# Cycle Intersection forSOp,q-Flag Domains

Authors: Faten Abu Shoga
Journal: International Journal of Mathematics and Mathematical Sciences
 (2020)

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## Abstract

A real formG0 of a complex semisimple Lie group G has only finitely many orbits in any given compact G-homogeneous projective algebraic manifold Z=G/Q. A maximal compact subgroup K0 of G0 has special orbits C which are complex submanifolds in the open orbits of G0. These special orbits C are characterized as the closed orbits in Z of the complexification K of K0. These are referred to as cycles. The cycles intersect Schubert varieties S transversely at finitely many points. Describing these points and their multiplicities was carried out for all real forms of SLn,ℂ by Brecan (Brecan, 2014) and (Brecan, 2017) and for the other real forms by Abu-Shoga (Abu-Shoga, 2017) and Huckleberry (Abu-Shoga and Huckleberry). In the present paper, we deal with the real form SOp,q acting on the SO (2n, C)-manifold of maximal isotropic full flags. We give a precise description of the relevant Schubert varieties in terms of certain subsets of the Weyl group and compute their total number. Furthermore, we give an explicit description of the points of intersection in terms of flags and their number. The results in the case of G/Q for all real forms will be given by Abu-Shoga and Huckleberry.


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## Body

## 1. Background

LetG be a complex semisimple Lie group and B a Borel subgroup. The compact algebraic homogeneous space Z=G/B is called a complex flag manifold. Let G0 be a real form of G in the sense that there is an antiholomorphic involution τ:G⟶G such that G0 is the fixed point set of τ. It is well known [1] that there are only finitely many G0-orbits on Z. Therefore, at least one of them must be open. Such open orbits are called flag domains.Fix a Cartan involutionθ:G0⟶G0. The fixed point set of θ in G0 is a maximal compact subgroup of G0 denoted by K0, and K is its complexification.IfD is a flag domain on Z, then K0 has exactly one open orbit in D that is a complex submanifold and denoted by C0. The complex submanifold C0 is both the unique K-orbit in D that is compact and the unique K0-orbit in D that is complex. This is the origin of what is known as “Matsuki duality.”Consider an Iwasawa decompositionG0=K0A0N0 of the real form G0. We refer to a Borel subgroup B⊂G as an Iwasawa–Borel subgroup of G if it contains an Iwasawa factor A0N0. Of course, these are just the Borel subgroups which occur as the isotropy groups at points of the closed G0-orbit in G/B.In general, given any Borel subgroupB in G and a point z∈G/B, the closure S=clO of the orbit O=B.z is called the Schubert variety or Schubert cycle. An Iwasawa–Schubert variety S=clO is transversal (relative to an open G0-orbit D) if S∩D is nonempty and contained in the open B-orbit O, and codimS=q−r where r=dimC0 and the intersection S∩C0 is transversal at each of its points.Thus, although the determination of ther-codimensional Iwasawa–Schubert varieties which have nonempty intersection with D, along with their points of intersection with C0, is apparently a problem of a combinatorial nature, such information has complex analytic significance.SinceS∩D≠∅ implies that S∩C0≠∅, the first goal of this paper is to determine which Schubert varieties S have nonempty intersection with C0. After doing so, we then describe this intersection in the case where S is of complementary dimension to C0. The Schubert varieties are determined by the elements of the Weyl group WI of a distinguished maximal torus TI in the Borel subgroup BI which fixes a certain base point in γcℓ. These Schubert varieties are denoted by Sw where w∈WI. Consequently, our results are formulated in terms of conditions on elements w in the Weyl group WI. In this paper, we discussed both even and odd cases for p and q. (In some steps, we discuss only the even case because the odd case is included in the even case.)


## 2. Flag Intersection forSOp,q

### 2.1. Preliminaries

Consider the semisimple Lie groupG=SOm,ℂ with the real form SOp,q. In this case, it is convenient to choose the bilinear form b on C2n depending on p and q. If p or q is even, then we choose it in the usual way:(1)bv,w=−∑i=1qviwi+∑i=q+1mviwi.Letσ be the complex conjugation σv=v¯, then the Hermitian form hv,w of signature p,q which defines the real form is defined by(2)hv,w=bv,σw=−∑i=1qviσwi+∑i=q+1mviσwi.If bothp and q are odd, then the complex bilinear form is(3)bv,w=−∑i=1q−1viwi+vqwq+1+vq+1wq+∑i=q+2mviwi,and the Hermitian form hv,w is defined by(4)hv,w=−∑i=1qviw¯i+∑i=q+1mviw¯i.The real form isG0=SOp,q=SO2n,ℂ∩SUp,q.A Cartan involutionθ=sop,q⟶sop,q in the Lie algebra level is given by θg=−gt, and the maximal compact subgroup associated with it is K0≔SOp×Oq⊂SUp×Uq.


### 2.2. Introduction to the Flags inZ

Recall that the (indefinite metric) orthogonal groupG=SOp,q acts on the space of full flags by two orbits: closed orbit and open orbit. The closed orbit is the set of all maximally b-isotropic full flags F with respect to b, where the maximally b-isotropic full flag F is defined as a sequence of 2n+1-vector spaces:(5)F=0=V0⊂V1⊂…⊂V2n=ℂ2n,such that dimVi=i,for all0≤i≤2n, and V2n−i=Vi⊥,for1≤i≤n. In particular, Vi is isotropic for 1≤i≤n. The set of all maximal isotropic flags is denoted by Z. The manifold structure of Z arises from the fact that the Lie group G acts transitively on it with Z=G/B where B is the stabilizer of any particular maximally b-isotropic flag in Z.The signature of a flagF=0⊂V1⊂…⊂V2n∈Z with respect to h consists of three sequences:(6)a:0≤a1≤a2≤…≤a2n,b:0≤b1≤b2≤…≤b2n,d:0≤d1≤d2≤…≤d2n,with signVi=ai,bi,di where ai (resp., bi) denotes the dimension of a maximal negative (positive) subspace and di the degeneracy Vi of the restriction of h to Vi and ai+bi+di=i∀1≤i≤2n. The 3-tuple a,b,d is called the signature of the flag F. If d=0, we refer to F as being nondegenerate and write signF=a,b.


### 2.3. Base Points

To parametrize the Schubert cells, it is important to focus on a maximal TorusTI in an Iwasawa–Borel subgroup BI of G which fixes the associated flag FI. On the contrary, we need another maximal torus TS which defines a basis to determine the base cycles called the standard basis. These two tori are conjugate.In the following, we define bases depending onp and q as well as the positive and negative spaces which will be essential for understanding the base cycle. Although the goal of the paper is to handle the case where m is even, since in this paragraph the discussion for the odd case is essentially the same, we include it. Also, the fundamental maximal torus TS is defined in each case by requiring that each basis vector is a TS-eigenvector.(i)
Ifm=2n and q are even, the ordered b-isotropic basis is(7)e1+ie2,e3+ie4,…,e2n−1+ie2n,e2n−1−ie2n,…,e1−ie2,where(8)E−≔e1+ie2,e3+ie4,…,eq−1+ieq,e1−ie2,e3−ie4,…,eq−1−ieq,E+≔eq+1+ieq+2,…,e2n−1+ie2n,eq+1−ieq+2,…,e2n−1−ie2n.(ii)
Ifm=2n and q are odd, the ordered b-isotropic basis is(9)e1+ie2,e3+ie4,…,e2n−1+ie2n,eq,eq+1,e2n−1−ie2n,…,e1−ie2,where(10)E−=e1+ie2,e3+ie4,…,eq,e1−ie2,e3−ie4,…,eq−1−ieq,E+=eq+1,eq+2+ieq+3,…,e2n−1+ie2n,eq+2−ieq+3,…,e2n−1−ie2n.Note that the vectorseq and eq+1 are isotropic and beq,eq+1=−1. Moreover, he3,e3=−1 and he4,e4=1 which means that eq∈E− and eq+1∈E+.(iii)
Ifm=2n+1 and q are even, the ordered b-isotropic basis is(11)e1+ie2,e3+ie4,…,e2n+ie2n+1,eq+1,e2n−ie2n+1,…,e1−ie2,where(12)E−≔e1+ie2,e3+ie4,…,eq−1+ieq,e1−ie2,e3−ie4,…,eq−1−ieq,E+=eq+1,eq+2+ieq+3,…,e2n−1+ie2n,eq+2−ieq+3,…,e2n−1−ie2n.(iv)
Ifm=2n+1 and q are odd, the ordered b-isotropic basis is(13)e1+ie2,e3+ie4,…,e2n+ie2n+1,eq,e2n−ie2n+1,…,e1−ie2,where(14)E−=e1+ie2,e3+ie4,…,eq,e1−ie2,e3−ie4,…,eq−2−ieq−1,E+=eq+1+ieq+2,…,e2n+ie2n+1,eq+1−ieq+2,…,e2n−ie2n+1.Remark 1.
Ifm=2n+1, we have the vector eq or eq+1 in the ordered b-isotropic basis. In this case, the vector sits in a fixed position in the middle of the basis and the terms at the beginning skip over eq or eq+1.
The above bases define the standard maximal torusTS in each case as the subgroup of diagonal matrices, i.e.,(15)TS=g∈G0:g=diagt1,t2,…,tn,−tn,…,−t2,−t1,ti∈ℂifm=2n,orTS=g∈G0:g=diagt1,t2,…,tn,0,−tn,…,−t2,−t1,ti∈ℂifm=2n+1.
For any of the ordered bases above, we denote byFS the associated flag in Z, where the reordering on these bases will determine the flag domains, the base cycles, and the intersection points.
Just as in the case ofTS, the maximal torus TI is defined to have a certain basis of eigenvectors which depends on m being odd or even.(i)
Ifm is even, then the basis is(16)e1+e2q,…,eq+eq+1,e2q+1+ie2q+2,e2q+3+ie2q+4,…,e2n−1+ie2n,e2n−1−ie2n,…,e2q+3−ie2q+4,e2q+1−ie2q+2,eq−eq+1,…,e1−e2q.(ii)
Ifm is odd, then the basis is(17)e1+e2q,e2+e2q−1,…,eq+eq+1,e2q+1,e2q+2+ie2q+3,e2q+4+ie2q+5,…,e2n+ie2n+1,e2q+1,e2n−ie2n+1,…,e2q+4−ie2q+5,e2q+2−ie2q+3,eq−eq+1,…,e2−e2q−1,e1−e2q.We denote byBI the Iwasawa–Borel subgroup of G which fixes the associated flag FI.


### 2.4. Weyl Groups, Flag Domains, and Base Cycles

Let us use shorthand notation for the bases used above. Byr,s−form, we mean a basis r1,…,rn,sn,…,s1 where bri,rj=bsi,sj=0 and bri,sj=δij for all i and j. This occurs in the even-dimensional case. By r,t,s-form, which occurs in the odd-dimensional case, ri and si satisfy the same conditions and t is a single vector with bt,t=±1 and bt,ri=bt,si=0 for all i.Here, we discuss the Weyl groupsWTS and WTI by their actions on these bases. If T is either of these tori, then it stabilizes the spaces spanned by ri and si for i=1,…,n for both kinds of bases. In this, we regard T as a product T=T1⋅…⋅Tn, in the second case acting trivially on the space spanned by t. In both cases, an arbitrary permutation in Sn acting diagonally on these spaces by ri,si⟼rπi,sπi normalizes the T-action and is in the given orthogonal group G.A simple reflectionri,si⟼si,ri for one such i, and being the identity on the other 2-dimensional spaces, is denoted by a sign change. This normalizes T but has negative determinant. In the first case, this means that only an even number of sign changes are allowed. In the second case, we may couple the sign change with the map t⟼−t so that this slightly modified sign change has positive determinant. Thus, in both cases, we have the action of Sn⋉ℤ2n normalizing the T-representation on the basis at hand. This is in fact the action of the full Weyl group.As in the cases ofSp2n,ℝ and SO∗2n, to describe the flag domains and their K0-base cycles, we use the Weyl group WTS of the standard torus TS. The base point for α is the flag FS defined by the standard basis. Then, all base points are Fα≔wFS where w∈ℤ2n−1 if m=2n and w∈ℤ2n if m=2n+1 is associated with the sign-change vector α. As before, the base cycles and flag domains are the orbits K.Fα and G0.Fα, respectively, and α defines a nondegenerate sign with signFα=a,b and then Dα=Da,b.In our particular case, there are two maximal tori, the maximal torusTI and the maximal torus TS, which were defined earlier. Recall that these two maximal tori TIandTS are conjugates, and a conjugation induces an isomorphism ψ of the associated Weyl groups. In this case, the bijective map ψ is described, depending on the case, as follows:(i)
Ifm=2n and q are even, define WTS to be the Weyl group with respect to the basis (7) and let WI be the Weyl group with respect to the basis (16). In this case, the bijective map between WI and WTS is ψ∓2i−1=±i,ψ∓2i=±q−i+1 if 1≤i≤q/2 and ψ±i=±i if i>q.(ii)
Ifm=2n and q are odd, define WTS to be the Weyl group with respect to the basis (9) and let WI be the Weyl group with respect to the basis (16); then, the bijective map in this case is ψ:WI⟶WTo defined by ψ∓2i−1=±i,ψ∓2i=±q−i if 1≤i≤q−1/2 and ψ±i=±i−1 if i>q and ψ−q=−n.(iii)
Ifm=2n+1 and q are even, define WTe to be the Weyl group with respect to the basis (11) and let WI be the Weyl group with respect to the basis (17); then, we can define the bijective map between them to be ψ∓2i−1=±i,ψ∓2i=±q−i+1 if 1≤i≤q/2 and ψ±i=±i if i>q.(iv)
Ifm=2n+1 and q are odd, define WTo to be the Weyl group with respect to the basis (13). Let WI be the Weyl group with respect to the basis (17). The bijective map ψ between WI and WTo can be defined as ψ∓2i−1=±i,ψ∓2i=±q−i if 1≤i≤q−1/2 and ψ±i=±i if i>q.Moreover, through an argument quite similar to Proposition1, the base cycle Cα is the set(18)Cα=F∈Z:Vi=Vi∩E−⊕Vi∩E+,1≤i≤m,where E− and E+ defined in Section 2.3 and the intersection dimensions are determined by α (resp., a,b). The flag domains are parametrized by the signature of their base cycles.


### 2.5. The Length of the Elements ofSn⋉ℤ2n

Recall that everyB-Schubert cell S contains exactly one T-fixed point. Hence, at least in Z=G/B where the Weyl group acts freely and transitively on FixT, we should be able to compute dimℂS in terms of a corresponding Weyl element. In fact, if z0 is the B-fixed point in Z=G/B and w∈W, then the complex dimension of the Schubert cell Sw=B.wz0 is the length ℓw.As indicated above, we used a special way of writing the Weyl elements which is defined in the paper [2]. For dimension computations, let us state how we can compute the length of elements w∈Sn⋉ℤ2n relative to the notation for the Weyl group elements used in [2].Lemma 1 (see [2]).
Fixw∈Sn⋉ℤ2n and construct w˜∈Sn⋉ℤ2n by the following algorithm:(1)
Starting from left to right inw, using simple reflections, place all positive numbers in w step by step in a sequence of n−empty boxes beginning from the first one in w˜, in the same order as they appeared in w.(2)
From left to right inw, replace a negative number with its absolute value in the n−empty boxes starting from right to left.Ifw˜=w˜1,w˜2,…,w˜n, then define Lw˜=n2−n/2−number ofw˜i:i<kandw˜i<w˜k, and if we have m negative signs in w in the following positions j1,j2,…,jm, then define fw=∑i=1mn−ji. It follows that the length of w is(19)lw=Lw˜+fw+m.Example 1.
Letw=5−34−12∈S5⋉ℤ25, then by following the above remark, we have(20)5−34−12⟹542−3−1⟹54213,so w=5−34−12 and w˜=54213, and then Lw˜=8 and fw=4.
Hence,lw=8+4+2=14.


### 2.6. Description of theBI-Orbits

In the previous section, we have chosen a flagFI defined by the basis(21)e1−e2n,e2−e2n−1,…,en−en+1,en+en+1,…,e2+e2n−1,e1+e2n,which belongs to the G0-closed orbit γcl. Given a Weyl element w∈W, we consider the Schubert cell BI.Fw and an element F=bFw in it which is defined by an ordered basis b.vw1,…,b.vw2n, b∈BI, where vi=ei−e2n−i+1,1≤i≤n, and vi=e2n−i+1+ei,n+1≤i≤2n. If F∈Cα, then it can be defined by an ordered basis ε1,…,ε2n where for all k the vector εk is either in V− or in V+ and is a linear combination(22)εk=∑j≤kcjb.vwj.In this case, we call the basis asplit basis. According to the above discussion for Sp2n,ℂ, we can write the formulas for b.vwj for any element b∈BI as follows:(23)b.vwj=ηje−wj+e2n+wj+1+ζje−wj−e2n+wj+1+Bj,−n≤wj≤−1,b.vwj=ζjewj−e2n−wj+1+Bj,1≤wj≤n,where Bj does not involve the basis vectors ewj and e2n−wj+1.Remark 2.
When discussing theBI-orbit of the base point FI, it is important to explicitly understand the orbits BI.ei−e2n−i+1 and BI.ei+e2n−i+1. As we explained above, we have the following facts.
IfFI and BI are as above, then the orbits of interest are BI.ei±e2n−i+1, i=1,…,n. In this case, the orbits BI.ei+e2n−i+1 and BI.ei−e2n−i+1 have points of the forms(24)bei+e2n−i+1=λei+e2n−i+1+⋯+bnen+en+1+anen−en+1+⋯+a1e1−e2n,bei−e2n−i+1=λei−e2n−i+1+ai−1ei−1−e2n−i+⋯+a1e1−e2n,respectively, with λ≠0. Note that, in the above orbits, if b is chosen appropriately, then we can arrange bei+e2n+i−1=ei or bei+e2n+i−1=e2n−i+1 for all i. Does this play a role in the proof of Theorem 1.


### 2.7. Flag Domains and Base Cycles

In order to parametrize the flag domains, it is enough to parametrize the closedK-orbits (see [3], Sections 4.2 and 4.3). For this, we choose a base point FS∈FixTS. It follows that the set C of closed K-orbits can be identified with the Weyl group orbit WGTS.FS (see [3], Corollary 4.2.4). As we have observed above, WGTS≅Sn⋉ℤ2n where the Sn factor can be identified with WKTS. In fact, it is exactly the stabilizer of K.FS in C. Consequently, C can be identified with ℤ2n.FS. We regard an element of ℤ2n as a vector α of length n consisting of plus and minus signs, and thus, C can be identified with the set of such vectors. In concrete terms, the K-orbits (resp., G0-orbits) of base points ±e1,…,±en are exactly the base cycles (resp., flag domains) in Z. The respective flag domains are denoted by Dα.Observe thatα defines a nondegenerate signature a,b and that Dα⊂Da,b≔F:signF=a,b. Conversely, given F with signF=a,b, we may choose a basis v,w≔v1,…,vn,wn,…,w1 which defines F and which has the following properties:(1)
b is in canonical form(2)
v,w is h-orthogonal(3)
vjh2=±1 with sgnvjh=αj(4)
wjh2=−vjh2IfsignF=signF^, then we choose bases v,w and v^,w^ for F and F^, respectively, and note that the transformation which takes the one basis to the other is in G0. This argument shows that Da,b=Dα.DenoteE+=e1,…,en and E−=en+1,…,e2n, where ℂ2n=E+⊕E−.Proposition 1.
For a fixed open orbitDα, the base cycle Cα is the set(25)Cα=F∈Z:Vi=Vi∩E−⊕Vi∩E+,1≤i≤2n.Proof.
Firstly, it was shown above that there is a base pointFα with the splitting condition. Since K0 is the product K0=K0+×K0−, it acts transitively on all such flags. Let F∈Cα where Vi−=Vi∩E− and Vi+=Vi∩E+. Define F+ and F− to be the sets of maximally isotopic flags in E+ and E−, respectively. Recall that K0=K0+×K0− where K0+ and K0− act transitively on the sets F+ and F−, respectively, which implies that K0 acts transitively on the set of maximal isotropic flags obtained by put flags from F+ and F− in a way such that the new flag has signature α. Hence, Cα is a complex manifold. But K0 has a unique orbit in Dα which is a complex manifold. Therefore, Cα is the base cycle.


## 2.1. Preliminaries

Consider the semisimple Lie groupG=SOm,ℂ with the real form SOp,q. In this case, it is convenient to choose the bilinear form b on C2n depending on p and q. If p or q is even, then we choose it in the usual way:(1)bv,w=−∑i=1qviwi+∑i=q+1mviwi.Letσ be the complex conjugation σv=v¯, then the Hermitian form hv,w of signature p,q which defines the real form is defined by(2)hv,w=bv,σw=−∑i=1qviσwi+∑i=q+1mviσwi.If bothp and q are odd, then the complex bilinear form is(3)bv,w=−∑i=1q−1viwi+vqwq+1+vq+1wq+∑i=q+2mviwi,and the Hermitian form hv,w is defined by(4)hv,w=−∑i=1qviw¯i+∑i=q+1mviw¯i.The real form isG0=SOp,q=SO2n,ℂ∩SUp,q.A Cartan involutionθ=sop,q⟶sop,q in the Lie algebra level is given by θg=−gt, and the maximal compact subgroup associated with it is K0≔SOp×Oq⊂SUp×Uq.


## 2.2. Introduction to the Flags inZ

Recall that the (indefinite metric) orthogonal groupG=SOp,q acts on the space of full flags by two orbits: closed orbit and open orbit. The closed orbit is the set of all maximally b-isotropic full flags F with respect to b, where the maximally b-isotropic full flag F is defined as a sequence of 2n+1-vector spaces:(5)F=0=V0⊂V1⊂…⊂V2n=ℂ2n,such that dimVi=i,for all0≤i≤2n, and V2n−i=Vi⊥,for1≤i≤n. In particular, Vi is isotropic for 1≤i≤n. The set of all maximal isotropic flags is denoted by Z. The manifold structure of Z arises from the fact that the Lie group G acts transitively on it with Z=G/B where B is the stabilizer of any particular maximally b-isotropic flag in Z.The signature of a flagF=0⊂V1⊂…⊂V2n∈Z with respect to h consists of three sequences:(6)a:0≤a1≤a2≤…≤a2n,b:0≤b1≤b2≤…≤b2n,d:0≤d1≤d2≤…≤d2n,with signVi=ai,bi,di where ai (resp., bi) denotes the dimension of a maximal negative (positive) subspace and di the degeneracy Vi of the restriction of h to Vi and ai+bi+di=i∀1≤i≤2n. The 3-tuple a,b,d is called the signature of the flag F. If d=0, we refer to F as being nondegenerate and write signF=a,b.


## 2.3. Base Points

To parametrize the Schubert cells, it is important to focus on a maximal TorusTI in an Iwasawa–Borel subgroup BI of G which fixes the associated flag FI. On the contrary, we need another maximal torus TS which defines a basis to determine the base cycles called the standard basis. These two tori are conjugate.In the following, we define bases depending onp and q as well as the positive and negative spaces which will be essential for understanding the base cycle. Although the goal of the paper is to handle the case where m is even, since in this paragraph the discussion for the odd case is essentially the same, we include it. Also, the fundamental maximal torus TS is defined in each case by requiring that each basis vector is a TS-eigenvector.(i)
Ifm=2n and q are even, the ordered b-isotropic basis is(7)e1+ie2,e3+ie4,…,e2n−1+ie2n,e2n−1−ie2n,…,e1−ie2,where(8)E−≔e1+ie2,e3+ie4,…,eq−1+ieq,e1−ie2,e3−ie4,…,eq−1−ieq,E+≔eq+1+ieq+2,…,e2n−1+ie2n,eq+1−ieq+2,…,e2n−1−ie2n.(ii)
Ifm=2n and q are odd, the ordered b-isotropic basis is(9)e1+ie2,e3+ie4,…,e2n−1+ie2n,eq,eq+1,e2n−1−ie2n,…,e1−ie2,where(10)E−=e1+ie2,e3+ie4,…,eq,e1−ie2,e3−ie4,…,eq−1−ieq,E+=eq+1,eq+2+ieq+3,…,e2n−1+ie2n,eq+2−ieq+3,…,e2n−1−ie2n.Note that the vectorseq and eq+1 are isotropic and beq,eq+1=−1. Moreover, he3,e3=−1 and he4,e4=1 which means that eq∈E− and eq+1∈E+.(iii)
Ifm=2n+1 and q are even, the ordered b-isotropic basis is(11)e1+ie2,e3+ie4,…,e2n+ie2n+1,eq+1,e2n−ie2n+1,…,e1−ie2,where(12)E−≔e1+ie2,e3+ie4,…,eq−1+ieq,e1−ie2,e3−ie4,…,eq−1−ieq,E+=eq+1,eq+2+ieq+3,…,e2n−1+ie2n,eq+2−ieq+3,…,e2n−1−ie2n.(iv)
Ifm=2n+1 and q are odd, the ordered b-isotropic basis is(13)e1+ie2,e3+ie4,…,e2n+ie2n+1,eq,e2n−ie2n+1,…,e1−ie2,where(14)E−=e1+ie2,e3+ie4,…,eq,e1−ie2,e3−ie4,…,eq−2−ieq−1,E+=eq+1+ieq+2,…,e2n+ie2n+1,eq+1−ieq+2,…,e2n−ie2n+1.Remark 1.
Ifm=2n+1, we have the vector eq or eq+1 in the ordered b-isotropic basis. In this case, the vector sits in a fixed position in the middle of the basis and the terms at the beginning skip over eq or eq+1.
The above bases define the standard maximal torusTS in each case as the subgroup of diagonal matrices, i.e.,(15)TS=g∈G0:g=diagt1,t2,…,tn,−tn,…,−t2,−t1,ti∈ℂifm=2n,orTS=g∈G0:g=diagt1,t2,…,tn,0,−tn,…,−t2,−t1,ti∈ℂifm=2n+1.
For any of the ordered bases above, we denote byFS the associated flag in Z, where the reordering on these bases will determine the flag domains, the base cycles, and the intersection points.
Just as in the case ofTS, the maximal torus TI is defined to have a certain basis of eigenvectors which depends on m being odd or even.(i)
Ifm is even, then the basis is(16)e1+e2q,…,eq+eq+1,e2q+1+ie2q+2,e2q+3+ie2q+4,…,e2n−1+ie2n,e2n−1−ie2n,…,e2q+3−ie2q+4,e2q+1−ie2q+2,eq−eq+1,…,e1−e2q.(ii)
Ifm is odd, then the basis is(17)e1+e2q,e2+e2q−1,…,eq+eq+1,e2q+1,e2q+2+ie2q+3,e2q+4+ie2q+5,…,e2n+ie2n+1,e2q+1,e2n−ie2n+1,…,e2q+4−ie2q+5,e2q+2−ie2q+3,eq−eq+1,…,e2−e2q−1,e1−e2q.We denote byBI the Iwasawa–Borel subgroup of G which fixes the associated flag FI.


## 2.4. Weyl Groups, Flag Domains, and Base Cycles

Let us use shorthand notation for the bases used above. Byr,s−form, we mean a basis r1,…,rn,sn,…,s1 where bri,rj=bsi,sj=0 and bri,sj=δij for all i and j. This occurs in the even-dimensional case. By r,t,s-form, which occurs in the odd-dimensional case, ri and si satisfy the same conditions and t is a single vector with bt,t=±1 and bt,ri=bt,si=0 for all i.Here, we discuss the Weyl groupsWTS and WTI by their actions on these bases. If T is either of these tori, then it stabilizes the spaces spanned by ri and si for i=1,…,n for both kinds of bases. In this, we regard T as a product T=T1⋅…⋅Tn, in the second case acting trivially on the space spanned by t. In both cases, an arbitrary permutation in Sn acting diagonally on these spaces by ri,si⟼rπi,sπi normalizes the T-action and is in the given orthogonal group G.A simple reflectionri,si⟼si,ri for one such i, and being the identity on the other 2-dimensional spaces, is denoted by a sign change. This normalizes T but has negative determinant. In the first case, this means that only an even number of sign changes are allowed. In the second case, we may couple the sign change with the map t⟼−t so that this slightly modified sign change has positive determinant. Thus, in both cases, we have the action of Sn⋉ℤ2n normalizing the T-representation on the basis at hand. This is in fact the action of the full Weyl group.As in the cases ofSp2n,ℝ and SO∗2n, to describe the flag domains and their K0-base cycles, we use the Weyl group WTS of the standard torus TS. The base point for α is the flag FS defined by the standard basis. Then, all base points are Fα≔wFS where w∈ℤ2n−1 if m=2n and w∈ℤ2n if m=2n+1 is associated with the sign-change vector α. As before, the base cycles and flag domains are the orbits K.Fα and G0.Fα, respectively, and α defines a nondegenerate sign with signFα=a,b and then Dα=Da,b.In our particular case, there are two maximal tori, the maximal torusTI and the maximal torus TS, which were defined earlier. Recall that these two maximal tori TIandTS are conjugates, and a conjugation induces an isomorphism ψ of the associated Weyl groups. In this case, the bijective map ψ is described, depending on the case, as follows:(i)
Ifm=2n and q are even, define WTS to be the Weyl group with respect to the basis (7) and let WI be the Weyl group with respect to the basis (16). In this case, the bijective map between WI and WTS is ψ∓2i−1=±i,ψ∓2i=±q−i+1 if 1≤i≤q/2 and ψ±i=±i if i>q.(ii)
Ifm=2n and q are odd, define WTS to be the Weyl group with respect to the basis (9) and let WI be the Weyl group with respect to the basis (16); then, the bijective map in this case is ψ:WI⟶WTo defined by ψ∓2i−1=±i,ψ∓2i=±q−i if 1≤i≤q−1/2 and ψ±i=±i−1 if i>q and ψ−q=−n.(iii)
Ifm=2n+1 and q are even, define WTe to be the Weyl group with respect to the basis (11) and let WI be the Weyl group with respect to the basis (17); then, we can define the bijective map between them to be ψ∓2i−1=±i,ψ∓2i=±q−i+1 if 1≤i≤q/2 and ψ±i=±i if i>q.(iv)
Ifm=2n+1 and q are odd, define WTo to be the Weyl group with respect to the basis (13). Let WI be the Weyl group with respect to the basis (17). The bijective map ψ between WI and WTo can be defined as ψ∓2i−1=±i,ψ∓2i=±q−i if 1≤i≤q−1/2 and ψ±i=±i if i>q.Moreover, through an argument quite similar to Proposition1, the base cycle Cα is the set(18)Cα=F∈Z:Vi=Vi∩E−⊕Vi∩E+,1≤i≤m,where E− and E+ defined in Section 2.3 and the intersection dimensions are determined by α (resp., a,b). The flag domains are parametrized by the signature of their base cycles.


## 2.5. The Length of the Elements ofSn⋉ℤ2n

Recall that everyB-Schubert cell S contains exactly one T-fixed point. Hence, at least in Z=G/B where the Weyl group acts freely and transitively on FixT, we should be able to compute dimℂS in terms of a corresponding Weyl element. In fact, if z0 is the B-fixed point in Z=G/B and w∈W, then the complex dimension of the Schubert cell Sw=B.wz0 is the length ℓw.As indicated above, we used a special way of writing the Weyl elements which is defined in the paper [2]. For dimension computations, let us state how we can compute the length of elements w∈Sn⋉ℤ2n relative to the notation for the Weyl group elements used in [2].Lemma 1 (see [2]).
Fixw∈Sn⋉ℤ2n and construct w˜∈Sn⋉ℤ2n by the following algorithm:(1)
Starting from left to right inw, using simple reflections, place all positive numbers in w step by step in a sequence of n−empty boxes beginning from the first one in w˜, in the same order as they appeared in w.(2)
From left to right inw, replace a negative number with its absolute value in the n−empty boxes starting from right to left.Ifw˜=w˜1,w˜2,…,w˜n, then define Lw˜=n2−n/2−number ofw˜i:i<kandw˜i<w˜k, and if we have m negative signs in w in the following positions j1,j2,…,jm, then define fw=∑i=1mn−ji. It follows that the length of w is(19)lw=Lw˜+fw+m.Example 1.
Letw=5−34−12∈S5⋉ℤ25, then by following the above remark, we have(20)5−34−12⟹542−3−1⟹54213,so w=5−34−12 and w˜=54213, and then Lw˜=8 and fw=4.
Hence,lw=8+4+2=14.


## 2.6. Description of theBI-Orbits

In the previous section, we have chosen a flagFI defined by the basis(21)e1−e2n,e2−e2n−1,…,en−en+1,en+en+1,…,e2+e2n−1,e1+e2n,which belongs to the G0-closed orbit γcl. Given a Weyl element w∈W, we consider the Schubert cell BI.Fw and an element F=bFw in it which is defined by an ordered basis b.vw1,…,b.vw2n, b∈BI, where vi=ei−e2n−i+1,1≤i≤n, and vi=e2n−i+1+ei,n+1≤i≤2n. If F∈Cα, then it can be defined by an ordered basis ε1,…,ε2n where for all k the vector εk is either in V− or in V+ and is a linear combination(22)εk=∑j≤kcjb.vwj.In this case, we call the basis asplit basis. According to the above discussion for Sp2n,ℂ, we can write the formulas for b.vwj for any element b∈BI as follows:(23)b.vwj=ηje−wj+e2n+wj+1+ζje−wj−e2n+wj+1+Bj,−n≤wj≤−1,b.vwj=ζjewj−e2n−wj+1+Bj,1≤wj≤n,where Bj does not involve the basis vectors ewj and e2n−wj+1.Remark 2.
When discussing theBI-orbit of the base point FI, it is important to explicitly understand the orbits BI.ei−e2n−i+1 and BI.ei+e2n−i+1. As we explained above, we have the following facts.
IfFI and BI are as above, then the orbits of interest are BI.ei±e2n−i+1, i=1,…,n. In this case, the orbits BI.ei+e2n−i+1 and BI.ei−e2n−i+1 have points of the forms(24)bei+e2n−i+1=λei+e2n−i+1+⋯+bnen+en+1+anen−en+1+⋯+a1e1−e2n,bei−e2n−i+1=λei−e2n−i+1+ai−1ei−1−e2n−i+⋯+a1e1−e2n,respectively, with λ≠0. Note that, in the above orbits, if b is chosen appropriately, then we can arrange bei+e2n+i−1=ei or bei+e2n+i−1=e2n−i+1 for all i. Does this play a role in the proof of Theorem 1.


## 2.7. Flag Domains and Base Cycles

In order to parametrize the flag domains, it is enough to parametrize the closedK-orbits (see [3], Sections 4.2 and 4.3). For this, we choose a base point FS∈FixTS. It follows that the set C of closed K-orbits can be identified with the Weyl group orbit WGTS.FS (see [3], Corollary 4.2.4). As we have observed above, WGTS≅Sn⋉ℤ2n where the Sn factor can be identified with WKTS. In fact, it is exactly the stabilizer of K.FS in C. Consequently, C can be identified with ℤ2n.FS. We regard an element of ℤ2n as a vector α of length n consisting of plus and minus signs, and thus, C can be identified with the set of such vectors. In concrete terms, the K-orbits (resp., G0-orbits) of base points ±e1,…,±en are exactly the base cycles (resp., flag domains) in Z. The respective flag domains are denoted by Dα.Observe thatα defines a nondegenerate signature a,b and that Dα⊂Da,b≔F:signF=a,b. Conversely, given F with signF=a,b, we may choose a basis v,w≔v1,…,vn,wn,…,w1 which defines F and which has the following properties:(1)
b is in canonical form(2)
v,w is h-orthogonal(3)
vjh2=±1 with sgnvjh=αj(4)
wjh2=−vjh2IfsignF=signF^, then we choose bases v,w and v^,w^ for F and F^, respectively, and note that the transformation which takes the one basis to the other is in G0. This argument shows that Da,b=Dα.DenoteE+=e1,…,en and E−=en+1,…,e2n, where ℂ2n=E+⊕E−.Proposition 1.
For a fixed open orbitDα, the base cycle Cα is the set(25)Cα=F∈Z:Vi=Vi∩E−⊕Vi∩E+,1≤i≤2n.Proof.
Firstly, it was shown above that there is a base pointFα with the splitting condition. Since K0 is the product K0=K0+×K0−, it acts transitively on all such flags. Let F∈Cα where Vi−=Vi∩E− and Vi+=Vi∩E+. Define F+ and F− to be the sets of maximally isotopic flags in E+ and E−, respectively. Recall that K0=K0+×K0− where K0+ and K0− act transitively on the sets F+ and F−, respectively, which implies that K0 acts transitively on the set of maximal isotropic flags obtained by put flags from F+ and F− in a way such that the new flag has signature α. Hence, Cα is a complex manifold. But K0 has a unique orbit in Dα which is a complex manifold. Therefore, Cα is the base cycle.


## 3. Conditions forSw∩Cα≠ø

Here, we deal with the homogeneous spaceZ of maximally isotropic full flags of the complex orthogonal symmetric group G=SO2n,ℂ equipped with the action of the real form G0=SOp,q. The general results here are stated in terms of algorithms (See Definitions 1 and 2); in fact, it seems impossible to avoid this. In Corollary 1, we give concrete formulas for the intersection points in Sw∩Cα if the intersection is nonempty and Sw is of complementary dimension. Also, the number of intersection points with C0 is explicitly computed in Theorem 2.Since the Schubert variety Sw is determined by the Weyl element w, we will describe in this section the conditions for an element w of the Weyl group which parametrize the Schubert variety Sw that satisfiesSw∩Dα≠∅ for some flag domain Dα. As would be expected, a special type of permutation plays a fundamental role.Definition 1.
An elementw∈W is called a harmonic permutation if it satisfies the following conditions:Ifq is even, the number −2i−1,1≤i≤q/2, sits in any place to the left of the number 2i or −2i,1≤i≤q/2, in the one line notation of the permutation and the order of the numbers or −q+i, where 1≤i≤p−q is arbitrary.Ifq is odd, the number −2i−1,1≤i≤q−1/2, sits in any place to the left of the number 2i,1≤i≤q−1/2, and the number −q sits in the last position in the one line notation of the permutation and the order of the numbers q+i or −q+i, where 1≤i≤p−q/2 is arbitrary.Example 2.
InSO4,2, the relevant pairs are −12 and −1−2. As a result, we have 6 harmonic permutations. They are −12−3,−1−32,−3−12,−1−23,−13−2,and3−1−2.
Recall that the fixed point in the closed orbitγcl is the flag associated with the following ordered basis:(i)
Ifm is even, then the basis is(26)e1+e2q,…,eq+eq+1,e2q+1+ie2q+2,e2q+3+ie2q+4,…,e2n−1+ie2n,e2n−1−ie2n,…,e2q+3−ie2q+4,e2q+1−ie2q+2,eq−eq+1,…,e1−e2q.(ii)
Ifm is odd, then the basis is(27)e1+e2q,e2+e2q−1,…,eq+eq+1,e2q+1,e2q+2+ie2q+3,e2q+4+ie2q+5,…,e2n+ie2n+1,e2q+1,e2n−ie2n+1,…,e2q+4−ie2q+5,e2q+2−ie2q+3,eq−eq+1,…,e2−e2q−1,e1−e2q.Remark 3.
Forvi, any such basis vector, and b∈BI, the form of b.vi is given as follows:(1)
b.ei−e2q−i+1=λei−e2q−i+1+⋯+bneq−eq+1+aneq+eq+1+⋯+a1e1+e2q(2)
b.ei+e2q−i+1=λei+e2q−i+1+ai−1ei−1−e2q−i+⋯+a1e1−e2q(3)
b.e2q+2j−1−ie2q+2j=λe2q+2j−1−ie2q+2j+⋯+bmem−1−em+amem−1+em+⋯+a1e1+e2q where m=2n or 2n+1(4)
b.e2q+2j−1+ie2q+2j=λe2q+2j−1+ie2q+2j+ai−1e2q+2j+1+ie2q+2j+2+⋯+a1e1+e2q(5)
b.e2q+1=λe2q+1+ane2n+ie2n+1+⋯+a1e1+e2q if m=2n+1λ≠0 in all cases above. Note that, in the above orbits, if b is chosen appropriately, then we can arrange all linear combination for every one of the above vectors to be in the standard basis of TS-eigenvectors. See Theorem 1.
In the following result,ψ:WI⟶WS denotes the bijective map between Weyl groups which was introduced in Section 2.Proposition 2.
Ifw is a harmonic permutation, then the flag Fψw belongs to the orbit BIFw.Proof.
We handle the case wherem=2n. The proof for m=2n+1 goes analogously. Let us first prove the theorem for the case that q is even. For this, let w∈WI be a harmonic permutation and define Fw=w.FI to be the isotropic full flag associated with w. Denote by(28)0⊂uw1⊂uw1,uw2⊂…⊂uw1,…,uwn,the first n subspaces of Fw=wFI, where uwi is a vector from the basis above. Let w˜∈WS be the image of w under the bijective map ψ, and let Yw˜ be the isotropic flag associated with w˜ such that the first half of Yw˜ is(29)0⊂εw˜1⊂εw˜1,εw˜2⊂…⊂εw˜1,…,εw˜n.
Our claim here is that this flag is an intersection point inB.Fw∩Cα. To prove this, we will construct b with bFw=Yw˜. From the definition of harmonic permutation, there are two possibilities for w1: w1=2i−1≤q or w1>q.Case 1: ifw1>q, then uw1=vw˜1, so the orbit BI.uw1 contains the point εw˜1.Case 2: ifw1=2i−1≤q, then we must consider the orbit BI.e2w˜1−1−e2q+w1+1. By Remark 3 above, the orbit BI.e2w˜1−1−e2q+w1+1 contains points of the form(30)y=α1e2w˜1−1+e2q+w1+1+α2e2w˜1+e2q+w1+α3e2w˜1−e2q+w1+α4e2w˜1−1−e2q+w1+1,where α1=±α4 and α2=±α3. By taking α1=α4=1/2 and α2=α3=1/2i, it follows that y=e2w˜1−1+ie2w˜1=vw˜1.
To construct thej-vector of bvwj to obtain the subspace Vw˜j, we must consider three cases:Case 1: ifwj>q, then wj=w˜j and εw˜j=uwj, so the orbit BI.uwj contains the point εw˜j.Case 2: ifwj=2i−1≤q, then our job goes through the orbit BI.e2w˜j−1−e2q+wj+1. By using Remark 3, we see that the orbit BI.e2w˜j−1−e2q+wj+1 contains points of the form(31)y=α1e2w˜j−1+e2q+wj+1+α2e2w˜j+e2q+wj+α3e2w˜j−e2q+wj+α4e2w˜j−1−e2q+wj+1,where α1=±α4 and α2=±α3. By taking α1=α4=1/2 and α2=α3=1/2i, it follows that y=e2w˜j−1+ie2w˜j=εw˜j. Therefore, b is constructed to obtain the flag(32)0⊂εw˜1⊂εw˜1,εw˜2⊂…⊂εw˜1,…,εw˜j.Case 3: ifwj=2i, then the orbit is BI.ewj+e2w˜j−1 which is relevant. In this case, the points(33)y=α1ewj−1+e2w˜j+α2ewj+e2w˜j−1+α3ewj−1+iewjbelong to the orbit BI.ewj+e2w˜j−1. For α1=−i,α2=1, and α3=i, we have y=e2w˜j−1+ie2w˜j. Therefore, the j-vector of b is constructed in this case as well.
Thus, by induction, we observe thatb∈BI can be constructed with bFw=Yw˜.
We complete the proof by handling the case whereq is odd. For this point, we can repeat the steps of the proof above for wj>q and wj=2i−1. For the case where wj=2i, we apply the same method as above, only changing 2w˜j by 2w^j+1. If wn=−q, then w^j=−n. In this case, the orbit of relevance is BI.eq−eq+1. As we see from Remark 3, y=α1eq+eq+1+α2eq−eq+1 belongs to this orbit. The desired result is then obtained by taking α1=1,α2=−1, and the point y=eq+1 belongs to the orbit BI.eq−eq+1. Therefore, the element of BI is also constructed in the case where q is odd.Theorem 1 (harmonic permutation theorem).
The following are equivalent:(i)
w is harmonic(ii)
BIFw∩Dα≠∅ for some α
Under either of these conditions for everyα, the nonempty intersection BIFw∩Cα contains a TS-fixed point.Proof.
(i)⟹ (ii) is exactly the statement of Proposition 2.
(ii)⟹ (i) Assuming that w is not harmonic permutation, we will show that Sw∩Dα=∅, for all α, i.e., Sw has no TS-fixed points. Assume on the contrary that there exists b∈BI such that b.Fw∈Sw∩Cα. For bFw to be fixed, b has to have a certain shape, and at each stage where the condition of harmonicity is violated, we should prove that there is no such b. For this, recall that the complex bilinear form b has been defined to satisfy the following orthogonality condition:(34)bej−e2q−j+1,ej+e2q−j+1=1,1≤j≤q,bej±e2q−j+1,ek±e2q−k+1=0,fork≠j.
Sincew is not a harmonic permutation, there exists a pair of the form −2i−1,2i or of the form −2i−1,−2i, where 1≤i≤q/2, such that ∓2i sits to the left of −2i−1. Assume that wj=2i is the first even number which appears in w such that 2i sits to the left of −2i−1. Then, b does in fact yield a TS-fixed partial flag, i.e.,(35)0⊂εw˜1⊂εw˜1,εw˜2⊂…⊂εw˜1,…,vw˜j−1.
Now, we check that there is nob∈BI so that b.vwj defines an extended partial flag which is TS-fixed. For this, we consider the orbit BI.ewj+e2w˜j−1 which contains points of the form(36)α1e1+e2q+α2e2+e2q−1+⋯+αjewj+e2w˜j−1.
This is a linear combination ofh-isotropic vectors for all αi. Recall that the flag of intersection has non-h-isotropic vectors, and to have one of these vectors in this step which is linearly independent of all vectors in the flag (35) above, we should add the point ewj−1+iewj from the flag (35) to the linear combination in (36), but this vector is not in the flag (35). Thus, as was claimed, no b∈BI has the property that bFw is TS-fixed.
Ifwj=−2i is the first even number such that −2i sits to the left of −2i−1, then the relevant orbit is BI.ewj+e2w˜j−1 which contains points of the form(37)α1e1+e2q+⋯+αqeq+eq+1+αq+1eq−eq+1+⋯+αsewj−e2w˜j−1.
To have the nonisotropic point which is linearly independent of all points in the flag (35), we should add any of the points ewj−1+iewj>, ewj−1−iewj>, e2q−wj+1+ie2q−wj+2>, or e2q−wj+1−ie2q−wj+2> from the flag (35) to the linear combination in (36). But the flag (35) does not contain any of these points. So again, for all b∈BI, the flag bFw is not TS-fixed, and consequently, Sα∩Dα=∅.


## 4. Introduction to the Combinatorics

For the remainder of this paper, we only discuss the intersection properties of the Iwasawa–Schubert cells which are of complementary dimension toC0. Recall that the maximal compact subgroup of SOp,q is K0≔SOp×Oq⊂SUp×Uq. For E+ and E− as in Section 2.4, the base cycle in the flag domain Da,b is given by(38)C0=F∈Z:dimVi∩E−=∑j=1iajanddimVi∩E+=∑j=1ibj,1≤i≤m.Ifm=2n, we have two cases: If q is even, the dimension of C0 is pp−2/4+qq−2/4. If q is odd, then C0 has the dimension p−12/4+q−12/4. If m=2n+1, then the dimension of the base cycle is p−12/4+qq−2/4 if q is even and is pp−2/4+q−12/4 if q is odd. Since we have restricted to the case where Z=G/B is the manifold of complete flags, it follows that dimSw=pq/2 if q and n are even and dimSw=pq−1/2 if q or n is odd.Definition 2.
A harmonic permutationw∈Sn⋉ℤ2n−1 is called a perfect harmonic permutation if it is constructed by the following algorithm:(A)Start with a sequence ofn empty boxes which are to be filled in order to construct w.(B)Consider the pairs−2j−1,2j and −2j−1,−2j for all 1≤j≤q/2.(C)Ifq is even,(1)
The pairs in step B are−1,2,−1,−2,−3,4,…,−q−1,q,−q−1,−q.(2)
Step by step, starting from−1,2 until −q−1,−q, for each 1≤j≤q/2, we have two pairs of the forms −2j−1,2j and −2j−1,−2j, so choose only one pair of them for each step and omit the other from the above list.(3)
If we choose the pair of the form−2j−1,2j, place this pair in any box in w such that the components −2j−1 and 2j of this pair sit as close as possible to each other.(4)
If we choose the pair of the form−2j−1,−2j, place this pair in any box in w such that the components −2j−1 and −2j of this pair sit as close as possible to each other and all pairs −2i−1,2i or −2i−1,−2i with i>j are sitting to the left of this pair and the pairs of the form −2i−1,−2i with i<j are sitting in a decreasing order with respect to i to the right of −2j−1,−2j.(5)
Once a pair is placed, its position can be ignored so that the places at the immediate left and right of this pair become adjacent.(6)
After all pairs are placed, the remaining numbers±q+i for 1≤i≤p−q/2 are placed in the remaining spots in the strictly increasing order with respect to wi such that all numbers ±q+i for 1≤i≤p−q/2 are sitting to the left of the pairs −2i−1,−2i1≤i≤q/2. If the number of negative signs in all pairs from steps 2 and 3 is even, then all the remaining numbers are positive, and if the number of negative signs is odd, then all the remaining numbers are positive except the number n is negative.(D)Ifq is odd,(1)
The pairs in step B are−1,2,−3,4,−5,6,…,−q−2,q−1,−q.(2)
In the last box ofw, put the number −q.(3)
Step by step, starting from−1,2 until −q−2,q−1, choose a pair and place this pair in any box of the first n−1 boxes in w such that the components −2j−1 and 2j of this pair sit as close as possible to each other. This means that once a pair is placed, it can be ignored so that the places at the immediate left and right of this pair become adjacent.(4)
After all pairs are placed, the numbers±q+i for 1≤i≤p−q/2 are placed in the remaining spots in the strictly increasing order with respect to wi. If the number of negative signs in all pairs from steps 1 and 2 is even, then all the remaining numbers are positive, and if the number of negative signs is odd, then all the remaining numbers are positive except the number n is negative.Remark 4.
Ifq is odd, then the signature of the flag domain Dα is + in the last position.Remark 5.
IfW=Sn⋉ℤ2n, then a perfect harmonic permutation consists only of pairs of the form −2j−1,2j for all 1≤j≤q/2. It is constructed as above, in particular such that the sign of n is +.Example 3.
Ifp=6,q=4, then the perfect harmonic permutations are(39)−12−345,−125−34,5−12−34,−34−125,−345−12,5−34−12,−12−5−3−4,−5−12−3−4,−5−3−4−12,−34−5−1−2,−5−34−1−2,5−3−4−1−2,−3−1245,5−3−124,−5−3−12−4.Example 4.
Ifp=10,q=6, then the element 78−56−3−4−12 is a perfect harmonic permutation, while the element 78−3−4−12−56 is not a perfect harmonic permutation.
Recall that the dimension of the cells corresponds to the length of the wordw in the Weyl group, i.e., if Fw=w.FI, then dimB.Fw=lw. So if we want to discuss the dimension of the Schubert cell, it is enough to discuss the length of Weyl elements.Proposition 3.
Every perfect harmonic permutationw∈Sn⋉ℤ2n−1 has dimension pq/2 if q is even and pq−1/2 if q is odd.Proof.
Given a perfect harmonic permutationw, we consider three cases. These depend on which of the pairs −12,−34,−1−2, or −3−4 is contained in w. In each of these cases, our proof goes by induction on the dimension of the flag manifold. Without loss of generality, let p>q≥6 because if q≤6, then the permutation has only one or two pairs and the proof becomes trivial.Case 1.
The permutationw contains the pairs −12 and −34.
First, we remove the pairs−12 and −34 from w to have a new permutation v consisting of the numbers 5,6,…,n. We define a function f:5,…,n⟶1,…,n−4 by fi=i−4 for all 1≤i≤n−4. This is a bijective map which sends v to w^∈Sn−4⋉ℤ2n−5. Note that w^ is a perfect harmonic permutation in Sn−4⋉ℤ2n−5. Thus by the induction assumption, lw^=p−4q−4/2. Since v=f−1w^, it follows that v has the same length as w^. So we put the numbers 1234 to the left of v to have an element w˜∈Sn⋉ℤ2n−1 with length p−4q−4/2. To split the sign of 1 and 3 (i.e., to change the positive sign to the negative sign), we add n−4+1 to p−4q−4/2 to send 3 to the last position and add n−4+2 to send 1 to the position before the last position. Consequently, we have the element 24v13 with length p−4q−4/2+n−4+1+n−4+2=p−4q−4/2+2n−5, and it follows that the element 24v−3−1 has length p−4q−4/2+2n−4. We then return to the original w and remove only the pair −12. Then, we define g to be the number of positions to the left of −34. As a result, we have n−4−g positions to the left of −34, and −3 should cross n−4−g+1 positions to end up in the last position and 4 should cross g boxes to end up in the first position.
Finally, we return to the originalw and define h to be the number of positions to the left of −12 and n−2−f to the right. In this situation, −1 must cross n−2−h+1 positions to end up in the last position and 2 must cross g positions to end up in the first position. It follows that the length of w is p−4q−4/2+2n−4+n−4−g+1+g+n−2−h+1+h=pq/2.
Ifq is odd, then the length of w^∈Sn−4⋉ℤ2n−5 is lw^=p−4q−4−1/2. After applying the same steps as above, it follows that lw=pq−1/2.Case 2.
The permutationw contains the pair −1−2.
Note that this case only occurs ifq is even. Since w is a perfect harmonic permutation, the pair −1−2 sits in the last 2 positions of w. We remove the pair −1−2 from w to obtain v^. Since q is even, a similar argument to that shown above shows that lv^=p−2q−2/2 because q is even. We put the pair 12 to the left of v^. It follows that v˜=12v^ has the same length as v^. To split the sign of 1 and 2, each of them must cross n−2 positions. Having made this move, we then apply 2 reflections to obtain the pair −1−2 in the last two positions. Hence, the length of w becomes lw=p−2q−2/2+2n−2+2=pq/2.Case 3.
The permutationw contains the pairs −12 and −3−4.
Note that this case appears only ifq is even. In this case, these two pairs appear in w in the following form: −3−4−12 or −3−12−4, or the pair −3−4 sits in the last two positions of w. If the pair −3−4 sits in the last two positions of w, then the argument goes as in Case 2 above.
If we have the form−3−12−4, then we must add n−4 to p−2q−2/2 to put 3 in the last position and add n−2 to p−2q−2/2 to put 4 in the last position. Then, we must add 1 to split the signs and 3 to send −3 to its position in the originalw; then, the length of w becomes p−2q−2/2+n−4+n−2+1+3=pq/2.
If we have the form−3−4−12, then we must add n−4 to p−2q−2/2 to put 3 in the last position and add n−1 to p−2q−2/2 to put 4 in the position before the last position. Then, we must add 1 to split the signs, 3 to send −3 to its position in the original w, and 1 to send −4 to its position in the original w. It follows that the length of w is p−2q−2/2+n−4+n−3+1+3+1=pq/2.Proposition 4.
Every perfect harmonic permutationw∈Sn⋉ℤ2n has dimension pq/2.Proof.
The argument goes exactly along the lines as that of the above proposition. Here, it is in fact simpler because only the pairs−12 and −34 appear.


## 5. Intersection Points of Schubert Duality

Letw∈WI be a perfect harmonic permutation, in particular so that Sw is of complementary dimension to the cycles. Recall that, in this case, either Sw∩Cα=∅ or is pointwise TS-fixed (see Section 2.3). The main goal in this section is to compute all such intersection points. The argument in the case of SOp,q is carried out by means of algorithms. Nevertheless, we are able to provide formulas for the cardinality of Sw∩Cα when it is nonzero and the total number of cycles Cα for which this intersection is nonempty (see Theorem 2).Proposition 5.
Ifw is a perfect harmonic permutation so that BI.Fw intersects a cycle Cα at a point given by the ε-basis ε1,…,εm of TS-eigenvectors, then for any such eigenvector εk, it follows that the ε-basis is given by εk=e2j−1+ie2j or e2j−1−ie2j if q and p are even and εk=e2j−1+ie2j, e2j−1−ie2j, eq, eq+1, or e2q+1 if q or p is odd, depending on the dimension m and the signature α.Proof.
This is a consequence of the following:(1)
w is a perfect harmonic permutation and the flag basis is that of wFI.(2)
We have the following cases:(i)
b.vj=ηjer+e2q−r+1+ζjer−e2q−r+1+η˜jer+1+e2q−r+ζ˜jer+1−e2q−r+Bj=Kj+Bj(ii)
b.vj=ηje2q+r+ie2q+r+1+Bj=Kj+BjHere,η˜j=±iηj and ζ˜=±iζj and Bj does not involve the basis vectors er, er+1, e2q−r+1, and e2q−r.(iii)
b.vj=ηje2q+1+Bj=Kj+Bj,ifm=2n+1(3)
ηj≠0 and η˜j≠0.(4)
The intersectionSw∩Cα is a flag defined by TS-eigenvectors.
From the expression forvj, it is obvious that all of the possibilities in the statement occur. Furthermore, since ηj≠0 and η˜j≠0, for every j, a nonzero contribution from Kj occurs in the sum(40)vj=∑k≤jckjb.vwk.
Sincee2j−1+ie2j and e2j−1−ie2j and eq or eq+1 do not occur in b.vwk for k<j, it follows that εj=Kj+Ej in the standard basis. Finally, since εj is a TS-eigenvector, it follows that εj=Kj where Kj of the εj forms indicated in the proposition.
As a result of the above proposition, the following corollary gives us all intersection points ofSw∩Cα.Corollary 1.
LetDα be a flag domain parametrized by a sequence α and w∈WI be a perfect harmonic permutation such that Sw∩Dα≠ϕ. Then, the following algorithm produces us all intersection points of Sw∩Cα:Ifq is even,(i)
Consider a copy of α denoted by β.(ii)
For any pair−2j−1,2j,1≤j≤q/2, in w, if the corresponding signature of it in β is +−, then replace this +− in β by(41)e2q−2j+1+ie2q−2j+2,e2j−1−ie2jore2q−2j+1−ie2q−2j+2,e2j−1+ie2j,
and if the corresponding signature of it inβ is −+, then replace this −+ in β by
(42)
e
2
j
−
1
+
i
e
2
j
,
e
2
q
−
2
j
+
1
−
i
e
2
q
−
2
j
+
2
or
e
2
j
−
1
−
i
e
2
j
,
e
2
q
−
2
j
+
1
+
i
e
2
q
−
2
j
+
2
.
(iii)
For any pair−2j−1,−2j,1≤j≤q/2, in w, if the corresponding signature of it in β is +−, then replace this +− in β by
(43)
e
2
q
−
2
j
+
1
+
i
e
2
q
−
2
j
+
2
,
e
2
j
−
1
+
i
e
2
j
or
e
2
q
−
2
j
+
1
−
i
e
2
q
−
2
j
+
2
,
e
2
j
−
1
−
i
e
2
j
,
and if the corresponding signature of it inβ is −+, then replace this −+ in β by
(44)
e
2
j
−
1
+
i
e
2
j
,
e
2
q
−
2
j
+
1
+
i
e
2
q
−
2
j
+
2
or
e
2
j
−
1
−
i
e
2
j
,
e
2
q
−
2
j
+
1
−
i
e
2
q
−
2
j
+
2
.
(iv)
For the remaining numbers, for eachq+1≤j≤n−1, replace the corresponding + in β by e2j−1+ie2j, and for ±n, replace the corresponding + by e2n−1±ie2n.
Ifq is odd,
(i)
For any pair−2j−1,2j,1≤j≤q−1/2, in w, if the corresponding signature of it in β is +−, then replace this +− in β by
(45)
e
2
q
−
2
j
+
1
+
i
e
2
q
−
2
j
+
2
,
e
2
j
−
1
−
i
e
2
j
or
e
2
q
−
2
j
+
1
−
i
e
2
q
−
2
j
+
2
,
e
2
j
−
1
+
i
e
2
j
,
and if the corresponding signature of it inβ is −+, then replace this −+ in β by
(46)
e
2
j
−
1
+
i
e
2
j
,
e
2
q
−
2
j
+
1
−
i
e
2
q
−
2
j
+
2
or
e
2
j
−
1
−
i
e
2
j
,
e
2
q
−
2
j
+
1
+
i
e
2
q
−
2
j
+
2
.
(ii)
For the number−q, replace the corresponding + in β by eq+1.
(iii)
For the remaining numbers, for eachq+1≤j≤n−1, replace the corresponding + in β by e2j−1+ie2j, and for ±n, replace the corresponding + by e2n−1±ie2n.
Each point obtained from this algorithm is a point of the intersection ofSw∩Dα.Theorem 2.
A Schubert varietySw which is parametrized by a perfect harmonic permutation w intersects 2q/2 flag domains if q is even and intersects the base cycles of these flag domains in 2q points. If q is odd, it intersects 2q−1/2 flag domains and intersects the base cycles of these flag domains in 2q−1 points.Proof.
Letw∈WI be a perfect harmonic permutation. We first show that if q is even, then Sw∩Cα consists of 2q points and 2q−1 points if q is odd. Since w is a perfect harmonic permutation, we have two cases. In the first case, if w contains the pair −12, then the pair −12 sits inside consecutive boxes in w. The goal here is to show that there are exactly 4 possibilities for this pair which can be completed to the maximal isotropic flag. We begin by considering the BI-orbit of e1−e2q. By Remark 3, the elements in this orbit have the form(47)β1e1+e2q+⋯+β2ne1−e2q.
The question is how many 1-dimensional subspaces (spanned by vectors in Proposition5) do we have such that these subspaces can be completed to the maximal isotropic flag. To compute this number, we denote by v1 the vector we have from the first step which spans the 1-dimensional subspace; in the second step, we consider the orbit BI.e2+e2q−1, which has points of the form(48)β1e1+e2q+β2e2+e2q−1.
The 2-dimensional subspace in the flag is spanned by linear combinations of the form(49)v2=β1e1+e2q+β2e2+e2q−1+γv1.
Note thatv2 should be in E+ or in E− and of the form stated in Proposition 5, and therefore, v1 should contain the terms e1 and e2 or the terms e2q−1 and e2q. Thus, we have the following 4 possibilities of v1 as follows: e1−ie2,e1+ie2,e2q−1+ie2q,ande2q−1−ie2q.
Ifv1=e1∓ie2, then for β1=±i,β2=1, and γ=∓i, the vector v2 is v2=e2q−1±ie2q, so the 2-dimensional subspace corresponding to the pair −12 is spanned by(50)e1∓ie2,e2q−1±ie2q.
Ifv1=e2q−1∓ie2q, then for β1=1,β2=±i, and γ=∓i, it follows that v2 is v2=e1±ie2. Thus, the 2-dimensional subspace corresponding to the pair −12 is spanned by(51)e2q−1∓ie2q,e1±ie2.
Having constructed the 2-dimensional space, we ignore the pair−12 in w and repeat the same steps for the next pairs step by step. So if w contains the pair −2j−1,2j,1≤j≤q/2, then in the same way, the only possible 2-dimensional subspaces which can be completed to maximal isotropic flags are the subspaces spanned by(52)e2j−1∓ie2j,e2q−2j+1±ie2q−2j+2ore2q−2j+1±ie2q−2j+2,e2j−1∓ie2j.
The second case is wherew contains the pair −1−2 which sits inside consecutive boxes in w. Look at the orbit BI.e1−e2q, then this orbit has points of the form(53)β1e1+e2q+⋯+β2n−1e2−e2q−1+β2ne1−e2q.
We also consider the orbitBI.e2−e2q−1 which has points of the form(54)β1e1+e2q+⋯+β2n−1e2−e2q−1.
Then, if we have 1-dimensional subspace spanned byv1 from the first step, the 2-dimensional subspace in our flag spanned by v1 and a vector v2 is of the form v2=β1e1+e2q+β2e2+e2q−1+β3e2−e2q−1+γv1. So we have 4 possibilities of v1 which can be extended to the maximal isotropic flag. They are e1−ie2,e1+ie2,e2q−1+ie2q,ande2q−1−ie2q.
Ifv1=e1∓ie2, then for β1=∓i,β2=0,β3=−1, and γ=±i, the vector v2 is v2=e2q−1∓ie2q. Thus, the 2-dimensional subspace corresponding to the pair −1−2 is spanned by(55)e1∓ie2,e2q−1∓ie2q.
Ifv1=e2q−1∓ie2q, then for β1=1,β2=0,β3=∓i, and γ=∓i, the vector v2 is v2=e1±ie2. As a result, the 2-dimensional subspace corresponding to the pair −1−2 is spanned by(56)e2q−1∓ie2q,e1∓ie2.
Having determined the 2-dimensional subspace, we ignore the pair−1−2 from w and repeat the same steps for the next pairs step by step. More generally, if w contains the pair −2j−1,−2j,1≤j≤q/2, then by the same method, the only possible 2-dimensional subspaces which can be completed to the maximal isotropic flag are(57)e2j−1∓ie2j,e2q−2j+1∓ie2q−2j+2ore2q−2j+1∓ie2q−2j+2,e2j−1∓ie2j.
Therefore, for each pair ofw, we have 4 possibilities.
For the remaining numbers, recall thatw is a perfect harmonic permutation. In particular, all numbers in the remaining boxes sit in an increasing order. Thus, the only possibility for these which can be completed to the maximal isotropic flag is the following point: if n is positive, the point is the flag associated with the ordered basis(58)e2q+1+ie2q+2,…,e2n−1+ie2n,and if n is negative, then the point is the flag associated with the ordered basis(59)e2q+1+ie2q+2,…,e2n−3+ie2n−2,e2n−1−ie2n.
Hence, ifq is even, then w contains q/2 pairs and each pair has 4 possibilities. Therefore, in this case, the number of possible intersection points is 4q/2.1=2q. If q is odd, then w contains q−1/2 pairs and each pair has 4 possibilities; then, the number of possible intersection points is 4q−1/2.1=2q−1.
Finally, we show that the points described above belong to exactlyq/2 flag domains if q is even and to q−1/2 flag domains if q is odd. For this, recall that, for each pair in w, we have 4 possibilities of 2-dimensional subspaces and note that the signature of these subspaces is +− and −+, and the signature of the remainder is ++…++. Thus, the number of flag domains which have nonempty intersection with Sw for the given w is 2q/2 if q is even and is 2q−1/2 if q is odd.
Also, since each two of these 4 possibilities of 2-dimensional subspaces have the same signature, then for any fixed signature for a flag domain, there are2q/2 points belonging to the base cycle of that flag domain if q is even and 2q−1/2 points belonging to the same base cycle of that flag domain if q is odd.Remark 6.
In the case of the groupSL2nC with the real form SUp,q, Brecan [4, 5] shows that the number of Iwasawa–Schubert varieties which intersect at least one base cycle and have the minimal dimension pq is 2n−1.2n−3…2n−2q+1. But for the group SO2n,ℂ with the real form SOp,q, if q is even, the number of Schubert varieties Sw which have the minimal dimension pq/2 and intersect at least one base cycle is n.n−2…n−q+2, and if q is odd, the number of Schubert varieties Sw which have the minimal dimension pq−1/2 and intersect at least one base cycle is n−2.n−4…n−q+1.
In the case ofSO2n+1,ℂ with the real form SOp,q, if q is even, the number of Schubert varieties Sw which have the minimal dimension pq/2 and intersect at least one base cycle is n−1.n−3…n−q+1, and if q is odd, the number of Schubert varieties Sw which have the minimal dimension pq/2 and intersect at least one base cycle is n−2.n−4…n−q+1.
The following remark, which is a consequence of the proof of Theorem2, describes all intersection points of Sw with the base cycles Cα.Remark 7.
To determine all intersection points between the base cycles and the Iwasawa–Schubert varietySw of complementary dimension, we will define a set for each case of q:Ifq is even, let w∈WI be a perfect harmonic permutation and define the following:Switew≔ψwr:wr is obtained from w by switching none, some, or all pairs −2i−1,2iby2i,−2i−1or−2i,2i−1or2i−1,−2i and switching none, some, or all pairs −2i−1,−2i by −2i,−2i−1or2i,2i−1or2i−1,2i,1≤i≤q/2. Define FeFixTS to be the set of all maximally b-isotropic flags associated with the basis(60)e1+ie2,e3+ie4,…,e2n−1+ie2n,e2n−1−ie2n,…,e1−ie2.LetMw⊂FeFixTS be the set of all maximally b-isotropic flags associated with all elements in Switew. Note that we have q/2 of the pairs −2i−1,2i and −2i−1,−2i,1≤i≤q/2, in any w∈WI, and for each pair, we have 4 possibilities to switch it, so the cardinality of Switew is 4q/2=2q. The set Switew gives us all intersection points of Sw, and each 2q/2 of these points belongs to only one flag domain where these points of intersection sit in the base cycle of that flag domain.Ifq is odd, let w∈WI be a perfect harmonic permutation and define the following:Switow≔ψwr:wr is obtained from w by switching none, some, or all pairs−2i−1,2iby2i,−2i−1or−2i,2i−1or2i−1,−2i,1≤i≤q/2. Define FoFixTS to be the set of all maximal b-isotropic flags associated with the basis(61)e1+ie2,e3+ie4,…,eq−2+ieq−1,eq+2+ieq+3,…,e2n−1+ie2n,eq,eq+1,e2n−1−ie2n,…,eq+2−ieq+3,eq−2−ieq−1,…,e1−ie2.
LetMw⊂FoFixTS be the set of all maximally b-isotropic flags associated with all elements in Switow. Note that we have q−1/2 of the pairs −2i−1,2i,1≤i≤q/2, in any w∈WI, and for each pair, we have 4 possibilities for switching it. Hence, the cardinality of Switow is 4q−1/2=2q−1. The set Switow gives us all intersection points of Sw, and each 2q−1/2 of these points belongs to only one flag domain where these points of intersection sit in the base cycle of that flag domain.Example 5.
InG0=SO6,4, fix w=−35−142 as a perfect harmonic permutation, then(62)Switew=251−3−4,25−1−34,−2513−4,−25−134,−3512−4,351−2−4,−35−124,35−1−24,25−4−31,−25−431,254−3−1,−2543−1,−35−421,35−4−21,−3542−1,354−2−1.Example 6.
InG0=SO5,3, fix w=−124−3 as a perfect harmonic permutation, then(63)Switow=1−23−4,−213−4,2−13−4,−123−4.
Recall that, in the cases ofSP2n,ℝ and SO∗2n, every flag domain intersects all Schubert varieties of complementary dimension. But in the case of SOp,q, we do not have this property except in a very special case. We explain this case in the following example.Example 7.
Ifn=q+1, then the flag domain parametrized by the sequence(64)α=+−+−…+−+−+intersects all Schubert varieties of dimension pq/2 if q is even. And the flag domain parametrized by the sequence(65)β=+−+−…+−+−++intersects all Schubert varieties of dimension pq−1/2 if q is odd.



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