\documentclass[12pt]{amsart} \headheight=8pt \topmargin=0pt \textheight=624pt \textwidth=432pt \oddsidemargin=18pt \evensidemargin=18pt \usepackage{amsfonts} \usepackage{amsmath} \def\frak{\mathfrak} \newtheorem{theorem}{Theorem}[section] \newtheorem{corollary}[theorem]{Corollary} \newtheorem{remark}[theorem]{Remark} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{example}[theorem]{Example} \newtheorem{definition}[theorem]{Definition} \newtheorem{problem}[theorem]{Problem} \newtheorem{conjecture}[theorem]{Conjecture} \def\bc{\mathbb C} \def\br{\mathbb R} \def\bh{\mathbb H} \def\mand{\quad\mbox{and}\quad} \def\Ad{{\rm Ad}} \def\ad{{\rm ad}} \def\ggc{{\frak g}_\bc} \def\mbx#1{\makebox[#1cm]{}} \def\sdi{\,\mbox{$\rhd\kern-.55em<$}\,} \def\Aut{\,\mbox{Aut}\,} \def\Int{\,\mbox{Int}\,} \def\Im{\,\mbox{Im}\,} \def\Re{\,\mbox{Re}\,} \def\Lie{\,\mbox{Lie}\,} \begin{document} \bibliographystyle{alpha} \title[Stein extensions of Riemann symmetric spaces]{Stein extensions of Riemann symmetric spaces and dualities of orbits on flag manifolds} \author[S. Gindikin]{Simon Gindikin{*} } \address{\hskip-\parindent Simon Gindikin\\ Department of Mathematics\\ Hill Center, Rutgers University\\ 110 Frelinghysen Road\\ Piscataway, NJ 08854-8019, U.S.A.} \email{gindikin@math.rutgers.edu} \thanks{{*} The first author is partly supported by NSF grant DMS 00-70816.} \author[T. Matsuki]{Toshihiko Matsuki} \address{\hskip-\parindent Toshihiko Matsuki\\ Faculty of Integrated Human Studies\\ Kyoto University\\ %Street address or POBox\\ Kyoto 606-8501, Japan} \email{matsuki@math.h.kyoto-u.ac.jp} %\thanks{Research at MSRI is supported in part by NSF grant DMS-9701755. %More thanks...} \begin{abstract} It is known that $K_\bc$-orbits $S$ and $G_\br$-orbits $S'$ on a complex flag manifold are in one-to-one correspondence by the condition that $S\cap S'$ is non-empty and compact. We may replace $K_\bc$ by some conjugate $xK_\bc x^{-1}$ so that the correspondence is preserved. We will show that this replacement is related to the domain introduced by Akhiezer and Gindikin. \end{abstract} \maketitle \section{Introduction} Let $G_\bc$ be a connected complex semisimple Lie group and $G_\br$ a connected real form of $G_\bc$. Let $K_\bc$ be the complexification in $G_\bc$ of a maximal compact subgroup $K$ of $G_\br$. Let $X=G_\bc/P$ be a flag manifold of $G_\bc$ where $P$ is an arbitrary parabolic subgroup of $G_\bc$. Then there exists a natural one-to-one correspondence between the set $K_\bc\backslash X$ of $K_\bc$-orbits $S$ and the set $G_\br\backslash X$ of $G_\br$-orbits $S'$ on $X$ given by the condition: \begin{equation} S\leftrightarrow S'\Longleftrightarrow S\cap S'\mbox{ is non-empty and compact} \tag{1.1} \end{equation} (\cite{M3}, c.f. \cite{MUV} for an extension to sheaf theory). It is possible to interpret this correspondence as the duality between $K_\bc$-orbits and $G_\br$-orbits on flag manifolds $X=G_\bc/P$: dual orbits have non-empty compact intersections. It turns out that for a fixed group $G_\br$, this duality between its orbits with $K_\bc$-orbits is inherited by some conjugate subgroups $xK_\bc x^{-1}$ in $G_\bc$. In other words, let $M_\br=G_\br/K$ be the Riemann symmetric space, canonically embedded in its complexification $M_\bc=G_\bc/K_\bc$. Then in (1.1), instead of $K_\bc$, we can, of course, take $K_\bc(z)$ the isotropy subgroup of any point $z \in M_\br$. Our aim is to find the set $\widetilde M\supset M_\br$ of such $z \in M_\bc$ that (1.1) holds for $K_\bc(z)$. This problem is one of several problems about Stein extensions $\widetilde M$ of Riemann symmetric spaces $M_\br$, on which can be extended different analytic or geometrical constructions on $M_\br$. There is an informal conjecture that is an universal Stein extension for many such problems which in \cite{G} was suggested to call by complex crowns of Riemann symmetric spaces. For every $K_\bc$-orbit $S$ on $X=G_\bc/P$, define a subset $$C(S)=\{x\in G_\bc\mid xS\cap S'\mbox{ is non-empty and compact}\}$$ of $G_\bc$ where $S'$ is the $G_\br$-orbit on $X$ given by (1.1). \begin{remark} {\rm \ (1) \ Let $h\in G_\br$ and $k\in K_\bc$. Then we have $$hxkS\cap S'=hxS\cap hS'=h(xS\cap S').$$ Hence the sets $C(S)$ are left $G_\br$-invariant and right $K_\bc$-invariant. So we can also define a $G_\br$-invariant subset $\widetilde{M}(S)=C(S)/K_\bc$ of $M_\bc$ (2) \ Let $Z$ denote the center of $G_\bc$. Then the sets $C(S)$ are $Z$-invariant because $Z\subset P$. Hence the sets $C(S)$ depend only on Lie algebras. Note that the $K_\bc$-orbit structure and $G_\br$-orbit structure on $X$ depend only on Lie algebras since we assume that $K_\bc$ and $G_\br$ are connected. }\end{remark} Now we can propose the following problem. \begin{problem} \ What is the set $$C=\bigcap C(S)\ ?$$ Here we take the intersection for all $K_\bc$-orbits $S$ on $X$ and all flag manifolds $X=G_\bc/P$ of $G_\bc$. Clearly, $\displaystyle{\widetilde M =\bigcap \widetilde{M}(S)=C/K_\bc}$. \end{problem} Our conjecture is that the sets $C$ and $\widetilde{M}$ correspond to the domain $D_0$ from \cite{AG}. This domain includes $M_\br$ and is the maximal domain which unifies $G_\br$-orbits with compact isotropy subgroups. These orbits admit the next description. Let ${\frak g}_\br={\frak k}\oplus{\frak p}$ denote the Cartan decomposition of ${\frak g}_\br={\rm Lie}(G_\br)$ with respect to $K$. Let ${\frak t}$ be a maximal abelian subspace in $i{\frak p}$. Put $${\frak t}^+=\{Y\in{\frak t}\mid |\alpha(Y)|<{\pi\over 2} \mbox{ for all }\alpha\in\Sigma\}$$ where $\Sigma$ is the restricted root system of ${\frak g}_\bc$ with respect to ${\frak t}$. Then we can define a left $G_\br$-invariant right $K_\bc$-invariant open subset in $G_\bc$ by $$\widetilde{D_0}=G_\br(\exp{\frak t}^+)K_\bc$$ and a $G_\br$-invariant open subset $D_0=\widetilde{D_0}/K_\bc$ in $M_\bc$. \begin{conjecture} \ $C=\widetilde{D_0}Z$. (Equivalently, $\widetilde{M}=ZD_0$.) \end{conjecture} The aim of this paper is to prove: \begin{theorem} \ $C=\widetilde{D_0}Z$ for all classical cases and exceptional Hermitian cases. \end{theorem} We will prove this theorem by case-by-case checkings in Section 2-7 and by a general trick in Section 8. We will deal with the case where $G_\br$ is Hermitian type in Section 2, $SL(n,\br)$-case in Section 3, $SO(p,q)$-case in Section 4, cases of complex classical Lie groups in Section 5, $GL(n,\bh)$-case in Section 6 and $Sp(p,q)$-case in Section 7. (The essential part in $SL(n,\br)$-case was already proved in \cite{AG}, Example 1.) Let us explain a structure of these results. Let $B$ denote a Borel subgroup of $G_\bc$ and let $C_B$ denote the intersection of all $C(S)$ such that $S$ are closed $K_\bc$-orbits on $G_\bc/B$. In Section 2, we assume that $G_\br$ is Hermitian type. This case in many relations is a special one. Firstly, we consider the natural embedding $$\iota:M_\bc\to G_\bc/P\times G_\bc/\overline{P}$$ (\cite{WZ}) where $P$ and $\overline{P}$ are the usual maximal parabolic subgroups of $G_\bc$ containing $K_\bc$. Then we will prove in Proposition 2.2 that \begin{equation} \iota(D_0)=G_\br P/P\times G_\br\overline{P}/\overline{P} \ (\cong M_\br\times \overline{M_\br}). \tag{1.2} \end{equation} Let $B$ be a Borel subgroup of $G_\bc$ contained in $P$ and let consider two special closed $K_\bc$-orbits $S_1=K_\bc B/B=P/B$ and $S_2=K_\bc w_0B/B=\overline{P}w_0/B$ on $X=G_\bc/B$ where $w_0$ is the longest element in the Weyl group. (The corresponding open $G_\br$-orbits $S'_1$ and $S'_2$ are used to realize holomorphic and anti-holomorphic discrete series representations of $G_\br$.) Then the above realization (1.2) of $D_0$ clearly implies that $$\widetilde{D_0}=C(S_1)\cap C(S_2).$$ The second part of the section is dedicated to the proof of $$C_B=C(S_1)\cap C(S_2)$$ (Proposition 2.4) which was already proved in \cite{WZ} for classical cases by case-by-case checkings. But we will give, in this paper, a simple general proof. \begin{remark} {\rm Let $Q$ be a parabolic subgroup of $G_\bc$ containing $B$ and $\pi: G_\bc/B\to G_\bc/Q$ the canonical projection. Then a $K_\bc$-orbit $S$ (resp. $G_\br$-orbit $S'$) is called to be of holomorphic type if $$S=\pi(S_1)\quad\mbox{or}\quad \pi(S_2)\qquad(\mbox{resp. }S'=\pi(S'_1)\quad\mbox{or}\quad \pi(S'_2)).$$ Clearly, this definition is equal to the one in \cite{WZ} when $S'$ is open in $G_\bc/Q$. By this definition, $S$ is automatically closed and $S'$ is open in $G_\bc/Q$. By the result in \cite{WZ}, we have $$\widetilde{M}(S)=\begin{cases} G_\br P/K_\bc & \text{if $S=\pi(S_1)$,} \\ G_\br \overline{P}/K_\bc & \text{if $S=\pi(S_2)$.} \end{cases}$$ }\end{remark} All five sections from 3 to 7 have the same structure. We first take a minimal flag manifold $X_0$ such that the $K_\bc$-orbit structure on $X_0$ is nontrivial and take a closed $K_\bc$-orbit $S_0$ \ ($S_+$ in Section 7). Then we will show that $C(S_0)$ is essentially equal to $\widetilde{D_0}Z$. (In Section 5, 6 and 7, they are equal. In Section 3 and 4, they are not equal in general but \begin{equation} C(S_0)_0=\widetilde{D_0}. \tag{1.3} \end{equation} where $C(S)_0$ denote the connected component of $C(S)$ containing the identity.) The most essential idea in these sections is that the problem is reduced to some nonstandard problems of linear algebra in a classical style as in \cite{AG}, Example 1 which we will solve by elementary calculus. Here are samples of these problems. For $SL(n,\br)$, we need to describe such (complex) nondegenerate symmetric forms $m$ that corresponding quadrics $(^txmx=0)$ in the projective space $P^{n-1}(\bc)$ have no real points. Of course this set contains the symmetric space $M_\br$ of real positive matrixes. For $SL(n,\bc)$, we need to describe such $m\in SL(n,\bc)$ that $(^t\overline{x}mx=0)$ has no real points in the complex projective space. Again hermitian positive $m$ (Riemann symmetric space $M_\br$) satisfy these condition but we want to describe all of them. One more example: for $SO(p,q)$ we take the quadric corresponding to the canonical $(p,q)$-form in the complex projective space and describe $(p-1)$-sections on which the corresponding hermitian form will be positive. Evidently this set includes the real Grassmannian $Gr_{p}(\br)$. In all cases these sets are $G_\br$-invariant and we describe them by a parameterization of orbits, which has a similar structure in all these and some other cases. Next we consider full flag manifolds $X=G_\bc/B$ and show that \begin{equation} C(S)=\widetilde{D_0}Z \tag{1.4} \end{equation} for all closed $K_\bc$-orbits $S$ on $X$. In Section 8, we will show in general (Proposition 8.1) that $$C=C_B.$$ So we have proved Theorem 1.4 for all classical groups and exceptional groups of Hermitian type. Moreover we will show in Proposition 8.3 that \begin{equation} C(S)_0=C_0 \tag{1.5} \end{equation} for the open $K_\bc$-orbit $S$ on $X=G_\bc/B$ for any $G_\br$. This result has an interesting application (see below). In the appendix, we will discuss some geometric consideration for $SO(p,q)$-case: We may embed $D_0$ into the Grassmann manifold $X_1=Gr_{p,p+q}$ of $SL(p+q,C)$ in this case. It turns out that the image is bigger than the expected Hermitian symmetric domain which is an open $SU(p,q)$-orbit on $X_1$. Though this consideration is independent of our conjecture, it seems to be important for geometric studies on $D_0$. \bigskip We have seen in (1.3), (1.4) and (1.5) that for typical orbits $S$ we have \begin{equation} C(S)_0=\widetilde{D_0}. \tag{1.6} \end{equation} We can also prove it for all closed $K_\bc$-orbits $S$ on $X=G_\bc/P$ ($P$ is arbitrary) of nonholomorphic type (c.f. \cite{WZ}) for all classical cases. It will be published in a subsequent paper. Of course, we assume that $X$ is not $K_\bc$-homogeneous. For non-Hermitian cases, every closed $K_\bc$-orbit (open $G_\br$-orbit) is nonholomorphic type. By (1.6), we expect that \begin{conjecture} \ $C(S)_0=\widetilde{D_0}$ for all $K_\bc$-orbits $S$ on $X=G_\bc/P$ of nonholomorphic type. (Every non-closed $K_\bc$-orbit on $X$ and each orbit for groups of non Hermitian type is defined to be of nonholomorphic type by Remark 1.5.) \end{conjecture} Apparently, the new element which we need for proving of this conjecture is a proof of the inclusions $C(S)_0 \subset \widetilde{D_0}$. (The opposite inclusion is already proved in Theorem 1.4 for all classical cases and exceptional Hermitian cases.) This problem (Conjecture 1.6) for different types of orbits corresponds to different geometrical problems, some of which were studied extensively. Open $G_\br$-orbits $S'$ are called by flag domains and they are connected with realizations of discrete series and some other unitary representations. Then $xS$ are compact complex cycles and we have the problem on the parameterization of cycles in flag domains. Wolf proved that for these cases $C(S)$ are Stein manifolds (\cite{W}). Wolf and Zierau proved that $\widetilde{M}(S)_0\supset M_\br\times \overline{M_\br}$ for Hermitian classical groups and orbits $S$ of nonholomorphic type (\cite{WZ}). By our computation, it seems that their proof of the opposite inclusion contains a gap. Another interesting special case is when $S$ is open. There is only one such orbit and it is Zariski open. Correspondingly, there is a unique closed $G_\br$-orbit $S'$ which is contained in $S$. The supplement $\widetilde S$ to $S$ can be defined by an polynomial equation $D(g)=0$, where the function $D$ satisfies the condition: $$ D(kgp)=D(g) \chi(p)$$ for a character $\chi$ of $P$. (This function $D$ is a $K_\bc$-invariant element of a class-one finite dimensional representation of $G_\bc$ which is realized, by the Borel-Weil theorem, on the space of holomorphic cross sections of the holomorphic line bundle over $G_\bc/P$ with some strictly dominant class-one integral character of $P$.) In the case $P=B$, this condition is the obstruction to the existence of the Iwasawa decomposition on $G_\bc$. Then we can define $C(S)$ as $$ C(S) =\{x\mid D(x^{-1}g) \neq 0, \ g \in S'\}$$ which is clearly a Stein manifold because $C(S)$ is obtained by deleting from $G_\bc$ an infinite family of analytic hypersurfaces. If $P=B$, then we proved in Proposition 8.3 that $C(S)_0=C_0$ and we proved $C_0=\widetilde{D_0}$ for all classical groups. Hence $D_0=\widetilde{D_0}/K_\bc$ is a Stein manifold for all these groups. In such a way, we prove the conjecture from \cite{AG} for all these groups. The functions $D$ (let us call them determinant functions) must play essential role in the analysis on flag manifolds and symmetric spaces and deserve a special investigation. In simplest cases they reduce to the norm-functions on symmetric cones (Jordan algebras); for $SU(p,q)$ they constructed in \cite{BGW}. Finally, we want to explain, by the following simplest example, the relation to the structure of the double coset decomposition $G_\br\backslash G_\bc/K_\bc$ which was studied in \cite{M4}. \begin{example} \ {\rm Let $G_\bc=SL(2,\bc), G_\br=SL(2,\br)$ and $K_\bc=SO(2,\bc)$. Then the flag manifold $X=G_\bc/P\cong P^1(\bc)=\bc\cup\{\infty\}$ is decomposed to $K_\bc$-orbits as $$X=S_1\sqcup S_2\sqcup S_3$$ where $S_1=\{i\}$ and $S_2=\{-i\}$. $X$ is also decomposed to $G_\br$-orbits as $$X=S'_1\sqcup S'_2\sqcup S'_3$$ where $S'_1=\{z\in X\mid \Im z>0\}$ and $S'_2=\{z\in X\mid \Im z<0\}$. By the results of \cite{M4}, we can take representatives $x$ of the double coset decomposition $G_\br\backslash G_\bc/K_\bc$ as follows. (Then $xS_1$ and $xS_2$ are as in the remaining columns.) \bigskip \centerline{ \vbox{\offinterlineskip \hrule \halign{&\vrule#&\strut\quad$\hfil#\hfil$\quad\cr height2pt&\omit&&\omit&&\omit&&\omit&&\omit&\cr & x && xS_1 && xS_2 && xS_1\subset && xS_2\subset &\cr height2pt&\omit&&\omit&&\omit&&\omit&&\omit&\cr \noalign{\hrule} height4pt&\omit&&\omit&&\omit&&\omit&&\omit&\cr & a_\theta \ (0\le\theta<\displaystyle{\pi\over 4}) && \left\{i\tan\left(\theta+\displaystyle{\pi\over 4}\right)\right\} && \left\{-i\cot\left(\theta+\displaystyle{\pi\over 4}\right)\right\} && S'_1 && S'_2 &\cr height4pt&\omit&&\omit&&\omit&&\omit&&\omit&\cr & a_{\pi/4} && \{\infty\} && \{0\} && S'_3 && S'_3 &\cr height4pt&\omit&&\omit&&\omit&&\omit&&\omit&\cr & a_\theta \ (\displaystyle{\pi\over 4}<\theta\le\displaystyle{\pi\over 2}) && \left\{i\tan\left(\theta+\displaystyle{\pi\over 4}\right)\right\} && \left\{-i\cot\left(\theta+\displaystyle{\pi\over 4}\right)\right\} && S'_2 && S'_1 &\cr height4pt&\omit&&\omit&&\omit&&\omit&&\omit&\cr & b_ta_{\pi/4} \ (t>0) && \{i\coth t\} && \{i\tanh t\} && S'_1 && S'_1 &\cr height4pt&\omit&&\omit&&\omit&&\omit&&\omit&\cr & b_ta_{\pi/4} \ (t<0) && \{i\coth t\} && \{i\tanh t\} && S'_2 && S'_2 &\cr height4pt&\omit&&\omit&&\omit&&\omit&&\omit&\cr & c_+a_{\pi/4} && \{\infty\} && \{i\} && S'_3 && S'_1 &\cr height4pt&\omit&&\omit&&\omit&&\omit&&\omit&\cr & c_-a_{\pi/4} && \{\infty\} && \{-i\} && S'_3 && S'_2 &\cr height4pt&\omit&&\omit&&\omit&&\omit&&\omit&\cr & d_+a_{\pi/4} && \{-i\} && \{0\} && S'_2 && S'_3 &\cr height4pt&\omit&&\omit&&\omit&&\omit&&\omit&\cr & d_-a_{\pi/4} && \{i\} && \{0\} && S'_1 && S'_3 &\cr height4pt&\omit&&\omit&&\omit&&\omit&&\omit&\cr} \hrule} } \bigskip \noindent Here \begin{gather*} a_\theta=\begin{pmatrix} \cos\theta & i\sin\theta \\ i\sin\theta & \cos\theta \end{pmatrix}, \ b_t=\begin{pmatrix} \cosh t & i\sinh t \\ -i\sinh t & \cosh t \end{pmatrix}, \ c_\pm=\begin{pmatrix} 1 & \pm i \\ 0 & 1 \end{pmatrix} \mbox{ and } d_\pm=\begin{pmatrix} 1 & 0 \\ \pm i & 1 \end{pmatrix}. \end{gather*} By the above list, we can see $$C=C(S_1)\cap C(S_2)=\bigsqcup_{0\le\theta<\pi/4} G_\br a_\theta K_\bc=\widetilde{D_0}.$$ On the other hand, $C(S_3)$ consists of four connected components $$\bigsqcup_{0\le\theta<\pi/4} G_\br a_\theta K_\bc,\qquad \bigsqcup_{\pi/4<\theta\le\pi/2} G_\br a_\theta K_\bc,\qquad \bigsqcup_{t>0} G_\br b_ta_{\pi/4} K_\bc \mand \bigsqcup_{t<0} G_\br b_ta_{\pi/4} K_\bc.$$ Note that there are four non-closed $G_\br$-$K_\bc$ double cosets represented by $c_\pm a_{\pi/4}$ and $d_\pm a_{\pi/4}$ which have common boundary represented by $a_{\pi/4}$. }\end{example} Acknowledgement: The authors are grateful to H. Ochiai who pointed out an elementary mistake in our first draft. While we were correcting it, the proofs were very much improved. \section{Hermitian cases} Suppose that $G_\br$ is simple and of Hermitian type. Then ${\frak k}=\Lie(K)$ has a one-dimensional center and a nontrivial central element in $i{\frak k}$ defines a maximal parabolic subgroup $$P=K_\bc\exp{\frak n}$$ of $G_\bc$ where $${\frak n}=\bigoplus_{\alpha\in\Delta_n^+} \ggc({\frak j},\alpha)$$ is the nilpotent radical of $\Lie(P)$. Here ${\frak j}$ is a maximal torus of ${\frak k}$ and $\ggc({\frak j},\alpha)$ denote the root space for a root $\alpha$ in the root system $\Delta=\Delta(\ggc,{\frak j})$. Let $x\mapsto \overline{x}$ denote the conjugation of $G_\bc$ with respect to $G_\br$. Then we have $$\overline{P}=K_\bc\exp\overline{\frak n}\mand \overline{\frak n}=\bigoplus_{\alpha\in\Delta_n^+} \ggc({\frak j},-\alpha).$$ Note that $${\frak p}_\bc={\frak n}\oplus\overline{\frak n}.$$ Let $\{\beta_1,\ldots,\beta_\ell\}$ be a maximal strongly othogonal system in $\Delta_n^+$ and $X_1,\ldots,X_\ell$ be nonzero elements of $\ggc({\frak j},\beta_1),\ldots,\ggc({\frak j},\beta_\ell)$, respectively. Then we can define maximal abelian subspaces $${\frak a}= \br(X_1+\overline{X_1})\oplus\cdots\oplus \br(X_\ell+\overline{X_\ell})$$ of ${\frak p}$ and $${\frak t}= \br(X_1-\overline{X_1})\oplus\cdots\oplus \br(X_\ell-\overline{X_\ell})$$ of $i{\frak p}$. Put $A=\exp{\frak a}, \ T=\exp{\frak t}$ and $T^+=\exp{\frak t}^+$. Though the following fact seems to be known, we will give a proof for the sake of completeness. \begin{lemma} \ {\rm (1)} \ $AP=T^+P$. {\rm (2)} \ $G_\br P=KT^+P$. \end{lemma} Proof. \ (1) Let ${\frak l}_\bc$ be the complex Lie subalgebra of ${\frak g}_\bc$ generated by $\{X_1,\overline{X_1},X_2,\overline{X_2},\break \ldots, X_\ell,\overline{X_\ell}\}$. Then ${\frak l}_\bc$ is isomorphic to the direct sum of $\ell$-copies of ${\frak sl}(2,\bc)$. Note that the set ${\frak t}^+$ is characterized by the subset $\Delta({\frak l}_\bc,{\frak t})$ of $\Sigma=\Delta({\frak g}_\bc,{\frak t})$ because $\Sigma$ is of type ${\rm C}_\ell$ or ${\rm BC}_\ell$ and $\Delta({\frak l}_\bc,{\frak t})$ is the set of longest roots in $\Sigma$. Let $L_\bc$ denote the analytic subgroup of $G_\bc$ for ${\frak l}_\bc$. Then by a computation in $SL(2,\bc)$, we have $$A(L_\bc\cap P)=T^+(L_\bc\cap P).$$ Hence we have $$AP=T^+P.$$ (2) is clear by the Cartan decomposition $G_\br=KAK$. \hfill q.e.d. \bigskip Consider the natural embedding $$\iota: G_\bc/K_\bc \hookrightarrow G_\bc/P\times G_\bc/\overline{P}$$ given by $$xK_\bc\mapsto (xP,x\overline{P})$$ (\cite{WZ}). \begin{proposition} \ $\iota(D_0)=G_\br P/P\times G_\br \overline{P}/\overline{P}$. \ $($Here $D_0=\widetilde{D_0}/K_\bc.)$ \end{proposition} Proof. \ By Lemma 2.1, we have $$\iota(T^+K_\bc/K_\bc)\subset G_\br P/P\times G_\br \overline{P}/\overline{P}$$ and hence $$\iota(D_0)\subset G_\br P/P\times G_\br \overline{P}/\overline{P}.$$ Conversely, let $g$ and $h$ be elements of $G_\br$. Then we have only to find an $x\in D_0$ such that $\iota(x)=(gP,h\overline{P})$. By $G_\br$-action, we have only to consider $(h^{-1}gP,\overline{P})$. Write $h^{-1}g=k_1ak_2$ with $k_1,k_2\in K$ and $a\in A$. Then we have only to consider $(aP,\overline{P})$ since $$(k_1^{-1}h^{-1}gP,k_1^{-1}\overline{P})=(aP,\overline{P}).$$ Furthermore we have only to consider $(a^{1/2}P,a^{-1/2}\overline{P})$. Let $\tau$ denote the conjugation in $G_\bc$ with respect to the compact real form $U=K\exp i{\frak p}$. Then we have $\tau(P)=\overline{P}$ since $P$ is invariant by the holomorphic involution defining $K_\bc$ and we also have $\tau(a^{1/2})=a^{-1/2}$. Write $$a^{1/2}=tp$$ with some $t\in T^+$ and $p\in P$ by Lemma 2.1 (1). Then applying $\tau$ on the both sides, we get $$a^{-1/2}=t\tau(p).$$ Hence we have $$\iota(tK_\bc)=(a^{1/2}P,a^{-1/2}\overline{P}).$$ Thus we have proved $\iota(D_0)=G_\br P/P\times G_\br \overline{P}/\overline{P}$. \hfill q.e.d. \bigskip Let $\Delta^+$ be a positive system of $\Delta(\ggc,{\frak j})$ containing $\Delta_n^+$. Then $\Delta^+$ defines a Borel subgroup $B$ of $G_\bc$ such that $B\subset P$. Let $w_0$ be the longest element (with respect to $\Delta^+$) of the Weyl group $W=W({\frak j})$. Then the two closed sets $$S_1=P=K_\bc P=K_\bc B \mand S_2=\overline{P}w_0=K_\bc \overline{P}w_0 =K_\bc(w_0Bw_0^{-1})w_0=K_\bc w_0 B$$ are contained in the open sets $$S'_1=G_\br P=G_\br B \mand S'_2=G_\br\overline{P}w_0=G_\br w_0 B,$$ respectively (c.f. \cite{M1}). Let $x$ be an element of $G_\bc$. Then Proposition 2.2 implies that \begin{equation} x\in\widetilde{D_0}\Longleftrightarrow xS_1\subset S'_1 \mbox{ and } xS_2\subset S'_2 \notag \end{equation} Let $W_K$ be the subgroup of $W$ defined by $W_K=N_{K_\bc}({\frak j})/Z_{K_\bc}({\frak j})$. Then the set of closed $K_\bc$-$B$ double cosets in $G_\bc$ is parametrized by $W_K\backslash W$ (\cite{M0}, \cite{R}). \begin{lemma} \ For $w\in W$, we have $$K_\bc(BwB)^{cl} \cap K_\bc w_0(Bw_0^{-1}wB)^{cl} =K_\bc wB.$$ \end{lemma} Proof. \ We can choose a $w'=w_Kw \in W_Kw$ such that $$K_\bc(BwB)^{cl}=(Bw'B)^{cl}$$ since $P=K_\bc B$ and $(BwB)^{cl}$ are both expressed as products of parabolic subgoups $P_\alpha=B\cup Bw_\alpha B$ with $\alpha\in\Psi$ where $\Psi$ denote the set of simple roots in $\Delta^+$. It is known that $$(Bw'B)^{cl}\cap(B^-w'B)^{cl}=w'B$$ where $B^-=w_0Bw_0^{-1}$. So we have \begin{align*} K_\bc(BwB)^{cl} \cap K_\bc w_0(Bw_0^{-1}wB)^{cl} & = (Bw'B)^{cl}\cap K_\bc (B^-wB)^{cl} \\ & = (Bw'B)^{cl}\cap K_\bc (B^-w'B)^{cl} \\ & = K_\bc w'B \\ & = K_\bc wB \end{align*} since $K_\bc (B^-wB)^{cl}=(\overline{P}wB)^{cl}=(\overline{P}w'B)^{cl}=K_\bc (B^-w'B)^{cl}$ and since $w_K$ is contained in $K_\bc$. \hfill q.e.d. \begin{proposition} \ Let $x$ be an element of $G_\bc$. If $xS_1\subset S'_1$ and $xS_2\subset S'_2$, then $xS\subset S'$ for all closed $K_\bc$-$B$ double cosets $S$ in $G_\bc$. \end{proposition} Proof. \ Let $w\in W$. Then by Lemma 2.3, we have $$S_1(BwB)^{cl} \cap S_2(Bw_0^{-1}wB)^{cl} =K_\bc wB.$$ We also have $$S'_1(BwB)^{cl} \cap S'_2(Bw_0^{-1}wB)^{cl} =G_\br wB$$ since the correspondence between $K_\bc$-orbits and $G_\br$-orbits is compatible with the right $P_\alpha$-action for $\alpha\in\Psi$ (\cite{M1}). Since $xS_1\subset S'_1$ and $xS_2\subset S'_2$, we have $$xS_1(BwB)^{cl}\subset S'_1(BwB)^{cl} \mand xS_2(Bw_0^{-1}wB)^{cl}\subset S'_2(Bw_0^{-1}wB)^{cl}.$$ Hence we have \begin{equation*} \mbx5 xK_\bc wB\subset G_\br wB.\mbx5\mbox{q.e.d.} \end{equation*} \section{$SL(n,\br)$-case} Let $G_\bc=SL(n,\bc), \ K_\bc=SO(n,\bc)$ and $G_\br=SL(n,\br)$ with $n\ge 3$. First consider the flag manifold $X_0=P^{n-1}(\bc)=(\bc^n-\{0\})/\bc^\times$. Then there are two $K_\bc$-orbits and two $G_\br$-orbits on $X_0$. The closed $K_\bc$-orbit is defined by $$S_0: z_1^2+\cdots+z_n^2=0$$ on $X_0$ and the open $G_\br$-orbit is $$S'_0=X_0-P^{n-1}(\br).$$ We can take $${\frak t}^+=\left\{\begin{pmatrix} i\theta_1 && 0 \\ & \ddots & \\ 0 && i\theta_n \end{pmatrix} \in{\frak s}{\frak l}(n,\bc) \Bigm| |\theta_j-\theta_k|<{\pi\over 2} \mbox{ for all }j\mbox{ and } k \right\}.$$ Put $$y=\begin{pmatrix} e^{i\pi/n}I_{n-1} & 0 \\ 0 & -e^{i\pi/n} \end{pmatrix}= e^{i\pi/n}\begin{pmatrix} I_{n-1} & 0 \\ 0 & -1 \end{pmatrix}$$ and let $Y$ denote the subgroup of $G_\bc$ generated by $y$. Note that $y$ normalizes the sets $G_\br$, \ $K_\bc$ and $T^+$ and that $y^2$ generates the center $Z$ of $G_\bc$. (If $n$ is odd, then $Y=Z$.) Though the following fact was already proved in Example 1 of \cite{AG}, we will give a detailed proof here because an elementary calculus used in this proof is the key idea of this paper. \begin{proposition} \ $xS_0\subset S'_0\Longleftrightarrow x\in G_\br T^+YK_\bc$. \end{proposition} Proof. \ Note that the projective cone $xS_0\subset X_0$ is defined by $${}^tzmz=0$$ where $m={}^tx^{-1}x^{-1}$. So the condition $xS_0\subset S'_0$ implies that \begin{equation} {}^tzmz\ne 0\quad\mbox{for all }z\in \br^n-\{0\}. \tag{$3.1$} \end{equation} Assume (3.1). Since $F=\br^n-\{0\}$ is simply connected, we can define a smooth function $\xi:F\to \br$ by $$\xi(z)=\arg({}^tzmz)$$ taking some branch of $\arg$. Since $P^{n-1}(\br)=F/\br^\times$ is compact, the image $R$ of $\xi$ is a closed interval in $\br$. So we can take a $v_1\in F$ such that $\xi(v_1)$ is on the boundary of $R$ and that $|{}^tv_1mv_1|=1$. Let $u$ be an element of $\br^n$ and consider the function $$\eta(t)=\xi(v_1+tu)=\arg({}^tv_1mv_1+2t({}^tv_1mu)+t^2({}^tumu))$$ of $t\in\br$. Then we have $${}^tv_1mu\in \br\,{}^tv_1mv_1\quad\mbox{for all }u\in\br^n$$ since $\eta(0)=\xi(v_1)$ is on the boundary of $R$. Put $$U_1=\{u\in\br^n\mid {}^tv_1mu=0\}.$$ Then $U_1$ is an $n-1$-dimensional subspace of $\br^n$ which does not contain $v_1$. Next consider the restriction of $\xi$ on $U_1-\{0\}$ and take a $v_2\in U_1-\{0\}$ such that $\xi(v_2)$ is on the boundary of $\xi(U_1-\{0\})$ and that $|{}^tv_2mv_2|=1$. Repeating this procedure, we can take basis $v_1,\ldots,v_n$ of $\br^n$ such that $${}^tv_jmv_k=0 \quad\mbox{if }j\ne k$$ and that $|{}^tv_jmv_j|=1$ for $j=1,\ldots,n$. Hence if we put $g=(v_1\cdots v_n)\in G_\br$, then we have $${}^tgmg=\begin{pmatrix} \lambda_1 && 0 \\ & \ddots & \\ 0 && \lambda_n \end{pmatrix}$$ where $\lambda_j={}^tv_jmv_j\in U(1)$. Write $z=gc$ with $c\in\br^n$. Then we have $${}^tzmz={}^tc\,{}^tgmgc=\lambda_1c_1^2+\cdots+\lambda_nc_n^2.$$ Considering the convex real cone in $\bc$ generated by $\lambda_1,\ldots,\lambda_n$, we see that \begin{equation} \mbox{(3.1)}\Longleftrightarrow \{\lambda_1,\ldots,\lambda_n\}\subset\{z\in\bc \mid \Re\mu z>0\}\mbox{ for some }\mu\in\bc.\tag{$3.2$} \end{equation} Hence we can write $${}^tg\,{}^tx^{-1}x^{-1}g={}^tgmg=t^2={}^ttt$$ with some $t\in T^+Y$. Thus we have $$x\in gt^{-1}K_\bc \subset G_\br T^+YK_\bc.$$ Converse assertion is clear by the above arguments. \hfill q.e.d. \bigskip Now consider the full flag manifold $X$ of $G_\bc$ consisting of full flags \begin{equation} V_1\subset\cdots\subset V_{n-1} \tag{$3.3$} \end{equation} where $\dim_\bc V_j=j$. Then such a flag is contained in a closed $K_\bc$-orbit on $X$ if and only if $$Q(V_j,V_{n-j})=\{0\}\quad\mbox{for all }j.$$ Here $Q(\ ,\ )$ denote the canonical quadratic form on $\bc^n$ defined by $Q(z,w)=z_1w_1+\cdots+z_nw_n$. On the other hand, a flag (3.3) is contained in an open $G_\br$-orbit on $X$ if and only if $$V_j\cap\overline{V_{n-j}}=\{0\}\quad\mbox{for all }j.$$ Let $S$ be a closed $K_\bc$-orbit on $X$. Then every flag (3.3) contained in $xS$ satisfies \begin{equation} Q_x(V_j,V_{n-j})=\{0\}\quad\mbox{for all }j \tag{$3.4$} \end{equation} where $Q_x(z,w)=Q(x^{-1}z,x^{-1}w)={}^tzmw$. Assume (3.1). Then we can show that $xS$ is contained in an open $G_\br$-orbit as follows. Let (3.3) be an element of $xS$ and $z$ an element of $V_j\cap\overline{V_{n-j}}$. Then by (3.4), we have $$Q_x(z,\overline{z})=0.$$ Write $z=c_1v_1+\cdots+c_nv_n$ with the basis $v_1,\ldots,v_n$ of $\br^n$ given in the proof of Proposition 3.1. Then we have $$0=Q_x(z,\overline{z})={}^tzm\overline{z}=|c_1|^2\lambda_1+\cdots +|c_n|^2\lambda_n.$$ By (3.2), this implies $c_1=\cdots=c_n=0$ and $z=0$. Hence $xS$ is contained in an open $G_\br$-orbit on $X$. Let ${\frak j}$ be a maximal torus of ${\frak k}={\frak o}(n)$. Then the set of closed $K_\bc$-orbits on $X$ are parametrized by the left coset $$W_{K_\bc}({\frak j})\backslash W_{G_\bc}({\frak j})$$ where $W_*({\frak j})=N_*({\frak j})/Z_*({\frak j})$ (\cite{M0}, \cite{R}). First suppose $n$ is odd. Then we have $W_{K_\bc}({\frak j})=W_{G_\bc}({\frak j})$ \ (${\rm B}_{[n/2]}$-type) and hence $X$ has only one closed $K_\bc$-orbit. Since $Y=Z$ in this case, we have $G_\br T^+YK_\bc=\widetilde{D_0}Z$. Next suppose that $n$ is even. Then we have $|W_{K_\bc}({\frak j})\backslash W_{G_\bc}({\frak j})|=2$ since $W_{K_\bc}({\frak j})$ is ${\rm D}_{n/2}$-type and $W_{G_\bc}({\frak j})$ is ${\rm B}_{n/2}$-type. But since $W_{O(n)}({\frak j})=W_{G_\bc}({\frak j})$ and since $y$ can be replaced by an element $$\begin{pmatrix} I_{n-1} & 0 \\ 0 & -1 \end{pmatrix}$$ of $O(n)$, the two closed $K_\bc$-orbits on $X$ are written as $$S\mand yS.$$ Since $y^2\in Z$, we have $Y=Z\sqcup Zy$ and hence $$G_\br T^+YK_\bc=\widetilde{D_0}Z\sqcup \widetilde{D_0}Zy.$$ Thus we have $$xS\subset S'\mbox{ and }xyS\subset yS'\quad\mbox{if }x\in\widetilde{D_0}Z$$ and $$xS\subset yS'\mbox{ and }xyS\subset S'\quad\mbox{if }x\in\widetilde{D_0}Zy.$$ Conversely if $xS\subset S'$, then we have $xS_0\subset S'_0$ by a general argument as in Section 8. By the above arguments, we get: \begin{proposition} \ $x\in\widetilde{D_0}Z \Longleftrightarrow xS\subset S'\mbox{ for a closed $K_\bc$-orbit $S$ on }X.$ \end{proposition} \section{$SO(p,q)$-case} Let $Q$ denote the quadratic form on $\bc^n \ (n=p+q)$ defined by $$Q(z,w)=z_1w_1+\cdots+z_pw_p-z_{p+1}w_{p+1}-\cdots-z_nw_n$$ and put \begin{gather*} G_\bc=\{g\in SL(n,\bc)\mid Q(gz,gw)=Q(z,w)\mbox{ for all }z,w\in\bc^n\}\cong SO(n,\bc), \\ \widetilde{K_\bc}=\{g\in G_\bc\mid gV_+=V_+\}=\{g\in G_\bc\mid gV_-=V_-\}\\ \mbox{ and } \widetilde{G_\br}=G_\bc\cap SL(n,\br)=SO(p,q) %=\left\{\begin{pmatrix} A & 0 \\ 0 & B \end{pmatrix} \Bigm| A\in O(p,\bc), \ B\in O(q,\bc)\right\} \end{gather*} where $V_+=\bc e_1\oplus\cdots\oplus\bc e_p$ and $V_-=\bc e_{p+1}\oplus\cdots\oplus\bc e_n$ with the canonical basis $e_1,\ldots,e_n$ of $\bc^n$. Let $K_\bc$ and $G_\br$ denote the connected components of $\widetilde{K_\bc}$ and $\widetilde{G_\br}$, respectively, containing the identity. Let $F$ be the cone in $\bc^n$ defined by $$F=\{z\in\bc^n\mid Q(z,z)=0, \ z\ne 0\}.$$ Then $X_0=F/\bc^\times$ is a flag manifold of $G_\bc$. Suppose that $p\ge 3$ and $q\ge 1$. Then $$S_0=(V_+\cap F)/\bc^\times$$ is a closed $\widetilde{K_\bc}$-orbit on $X_0$. Since $S_0$ is connected, it is also a closed $K_\bc$-orbit on $X_0$. Put $$S'_0=\{z\in F\mid Q(z,\overline z)>0\}/\bc^\times.$$ Then $S'_0$ is an open $\widetilde{G_\br}$-orbit ($G_\br$-orbit) on $X_0$ containing $S_0$. Let $\ell=\min(p,q)$ and write $$Y(\theta_1,\ldots,\theta_\ell)=i\theta_1(E_{1,n}+E_{n,1})+\cdots +i\theta_\ell(E_{\ell,n-\ell+1}+E_{n-\ell+1,\ell})$$ where $E_{j,k}$ are the matrix units. Then $${\frak t}=\{Y(\theta_1,\ldots,\theta_\ell)\mid \theta_1,\ldots,\theta_\ell\in\br\}$$ is a maximal abelian subspace of $i{\frak p}$ and the set ${\frak t}^+$ is written as $${\frak t}^+=\{Y(\theta_1,\ldots,\theta_\ell)\mid |\theta_j|+|\theta_k|<{\pi\over 2} \mbox{ for all }j\ne k\}.$$ Put $$y=\begin{pmatrix} -1 && 0 \\ & I_{n-2} & \\ 0 && -1 \end{pmatrix}.$$ Then we have $$\widetilde{K_\bc}=K_\bc\sqcup K_\bc y \mand \widetilde{G_\br}=G_\br \sqcup G_\br y.$$ Since $yT^+y^{-1}=T^+$, we have $$\widetilde{G_\br}T^+\widetilde{K_\bc}=G_\br T^+K_\bc \sqcup G_\br T^+K_\bc y=\widetilde{D_0}\sqcup \widetilde{D_0}y.$$ \begin{proposition} \ Let $x$ be an element of $G_\bc$. Then $$x\in\widetilde{G_\br}T^+\widetilde{K_\bc} \Longleftrightarrow xS_0\subset S'_0.$$ \end{proposition} Proof. \ Suppose that $xS_0\subset S'_0$. This implies that $$Q(z,\overline{z})>0 \mbox{ for any } z\in xV_+\cap F.$$ We will find good basis $v_1,\ldots,v_p$ of $V=xV_+$ as follows. Let $f(z)$ be the restriction of the function $Q(z,\overline{z})$ on the quadratic hypersurface $H=\{z\in V\mid Q(z,z)=1\}$ in $V$. Then by the above assumption, $f(z)$ goes to $+\infty$ when $z$ tends to the boundary of $H$. So $f(z)$ has its minimum at some point $v_1$ in $H$. Hence by elementary calculus, we have $$v\in V, \ Q(v,v_1)=0 \Longrightarrow {\rm Re}\,Q(v,\overline{v_1})=0.$$ Since $Q(v,v_1)=0$ implies $Q(iv,v_1)=0$, we also have $$v\in V, \ Q(v,v_1)=0 \Longrightarrow {\rm Re}\,Q(iv,\overline{v_1})=0.$$ Hence we get $$v\in V, \ Q(v,v_1)=0 \Longrightarrow Q(v,\overline{v_1})=0.$$ Next consider the orthogonal compliment $U_1$ of $v_1$ in $V$ and take a $v_2\in U_1\cap H$ by the same procedure. Repeating this procedure, we get basis $v_1,\ldots,v_p$ of $V$ such that \begin{equation} Q(v_j,v_k)=\delta_{jk}\quad\mbox{and that}\quad Q(v_j,\overline{v_k})=0\mbox{ if } j\ne k.\tag{$4.1$} \end{equation} Write $v_j=u_j+iu'_j$ with $u_j, u'_j\in\br^n$. Then the $2p$-vectors $u_1,\ldots,u_p,u'_1,\ldots,u'_p$ in $\br^n$ are mutually orthogonal by (4.1) and $$Q(u_j,u_j)-Q(u'_j,u'_j)=Q(v_j,v_j)=1,$$ $$Q(u_j,u_j)+Q(u'_j,u'_j)=Q(v_j,\overline{v_j}).$$ If $Q(v_2,\overline{v_2})>1$, then since $Q(v_2,\overline{v_2})\le\cdots\le Q(v_p,\overline{v_p})$, we have $$Q(u_j,u_j)>0\mand Q(u'_j,u'_j)>0\quad\mbox{for }j=2,\ldots,p,$$ a contradiction to the signature of $Q$. Hence we have $$Q(v_2,\overline{v_2})\le 1.$$ Since $v_1+iv_2\in F$, we have \begin{equation} 0-1$$ and hence $$Q(u_1,u_1)>0.$$ Since \begin{equation} Q(v_1,\overline{v_1})\le\cdots\le Q(v_p,\overline{v_p}), \tag{$4.3$} \end{equation} we have $$0\theta_1\ge\cdots\ge\theta_p\ge 0 \mand \theta_1+\theta_2<{\pi\over 2} \tag{$4.4$} \end{gather} by (4.2). If $q & 0 \end{eqnarray*} by (4.2). \hfill q.e.d. \bigskip Now let us consider the full flag manifold. The full flag manifold $X=G_\bc/B$ consists of full flags $$V_1\subset\cdots\subset V_{[n/2]}$$ where $V_j$ are $j$-dimensional complex linear subspaces $V_j$ contained in $F\cup\{0\}$. Such a flag is contained in a closed $\widetilde{K_\bc}$-orbit if and only if $$V_j=(V_j\cap V_+)\oplus(V_j\cap V_-)\quad\mbox{for all }j=1,\ldots,n'$$ where $n'=[p/2]+[q/2]$. It is clear that $$p_j=\dim(V_j\cap V_+)\mand q_j=\dim(V_j\cap V_-)\quad(j=1,\ldots,n')$$ are the invariants of closed $\widetilde{K_\bc}$-orbits. Hence there are ${n'\choose [p/2]}$ closed $\widetilde{K_\bc}$-orbits on $X$. On the other hand, a flag is contained in an open $\widetilde{G_\br}$-orbit if and only if the hermitian form $Q(z,\overline z)$ is nondegenerate on each $V_j \ (j=1,\ldots,n')$. So the signatures $(p_j,q_j)$ of $Q(z,\overline z)$ on $V_j \ (j=1,\ldots,n')$ are the invariants of open $\widetilde{G_\br}$-orbits on $X$. We can see that closed $\widetilde{K_\bc}$-orbits and open $\widetilde{G_\br}$-orbits on $X$ are in one-to-one correspondence by these numbers $(p_j,q_j) \ (j=1,\ldots,n')$. Suppose that $x\in\widetilde{G_\br}T^+\widetilde{K_\bc}$. Then we want to show that $xS\subset S'$ for all closed $\widetilde{K_\bc}$-orbits $S$ on $X$. By Proposition 4.1, we have $$Q(z,\overline{z})>0 \mbox{ for any } z\in xV_+\cap F.$$ We also have $$Q(z,\overline{z})<0 \mbox{ for any } z\in xV_-\cap F.$$ by the same arguments as in the proof of Proposition 4.1. Let $$V_1\subset\cdots\subset V_{[n/2]}$$ be a flag contained in $S$. Then since $x(V_j\cap V_+)=xV_j\cap xV_+\subset xV_+\cap F$, we have $$Q(z,\overline{z})>0\quad\mbox{for all }z\in x(V_j\cap V_+).$$ Similarly, we also have $$Q(z,\overline{z})<0\quad\mbox{for all }z\in x(V_j\cap V_-).$$ Since $xV_j=x(V_j\cap V_+)\oplus x(V_j\cap V_-)$ for $j=1,\ldots,n'$, the signature of $Q(z,\overline{z})$ on $xV_j$ is $(\dim(V_j\cap V_+),\dim(V_j\cap V_-))$ for $j=1,\ldots,n'$. Thus we have proved $$xS\subset S'.$$ Now let us consider $K_\bc$-orbits on $X$. Let ${\frak j}$ be a maximal torus of ${\frak k}$. Then the set of closed $K_\bc$-orbits and $\widetilde{K_\bc}$-orbits on $X$ are parametrized by the left cosets $$W_{K_\bc}({\frak j})\backslash W_{G_\bc}({\frak j}) \mand W_{\widetilde{K_\bc}}({\frak j})\backslash W_{G_\bc}({\frak j}),$$ respectively, where $W_*({\frak j})=N_*({\frak j})/Z_*({\frak j})$. First suppose that either $p$ or $q$ is even. Then we can see that $W_{\widetilde{K_\bc}}({\frak j})$ is bigger than $W_{K_\bc}({\frak j})$. Hence in this case, every closed $\widetilde{K_\bc}$-orbit $S$ on $X$ is decomosed into two closed $K_\bc$-orbits as $$S=S_1\sqcup yS_1.$$ Since $xS\subset S'$, we see that $$xS_1\subset S'_1\mand xyS_1\subset yS'_1\quad\mbox{if }x\in\widetilde{D_0},$$ and that $$xS_1\subset yS'_1\mand xyS_1\subset S'_1\quad\mbox{if }x\in\widetilde{D_0}y.$$ On the other hand, if $p$ and $q$ are both odd, then we can replace $y$ by the central element $-I_n$. So the $K_\bc$-orbits and $\widetilde{K_\bc}$-orbits on $X$ coincide. Conversely, suppose that $xS_1\subset S'_1$ for some closed $K_\bc$-orbit $S_1$ on $X$. Let $$V_1\subset\cdots\subset V_{[n/2]}$$ be a flag contained in $S_1$. Then every flag contained in $S_1$ is of the form $$kV_1\subset\cdots\subset kV_{[n/2]}$$ with some $k\in K_\bc$ and $$V_1\subset V_+\quad\mbox{or}\quad V_1\subset V_-.$$ If $V_1\subset V_+$, then by the property of $S'_1$, we have $$Q(z,\overline{z})\mbox{ is positive definite on }xkV_1\mbox{ for every }k\in K_\bc$$ and therefore $$Q(z,\overline{z})>0\quad\mbox{for all }z\in xV_+\cap F$$ since $V_+\cap F$ is $K_\bc$-homogeneous ($p\ge 3$). Hence by Proposition 4.1, we have $$x\in\widetilde{G_\br}T^+\widetilde{K_\bc}$$ and $$x\in\widetilde{D_0}Z$$ by the preceding arguments. When $V_1\subset V_-$ and $q\ge 3$, we also get $x\in\widetilde{D_0}Z$ by the same arguments. Summarizing the above arguments, we have: \begin{proposition} \ {\rm (1)} Suppose $p\ne 2$ and $q\ne 2$. Then the following three conditions on $x\in G_\bc$ are equivalent: {\rm (a)} \ $x\in \widetilde{D_0}Z$. {\rm (b)} \ $xS\subset S'$ for a closed $K_\bc$-orbit $S$ on $X$. {\rm (c)} \ $xS\subset S'$ for all closed $K_\bc$-orbits $S$ on $X$. {\rm (2)} Suppose $p\ge 3$ and $q=2$. Then the following three conditions on $x\in G_\bc$ are equivalent: {\rm (a)} \ $x\in \widetilde{D_0}Z$. {\rm (b)} \ $xS\subset S'$ for a closed $K_\bc$-orbit $S$ on $X$ consisting of flags $V_1\subset\cdots\subset V_{[n/2]}$ such that $V_1\subset V_+$. {\rm (c)} \ $xS\subset S'$ for all closed $K_\bc$-orbits $S$ on $X$. \end{proposition} \begin{remark}{\rm \ When $p\ge 3$ and $q=2$, $G_\br$ is a Lie group of Hermitian type. Open $G_\br$-orbits $S'$ on $X$ in the above condition (b) are called ``nonholomorphic type'' in \cite{WZ}. On the other hand, the two open $G_\br$-orbits $S'_1$ and $S'_2$ on $X$ which we considered in Section 2 are called ``holomorphic type''. %In \cite{WZ}, they considered correspondence between closed $K_\bc$-orbits and open $G_\br$-orbits on an arbitrary flag manifold of $G_\bc$ when $G_\br$ is Hermitian type. }\end{remark} \section{Cases of complex classical Lie groups} Let $G$ be a connected complex simple Lie group and $G_\bc=G\times G$. Take a compact real form $K$ of $G$ and let $\tau$ be the conjugation of $G$ with respect to $K$. Then $$G_\br=\{(g,\tau(g))\mid g\in G\}\cong G$$ is a real form of $G_\bc$. We can take $$K_\bc=\{(g,g)\mid g\in G\}\cong G.$$ Let $X=G/B$ be the full flag manifold of $G$. Then by the correspondence $(g_1,g_2)\mapsto g_1^{-1}g_2$, we have a natural bijection $$K_\bc\backslash G_\bc/B\times B\cong B\backslash G/B.$$ Since the right hand side has only one closed double coset $B=BeB$ by the Bruhat decomposition, the full flag manifold $X\times X$ of $G_\bc$ has only one closed $K_\bc$-orbit $$S=\{(m,m)\mid m\in X\}.$$ When $G$ is classical, we can take a minimal flag manifold $X_0=F/\bc^\times$ of $G$ where $F$ is given as follows. \bigskip \centerline{ \vbox{\offinterlineskip \hrule \halign{&\vrule#&\strut\quad$\hfil#\hfil$\quad\cr height2pt&\omit&&\omit&\cr & G && F &\cr height2pt&\omit&&\omit&\cr \noalign{\hrule} height4pt&\omit&&\omit&\cr & SL(n,\bc) && \bc^n-\{0\} &\cr height4pt&\omit&&\omit&\cr & SO(n,\bc) && \{z\in \bc^n-\{0\} \mid z_1^2+\cdots+z_n^2=0\} &\cr height4pt&\omit&&\omit&\cr & Sp(n,\bc) && \bc^{2n}-\{0\} &\cr height4pt&\omit&&\omit&\cr} \hrule} } \bigskip \noindent If we write $X_0=G/P$, then we have a natural bijection $$K_\bc\backslash G_\bc/P\times P\cong P\backslash G/P\cong W_P\backslash W/W_P.$$ When $W$ is ${\rm *}_n$-type (${\rm *}=$ A, B, C or D), we can see that $W_P$ is ${\rm *}_{n-1}$-type. Hence we have $$|W_P\backslash W/W_P|=\begin{cases} 2 & \text{if $G=SL(n,\bc)$,} \\ 3 & \text{if $G=SO(n,\bc) \ (n\ge 5)$ or $G=Sp(n,\bc) \ (n\ge 2)$,} \\ 4 & \text{if $G=SO(4,\bc)$.} \end{cases}$$ The flag manifold $X_0\times X_0$ of $G_\bc$ has only one closed $K_\bc$-orbit $$S_0=\{(m,m)\mid m\in X_0\}=p\{(z,z)\mid z\in F\}$$ where $p:F\times F\to X_0\times X_0$ is the canonical projection. When $G=SO(n,\bc)$ with $n\ge 5$, the $K_\bc$-orbit decomposition of $X_0\times X_0$ is $$X_0\times X_0=S_0\sqcup \{p(z,w)\mid \bc z\ne \bc w, \ Q(z,w)=0\} \sqcup \{p(z,w)\mid Q(z,w)\ne 0\}$$ where $Q(z,w)=z_1w_1+\cdots+z_nw_n$. When $G=Sp(n,\bc)$ with $n\ge 2$, the $K_\bc$-orbit decomposition of $X_0\times X_0$ is $$X_0\times X_0=S_0\sqcup \{p(z,w)\mid \bc z\ne \bc w, \ {}^tzJw=0\} \sqcup \{p(z,w)\mid {}^tzJw\ne 0\}$$ where $J$ is the canonical alternating matrix defining $G$. We will prove the following proposition in this section. \begin{proposition} \ Let $G=SL(n,\bc),\ SO(n,\bc)$ or $Sp(n,\bc)$. Then the following three conditions on $x\in G_\bc$ are equivalent: {\rm (i)} \ $x\in\widetilde{D_0}Z$,\qquad {\rm (ii)} \ $xS_0\subset S'_0$, \qquad {\rm (iii)} \ $xS\subset S'$. \noindent {\rm (Remark. (iii) implies (ii) by a general argument as in Section 8. Since $SO(3,\bc)$ and $SO(4,\bc)$ are locally isomorphic to $SL(2,\bc)$ and $SL(2,\bc)\times SL(2,\bc)$, respectively, we may assume $n\ge 5$ for $SO(n,\bc)$-case.)} \end{proposition} Proof. \ Since $G=SL(n,\bc),\ SO(n,\bc)$ or $Sp(n,\bc)$, we can take $$\tau(g)={}^t\overline{g}^{-1}\quad\mbox{for }g\in G$$ as usual. Then the open $G_\br$-orbit on $X_0\times X_0$ is $$S'_0=p\{(w,z)\in F\times F \mid {}^t\overline{z}w\ne 0\}.$$ The space $G_\bc/K_\bc$ is identified with $G$ by the map $$\varphi: (g,h)K_\bc\mapsto gh^{-1}.$$ By this identification, the left $G_\br$-action on $G_\bc/K_\bc$ corresponds to the $G_\br$-action on $G$ given by \begin{equation} ((h, {}^t\overline{h}^{-1}), g)\mapsto hg\,{}^t\overline{h}. \tag{$5.1$} \end{equation} Write $x=(g_1,g_2)$ with $g_1, g_2\in G$. Then we have $$xS_0=(g_1,g_2)S_0=(g,e)S_0=p\{(gz,z)\mid z\in F\}$$ where $g=\varphi(x)=g_1g_2^{-1}$. So the condition $xS_0\subset S'_0$ implies that \begin{equation} {}^t\overline{z}gz\ne 0\quad\mbox{for all }z\in F. \tag{$5.2$} \end{equation} \bigskip \noindent {\bf A. $SL(n,\bc)$-case} \bigskip We can take $${\frak t}^+=\left\{\left(\begin{pmatrix} i\theta_1 && 0 \\ & \ddots & \\ 0 && i\theta_n \end{pmatrix}, -\begin{pmatrix} i\theta_1 && 0 \\ & \ddots & \\ 0 && i\theta_n \end{pmatrix} \right) \Bigm| |\theta_j-\theta_k|<{\pi\over 2}\mbox{ for all }j\mbox{ and }k\right\}.$$ (ii) $\Longrightarrow$ (i). \ Assume (5.2). Then we can diagonalize $g$ by the $G_\br$-action (5.1) by an argument similar to Proposition 3.1 as follows. Since $F=\bc^n-\{0\}$ is simply connected, we can define a smooth function $\xi:F\to \br$ by $$\xi(z)=\arg({}^t\overline{z}gz)$$ for $z\in F$. Then the image $R$ of $\xi$ is a bounded closed interval in $\br$ since $X_0=P^{n-1}(\bc)=F/\bc^\times$ is compact. Take $v_1$ so that $\xi(v_1)$ is on the boundary of $R$ and that $|{}^t\overline{v_1}gv_1|=1$. Let $u$ be an element of $\bc^n$ and consider the function $$\eta(t)=\xi(v_1+tu)=\arg({}^t\overline{v_1}gv_1+t({}^t\overline{v_1}gu+{}^t\overline{u}gv_1)+t^2({}^t\overline{u}gu))$$ of $t\in\br$. Then we have \begin{equation} {}^t\overline{v_1}gu+{}^t\overline{u}gv_1\in \br\,{}^t\overline{v_1}gv_1 \quad\mbox{for all }u\in\bc^n \tag{5.3} \end{equation} since $\eta(0)=\xi(v_1)$ is on the boundary of $R$. Put $$U_1=\{u\in\bc^n\mid {}^t\overline{u}gv_1=0\}.$$ Then $U_1$ is an $n-1$-dimensional complex subspace of $\bc^n$ which does not contain $v_1$. By (5.3), we have $${}^t\overline{v_1}gU_1\subset \br\,{}^t\overline{v_1}gv_1.$$ Hence we have $${}^t\overline{v_1}gU_1=\{0\}$$ because the left hand side is a complex subspace of $\bc$. Next consider the restriction of $\xi$ on $U_1-\{0\}$ and take a $v_2\in U_1-\{0\}$ such that $\xi(v_2)$ is on the boundary of $\xi(U_1-\{0\})$ and that $|{}^t\overline{v_2}gv_2|=1$. Repeating this procedure, we can take basis $v_1,\ldots,v_n$ of $\bc^n$ such that $${}^t\overline{v_j}gv_k=0 \quad\mbox{if }j\ne k$$ and that $|{}^t\overline{v_j}gv_j|=1$ for $j=1,\ldots,n$. Hence if we put $h=(v_1\cdots v_n)\in G$, then we have $${}^t\overline{h}gh=\begin{pmatrix} \lambda_1 && 0 \\ & \ddots & \\ 0 && \lambda_n \end{pmatrix}$$ where $\lambda_j={}^t\overline{v_j}gv_j\in U(1)$. By (5.2), $\lambda_1,\ldots,\lambda_n$ satisfy the condition in (3.2). Hence we have $${}^t\overline{h}gh\in \varphi(t)Z'=\varphi(tZ)$$ with some $t\in T^+$. Here $Z'=\varphi(Z)$ is the center of $G$. This implies that $$x=(g_1,g_2)\in ({}^t\overline{h}^{-1},h)tZK_\bc\subset G_\br T^+K_\bc Z=\widetilde{D_0}Z.$$ (i) $\Longrightarrow$ (ii) is clear by the above arguments. (ii) $\Longrightarrow$ (iii). \ Elements of the full flag manifold $X\times X$ of $G_\bc$ are pairs of sequences \begin{equation} (V_1\subset\cdots\subset V_{n-1},\ V'_1\subset\cdots\subset V'_{n-1}) \tag{5.4} \end{equation} where $V_j$ and $V'_j$ are $j$-dimensional complex subspaces of $\bc^n$. The open $G_\br$-orbit $S'$ on $X\times X$ consists of flags (5.4) such that the canonical hermitian form is nondegenerate on the pair $(V_j,V'_j)$ for every $j=1,\ldots,n-1$. \ (i.e. For every $v\in V_j$, there exists a $u\in V'_j$ such that ${}^t\overline{u}v\ne 0$.) For $x=(g_1,g_2)\in G_\bc$, we can see that the set $xS$ consists of flags of the form $$(gV_1\subset\cdots\subset gV_{n-1},\ V_1\subset\cdots\subset V_{n-1})$$ ($g=\varphi(x)=g_1g_2^{-1}$). Suppose that $xS_0\subset S'_0$. Then by (5.2), ${}^t\overline{z}gz\ne 0$ for every $z\in\bc^n-\{0\}$. So the canonical hermitian form is clearly nondegenerate on the pair $(gV_j,V_j)$ for $j=1,\ldots,n-1$. Hence we have $xS\subset S'$. \bigskip \noindent {\bf B. $SO(n,\bc)$-case ($n\ge 5$)} \bigskip We can take $${\frak t}^+=\{(Y(\theta_1,\ldots,\theta_m), -Y(\theta_1,\ldots,\theta_m))\mid |\theta_j|+|\theta_k|<{\pi\over 2} \mbox{ if }j\ne k\}.$$ Here $$Y(\theta_1,\ldots,\theta_m)=\begin{cases} \begin{pmatrix} \theta_1J_1 && 0 \\ & \ddots & \\ 0 && \theta_mJ_1 \end{pmatrix} & \text{if $n=2m$,} \\ \noalign{\medskip} \begin{pmatrix} \theta_1J_1 &&& 0 \\ & \ddots && \\ && \theta_mJ_1 & \\ 0 &&& 1 \end{pmatrix} & \text{if $n=2m+1$} \end{cases}$$ with $$J_1=\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}.$$ Let $Q$ denote the canonical quadratic form on $\bc^n$ defined by $$Q(z,w)=z_1w_1+\cdots+z_nw_n.$$ Then (5.2) is written as \begin{equation} Q(gz,\overline z)\ne 0\quad\mbox{for all }z\in F. \tag{$5.2'$} \end{equation} (ii) $\Longrightarrow$ (i). \ Assume (5.2'). Since $F=\{z\in\bc^n-\{0\}\mid Q(z,z)=0\}$ is simply connected, we can define a smooth function $\xi_g:F\to\br$ by $$\xi_g(z)=\arg Q(gz,\overline z)\quad\mbox{for }z\in F.$$ Since $X_0=F/\bc^\times$ is compact, the image $R$ of $\xi_g$ is a bounded closed interval in $\br$. If $g'=\overline{h}^{-1}gh$ with an $h\in G$, then we can put \begin{equation} \xi_{g'}(z)=\xi_g(hz)\tag{5.5} \end{equation} because $Q(g'z,\overline{z})=Q(\overline{h}^{-1}ghz,\overline{z})= Q(ghz,\overline{hz})$. Let $v$ be an element of $F$ such that $\xi_g(v)=\theta_1$ is on the boundary of $R$. Replacing $g$ by a twisted conjugate $\overline{h_0}^{-1}gh_0$ with an element $h_0\in G$ such that $v=h_0(e_1+ie_2)$, we may assume $$v=e_1+ie_2,$$ by (5.5), where $\{e_1,\ldots,e_n\}$ is the canonical basis of $\br^n$. Since $Q(v,\overline{v})=2\ne 0$ and since $Q(gv,\overline{v})\ne 0$, two points $(\bc v,\bc\overline{v})$ and $(\bc gv,\bc\overline{v})$ are in the open $K_\bc$-orbit on $X_0\times X_0$. So we can take an $h\in G$ such that $$\overline{h}\bc\overline{v}=\bc\overline{v} \quad\mbox{and that}\quad \overline{h}\bc v=\bc gv.$$ Hence $$\overline h^{-1}ghv\in \overline h^{-1}g\bc v=\bc v$$ and so we can write $$\overline h^{-1}ghv=\mu v$$ with some $\mu\in\bc^\times$. Take an element $h'$ of $G$ such that $$h'v=|\mu|^{-1/2}v \quad\mbox{and that}\quad h'\overline{v}=|\mu|^{1/2}\overline{v}.$$ Then we have $$\overline{h'}^{-1}\overline h^{-1}ghh'v={\mu\over|\mu|}v.$$ Rewrite $\overline{h'}^{-1}\overline h^{-1}ghh'$ as $g$. Then we have $$\xi_g(v)=\theta_1$$ by (5.5) since $hh'v\in\bc v$. Hence we have \begin{equation} gv={\mu\over|\mu|}v=e^{i\theta_1}v.\tag{5.6} \end{equation} Let $u$ be an element of $U_1=\bc e_3\oplus\cdots\oplus \bc e_n$ and consider the function $$\eta(t)=\xi_g(v+tu)=\arg(Q(gv,\overline{v})+tQ(gv,\overline{u})+ tQ(gu,\overline{v})+t^2Q(gu,\overline{u}))$$ of $t\in\br$. Then we have $${d\eta\over dt}(0)=0$$ since $\eta(0)=\xi_g(v)$ is on the boundary of $R$ and since the line $v+tu$ is tangent to $F$ at $t=0$. Since $gv=e^{i\theta_1}v$, we have $$Q(gv,\overline{u})=0.$$ Hence $$Q(gu,\overline{v})\in \br Q(gv,\overline{v})$$ for all $u\in U_1$. Since $U_1$ is a complex subspace of $\bc^n$, this implies $$Q(gu,\overline{v})=0$$ and therefore $$g^{-1}\overline{v}\in \bc v\oplus \bc\overline{v}.$$ Since $Q(v,g^{-1}\overline v)=Q(gv,\overline v)\ne 0$ and since $g^{-1}\overline v\in F$, we have $$g^{-1}\overline v\in \bc\overline v.$$ Hence $$g\overline v=\beta\overline v$$ with some $\beta\in\bc^\times$. Since $2e^{i\theta_1}\beta=Q(gv,g\overline v)=Q(v,\overline v)=2$, we have $\beta=e^{-i\theta_1}$. Thus we get \begin{equation} g\overline v=e^{-i\theta_1}\overline v. \tag{5.7} \end{equation} By (5.6) and (5.7), $g$ is of the form $$\begin{pmatrix} B_{\theta_1} & 0 \\ 0 & g' \end{pmatrix}$$ with $g'\in SO(n-2,\bc)$ where $$B_\theta=\begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}.$$ So by induction, we can show that a twisted conjugate $g_0$ of $g$ is of the form $$g_0=a(\theta_1,\ldots,\theta_m)=\begin{cases} \begin{pmatrix} B_{\theta_1} && 0 \cr & \ddots & \cr 0 && B_{\theta_m} \end{pmatrix} & \text{if $n=2m$,} \\ \noalign{\medskip} \begin{pmatrix} B_{\theta_1} &&& 0 \cr & \ddots && \cr && B_{\theta_m} & \cr 0 &&& 1 \end{pmatrix} & \text{if $n=2m+1$}. \end{cases}$$ By the action of $K=SO(n)$, we may assume $$\pi\ge|\theta_1|\ge\cdots\ge|\theta_m|\ge 0.$$ First consider the case where $n$ is odd. Suppose $|\theta_1|+|\theta_2|\ge\pi$. Then we have $$Q(g_0(e_1+i\alpha e_3+i\beta e_n),e_1-i\alpha e_3-i\beta e_n)= \cos\theta_1+\alpha^2\cos\theta_2+\beta^2=0$$ for some $(\alpha,\beta)\in[0,1]\times [0,1]$ satisfying $\alpha^2+\beta^2=1$ because $$\cos\theta_1+\cos\theta_2\le 0 \mand \cos\theta_1+1\ge 0.$$ Since $e_1+i\alpha e_3+i\beta e_n\in F$, this contradicts to (5.2'). Hence $$|\theta_1|+|\theta_2|<\pi$$ and so we have $$g_0\in\varphi(T^+),$$ which implies (i). Next consider the case where $n$ is even ($n=2m\ge 6$). Suppose $|\theta_1|+|\theta_m|\le \pi$. Then we can show $|\theta_1|+|\theta_2|<\pi$ as follows. If $|\theta_1|+|\theta_2|\ge \pi$, then we have $$Q(g_0(e_1+i\alpha e_3+i\beta e_{2m-1}),e_1-i\alpha e_3-i\beta e_{2m-1})= \cos\theta_1+\alpha^2\cos\theta_2+\beta^2\cos\theta_m=0$$ for some $(\alpha,\beta)\in[0,1]\times [0,1]$ satisfying $\alpha^2+\beta^2=1$ because $$\cos\theta_1+\cos\theta_2\le 0 \mand \cos\theta_1+\cos\theta_m\ge 0.$$ Since $e_1+i\alpha e_3+i\beta e_{2m-1}\in F$, this contradicts to (5.2'). Hence $$|\theta_1|+|\theta_2|<\pi$$ and so we have $g_0\in\varphi(T^+)$. On the other hand, suppose $|\theta_1|+|\theta_m|\ge \pi$. Then we can show $|\theta_{m-1}|+|\theta_m|>\pi$ by the same argument as above. Hence we have $$g_0=-a(\theta_1+\pi,\ldots,\theta_m+\pi)\in -\varphi(T^+)\subset \varphi(T^+Z).$$ \bigskip (i) $\Longrightarrow$ (ii). \ Let $g=\varepsilon a(\theta_1,\ldots,\theta_m)$ with $$\varepsilon=\pm 1,\quad |\theta_j|+|\theta_k|<\pi\mbox{ if }j\ne k.$$ Then we have only to show (5.2'). Clearly we may assume $\varepsilon=1$. By the action of $K=SO(n)$, we may assume $$|\theta_1|\ge\cdots\ge|\theta_m|\ge 0.$$ Let $z=z_1e_1+\cdots+z_ne_n$ be an element of $F$. Then we have \begin{eqnarray*} Q(gz,\overline z) & = & \sum_{j=1}^m((z_{2j-1}\cos\theta_j+z_{2j}\sin\theta_j)\overline{z_{2j-1}} + (-z_{2j-1}\sin\theta_j+z_{2j}\cos\theta_j)\overline{z_{2j}}) + \delta z_n\overline{z_n} \\ & = & \sum_{j=1}^m(|z_{2j-1}|^2+|z_{2j}|^2)\cos\theta_j + \delta |z_n|^2 + 2i\sum_{j=1}^m{\rm Im}(\overline{z_{2j-1}}z_{2j})\sin\theta_j \end{eqnarray*} where $\delta=n-2m$. If $|\theta_1|<\pi/2$, then ${\rm Re}\,Q(gz,\overline z)>0$. So we may assume that $|\theta_1|\ge \pi/2$. Suppose ${\rm Im}\,Q(gz,\overline z)=0$. Since $|\sin\theta_1|>|\sin\theta_2|\ge\cdots\ge|\sin\theta_m|\ge 0$, we have $$|{\rm Im}(\overline{z_1}z_2)|\le |{\rm Im}(\overline{z_3}z_4)| +\cdots+ |{\rm Im}(\overline{z_{2m-1}}z_{2m})|.$$ On the other hand, since $z\in F$, we have $$|z_1^2+z_2^2|=|z_3^2+\cdots+z_n^2|\le |z_3^2+z_4^2|+\cdots +|z_{2m-1}^2+z_{2m}^2|+ \delta |z_n^2|.$$ Since $|z_{2j-1}^2+z_{2j}^2|^2+4|{\rm Im}(\overline{z_{2j-1}}z_{2j})|^2 =(|z_{2j-1}|^2+|z_{2j}|^2)^2$, we get $$|z_1|^2+|z_2|^2\le |z_3|^2+\cdots+|z_n|^2.$$ Hence \begin{eqnarray*} {\rm Re}\,Q(gz,\overline z) & \ge & (|z_1|^2+|z_2|^2)\cos\theta_1+(|z_3|^2+\cdots+|z_n|^2)\cos\theta_2 \\ & = & (|z_1|^2+|z_2|^2)(\cos\theta_1+\cos\theta_2) + (|z_3|^2+\cdots+|z_n|^2-|z_1|^2-|z_2|^2)\cos\theta_2 \\ & > & 0 \end{eqnarray*} because $\cos\theta_2\ge \cos\theta_1+\cos\theta_2 >0$ and $z\ne 0$. Thus the assertion is proved. (ii) $\Longrightarrow$ (iii). \ Elements of the full flag manifold $X\times X$ of $G_\bc$ are pairs of sequences \begin{equation} (V_1\subset\cdots\subset V_{[n/2]},\ V'_1\subset\cdots\subset V'_{[n/2]}) \tag{5.8} \end{equation} where $V_j$ and $V'_j$ are $j$-dimensional complex subspaces contained in $F\cup\{0\}$. The open $G_\br$-orbit $S'$ on $X\times X$ consists of flags (5.8) such that the canonical hermitian form is nondegenerate on the pair $(V_j,V'_j)$ for every $j=1,\ldots,[n/2]$. So we can prove (ii) $\Longrightarrow$ (iii) by the same argument as in the case of $SL(n,\bc)$. \bigskip \noindent {\bf C. $Sp(n,\bc)$-case} \bigskip Define $G=Sp(n,\bc)$ by $$G=\{g\in GL(2n,\bc)\mid {}^tgJg=J\}$$ where $$J=\begin{pmatrix} J_1 && 0 \\ & \ddots & \\ 0 && J_1 \end{pmatrix} \mand J_1=\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}.$$ Then we can take $${\frak t}^+=\{(Y(\theta_1,\ldots,\theta_n), -Y(\theta_1,\ldots,\theta_n))\mid |\theta_j|<{\pi\over 4}\mbox{ for }j=1,\ldots,n\}$$ where $$Y(\theta_1,\ldots,\theta_n)=\begin{pmatrix} \theta_1 &&&& 0 \\ & -\theta_1 &&& \\ && \ddots && \\ &&& \theta_n & \\ 0 &&&& -\theta_n \end{pmatrix}.$$ (ii) $\Longrightarrow$ (i). \ Assume (5.2). Since $F=\bc^{2n}-\{0\}$ is simply connected, we can define a smooth function $\xi_g:F\to\br$ by $$\xi_g(z)=\arg {}^t\overline{z}gz\quad\mbox{for }z\in F.$$ Since $X_0=F/\bc^\times$ is compact, the image $R$ of $\xi_g$ is a bounded closed interval in $\br$. If $g'={}^t\overline{h}gh$ with an $h\in G$, then we can put \begin{equation} \xi_{g'}(z)=\xi_g(hz). \tag{5.9} \end{equation} Let $v$ be an element of $F$ such that $\xi_g(v)=\theta_1$ is on the boundary of $R$. Replacing $g$ by a conjugate ${}^t\overline{h_0}gh_0$ with an element $h_0\in G$ such that $v=h_0 e_1$, we may assume $$v=e_1$$ by (5.9). Since ${}^te_2Je_1=-1\ne 0$ and since $${}^te_2Jge_1=-\,{}^te_1ge_1=-\,{}^t\overline{e_1}ge_1\ne 0,$$ two points $(\bc e_1,\bc e_2)$ and $(\bc ge_1,\bc e_2)$ are in the open $K_\bc$-orbit on $X_0\times X_0$. So we can take an $h\in G$ such that $${}^t\overline{h}\bc e_2=\bc e_2 \quad\mbox{and that}\quad {}^t\overline{h}\bc ge_1=\bc e_1.$$ Hence we have $$he_1=J^{-1}{}^th^{-1}Je_1=-J^{-1}{}^th^{-1}e_2\in J^{-1}\bc e_2=\bc e_1$$ and therefore $${}^t\overline{h}ghe_1\in {}^t\overline{h}g\bc e_1=\bc e_1.$$ Write ${}^t\overline{h}ghe_1=\mu e_1$ with $\mu\in\bc^\times$ and take an element $$h'=\begin{pmatrix} |\mu|^{-1/2} && 0 \\ & |\mu|^{1/2} & \\ 0 && I_{2n-2} \end{pmatrix}$$ of $G$. Then we have $${}^t\overline{h'}\,{}^t\overline{h}ghh'e_1={\mu\over |\mu|}e_1.$$ Rewrite ${}^t\overline{h'}\,{}^t\overline{h}ghh'$ as $g$. Then we have $$\xi_g(e_1)=\theta_1$$ by (5.9) since $hh'e_1\in\bc e_1$. Hence we have \begin{equation} ge_1={\mu\over|\mu|}e_1=e^{i\theta_1}e_1.\tag{5.10} \end{equation} Let $u$ be an element of $U_1=\bc e_2\oplus\cdots\oplus \bc e_{2n}$ and consider the function $$\eta(t)=\xi_g(e_1+tu)=\arg ({}^t\overline{e_1}ge_1+ t\,{}^t\overline{u}ge_1+t\,{}^t\overline{e_1}gu+ t^2\,{}^t\overline{u}gu)$$ of $t\in\br$. Then we have $${d\eta\over dt}(0)=0$$ since $\eta(0)=\xi_g(e_1)$ is on the boundary of $R$. Since $ge_1=e^{i\theta_1}e_1$, we have $${}^t\overline{u}ge_1=0.$$ Hence $${}^t\overline{e_1}gu\in \br\,{}^t\overline{e_1}ge_1$$ for all $u\in U_1$. Since $U_1$ is a complex subspace of $\bc^n$, this implies $${}^t\overline{e_1}gu=0$$ and therefore $${}^tge_1\in\bc e_1.$$ Hence we have $$g^{-1}e_2 =J^{-1}{}^tgJe_2 =J^{-1}{}^tge_1 \in J^{-1}\bc e_1 =\bc e_2.$$ So we can write $ge_2=\beta e_2$ and we can compute $\beta$ as $$\beta={}^te_1Jge_2={}^te_1\,{}^tg^{-1}Je_2={}^t(g^{-1}e_1)Je_2= e^{-i\theta_1}.$$ Thus we have \begin{equation} ge_2=e^{-i\theta_1}e_2. \tag{5.11} \end{equation} By (5.10) and (5.11), $g$ is of the form $$\begin{pmatrix} e^{i\theta_1} && 0 \\ & e^{-i\theta_1} & \\ 0 && g' \end{pmatrix}$$ with $g'\in Sp(n-1,\bc)$. So by induction, we can show that $g$ is conjugate to a diagonal matrix $$g_0=a(\theta_1,\ldots,\theta_n)=\exp Y(\theta_1,\ldots,\theta_n).$$ Since $${}^t\overline{z}g_0z= e^{i\theta_1}|z_1|^2 +e^{-i\theta_1}|z_2|^2 +\cdots +e^{i\theta_n}|z_{2n-1}|^2 +e^{-i\theta_n}|z_{2n}|^2,$$ it is clear that ${}^t\overline{z}g_0z\ne 0$ for all $z\in F$ if and only if $$\Re e^{i\theta_1},\ldots,\Re e^{i\theta_n}\mbox{ are all positive or all negative.}$$ Hence it is equivalent to $$g_0\in \varphi(T^+Z).$$ \bigskip (i) $\Longrightarrow$ (ii) is included in the above argument. \bigskip (ii) $\Longrightarrow$ (iii). \ Elements of the full flag manifold $X\times X$ of $G_\bc$ are pairs of sequences \begin{equation} (V_1\subset\cdots\subset V_n,\ V'_1\subset\cdots\subset V'_n) \tag{5.12} \end{equation} where $V_j$ and $V'_j$ are $j$-dimensional complex subspaces on which the canonical symplectic form for $G=Sp(n,\bc)$ vanish. The open $G_\br$-orbit $S'$ on $X\times X$ consists of flags (5.12) such that the canonical hermtian form is nondegenerate on the pair $(V_j,V'_j)$ for every $j=1,\ldots,n$. So we can prove (ii) $\Longrightarrow$ (iii) by the same argument as in the case of $SL(n,\bc)$. \hfill q.e.d. \section{$GL(n,\bh)$-case} Let $G_\bc=SL(2n,\bc), \ K_\bc=\{g\in G_\bc\mid {}^tgJg=J\}\cong Sp(n,\bc)$ and $G_\br=\{g\in G_\bc\mid J\overline{g}J^{-1}=g\} \ (\cong GL(n,\bh) \mbox{ modulo center})$. Here $$J=\begin{pmatrix} J_1 && 0 \\ & \ddots & \\ 0 && J_1 \end{pmatrix} \mand J_1=\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}.$$ As a flag manifold $X_0$, we consider the Grassmann manifold $$X_0=\{V\subset \bc^{2n}\mid \dim_\bc V=2\}.$$ (Note that $P^{2n-1}(\bc)$ is $K_\bc$-homogeneous.) There are two $K_\bc$-orbits and two $G_\br$-orbits on $X_0$. The closed $K_\bc$-orbit $S_0$ on $X_0$ is $$S_0=\{V\in X_0\mid {}^tuJv=0\mbox{ for all }u, v\in V\}$$ and the open $G_\br$-orbit $S'_0$ on $X_0$ is $$S'_0=\{V\in X_0\mid J\overline{V}\cap V=\{0\}\}.$$ We can take $${\frak t}^+=\left\{\begin{pmatrix} i\theta_1 &&&& 0 \\ & i\theta_1 &&& \\ && \ddots && \\ &&& i\theta_n & \\ 0 &&&& i\theta_n \end{pmatrix} \in{\frak s}{\frak l}(2n,\bc) \Bigm| |\theta_j-\theta_k|<{\pi\over 2} \mbox{ for all }j\mbox{ and } k \right\}.$$ \begin{proposition} \ $xS_0\subset S'_0\Longleftrightarrow x\in \widetilde{D_0}Z$. \end{proposition} Proof. \ Suppose that $V\in S_0$ and $xV\in X_0-S'_0 \ (=\mbox{the closed $G_\br$-orbit})$. Note that $$xV\in X_0-S'_0\Longleftrightarrow J\overline{xV}=xV \Longleftrightarrow xV=\bc v\oplus \bc J\overline{v}\mbox{ for }v\in xV-\{0\}.$$ Since $x^{-1}v, \ x^{-1}J\overline{v}\in V\in S_0$, we have $${}^t(x^{-1}J\overline{v})Jx^{-1}v ={}^t\overline{v}\,{}^tJ\,{}^tx^{-1}Jx^{-1}v=0.$$ Hence the condition $xS_0\subset S'_0$ is equivalent to \begin{equation} {}^t\overline{v}gv\ne 0\quad\mbox{for all }v\in\bc^{2n}-\{0\} \tag{6.1} \end{equation} where $g={}^tJ\,{}^tx^{-1}Jx^{-1}$. Note that $gJ$ and ${}^tJg$ are skew symmetric. Assume (6.1). Since $\bc^{2n}-\{0\}$ is simply connected, we can define a smooth function $$\xi(v)=\arg {}^t\overline{v}gv$$ of $v\in\bc^{2n}-\{0\}$. The range $R$ of $\xi$ is a bounded closed interval in $\br$ since $P^{2n-1}(\bc)=(\bc^{2n}-\{0\})/\bc^\times$ is compact. Take a $v_1\in\bc^{2n}-\{0\}$ such that $\xi(v_1)$ is on the boudary of $R$ and that ${}^t\overline{v_1}gv_1=\lambda_1\in U(1)$. Put $$U_1=\{u\in\bc^{2n}\mid {}^t\overline{v_1}gu= {}^t\overline{v_1}gJ\overline{u}=0\}.$$ Then $U_1$ is a complex subspace of $\bc^{2n}$ such that \begin{equation} J\overline{U_1}=U_1. \tag{6.2} \end{equation} Since ${}^t\overline{v_1}gv_1\ne 0$ by (6.1), $U_1$ is a $2n-2$-dimensional complex subspace of $\bc^{2n}$ satisfying $$U_1\cap(\bc v_1\oplus \bc J\overline{v_1})=\{0\}.$$ Let $u$ be an element of $U_1$ and define a function $$\eta(t)=\xi(v_1+tu)=\arg ({}^t\overline{v_1}gv_1+ t\,{}^t\overline{v_1}gu+t\,{}^t\overline{u}gv_1 +t^2\,{}^t\overline{u}gu)$$ of $t\in\br$. Since $\eta(0)=\xi(v_1)$ is on the boundary of $R$ and since ${}^t\overline{v_1}gu=0$, we have $${}^t\overline{u}gv_1\in \br\,{}^t\overline{v_1}gv_1$$ for all $u\in U_1$. Since $U_1$ is a complex subspace of $\bc^{2n}$, this implies $${}^t\overline{u}gv_1=0.$$ By (6.2), we also have $${}^t\overline{J\overline{u}}gv_1=0.$$ Since ${}^tJg$ is skew symmetric, we have $${}^t\overline{J\overline{v_1}}gu={}^tv_1\,{}^tJgu= -{}^tu\,{}^tJgv_1 =-{}^t\overline{J\overline{u}}gv_1=0.$$ On the other hand, since $gJ$ is skew symmetric, we also have $${}^t\overline{u}gJ\overline{v_1}=-{}^t\overline{v_1}gJ\overline{u}= 0$$ by the definition of $U_1$. Thus we have proved \begin{equation} {}^t\overline{v_1}gu= {}^t\overline{J\overline{v_1}}gu= {}^t\overline{u}gv_1={}^t\overline{u}gJ\overline{v_1}=0 \notag \end{equation} for all $u\in U_1$. Since $gJ$ and ${}^tJg$ are skew symmetric, we have $${}^t\overline{v_1}gJ\overline{v_1}=0,\quad {}^t\overline{J\overline{v_1}}gv_1={}^tv_1\,{}^tJgv_1=0$$ and $${}^t\overline{J\overline{v_1}}gJ\overline{v_1}= {}^tv_1\,{}^tJgJ\overline{v_1} =-{}^tv_1\,{}^tgJ^2\overline{v_1} ={}^tv_1\,{}^tg\overline{v_1} ={}^t\overline{v_1}gv_1=\lambda_1.$$ Hence we have \begin{equation} \begin{pmatrix} {}^t\overline{v_1}gv_1 & {}^t\overline{v_1}gJ\overline{v_1} \\ \noalign{\smallskip} {}^t\overline{J\overline{v_1}}gv_1 & {}^t\overline{J\overline{v_1}}gJ\overline{v_1} \end{pmatrix} = \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_1 \end{pmatrix} \notag \end{equation} Next we take an element $v_2\in U_1-\{0\}$ such that $\xi(v_2)$ is on the boundary of $\xi(U_1-\{0\})$ and that ${}^t\overline{v_2}gv_2 =\lambda_2\in U(1)$. Repeating this procedure, we can take basis $v_1,J\overline{v_1},\ldots,v_n,J\overline{v_n}$ of $\bc^{2n}$ such that $${}^t\overline{v_j}gv_k={}^t\overline{v_j}gJ\overline{v_k} ={}^t(\overline{J\overline{v_j}})gv_k ={}^t(\overline{J\overline{v_j}})gJ\overline{v_k}=0 \quad\mbox{if }j\ne k$$ and that $$\begin{pmatrix} {}^t\overline{v_j}gv_j & {}^t\overline{v_j}gJ\overline{v_j} \\ \noalign{\smallskip} {}^t\overline{J\overline{v_j}}gv_j & {}^t\overline{J\overline{v_j}}gJ\overline{v_j} \end{pmatrix} = \begin{pmatrix} \lambda_j & 0 \\ 0 & \lambda_j \end{pmatrix}$$ for $j=1,\ldots,n$. Let $h$ be an element of $GL(2n,\bc)$ such that $$he_1=v_1, he_2=-J\overline{v_1}, \ldots, he_{2n-1}=v_n, he_{2n}=-J\overline{v_n}.$$ Then $h$ is an element of $\widetilde{G_\br}=\{h\in GL(2n,\bc)\mid J\overline{h}J^{-1}=h\}\cong GL(n,\bh)$ and $${}^t\overline{h}gh=\begin{pmatrix} \lambda_1 &&&& 0 \\ & \lambda_1 &&& \\ && \ddots && \\ &&& \lambda_n & \\ 0 &&&& \lambda_n \end{pmatrix}.$$ We can see that $\det h\in\br$ and moreover that $\det h>0$ since $\widetilde{G_\br}$ is connected. Since $g={}^tJ\,{}^tx^{-1}Jx^{-1}\in SL(n,\bc)$, we have $$(\det h)^2=\lambda_1^2\cdots\lambda_n^2\in U(1).$$ So we have $\det h=1$ which implies $h\in G_\br$ and we also have $$\lambda_1^2\cdots\lambda_n^2=1.$$ We have $${}^th\,{}^tx^{-1}Jx^{-1}h =J\,{}^t\overline{h}\,{}^tJ\,{}^tx^{-1}Jx^{-1}h =J\,{}^t\overline{h}gh =\begin{pmatrix} \lambda_1J_1 && 0 \\ & \ddots & \\ 0 && \lambda_nJ_1 \end{pmatrix}.$$ We can take the pfaffian as a polynomial function of degree $n$ on the space of skew symmetric matrices such that $${\rm Pf}\,J=1.$$ Then we have \begin{equation} {\rm Pf}\,\begin{pmatrix} \lambda_1J_1 && 0 \\ & \ddots & \\ 0 && \lambda_nJ_1 \end{pmatrix} =\lambda_1\cdots\lambda_n=1 \tag{6.3} \end{equation} because ${\rm Pf}({}^tmJm)=1$ for any $m\in SL(2n,\bc)$. Let $c$ be an element of $\bc^{2n}$. Then we have $${}^t\overline{c}\,{}^t\overline{h}ghc =\lambda_1(|c_1|^2+|c_2|^2)+\cdots+ \lambda_n(|c_{2n-1}|^2+|c_{2n}|^2).$$ Hence the condition (6.1) is equivalent to the condition (3.2). So we can write $$\begin{pmatrix} \lambda_1J_1 && 0 \\ & \ddots & \\ 0 && \lambda_nJ_1 \end{pmatrix} =t^2z^2J={}^t(tz)Jtz$$ with some $t\in T^+$ and $z\in Z$ by using the condition (6.3). Hence $x^{-1}hz^{-1}t^{-1}\in K_\bc$ and therefore $$x\in hz^{-1}t^{-1}K_\bc\subset \widetilde{D_0}Z.$$ Converse assertion is clear by the above arguments. Now let us consider the full flag manifold $X$ consisting of full flags $$V_1\subset\cdots\subset V_{2n-1}$$ where $V_j$ is a $j$-dimensional complex subspace in $\bc^{2n}$. There is only one closed $K_\bc$-orbit $S$ on $X$ consisting of flags satisfying $$u\in V_j, \ v\in V_{2n-j} \Longrightarrow {}^tuJv=0$$ for all $j$. On the other hand, the corresponding open $G_\br$-orbit $S'$ consists of flags satisfying $$J\overline{V_j}\cap V_{2n-j}=\{0\}$$ for all $j$. Suppose $xS_0\subset S'_0$. Then we can show $xS\subset S'$ as follows. Any flag contained in $xS$ is written as $$xV_1\subset\cdots\subset xV_{2n-1}$$ with some $$V_1\subset\cdots\subset V_{2n-1}$$ in $S$. Let $v$ be an element of $J\overline{xV_j}\cap xV_{2n-j}$. Then since $x^{-1}\overline{J^{-1}v}\in V_j$ and $x^{-1}v\in V_{2n-j}$, we have $$0={}^t(x^{-1}\overline{J^{-1}v})Jx^{-1}v ={}^t\overline{v}\,{}^tJ\,{}^tx^{-1}Jx^{-1}v={}^t\overline{v}gv.$$ Hence $v=0$ by (6.1). Thus we have proved $S_0\subset S'_0 \Longrightarrow xS\subset S'$. The assertion $xS\subset S' \Longrightarrow xS_0\subset S'_0$ is proved by a general argument as in Section 8. Thus we have shown: \begin{proposition} \ $xS_0\subset S'_0 \Longleftrightarrow xS\subset S'$. \end{proposition} \section{$Sp(p,q)$-case} Let $G_\bc=\{g\in GL(2n,\bc)\mid {}^tgJg=J\}\cong Sp(n,\bc)$ where $J$ is defined as in Section 5 and 6. Put $V_+=\bc e_1\oplus\cdots\oplus\bc e_{2p}$ and $V_-=\bc e_{2p+1}\oplus\cdots\oplus\bc e_n$ with the canonical basis $e_1,\ldots,e_{2n}$ of $\bc^{2n}$ and let $Q$ be the hermitian form on $\bc^{2n}$ defined by $$Q(z,w)={}^tzI_{2p,2q}\overline{w}=z_1\overline{w_1}+\cdots+ z_{2p}\overline{w_{2p}}-z_{2p+1}\overline{w_{2p+1}}-\cdots -z_{2n}\overline{w_{2n}}.$$ Then we define subgroups $$K_\bc=\{g\in G_\bc\mid gV_+=V_+\}=\{g\in G_\bc\mid gV_-=V_-\} \cong Sp(p,\bc)\times Sp(q,\bc)$$ ($q=n-p$) and $$G_\br=\{g\in G_\bc \mid Q(gz,gw)=Q(z,w)\mbox{ for all }z,w\in\bc^{2n}\}\cong Sp(p,q)$$ of $G_\bc$. We can take a flag manifold $$X_0=P^{2n-1}(\bc)=(\bc^{2n}-\{0\})/\bc^\times$$ of $G_\bc$. There are three $K_\bc$-orbits on $X_0$ consisting of two closed orbits $S_\pm=(V_\pm-\{0\})/\bc^\times$ and the compliment $S_0$. There are also three $G_\br$-orbits $S'_\pm$ and $S'_0$ on $X_0$ defined by the signature of $Q$. Let $\ell=\min(p,q)$ and write $$Y(\theta_1,\ldots,\theta_\ell) =\theta_1(E_{2p+1,1}+E_{2p+2,2}-E_{1,2p+1}-E_{2,2p+2})+\cdots$$ $$\mbx5 +\theta_\ell(E_{2p+2\ell-1,2\ell-1}+E_{2p+2\ell,2\ell} -E_{2\ell-1,2p+2\ell-1}-E_{2\ell,2p+2\ell}).$$ Then we can take $${\frak t}^+=\{Y(\theta_1,\ldots,\theta_\ell) \mid |\theta_j|<\pi/4\mbox{ for all }j=1,\ldots,\ell\}.$$ \begin{proposition} \ $x\in G_\br T^+K_\bc \Longleftrightarrow xS_+\subset S'_+ \Longleftrightarrow xS_-\subset S'_-$. \end{proposition} Proof. \ The condition $xS_+\subset S'_+$ is equivalent to \begin{equation} Q\mbox{ is positive definite on }V=xV_+ \tag{7.1} \end{equation} Assume (7.1). Let $H$ denote the hypersurface in $V\times V$ defined by $$H=\{(u,v)\in V\times V\mid {}^tuJv=1\}$$ and consider the function $$f(u,v)=Q(u,u)Q(v,v)$$ of $(u,v)\in V\times V$. Then the range of $f|_H$ is of the form $$[M,\infty)$$ with some $M>0$ because we can write $$\{|{}^tuJv|\mid u,v\in V,\ Q(u,u)=Q(v,v)=1\}=\left[0,{1\over M}\right]$$ by (7.1). Let $(u_1,v_1)$ be an element of $H$ such that $$f(u_1,v_1)=M.$$ Clearly, we may choose $u_1$ and $v_1$ so that $$Q(u_1,u_1)=Q(v_1,v_1)=\sqrt{M}.$$ Put $\lambda_1=\sqrt{M}$. By elementary calculus, we have $${}^tu_1Jw=0\Longrightarrow \Re Q(v_1,w)=0$$ for $w\in\bc^{2n}$. Since $w$ can be replaced by $iw$, we have $${}^tu_1Jw=0\Longrightarrow \Im Q(v_1,w)=0$$ and therefore $${}^tu_1Jw=0\Longrightarrow Q(v_1,w)=0.$$ In the same way, we also have $${}^tv_1Jw=0\Longrightarrow Q(u_1,w)=0.$$ Put $U_1=\{w\in V\mid {}^tu_1Jw={}^tv_1Jw=0\}$. Then the alternating form ${}^tzJw$ is nondegenerate on $U_1$ since it is nondegenerate on $V$. By the same argument as above, we can take $(u_2,v_2)\in H_1=(U_1\times U_1)\cap H$ such that $f|_{H_1}$ attains its minimum at $(u_2,v_2)$ and that $$Q(u_2,u_2)=Q(v_2,v_2)=\lambda_2\ge\lambda_1.$$ Repeating this procedure, we can get basis $u_1,v_1,\ldots,u_p,v_p$ of $V$ satisfying $${}^tu_jJv_j=1,\qquad Q(u_j,u_j)=Q(v_j,v_j)=\lambda_j,\qquad Q(u_j,v_j)=0$$ and \begin{equation} {}^tu_jJu_k={}^tu_jJv_k={}^tv_jJv_k=Q(u_j,u_k)=Q(u_j,v_k)= Q(v_j,v_k)=0\quad\mbox{if }j\ne k \tag{7.2} \end{equation} where $0<\lambda_1\le\cdots\le\lambda_p$. Put $J'=I_{2p,2q}J$. Then we have \begin{equation} Q(u,J'\overline{v})={}^tuJv,\qquad Q(J'\overline{u},J'\overline{v}) =Q(v,u)\mand {}^t(J'\overline{u})JJ'\overline{u}=\overline{{}^tuJv} \tag{7.3} \end{equation} for any $u,v\in\bc^{2n}$. Put $$w_j^{(1)}=u_j+J'\overline{v_j},\quad w_j^{(2)}=-J'\overline{w_j^{(1)}},\quad w_j^{(3)}=u_j-J'\overline{v_j},\quad w_j^{(4)}=-J'\overline{w_j^{(3)}}$$ for $j=1,\ldots,p$. Then by (7.2) and (7.3), we have \begin{equation} Q(w_j^{(r)},w_k^{(s)})={}^tw_j^{(r)}Jw_k^{(s)}=0\quad\mbox{if }j\ne k \tag{7.4} \end{equation} for every $r,s=1,2,3,4$. On the other hand, we have \begin{align*} Q(w_j^{(1)},w_j^{(1)}) & =Q(u_j,u_j)+Q(u_j,J'\overline{v_j}) +Q(J'\overline{v_j},u_j)+Q(J'\overline{v_j},J'\overline{v_j}) \\ & =Q(u_j,u_j)+{}^tu_jJv_j+\overline{{}^tu_jJv_j}+Q(v_j,v_j) \\ & =2\lambda_j+2, \end{align*} $Q(w_j^{(3)},w_j^{(3)})=2\lambda_j-2, \ Q(w_j^{(1)},w_j^{(3)})=0$ and $${}^tw_j^{(1)}Jw_j^{(3)} ={}^t(u_j+J'\overline{v_j})J(u_j-J'\overline{v_j}) =-2\,{}^tu_jJJ'\overline{v_j} =2Q(u_j,v_j)=0.$$ Hence we have \begin{equation} Q(w_j^{(r)},w_j^{(s)})=\begin{cases} 2\lambda_j+2 & \text{if $r=s=1$ or $r=s=2$} \\ 2\lambda_j-2 & \text{if $r=s=3$ or $r=s=4$} \\ 0 & \text{otherwise} \end{cases} \tag{7.5} \end{equation} and \begin{equation} {}^tw_j^{(r)}Jw_j^{(s)}=\begin{cases} 2\lambda_j+2 & \text{if $(r,s)=(1,2)$} \\ 2\lambda_j-2 & \text{if $(r,s)=(3,4)$} \\ 0 & \text{if $r\le s$ and $(r,s)\ne (1,2),(3,4)$} \end{cases} \tag{7.6} \end{equation} by (7.3). By (7.4) and (7.5), the $4p$ vectors $w_j^{(r)}$ are mutually orthogonal with respect to $Q$. Since $2\lambda_j+2>2>0$ and since the signature of $Q$ is $(2p,2q)$, we have $$2\lambda_j-2\le 0$$ and hence $\lambda_j\le 1$. Moreover if $\lambda_j=1$, then $$w_j^{(3)}=w_j^{(4)}=0$$ since $Q$ is negative definite on the orthogonal compliment of the $2p$-dimensional space spanned by $w_1^{(1)},\ldots,w_p^{(1)}, w_1^{(1)},\ldots,w_p^{(2)}$. (If $p>q$, then we have automatically $\lambda_j=1$ and $w_j^{(3)}=w_j^{(4)}=0$ for $j=q+1,\ldots,p$ since $\lambda_1\le\cdots \le\lambda_p$.) Since $0<\lambda_1\le\cdots \le\lambda_p\le 1$, we can write $\lambda_j=\cos^2 \theta_j-\sin^2 \theta_j$ with $${\pi\over 4}\ge\theta_1\ge\cdots\ge\theta_p\ge 0.$$ Then $2\lambda_j+2=4\cos^2 \theta_j$ and $2\lambda_j-2=-4\sin^2 \theta_j$. By (7.4), (7.5) and (7.6), we can take an element $g\in G_\br$ such that $$gw_j^{(1)}=2\cos\theta_je_{2j-1},\quad gw_j^{(2)}=2\cos\theta_je_{2j}$$ for $j=1,\ldots,p$ and that $$gw_j^{(3)}=2\sin\theta_je_{2p+2j-1},\quad gw_j^{(4)}=2\sin\theta_je_{2p+2j}$$ for $j=1,\ldots,\ell'$ where $\ell'=\max\{j\mid \theta_j\ne 0\}$. So we have \begin{align*} gu_j & ={1\over 2}g(w_j^{(1)}+w_j^{(3)})= \cos\theta_je_{2j-1}+\sin\theta_je_{2p+2j-1} \\ gv_j & ={1\over 2}g(w_j^{(2)}+w_j^{(4)})= \cos\theta_je_{2j}+\sin\theta_je_{2p+2j} \end{align*} for $j=1,\ldots,p$. This implies $$gxV_+=gV=\exp Y(\theta_1,\ldots,\theta_\ell)V_+$$ and therefore $$x\in g^{-1}\exp Y(\theta_1,\ldots,\theta_\ell) K_\bc \subset G_\br T^+K_\bc.$$ Conversely, let $v=\sum_{j=1}^{2p}c_je_j\in V_+$ and $x= \exp Y(\theta_1,\ldots,\theta_\ell)$ with $|\theta_j|<\pi/4$. Then we have $$Q(xv,xv)=(|c_1|^2+|c_2|^2)(\cos^2 \theta_1-\sin^2 \theta_1)+\cdots +(|c_{2p-1}|^2+|c_{2p}|^2)(\cos^2 \theta_p-\sin^2 \theta_p)>0$$ if $v\ne 0$. (Here $\theta_{\ell+1}=\cdots=\theta_p=0$ if $p>q$.) Hence $xS_+\subset S'_+$. We also have $x\in G_\br T^+K_\bc \Longleftrightarrow xS_-\subset S'_-$ by the same argument. \hfill q.e.d. \bigskip Now let us consider the full flag manifold. The full flag manifold $X=G_\bc/B$ consists of full flags $$V_1\subset\cdots\subset V_n$$ where $V_j$ are $j$-dimensional complex linear subspaces $V_j$ such that ${}^tzJw=0$ for all $z,w\in V_j$. Such a flag is contained in a closed $K_\bc$-orbit if and only if $$V_j=(V_j\cap V_+)\oplus(V_j\cap V_-)\quad\mbox{for all }j=1,\ldots,n.$$ It is clear that $$p_j=\dim(V_j\cap V_+)\mand q_j=\dim(V_j\cap V_-)\quad(j=1,\ldots,n)$$ are the invariants of closed $K_\bc$-orbits. Hence there are ${n\choose p}$ closed $K_\bc$-orbits on $X$. On the other hand, a flag is contained in an open $G_\br$-orbit if and only if the hermitian form $Q$ is nondegenerate on each $V_j \ (j=1,\ldots,n)$. So the signatures $(p_j,q_j)$ of $Q$ on $V_j \ (j=1,\ldots,n)$ are the invariants of open $G_\br$-orbits on $X$. We can see that closed $K_\bc$-orbits and open $G_\br$-orbits on $X$ are in one-to-one correspondence by these numbers $(p_j,q_j) \ (j=1,\ldots,n)$. Suppose that $x\in G_\br T^+ K_\bc$. Then we want to show that $xS\subset S'$ for all closed $K_\bc$-orbits $S$ on $X$. By Proposition 7.1, $Q$ is positive defnite on $xV_+$ and negative definite on $xV_-$. Let $$V_1\subset\cdots\subset V_n$$ be a flag contained in $S$. Then $Q$ is positive definite on $xV_j\cap xV_+=x(V_j\cap V_+)$ and negative definite on $xV_j\cap xV_-=x(V_j\cap V_-)$ for every $j=1,\ldots, n$. So the signature of $Q$ on $xV_j$ is $(\dim(V_j\cap V_+),\dim(V_j\cap V_-))$ for $j=1,\ldots,n$. Thus we have proved $xS\subset S'$. Conversely, if $xS\subset S'$ for some closed $K_\bc$-orbit on $X$, then we have $xS_+\subset S'_+$ or $xS_-\subset S'_-$ by a general argument as in Section 8 and hence $x\in G_\br T^+ K_\bc$ by Proposition 7.1. Summarizing the above arguments, we have: \begin{proposition} \ $x\in G_\br T^+ K_\bc \Longleftrightarrow xS\subset S'$ for a closed $K_\bc$-orbit on $X$. \end{proposition} \section{Arguments on full flag manifolds} Let $B$ be a Borel subgroup of $G_\bc$. Let $\Psi$ denote the set of simple roots corresponding to $B$. For $\alpha\in\Psi$, we can take a parabolic subgroup $P_\alpha=B\cup Bw_\alpha B$ of $G_\bc$. \begin{proposition} \ Let $x$ be an element of $G_\bc$. If $xS\subset S'$ for all closed $K_\bc$-$B$ double cosets in $G_\bc$, then $$xS\cap S'\mbox{ is non-empty and closed in }G_\bc\mbox{ for all }S\in K_\bc\backslash G_\bc/P$$ for an arbitrary parabolic subgroup $P$ of $G_\bc$. \end{proposition} Proof. \ First assume that $P=B$. We will prove this by induction on $\dim S$. If $S\in K_\bc\backslash G_\bc/B$ is not closed, then there exists an $\alpha\in\Psi$ such that (1) $SP_\alpha=S\cup S_1\cup S_2, \ \dim S_1=\dim S_2=\dim S-2$, \noindent or (2) $SP_\alpha=S\cup S_1, \ \dim S_1=\dim S-2$. Consider the case (1). (The case (2) is easier than (1).) It is known (\cite{M2}, \cite{Sp}) that the closure $S_1^{cl}$ of $S_1$ is written as $$S_1^{cl}=S_0P_{\alpha_1}\cdots P_{\alpha_\ell}$$ with some closed $K_\bc$-$B$ double coset $S_0$ and $\alpha_1,\ldots,\alpha_\ell\in\Psi$. So by the assumption, we have \begin{equation} xS_1\subset xS_1^{cl}=xS_0P_{\alpha_1}\cdots P_{\alpha_\ell} \subset S'_0P_{\alpha_1}\cdots P_{\alpha_\ell} =\bigcup\{T'\in G_\br\backslash G_\bc/B\mid T\subset S_1^{cl}\}.\tag{$8.1$} \end{equation} (Note that the right hand side is open in $G_\bc$ since $S'_0$ is open.) Hence we have \begin{equation} xS_1\cap S'=\phi \mand xS_1\cap S'_2=\phi.\tag{$8.2$} \end{equation} because $S$ and $S_2$ are not contained in $S_1^{cl}$. In the same way, we also have \begin{equation} xS_2\cap S'=\phi \mand xS_2\cap S'_1=\phi.\tag{$8.3$} \end{equation} By (8.2), we have $$(xS_1\cap S'_1)P_\alpha=(xS_1\cap(S'_1\cup S'_2\cup S'))P_\alpha =(xS_1\cap S'_1P_\alpha)P_\alpha =xS_1P_\alpha\cap S'_1P_\alpha.$$ By the assumption of induction, this set is closed in $G_\bc$ (compact in $G_\bc/B$). On the other hand, we have $$xS_1P_\alpha\cap S'_1P_\alpha = (xS_1\cup xS_2\cup xS)\cap (S'_1\cup S'_2\cup S') = ((xS_1\cup xS_2\cup xS)\cap (S'_1\cup S'_2)) \cup (xS\cap S')$$ since \begin{equation} (xS_1\cup xS_2\cup xS)\cap S'=xS\cap S'\tag{$8.4$} \end{equation} by (8.2) and (8.3). Hence the set $xS\cap S'$ is closed in $G_\bc$ because it is the compliment of the open set $$\bigcup\{T'\in G_\br\backslash G_\bc/B\mid T\subset(S_1)^{cl}\} \cup \bigcup\{T'\in G_R\backslash G_\bc/B\mid T\subset(S_2)^{cl}\}$$ in the closed set $xS_1P_\alpha\cap S'_1P_\alpha$. By (8.4), we have $$(xS\cap S')P_\alpha=(xSP_\alpha\cap S')P_\alpha=xSP_\alpha\cap S'P_\alpha=xS_1P_\alpha\cap S'_1P_\alpha.$$ This set is non-empty since $xS_1\cap S'_1$ is non-empty by the assumption of induction. Hence $xS\cap S'$ is non-empty. Now consider an arbitrary parabolic subgroup $P$. We may assume that $B\subset P$. Take a $K_\bc$-$B$ double coset $S_1$ contained in $S$ such that $S_1$ is relatively closed in $S$. Then by (8.1), we have $$xS_1\cap T'=\phi$$ for $K_\bc$-$B$ double cosets $T$ contained in $S-S_1$ because $T\not\subset (S_1)^{cl}$. Hence we have $$xS_1\cap S'=xS_1\cap S'_1$$ since $S'=\bigcup_{T\subset S} T'$ as in \cite{M1}. Thus the set $$xS\cap S'=xS_1P\cap S'=(xS_1\cap S')P=(xS_1\cap S'_1)P$$ is closed in $G_\bc$ since $(xS_1\cap S'_1)/B$ is compact. \hfill q.e.d. \begin{remark} \ {\rm (More precise arguments when $P=B$) \ First consider the case (1). Let $y$ be an element of $xS\cap S'$. Then we have \begin{eqnarray*} yP_\alpha & = & (yP_\alpha\cap xS_1)\cup (yP_\alpha\cap xS_2)\cup (yP_\alpha\cap xS) \\ & = & (yP_\alpha\cap S'_1)\cup (yP_\alpha\cap S'_2)\cup (yP_\alpha\cap S') \end{eqnarray*} Here $yP_\alpha/B\cong P^1(\bc), \ (yP_\alpha\cap xS_1)/B\cong\{\mbox{a point}\}, \ (yP_\alpha\cap xS_2)/B\cong\{\mbox{a point}\}$ and $(yP_\alpha\cap S')/B\cong S^1$. By (8.2) and (8.3), we have $$yP_\alpha\cap xS_1 \subset yP_\alpha\cap S'_1 \mand yP_\alpha\cap xS_2 \subset yP_\alpha\cap S'_2.$$ We also have the same formulas when $y\in xS_1\cap S'_1$. Thus we can construct a smooth surjection of $(xS\cap S')/B$ onto $(xS_1\cap S'_1)/B$ with fibers diffeomorphic to $S^1$. Next consider the case (2). Let $y$ be an element of $xS\cap S'$. Then we have \begin{eqnarray*} yP_\alpha & = & (yP_\alpha\cap xS_1)\cup (yP_\alpha\cap xS) \\ & = & (yP_\alpha\cap S'_1)\cup (yP_\alpha\cap S') \end{eqnarray*} There are two cases: (2a) $(yP_\alpha\cap xS_1)/B\cong\{\mbox{a point}\}$ and $(yP_\alpha\cap S')/B\cong\{\mbox{a point}\}$, (2b) $(yP_\alpha\cap xS_1)/B\cong\{\mbox{two points}\}$ and $(yP_\alpha\cap S')/B\cong S^1$. In (2a)-case, we get a diffeomorphism of $(xS\cap S')/B$ onto $(xS_1\cap S'_1)/B$. On the other hand, in (2b)-case, we have a $S^1$-to-$\{\mbox{2 points}\}$ correspondence. }\end{remark} \begin{proposition} \ Let $S$ be the open $K_\bc$-orbit on $X$. Then $$C(S)_0=C_0.$$ Here $*_0$ denote the connected component of $*$ containing the identity $e$. \end{proposition} Proof. \ Let $[0,1]\ni t\mapsto x_t$ be a continuous curve in $C(S)$ such that $x_0=e$. Then by Proposition 8.1, we have only to show that $x_t\in C(S_1)$ for all closed $K_\bc$-orbits $S_1$ on $X$. Suppose $x_t\notin C(S_1)$ for some $t$ and $S_1$. Then we will get a contradiction as follows. We may assume that $x_tS_1\cap S'_2\ne\phi$ for some $G_\br$-orbit $S'_2$ contained in the boundary of $S'_1$. Put $\ell=\dim_\bc S-\dim_\bc S_2$ and let $\alpha_1,\ldots,\alpha_\ell$ be a sequence of simple roots such that $$\dim_\bc S_2P_{\alpha_1}\cdots P_{\alpha_k}=\dim_\bc S_2+k$$ for all $k=1,\ldots,\ell$. Since $S_2P_{\alpha_1}\cdots P_{\alpha_k} \supset S$, we have $$S'_2P_{\alpha_1}\cdots P_{\alpha_k}\supset S'$$ (cf. [M2]) and therefore $$S'_2\subset S'P_{\alpha_k}\cdots P_{\alpha_1}.$$ Since $x_t\in C(S)$, we have $S'\subset x_tS$. So we have $$S'_2\subset x_tSP_{\alpha_k}\cdots P_{\alpha_1}$$ and hence $$\phi\ne x_tS_1\cap x_tSP_{\alpha_k}\cdots P_{\alpha_1} = x_t(S_1\cap SP_{\alpha_k}\cdots P_{\alpha_1}).$$ But since the set $SP_{\alpha_k}\cdots P_{\alpha_1}$ does not contain $K_\bc$-orbits whose dimensions are smaller than $\dim_\bc S_2$, the right hand side of the above formula must be $\phi$. Thus we have got a contradiction. \hfill q.e.d. \section{Appendix for $SO(p,q)$-case} Retain the notations in Section 4. Assume $p\le q$. The orthogonal compliment ${\frak m}$ of ${\frak s}{\frak o}(p,q)+{\frak s}({\frak u}(p)\oplus{\frak u}(q))$ in ${\frak s}{\frak u}(p,q)$ is written as $${\frak m}=\left\{\begin{pmatrix} 0 & -i\,{}^tB \\ iB & 0 \end{pmatrix}\Bigm| B\mbox{ is a $q\times p$ real matrix}\right\}.$$ So if we write $$Z(t_1,\ldots,t_p)=it_1(E_{n,1}-E_{1,n})+\cdots +it_p(E_{n-p+1,p}-E_{p,n-p+1}),$$ then we can take $${\frak a}=\{Z(t_1,\ldots,t_p)\mid t_1,\ldots,t_p\in\br\}$$ as a maximal abelian subspace of ${\frak m}$. By a fundamental decomposition theorem (c.f. \cite{R} Theorem 10), we have \begin{equation} SU(p,q)=SO(p,q)_0AS(U(p)\times U(q)) \tag{9.1} \end{equation} where $A=\exp{\frak a}$. Let $X$ be the Grassmann manifold consisting of $p$-dimensional complex subspaces of $\bc^n$. Let $D'$ denote the $SU(p,q)$-orbit on $X$ containing $V_+=\bc e_1\oplus\cdots\oplus\bc e_p$. Then by (9.1), we have $$D'=SO(p,q)_0AV_+.$$ Let ${\frak t}$ be as in Section 4 and define $${{\frak t}^+}'=\{Y(\theta_1,\ldots,\theta_p)\mid |\theta_j|<\pi/4 \mbox{ for all }j=1,\ldots,p\}$$ which is a subset of ${\frak t}^+$. Put ${T^+}'=\exp{{\frak t}^+}'$. Then we have $$AV_+={T^+}'V_+$$ because $$(\exp Z(t_1,\ldots,t_p))e_j=(\cosh t_j)e_j+i(\sinh t_j)e_{n-j+1}$$ and $$(\exp Y(\theta_1,\ldots,\theta_p))e_j= (\cos \theta_j)e_j+i(\sin \theta_j)e_{n-j+1}.$$ (This is a typical example of Lemma 2.1 (1).) Hence $D_0=SO(p,q)_0T^+ V_+$ is bigger than $D'=SO(p,q)_0{T^+}'V_+$. \begin{thebibliography}{BGW} \bibitem[AG]{AG} D. N. Akhiezer and S. G. Gindikin. \newblock On Stein extensions of real symmetric spaces. \newblock {\it Math. Ann.}, 286:1--12, 1990. \bibitem[BGW]{BGW} L. Barchini, S. G. Gindikin and H. W. Wong. \newblock Determinant functions and geometry of the flag manifold for $SU(p,q)$. \newblock {\it J. of Lie Theory}, 6:191-206, 1996. \bibitem[G]{G} S. 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