Counting Borel Orbits in Symmetric Spaces of Types BI and CII

Abstract

This is a continuation of our combinatorial program on the enumeration of Borel orbits in symmetric spaces of classical types. Here, we determine the generating series the numbers of Borel orbits in ${\mathbf {SO}}_{2n+1}/{\mathbf {S(O}}_{2p}\times {\mathbf {O}}_{2q+1} \mathbf {)}$ (type BI) and in ${\mathbf {Sp}}_n/{\mathbf {Sp}}_p \times {\mathbf {Sp}}_q$ (type CII). In addition, we explore relations to lattice path enumeration.

Appendix

In this appendix, as we promised in the introduction, we outline a method for approximating the number of symmetric $(2p,2q+1)$ clans with k pairs, $\beta _{k,p,q}$. Recall our notation that $A_e(x) = \sum \nolimits _{l=0}^{q} \beta _{2l,p,q} x^{2l}$ and $A_o(x) = \sum \nolimits _{l=0}^{q} \beta _{2l+1,p,q} x^{2l+1}$.

First of all, by multiplying both sides of the recurrence relation (3.8) by $x^{2l}$ and summing over l lead us to the following integral/differential equation

$$\begin{aligned} A_e(x)-\beta _{0,p,q}= & {} (q+1) \int (A_o(x) - \beta _{2q+1,p,q} x^{2q+1})dx \\&- \frac{x}{2}(A_o(x) - \beta _{2q+1,p,q} x^{2q+1}) \\&+ 2(pq+p+q+1)\int x(A_e(x)- \beta _{2q,p,q} x^{2q})dx \\&- (p+q+1)x^2 (A_e(x) - \beta _{2q,p,q} x^{2q}) \\&+\frac{x^3}{2}A_e^{'}(x) - q\beta _{2q,p,q} x^{2q+2}. \end{aligned}$$

We get rid of the integrals by taking the derivative with respect to x and then we reorganize our equation which is now a second order ODE as in

$$\begin{aligned}&x^3A_e^{''}(x) - \biggr ((2p+2q-1)x^2 +2\biggr )A_e^{'}(x) +4pq x A_e(x) \\&\quad - x A_o^{'}(x) + (2q+1)A_0(x) =0. \end{aligned}$$

By applying a similar procedure to the recurrence relation (3.9) and also by using the fact that $\beta _{1,p,q} = p \beta _{0,p,q}$, we obtain our second second order ODE:

$$\begin{aligned}&x^3A_o^{''}(x) - \biggr ((2p+2q-1)x^2 + 2\biggr )A_o^{'}(x) +(4pq+2p-2q-1)x A_o(x) \\&\quad - xA_e^{'}(x) + 2p A_e(x) =0. \end{aligned}$$

Note that we the following initial conditions that follow from the definitions of $A_e(x)$ and $A_o(x)$:

$$\begin{aligned} A_e(0)= & {} \beta _{0,p,q} \; \text {and}\; A_o(0) = 0, \\ A_e^{'}(0)= & {} 0 \; \text {and} \; A_o^{'}(0) = p\beta _{0,p,q} = \beta _{1,p,q}. \end{aligned}$$

We will reduce our second order system to a first order ODE by setting $u(x):=A_e^{'}(x)$ and $v(x):=A_o^{'}(x)$. Then

$$\begin{aligned} x^3 u'(x)= & {} ((2p+2q-1)x^2 +2 )u(x)-4pqx A_e(x) - (2q+1)A_0(x) + xv(x) \\ x^3 v'(x)= & {} ((2p+2q-1)x^2 +2) v(x)- (4pq+2p-2q-1)x A_o(x) \\&-2p A_e(x) + x u(x) \\ x^3 A_e^{'}(x)= & {} x^3 u(x) \\ x^3 A_o^{'}(x)= & {} x^3 v(x). \end{aligned}$$

We write this system in matrix form

$$\begin{aligned} x^3 X' = A(x) X, \end{aligned}$$

where A(x) the $4\times 4$ matrix as in (1.5). Note that our initial conditions become

$$\begin{aligned} X(0) = \begin{bmatrix} u(0)\\ v(0) \\ A(0)\\ A(0) \end{bmatrix} = \begin{bmatrix} 0\\ p\beta _{0,p,q} \\ \beta _{0,p,q}\\ 0 \end{bmatrix}. \end{aligned}$$

(6.1)

Once a system of first order ordinary differential equations of this type is given, formal series solutions can always be obtained by carrying out the computational procedure, which is outlined in Turrittin (1955). We will use those techniques to solve the above system of first order ordinary differential equations.

Before proceeding any further let us define the matrices $A_0,A_1,\ldots$ by decomposing the coefficient matrix A(x):

$$\begin{aligned} A(x)= & {} \sum _{k= 0}^{\infty } A_k x^k = \begin{bmatrix} 2&0&0&-(2q+1) \\ 0&2&-2p&0 \\ 0&0&0&0 \\ 0&0&0&0 \end{bmatrix} x^0 \\&+ \begin{bmatrix} 0&1&-4pq&0 \\ 1&0&0&-(4pq+2q-2q-1) \\ 0&0&0&0 \\ 0&0&0&0 \end{bmatrix} x^1 \\&+ \begin{bmatrix} (2q+2q-1)&0&0&0 \\ 0&(2p+2q-1)&0&0 \\ 0&0&0&0 \\ 0&0&0&0 \end{bmatrix} x^2 \\&+ \begin{bmatrix} 0&0&0&0\\ 0&0&0&0 \\ 1&0&0&0 \\ 0&1&0&0 \end{bmatrix} x^3 + {\varvec{0}}x^4 + \cdots \end{aligned}$$

Since the eigenvalues of the leading matrix $A_0$ fall into two groups, namely $\lambda _1 = \lambda _2 = 0$ and $\lambda _3 = \lambda _4 = 2$, there exists a normalizing transformation matrix P obtained from the Jordan canonical form of $A_0$. More precisely, since

$$\begin{aligned} B_0= & {} \begin{bmatrix} 0&0&0&0 \\ 0&0&0&0 \\ 0&0&2&0 \\ 0&0&0&2 \end{bmatrix}\\= & {} \begin{bmatrix} 0&0&0&1 \\ 0&0&1&0 \\ 0&1&-p&0 \\ 1&0&0&-\frac{2q+1}{2} \end{bmatrix} \begin{bmatrix} 2&0&0&-(2q+1) \\ 0&2&-2p&0 \\ 0&0&0&0 \\ 0&0&0&0 \end{bmatrix} \begin{bmatrix} \frac{2q+1}{2}&0&0&1 \\ 0&p&1&0 \\ 0&1&0&0 \\ 1&0&0&0 \\ \end{bmatrix} \\= & {} P^{-1} A_0 P. \end{aligned}$$

the normalizing transformation $X = PY$ turns our system into

$$\begin{aligned} x^3 Y^{'} = B(x) Y; \quad \text {with} \quad Y(0) = \begin{bmatrix} 0 \\ \beta _{0,p,q} \\ 0 \\ 0 \end{bmatrix} \end{aligned}$$

(6.2)

where

$$\begin{aligned} B(x)= & {} P^{-1} A(x) P \\= & {} \begin{bmatrix} 0&px^3&x^3&0 \\ \frac{2q+1}{2}x^3&0&0&x^3 \\ -\frac{2q + 1}{2} (px^3 +4px - 3x)&p(2p+2q-1)x^2&(2p+2q-1)x^2+2&x- px^3 \\ \frac{(2q+1)(2p+2q-1)}{2}x^2&(p-4pq)x - \frac{p(2q+1)}{2}x^2&x- \frac{2q+1}{2}x^3&(2p+2q-1)x^2+2 \end{bmatrix}\\= & {} \begin{bmatrix} 0&0&0&0\\ 0&0&0&0\\ 0&0&2&0\\ 0&0&0&2 \end{bmatrix} + \begin{bmatrix} 0&0&0&0 \\ 0&0&0&0 \\ -\frac{(2q+1)(4p-3)}{2}&0&0&1 \\ 0&p(1- 4q)&1&0 \end{bmatrix} x \\&+ \begin{bmatrix} 0&0&0&0 \\ 0&0&0&0\\ 0&p(2p+2q-1)&2p+2q-1&0\\ \frac{(2q+1)(2p+2q-1)}{2}&0&0&2p+2q-1 \end{bmatrix} x^2\\&+ \begin{bmatrix} 0&p&1&0 \\ \frac{2q+1}{2}&0&0&1 \\ -\frac{p(2q+1)}{2}&0&0&-p \\ 0&-\frac{p(2q+1)}{2}&-\frac{2q+1}{2}&0 \end{bmatrix} x^3 + {\varvec{0}}x^4 + {\varvec{0}} x^5 + \cdots \end{aligned}$$

We denote the coefficient matrix of $x^i$ ($i=0,1,2,\ldots$) in B(x) by $B_i$. Thus,

$$\begin{aligned} B(x) = B_0 + B_1 x + B_2 x^2 +B_3 x^3. \end{aligned}$$

We will work with a system that is obtained from B(x) by a “shearing” transformation. Let Q be a formal power series of the form $Q=\sum Q_r x^r$ with $Q_r$’s are some constant matrices of order 4. We assume that our desired solution $Y=Y(x)$ for $x^3 Y' = B Y$ is of the form $Y=QZ$ for some $4\times 1$ column matrix $Z=Z(x)$. Formally, substituting QZ into $x^3 Y' = B(x) Y$ will give us a new ODE:

$$\begin{aligned} x^3 (QZ)' = B QZ \Rightarrow x^3(Q'Z+QZ') = BQZ\ \text { or } \ x^3 Z' = (Q^{-1} B Q + x^3 Q^{-1}Q')Z. \end{aligned}$$

Let C denote the formal power series $\sum C_r x^r$ that is defined by

$$\begin{aligned} Q^{-1} B Q+ x^3 Q^{-1}Q' = C = \sum C_r x^r, \end{aligned}$$

(6.3)

hence our ODE is equivalent to

$$\begin{aligned} x^3 Z'&= C Z. \end{aligned}$$

(6.4)

By multiplying both sides of (6.3) with Q and rearranging we obtain a new ODE whose solution will lead to a solution of (6.5):

$$\begin{aligned} x^3 Q' = Q C - B Q. \end{aligned}$$

(6.5)

To solve (6.5) we simply substitute $B=\sum B_r x^r$, $Q=\sum Q_r x^r$ and $C=\sum C_r x^r$ and equate the coefficients. Then we get the following relations which we call as our fundamental recurrences:

• (i) $0= Q_0 C_0 - B_0 Q_0$;

• (ii) $0= (Q_0 C_1 - B_1 Q_0)+(Q_1C_0 - B_0Q_1)$;

• (iii) $(r-2)Q_{r-2} = \sum \nolimits _{i=0}^r (Q_i C_{r-i} - B_{r-i} Q_i)$ for $r\ge 2$.

We will recursively assign specific values to the matrices $Q_i$, $i=0,1,2,\ldots$ which will allow us to solve (6.5). Along the way we will determine the series $C(x)=\sum x^i C_i$, which is what we want to solve in the first place. Indeed, our goal is to choose $Q_i$’s in such a way that $C_i$’s become block diagonal. To this end, we assume that $Q_i$ ($i=0,1,2,\ldots$) is a block anti-diagonal matrix:

$$\begin{aligned} Q_i = \begin{bmatrix} 0&Q_i^{12} \\ Q_i^{21}&0 \end{bmatrix} \end{aligned}$$

for some $2\times 2$ matrices $Q_i^{12},Q_i^{21}$ ($i=0,1,2,\ldots$).

Step 1. We choose $Q_0=I_4$, the $4\times 4$ identity matrix. It follows from (i) that

$$\begin{aligned} C_0 = B_0 = \begin{bmatrix} 0&0&0&0\\ 0&0&0&0\\ 0&0&2&0\\ 0&0&0&2 \end{bmatrix}. \end{aligned}$$

We have a remark in order.

Remark 6.1

Let us point out that, since

$$\begin{aligned} Q_i B_0 -C_0 C_i&= \begin{bmatrix} 0&2 Q_i^{12} \\ -2Q_i^{21}&0 \end{bmatrix} \ \text { for } i=1,2,\ldots \end{aligned}$$

(6.6)

by using the fundamental recurrences (ii) and (iii) we will always be able to choose $Q_i^{12}$ and $Q_i^{21}$ so that $C_i$ is of the form

$$\begin{aligned} C_i = \begin{bmatrix} C_i^{11}&0 \\ 0&C_i^{22} \end{bmatrix}, \end{aligned}$$

where $C_i^{11}$ and $C_i^{22}$ are $2\times 2$ matrices.

Step 2. By (ii) and Step 1, $C_1 = B_1 -Q_1 C_0+ B_0 Q_1$, so we set

$$\begin{aligned} Q_1 = \begin{bmatrix} 0&0&0&0 \\ 0&0&0&0 \\ -\frac{(2q+1)(4p-1)}{4}&0&0&0 \\ 0&\frac{p(4q-1)}{2}&0&0 \end{bmatrix} \implies C_1 = \begin{bmatrix} 0&0&0&0 \\ 0&0&0&0 \\ 0&0&0&1 \\ 0&0&1&0 \end{bmatrix}. \end{aligned}$$

Step 3. By (iii) and Steps 1,2, $C_2 = B_2 -Q_1 C_1+ B_1 Q_1 -Q_2C_0 + B_0 Q_2$, so we set

$$\begin{aligned} Q_2&= \begin{bmatrix} 0&0&0&0 \\ 0&0&0&0 \\ 0&-\frac{p(4p+8q-3)}{4}&0&0 \\ -\frac{(2q+1)(4q-1)}{8}&0&0&0 \end{bmatrix} \\&\quad \implies C_2 = \begin{bmatrix} 0&0&0&0 \\ 0&0&0&0 \\ 0&0&2p+2q-1&0 \\ 0&0&0&2p+2q-1 \end{bmatrix}. \end{aligned}$$

In a similar manner, we put

$$\begin{aligned} Q_3&= \begin{bmatrix} 0&0&\frac{1}{2}&0\\ 0&0&0&\frac{1}{2} \\ \frac{(2q+1)(16p^2+16pq-1)}{16}&0&0&0 \\ 0&\frac{-p(16pq+16q^2-8p-16q+1)}{8}&0&0 \end{bmatrix} \\&\quad \implies C_3 = \begin{bmatrix} 0&p&0&0 \\ \frac{2q+1}{2}&0&0&0 \\ 0&0&0&-p \\ 0&0&-\frac{2q+1}{2}&0 \end{bmatrix}. \end{aligned}$$

The above computations are in some sense are our initial conditions. To get a better understanding of the general case we make a few more preliminary observations and formal computations.

$$\begin{aligned} C_i^{jj}&= B_i^{jj} \ \text { for } i =0,1,2,3 \ \text { and } \ j=1,2 \end{aligned}$$

(6.7)

$$\begin{aligned} Q_iC_1&= \begin{bmatrix} 0&Q_i^{12} C_1^{22} \\ 0&0 \end{bmatrix} \qquad \text { and } \qquad B_1Q_i = \begin{bmatrix} 0&0 \\ B_1^{22}Q_i^{21}&B_1^{21}Q_i^{12} \end{bmatrix} \end{aligned}$$

(6.8)

$$\begin{aligned} Q_iC_2&= \begin{bmatrix} 0&Q_i^{12} C_2^{22} \\ 0&0 \end{bmatrix} \qquad \text { and } \qquad B_2Q_i = \begin{bmatrix} 0&0 \\ B_2^{22}Q_i^{21}&B_2^{21}Q_i^{12} \end{bmatrix} \end{aligned}$$

(6.9)

$$\begin{aligned} Q_iC_3&= \begin{bmatrix} 0&Q_i^{12} C_3^{22} \\ Q_i^{21} C_3^{11}&0 \end{bmatrix} \qquad \text { and } \qquad B_3Q_i = \begin{bmatrix} B_3^{12}Q_i^{21}&B_3^{11} Q_i^{12} \\ B_3^{22}Q_i^{21}&B_3^{21}Q_i^{12} \end{bmatrix} \end{aligned}$$

(6.10)

$$\begin{aligned} Q_iC_j&= \begin{bmatrix} 0&Q_i^{12} C_j^{22} \\ Q_i^{21} C_j^{11}&0 \end{bmatrix}. \end{aligned}$$

(6.11)

Finally, since $B_r = 0$, the fundamental recurrence (iii) simplifies to

$$\begin{aligned} C_r = (r-2)Q_{r-2} -\left( \sum _{i=0}^3 (Q_{r-i} C_i - B_i Q_{r-i})\right) - \left( \sum _{i=4}^{r-1} Q_{r-i} C_i \right) . \end{aligned}$$

(6.12)

Recall that we started with the system $x^3 X' = A(x) X$ which is transformed into $x^3 Y' = B(x) Y$ by conjugating with a constant matrix, and the latter system is transformed into $x^3 Z' = C(x) Z$ by the shearing transformation $Y= Q(x) Z$.

Proposition 6.2

Let $C(x) = \sum \nolimits _r C_r x^r$ and $Q(x) = \sum \nolimits _r Q_r x^r$ be as in the previous paragraph. If $r\ge 4$, then we have

$$\begin{aligned} C_r = \begin{bmatrix} Q_{r-3}^{21}&0 \\ 0&B_1^{21} Q_{r-1}^{12} + B_2^{21} Q_{r-2}^{12} + B_3^{21} Q_{r-3}^{12} \end{bmatrix}. \end{aligned}$$

In particular, the system $x^3 Z' = C(x) Z$ decomposes into two $2\times 2$ systems of ODE’s

$$\begin{aligned} x^3 K'&= R(x) K, \end{aligned}$$

(6.13)

$$\begin{aligned} x^3 L'&= T(x) L, \end{aligned}$$

(6.14)

where

$$\begin{aligned} R(x)&= C_3^{11} x^3 + \sum _{r\ge 4} Q_{r-3}^{21} x^r; \\ T(x)&= \sum _{i=0}^3 C_i^{22}x^3 + \sum _{r\ge 4} (B_1^{21} Q_{r-1}^{12} + B_2^{21} Q_{r-2}^{12} + B_3^{21} Q_{r-3}^{12}) x^r. \end{aligned}$$

Proof

Since $C_r$ is a block diagonal matrix, recurrence (6.12) combined with equations (6.7)–(6.11) gives us the desired result. $\square$

What remains is to solving the systems (6.13) and (6.14). The former ODE is relatively easy since it does not have a singularity anymore. However, the second ODE (6.14) does have a singularity. Moreover, we still do not know the exact forms of neither $Q^{12}(x)$ nor $Q^{21}(x)$. On the positive side, by taking advantage of the particular structure of B(x)’s we are able to find recurrences for R(x) and T(x).

To find a recurrence for the blocks of $Q_r$’s, we use Proposition 6.2 as well as the simplified fundamental recurrence (6.12) as follows:

$$\begin{aligned} C_r =&\begin{bmatrix} Q_{r-3}^{21}&0 \\ 0&B_1^{21} Q_{r-1}^{12} + B_2^{21} Q_{r-2}^{12} + B_3^{21} Q_{r-3}^{12} \end{bmatrix}\\ =&\begin{bmatrix} 0&(r-2)Q_{r-2}^{12} \\ (r-2) Q_{r-2}^{21}&0 \end{bmatrix} - \begin{bmatrix} 0&2Q_{r}^{12} \\ -2Q_{r}^{21}&0 \end{bmatrix} - \begin{bmatrix} 0&Q_{r-1}^{12}B_{1}^{22} \\ -B_{1}^{22}Q_{r-1}^{21}&-B_{1}^{21}Q_{r-1}^{12} \end{bmatrix}\\&- \begin{bmatrix} 0&Q_{r-2}^{12}B_{2}^{22} \\ -B_{2}^{22}Q_{r-2}^{21}&-B_{2}^{21}Q_{r-2}^{12} \end{bmatrix}- \begin{bmatrix} -B_{3}^{12}Q_{r-3}^{21}&Q_{r-3}^{12}B_{3}^{22} - B_{3}^{11}Q_{r-3}^{12} \\ Q_{r-3}^{21}B_{3}^{11} -B_{3}^{22} Q_{r-3}^{21}&-B_{3}^{21}Q_{r-3}^{12} \end{bmatrix}\\&- \sum _{i=4} \begin{bmatrix} 0&Q_{r-i}^{12} \\ Q_{r-i}^{21}&0 \end{bmatrix} \begin{bmatrix} Q_{i-3}^{21}&0 \\ 0&B_1^{21} Q_{i-1}^{12} + B_2^{21} Q_{i-2}^{12} + B_3^{21} Q_{i-3}^{12} \end{bmatrix} \\ =&\begin{bmatrix} Q_{i-3}^{21}&\begin{matrix} (r-2)Q_{r-2}^{12}-2Q_{r}^{12}-Q_{r-1}^{12}B_1^{22} \\ {}-Q_{r-2}^{12}B_2^{22}-Q_{r-3}^{12}B_3^{22}+B_3^{11}Q_{r-3}^{12}\\ {}- \sum _{i=4} Q_{r-i}^{12}(B_1^{21} Q_{i-1}^{12} + B_2^{21} Q_{i-2}^{12} + B_3^{21} Q_{i-3}^{12}) \end{matrix} \\ \begin{matrix} (r-2)Q_{r-2}^{21}+2Q_{r}^{21}+B_1^{22}Q_{r-1}^{21}+B_2^{22}Q_{r-2}^{21}\\ {}-Q_{r-3}^{21}B_3^{11}+B_3^{22}Q_{r-3}^{21}- \sum _{i=4} Q_{r-i}^{21}Q_{i-3}^{21} \end{matrix}&B_1^{21} Q_{i-3}^{12} + B_2^{21} Q_{i-2}^{12} + B_3^{21} Q_{i-3}^{12} \end{bmatrix}. \end{aligned}$$

Observe that the diagonal blocks do not give us any new information, however, the anti-diagonal blocks do. By the equality of the bottom left blocks, we have

$$\begin{aligned} 2Q_{r}^{21}&=-(r-2)Q_{r-2}^{21} -B_1^{22}Q_{r-1}^{21}- B_2^{22}Q_{r-2}^{21}+Q_{r-3}^{21}B_3^{11}-B_3^{22}Q_{r-3}^{21} \nonumber \\&\quad + \sum _{i=4}^{r-1} Q_{r-i}^{21}Q_{i-3}^{21}. \end{aligned}$$

(6.15)

Similarly, the equality of the top right blocks give

$$\begin{aligned} \begin{aligned} 2Q_r^{12} =&(r-2)Q_{r-2}^{12}-Q_{r-1}^{12} B_1^{22}-Q_{r-2}^{12}B_2^{22}- Q_{r-3}^{12}B_3^{22}+B_3^{11}Q_{r-3}^{12}\\&- \sum _{i=4}^{r-1} Q_{r-i}^{12}(B_1^{21} Q_{i-1}^{12} + B_2^{21} Q_{i-2}^{12} + B_3^{21} Q_{i-3}^{12}). \end{aligned} \end{aligned}$$

(6.16)

Obviously, these recurrences enable us to write the precise forms of the ODE’s (6.13) and (6.14). Both of these ODE’s can now be solved by applying suitable shearing transformations leading to a solution of our original equation $x^3 X'=A(x) X$. However, due to its high computational cost the result is still not better than the expressions for $\beta _{k,p,q}$’s that we recorded in Theorem 1.2.