ; 1] + [1; 2 ; 2] [2; 2;2] + [1;3; 3]18 2A2 [1; 1; 1] + [2;2; 2] + [2;1; 3] + [1;2; 4]19 A3 + A1 [1; 1; 1] + [2;2; 2] + [1;3; 3] + [1;2; 4]20 A3 + A2 [1; 1; 1] + [2; 2; 2] + [1; 3; 3] + [2; 1;3] + [1; 2;4]21 A4 [1; 1; 1] + [2;2; 2] + [1;3; 3] + [2;1; 4]22 A4 + A1 [1; 1; 1] + [2; 2; 2] + [1; 3; 3] + [2; 1;4] + [1; 2;4]23 A4 + A2 [1; 1; 1] + [2; 2; 2] + [1; 3; 3] + [2; 1;4] + [1; 2;4] + [2; 3;1]24 A5 [1; 1; 1] + [2; 2; 2] + [1; 3; 3] + [2; 1;4] + [1; 2;5]25 A6 [1; 1; 1] + [2; 2; 2] + [1; 3; 3] + [2; 1;4] + [1; 2;5] + [2; 3;1]

Exercises for section 5. Analyze one of the smaller cases, for example (D5, 2).

6. Triple tensor products: quiver approach.

In the case of (E6, 4), (E7, 4), (E8, 4) we are mostly interested in, there isanother approach using representations of Kronecker quiver.

The Kronecker quiver is the quiver

Q :

a

1 2b


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18 JERZY WEYMAN

A representation V ofQ is just a pair of vector spaces {V(1), V(2)} and a pair oflinear maps. Each representation V ofQ has dimension vector = (dimV(1),dimV(2)) Z2. The representations of Q form an Abelian category (in fact a category of KQ-

modules where KQ is a path algebra of Q). By Krull-Remak-Schmidt theoremeach representation decomposes uniquely (up to permutation of factors) to a directsum of indecomposable representations.

In the case of Kronecker quiver one can list all indecomposable representations.The are as follows.

In each dimension vector (n, n) there is P1 worth of representations, with thestandard A1 is given by

V(1) = Kn, V(2) = Kn, V(a) = Id,V(b) = Jn

where Jn is an n n Jordan block of dimension n. We denote these representationsby Vn .

Moreover, there are two discrete infinite families: {Pn} of dimension vector(n, n + 1), n 0, and {Qn} of dimension vector (n + 1, n), n 0), defined asfollows

Pn(a) =

1 0 . . . 0 00 1 . . . 0 0

. . . . . . . . . . . . . . .0 0 . . . 0 10 0 . . . 0 0

Pn(b) =

0 0 . . . 0 01 0 . . . 0 0

. . . . . . . . . . . . . . .0 0 . . . 1 00 0 . . . 0 1

The matrices of Qn are just transposes of the matrices of Pn.Proof. The idea is to introduce reflection functors C+(2) and C(1), and their

compositions and 1. They take Q into itself. Thus they take indecomposablesinto indecomposables. C+(2) takes dimension vector (m, n) into (2m n, m). So,the vector (p+1, p) goes to (p+2, p+1) and (p, p+1) goes to (p1, p). Respectively,C(1) takes (m, n) into (n, 2n m). This means (q, q +1) goes to (q + 1, q +2) and(q + 1, q) goes to (q, q 1).

Notice that the action of both functors on dimenson vectors (m, n) preservem n. This means there are no indecomposables in dimension vectors (m, n) with|m n| 2, as they would imply the existence of indecomposables in dimensionvectors (|m n|, 0) or (0, |m n|). In the dimension vectors (p + 1, p) and (p, p + 1)there is one indecomposable, as it is the case in dimension vectors (1 , 0) and (0, 1).Finally, in dimension vector (n, n), assuming one of the matrices is nonsingular, wecan use it to identify V(1) and V(2) so the description of indecomposables follows


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from Jordan canonical form. This is the case if any linear combination of V(a) andV(b) is nonsingular. If not, then there has to be subspace of dimension p + 1 inV(1) that is taken to a subspace of dimension p in V(2) by both V(a) and V(b)(apply Hilbert-Mumford criterion-exercise), which means V s not indecomposable,as , after applying reflection functors we reduce to a representation of dimension(n, n) having a subrepresentation of dimension (1, 0) which is injective, so it splits.

To a tensor in C2 Cp Cq we can associate the representation V() bychoosing a fixed basis {e1, e2} of C

2, write = e1 (a) + e2 (b) and takeV()(a) = (a), V((b) = (b). Alternatively, the group GL(2) acts on the set ofrepresentations ofQ of dimension vector (p, q) by replacing V(a), V(b) by the linearcombinations of these maps, and the orbits in C2 Cp Cq are the GL(2)-orbitsof this action. In particular the P1 families become one GL(2)-orbit, and we getfinitely many possibilities, as long as the numer of summands of that type is 3.

In this language our three tables have the following meaning.

(E6, 4).

number representative quiver0 0 3P0 3Q0

1 [1; 1; 1] V1 2P0 2Q0

2 [1; 1; 1] + [2; 2; 1] P1 P0 2Q0

3 [1; 1; 1] + [2; 1; 2] Q1 2P0 Q0

4 [1; 1; 1] + [1; 2; 2] 2V1 P0 Q0

5 [1; 1; 1] + [1; 2; 2] + [2; 1; 2]6 [1; 1; 1] + [1; 2; 2] + [1; 3; 3]7 [1; 1; 1] + [2; 2; 2]8 [1; 1; 1] + [2; 2; 2] + [1; 3; 2]9 [1; 1; 1] + [2; 2; 2] + [1; 2; 3]

10 [1; 1; 1] + [2; 2; 2] + [1; 3; 2] + [2; 3; 1]11 [1; 1; 1] + [2; 2; 2] + [1; 2; 3] + [2; 1; 3]

12 [1; 1; 1] + [2; 2; 2] + [1; 2; 3] + [1; 3; 2]13 [1; 1; 1] + [2; 2; 2] + [1; 2; 3] + [2; 3; 1]14 [1; 1; 1] + [2; 2; 2] + [1; 3; 3]15 [1; 1;1] + [2; 2;2] + [1; 2;3] + [2;3;1] + [1;3;2]16 [1; 1; 1] + [2; 2; 2] + [1; 3; 3] + [2; 1; 3]17 [1; 1;1] + [2; 1;1] + [1; 2;2] [2; 2;2] + [1; 3;3]

We end this section with some remarks on the castling transform. This is aversion of the reflection functors, suitable for the triple tensor products. The methodallows to classify the cases when the triple tensor product has an open orbit. Thismethod goes back at least to Sato-Kimura [SK77] who classified the irreduciblerepresentations of reductive groups with an open orbit.


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When p = 2 we have

N(p, q, r) = 4 + q2 + r2 2qr = 4 + (q r)2 4.

Detailed case by case analysis shows that in the cases 3 p q r the value ofN is never 2.In the cases with p = 1 example 5 shows that the algebra A3(1, q , r) is finite

dimensional.Let us look at the examples with p = 2. We will think of C2 Cq Cr as of

the set of representations of Kronecker quiver of dimension vector ( q, r) this willallow us to identify the generic tensors. The generic tensors can be determined bythe generic decomposition. The indecomposables occur in dimensions (n, n) (oneparameter families) and (n, n + 1) and (n + 1, n). Thus canonical decompositionof (n, n + k) is (k v)(u, u + 1) + v(u + 1, u + 2) where n = uk + v is divisionwith remainder. Similarly (n + k, n) = (k v)(u + 1, u) + v(u + 2, u + 1). Thisdecomposition allows to write easily the generic tensors.

Exercises for section 6. 1. Apply Hilbert-Mumford criterion to prove that if arepresentation V of the Kronecker quiver has the property that for each , thelinear combination V(a) + V(b) is singular, then there exist p, 0 p n 1and a subspace W of dimension p + 1 in V(1) such that

dim (V(a)W + V(b)W) p.

2. Fill the rest of the entries in the table (E6, 4)

7. Geometric technique.In this section we provide a quick description of main facts related to geometric

technique of calculating syzygies. We work over an algebraically closed field K.Let us consider the projective variety V of dimension m. Let X = AN

Cbe the

affine space. The space XV can be viewed as a total space of trivial vector bundleE of dimension n over V. Let us consider the subvariety Z in X V which is thetotal space of a subbundle S in E. We denote by q the projection q : X V Xand by q the restriction of q to Z. Let Y = q(Z). We get the basic diagram

Z X V q q

Y X


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22 JERZY WEYMAN

The projection from X V onto V is denoted by p and the quotient bundle E/Sby T. Thus we have the exact sequence of vector bundles on V

0 S E T 0

The dimensions of S and T will be denoted by s, t respectively. The coordinatering of X will be denoted by A. It is a polynomial ring in N variables over C. Wewill identify the sheaves on X with A-modules.

The locally free resolution of the sheaf OZ as an OXV-module is given by theKoszul complex

K() : 0 t

(p) . . . 2

(p) p() OXV

where = T. The differentials in this complex are homogeneous of degree 1 in thecoordinate functions on X. The direct image p(OZ) can be identified with the thesheaf of algebras Sym() where = S.

The idea of the geometric technique is to use the Koszul complex K() toconstruct for each vector bundle V on V the free complex F(V) ofA-modules withthe homology supported in Y. In many cases the complex F(OV) gives the freeresolution of the defining ideal of Y.

For every vector bundle V on V we introduce the complex

K(, V) := K() OXV pV

This complex is a locally free resolution of the OXV-module M(V) := OZ pV.

Now we are ready to state the basic theorem (Theorem (5.1.2) in [W]).

Theorem 7.1. . For a vector bundle V on V we define a free graded A-modules

F(V)i =j0

Hj(V,

i+j V) k A(i j)

a) There exist minimal differentials

di(V) : F(V)i F(V)i1

of degree 0 such thatF(V) is a complex of graded free A-modules with

Hi(F(V)) = RiqM(V)

In particular the complexF(V) is exact in positive degrees.


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b) The sheaf RiqM(V) is equal to Hi(Z, M(V)) and it can be also identified with

the graded A-module Hi(V,Sym() V).c) If : M(V) M(V)(n) is a morphism of graded sheaves then there exists a

morphism of complexes

f() : F(V) F(V

)(n)Its induced map Hi(f()) can be identified with the induced map

Hi(Z, M(V)) Hi(Z, M(V))(n).

IfV is a one dimensional trivial bundle on V then the complex F(V) is denotedsimply by F.

The next theorem gives the criterion for the complex F to be the free resolutionof the coordinate ring of Y.

Theorem 7.2.. Let us assume that the map q : Z Y is a birational isomor-phism. Then the following properties hold.

a) The module qOZ is the normalization ofC[Y].

b) IfRi

qOZ = 0 for i > 0, thenF is a finite free resolution of the normalization

ofC[Y] treated as an A-module.

c) If RiqOZ = 0 for i > 0 andF0 = H0(V,0

) A = A then Y is normal andit has rational singularities.

This is Theorem (5.1.3) in [W].The complexes F(V) satisfy the Grothendieck type duality. Let V denote the

canonical divisor on V.

Theorem 7.3. LetV be a vector bundle on V. Let us introduce the dual bundle

V = V t

V.

ThenF

(V

) =F

(V)

[m t]This is Theorem (5.1.4) in [W].In all our applications the projective variety V will be a Grassmannian. To fix the

notation, let us work with the Grassmannian Grass(r, E) of subspaces of dimensionr in a vector space F of dimension n. Let

0 R E Grass(r, E) Q 0

be a tautological sequence of the vector bundles on Grass(r, E). The vector bundle will be a direct sum of the bundles of the form S1,...,nrQ S1,...,rR. Thusall the exterior powers of will also be the direct sums of such bundles. We willapply repreatedly the following result to calculate cohomology of vector bundlesS1,...,nrQ S1,...,rR.


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24 JERZY WEYMAN

Proposition 7.4 (Botts algorithm). . The cohomology of the vector bundleS1,...,nrQ S1,...,rR on Grass(r, E) is calculated as follows. We look at theweight

(, ) = (1, . . . , nr, 1, . . . , r)

and add to it = (n, n1, . . . , 1). Then one of two mutually exclusive cases occurs.a) The resulting sequence

(, ) + = (1 + n , . . . nr + r + 1, 1 + r, . . . , r + 1)

has repetitions. In such case

Hi(Grass(r, E), SQ SR) = 0

for all i 0.b) The sequence (, ) + has no repetitions. Then there is a unique permutation

w n that makes this sequence decreasing. The sequence = w((, ) + ) is a non-increasing sequence. Then the only non-zero cohomology group of the

sheaf SQ SR is the group Hl

where l = l(w) is the length of w. We have

Hl(Grass(r, E), SQ SR) = SE.

Here SE denotes the highest weight representationSE ofGL(E) correspondingto the highest weight (so-called Schur module).

8. Singularities, defining ideals and syzygies.

The geometric method of calculating syzygies [W03] is applicable to analyze thesingularities and defining ideals of orbit closures. We will look more closely at thethree cases (E6, 4), (E7, 4) and (E8, 4).

Let us recall our basic notions. The coordinate ring of the triple tensor productis A = Sym(E F G) = C[Xi,j,k]. For an orbit closure Y = O(t) we defineits coordinate ring

C[Y] = A/J, J = {f A | f|O(t) = 0}.

The ideal J is called the defining ideal of Y.In order to find the data for some orbit closure we need to find a desingularization

of each orbit closure and then use the push-down of the Koszul complex to calculatethe terms of a minimal free resolution. It is possible in most of the cases (and allcases for triple tensor products).

Before we start talking about orbit closures for triple tensor products let usintroduce the concept of a degenerate orbit.


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Definition 8.1. The GL(E) GL(F) GL(G)-orbit of a tensor t E F Gis E-degenerate (resp. F, G-degenerate) if there exists a proper subspace E E(resp. F F, G G) such that t E F G (resp. E F G, E F G).

The point is that for an E (resp. F, G)- degenerate obit closure we can get an

estimate on the free resolution of the coordinate ring of the orbit closure as follows.Without loss of generality we can assume that dim E = dim E 1.

Assume that we know the terms F of the free resolution of the coordinate ring

of the orbit closure Y = O(t) where O(t) is the closure of the orbit of a tensor ttreated as an element of E F G.

Consider the desingularization of the subspace variety

Z = {(t, R) E F G Grass(m 1, E) | r R F G}

The tautological sequence associated to the projective space is

0 R E Grass(m 1, E) Q 0

Working over the sheaf of rings B = Sym(R

F

G

) we can produce the exactsequence F(R, F , G) of B modules modeled after F, whose terms are SR

SF SG B(s). Taking sections we get an exact complex of A-modules

supported in the subspace variety q(Z). Each of the modules

H0(Grass(m 1, E), SR SF

SG B)

has a resolution with the terms

F(, , ) = H(Grass(m 1, E),

)

where = Q F G. We can use the cone construction to find a (non-minimal)resolution of the coordinate ring of O(t) in E F G.

Example 8.2. Look at the orbitO10 in the case (E6, 4). It is clearlyG-degenerate.In fact this is a general G-degenerate orbit, so we get its desingularization usingthe Grassmannian Grass(2, G) with the tautological sequence

0 R G Grass(2, G) Q 0

and the bundles = E F R, = E F Q. the resolutionF has theterms

F0 = A,F1 =3

(E F) S1,1,1G A(3),

F2 =

4(E F) S2,1,1G

A(4),


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26 JERZY WEYMAN

F3 =

5(E F) S3,1,1G

A(5),

F4 =6

(E F) S4,1,1G A(6),

and our complex is just the Eagon-Northcott complex asscociated to the 33 minorsof a 3 6 matrix we get when we flatten our 3-dimensional matrix

Example 8.3. Consider the orbitO8 in the case (E6, 4). This is a G-degenerateorbit which corresponds to an orbit of codimension one inEFG. The resolutionof this orbit closure is given by an invariant of degree 6, i.e. the complex

0 S3,3E S2,2,2F

S3,3G A(6) A.

We construct a complex of sheaves F over B = Sym(E F R) with

F1 = S3,3E S2,2,2F

S3,3R B, F0 = B.

Bott theorem implies both sheaves have no higher cohomology, so we get the exact

sequence

0 H0(Grass(2, G), F1) H0(Grass(2, G), F0) C[O8].

Now we can find the resolutions of both modules using the geometric construc-tion from the previous example. The resolution of H0(Grass(2, G), F0) was workedout in the previous example. The resolution of H0(Grass(2, G), F1) is (use V =S3,3E

S2,2,2F S3,3R and from previous example).

G0 = S3,3E S2,2,2F

S3,3,0G A(6),

G1 = S3,3E S2,2,2F

(E F) S3,3,1G A(7),

G2 = S3,3E

S2,2,2F

2

(E

F

) S3,3,2G

A(8),

G3 = S3,3E S2,2,2F

3

(E F) S3,3,3G A(9),

G4 = S3,3E S2,2,2F

6

(E F) S4,4,4G A(10).

The mapping coneG F

gives a (possibly non-minimal) resolution of the coordinate ringC[O8]. In his casethe length of our resolution is 5, i.e. equal to codimension of O8 so the orbit closureis Cohen-Macaulay.


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28 JERZY WEYMAN

in fact a proof that the incidence space Z(15) is a desingularization. Dividing bythe regular sequence of 16 generic linear forms gives a resolution of the ring withHilbert function

1 3t6 + 2t9

(1 t)2 = 1 + 2t + 3t

2

+ 4t

3

+ 5t

4

+ 6t

5

+ 4t

6

+ 2t

7

.

This means the degree of the orbit closure is 21.

Example 8.6. The codimension 3 orbitO14. This orbit closure has a nice geomet-ric description. It consists of pencils of3 3 matrices of linear forms containing amatrix of rank 1.

The bundle is the submodule with the weights

(0, 1; 0, 0, 1; 0, 0, 1), (0, 1; 0, 1, 0; 0, 0, 1), (0, 1; 0, 0, 1; 0, 1, 0),

(0, 1; 0, 1, 0; 0, 1, 0), (0, 1; 1, 0, 0; 0, 0, 1), (0, 1; 0, 0, 1; 1, 0, 0),

(0, 1; 1, 0, 0; 0, 1, 0), (0, 1; 0, 1, 0; 1, 0, 0).

This bundle can be thought of as the set of weights in the graphic form as follows.

X X XX X XX X X

X O OO O OO O O

Here the first matrix represents the weights with(1, 0) on the first coordinate, andthe second matrix represents the weghts with (0, 1) on the first coordinate. Thesymbol X denotes the weight in , the symbol O-the weight in . Our incidencespace has dimension 10 + 2 + 2 + 1 = 15, so it projects on the orbit of codimension3. It is clearly O14.

The calculation of cohomology reveals that we get a complexF(14)

0 (5, 1; 2, 2, 2; 2, 2, 2) A(6) (3, 1; 2, 1, 1; 2, 1, 1) A(4)

(2, 1; 1, 1, 1; 2, 1, 0) A(3) (2, 1; 2, 1, 0; 1, 1, 1) A(3)

(1, 1; 1, 1, 0; 1, 1, 0) A(2) A.

The terms of the complex F(14) can be understood by looking at the partitionson two last coordinates. They give terms of the Gulliksen-Nagard complex resolving2 2 minors of the generic 3 3 matrix. The corresponding term on the firstcoordnate comes from the symmetric power on the bundle QE . This means theterm comes from H1 on the P(E) coordinate (except the trivial term), hence theshift in homological degree. The orbit is not normal. One still needs to resolve thecokernel module. This is an important problem, as similar situation occurs in severalother cases.


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Example 8.7. The codimension 4 orbit O13. The geometric description of thisorbit can be understood from the quiver point of view. Our representation can bethought of as the set of representations of Kronecker quiver of dimension vector(3, 3).

C3

C3.

Our orbit closure then is the set of representations having a subrepresentationof dimension vector (2, 1). The desingularization Z(13) lives on Grass(1, F) Grass(1, G) and the bundle is

= E QF QG .

Graphically we have

X X XX O O

X O O

X X XX O O

X O O

.

Our incidence space has dimension 10 + 2 + 2 = 14, so it projects on the orbitof codimension4. In this case one can see directly that Z(13) is a desingularizationof O13.

The calculation of cohomology reveals that we get a complexF(13)

0 (4, 4; 3, 3, 2; 3, 3, 2) (4, 3; 3, 2, 2; 3, 2, 2)

(2, 2; 2, 1, 1; 2, 1, 1) (3, 1; 2, 1, 1; 2, 1, 1) (3, 2; 2, 2, 2; 2, 2, 2)

(2, 1; 1, 1, 1; 1, 1, 1) (2, 1; 1, 1, 1; 2, 1, 0) (2, 1; 2, 1, 0; 1, 1, 1) (3, 0; 1, 1, 1; 1, 1, 1)

(1, 1; 1, 1, 0; 1, 1, 0)) (0, 0; 0, 0, 0; 0, 0, 0).

The orbit closure is obviously not normal. The complex F(13) gives a minimalresolution of the normalization N(O13). We have an exact sequence

0 C[O13] C[N(O13)] C(13) 0.

The complexF(13) reveals that the A-module C(13) has the presentation

(2, 1; 1, 1, 1; 1, 1, 1) (2, 1; 1, 1, 1; 2, 1, 0) (2, 1; 2, 1, 0; 1, 1, 1) (1, 1; 1, 1, 0; 1, 1, 0).

Indeed, the representation (3, 0; 1, 1, 1; 1, 1, 1) can map only into the trivial term inthe complexF(13).


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30 JERZY WEYMAN

We can look for the module with the above presentation among twisted modulessupported in smaller orbits. It turns out the right choice is the orbit closure O7. Itsbundle has a diagram

X X O

X X OO O O

X X O

X X OO O O

.

This bundle lives on the space Grass(2, F) Grass(2, G). Consider the twisted

complexF(2

E 2 RF

2 RG)(7). Its terms are:

0 (6, 6; 4, 4, 4; 4, 4, 4)

(5, 4; 3, 3, 3; 4, 3, 2) (5, 4; 4, 3, 2; 3, 3, 3) (6, 3; 3, 3, 3; 3, 3, 3)

(4, 4; 3, 3, 2; 3, 3, 2)(4, 4; 3, 3, 2; 4, 2, 2)(4, 4; 4, 2, 2; 3, 3, 2)2(5, 3; 3, 3, 2; 3, 3, 2)

(4, 3; 3, 2, 2; 3, 2, 2) (4, 3; 3, 2, 2; 3, 3, 1) (4, 3; 3, 3, 1; 3, 2, 2) (5, 1; 2, 2, 2; 2, 2, 2)

(2, 2; 2, 1, 1; 2, 1, 1) (3, 1; 2, 1, 1; 2, 1, 1) (3, 3; 2, 2, 2; 3, 3, 0) (3, 3; 3, 3, 0; 2, 2, 2)

(2, 1; 1, 1, 1; 1, 1, 1) (2, 1; 1, 1, 1; 2, 1, 0) (2, 1; 2, 1, 0; 1, 1, 1) (1, 1; 1, 1, 0; 1, 1, 0).

The resolution of he A-module C[O13] can be constructed as a mapping cone ofthe map

F(13) F(2

E2

QF 2

QH)(7)

covering the natural epimorphism of A-modules. The mapping cone is not a min-imal resolution but the repeating representations might cancel out. Let us see thatthe pairs of representations (2, 2; 2, 1, 1; 2, 1, 1) and (3, 1; 2, 1, 1; 2, 1, 1) cancel out.From this we can deduce that the defining ideal of O13 is generated by the repre-sentations (3, 0; 1, 1, 1; 1, 1, 1) in degree 3 and the representations (3, 3; 2, 2, 2; 3, 3, 0)and (3, 3; 3, 3, 0; 2, 2, 2) in degree 6.

Indeed, if (3, 1; 2, 1, 1; 2, 1, 1) would not cancel out, it would contribute to theminimal generators of the defining ideal. However it occurs once in S4(E

F G) so that representation is already in the ideal generated by (3, 0; 1, 1, 1; 1, 1, 1).Regarding representation (2, 2; 2, 1, 1; 2, 1, 1), if it would occur in the defining idealof O13 the analysis of the next section will show that then the defining ideal ofO13 would contain the defining ideal of O12. But this is impossible since both orbitclosures have the same dimension. The homological dimension ofC[O13] as an A-module equals 5 because the top of the resolution of C(13) does not cancel out. Sothis coordinate ring is not Cohen-Macaulay. Also the non-normality locus of O13equals to O7.


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2010 summer school

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