$L^2$-Poisson integral representations of eigensections of invariant differential operators on a homogeneous line bundle over the complex Grassmann manifold $SU(r,r+b)/S( U(r)\times U(r+b))$

Abstract

Let $E_l=G\times _K{{\mathbb {C}}}$ be the associated homogeneous line bundle to a one-dimensional $K$-representation $\tau _l$ ($l\in {{\mathbb {Z}}}$) over the noncompact complex Grassmann manifold $G/K$; $G=SU(r,r+b)$ and $K=S(U(r)\times U(r+b))$. Let ${\mathbb {D}}(E_l)$ be the algebra of $G$-invariant differential operators on $E_l$. Let $\lambda$ be a real and regular spectral parameter in ${\mathfrak {a}}^*$, and let $F$ be a solution of the system differential equations on $E_l$: $DF=\chi _{\lambda ,l}(D)F$ for all $D$ in ${\mathbb {D}}(E_l)$. In this article, we obtain a necessary and sufficient condition for this $F$ to be represented by the Poisson transform of $f$ in the section space $L^2(K\times _M {{\mathbb {C}}})$.

Appendix

We first review some facts on Jacobi functions. We refer to [17] for more details.

For $\alpha ,\beta , \lambda \in {{\mathbb {C}}}$, $\alpha \ne -1-2,\ldots$, the Jacobi functions of the first kind $\displaystyle \phi ^{\alpha ,\beta }_\lambda (t)$ is the solution of the differential equation

$$\begin{aligned} \frac{d^2}{dt^2}\phi _\lambda ^{\alpha ,\beta }+[2(\alpha +1)\coth t+(2\beta +1)\tanh t]\frac{d}{dt}\,\phi _\lambda ^{\alpha ,\beta }=-(\lambda ^2+(\alpha +\beta +1)^2)\phi _\lambda ^{\alpha ,\beta },\nonumber \\ \end{aligned}$$

(7.1)

which satisfied $\phi _\lambda ^{\alpha ,\beta }(0)=1$ and $\frac{d}{dt}\phi _\lambda ^{\alpha ,\beta }(t)_{\mid t=0} =0$. The Jacobi functions can be expressed by the Gauss hypergeometric functions as

$$\begin{aligned} \phi ^{(\alpha ,\beta )}_\lambda (t)=F\left( \frac{i\lambda +\alpha +\beta }{2},\frac{-i\lambda +\alpha +\beta }{2}; \alpha +1; -\sinh ^2 t\right) . \end{aligned}$$

For $\lambda \notin -i{{\mathbb {N}}}$, another solution $\Phi _\mu ^{\alpha ,\beta }$ of (7.1) such that

$$\begin{aligned} \Phi _\mu ^{\alpha ,\beta }(t)=\mathrm{e}^{(i\mu -\alpha -\beta -1)t}(1+\circ (1)) \quad \textit{as} \,\, t\rightarrow +\infty , \end{aligned}$$

(7.2)

is given by (see [17, (2.4)]

$$\begin{aligned} \Phi _\lambda ^{\alpha ,\beta }(t)=(2\sinh t)^{i\lambda -\alpha -\beta -1}F\left( \frac{-i\lambda +\alpha +\beta +1}{2},\frac{-i\lambda -\alpha +\beta +1}{2}, 1-i\lambda ; -(\sinh t)^{-2} \right) . \end{aligned}$$

For $\lambda \notin {{\mathbb {Z}}}$, $\Phi _\lambda ^{\alpha ,\beta }$ and $\Phi _{-\lambda }^{\alpha ,\beta }$ are linearly independent solutions of (7.1), so we have

$$\begin{aligned} \phi _\lambda ^{\alpha ,\beta }(t)=c_{\alpha ,\beta }(\lambda )\Phi _\lambda ^{\alpha ,\beta }(t)+c_{\alpha ,\beta }(-\lambda )\Phi _{-\lambda }^{\alpha ,\beta }(t), \end{aligned}$$

where

$$\begin{aligned} c_{\alpha ,\beta }(\lambda )=\frac{2^{\alpha +\beta +1-i\lambda }\Gamma (\alpha +1)\Gamma (i\lambda )}{\Gamma \left( \frac{\alpha +\beta +1+i\lambda }{2}\right) \Gamma \left( \frac{\alpha -\beta +1+i\lambda }{2}\right) }. \end{aligned}$$

We shall also need some estimates on the Jacobi functions from [15], but which will be stated here for $\alpha ,\beta , \lambda \in {{\mathbb {R}}}, \alpha >-\frac{1}{2}$.

Lemma 7.1

[15] Assume $\alpha ,\beta$ with $\alpha >-\frac{1}{2}$.

• (i) For each and $\delta >0$, there exists a positive constant $C$ such that for all $t\ge \delta$ and all $\lambda \in {{\mathbb {R}}}$

  $$\begin{aligned} \mid \Phi _{\lambda }^{\alpha ,\beta }(t)\mid \le C\mathrm{e}^{-(\alpha +\beta +1)t}. \end{aligned}$$

  (7.3)

• (ii) There exists a positive constant $C$ such that for all $t\ge 0$ and all $\lambda \in {{\mathbb {R}}}$

  $$\begin{aligned} \mid \frac{d^n}{dt^n}\phi _\lambda ^{\alpha ,\beta }(t)\mid \le C(1+\mid \lambda \mid )^n(1+t)\mathrm{e}^{-(\alpha +\beta +1)t}. \end{aligned}$$

  (7.4)

Lemma A1

For $\alpha ,\beta \in {\mathbb {R}}$ with $\alpha >-\frac{1}{2}$ and $n\in {\mathbb {N}}$, there exists a positive constant $C$ such that for all $t\ge 0$ and $\lambda \in {{\mathbb {R}}}{\setminus }\{0\}$

$$\begin{aligned} \mid \lambda \dfrac{d^{n}}{dt^{n}}\phi _{\lambda }^{\alpha ,\beta }(t)\mid \le C(1+\lambda ^{2})^{n+1}e^{-(\alpha +\beta +1) t} \end{aligned}$$

Proof

Consider first the case $n=0$. Since $\alpha >-\frac{1}{2}$ it follows from the Stirling formula that there exists a positive constant $C$ such that for all $\lambda \in {{\mathbb {R}}}{\setminus }\{0\}$ $\displaystyle \mid \lambda c(\lambda )\mid \le C(1+\mid \lambda \mid )$.

It follows from (7.3) that there exists $C>0$ such that for all $t\ge 1$

$$\begin{aligned} \mid \lambda \phi _\lambda ^{\alpha ,\beta }(t)\mid\le & {} C(1+\mid \lambda \mid )\mathrm{e}^{-(\alpha +\beta +1)t}\\\le & {} 2C(1+\lambda ^{2})\mathrm{e}^{-(\alpha +\beta +1)t}. \end{aligned}$$

We also have from (7.4) that there exists $C>0$ such that for all $t\le 1$

$$\begin{aligned} \mid \lambda \phi _\lambda ^{\alpha ,\beta }(t)\mid\le & {} C(1+\mid \lambda \mid )\mathrm{e}^{-(\alpha +\beta +1)t}\\\le & {} 2C(1+\lambda ^{2})\mathrm{e}^{-(\alpha +\beta +1)t}. \end{aligned}$$

Combining these inequalities, the lemma follows for $n=0$.

We prove the case $n\ge 1$ by induction with respect to $n$ using the formula

$$\begin{aligned} \dfrac{d^{n}}{dt^{n}}\phi ^{\alpha ,\beta }_{\lambda }(t)=\dfrac{-\Gamma (\alpha +1)}{4}((\alpha +\beta +1)^{2}+\lambda ^{2})\sum _{k=0}^{n-1}2^{k}\dfrac{d^{k}}{dt^{k}}(\sinh (2t))\dfrac{d^{n-1-k}}{dt^{n-1-k}}\phi ^{\alpha +1,\beta +1}_{\lambda }(t). \end{aligned}$$

This last formula follows from the identity obtained by a simple calculation:

$$\begin{aligned} \dfrac{d}{dt}\phi ^{\alpha ,\beta }_{\lambda }(t)= & {} \dfrac{-\Gamma (\alpha +1)}{4}((\alpha +\beta +1)^{2}+\lambda ^{2})\sinh (2t)\phi ^{\alpha +1,\beta +1}_{\lambda }(t). \end{aligned}$$

$\square$

Lemma 7.2

Let $f$ be a ${{\mathbb {C}}}$-valued function on ${{\mathbb {R}}}$ whose $n$th-order derivative is continuous. Assume $f(t_1)=f(t_2)=\cdots =f(t_n)=0$, $t_1,..,t_n$ being real numbers such that $t_{1}>t_{2}>\cdots > t_{n}$. Then for any $t\in {{\mathbb {R}}}$, we have

$$\begin{aligned} |\dfrac{f(t)}{\prod \limits _{i=1}^{n}(t-t_{i})}|\le & {} \sup _{s\in [\min (t,t_{n}),\max (t,t_{1})]}|f^{(n)}(s)|. \end{aligned}$$

Proof

For $n=1$, the result is obvious. We prove the case $n\ge 1$ by induction on $n$ using the fundamental theorem of calculus.$\square$

Recall that by the Cartan decomposition $G=K\exp {\mathfrak {p}}$ each $g\in G$ can be uniquely written $\displaystyle g=\pi _0(g)\exp X(g)$ where $\pi _0(g)\in K$ and $X(g)\in {\mathfrak {p}}$. For $g=\kappa _{1}(g)\mathrm{e}^{A^+(g)})\kappa _{2}(g)$ we know that $\pi _0(g)=\kappa _{1}(g)\kappa _{2}(g)$.

Lemma 7.3

Let $g,h\in SU(r,r+b)$ and $H_T\in {\mathfrak {a}}^+$. Then

$$\begin{aligned}&\displaystyle \kappa _1(h^{-1}g)\kappa _2(h^{-1}g)=\kappa _1(h^{-1}\kappa _1(g)\mathrm{e}^{A^+(g)})\kappa _2(h^{-1}\kappa _1(g)\mathrm{e}^{A^+(g)})\kappa _2(g). \end{aligned}$$

(7.5)

$$\begin{aligned}&\lim _{R\rightarrow +\infty }\tau _l(\kappa _{1}(g\mathrm{e}^{RH_T})\kappa _{2}(g\mathrm{e}^{RH_T}))=\tau _l(\kappa (g)) \end{aligned}$$

(7.6)

Proof

The identity (7.5) is obvious.

We write $g=\kappa (g)e^{H(g)}n(g)$ with respect to the Iwasawa decomposition $G=KAN$. Then we have

$$\begin{aligned} \pi _{0}(g\mathrm{e}^{RH_{T}})=\kappa (g)\pi _{0}(e^{H(g)}n(g)\mathrm{e}^{RH_T}) \qquad \forall R>0. \end{aligned}$$

Therefore, it is enough to prove

$$\begin{aligned} \lim _{R\rightarrow +\infty }\tau _l(\pi _0(\mathrm{e}^{H(g)}n(g)\mathrm{e}^{RH_T}))=1. \end{aligned}$$

We first observe that if $g=\begin{pmatrix} A&{}B \\ C&{}D\\ \end{pmatrix}$, then

$$\begin{aligned} \tau _l(\pi _0(g))=\left( \frac{\det D}{\mid \det D\mid }\right) ^{-l} \end{aligned}$$

(7.7)

We write $e^{-RH_T}n(g)e^{RH_T}$ in $r\times r+b$-block notation as

$$\begin{aligned} e^{-RH_T}n(g)e^{RH_T}=\begin{pmatrix} A_{1}(R)&{}B_{1}(R) \\ A_{2}(R)&{}B_{2}(R)\\ \end{pmatrix}. \end{aligned}$$

Then, a straightforward computation shows

$$\begin{aligned}&\tau _{l}(\pi _{0}(e^{H(g)+RH_T}e^{-RH_T}n(g)e^{RH_T}))\\&\quad = \left( \dfrac{det\left( \begin{pmatrix}\tanh (RT+S)\\ 0_{b\times r} \end{pmatrix}B_{1}(R)+B_{2}(R)\right) }{|det(\begin{pmatrix}\tanh (RT+S)\\ 0_{b\times r} \end{pmatrix}B_{1}(R)+B_{2}(R)|}\right) ^{-l}, \quad H(g)=H_S \end{aligned}$$

Since $\lim _{R\rightarrow +\infty }\mathrm{e}^{-RH_T}n(g)\mathrm{e}^{RH_T}=I_{2r+b}$, we get

$$\begin{aligned} \lim _{R\rightarrow +\infty }\tau _{l}(\pi _{0}(e^{H(g)+RH_T})=1, \end{aligned}$$

as was to be shown. $\square$