Cycle Intersection for SO(p, q)-Flag Domains

A real form G0 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 ofG0.+ese special orbitsC are characterized as the closed orbits in Z of the complexificationK ofK0. +ese are referred to as cycles. +e cycles intersect Schubert varieties S transversely at finitely many points. Describing these points and their multiplicities was carried out for all real forms of SL(n,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 SO(p, 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 theWeyl group and compute their total number. Furthermore, we give an explicit description of the points of intersection in terms of flags and their number.+e results in the case of G/Q for all real forms will be given by Abu-Shoga and Huckleberry.


Background
Let G be a complex semisimple Lie group and B a Borel subgroup. e compact algebraic homogeneous space Z � G/B is called a complex flag manifold. Let G 0 be a real form of G in the sense that there is an antiholomorphic involution τ : G ⟶ G such that G 0 is the fixed point set of τ. It is well known [1] that there are only finitely many G 0 -orbits on Z.
erefore, at least one of them must be open. Such open orbits are called flag domains.
Fix a Cartan involution θ : G 0 ⟶ G 0 . e fixed point set of θ in G 0 is a maximal compact subgroup of G 0 denoted by K 0 , and K is its complexification.
If D is a flag domain on Z, then K 0 has exactly one open orbit in D that is a complex submanifold and denoted by C 0 .
e complex submanifold C 0 is both the unique K-orbit in D that is compact and the unique K 0 -orbit in D that is complex. is is the origin of what is known as "Matsuki duality." Consider an Iwasawa decomposition G 0 � K 0 A 0 N 0 of the real form G 0 . We refer to a Borel subgroup B ⊂ G as an Iwasawa-Borel subgroup of G if it contains an Iwasawa factor A 0 N 0 . Of course, these are just the Borel subgroups which occur as the isotropy groups at points of the closed us, although the determination of the r-codimensional Iwasawa-Schubert varieties which have nonempty intersection with D, along with their points of intersection with C 0 , is apparently a problem of a combinatorial nature, such information has complex analytic significance.
Since S ∩ D ≠ ∅ implies that S ∩ C 0 ≠ ∅, the first goal of this paper is to determine which Schubert varieties S have nonempty intersection with C 0 . After doing so, we then describe this intersection in the case where S is of complementary dimension to C 0 .
e Schubert varieties are determined by the elements of the Weyl group W I of a distinguished maximal torus T I in the Borel subgroup B I which fixes a certain base point in c cℓ .
ese Schubert varieties are denoted by S w where w ∈ W I . Consequently, our results are formulated in terms of conditions on elements w in the Weyl group W I . 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.)

Flag Intersection for SO(p, q)
2.1. Preliminaries. Consider the semisimple Lie group G � SO(m, C) with the real form SO(p, q). In this case, it is convenient to choose the bilinear form b on C 2n depending on p and q. If p or q is even, then we choose it in the usual way: Let σ be the complex conjugation σ(v) � v, then the Hermitian form h(v, w) of signature (p, q) which defines the real form is defined by If both p and q are odd, then the complex bilinear form is (3) and the Hermitian form h(v, w) is defined by e real form is G 0 � SO(p, q) � SO(2n, C) ∩ SU(p, q). A Cartan involution θ � so(p, q) ⟶ so(p, q) in the Lie algebra level is given by θ(g) � − g t , and the maximal compact subgroup associated with it is K 0 ≔ S(O(p) × O(q)) ⊂ S(U(p) × U(q)).

Introduction to the Flags in Z.
Recall that the (indefinite metric) orthogonal group G � SO(p, q) acts on the space of full flags by two orbits: closed orbit and open orbit. e 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: In particular, V i is isotropic for 1 ≤ i ≤ n. e set of all maximal isotropic flags is denoted by Z. e 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. e signature of a flag F � (0 ⊂ V 1 ⊂ . . . ⊂ V 2n ) ∈ Z with respect to h consists of three sequences: where a i (resp., b i ) denotes the dimension of a maximal negative (positive) subspace and d i the degeneracy V i of the restriction of h to V i and is called the signature of the flag F. If d � 0, we refer to F as being nondegenerate and write sign(F) � (a, b).

Base Points.
To parametrize the Schubert cells, it is important to focus on a maximal Torus T I in an Iwasawa-Borel subgroup B I of G which fixes the associated flag F I . On the contrary, we need another maximal torus T S which defines a basis to determine the base cycles called the standard basis. ese two tori are conjugate.
In the following, we define bases depending on p 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 T S is defined in each case by requiring that each basis vector is a T S -eigenvector.
(i) If m � 2n and q are even, the ordered b-isotropic basis is where (ii) If m � 2n and q are odd, the ordered b-isotropic basis is where where (iv) If m � 2n + 1 and q are odd, the ordered b-isotropic basis is where Remark 1. If m � 2n + 1, we have the vector e q or e q+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 e q or e q+1 . e above bases define the standard maximal torus T S in each case as the subgroup of diagonal matrices, i.e., For any of the ordered bases above, we denote by F S 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 of T S , the maximal torus T I is defined to have a certain basis of eigenvectors which depends on m being odd or even.

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We denote by B I the Iwasawa-Borel subgroup of G which fixes the associated flag F I .

Weyl Groups, Flag Domains, and Base
Cycles. Let us use shorthand notation for the bases used above. By (r, s) − form, we mean a basis (r 1 , . . . , r n , s n , . . . , s 1 ) where b(r i , r j ) � b(s i , s j ) � 0 and b(r i , s j ) � δ ij for all i and j. is occurs in the even-dimensional case. By (r, t, s)-form, which occurs in the odd-dimensional case, r i and s i satisfy the same conditions and t is a single vector with b(t, t) � ± 1 and b(t, r i ) � b(t, s i ) � 0 for all i.
Here, we discuss the Weyl groups W(T S ) and W(T I ) by their actions on these bases. If T is either of these tori, then it stabilizes the spaces spanned by r i and s i for i � 1, . . . , n for both kinds of bases. In this, we regard T as a product T � T 1 · . . . · T n , in the second case acting trivially on the space spanned by t. In both cases, an arbitrary permutation in S n acting diagonally on these spaces by (r i , s i ) ⟼ (r π(i) , s π(i) ) normalizes the T-action and is in the given orthogonal group G.
A simple reflection (r i , s i ) ⟼ (s i , r i ) for one such i, and being the identity on the other 2-dimensional spaces, is denoted by a sign change. is 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.
us, in both cases, we have the action of S n ⋉ Z n 2 normalizing the T-representation on the basis at hand. is is in fact the action of the full Weyl group.
As in the cases of Sp(2n, R) and SO * (2n), to describe the flag domains and their K 0 -base cycles, we use the Weyl group W(T S ) of the standard torus T S . e base point for α is the flag F S defined by the standard basis. en, all base points are F α ≔ w(F S ) where w ∈ Z n− 1 2 if m � 2n and w ∈ Z n 2 if m � 2n + 1 is associated with the sign-change vector α. As before, the base cycles and flag domains are the orbits K.F α and G 0 .F α , respectively, and α defines a nondegenerate sign with sign(F α ) � (a, b) and then D α � D a,b .
In our particular case, there are two maximal tori, the maximal torus T I and the maximal torus T S , which were International Journal of Mathematics and Mathematical Sciences defined earlier. Recall that these two maximal tori T I and T S 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) If m � 2n and q are even, define W T S to be the Weyl group with respect to the basis (7) and let W I be the Weyl group with respect to the basis (16). In this case, the bijective map between W I and W T S is If m � 2n and q are odd, define W T S to be the Weyl group with respect to the basis (9) and let W I be the Weyl group with respect to the basis (16); then, the bijective map in this case is ψ : (iii) If m � 2n + 1 and q are even, define W T e to be the Weyl group with respect to the basis (11) and let W I be the Weyl group with respect to the basis (17); then, we can define the bijective map between them to be ψ( Weyl group with respect to the basis (13). Let W I be the Weyl group with respect to the basis (17). e bijective map ψ between W I and W T o can be defined Moreover, through an argument quite similar to Proposition 1, the base cycle C α is the set where E − and E + defined in Section 2.3 and the intersection dimensions are determined by α (resp., (a, b)). e flag domains are parametrized by the signature of their base cycles.

e Length of the Elements of
Recall that every B-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 Fix(T), we should be able to compute di m C S in terms of a corresponding Weyl element. In fact, if z 0 is the Bfixed point in Z � G/B and w ∈ W, then the complex dimension of the Schubert cell S w � B.w(z 0 ) is the length ℓ(w).
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 ∈ S n ⋉ Z n 2 relative to the notation for the Weyl group elements used in [2].
Lemma 1 (see [2]). Fix w ∈ S n ⋉ Z n 2 and construct w ∈ S n ⋉ Z n 2 by the following algorithm: (1) Starting from left to right in w, 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 in w, replace a negative number with its absolute value in the n− empty boxes starting from right to left.

Description of the B I -Orbits.
In the previous section, we have chosen a flag F I defined by the basis which belongs to the G 0 -closed orbit c cl . Given a Weyl element w ∈ W, we consider the Schubert cell B I .F w and an element F � b(F w ) in it which is defined by an ordered basis If F ∈ C α , then it can be defined by an ordered basis (ε 1 , . . . , ε 2n ) where for all k the vector ε k is either in V − or in V + and is a linear combination In this case, we call the basis a split basis. According to the above discussion for Sp(2n, C), we can write the formulas for b.v w j for any element b ∈ B I as follows: where B j does not involve the basis vectors e w j and e 2n− w j +1 . International Journal of Mathematics and Mathematical Sciences b e i + e 2n− i+1 � λ e i + e 2n− i+1 + · · · + b n e n + e n+1 + a n e n − e n+1 + · · · + a 1 e 1 − e 2n , respectively, with λ ≠ 0. Note that, in the above orbits, if b is chosen appropriately, then we can arrange b(e i + e 2n+i− 1 ) � e i or b(e i + e 2n+i− 1 ) � e 2n− i+1 for all i. Does this play a role in the proof of eorem 1.

Flag Domains and Base Cycles.
In order to parametrize the flag domains, it is enough to parametrize the closed Korbits (see [3], Sections 4.2 and 4.3). For this, we choose a base point F S ∈ Fix(T S ). It follows that the set C of closed Korbits can be identified with the Weyl group orbit W G (T S ).F S (see [3], Corollary 4.2.4). As we have observed above, W G (T S ) � S n ⋉ Z n 2 where the S n factor can be identified with W K (T S ). In fact, it is exactly the stabilizer of K.F S in C. Consequently, C can be identified with Z n 2 .F S . We regard an element of Z n 2 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., G 0 -orbits) of base points ( ± e 1 , . . . , ± e n ) are exactly the base cycles (resp., flag domains) in Z. e respective flag domains are denoted by D α .
Observe that α defines a nondegenerate signature (a, b) and that D α ⊂ D a,b ≔ F : sign(F) � (a, b) . Conversely, given F with sign(F) � (a, b), we may choose a basis (v, w) ≔ (v 1 , . . . , v n , w n , . . . , w 1 ) which defines F and which has the following properties: If sign(F) � sign(F), 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

Proposition 1.
For a fixed open orbit D α , the base cycle C α is the set Proof. Firstly, it was shown above that there is a base point F α with the splitting condition. Since K 0 is the product Define F + and F − to be the sets of maximally isotopic flags in E + and E − , respectively. Recall that K 0 � K + 0 × K − 0 where K + 0 and K − 0 act transitively on the sets F + and F − , respectively, which implies that K 0 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, C α is a complex manifold. But K 0 has a unique orbit in D α which is a complex manifold. erefore, C α is the base cycle.

Conditions for
Here, we deal with the homogeneous space Z of maximally isotropic full flags of the complex orthogonal symmetric group G � SO(2n, C) equipped with the action of the real form G 0 � SO(p, q). e 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 S w ∩ C α if the intersection is nonempty and S w is of complementary dimension. Also, the number of intersection points with C 0 is explicitly computed in eorem 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 satisfies S w ∩ D α ≠ ∅ for some flag domain D α . As would be expected, a special type of permutation plays a fundamental role. Definition 1. An element w ∈ W is called a harmonic permutation if it satisfies the following conditions: If q 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. If q 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. Recall that the fixed point in the closed orbit c cl is the flag associated with the following ordered basis: (i) If m is even, then the basis is For v i , any such basis vector, and b ∈ B I , the form of b.v i is given as follows: + a n (e q + e q+1 ) + · · · + a 1 (e 1 + e 2q ) ) + a n (e 2n + ie 2n+1 ) + · · · + a 1 (e 1 + e 2q ) if m � 2n + 1 λ ≠ 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 T S -eigenvectors. See eorem 1.
In the following result, ψ : W I ⟶ W S denotes the bijective map between Weyl groups which was introduced in Section 2.

Proposition 2.
If w is a harmonic permutation, then the flag F ψ(w) belongs to the orbit B I (F w ).
Proof. We handle the case where m � 2n. e proof for m � 2n + 1 goes analogously. Let us first prove the theorem for the case that q is even. For this, let w ∈ W I be a harmonic permutation and define F w � w.(F I ) to be the isotropic full flag associated with w. Denote by the first n subspaces of F w � w(F I ), where u w i is a vector from the basis above. Let w ∈ W S be the image of w under the bijective map ψ, and let Y w be the isotropic flag associated with w such that the first half of Y w is Our claim here is that this flag is an intersection point in B.F w ∩ C α . To prove this, we will construct b with b(F w ) � Y w . From the definition of harmonic permutation, there are two possibilities for w 1 : |w 1 | � 2i − 1 ≤ q or |w 1 | > q.
, so the orbit B I .〈u w 1 〉 contains the point 〈ε w 1 〉.
us, by induction, we observe that b ∈ B I can be constructed with b(F w ) � Y w .
We complete the proof by handling the case where q is odd. For this point, we can repeat the steps of the proof above for w j > q and |w j | � 2i − 1. For the case where |w j | � 2i, we apply the same method as above, only changing 2|w j | by 2(|w j | + 1). If w n � − q, then w j � − n. In this case, the orbit of relevance is B I .〈e q − e q+1 〉. As we see from Remark 3, y � α 1 (e q + e q+1 ) + α 2 (e q − e q+1 ) belongs to this orbit. e desired result is then obtained by taking α 1 � 1, α 2 � − 1, and the point y � e q+1 belongs to the orbit B I .〈e q − e q+1 〉. erefore, the element of B I is also constructed in the case where q is odd.
International Journal of Mathematics and Mathematical Sciences Theorem 1 (harmonic permutation theorem).
e following are equivalent: Under either of these conditions for every α, the nonempty intersection B I (F w ) ∩ C α contains a T S -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 S w ∩ D α � ∅, for all α, i.e., S w has no T S -fixed points. Assume on the contrary that there exists b ∈ B I such that b.(F w ) ∈ S w ∩ C α . For b(F w ) 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: b e j − e 2q− j+1 , e j + e 2q− j+1 � 1, 1 ≤ j ≤ q, b e j ± e 2q− j+1 , e k ± e 2q− k+1 � 0, for k ≠ j.

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Since w 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 w j � 2i is the first even number which appears in w such that 2i sits to the left of − (2i − 1). en, b does in fact yield a T S -fixed partial flag, i.e., Now, we check that there is no b ∈ B I so that b.(v w j ) defines an extended partial flag which is T S -fixed. For this, we consider the orbit B I .〈e |w j | + e |2 w j |− 1 〉 which contains points of the form α 1 e 1 + e 2q + α 2 e 2 + e 2q− 1 + · · · + α j e w j + e 2w j |− 1 .
is is a linear combination of h-isotropic vectors for all α i . 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 〈(e |w j |− 1 + ie |w j | )〉 from the flag (35) to the linear combination in (36), but this vector is not in the flag (35). us, as was claimed, no b ∈ B I has the property that b(F w ) is T S -fixed.

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 to C 0 . Recall that the maximal U(q)). For E + and E − as in Section 2.4, the base cycle in the flag domain D a,b is given by If m � 2n, we have two cases: If q is even, the dimension of C 0 is (p(p − 2)/4) + (q(q − 2)/4). If q is odd, then C 0 has the dimension ((p − 1) 2 /4) + ((q − 1) 2 /4). If m � 2n + 1, then the dimension of the base cycle is ((p − 1) 2 /4) + (q(q − 2)/4) if q is even and is (p(p − 2)/4) + ((q − 1) 2 /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 dim S w � pq/2 if q and n are even and dim S w � (pq − 1)/2 if q or n is odd.    Remark 5. If W � S n ⋉ Z n 2 , 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 +.  Recall that the dimension of the cells corresponds to the length of the word w in the Weyl group, i.e., if F w � w.F I , then dim(B.F w ) � l(w). 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 permutation w ∈ S n ⋉ Z n− 1 2 has dimension pq/2 if q is even and (pq − 1)/2 if q is odd.
Proof. Given a perfect harmonic permutation w, we consider three cases.
ese depend on which of the pairs 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. We define a function f : 5, . . . , n is is a bijective map which sends v to w ∈ S n− 4 ⋉ Z n− 5 2 . Note that w is a perfect harmonic permutation in S n− 4 ⋉ Z n− 5 2 . us by the induction assumption, l(w) � (p − 4)(q − 4)/2. Since v � f − 1 (w), 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 ∈ S n ⋉ Z n− 1 2 with length (p − 4)(q − 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 − 4)(q − 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 (24)v(13) with length ((p − 4)(q − 4)/2) + n − 4 + 1 + n − 4 + 2 � ((p − 4)(q − 4)/ 2) + 2n − 5, and it follows that the element (24)v(− 3 − 1) has length ((p − 4)(q − 4)/2) + 2n − 4. We then return to the original w and remove only the pair (− 1 2). en, 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 (− 3 4), 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 original w and define h to be the number of positions to the left of (− 1 2) 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 If q is odd, then the length of w ∈ S n− 4 ⋉ Z n− 5 2 is l(w) � (p − 4)(q − 4) − 1/2. After applying the same steps as above, it follows that l(w) � (pq − 1)/2.

Case 2.
e permutation w contains the pair (− 1 − 2). Note that this case only occurs if q is even. Since w is a perfect harmonic permutation, the pair (− 1 − 2) sits in the 8 International Journal of Mathematics and Mathematical Sciences 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 l(v) � (p − 2)(q − 2)/2 because q is even. We put the pair (1 2) to the left of v. It follows that v � (1 2)v 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 l(w) � ((p − 2)(q − 2)/2) + 2(n − 2) + 2 � pq/2. Note that this case appears only if q is even. In this case, these two pairs appear in w in the following form: If we have the form (− 3 − 1 2 − 4), then we must add n − 4 to (p − 2)(q − 2)/2 to put 3 in the last position and add n − 2 to (p − 2)(q − 2)/2 to put 4 in the last position. en, 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 − 2)(q − 2)/2) + (n − 4) + (n − 2) + 1 + 3 � pq/2.
If we have the form (− 3 − 4 − 1 2), then we must add n − 4 to (p − 2)(q − 2)/2 to put 3 in the last position and add n − 1 to (p − 2)(q − 2)/2 to put 4 in the position before the last position. en, 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 − 2)(q − 2)/2) + (n − 4) + (n − 3) + 1 + 3 + 1 � pq/2. Proposition 4. Every perfect harmonic permutation w ∈ S n ⋉ Z n 2 has dimension pq/2.
Proof. e argument goes exactly along the lines as that of the above proposition. Here, it is in fact simpler because only the pairs (− 1 2) and (− 3 4) appear.

Intersection Points of Schubert Duality
Let w ∈ W I be a perfect harmonic permutation, in particular so that S w is of complementary dimension to the cycles. Recall that, in this case, either S w ∩ C α � ∅ or is pointwise T S -fixed (see Section 2.3). e main goal in this section is to compute all such intersection points. e argument in the case of SO(p, q) is carried out by means of algorithms. Nevertheless, we are able to provide formulas for the cardinality of S w ∩ C α when it is nonzero and the total number of cycles C α for which this intersection is nonempty (see eorem 2).

Proposition 5.
If w is a perfect harmonic permutation so that B I .F w intersects a cycle C α at a point given by the ε-basis (ε 1 , . . . , ε m ) of T S -eigenvectors, then for any such eigenvector ε k , it follows that the ε-basis is given by ε k � e 2j− 1 + ie 2j or e 2j− 1 − ie 2j if q and p are even and ε k � e 2j− 1 + ie 2j , e 2j− 1 − ie 2j , e q , e q+1 , or e 2q+1 if q or p is odd, depending on the dimension m and the signature α.
Proof. is is a consequence of the following: (1) w is a perfect harmonic permutation and the flag basis is that of w(F I ).
(2) We have the following cases: Here, η j � ± iη j and ζ � ± iζ j and B j does not involve the basis vectors e r , e r+1 , e 2q− r+1 , and e 2q− r .
From the expression for v j , it is obvious that all of the possibilities in the statement occur. Furthermore, since η j ≠ 0 and η j ≠ 0, for every j, a nonzero contribution from K j occurs in the sum Since e 2j− 1 + ie 2j and e 2j− 1 − ie 2j and e q or e q+1 do not occur in b.v w k for k < j, it follows that ε j � K j + E j in the standard basis. Finally, since ε j is a T S -eigenvector, it follows that ε j � K j where K j of the ε j forms indicated in the proposition.
As a result of the above proposition, the following corollary gives us all intersection points of S w ∩ C α . □ Corollary 1. Let D α be a flag domain parametrized by a sequence α and w ∈ W I be a perfect harmonic permutation such that S w ∩ D α ≠ ϕ. en, the following algorithm produces us all intersection points of S w ∩ C α : If q 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 and if the corresponding signature of it in β is − +, then replace this − + in β by (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 International Journal of Mathematics and Mathematical Sciences and if the corresponding signature of it in β is − +, then replace this − + in β by (iv) For the remaining numbers, for each q + 1 ≤ j ≤ n − 1, replace the corresponding + in β by (e 2j− 1 + ie 2j ), and for ± n, replace the corresponding + by (e 2n− 1 ± ie 2n ).
If q 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 and if the corresponding signature of it in β is − +, then replace this − + in β by (ii) For the number − q, replace the corresponding + in β by e q+1 . (iii) For the remaining numbers, for each q + 1 ≤ j ≤ n − 1, replace the corresponding + in β by (e 2j− 1 + ie 2j ), and for ± n, replace the corresponding + by (e 2n− 1 ± ie 2n ).
Each point obtained from this algorithm is a point of the intersection of S w ∩ D α .

Theorem 2.
A Schubert variety S w which is parametrized by a perfect harmonic permutation w intersects 2 q/2 flag domains if q is even and intersects the base cycles of these flag domains in 2 q points. If q is odd, it intersects 2 (q− 1)/2 flag domains and intersects the base cycles of these flag domains in 2 q− 1 points.
Proof. Let w ∈ W I be a perfect harmonic permutation. We first show that if q is even, then S w ∩ C α consists of 2 q points and 2 q− 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 (− 1 2), then the pair (− 1 2) sits inside consecutive boxes in w. e 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 B I -orbit of 〈e 1 − e 2q 〉. By Remark 3, the elements in this orbit have the form β 1 e 1 + e 2q + · · · + β 2n e 1 − e 2q .
(47) e question is how many 1-dimensional subspaces (spanned by vectors in Proposition 5) do we have such that these subspaces can be completed to the maximal isotropic flag. To compute this number, we denote by v 1 the vector we have from the first step which spans the 1-dimensional subspace; in the second step, we consider the orbit B I .〈e 2 + e 2q− 1 〉, which has points of the form β 1 e 1 + e 2q + β 2 e 2 + e 2q− 1 .
(48) e 2-dimensional subspace in the flag is spanned by linear combinations of the form v 2 � β 1 e 1 + e 2q + β 2 e 2 + e 2q− 1 + cv 1 . (49) Note that v 2 should be in E + or in E − and of the form stated in Proposition 5, and therefore, v 1 should contain the terms e 1 and e 2 or the terms e 2q− 1 and e 2q . us, we have the following 4 possibilities of v 1 as follows: e 1 − ie 2 , e 1 + ie 2 , e 2q− 1 + ie 2q , and e 2q− 1 − ie 2q .

(61)
Let M w ⊂ F o (Fix T S ) be the set of all maximally b-isotropic flags associated with all elements in Swito w . Note that we have (q − 1)/2 of the pairs (− (2i − 1), 2i), 1 ≤ i ≤ q/2, in any w ∈ W I , and for each pair, we have 4 possibilities for switching it. Hence, the cardinality of Swito w is 4 (q− 1)/2 � 2 q− 1 . e set Swito w gives us all intersection points of S w , and each 2 (q− 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. (63) Recall that, in the cases of SP(2n, R) and SO * (2n), every flag domain intersects all Schubert varieties of complementary dimension. But in the case of SO(p, q), we do not have this property except in a very special case. We explain this case in the following example.

Data Availability
e research data used to support this study are included within the article.

Disclosure
Parts of this work are from the author's PhD thesis at Ruhr-Universität, Bochum, Germany [6].

Conflicts of Interest
e author declares that there are no conflicts of interest regarding the publication of this paper.