DOI 10.1007/s10801-009-0175-2
Simple SL(n)-modules with normal closures of maximal torus orbits
Karine Kuyumzhiyan
Received: 16 August 2008 / Accepted: 24 February 2009 / Published online: 16 March 2009
© Springer Science+Business Media, LLC 2009
Abstract LetT be the subgroup of diagonal matrices in the groupSL(n). The aim of this paper is to find all finite-dimensional simple rationalSL(n)-modulesV with the following property: for each pointv∈V the closureT vof itsT-orbit is a normal affine variety. Moreover, for anySL(n)-module without this property aT-orbit with non-normal closure is constructed. The proof is purely combinatorial: it deals with the set of weights of simpleSL(n)-modules. The saturation property is checked for each subset in the set of weights.
Keywords Toric variety·Normality·Saturation
Introduction
LetT be an algebraic torus defined over an algebraically closed fieldkof characteris- tic zero. Recall that an irreducible algebraicT-varietyXis called toric ifXis normal andT acts onXwith an open orbit. This class of varieties plays an important role in algebraic geometry, topology and combinatorics due to its remarkable description in terms of convex geometry, see [6]. Assume that the torusT acts on a varietyY. Then the closureX=T yof theT-orbit of a pointy∈Y is a natural candidate to be a toric variety. To verify it, one should check thatXis normal.
During last decades, normality of torus orbits’ closures was an object of numer- ous investigations. For example, letGbe a semisimple algebraic group with a Borel subgroupBand a maximal torusT ⊂B. In [10], it was proved that the closure of a generalT-orbit on the flag varietyG/Bis normal. Later it was shown that the closure of a generalT-orbit inG/P, whereP ⊂Gis a parabolic subgroup, is also normal,
K. Kuyumzhiyan (
)Department of Higher Algebra, Faculty of Mechanics and Mathematics, Moscow State University, 119991 Moscow, Russia
e-mail:[email protected]
see [5]. Examples of non-normal closures of non-general torus orbits can be found in [3].
Now let us consider a finite-dimensional rationalT-moduleV. There exists an easy combinatorial criterion of normality ofT vfor a vectorv∈V. Namely, letv= vχ1+ · · · +vχm,vχi=0, be the weight decomposition of the vectorv. Consider the corresponding set ofT-weightsχ1, . . . , χm. If we takeχ1, . . . , χmas elements of the character latticeX(T ), we can generate a semigroupZ+(χ1, . . . , χm), a sublattice Z(χ1, . . . , χm), and a rational polyhedral coneQ+(χ1, . . . , χm). The setχ1, . . . , χm is called saturated ifZ+(χ1, . . . , χm)=Z(χ1, . . . , χm)∩Q+(χ1, . . . , χm). It is well known (see [9, page 5]) that the following two conditions are equivalent: the set {χ1, . . . , χm}is saturated and the closureT vof theT-orbitT vis normal. There is an analogous criterion for theT-action on the projectivisationP(V ), see [3].
The saturation property occurs in many algebraic and geometric problems. In [17], it was proved that the set of incidence vectors of the bases of a realizable matroid is saturated. The geometric conclusion of this fact is that for any pointy in the affine cone over the classical Grassmannian Gr(k, n)the closureT yis normal.
Taken a finite graphwithnvertices, one can associate a finite collectionM() of vectors in the latticeZnwith it:
M()= {εi+εj:(ij )is an edge of},
whereε1, ε2, . . . , εnis the standard basis ofZn. The saturation property for this set is equivalent to the fact that for two arbitrary minimal odd cyclesC andC in, eitherC andChave a common vertex or there exists an edge ofjoining a vertex ofC with a vertex ofC (see [12] and [13]). Algebraically, the saturation property for M()is equivalent to the integral closureness for the subalgebra A() of the polynomial algebrak[x1, x2, . . . , xn],
A()=k[xixj:(ij )is an edge of], in its field of fractionsQA().
Some general results concerning quivers and the saturation property were obtained in [4]. It was shown that a finite, connected quiverQ without oriented cycles is a Dynkin or Euclidean quiver if and only if all orbit semigroups of representations of Qare saturated.
In the paper [11], the following problem is solved. LetGbe a semisimple algebraic group with a maximal torusT andV be its adjoint module. For whichGis the closure T vnormal for allv∈V? The surprising fact is that forG=SL(n)this is always the case (see [11, Thm.1], and also [15, Ex. 3.7], [14] and [2, Prop.2.1]). In [2], this combinatorial result is interpreted in terms of representations of quivers.
The aim of this paper is to classify all simple finite-dimensional rationalSL(n)- modulesV such that for anyv∈V the closureT vis normal.
Main Theorem The representations below, together with their dual, form the list of all irreducible representations ofSL(n)where all maximal torus orbits’ closures are normal:
(1) the tautological representation ofSL(n);
(2) the adjoint representation ofSL(n);
(3) exceptional cases:
Group Highest weight G-module
SL(2) 3π1 S3k2
SL(2) 4π1 S4k2
SL(3) 2π1 S2k3
SL(4) π2 2k4
SL(5) π2 2k5
SL(6) π2 2k6
SL(6) π3 3k6
The paper is organized as follows. In Section1we give some algebraic definitions and reformulate the problem in combinatorial terms. From that point, it remains to check the saturation property for any subset in the system ofT-weights of a simple SL(n)-module. In Section 2 we prove that the saturation property holds for each subset in the set of weights of the representations listed in the Main Theorem. The most powerful methods used here are two criteria given in [15, Thm. 3.5] and [15, Thm. 3.8]. Some reasoning uses the graph theory language. Our reference for graph theory is [7]. In Section3 we produce non-saturated subsets in sets of weights for all other representations. If the set of weights of the representation with the highest weightλis a subset in the set of weights of the representation with the highest weight μ, and a non-saturated subset forλis known, then one can use it as a non-saturated subset for μ. Fundamental representations form the most difficult class. To work with them, we use the following observation. If a non-saturated subset in the set of weights of thekth fundamental representation ofSL(n)is found, then the analogous non-saturated subset exists in the set of weights of thekth fundamental representation ofSL(n+k).
Another motivation to study normality of torus orbits’ closures comes from a more general problem. LetGbe a connected reductive group with a maximal unipotent subgroupUnormalized by a maximal torusT. Assume thatGacts on an affine va- rietyX. Then the passage toU-invariants allows to reduce the question of normality of a sphericalG-orbit closure onXto the question of normality of aT-orbit closure onX//U, see [1] for a partial realization of this approach.
In further publications, we plan to give a classification of simpleG-modules with normalT-orbit closures for other simple algebraic groupsG.
1 Algebraic background and notation
LetV be a finite-dimensional rational T-module. Given any character χ from the character latticeX(T ), define a weight subspaceVχ asVχ= {v∈V :t·v=χ (t )v}. It is well known thatV =
χ∈X(T )
Vχ, and only finitely manyVχ are nonzero. The set {χ1, χ2, . . . , χk}of thoseχi for whichVχi=0 is called the system of weights of the T-moduleV.
LetZ+andQ+denote the sets of integer and rational non-negative numbers, re- spectively; and letv1, v2, . . . , vm∈Qn. Consider the semigroupZ+(v1, v2, . . . , vm)= {n1v1+n2v2+. . .+nmvm:ni ∈Z+}, the sublatticeZ(v1, v2, . . . , vm)= {z1v1+ z2v2+. . .+zmvm:zi∈Z}, and the rational polyhedral coneQ+(v1, v2, . . . , vm)= {q1v1+q2v2+. . .+qmvm: qi ∈Q+}. Define the following important property of the set{v1, v2, . . . , vm}.
Definition The set of points{v1, v2, . . . , vm} ⊂Qnis called saturated if Z+(v1, v2, . . . , vm)=Z(v1, v2, . . . , vm)∩Q+(v1, v2, . . . , vm).
The following result provides a well-known combinatorial criterion of normality of the torus orbit closure, see [9]:
Theorem 1.1 Consider a rational linear action of a torusT on a vector spaceV. Letv∈V, andv=vλ1 + · · · +vλs,vλi=0, be its weight decomposition. Then the closureT vis normal if and only if the set of characters{λ1, . . . , λs}is saturated.
Definition The set of points{λ1, λ2, . . . , λm} ⊂Qn is called hereditary normal if each its subset is saturated.
Corollary 1.2 Given a rational linear action of a torusT on a vector spaceV; let {λ1, . . . , λs}be the set of weights of this action. Then the closureT v is normal for eachv∈V if and only if{λ1, . . . , λs}is hereditary normal.
Remark The weight system is multiplied by−1 while changing a representationV of the torusT with its dual. Hence the property of normality of allT-orbits’ closures is preserved.
LetG=SL(n). We fix a maximal torus T ⊂Gconsisting of all diagonal ma- trices. An element a =(a1, a2, . . . , an) of the lattice Zn can be interpreted as a character χa of the torus T in the following way: χa(t )=t1a1t2a2. . . tnan, where t=diag(t1, t2, . . . , tn). Sincet1t2. . . tn=1, the pointsaandbdefine the same char- acter if and only ifa−b=α(1,1, . . . ,1). EachafromZnhas a unique representation a= ˜a+α(1,1, . . . ,1), wherea˜∈Qn,α∈Q, and
˜ ai=0.
Letε1, ε2, . . . , εnbe the standard basis of the latticeZn, and
ei=εi=
⎛
⎝−1 n,−1
n, . . . ,−1 n,n−1
ith placen ,−1
n, . . . ,−1 n
⎞
⎠.
Notice thate1, e2, . . . , en(further referred to as a quasi-basis) satisfy the only linear relation
e1+e2+. . .+en=0. (∗)
IdentifyX(T )with theZ-lattice generated bye1, e2, . . . , en. Recall that a weightχa is called dominant if and only ifa1a2. . .an. The root lattice forSL(n)is a
lattice generated by the vectorse1−e2,e2−e3, . . . , en−1−en. Due to the ambiguity of notation,
= {a1e1+a2e2+. . .+anen : n|(a1+a2+. . .+an)}, where|stands for divisibility, i.e.n|m ⇐⇒ ∃z∈Z, m=nz.
For a positive integers|n, define Z≡0(s)(e1, . . . , en)= n
i=1xiei:xi∈Z, s|n i=1xi
. In this notation the root lattice coincides withZ≡0(n)(e1, . . . , en).
LetV be a finite-dimensional simple rationalSL(n)-module, andM(V )be the system of weights of the moduleV with respect to the restricted actionT: V. In- troduce a partial order onM(V ):μν if and only if forξ=μ−ν the following conditions hold:ξ10,ξ1+ξ20,. . . , ξ1+ξ2+. . .+ξn−10. It is well known thatM(V )contains the only maximal elementλwith respect to , it is called the highest weight of the module. The weightλis dominant; moreover, for any dominant weightλ∈X(T )there exists a unique simpleSL(n)-moduleV (λ)with the highest weightλ(see [16, Chapter 4, §3, Thm.11] or [8, §§20–21]). The role of the Weyl groupW is played here by the permutation groupSn, which acts onZnby permuta- tions of coordinates. It is known that
M(λ):=M(V (λ))=conv{wλ:w∈W} ∩(λ+ ), where conv(L)denotes the convex hull of the setL⊂Rn; see [8, §21.3].
In our situation, Corollary1.2can be reformulated in the following way:
Proposition 1.3 LetV (λ) be a simple module of a semisimple groupG with the highest weightλ. Then the closure of eachT-orbit inV (λ)is normal if and only if M(λ)is hereditary normal.
2 Positive results
In this section we prove that certain sets of weights are hereditary normal. First, let us present some machinery which proves that a given set of pointsLis hereditary normal. Choose a basis in which for anyv∈Lits coordinates are integer. Represent each point as a column vector of its coordinates in this basis, and letK=K(L)be an integerp×q matrix formed by all these column vectors. The first algorithm deals with the objects which we had in the very beginning – with the monomial algebras.
Definition To each columnKi=(k1i, . . . , kpi)T we put into correspondence a Lau- rent monomialtKi=t1k1i. . . tpkpi. The toric idealIK associated withK is the kernel of thek-algebra homomorphism
C[x1, x2, . . . , xq] →C[t1, . . . , td, t1−1, . . . , td−1], xi→tKi.
Definition Suppose thatu+andu−are two vectors fromZp+with disjoint supports, and letf =xu+−xu−∈IK. We say thatf is a circuit inIK if the following two conditions hold:
• all the coordinates ofu+andu−have no common divisor greater than 1;
• the set of variables which actually occur inf is minimal with respect to inclusion among all the binomials ofIK.
Theorem 2.1 ([15, Thm. 3.8]) A set of pointsLis hereditary normal if and only if every circuit inIK(L)has at least one square-free monomial.
The next property ofK=K(L)deals only withK, which makes this method so easy.
Definition The matrixKis called unimodular if it has maximal rankp, and all non- zerop×pminors ofKhave the same absolute value.
If the rows ofKare linearly dependent, then one can omit the redundant rows and check unimodularity for the remaining matrix.
Theorem 2.2 ([15, Thm. 3.5]) IfK(L)is unimodular, thenLis hereditary normal.
2.1 The tautological representation
Its highest weight equals e1. The set M(e1)equals {e1, . . . , en}. It is unimodular, hence it is hereditary normal.
2.2 The adjoint representation
Its highest weightλ is equal toe1−en. Acting byW =Sn, we get all vectors of the formei−ej. Taking the convex hull adds only0 to this set. We get¯ M(λ)= {0} ∪ {ei−ej:1≤i, j≤n, i=j}. It was proved in [15, Ex. 3.7] or [11, Thm.1] that this set is hereditary normal.
2.3 The representation ofSL(2)with the highest weight 3π1
Here we haveM(λ)= 32,−32 ,
1 2,−12
, −12,12
,
−32,32
(in the usual ba- sis). After an appropriate change of basis,K(M(λ))=(3,1,−1,−3), the set of all circuits ofIK(M(λ))is{x1x33−1, x1x4−1, x1−x23, x2x3−1, x23x4−1, x33−x4}. By Theorem2.1, the setM(λ)is hereditary normal.
2.4 The representation ofSL(2)with the highest weight 4π1
One has to verify that the setM(λ)= {(2,−2), (1,−1), (0,0), (−1,1), (−2,2)}is hereditary normal. Here we can makeK(M(λ))=(2,1,0,−1,−2), the set of all circuits ofIK(M(λ))is{x1−x22, x1x24−1, x1x5−1, x2x4−1, x22x5−1, x42−x5}. By Theorem2.1, it is hereditary normal.
2.5 The representation ofSL(3)with the highest weight 2π1
Its highest weightλis equal to 2π1=2e1, and all the weights of this representation are pointed in the figure below.
After an appropriate change of basis, K(M(λ))=
1 0−1−2 0 2 0 1−1 0 −2 2
. The set of all circuits of IK(M(λ)) is{x1x2x3−1, x4x5x6−1, x12x4−1, x22x5−1, x32x6− 1, x12x22−x6, x12x32−x5, x22x32−x4, x12−x4x5, x22−x4x6, x32−x5x6}. By Theorem2.1, it is hereditary normal.
2.6 The representation ofSL(6)with the highest weightπ3 The highest weightλequals
1
2,12,12,−12,−12,−12
,M(λ)= {12(ε1, . . . , ε6):εi=
±1,
εi=0}. The corresponding matrixKis(K1| −K1), where
K1=1 2
⎛
⎜⎜
⎜⎜
⎜⎜
⎝
1 1 1 1 1 1 1 1 1 1
1 1−1 1−1−1 1−1−1−1 1−1 1−1 1−1−1 1−1−1
−1 1 1−1−1 1−1−1 1−1
−1−1−1 1 1 1−1−1−1 1
−1−1−1−1−1−1 1 1 1 1
⎞
⎟⎟
⎟⎟
⎟⎟
⎠ .
The rows ofK1are linearly dependent. One can exclude the first row, and it re- mains to check unimodularity ofK2,
K2=
⎛
⎜⎜
⎜⎜
⎝
1 1−1 1−1−1 1−1−1−1 1−1 1−1 1−1−1 1−1−1
−1 1 1−1−1 1−1−1 1−1
−1−1−1 1 1 1−1−1−1 1
−1−1−1−1−1−1 1 1 1 1
⎞
⎟⎟
⎟⎟
⎠.
An easy but cumbersome calculation shows thatK2is unimodular, all its non-zero minors are equal to±16, henceM(λ)is hereditary normal.
In the next case we use the following lemmas, their proof can be found in [11].
Lemma 2.3 LetM be a non-saturated set andαbe a vector such thatα∈Mand
−α∈M. Then eitherM\{α}orM\{−α}is non-saturated.
Lemma 2.4 Any set of linearly independent vectors is saturated.
Lemma 2.5 Letv=q1v1+ · · · +qmvm, wherev, vi are arbitrary vectors, andqi∈ Q+. Then one can choose a linearly independent subset{vi1, . . . , vis} ⊂ {v1, . . . , vm} and numbersqi
1, . . . , qi
s∈Q+such that v=qi
1vi1+ · · · +qi
svis.
Definition We mean by an NSS a non-saturated subset{v1, v2, . . . , vs}in the setM of weights of a representation. By an ENSS we mean the NSS together with a vector v, wherev∈Z(v1, v2, . . . , vm)∩Q+(v1, v2, . . . , vm), andv∈Z+(v1, v2, . . . , vm).
2.7 The representations ofSL(4),SL(5)andSL(6)with the highest weightπ2 The set of weightsM(λ)is equal to {ei +ej :1≤i < j ≤n}, where n=4, 5, 6.
Suppose that there exists an ENSS{w;v1, . . . , vm},vi∈M(λ):
w=z1v1+ · · · +zmvm=q1v1+ · · · +qmvm, zi∈Z, qi∈Q+, w∈Z+(v1, . . . , vm).
Consider allvi occurring in the right side of this equality with a nonzero coefficient qi. By Lemma2.5, we may assume that they are linearly independent. To simplify the reasoning, consider vectorv=w− q1v1− · · · − qmvm instead ofw. It is easy to see thatv belongs toZ(v1, . . . , vm), toQ+(v1, . . . , vm)and does not belong toZ+(v1, . . . , vm). We yield that{v;v1, . . . , vm}is also an ENSS. After this change all the coefficients of theQ+-combination belong to[0,1).
Letbe the quiver associated with{v1, . . . , vm}. Construct a subgraph⊂: take all the vertices of and all the edges ofentering into the Q+-combination above with nonzero coefficients. Write the coefficients of theQ+-combination at the edges of. The further proof consists of a search of all possible graphs . The following observations simplify the search.
(0.1) The number of edges in each connected component ofis not greater than the number of vertices. Otherwise, the vectors corresponding to the edges of this component are linearly dependent.
(0.2) The number of edges in is less than the number of vertices (it follows from(∗)that the dimension of the enveloping space equalsn−1).
(0.3) The graphdoes not contain even cycles. It follows from the fact that the edges of an even cycle are linearly dependent: their alternating sum is zero.
(0.4) It follows from (0.1) that each connected component ofeither is a tree or contains exactly one cycle. In the second case it follows from (0.3) that this cycle is odd.
(0.5) It follows from (0.2) thathas a vertex of degree 0 or 1.
At each vertex, count the sum of all coefficients on the incident edges, and for each sum take its fractional part. All these fractional parts are equal due to the fact that all the sums in vertices (they equal the coordinates ofv) become integer after subtracting(∗)with an appropriate coefficient. Now we conclude that
Fig. 1 Graph
(0.6)does not contain vertices of degree 0 and 1 simultaneously: if it does, the fractional parts of the sums in vertices are all equal to 0, but in the terminal vertex this sum has only one summand and is not an integer. We consider these two cases independently.
Case 1. Graphhas a vertex of degree 0.
(1.1) Any other connected component of is either a point or has no terminal vertices (it follows from (0.6)). Moreover, it follows from (0.4) that it is an odd cycle.
(1.2) We haven≤6, consequently, the number of edges inis≤5, but any odd cycle has≥3 edges, and we yield thathas at most one cycle.
Fulfill an exhaustive search within all graphshaving a vertex of degree 0:
n=4, graph is a cycle of length 3 and a point, n=5, graph is a cycle of length 3 and two points, n=6, graph is a cycle of length 3 and three points,
n=6, graph is a cycle of length 5 and a point.
The only possibleQ+-combination in these cases is12(v1+· · ·+vs). This means that v=e1+e2+. . .+es,s∈ {3,5},e1, . . . , es correspond to the vertices of the cycle.
But it does not lie inZ(v1, . . . , vm)whennis even. Whenn=5, consider also the graph. Sincev is aZ-combination of the edges of, has more than 3 edges:
⊃,=andhas 3 edges. In the representatione1+e2+. . .+es the sum of coefficients ofvis odd, hence we should apply(∗)to the existingZ-combination to get the representatione1+e2+. . .+es. For this purpose the edges ofshould touch all the vertices of(we name this property(∗∗)).
In Fig.1 the graph is drawn. To satisfy (∗∗), should contain at least the following edges (up to symmetry): see Fig.2a,2b or2c. The vertices corresponding toeiare calledVi. But in all cases we get a contradiction, sincee1+e2+e3is already aZ+-combination:
in Fig.2a: e1+e2+e3=V1V2+V2V3+V1V3+V4V5, in Fig.2b: e1+e2+e3=V4V2+V2V1+V1V3+V3V5, in Fig.2c: e1+e2+e3=V4V2+V2V5+2V1V3.
We have shown that the graphdoes not provide an NSS if has a vertex of degree 0.
Case 2. The graph has a vertexXof degree 1. LetXY be an edge incident to X. We need to subtract(∗)with the same multiplicity as atXY. We get:
(a) (b) (c)
Fig. 2 Graphshould contain one of these graphs as a subgraph
(2.1) SinceXis a terminal vertex of, eitherXY is a connected component of or the degree ofY is≥3. Indeed, suppose that the degree ofY is 2. LetY Zbe the second edge incident toY, and letqbe the value written atY Z. After subtracting(∗) the coefficient atY becomes equal toq, but it should be integer, and we know that q∈(0,1). This is a contradiction.
Find all possible connected components of. On 2 vertices:
On 3 vertices:
On 4 vertices: and
Ifhas a connected component on 5 or 6 vertices, then this component coincides with. Using this observation together with (0.2), we obtain thatis a tree. Taking into account (2.1), it remains to consider only the following trees: , and . But the edges of are linearly dependent (whenn=6, one should sum all the thin edges, then subtract the thick one, and obtain(∗)). Therefore, this graph should not be considered. The result is
On 5 vertices:
On 6 vertices:
Fulfill an exhaustive search within all graphs onn vertices satisfying all the conditions above. In the case when one of the connected components ofis a claw (i.e., all the edges are incident to one vertex) with at least 4 vertices, its central vertex will correspond toe1(it is easy to see thatcannot have more than one claw).
n Splitting into Admissible
connected components graphs
4 2+2
4 4 or
5 2+3
5 5
6 2+2+2
6 2+4 or
6 6
The graphs and do not satisfy our conditions: their edges are linearly de- pendent.
If we start with , we can obtain only e1 as the Q+-combination: all the three edges should appear in theQ+-combination with the same coefficient, saya, a∈(0,1). We sum these three vectors, obtain 3ae1+ae2+ae3+ae4, and sub- tract(∗)with an appropriate coefficient. Finally we obtain 2ae1. In this notation it already has integer coordinates (equal to zero), this means that all the other coordi- nates, 2a among them, should be integers,a=12,v=e1. Butv cannot be obtained as aZ-combination of vectors of typeei +ej. Indeed, eachvi has an even sum of coordinates,nis even, subtracting (∗)with an integer coefficient does not change parity of the sum of coordinates, and this proves that any vector fromZ{vi}mi=1has even sum of coordinates.
The edges of the graph are linearly dependent (heren=5) because (2·first edge + the sum of the edges of the cycle)=0.
The graph : using similar reasoning,v=e1or 2e1. But there exists an edge in\, hencee1is aZ+-combination of the edges of: take the sum of thick edges
of .
The edges of and are linearly dependent.
In the graph the vectorvmay be equal only toe1, bute1can not be obtained as aZ-combination: 6 is even, and the sum of coordinates ofe1is odd.
In the graph the vectorv has to be proportional toe1, moreover, the coeffi- cient should be even (we use the reasoning as above, from the fact that 6 is even it follows that the sum of coordinates is even for any vector fromZ(v1, . . . , vm)). But if we add any edge to this set, 2e1will be obtained as aZ+-combination: take the sum of thick edges of .
All the cases are considered, and this completes the proof.
3 Negative results
Letλbe a highest weight not listed in the Main Theorem. One has to construct an NSS inM(λ). There are two possibilities forλ: either the absolute values of all its usual coordinates are<1, orλ has a coordinate with the absolute value 1. Speaking informally, the second case is practically always the consequence of the first one (Lemma3.10), but the NSS in the first case is constructed recursively and its capacity increases whennincreases. The construction of the second case gives an NSS of only 4 vectors for anyn.
To prove that a set {v1, . . . , vm} is not saturated, we construct a so-called dis- criminating functionf (v) with the following properties: linearity and f (vi) >0, i=1,2, . . . , m. The discriminating function will be applied as follows. If one wants to show that{v0;v1, . . . , vm}is an ENSS, it suffices to present the correspondingQ+- andZ-combinations forv0and construct a discriminating functionf, such thatf (v0) cannot be composed as the sum off (vi)withZ+-coefficients.
Byxi we denote theith quasi-coordinate of a vector, if its quasi-basis represen- tation is fixed. Note thatf =a1x1+. . .+anxnin quasi-basis is well-defined if and only ifa1+. . .+an=0.
3.1 The fundamental weights In this caseλequals
πk=πk,n= n−k
n , . . . ,n−k n ,−k
n, . . . ,−k n
in the usual basis, 0< k < n,n3 (ifn=2, the corresponding representation is mentioned in the Main Theorem). In some proofs we consider πk for SL(n)’s of different dimensions simultaneously, so the second index in the notationπk,ncarries this data. HereM(λ)= {σ λ:σ∈Sn}. The highest weight is equal toe1+e2+. . .+ek in quasi-basis, all the points ofM(λ)have formei1+ei2+. . .+eik, 1≤i1< i2<
. . . < ik≤n.
Now we can reformulate the problem. Let{ei}be the quasi-basis,k < n, and the weightλ=πk is not listed in the Main Theorem. One has to find a non-saturated subset in the set
{ei1+ei2+. . .+eik:1≤i1< i2< . . . < ik≤n}.
The construction uses induction onn. In the next section we produce the NSSes which are the base of induction.
3.1.1 Important particular cases
Example 3.1 n=7, k=2. The NSS consists of those and only those vectors which are the sums of two quasi-basis vectors connected with an edge in the graph below.
We have
v=e1+e2+e3=1 2
(e1+e2)+(e2+e3)+(e1+e3)
, v= −(e4+e5+e6+e7)
=2(e2+e3)−(e2+e4)−(e2+e5)−(e3+e6)−(e3+e7).
Let f =5(x2+x3)−2(x1+x4+x5+x6+x7). Then f (e1+e2)=f (e1+e3)=f (e2+e4)=f (e2+e5)=f (e3+e6)
=f (e3+e7)=3,
f (e2+e3)=10, f (v)=f (e1+e2+e3)=5·2−2=8.
It is clear that 8 cannot be represented as a sum where each summand equals either 3 or 10.
Example 3.2 n=8, k=3. Consider the following vectors (in quasi-basis):
⎛
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎝ v1 v2 v3 v4
v5
v6
v7
v8
⎞
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎠
=
⎛
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎝
0 0 1 1 1 0 0 0 1 0 0 1 1 0 0 0 1 1 0 0 1 0 0 0 1 1 1 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 1 1 0 1 0 0 0 1 0 1 0 0 1 0 0 1 1 0 0 0 0 1
⎞
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎠ .
Takev=(1,1,1,1,1,0,0,0)=13(v1+v2+v3+v4+v5)=2v5−v6−v7−v8. Let f=x1+5(x2+x3+x4)+2x5−6(x6+x7+x8). Then
f (v1)=12, f (v2)=f (v3)=8, f (v4)=11, f (v5)=15, f (v6)=f (v7)=f (v8)=4, f (v)=18.
It is easy to see that 18 cannot be represented as a sum of 4, 8, 11, 12, or 15.
Example 3.3 n=2k, k4.
⎛
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎝ v1 v2 v3 ... vk−1
vk
vk+1
vk+2
⎞
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎠
=
⎛
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎝
1 0 0. . .0 0 0 1 1. . .1 1 0 1 0. . .0 0 1 0 1. . .1 1 0 0 1. . .0 0 1 1 0. . .1 1 ... ... ... . .. ... ... ... ... ... . .. ... ... 0 0 0. . .1 0 1 1 1. . .0 1 0 0 0. . .0 1 1 1 1. . .1 0 0 1 0. . .0 0 1 1 1. . .1 0 1 1 0. . .0 0 0 1 1. . .1 0
⎞
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎠
Let us show that it is an NSS. Setv=(0, . . . ,0
k
,1, . . . , 1
k
),v=v1+vk+1−vk+2,
1
k−2(v1+ · · · +vk)= 1
k−2(1, . . . ,1
k
, k−1, . . . , k −1
k
)=
= 1
k−2(0, . . . ,0
k
, k−2, . . . , k −2
k
)=(0, . . . ,0
k
,1, . . . , 1
k
)=v.
To explain whyvis not aZ+-combination of vectorsvi, consider two cases.
Case 1:k=4. Letf = −6x3−7x4+5(x5+x6+x7)−2x8. Then
f (v1)=f (v2)=8, f (v3)=2, f (v4)=8, f (v5)=15, f (v6)=10, f (v)=13.
But it is easy to see that 13 cannot be represented as a sum of 2, 8, 10, or 15.
Case 2:k5. Letf =(k−2)(xk+1+ · · · +x2k)−k(x3+ · · · +xk). Then f (v1)=f (v2)=(k−2)(k−1),
f (v3)=f (v4)= · · · =f (vk)=(k−1)(k−2)−k,
f (vk+1)=(k−2)(k−1), f (vk+2)=(k−2)2, f (v)=k(k−2).
Ifk6, then two least possible summands give too much: 2((k−1)(k−2)−k) >
k(k−2), ifk=5, then 15 should be represented as a sum of 12, 7, or 9, but this is impossible.
Example 3.4 n=2k+1, k3.
⎛
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎝ v1 v2 v3 ... vk vk+1
vk+2
vk+3
... v2k v2k+1
⎞
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎠
=
⎛
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎝
0 1 1 . . .1 1 0 0. . .0 0 1 0 1 . . .1 1 0 0. . .0 0 1 1 0 . . .1 1 0 0. . .0 0 ... ... ... . .. ... ... ... ... . .. ... ... 1 1 1 . . .0 1 0 0. . .0 0 1 1 1 . . .1 0 0 0. . .0 0 0 1. . . 1 1 0 1 0. . .0 0 1 0. . . 1 1 0 0 1. . .0 0 ... ... . .. ... ... ... ... ... . .. ... ... 1 1. . . 0 1 0 0 0. . .1 0 1 1. . . 1 0 0 0 0. . .0 1
⎞
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎠
Letv=(1, . . . , 1
k+1
,0, . . . ,0
k
). Then
v=1
k(v1+ · · · +vk+1)=1
k(k, . . . , k
k+1
,0, . . . ,0
k
)=(1, . . . , 1
k+1
,0, . . . ,0
k
),
(k−1)vk+1−vk+2− · · · −v2k+1=
=(k−1)(1, . . . , 1
k
,0, . . . , 0
k+1
)−(k−1, . . . , k −1
k
,0,1, . . . , 1
k
)=
=(0, . . . , 0
k+1
,−1, . . . , −1
k
)=(1, . . . ,1
k+1
,0, . . . , 0
k
)=v.
It suffices to show thatv does not belong to Z+(v1, v2, . . . , v2k+1). Letf =(k+ 1)(x1+ · · · +xk)−k(xk+1+ · · · +x2k+1). Then
f (v1)= · · · =f (vk)=k2−k−1, f (vk+1)=k(k+1), f (vk+2)= · · · =f (v2k+1)=k2−k−1, f (v)=k2.
But if k3, then k2<2(k2−k−1), so k2 cannot be represented as a sum where each summand equals either (k2−k−1) or k(k+1). This means that v∈Z+(v1, v2, . . . , v2k+1).
Example 3.5 n=8, k=2.
The NSS consists of vectors which are sums of two quasi-basis vectors connected with an edge in the graph above. Letv=e1+e2+e3+e5+e6+e7. Then
v=1
2((e1+e2)+(e2+e3)+(e1+e3)+(e5+e6)+(e6+e7)+(e5+e7)) , v=(e1+e2)+(e3+e4)−(e4+e5)+(e5+e6)+(e5+e7).
Check thate1+e2+e3+e5+e6+e7cannot be represented as aZ+-combination of the vectors of our set. Let f =x1+x2+x3+2(x5+x6+x7)+9x4−18x8. Then
f (e1+e2)=f (e2+e3)=f (e1+e3)=2, f (e5+e6)=f (e6+e7)=f (e5+e7)=4, f (e3+e4)=10, f (e4+e5)=11, f (v)=9.
But 9 cannot be represented as the sum of integers 2, 4, 10, or 11.
Example 3.6 n=9, k=3. Consider the following vectors:
v1=e1+e2+e4, v5=e1+e3+e8, v2=e1+e2+e5, v6=e1+e3+e9, v3=e2+e3+e6, v7=e2+e4+e6.
v4=e2+e3+e7,
Thenv=e1+e2+e3=13(v1+v2+v3+v4+v5+v6)=v1+v3−v7.
Check thatvis not aZ+-combination ofv1,v2,v3,v4,v5,v6, andv7. Letf = 5(x1+x2+x3+x4)−4(x5+x6+x7+x8+x9). Thenf (v1)=15,
f (v2)=f (v3)=f (v4)=f (v5)=f (v6)=f (v7)=6, f (v)=15.
Note thatv=v1andf (v1)=f (v), so we conclude that ifv∈Z+(v1, . . . , v7), then v1does not occur in this decomposition. But 615, and this means thatvcannot be obtained as aZ+-combination ofvi’s.
Example 3.7 n=10, k=4. Consider the following vectors:
v1=e1+e2+e3+e5, v4=e5+e6+e7+e8, v2=e1+e2+e4+e6, v5=e5+e7+e8+e9, v3=e3+e4+e5+e6, v6=e6+e7+e8+e10,
v=e1+e2+e3+e4+e5+e6=1
2(v1+v2+v3)=v4−v5−v6. Let us show thatv∈Z+(v1, . . . , v6). Letf =x1+x3+x4+6x7+6x8−7x9− 8x10. Then
f (v1)=f (v2)=f (v3)=2, f (v4)=12, f (v5)=5, f (v6)=4, f (v)=3.
But it is clear that 3 cannot be represented as a sum of 2, 4, 5, or 12.
3.1.2 Case whenkn, (n−k)n
It follows thatn5. The exceptional case nk ∈ {25,35}will be considered at the end of the section. Below we suppose thatkn∈ {25,35}, which givesn7.
Lemma 3.8 Assume that there exists an NSS for a pair(n, k), where(n, k)satisfies the conditions above. Then for eachr∈N, there exists an NSS for the pair(nr, kr).
Proof Consider an arbitrary vector fromM(πk,n). Write down its quasi-coordinates rtimes in succession. The result is a vector fromM(πkr,nr): it haskr1’s and(n−k)r 0’s. If one takes an NSS for(n, k)and performs this procedure on each vector, the
result will be an NSS for(nr, kr).
Thus, if we construct an NSS for all pairs(n, k)where gcd(n, k)=1, then the NSS for all other pairs will be constructed according to Lemma3.8.
Lemma 3.9 (The Step procedure) If there exists an NSS for a pair(n, k), then there exists an NSS for the pair(n+k, k).