Harish-Chandra highest weight representations

of semisimple Lie algebras and Lie groups

R. Fioresi, V.S. Varadarajan

###### Abstract.

In this expository paper we describe the theory of Harish-Chandra highest
weight representations and their explicit geometric realizations.

##  1\. Introduction

The theory initiated by Harish-Chandra in 1956, with his three seminal papers
in the American Journal of Mathematics [16] on representations of real
semisimple Lie groups, influenced researchers for the next decades and are
still a source of inspiration for many. He constructed these modules first
infinitesimally and then globally, in the space of sections on holomorphic
bundles on symmetric spaces and laid the foundation for further investigation
of unitary representations and harmonic analysis of real semisimple Lie groups
([14], [31]). His research also paved the road to the successes of geometric
quantization (orbit method) and models (see [28], [24], [12], [13] and refs.
therein) and later on to the SUSY generalizations [22], [4], [5] and refs.
therein. Despite his untimely death, his mathematical vision had a great
impact on generations of mathematicians; for a thorough walk through his
beautiful mathematical achievements, see recollections by Howe [20], Langlands
[29] and Varadarajan et al. [36], [37].

In the present work we want to give a self contained exposition of the theory
of the infinitesimal and global realizations of the highest weight Harish-
Chandra modules, elucidating Harish-Chandra arguments as in [16].

The first step (Sec. 2) is the infinitesimal theory, that is the study of
Harish-Chandra representations for the pair $(\mathfrak{g},\mathfrak{k})$,
where $\mathfrak{g}$ is a semisimple Lie algebra and $\mathfrak{k}$ the
complexification of the maximal compact subalgebra of a chosen real form
$\mathfrak{g}_{0}=\mathfrak{k}_{0}\oplus\mathfrak{p}_{0}$ (real Cartan
decomposition). Notice that the index $0$ always denotes real forms and when
we drop it we mean the complexification. Such pair
$(\mathfrak{g},\mathfrak{k})$ is called an Harish-Chandra pair (HC), and it is
uniquely determined by a real form $\mathfrak{g}_{0}$ of $\mathfrak{g}$ (see
Sec. 2.2).

By definition, an Harish-Chandra representation of the HC pair
$(\mathfrak{g},\mathfrak{k})$, is a representation of $\mathfrak{g}$ in which
$\mathfrak{k}$ acts finitely, that is the sum of irreducible representations
of $\mathfrak{k}$ with the same character is finite dimensional (see (2)). The
construction of the universal highest weight Harish-Chandra representation
$U^{\lambda}$, $\lambda\in\mathfrak{h}^{*}$ ($\mathfrak{h}$ a Cartan
subalgebra of both $\mathfrak{g}$ and $\mathfrak{k}$) is based on the
existence of an admissible system for $\mathfrak{g}$. These are positive
systems in which the adjoint representation of $\mathfrak{k}$ on
$\mathfrak{p}$ stabilizes $\mathfrak{p}^{\pm}$, the sum of the positive
(negative) non compact root spaces. The existence of such systems is
equivalent to the existence of center $\mathfrak{c}$ for $\mathfrak{k}$ (Prop.
2.18) and to a natural invariant complex structure on the real symmetric space
$G_{0}/K_{0}$, $\mathfrak{g}_{0}=\mathrm{Lie}(G_{0})$,
$\mathfrak{k}_{0}=\mathrm{Lie}(K_{0})$ (Rem. 2.19). The dimension of the
center is linked to the number of non equivalent invariant complex structures
on $G_{0}/K_{0}$ (see [1]). The main result for this part is Thm 2.8. Then,
based on our treatment on admissible systems, we proceed with the construction
of the infinitesimal universal highest weight HC module $U^{\lambda}$ and we
give a sufficient condition for its irreducibility (Thm 2.10).

Once the infinitesimal theory is fully elucidated, we proceed (Sec. 3),
following [16], to the geometric realization of the global Harish-Chandra
representation of the real supergroup $G_{0}$ in the space of holomorphic
sections of a line bundle on the symmetric superspace $G_{0}/K_{0}\subset
G/B$, $B$ the borel subgroup corresponding to the fixed admissible system.
These infinite dimensional representations of $G_{0}$ are the global
counterparts of the infinitesimal highest weight representations constructed
previously. We proceed as follows. First we consider the Fréchet space of
sections of the complex line bundle $L$ on the quotient $G/B$, associated with
the infinitesimal character $\lambda$ (see Def. 4). Then, on a neighbourhood
$U$ of the origin, we define a left $\ell$ and a right $\partial$ action of
$\mathcal{U}(\mathfrak{g})$ on $L(U)$ and establish a duality between a
subrepresentation of $L(U)$ (as left $\mathcal{U}(\mathfrak{g})$ module) and
the infinitesimal HC representations studied in Sec. 2 (Thm. 3.12). Choosing
$U=G_{0}B$, we can prove the main result of this part, namely obtain the
highest weight HC representation inside $L(G_{0}S)$, provided however that
$L(G_{0}S)\neq 0$ (Thm. 3.13). Then we give conditions (Thm. 3.17) for which
this occurs. In order to do so, we define the Harish-Chandra decomposition
$P^{-}\times K\times P^{+}\cong P^{-}KP^{+}\subset G$ open,
$\mathfrak{p}^{\pm}=\mathrm{Lie}(P^{\pm})$, where $P^{\pm}$ are abelian
subgroups (see Sec. 3.6 and Lemma 3.14). This is a remarkable result by
itself, based on the peculiar properties of admissible systems;
$\Omega=P^{-}KP^{+}$ is called the Harish-Chandra big cell.

Acknowledgements. The authors thank Prof. C. Carmeli for many illuminating
discussions. The authors also wish to thank the UCLA Dept. of Mathematics for
the kind hospitality, during the completion of this work.

##  2\. Theory on the Lie algebra

###  2.1. Highest weight modules

Let $\mathfrak{g}$ be a semisimple Lie algebra over $\mathbb{C}$. In the
theory of finite dimensional representations of $\mathfrak{g}$, a fundamental
role is played by the highest weight modules. These are defined with respect
to the choice of a Cartan subalgebra (CSA) $\mathfrak{h}$ of $\mathfrak{g}$
and a positive system $P$ of roots of $(\mathfrak{g},\mathfrak{k})$. They are
parametrized by their highest weights, namely the elements
$\lambda\in\mathfrak{h}^{*}$. The universal highest weight modules are known
as Verma modules and are infinite dimensional. The irreducible highest weight
modules are uniquely determined by their highest weights and are the unique
irreducible quotients of the Verma modules. The irreducible modules are finite
dimensional if and only if the highest weight is dominant integral, and one
obtains all irreducible finite dimensional representations of $\mathfrak{g}$
in this manner. These representations lift to the simply connected complex
group $G$ corresponding to $\mathfrak{g}$. For the basic theory of highest
weight modules see [34] Ch. 4, [21] Ch. 2, [2].

###  2.2. Harish-Chandra pairs and Harish-Chandra (HC) modules

In order to study the theory of representations for real semisimple Lie
groups, it is necessary to work in a more structured context. A Harish-Chandra
pair over a field $F$ of characteristic $0$ is a pair
$(\mathfrak{g},\mathfrak{k})$, where:

  1. (i)

$\mathfrak{g}$ is semisimple over $F$ and $\mathfrak{k}$ is the set of fixed
points of an involutive automorphism $\theta$ of $\mathfrak{g}$;

  2. (ii)

$\mathfrak{k}$ is reductive in $\mathfrak{g}$.

This implies that $\mathfrak{k}$ is reductive and

$\mathfrak{k}=\mathfrak{k}^{\prime}\oplus\mathfrak{c},\quad\mathfrak{k}^{\prime}=[\mathfrak{k},\mathfrak{k}],\qquad\mathfrak{c}=\,\hbox{center
of}\,\,\mathfrak{k}.$

(refer to [25], Ch. VI and [19] for the theory of Cartan involutions and their
complexifications). In what follows we work over $\mathbb{C}$. Since
$\theta^{2}=1$, $\mathfrak{g}$ is the direct sum of the subspaces where
$\theta=\pm 1$. We write $\mathfrak{p}$ for the eigenspace of $\theta$ for the
eigenvalue $-1$, $\mathfrak{k}$ by, its very definition, being the eigenspace
of eigenvalue $1$. Since the Cartan-Killing form is invariant under all
automorphisms of $\mathfrak{g}$ we have:

$\mathfrak{p}=\mathfrak{k}^{\perp},\qquad\mathfrak{g}=\mathfrak{k}\oplus\mathfrak{p}$

where $\perp$ refers to the Cartan-Killing form. The restrictions of this form
to $\mathfrak{k}$ and $\mathfrak{p}$ are therefore both non-degenerate. The
fact that $\theta$ is an involutive automorphism implies that

(1)
$[\mathfrak{k},\mathfrak{k}]\subset\mathfrak{k},\qquad[\mathfrak{k},\mathfrak{p}]\subset\mathfrak{p},\qquad[\mathfrak{p},\mathfrak{p}]\subset\mathfrak{k}.$

These properties of $(\mathfrak{g},\mathfrak{k})$ lead to calling the pairs
$(\mathfrak{g},\mathfrak{g}_{\overline{0}})$, in the theory of Lie super
algebras and Lie supergroups, super Harish-Chandra pairs [3], [4]
($\mathfrak{g}$ a Lie superalgebra, $\mathfrak{g}_{\overline{0}}$ its even
part). Nevertheless in this note, we limit ourselves to the ordinary theory of
Lie algebras, and we shall use, as customary, the suffix $0$ to indicate a
real form of a Lie algebra or a Lie group. Let $G_{0}$ be a real connected
semisimple group having finite center with Lie algebra $\mathfrak{g}_{0}$. Let
$K_{0}$ be a maximal compact subgroup of $G_{0}$ with Lie algebra
$\mathfrak{k}_{0}\subset\mathfrak{g}_{0}$. Then we say that
$(\mathfrak{g}_{0},\mathfrak{k}_{0})$ is a Harish-Chandra pair (HC pair) over
$\mathbb{R}$; its complexification $(\mathfrak{g},\mathfrak{k})$ is a HC pair
over $\mathbb{C}$. In this case, there is a unique involution of
$\mathfrak{g}_{0}$ such that $\mathfrak{k}_{0}$ is its set of fixed points,
called the Cartan involution (see [25] Ch. VI).

However, it must be noted that there are involutions of $\mathfrak{g}_{0}$
which are not Cartan involutions, which also can account for HC pairs. For
example, for $\mathfrak{g}_{0}=\mathfrak{sl}(n,\mathbb{R})$, we can take
$\theta$ to be $X\mapsto-FX^{t}F$, where $F$ is the matrix:

$F=\begin{pmatrix}I_{p}&0\\\ 0&I_{q}\end{pmatrix},\qquad n=p+q,\,\,p,q\leq 1.$

Then $\mathfrak{k}_{0}=\mathfrak{so}(p,q,\mathbb{R})$ and
$(\mathfrak{g}_{0},\mathfrak{k}_{0})$ is also a HC pair over $\mathbb{R}$, but
in this case $\mathfrak{k}_{0}$ corresponds to a non-compact subgroup of
$G_{0}$. The HC pairs arising from $(G_{0},K_{0})$, $K_{0}$ compact, are our
main interest, although much of the theory can be formulated in the more
general context of HC pairs over $\mathbb{C}$. When we write
$(\mathfrak{g},\mathfrak{k})$, it means the HC pair is defined over
$\mathbb{C}$; the suffix 0 indicates that it is defined over $\mathbb{R}$.

For a Harish-Chandra pair $(\mathfrak{g},\mathfrak{k})$, we have the class of
$(\mathfrak{g},\mathfrak{k})$-modules, which are $\mathfrak{g}$-modules $V$
such that $V$ splits as an algebraic direct sum of finite dimensional
irreducible $\mathfrak{k}$-modules. In this case we can write

$V=\oplus_{\theta\in E(\mathfrak{k})}V_{\theta}$

where $E(\mathfrak{k})$ is the set of equivalence classes of irreducible
finite dimensional representations of $\mathfrak{k}$ and $V_{\theta}$ is the
span of all $\mathfrak{k}$-irreducible subspaces of $V$ belonging to the class
$\theta$. The module $V$ is called a Harish-Chandra module if:

(2) $\hbox{dim}(V_{\theta})<\infty\qquad\hbox{for all}\,\theta.$

The significance of this class of modules can be seen from the following
example. Suppose that the Harish-Chandra pair $(\mathfrak{g},\mathfrak{k})$
arises from a pair $(G_{0},K_{0})$, i.e. by complexifying the Lie algebras.
Then, irreducible $(\mathfrak{g},\mathfrak{k})$-modules, which are HC-modules,
are the objects of interest from the point of view of representations of the
group $G_{0}$ because of the following fact: if $H$ is an irreducible Banach
space representation of $G_{0}$ and $V$ is the set of $K_{0}$-finite vectors
(see below), then $V$ is an irreducible HC-module. The converse is also true:
every irreducible HC-module arises in this manner. The generic HC module is
not of the highest weight type, but under certain circumstances highest weight
modules are also HC modules and it is the purpose of these notes to
investigate these closely.

###  2.3. Highest weight $(\mathfrak{g},\mathfrak{k})$-modules when rank of
$\mathfrak{g}$=rank of $\mathfrak{k}$

If $V$ is a $\mathfrak{g}$-module and $\mathfrak{a}\subset\mathfrak{g}$ is a
Lie subalgebra, a vector $v\in V$ is $\mathfrak{a}$-finite, if $v\in W$ for
some finite dimensional $\mathfrak{a}$-stable subspace $W$. Let
$V[\mathfrak{a}]$ be the subspace of all $\mathfrak{a}$-finite vectors. Then
$V[\mathfrak{a}]$ is a $\mathfrak{g}$-submodule of $V$; this follows from the
easily proved fact that if $W$ is finite dimensional and
$\mathfrak{a}$-stable, then $\mathfrak{g}[W]$ is again finite dimensional and
$\mathfrak{a}$-stable. Let us now assume that:

$(\mathfrak{g},\mathfrak{k})\quad\hbox{is a HC pair and rk
}\mathfrak{g}=\hbox{rk}\,\mathfrak{k}.$

Then we can choose a Cartan subalgebra (CSA) $\mathfrak{h}$ so that

$\mathfrak{h}\subset\mathfrak{k}\subset\mathfrak{g}$

and $\mathfrak{h}$ will be a CSA of both $\mathfrak{k}$ and $\mathfrak{g}$. We
fix a positive system $P$ of roots for $(\mathfrak{g},\mathfrak{h})$ and write
$\alpha>0$ interchangeably with $\alpha\in P$. We are interested in highest
weight modules (with respect to $P$) which are also HC modules.

######  Lemma 2.1.

If $U$ is a highest weight module with a highest weight vector $u$, the
following are equivalent:

  1. a)

dim$(\mathcal{U}(\mathfrak{k})u)<\infty$.

  2. b)

$U$ is $(\mathfrak{g},\mathfrak{k})$-module

  3. c)

$U$ is a HC module.

If these are satisfied, $\mathcal{U}(\mathfrak{k})u$ is an irreducible
$\mathfrak{k}$-module.

###### Proof.

Since $\mathfrak{k}$ is not semisimple, one must be a little careful. Let
$\mathfrak{c}$ denote the center of $\mathfrak{k}$. For instance, a finite
dimensional $\mathfrak{k}$-module is fully reducible if and only if the action
of $\mathfrak{c}$ is completely reducible. In the present case, since
$\mathfrak{c}\subset\mathfrak{h}$ and $U$ is a direct sum of weight spaces, it
follows that $\mathfrak{c}$ acts semisimply on $U$. In particular, any finite
dimensional submodule for $\mathfrak{k}$ is fully reducible. If $u$ is
$\mathfrak{k}$-finite, then $u\in U[\mathfrak{k}]$. As $U[\mathfrak{k}]$ is a
$\mathfrak{g}$-module, we see that, $U[\mathfrak{k}]=U$. So a) $\implies$ b).
c) $\implies$ a) trivially. If we assume b), we must prove that the spaces
$U_{\theta}$ are finite dimensional. If not, and if $\mu$ is a weight of
$\theta$, then $\mu$ occurs with infinite multiplicity, so that dim
$U[\mu]=\infty$, a contradiction. So b) $\implies$ c). Now
$\mathcal{U}(\mathfrak{k})u$ is a highest weight module for $\mathfrak{k}$ of
finite dimension on which $\mathfrak{c}$ acts through scalars, namely,
$Cw=\lambda(C)w$ where $C\in\mathfrak{c},w\in\mathcal{U}(\mathfrak{k})u$ and
$\lambda$ is the highest weight. Hence
$\mathcal{U}(\mathfrak{k})u=\mathcal{U}(\mathfrak{k}^{\prime})u$,
($\mathfrak{k}^{\prime}=[\mathfrak{k},\mathfrak{k}]$), and so it is a finite
dimensional highest weight module for the semisimple algebra
$\mathfrak{k}^{\prime}$. Hence it is irreducible. ∎

Since $\mathfrak{h}\subset\mathfrak{k}$, both $\mathfrak{k}$ and
$\mathfrak{p}$ are stable under ad $\mathfrak{h}$, and as the root spaces are
one-dimensional, each root space $\mathfrak{g}_{\alpha}$ is contained either
in $\mathfrak{k}$ or $\mathfrak{p}$; we then refer to $\alpha$ as a compact or
non-compact root respectively.

Write $P_{k}$, $P_{n}$ for the set of compact and non-compact roots in $P$. If
$\alpha,\beta,\alpha+\beta$ are roots, then the relation
$[\mathfrak{g}_{\alpha},\mathfrak{g}_{\beta}]=\mathfrak{g}_{\alpha+\beta}$
implies the following: if $\alpha,\beta$ are both compact or both non-compact,
then $\alpha+\beta$ is compact, while if one of them is compact and the other
non-compact, then $\alpha+\beta$ is non-compact; this is a straighforward
consequence of (1). We are interested in determining the highest weight
modules for $(\mathfrak{g},\mathfrak{h},P)$ which are
$(\mathfrak{g},\mathfrak{k})$-modules. Let $V$ be one of such modules, of
highest weight $\lambda\in\mathfrak{h}^{*}$ and $v\in V$ a highest weight
vector. Since $\mathcal{U}(\mathfrak{k})v$ is a finite dimensional highest
weight module for $(\mathfrak{k},\mathfrak{h},P_{k})$, it follows that the
highest weight $\lambda$ must be dominant integral for $P_{k}$, i.e., we must
have

$\lambda(H_{\alpha})\in\mathbb{Z}_{\geq 0},\quad(\alpha\in P_{k})$

We shall now show that $\lambda$ must satisfy additional conditions.

Let $\mathfrak{p}^{\pm}$ be the span of the root spaces corresponding to the
roots $\beta\in\pm P_{n}$. Neither of these is in general stable under the
adjoint action of $\mathfrak{k}$, but they may admit non-zero subspaces stable
under ad $\mathfrak{k}$; since $\mathfrak{h}\subset\mathfrak{k}$, such
subspaces are spans of root spaces for non-compact roots. A root $\beta$ is
said to be totally positive if $\mathfrak{g}_{\beta}$ is contained in a
subspace $\mathfrak{m}$ of $\mathfrak{p}^{+}$ stable under $\mathfrak{k}$; if
$\mathfrak{m}=\oplus_{\gamma\in R}\mathfrak{g}_{\gamma}$, then $\beta\in R$
and all roots in $R$ are also totally positive. Negatives of totally positive
roots are called totally negative. Their behavior is similar to the totally
positive roots because of the fact that there is an automorphism of
$\mathfrak{g}$ that is $-\mathrm{id}$ on $\mathfrak{h}$. Such an automorphism
will take $\mathfrak{g}_{\alpha}$ to $\mathfrak{g}_{-\alpha}$ for all roots
$\alpha$, in particular preserving $\mathfrak{k}$ and $\mathfrak{p}$. We write
$P_{t}$ for the set of totally positive roots and

$\mathfrak{p}^{\pm}_{t}=\oplus_{\beta\in P_{t}}\mathfrak{g}_{\pm\beta}$

Obviously $\mathfrak{p}_{t}$ is the largest ad $\mathfrak{k}$-stable subspace
of $\mathfrak{p}$; it may be $0$. We say that the positive system $P$ is
admissible if

$\mathfrak{p}^{\pm}_{t}\neq 0,\qquad\hbox{i.e.},\qquad\pm P_{t}\neq 0.$

For example in $A_{2}$, with root system
$\Delta=\pm\\{\alpha,\beta,\alpha+\beta\\}$, and compact roots $\pm\alpha$, we
have that $P=\\{\alpha,\beta,\alpha+\beta\\}$ is admissible
($P_{k}=\\{\alpha\\}$, $P_{n}=\\{\beta,\alpha+\beta\\}$), while
$P^{\prime}=\\{\alpha+\beta,-\beta,\alpha\\}$ is not admissible
($P^{\prime}_{k}=\\{\alpha\\}$, $P^{\prime}_{n}=\\{\alpha+\beta,-\beta\\}$),
see [25] Ch. VII, [27] and also [7], [8] for generalizations.

Our aim is to prove the following theorem, which reveals the significance of
total positivity for the problem of constructing infinite dimensional highest
weight $(\mathfrak{g},\mathfrak{k})$-modules.

######  Theorem 2.2.

Let $V$ be a non-zero highest weight $(\mathfrak{g},\mathfrak{k})$-module with
respect to a positive system $P$ of roots of $(\mathfrak{g},\mathfrak{h})$, of
highest weight $\lambda$. Then $\lambda(H_{\gamma})\in\mathbb{Z}_{\geq 0}$ for
all positive roots $\gamma$ which are not totally positive. In particular, if
$P$ is not admissible and $V$ is irreducible, then $V$ is finite dimensional.

The proof of this theorem is quite delicate and depends on the following
lemmas. Before stating them we need some preparation.

Let $v$ be a highest weight vector. For any positive root $\gamma$ let
$\mathfrak{a}_{\gamma}=\mathfrak{h}\oplus\mathfrak{g}_{\gamma}\oplus\mathfrak{g}_{-\gamma}$.
Let $Q$ be the set of all positive roots such that $v$ is
$\mathfrak{a}_{\gamma}$-finite and let $W_{Q}$ be the subgroup of the Weyl
group generated by the reflections $s_{\gamma}$ for $\gamma\in Q$. Let us also
write $W_{k}$ for the Weyl group of $\mathfrak{k}$. Since the vectors $w\in
V$, such that $w$ is $\mathfrak{a}_{\gamma}$-finite, form a
$\mathfrak{g}$-module, $Q$ is also the set of all roots $\gamma>0$ such that
$v$ is $\mathfrak{a}_{\gamma}$-finite. It is known that $v$ is
$\mathfrak{a}_{\gamma}$-finite if and only if $X^{r}v=0$ for $r>>0$, and that,
in this case, $\lambda(H_{\gamma})\in\mathbb{Z}_{\geq 0}$. If $\gamma\in Q$,
we can split $V$ as a direct sum of finite dimensional irreducible
$\mathfrak{a}_{\gamma}$-modules from which we conclude that the set of weights
of $V$ is stable under $W_{Q}$. Since $v$ is $\mathfrak{k}$-finite,
$P_{k}\subset Q$ and so $W_{k}\subset W_{Q}$. Suppose $\beta$ is totally
positive. Then, there is a subset $R$ of $P$ such that $R$ contains $\beta$
and $\oplus_{\gamma\in R}\mathfrak{g}_{\gamma}$ is stable under the adjoint
action of $\mathfrak{k}$. Hence $R$ is stable under $W_{k}$, showing that
$s\beta>0$ and in fact totally positive for all $s\in W_{k}$. The lemma below
goes in the other direction.

######  Lemma 2.3.

Let $\gamma$ be a positive non-compact root. If $s\gamma<0$ for some $s\in
W_{Q}$, then $\gamma\in Q$.

###### Proof.

The set of weights of $V$ is stable under $W_{Q}$ and so $s\lambda$ is a
weight of $V$. Hence

$\lambda-s\lambda=\sum_{1\leq i\leq\ell}n_{i}\beta_{i},\qquad\hbox{($n_{i}$
are integers $\geq 0$)}$

where $\\{\beta_{1},\dots,\beta_{\ell}\\}$ is the set of simple roots in $P$.
We shall show that $X^{p}_{-\gamma}v=0$ if $p>\mathrm{max}_{i}n_{i}$. Suppose
for some integer $p\geq 1$, we have $X^{p}_{-\gamma}v\neq 0$. Then
$\lambda-p\gamma$ is a weight of $V$. Write $s\gamma=-\beta$ where $\beta>0$.
Then $s\lambda+p\beta$ is a weight of $V$ so that
$\lambda-s\lambda-p\beta=\sum_{1\leq i\leq\ell}m_{i}\beta_{i}$ the
$m_{i}\in\mathbb{Z}_{\geq 0}$. Then $p\beta=\sum_{1\leq
i\leq\ell}(n_{i}-m_{i})\beta_{i}$; as $\beta>0$, we can also write
$\beta=\sum_{i}k_{i}\beta_{i}$ where the $k_{i}\in\mathbb{Z}_{\geq 0}$
$k_{i_{0}}>0$ for some $i_{0}$. We have $pk_{i_{0}}=n_{i_{0}}-m_{i_{0}}\leq
n_{i_{0}}$ giving $p\leq pk_{i_{0}}\leq n_{i_{0}}$. ∎

######  Lemma 2.4.

If $\gamma$ is a positive root that vanishes on the center $\mathfrak{c}$ of
$\mathfrak{k}$, then $s\gamma<0$ for some $s\in W_{k}$ and hence $\gamma\in
Q$. If $\gamma$ is totally positive, it cannot vanish on $\mathfrak{c}$ and
all $s\gamma>0$, even totally positive, for $s\in W_{k}$. In particular
$\mathfrak{c}\neq 0$ when totally positive roots exist.

###### Proof.

Let $\delta=\sum_{s\in W_{k}}s\gamma$ ($\delta$ need not be a root). Since
$\mathfrak{c}$ is fixed elementwise by $W_{k}$, we see that all $s\gamma$
vanish on $\mathfrak{c}$ and so $\delta$ must be $0$ on $\mathfrak{c}$. On the
other hand, $\delta$ is fixed by all elements of $W_{k}$ and so
$\delta(H_{\theta})=0$, for all $\theta\in P_{k}$. So $\delta$ must be 0 since
the $H_{\theta}$ and $\mathfrak{c}$ span $\mathfrak{h}$. But then a sum of
positive roots cannot be $0$ and so $s\gamma$ must be $<0$ for some $s\in
W_{k}$. By Lemma 2.3, we conclude that $\gamma\in Q$. If $\gamma$ is totally
positive, then $\mathfrak{g}_{\gamma}$ is contained in a sum $\mathfrak{m}$ of
root spaces contained in $\mathfrak{p}^{+}$ and stable under $\mathfrak{k}$,
so that all the root spaces $\mathfrak{g}_{s\gamma}$ are contained in
$\mathfrak{m}$; this shows that all $s\gamma>0$. The previous argument shows
that $\gamma$ cannot be 0 on $\mathfrak{c}$. ∎

Now we go to the proof. of Theorem 2.2.

###### Proof.

Let $v$ be a highest weight vector. We want to prove that if $\gamma>0$ is not
in $Q$, then $\gamma$ is totally positive. Since we know that $V$ is
$\mathfrak{k}$-finite, we have $P_{k}\subset Q$ and so we may assume that
$\gamma$ is a non-compact positive root. Let $\mathfrak{m}$ be the minimal ad
$\mathfrak{k}$-stable subspace of $\mathfrak{p}$ containing
$\mathfrak{g}_{\gamma}$. Now $\mathfrak{m}$ is spanned by the non-compact root
spaces, and as these are all one- dimensional, it follows from the complete
reducibility of the action of $\mathfrak{k}$ on $\mathfrak{p}$ that
$\mathfrak{m}$ is irreducible. Let $\beta_{0}$ and $\beta_{1}$ be the highest
and lowest roots belonging to $\mathfrak{m}$. If $\beta_{1}>0$ all the roots
belonging to $\mathfrak{m}$ will be $>0$, showing that $\gamma$ must be
totally positive. Hence it suffices to prove that $\beta_{1}>0$. Suppose to
the contrary that $\beta_{1}<0$. Now there is an element $s\in W_{k}$ such
that $s\beta_{0}=\beta_{1}<0$. So by Lemma 2.3, we know that $\beta_{0}\in Q$.
Let $\beta$ be a maximal positive root belonging to $\mathfrak{m}$, but which
is not in $Q$. Clearly $\beta\neq\beta_{0}$. Then there is a positive compact
root $\alpha$ such that
$[\mathfrak{g}_{\alpha},\mathfrak{g}_{\beta}]=\mathfrak{g}_{\alpha+\beta}\neq
0$ and so $\alpha+\beta$ is a root, which is positive. Moreover
$\alpha+\beta\in Q$. Clearly $\alpha+\beta$ is positive non-compact. Let
$\theta=\beta+(\alpha+\beta)=2\beta+\alpha$. We claim that $\theta$ is not a
root. If it were a root, it must be compact, and so $\theta$ and $\alpha$ are
both compact roots. Hence they both vanish on $\mathfrak{c}$, from which we
infer that $\beta=(1/2)(\theta-\alpha)$ must also vanish on $\mathfrak{c}$. By
Lemma 2.4 we have that $\beta\in Q$, a contradiction. Hence
$\beta+(\alpha+\beta)$ is not a root but $\beta-(\alpha+\beta)=-\alpha$ is a
root. Let $t\geq 1$ be the largest integer such that $\beta-t(\alpha+\beta)$
is a root. Then $s_{\alpha+\beta}(\beta-t(\alpha+\beta))=\beta$ or,
equivalently,

$s_{\alpha+\beta}\beta=\beta-t(\alpha+\beta)=-\alpha-(t-1)(\alpha+\beta)<0$

showing that $s_{\alpha+\beta}\beta<0$. But $\alpha+\beta\in Q$ and so, by
Lemma 2.3, $\beta\in Q$, a contradiction. This finishes the proof of the
theorem. ∎

###  2.4. Structure of the set of totally positive roots.

Recall that $P_{n}$ denotes the set of positive non compact roots in the
positive system $P=P_{k}\cup P_{n}$, while $\theta$ is the Cartan involution
of the complex semisimple Lie algebra $\mathfrak{g}$.

The basic result we want to prove is the following.

######  Theorem 2.5.

Let

$\mathfrak{g}_{t}=\mathfrak{k}_{t}\oplus\mathfrak{p}_{t},\quad\hbox{where}\quad\mathfrak{k}_{t}=[\mathfrak{p}_{t},\mathfrak{p}_{t}],\quad\mathfrak{g}_{1}=\mathfrak{g}^{\perp}_{t}.$

Then $\mathfrak{g}_{t}$, $\mathfrak{g}_{1}$ are ideals of $\mathfrak{g}$ which
are $\theta$-stable, and

$\mathfrak{g}=\mathfrak{g}_{t}\oplus\mathfrak{g}_{1}$

Moreover $P_{t}$ is precisely the set of positive non-compact roots of
$\mathfrak{g}_{t}$ and $P_{n}\setminus P_{t}$ is precisely the set of positive
non-compact roots of $\mathfrak{g}_{1}$.

Before its proof we need a lemma.

######  Lemma 2.6.

Let the notation be as above. Let $\mathfrak{q}_{1}$, $\mathfrak{q}_{2}$ be
two subspaces of $\mathfrak{p}$ stable under ad $\mathfrak{k}$. Suppose that
$\mathfrak{q}_{1}\perp\mathfrak{q}_{2}$. Then
$[\mathfrak{q}_{1},\mathfrak{q}_{2}]=0$.

###### Proof.

Let $B$ be the Cartan-Killing form of $\mathfrak{g}$. If $X\in\mathfrak{k}$,
$Y\in\mathfrak{q}_{1}$, $Z\in\mathfrak{q}_{2}$, then $B(X,[Y,Z])=B([X,Y],Z)=0$
because $[X,Y]\in\mathfrak{q}_{1}$, $Z\in\mathfrak{q}_{2}$. ∎

Now we go to the proof of Thm 2.5.

###### Proof.

Since $\mathfrak{p}_{t}$ is stable under ad $\mathfrak{k}$, it is immediate
that $\mathfrak{g}_{t}$ is stable under ad $\mathfrak{k}$. It is also obvious
that $\mathfrak{g}_{t}$ is stable under $\mathfrak{p}_{t}$. We must show that
it is stable under $\mathfrak{p}$. Let
$\mathfrak{q}=\mathfrak{p}^{\perp}_{t}$. Then $\mathfrak{q}$ is also stable
under ad $\mathfrak{k}$. By Lemma 2.3 we have
$[\mathfrak{q},\mathfrak{p}_{t}]=0$, hence also
$[\mathfrak{q},\mathfrak{k}_{t}]=0$, since
$\mathfrak{k}_{t}=[\mathfrak{p}_{t},\mathfrak{p}_{t}]$. So
$[\mathfrak{q},\mathfrak{g}_{t}]=0$. To prove that $\mathfrak{g}_{t}$ is
stable under ad $\mathfrak{p}_{t}$ is thus enough to show that
$\mathfrak{p}=\mathfrak{p}_{t}\oplus\mathfrak{q}$, or equivalently,
$\mathfrak{q}\cap\mathfrak{p}_{t}=0$. If $\mathfrak{g}_{\beta}$ belongs to
this intersection, then from $\mathfrak{g}_{\beta}\in\mathfrak{p}_{t}$, we get
$\beta=c\gamma$ where $c=\pm 1$ and $\gamma\in P_{t}$. On the other hand, as
$\mathfrak{g}_{\beta}\in\mathfrak{q}$, $\mathfrak{g}_{\beta}$ is orthogomal to
$\mathfrak{g}_{\pm\gamma}$. Hence
$\mathfrak{g}_{\beta}\perp\mathfrak{g}_{\pm\beta}$. This is impossible. Thus
we have proved that $\mathfrak{g}_{t}$ is an ideal. It is clearly
$\theta$-stable and so $\mathfrak{g}_{1}$ is stable under $\theta$ and is the
ideal complementary to $\mathfrak{g}_{t}$. The remaining assertions are now
obvious. ∎

######  Remark 2.7.

The condition $\mathfrak{k}_{t}=[\mathfrak{p}_{t},\mathfrak{p}_{t}]$ means
that $\mathfrak{g}_{t}$ has no ideal factors which are “compact”.

Because of this result, we can direct our attention to the case when
$P_{n}=P_{t}$, i.e., $\mathfrak{g}=\mathfrak{g}_{t}$; hence all non compact
positive roots are totally positive. Since $P_{n}=P_{t}$, we know that
$\mathfrak{p}^{+}_{t}$ is stable under ad $\mathfrak{k}$ and multiplicity free
(as the root spaces are one dimensional). Hence we have a unique (up to
ordering) decomposition

$\mathfrak{p}^{+}_{t}=\oplus_{1\leq i\leq s}\,\mathfrak{p}_{i}\qquad(1\leq
i\leq s)$

into irreducible modules for $\mathfrak{k}$. Define $\mathfrak{p}^{-}_{i}$ so
that the roots belonging $\mathfrak{p}_{i}^{-}$ are the negatives of the roots
belonging to $\mathfrak{p}_{i}^{+}$. Let $\beta_{i}$, $(1\leq i\leq s)$ be the
lowest root of $\mathfrak{p}^{+}_{i}$.

######  Theorem 2.8.

Let $P_{n}=P_{t}$ and let $\alpha_{1},\dots,\alpha_{r}$ be the simple roots of
$P_{k}$. Then, $\\{\alpha_{1},\dots,\alpha_{r},\beta_{1},\dots,\beta_{s}\\}$
is the set of simple roots of $P$. Moreover we have

  1. (a)

$r+s=$ rk $\mathfrak{h}$.

  2. (b)

$s$ = dim($\mathfrak{c}$) where $\mathfrak{c}$ is the center of
$\mathfrak{k}$, and the restrictions of the $\beta_{i}$ to $\mathfrak{c}$ are
linearly independent.

  3. (c)

If $\beta$ is a non-compact positive root, there is exactly one $i$, $(1\leq
i\leq s)$ and integers $m_{j}\geq 0$ such that

$\beta=\beta_{i}+m_{1}\alpha_{1}+\dots+m_{r}\alpha_{r}$

$i$, $m_{j}$ are uniquely determined by $\beta$. In particular
$[\mathfrak{p}^{+},\mathfrak{p}^{+}]=0$.

  4. (d)

$\mathfrak{g}$ decomposes as the sum of $s$ ideals
$\mathfrak{g}_{i}=\mathfrak{k}_{i}\oplus\mathfrak{p}_{i}$ which are
$\theta$-stable. Each of these have the property that all non-compact roots
are totally positive, $[\mathfrak{p}_{i},\mathfrak{p}_{i}]=\mathfrak{k}_{i}$,
and the dimension of the center of $\mathfrak{k}_{i}$ is $1$. In particular
each $\mathfrak{g}_{i}$ is simple.

  5. (e)

The integer $s$ is also the number of irreducible components of
$\mathfrak{p}^{+}_{t}$ as a $\mathfrak{k}$-module. In particular,
$\mathfrak{g}$ is simple if and only if $\mathfrak{p}^{+}_{t}$ is irreducible.

###### Proof.

Let $\beta_{i}$ be the lowest root (i.e. weight) of $\mathfrak{p}^{+}_{i}$.
Then, for any non-compact root $\beta$, there exits a unique $i$ such that
$\mathfrak{g}_{\beta}\subset\mathfrak{p}^{+}_{i}$, and $\mathfrak{g}_{\beta}$
can be reached by applying positive compact root vectors to
$\mathfrak{g}_{\beta_{i}}$. Hence, $\beta=\beta_{i}+m_{1}\alpha_{1}+\dots
m_{r}\alpha_{r}$, where the $m_{j}$ are integers $\geq 0$. Let
$S=\\{\alpha_{1},\dots,\alpha_{r},\beta_{1},\dots,\beta_{s}\\}$. Then, every
positive root is a non-negative integral linear combination of elements of
$S$. So $r+s\geq\ell$, where $\ell$ is the rank of $\mathfrak{g}$. On the
other hand, suppose $H\in\mathfrak{h}$ is an element such that all elements of
$S$ vanish at $H$. Then, $H$ must centralize $\mathfrak{k}$ and
$\mathfrak{p}$, hence also $\mathfrak{g}$. So $H=0$, showing that the elements
of $S$ are lineary independent. The same argument shows also that the
restrictions to $\mathfrak{c}$ of the $\beta_{i}$ are linearly independent. So
$S$ is the set of simple roots in $P$ and $i$ and the $m_{j}$ are unique. The
formula for the non-compact roots shows that the sum of two elements of
$P_{t}$ is never a root. The proofs of (a), (b), (c) are now clear.

We now take up (d) and (e). Let
$\mathfrak{p}_{i}=\mathfrak{p}^{+}_{i}\oplus\mathfrak{p}_{i}^{-}$. Clearly the
$\mathfrak{p}_{i}$ are mutually orthogonal and
$\mathfrak{p}=\oplus_{i}\mathfrak{p}_{i}$. Let
$\mathfrak{k}_{i}=[\mathfrak{p}_{i},\mathfrak{p}_{i}]$ and
$\mathfrak{g}_{i}=\mathfrak{k}_{i}\oplus\mathfrak{p}_{i}$. Then
$\mathfrak{g}_{i}$ is stable under $\theta$ as well as ad $\mathfrak{k}$ and
ad $\mathfrak{p}_{i}$. We now claim that for $i\neq j$,
$[\mathfrak{p}_{i},\mathfrak{p}_{j}]=0$. Certainly
$\mathfrak{p}_{i}\perp\mathfrak{p}_{j}$, because, if
$\beta_{r}\in\mathfrak{p}_{r}$, then $\beta_{i}\pm\beta_{j}\neq 0$ proving the
claim in view of Lemma 2.6. Hence $\mathfrak{g}_{i}$ is stable under all ad
$\mathfrak{p}_{j}$ $(j\neq i)$, hence under ad $\mathfrak{p}$. So the
$\mathfrak{g}_{i}$ are ideals in $\mathfrak{g}$. The results of (a)-(c) now
give (d) and (e). ∎

We also have the following result we use in the sequel.

######  Proposition 2.9.

Let the notation be as above.

  1. (a)

If $\beta$ is a totally positive root, then any root of the form
$\gamma=\beta+\sigma$, where $\sigma$ is an integral linear combination of the
compact roots is totally positive and lies in the irreducible ad
$\mathfrak{k}$-module generated by $\mathfrak{g}_{\beta}$.

  2. (b)

If $\gamma$ is totally positive and $\alpha$ is a compact root, the
$\alpha$-chain containing $\gamma$ has length $\leq 3$.

  3. (c)

If $\mathfrak{g}$ is simple then either no root is totally positive or all
positive roots are totally positive.

  4. (d)

If $\mathfrak{g}=\mathfrak{g}_{t}$ is simple, then $P$ and $P^{-}=P_{k}\cup-
P_{n}$ are the only two admissible positive systems containing $P_{k}$ a fixed
positive system for $\mathfrak{k}$. Otherwise, if $\mathfrak{g}_{i}$, $(1\leq
i\leq s)$ are the simple ideals of $\mathfrak{g}$, and
$\mathfrak{p}_{i}^{\pm}$ are as defined earlier, then the number of admissible
possible systems containing $P_{k}$ is $2^{s}$. A non-compact root belongs, to
exactly one of $\mathfrak{p}^{\pm}_{i}$.

###### Proof.

(a) We have the ideal decomposition
$\mathfrak{g}=\mathfrak{g}_{t}\oplus\mathfrak{g}_{1}$ and $\beta$ is a root of
$\mathfrak{g}_{t}$. If $\gamma$ is a root of $\mathfrak{g}_{1}$, it will
vanish on $\mathfrak{c}\cup\mathfrak{g}_{t}$ and so $\beta$ will also vanish
there, contradicting total positivity. Thus $\gamma$ is also a root of
$\mathfrak{g}_{t}$. Let us write $\sigma$ $(\sigma^{+})$ with or without
suffixes for integral (positive integral) linear combinations of the simple
compact roots of $\mathfrak{g}_{t}$. Then, using the notation of Theorem 2.8,
we have $\beta=\beta_{i}+\sigma_{1}^{+}$ so that $\gamma=\beta_{i}+\sigma_{2}$
on the one hand and $\gamma=\pm(\beta_{j}+\sigma_{3}^{+})$ on the other hand,
depending on whether $\gamma$ is positive or negative. Hence
$\beta_{i}=\pm(\beta_{j}+\sigma_{3}^{+})$. Restricting to the center of
$\mathfrak{k}_{t}$ and remembering that the restrictions of the $\beta_{m}$
are linearly independent, we see that we have to take the plus sign and $j=i$.
The conclusions of (a) now follow at once.

(b) We can take the $\alpha$-chain to be $\\{\gamma-p\alpha\\}$,
$(p=0,1,\dots,k)$ where $\gamma-k\alpha=s_{\alpha}\gamma$ so that
$k=\gamma(H_{\alpha})$. It is a question of proving that
$\gamma(H_{\alpha})\leq 2$. Suppose $\gamma(H_{\alpha})\geq 3$. Then
$\alpha(H_{\gamma})>0$ and so $\geq 1$, showing that
$m=\alpha(H_{\gamma})\gamma(H_{\alpha})\geq 3$. Consider the root
$\beta=s_{\gamma}s_{\alpha}\gamma=(m-1)\gamma-\gamma(H_{\alpha})\alpha$. Since
$m-1\geq 2$ this contradicts (c) of Theorem 2.8.

(c) If $\mathfrak{g}$ is simple, then $\mathfrak{g}=\mathfrak{g}_{t}$ or
$\mathfrak{g}=\mathfrak{g}_{1}$.

(d) Let $P^{\prime}$ be an admissible positive system. Then all roots of
$P^{\prime}_{n}$ belong to $\mathfrak{g}_{t}$. We may the assume that
$\mathfrak{g}=\mathfrak{g}_{t}$ and is simple. If $P_{t}^{\prime}$ contains an
element from $P^{\pm}_{n}$, then by the irreducibility of $\mathfrak{p}^{\pm}$
under $\mathfrak{k}$ we see that $P_{t}^{\prime}$ must contain all of
$P^{\pm}_{n}$. Thus $P^{\pm}$ are the only admissible positive systems
containing $P_{k}$. In the general case, let $\epsilon=(\epsilon_{i})$ be an
$s$-tuple of signs $\pm 1$ and let
$\mathfrak{q}^{\epsilon}=\oplus\mathfrak{p}_{i}^{\epsilon_{i}}$. Let
$P^{\epsilon}$ be the set of roots belonging to $\mathfrak{q}^{\epsilon}$. We
claim that $P^{\epsilon}$ is an admissible positive system. To prove that it
is a positive system, it is enough to find a point in $\mathfrak{h}$ at which
all the elements of $P^{\epsilon}$ are $>0$. Write $\beta_{i}^{+}=\beta_{i}$
for the lowest root in $\mathfrak{p}^{+}_{i}$; if $\gamma_{i}$ is the highest
root of $\mathfrak{p}^{+}_{i}$, then $\beta_{i}=-\gamma_{i}$ is the lowest
root of $\mathfrak{p}_{i}$, and it is a question of finding a point of
$\mathfrak{h}$ at which all of $P_{k}$ and all $\beta_{i}^{\epsilon_{i}}$ are
$>0$. Since the $\beta_{i}$ are linearly independent when restricted to
$\mathfrak{c}$ we can find a $C\in\mathfrak{c}$ such that
$\beta_{i}^{\epsilon_{i}}(C)>0$ for all $i$. On the other hand, we can find a
$U$ in the span $\mathfrak{h}^{\prime}$ of the $H_{\alpha}$ for the compact
roots such that $\alpha(U)>0$ for all $\alpha\in P_{k}$. Then, for $H=C+\eta
U$ for sufficiently small $\eta>0$ has the property that $\beta(H)>0$ for all
$\beta=\alpha\in P_{k}$, $\beta=\beta_{i}^{\epsilon_{i}}$. Obviously there are
no other admissible positive systems containing $P_{k}$. ∎

###  2.5. Harish-Chandra homomorphism

Let $\zeta$ be the center of $\mathcal{U}(\mathfrak{g})$ and $\mathcal{U}[0]$
the subalgebra of $\mathcal{U}(\mathfrak{g})$ commuting with $\mathfrak{h}$.
More generally, for any $\mu\in\mathfrak{h}$, let $\mathcal{U}[\mu]$ be the
subspace of $\mathcal{U}(\mathfrak{g})$ given by

$\mathcal{U}[\mu]=\\{a\in\mathcal{U}(\mathfrak{g})\,|\,[H,a]=\mu(H)a,\,\,\hbox{
for all}\,\,H\in h\\}$

Then $\mathcal{U}[0]$ is a subalgebra, $\zeta\subset\mathcal{U}[0]$, and
($\mathcal{U}[\mu]$) is a grading of $\mathcal{U}(\mathfrak{g})$; moreover
$\mathcal{U}[\mu]\neq 0$ if and only if $\mu$ is in the $\mathbb{Z}$-span of
the roots (the root lattice). If $\gamma_{1},\dots,\gamma_{t}$ is an
enumeration of the positive roots and $(H_{i})$ is a basis for $\mathfrak{h}$,
then elements of $\mathcal{U}[0]$ are linear combinations of

$X_{-\gamma_{1}}^{n_{1}}\dots H_{1}^{c_{1}}\dots
X_{\gamma_{1}}^{p_{1}}\dots\quad\hbox{with}\quad(p_{1}-n_{1})\gamma_{1}+\dots=0.$

It is then clear that every term occurring in such a linear combination must
necessarily have some $p_{i}>0$ except those that are just monomials in the
$H_{i}$ alone. So for any $u\in\mathcal{U}[0]$ we have an element
$\beta_{P}(u)=\beta(u)\in\mathcal{U}(\mathfrak{h})$ such that

(3)
$u\equiv\beta(u)\,(\hbox{mod}\mathcal{P}),\qquad\mathcal{P}:=\mathcal{U}(\mathfrak{g})\mathfrak{g}_{\gamma},\qquad\gamma\in
P$

The action of $u$ on the Verma module $V_{\lambda}$ must leave the weight
spaces stable since it commutes with $\mathfrak{h}$, and so applying it to the
highest weight vector $v_{\lambda}$ we see that
$uv_{\lambda}=\beta(u)(\lambda)v_{\lambda}$ where we are identifying
$\mathcal{U}(\mathfrak{h})$ with the algebra of all polynomials on
$\mathfrak{h}^{*}$ so that $\beta(u)(\lambda)$ makes sense. It follows from
this that, if $u\in\mathcal{U}(\mathfrak{h})\cap\mathcal{P}$, then
$u(\lambda)=0$ for all $\lambda$ and so $u=0$, i.e.,
$\mathcal{U}(\mathfrak{h})\cap P=0$. Hence $\beta(u)$ is uniquely determined
by the equation (3), and the map $u\mapsto\beta(u)$ is a homomorphism of
$\mathcal{U}[0]$ onto $\mathcal{U}(\mathfrak{h})$. Since
$\zeta\subset\mathcal{U}[0]$, we thus have a homomorphism of $\zeta$ into
$\mathcal{U}[\mathfrak{h}]$. This is the Harish-Chandra homomorphism (see [25]
Ch. VII). Harish-Chandra proved that $\beta$ is an isomorphism of $\zeta$ onto
the algebra of all elements of $\mathcal{U}(\mathfrak{h})$ invariant under a
certain (affine) action of the Weyl group $W$ on $\mathfrak{h}$. More
precisely, let $\delta=\delta_{P}=(1/2)\sum_{\alpha\in P}\alpha$, and for
$s\in W$ let $s_{A}\lambda=s(\lambda+\delta)-\delta$. Then $s\mapsto s_{A}$ is
an (affine) action of $W$ on $\mathfrak{h}^{*}$, and $\beta$ is an isomorphism
of $\zeta$ with $\mathcal{U}(\mathfrak{h})^{W}$. Now, for $z\in\zeta$,

$zv_{\lambda}=\beta(z)(\lambda)v_{\lambda}$

but since $z$ commutes with the actions of all elements of
$\mathcal{U}(\mathfrak{g})$ and so, as $v_{\lambda}$ is cyclic for
$V_{\lambda}$, we find that $zv=\beta(z)(\lambda)v$ on all $v\in V_{\lambda}$.
Thus

$z=\chi_{\lambda}(z)I,\quad\hbox{on}\quad
V_{\lambda},\qquad\chi_{\lambda}(z):=\beta(z)(\lambda).$

It follows from this that, if $\lambda,\mu\in\mathfrak{h}^{*}$ are such that
$\chi_{\lambda}(z)=\chi_{\mu}(z)$ for all $z\in\zeta$, then for some $s\in W$,
we must have $s_{A}\lambda=\mu$. A consequence of this is the following: if
$U_{1}$, $U_{2}$ are highest weight modules with highest weights $\mu_{1}$,
$\mu_{2}$ respectively, and if there is a non zero morphism
$U_{2}\longrightarrow U_{1}$, then there is an element $s\in W$ such that
$\mu_{2}+\delta=s(\mu_{1}+\delta)$. Indeed, if $z\in\zeta$, then $z$ acts as
$\chi_{\mu_{2}}(z)$ on $U_{2}$ and by $\chi_{\mu_{1}}(z)$ on $U_{1}$ and these
two numbers must be the same for all $z\in\zeta$. This gives the required
result. The Harish-Chandra homomorphism on $\zeta$ depends on the choice of
the positive system $P$. Let

$\gamma(z)(\lambda)=\beta_{P}(z)(\lambda-\delta_{P})\qquad(z\in\zeta).$

It is then easy to show that $\gamma(z)$ is independent of $P$, and that
$\gamma$ is an isomorphism of $\zeta$ with the algebra
$\mathcal{U}(\mathfrak{h})^{W}$ of all elements of $\mathcal{U}(\mathfrak{h})$
invariant under the usual linear action of $W$ on $\mathfrak{h}$ (see [25] Ch.
VII).

###  2.6. Converse to Theorem 2.2

We want to construct highest weight $(\mathfrak{g},\mathfrak{h})$-modules when
$\lambda$ satisfies the condition of Theorem 2.2. In view of the splitting
$\mathfrak{g}=\mathfrak{g}_{t}\times\mathfrak{g}_{1}$, it is enough to
consider the case of $\mathfrak{g}_{t}$, since we can tensor with finite
dimensional modules for $\mathfrak{g}_{1}$.

Let $\lambda\in\mathfrak{h}^{*}$ be such that $\lambda(H_{\alpha})$ is an
integer $\geq 0$ for all $\alpha\in P_{k}$. Let $F=F_{\lambda}$ be the
irreducible finite dimensional module for $\mathfrak{k}$ of highest weight
$\lambda$. Note that $\lambda(H_{\beta})$ can be arbitrary for positive non-
compact roots $\beta$. Write
$\mathfrak{q}=\mathfrak{k}\oplus\mathfrak{p}^{+}$. Recall that
$[\mathfrak{k},\mathfrak{p}^{+}]\subset\mathfrak{p}^{+}$ and so we can turn
$F$ into a left $\mathfrak{q}$-module by letting $\mathfrak{p}^{+}$ act
trivially.

Define

$U^{\lambda}=\mathcal{U}(\mathfrak{g})\otimes_{\mathcal{U}(\mathfrak{q})}F$

and view $U^{\lambda}$ as a $\mathcal{U}(\mathfrak{g})$-module by left action:

$a(b\otimes f)=ab\otimes f.$

Let

$\delta=(1/2)\sum_{\gamma\in P}\gamma.$

######  Theorem 2.10.

$U^{\lambda}$ is the universal HC module of highest weight $\lambda$. If

$(\lambda+\delta)(H_{\gamma})\quad\hbox{is real and}\quad\leq 0$

for all $\gamma\in P_{n}$, then $U^{\lambda}$ is irreducible.

###### Proof.

Let $M\subset U^{\lambda}$ be a nonzero submodule of $U^{\lambda}$. Since the
weights of $M$ are $\leq\lambda$ we can choose a maximal one, say $\mu$; if
$u$ is a corresponding weight vector, $X_{\gamma}u=0$ for all $\gamma\in P$
and so $\mathcal{U}(\mathfrak{g})u$ is a highest weight module of highest
weight $\mu$. From the properties of the infinitesimal character and the
Harish-Chandra homomorphism this implies that $\mu+\delta=s(\lambda+\delta)$
for some $s\in W$. The condition on $\lambda$ can be rewritten as
$(\lambda+\delta)(H_{\gamma})\geq 0$ for all $\gamma\in P=P_{k}\cup(-P_{n})$.
Let $s_{0}$ be the element of $W_{k}$ that takes $P_{k}$ to -$P_{k}$. Then
$s_{0}P=-P_{k}\cup P_{n}$ and so $P^{-}=-s_{0}P$ is also a positive system.
Let $\\{\alpha_{1},\dots,\alpha_{r},\beta_{1},\dots,\beta_{r}\\}$ be the
simple system of $P$and let $\gamma_{j}=s_{0}\beta_{j}$. Then the $\gamma_{j}$
are also in $P_{t}$ and in fact $\gamma_{j}$ is the highest root of the
$\mathfrak{k}$-module $\mathfrak{p}^{+}_{j}$ of which $\beta_{j}$ is the
lowest root. Hence $\gamma_{j}=\beta_{j}+\sigma_{j}$ where $\sigma_{j}$ is a
sum of positive compact roots.

We claim that
$\\{\alpha_{1},\dots,\alpha_{r},-\gamma_{1},\dots,-\gamma_{r}\\}$ is the
simple system of $P^{-}$. Let $\gamma\in P_{n}$. Then
$s_{0}\gamma=b_{1}\alpha_{1}+\dots+b_{r}\alpha_{r}+e_{1}\beta_{1}+\dots+e_{s}\beta_{s}$
where the $b_{i}$, $e_{j}$ are integers $\geq 0$. Hence applying $-s_{0}$, we
get
$-\gamma=b_{1}^{\prime}\alpha_{1}+\dots+b_{r}^{\prime}\alpha_{r}-e_{1}\gamma_{1}-\dots-
e_{s}\gamma_{s}$ where the $b_{j}^{\prime}$ are integers $\geq 0$. This proves
the claim.

Since $(\lambda+\delta)(H_{\gamma})\geq 0$ for all $\gamma\in P^{-}$ we can
write

$(\lambda+\delta)-s(\lambda+\delta)=\sum_{1\leq i\leq
r}c_{i}\alpha_{i}-\sum_{1\leq j\leq s}d_{j}\gamma_{j}$

where $c_{i},d_{j}\geq 0$. Thus

$\lambda-\mu=(\lambda+\delta)-s(\lambda+\delta)=\sum_{1\leq i\leq
r}c_{i}\alpha_{i}-\sum_{1\leq j\leq s}d_{j}\gamma_{j}$

But

$\lambda-\mu=\sum_{1\leq i\leq r}a_{i}\alpha_{i}+\sum_{1\leq j\leq
s}b_{j}\beta_{j}$

where the $a_{i},bj$ are integers $\geq 0$. Hence, writing
$\gamma_{j}=\beta_{j}+\sigma_{j}$ as above and restricting to $\mathfrak{c}$
we get

$\sum_{1\leq j\leq s}(d_{j}+f_{j})(\beta_{j})|_{\mathfrak{c}}=0$

Since the $(\beta_{j})|_{\mathfrak{c}}$ are linearly independent by (b) of
Theorem 2.8 we have $d_{j}+f_{j}=0$ for all $j$, and hence, as the
$d_{j},f_{j}$ are $\geq 0$, we must have $d_{j}=f_{j}=0$, for all $j$. Hence

$\lambda-\mu=\sum_{1\leq i\leq r}a_{i}\alpha_{i}$

where the $a_{i}$ are integers $\geq 0$.

Now $u$ is a linear combination of $X_{-\gamma_{1}}\dots X_{-\gamma_{m}}v$
where each $\gamma_{j}$ is in $\\{\alpha_{1},\dots,\alpha_{r},$
$\beta_{1},\dots\beta_{s}\\}$ and

$\lambda-\mu=\gamma_{1}+\dots+\gamma_{m}.$

Writing each $\gamma_{j}$ as a linear combination of $\alpha_{i}$ $(1\leq
i\leq r)$ and the $\beta_{j}$ $(1\leq j\leq s)$ with integer coefficients
$\geq 0$, and noting that $\lambda-\mu$ does not involve the $\beta_{j}$, we
conclude that each $\gamma_{j}$ does not involve any $\beta_{j}$. In other
words, $u\in\mathcal{U}(\mathfrak{k})v$. But then $u$ must be a multiple of
$v$, showing that $M=U^{\lambda}$. This proves that $U^{\lambda}$ is already
irreducible. ∎

We shall now study the structure of $U^{\lambda}$ as a $\mathfrak{q}$-module
for arbitrary $\lambda$ with $\lambda(H_{\alpha})$ an integer $\geq 0$, for
all $\alpha\in P_{k}$. For this we need a standard lemma.

######  Lemma 2.11.

Let $g$ be a field and $A$, $B$ algebras over $g$. Suppose $B\subset A$, $A$
is a free right $B$-module, $F$ a left $B$-module, and $V=A\otimes_{B}F$. If
$(a_{i})$ is a free basis for $A$ as a right $B$-module, and
$L=\sum_{i}g.a_{i}$, then the map taking $l\otimes_{g}f$ to $l\otimes_{B}f$ is
a linear isomorphism of $L\otimes_{g}F$ with $V$.

###### Proof.

This is standard but we give a proof. All symbols $\otimes$ without any suffix
mean tensor products over the field $g$. Let $(b_{j})$ be a $g$-basis for $B$
with $b_{0}=1$. Then $V$ is a quotient of $A\otimes F$ by the span $S$ of
elements of the form $ab\otimes f-a\otimes bf$ where $a\in A$, $b\in B$, $f\in
F$. Let $(f_{k})$ be a $g$-basis for $F$. We assert that $S$ is spanned by
$a_{i}b_{j}\otimes f_{k}-a_{i}\otimes b_{j}f_{k}$. Indeed, $S$ is spanned by
$a_{i}b_{j}b\otimes f_{k}-a_{i}b_{j}\otimes bf_{k}$. Now

$a_{i}b_{j}b\otimes f_{k}-a_{i}b_{j}\otimes bf_{k}=(a_{i}b_{j}b\otimes
f_{k}-a_{i}\otimes b_{j}bf_{k})-(a_{i}b_{j}\otimes bf_{k}-a_{i}\otimes
b_{j}bf_{k}).$

Expressing $b_{j}b$ in the terms of the first group as a linear combination of
the $br$ and the $bf_{k}$ of the second group in terms of the $f_{l}$, we see
that our assertion is proved. Note that we only need the terms with $j\neq 0$
as $a_{i}b_{j}\otimes f_{k}-a_{i}\otimes b_{j}f_{k}=0$ for $j=0$. Since the
map $L\otimes F\longrightarrow V$ is obviously surjective, it is enough to
show that the linear span of the $a_{i}\otimes f_{k}$ has $0$ intersection
with $S$. Suppose $\sum_{i,k}D_{ik}a_{i}\otimes f_{k}\in S$. Then we can write

$\sum_{i,j,k,j\neq 0}C_{ijk}(a_{i}b_{j}\otimes f_{k}-a_{i}\otimes
b_{j}f_{k})=\sum_{i,k}D_{ik}a_{i}\otimes f_{k}.$

Since $b_{j}f_{k}$ is a linear combination of the $f_{r}$ it follows that

$\sum_{i,j,k,j\neq 0}C_{ijk}a_{i}b_{j}\otimes f_{k}=\sum E_{ir}a_{i}\otimes
f_{r}.$

This means that $C_{ijk}=0$ for all $i,j,k$ with $j\neq 0$, hence that
$\sum_{i,k}D_{ik}a_{i}\otimes f_{k}=0$. This proves the lemma. ∎

We regard $\mathcal{U}(\mathfrak{p}^{-})\otimes F$ as a
$\mathcal{U}(\mathfrak{p}^{-})$-module by $a,b\otimes f\mapsto ab\otimes f$.
Since is stable under ad $\mathfrak{k}$ we may view
$\mathcal{U}(\mathfrak{p}^{-})\otimes F$ as a $\mathfrak{k}$-module also.

######  Corollary 2.12.

The map $\phi:a\otimes f\mapsto a\otimes U(\mathfrak{q})f$ is a linear
isomorphism of $\mathcal{U}(\mathfrak{p}^{-})\otimes F$ with $U^{\lambda}$
that intertwines the actions of $\mathcal{U}(\mathfrak{p}^{-})$ and
$\mathcal{U}(\mathfrak{k})$. In particular, $U^{\lambda}$ is a free
$\mathcal{U}(\mathfrak{p}^{-})$-module with basis
$1\otimes\mathcal{U}(\mathfrak{q})f_{j}$ where $(f_{j})$ is a basis for $F$.

###### Proof.

Since $\mathfrak{g}=\mathfrak{p}^{-}\oplus\mathfrak{q}$ it follows that
$a\otimes b\mapsto ab$ is a linear isomorphism of
$\mathcal{U}(\mathfrak{p}^{-})\otimes\mathcal{U}(\mathfrak{q})$ with
$\mathcal{U}(\mathfrak{g})$. It is clear from this that
$\mathcal{U}(\mathfrak{g})$ is a free right
$\mathcal{U}(\mathfrak{q})$-module, and that any basis of
$\mathcal{U}(\mathfrak{p}^{-})$ is a free right
$\mathcal{U}(\mathfrak{q})$-basis for $\mathcal{U}(\mathfrak{g})$.

Lemma 2.11 now applies and shows that $\phi$ is an isomorphism. It obviously
commutes with the action of $\mathcal{U}(\mathfrak{p}^{-})$. The verification
of the commutativity with respect to $\mathfrak{k}$ is also straightforward. ∎

######  Remark 2.13.

This gives a formula for the multiplicity for the weight spaces
$U^{\lambda}[\mu]$ of $U^{\lambda}$. Let $\lambda_{0}=\lambda$ and
$\lambda_{i}$, $(0\leq i\leq r)$ be the weights of $F$ with $k_{i}$ as the
multiplicity of $\lambda_{i}$. Let $\gamma_{1},\dots,\gamma_{q}$ be the
distinct totally positive roots. For any linear function $\nu$ on
$\mathfrak{h}$ let $N(\nu)$ be the number of distinct ways of writing
$\nu=m_{1}\gamma_{1}+\dots+m_{q}\gamma_{q}$ where the $m_{j}$ are integers
$\geq 0$. Then

$\mathrm{dim}U^{\lambda}[\mu]=\sum_{0\leq i\leq r}k_{i}\,N(\mu_{i}-\lambda).$

######  Remark 2.14.

There is a criterion for the Verma module $V^{\lambda}$ to be irreducible,
namely that

$(\lambda+\delta)(H_{\gamma})\in/\\{1,2,\dots\\}\qquad\hbox{for
all}\,\gamma\in P.$

This is due to M. Duflo [6] and it is a variant of the similar condition for
the spherical principal series for a complex group to be irreducible, due to
K. R. Parthasarathy, R. Ranga Rao and V. S. Varadarajan [30].

###  2.7. Totally positive roots, real HC pairs, and complex geometry

In practice the HC pairs arise by complexification of real HC pairs. Let
$(\mathfrak{g}_{0},\mathfrak{k}_{0})$ be a real HC pair, $\mathfrak{g}_{0}$
simple. We assume that $G_{0}$ is a connected real Lie group with Lie algebra
$\mathfrak{g}_{0}$ and $K_{0}$ is the analytic subgroup defined by
$\mathfrak{k}_{0}$. We also assume that:

  1. (1)

Ad $K_{0}$ is the maximal compact subgroup of Ad $G_{0}$;

  2. (2)

$G_{0}$ and $K_{0}$ have the same rank:

$\mathfrak{h}_{0}\subset\mathfrak{k}_{0}\subset\mathfrak{g}_{0}$

where $\mathfrak{h}_{0}$ is a CSA for both.

Let $A_{0}$ be the Cartan subgroup (CSG) of $\mathfrak{h}_{0}$ in $G_{0}$ so
that it is centralizer of $\mathfrak{h}_{0}$ in $G_{0}$. So Ad $A_{0}$ is
compact and the roots of $(\mathfrak{g},\mathfrak{k})$ are the eigencharacters
of Ad $A_{0}$. Thus all roots are pure imaginary on $\mathfrak{h}_{0}$. Now,
$\mathfrak{g}_{0}$ being a real form of $\mathfrak{g}$, there is a conjugation
$X\longrightarrow X^{\mathrm{conj}}$ on $\mathfrak{g}$, which is an antilinear
bijection of $\mathfrak{g}$ with itself preserving brackets, with
$\mathfrak{g}_{0}$ as the set of its fixed points. Since it is the identity on
$\mathfrak{h}_{0}$, it follows by conjugating
$[H,X_{\beta}]=\beta(H)X_{\beta}$ that
$\mathfrak{g}_{\beta}^{\mathrm{conj}}=\mathfrak{g}_{-\beta}$. In the above
setting $(\mathfrak{g}_{0},\mathfrak{k}_{0})$ is a HC pair but $K_{0}$ need
not be semisimple, i.e., the center $\mathfrak{c}_{0}$ may be $\neq 0$. If
$\mathfrak{k}_{0}$ has zero center, the group $K_{0}$ is compact even when
$G_{0}$ is the simply connected group corresponding to $\mathfrak{g}_{0}$ and
$G_{0}$ has finite center.

######  Lemma 2.15.

We have $[\mathfrak{p}_{0},\mathfrak{p}_{0}]=\mathfrak{k}_{0}$ and
$\mathfrak{p}_{0}$ is irreducible as a $\mathfrak{k}_{0}$ -module. Moreover
the $H_{\beta}$ for noncompact roots $\beta$ span $i\mathfrak{h}_{0}$ over
$\mathbb{R}$.

###### Proof.

The first statement follows from the fact that
$[\mathfrak{p}_{0},\mathfrak{p}_{0}]\oplus\mathfrak{p}_{0}$ is a nonzero ideal
in $\mathfrak{g}_{0}$ and so has to be $\mathfrak{g}_{0}$. Indeed, since
$\mathfrak{p}_{0}$ is stable under $\mathfrak{k}_{0}$, so is
$[\mathfrak{p}_{0},\mathfrak{p}_{0}]$, so that stability under
$\mathfrak{k}_{0}$ is clear; the stability under $\mathfrak{p}_{0}$ is
obvious. Since $[\mathfrak{p},\mathfrak{p}]=\mathfrak{k}$ is spanned by
$[X_{\beta},X_{\gamma}]$ for $\beta,\gamma$ noncompact, it follows that
$\mathfrak{h}$ is spanned by the $[X_{\beta},X_{-\beta}]$ for noncompact
$\beta$ which proves the last statement. Note that as all roots are pure
imaginary on $\mathfrak{h}_{0}$, $H_{\gamma}\in i\mathfrak{h}_{0}$ for all
roots $\gamma$. The irreducibility of $\mathfrak{p}_{0}$ is however less
trivial. The Killing form is positive definite on $\mathfrak{p}_{0}$ and
negative definite on $\mathfrak{k}_{0}$, and is $\mathfrak{g}_{0}$ invariant.
We write $\mathfrak{p}_{0}=\oplus_{1\leq j\leq r}\mathfrak{p}_{j}$ where the
$\mathfrak{p}_{j}$ are mutually orthogonal and irreducible for
$\mathfrak{k}_{0}$. Let
$\mathfrak{k}_{j}:=[\mathfrak{p}_{j},\mathfrak{p}_{j}]$,
$\mathfrak{g}_{j}:=\mathfrak{k}_{j}\oplus\mathfrak{p}_{j}$. We shall prove
that the $\mathfrak{g}_{j}$ are ideals in $\mathfrak{g}_{0}$ and that
$\mathfrak{g}_{0}=\oplus_{1\leq j\leq r}\mathfrak{g}_{j}$. If this is so, then
$j=1$ and we are done. Fix $j$, $k$ with $1\leq j,k\leq r$ and $j\neq k$. We
shall prove the following in succession.

A. $[\mathfrak{p}_{j},\mathfrak{p}_{k}]=0$,
$[\mathfrak{k}_{j},\mathfrak{p}_{k}]=0$,
$[\mathfrak{k}_{j},\mathfrak{k}_{k}]=0$, $j\neq k$.

Let $X\in\mathfrak{k}_{0}$ , $Y_{i}\in\mathfrak{p}_{i}$, $i=j,k$. Then
$B(X,[Y_{j},Y_{k}])=B([X,Y_{j}],Y_{k})=0$ as $[X,Y_{j}]\in\mathfrak{p}_{j}$.
Since this is true for all $X\in\mathfrak{k}_{0}$, we must have
$[Y_{j},Y_{k}]=0$. If $Y_{j}$, $Z_{j}\in\mathfrak{p}_{j}$,
$Y_{k}\in\mathfrak{p}_{k}$ and $X=[Y_{j},Z_{j}]$, then
$[X,Y_{k}]=-[[Z_{j},Y_{k}],Y_{j}]-[[Y_{k},Y_{j}],Z_{j}]=0$ by the previous
result.

If $Y_{j}$, $Z_{j}\in\mathfrak{p}_{j}$, $X_{k}\in\mathfrak{k}_{k}$, then

$[[Y_{j},Z_{j}],X_{k}]=-[[Z_{j},X_{k}],Y_{j}]-[[X_{k},Y_{j}],Z_{j}]=0$

by the previous result.

B. $\mathfrak{k}_{j}\perp\mathfrak{k}_{k}$

Let $Y_{j}$, $Z_{j}\in\mathfrak{p}_{j}$, $X_{k}\in\mathfrak{k}_{k}$. Then
$B([Y_{j},Z_{j}],X_{k})=-B(Z_{j},[Y_{j},X_{k}])=0$ since $[Y_{j},X_{k}]=0$. So
$\mathfrak{g}_{1},\dots,\mathfrak{g}_{r}$ are mutually orthogonal and
$[\mathfrak{g}_{j},\mathfrak{g}_{k}]=0$ for $j\neq k$. Since $B$ is separately
definite on $\mathfrak{k}_{0}$ and $\mathfrak{p}_{0}$, the $\mathfrak{k}_{j}$
and $\mathfrak{p}_{j}$ are linearly independent among themselves, from which
it follows easily that the $\mathfrak{g}_{j}$ are linearly independent. We now
claim that the $\mathfrak{g}_{j}$ are subalgebras. Since

$[\mathfrak{k}_{j},\mathfrak{p}_{j}]\subset[\mathfrak{k},\mathfrak{p}_{j}]\subset\mathfrak{p}_{j}$

and $[\mathfrak{p}_{j},\mathfrak{p}_{j}]\subset\mathfrak{k}_{j}$, it is enough
to verify that $[\mathfrak{k}_{j},\mathfrak{k}_{j}]\subset\mathfrak{k}_{j}$;
but $[\mathfrak{k}_{0},\mathfrak{p}_{j}]\subset\mathfrak{p}_{j}$ so that
$[\mathfrak{k}_{0},\mathfrak{k}_{j}]\subset\mathfrak{k}_{j}$, and
$[\mathfrak{k}_{j},\mathfrak{k}_{j}]\subset[\mathfrak{k}_{0},\mathfrak{k}_{j}]\subset\mathfrak{k}_{j}$.
As $[\mathfrak{p}_{0},\mathfrak{p}_{0}]=\mathfrak{k}_{0}$,
$[\mathfrak{p}_{j},\mathfrak{p}_{k}]=0$ $(j\neq k)$, we have
$\mathfrak{k}_{0}=\mathfrak{k}_{1}\oplus\dots\oplus\mathfrak{k}_{r}$, hence
$\mathfrak{g}_{0}=\mathfrak{g}_{1}\oplus\dots\oplus\mathfrak{g}_{r}$. The
$\mathfrak{g}_{j}$ are subalgebras and $[\mathfrak{g}_{j},\mathfrak{g}_{k}]=0$
for $j\neq k$. Hence the $\mathfrak{g}_{j}$ are ideals. ∎

######  Lemma 2.16.

Let $\mathfrak{g}_{0}$ be simple. Then $\mathfrak{c}_{0}$ is either 0 or has
dimension 1. In the latter case we can find a $J\in\mathfrak{c}_{0}$, unique
up to a sign, such that ad($J)^{2}=-I$ on $\mathfrak{p}_{0}$. Moreover
$\beta(iJ)=\pm 1$ for all non compact roots $\beta$.

###### Proof.

Suppose $\mathfrak{c}_{0}\neq 0$. Since $\mathfrak{c}_{0}$ is compact, the
action of $\mathfrak{c}_{0}$ on $\mathfrak{p}_{0}$ is completely reducible
with pure imaginary eigenvalues. The eigenvalues are of the form 0 or $\pm
im_{j}$ where $m_{j}$ are real non zero linear functions on
$\mathfrak{h}_{0}$. The eigenspaces are stable under $\mathfrak{k}_{0}$ and so
$\mathfrak{p}_{0}$ is a direct sum of $\mathfrak{p}_{j}$ where the
$\mathfrak{p}_{j}$ are stable under $\mathfrak{k}_{0}$ and the eigenvalues on
$\mathbb{C}\cdot\mathfrak{p}_{j}$ are either $0$ or $\pm im_{j}$. By Lemma
2.15, there is only one of the $\mathfrak{p}_{j}$. If 0 is an eigenvalue, then
$\mathfrak{c}_{0}$ centralizes $\mathfrak{p}_{0}$, hence $\mathfrak{g}_{0}$.
This is not possible and so the eigenvalues of $\mathfrak{c}_{0}$ are $\pm
im$. If dim$(\mathfrak{c}_{0})>1$ we can find a nonzero $H\in\mathfrak{c}_{0}$
such that $m(H)=0$ so that $H$ centralizes $\mathfrak{g}_{0}$, which is not
possible. So dim($\mathfrak{c}_{0}$ ) = 1. The existence of $J$ and its
uniqueness are now obvious. Since $J$ has eigenvalues $\pm i$ on
$\mathfrak{p}$, and $\mathfrak{p}$ is spanned by the $X_{\beta}$ for non
compact roots $\beta$, we have $\beta(iJ)=\pm 1$ for all $\beta$. ∎

######  Remark 2.17.

The tangent space to $S_{0}:=G_{0}/K_{0}$ at $\overline{1}$, the image of 1 in
the canonical map $G_{0}\longrightarrow G_{0}/K_{0}$, is isomorphic to
$\mathfrak{p}_{0}$ as a $K_{0}$ module. Thus $\pm
J\in{\mathrm{End}}(\mathfrak{p}_{0})$ and satisfies $(\pm J)^{2}=-I$ and is
fixed under the adjoint action of $K_{0}$ on $\mathfrak{p}_{0}$, hence defines
$K_{0}$ -invariant sections $\pm J$ of the endomorphism bundle of the tangent
bundle of $S_{0}$, satisfying $(\pm J)^{2}=-I$. Thus we have defined two
canonical almost complex structures on $S_{0}$. We shall see later that these
are actually integrable and so define $G_{0}$-invariant complex structures on
$S_{0}$.

We also notice that the two canonical complex structures are in bijection with
the admissible simple systems of $\mathfrak{g}$; this fact is true more in
general, for a non necessarily maximal compact $K_{0}$, as detailed in [1] and
refs. therein. In this case, in fact, the complex structures on $G_{0}/K_{0}$
are in bijection with simple systems of generalized root systems, as studied
by Kostant in [27] and later on in [8], [9].

We now state a result, which is a refinement of Theorem 2.8 in this context.

######  Proposition 2.18.

Let the notation be as above.

  1. (1)

If $\mathfrak{c}_{0}=0$ there are no admissible positive systems.

  2. (2)

If dim($\mathfrak{c}_{0}$ ) = 1, there are exactly two admissible positive
systems containing a given positive system $P_{k}$ of compact roots. For a
fixed choice of $J$ these are $P^{\pm}=P_{k}\cup\pm Q$ where $Q$ is the $P$
set of non compact roots taking the value 1 at $iJ$. The subspaces
$\mathfrak{p}^{\pm}:=\sum_{\beta\in\pm Q}\mathfrak{g}_{\beta}$ are stable
under $\mathfrak{k}$, abelian, and irreducible.

  3. (3)

Finally, if $\\{\alpha_{1},\dots,\alpha_{\ell-1}\\}$ are the simple roots in
$P_{k}$ , we can find a unique non compact root $\beta$ such that
$\\{\alpha_{1},\dots,\alpha_{\ell-1},\beta\\}$ is the set of simple roots of
$P^{+}$, $\\{\alpha_{1},\dots,\alpha_{\ell-1},-\beta\\}$ being then the simple
roots in $P^{-}$.

###### Proof.

(1) Assume first that $\mathfrak{c}_{0}=0$. Let $P$ be a positive system and
let $R\subset P$ be a non empty set such that $L=\sum_{\beta\in
R}\mathfrak{g}_{\beta}$ is $\mathfrak{k}$-stable. If $W_{k}$ is the subgroup
of the Weyl group generated by the compact roots, it is then clear that $R$,
the set of weights for $\mathfrak{k}$ in $L$, is stable under $W_{k}$. Fix a
$\beta\in R$ and let $\lambda=\sum_{s\in W_{k}}s\cdot\beta$. Then $\lambda$ is
invariant under $W_{k}$, and the condition $s_{\alpha}\lambda=\lambda$ for all
compact roots $\alpha$ gives $\lambda(H_{\alpha})=0$ for all compact roots
$\alpha$, so that $\lambda=0$, since the $H_{\alpha}$ span $\mathfrak{h}$.
This is impossible since all the $s\cdot\beta$ are $>0$ and so their sum
cannot be 0.

(2) Let us now suppose that dim($\mathfrak{c}_{0}$) = 1. If $\alpha$ is a
compact root and $\beta\in\pm P_{n}$ it is immediate that $\beta+\alpha$ and
$\beta$ have the same value $\pm 1$ at $iJ$. If $\beta+\alpha$ is a root, it
must be non compact and so is in $\pm P^{n}$. Hence $[X_{\alpha},X_{\beta}]$
is either 0 or in $\mathfrak{p}^{\pm}$, showing that $\mathfrak{p}^{\pm}$ are
stable under $\mathfrak{k}$. If $\beta,\gamma$ are both in $\pm P_{n}$,
$\beta+\gamma$ takes the value $\pm 2$ at $iJ$ and so is not a root, showing
that $\mathfrak{p}^{\pm}$ are abelian. Now
$\mathfrak{g}^{\mathrm{conj}}_{\beta}=\mathfrak{g}-\beta$ and so
$\mathfrak{p}^{-}=(\mathfrak{p}^{+})^{\mathrm{conj}}$. So, if there is a
proper subspace $\mathfrak{q}$ of $\mathfrak{p}^{+}$ stable under
$\mathfrak{k}$, then
$(\mathfrak{q}+\mathfrak{q}^{\mathrm{conj}})\cap\mathfrak{p}_{0}$ is a proper
subspace of $\mathfrak{p}_{0}$ stable under $\mathfrak{k}_{0}$ , which is not
possible since $\mathfrak{p}_{0}$ is irreducible. So $\mathfrak{p}^{\pm}$ are
irreducible.

(3) If the positive system $P^{\prime}$ contains $P_{k}$ and is admissible,
there is a non empty subset $R\subset P^{n}$ such that $L:=\sum_{\beta\in
R}\mathfrak{g}_{\beta}$ is stable under $\mathfrak{k}$. If $\beta\in R$, then
$\beta\in P^{\pm}$ and so $L\cap\mathfrak{p}^{\pm}\neq 0$ showing that $L$
contains one of $\mathfrak{p}^{\pm}$, hence equal to one of them. Hence
$P^{\prime}=P^{\pm}$. To find the simple roots of $\mathfrak{p}^{+}$, let
$\beta$ be the lowest weight (relative to $P_{k}$) for $\mathfrak{p}^{+}$ as a
$\mathfrak{k}$-module. If $\gamma$ is a non compact root in $P_{n}$, it is
then clear that $\gamma-\beta$ is a sum of positive compact roots and so
$\\{\alpha_{1},\dots,\alpha_{\ell-1},\beta\\}$ is the simple system for
$\mathfrak{p}^{+}$; changing $\beta$ to $-\beta$, we get the simple system for
$P^{-}$. The uniqueness of $\beta$ is obvious. ∎

######  Remark 2.19.

The relation
$\mathfrak{p}=\mathfrak{p}^{+}\oplus\mathfrak{p}^{-}=\mathfrak{p}^{+}\oplus(\mathfrak{p}^{+})^{\mathrm{conj}}$
shows that the action of $K_{0}$ on $\mathfrak{p}$ splits as
$\sigma\oplus\sigma^{\mathrm{conj}}$ where $\sigma$ is the irreducible action
of $K_{0}$ on $\mathfrak{p}^{+}$. If $L\in{\mathrm{End}}(\mathfrak{p}_{0})$
commutes with $K_{0}$, then $L$ is a scalar $\lambda$ on $\mathfrak{p}^{+}$
and $\lambda^{\mathrm{conj}}$ on $\mathfrak{p}^{-}$. If now in addition we
have $L^{2}=-I$ it is immediate that $\lambda=\pm i$. Thus $L=\pm J$. This
shows that the only $G_{0}$ -invariant almost complex structures on
$S_{0}=G_{0}/K_{0}$ are those defined by $\pm J$. This also shows that the
type of the $\mathbb{R}$-representation of $K_{0}$ on $\mathfrak{p}_{0}$ is
$\mathbb{R}$, namely, that the commutator is inside $\mathbb{C}_{\mathbb{R}}$,
the complex numbers viewed as a real algebra.

###  2.8. Unitarity

We extend the homomorphism
$\beta:\mathcal{U}[0]\longrightarrow\mathcal{U}(\mathfrak{h})$ constructed in
Sec. 2.5 to a linear map
$\mathcal{U}(\mathfrak{g})\longrightarrow\mathcal{U}(\mathfrak{h})$ by making
it 0 on $\mathcal{U}[q]$ for $q\neq 0$. Let $V$ be a $\mathfrak{g}$-module.
Then $V$ is said to be unitary if there is a positive definite hermitian
product $(,)$ for $V$ such that

$(Xu,v)+(u,Xv)=0,\qquad(u,v\in V,X\in\mathfrak{g}_{0})$

It is a theorem of Harish-Chandra that if $V$ is a unitary HC module and $H$
is the completion of $V$ in the norm $||x||=+\sqrt{(x,x)}$, and if the
$\mathfrak{k}_{0}$ -action on $V$ lifts to a $K_{0}$-action, there is a
unitary representation of $G_{0}$ in $H$ such that $V$ is the space of $K_{0}$
-finite vectors of $H$ as a $\mathfrak{g}$-module (see [16], [25]).

To discuss unitarity it is convenient to define the adjoint operation directly
on $\mathcal{U}(\mathfrak{g})$. The map $X\mapsto-X$ extends to an
antiautomorphism of $\mathcal{U}(\mathfrak{g}_{0})$; it can then be extended
to a conjugate linear antiautomorphism of $\mathcal{U}(\mathfrak{g})$. It is
denoted by $a\mapsto a^{*}$ and has the following properties:

  1. (i)

$a^{**}=a$

  2. (ii)

$(ab)^{*}=b^{*}a^{*}$

  3. (iii)

$a^{*}$ is conjugate linear in $a$

  4. (iv)

$X^{*}=-X$ for all $X\in\mathfrak{g}_{0}$.

It is uniquely determined by these requirements. The unitarity condition is
now

$(au,v)=(u,a^{*}v),\qquad(a\in\mathcal{U}(\mathfrak{g}),u,v\in V).$

######  Lemma 2.20.

We can choose the root vectors $X_{\gamma}\in\mathfrak{g}_{\gamma}$ in such a
way that $[X_{\gamma},X_{-\gamma}]=H_{\gamma}$ and

$X_{\gamma}^{*}=\begin{cases}X_{-\gamma},&\gamma\,\hbox{compact}\\\
-X_{-\gamma},&\gamma\,\hbox{ non compact}\end{cases}\qquad
X_{\gamma}^{\mathrm{conj}}=\begin{cases}-X_{-\gamma},&\gamma\,\hbox{compact}\\\
X_{-\gamma},&\gamma\,\hbox{ non compact}\end{cases}$

where $\mathrm{conj}$ is the conjugation of $\mathfrak{g}$ defined by
$\mathfrak{g}_{0}$.

###### Proof.

Let $0\neq X_{\gamma}\in\mathfrak{g}_{\gamma}$ be arbitrary to start with, but
satisfying $[X_{\gamma},X_{-\gamma}]=H_{\gamma}$. The relation
$[H,X_{\gamma}]=\gamma(H)X_{\gamma}$ gives, on applying $*$,
$[H,X_{\gamma}^{*}]=-\gamma(H)X_{\gamma}^{*}$, so that
$X_{\gamma}^{*}=c(\gamma)X_{-\gamma}$. As $*$ is involutive we get
$c(\gamma)^{\mathrm{conj}}c(-\gamma)=1$, where $\mathrm{conj}$ denotes complex
conjugation. On the other hand, from the standard theory,
$[X_{\gamma},X_{-\gamma}]=B(X_{\gamma},X_{-\gamma})H_{\gamma}^{\prime}$ where
$B$ is the Killing form and $H_{\gamma}^{\prime}$ is the image of $\gamma$
under the isomorphism $\mathfrak{h}^{*}\cong\mathfrak{h}$ induced by $B$.
Hence $H_{\gamma}=B(X_{\gamma},X_{-\gamma})H_{\gamma}^{\prime}$. As
$H_{\gamma}=cH_{\gamma}^{\prime}$ where $c>0$, we see that
$B(X_{\gamma},X_{-\gamma})$ is real and $>0$. On the other hand, we claim that
if $X\neq 0$ is in $\mathfrak{g}$, $B(X,X^{*})$ is $>0$ or $<0$ according as
$X\in\mathfrak{k}$ or $X\in\mathfrak{p}$. To see this, write $X=Y+iZ$ where
$Y,Z\in\mathfrak{g}_{0}$; then

$B(X,X^{*})=B(Y+iZ,-Y+iZ)=-B(Y,Y)-B(Z,Z)$

which proves our claim (since $Y,Z\in\mathfrak{k}_{0}$ or $\mathfrak{p}_{0}$
according as $X\in\mathfrak{k}$ or $\mathfrak{p}$). This proves the claim. But
now $B(X_{\gamma},X_{\gamma}^{*})=c(\gamma)B(X_{\gamma},X_{-\gamma})$, so
that, from our earlier remark we infer that $c(\gamma)$ is $>0$ or $<0$
according as $\gamma$ is compact or non compact. In any case $c(\gamma)$ is
real and so $c(\gamma)c(-\gamma)=1$. Write
$X_{\gamma}^{\prime}=|c(\gamma)|^{-1/2}X_{\gamma}$. Then
$[X_{\gamma}^{\prime},X_{-\gamma}]=H_{\gamma}$ still. On the other hand,

$\begin{array}[]{rl}(X_{\gamma}^{\prime})^{*}&=|c(\gamma)|^{-1/2}X_{\gamma}^{*}=|c(\gamma)|^{-1/2}c(\gamma)|c(-\gamma)|^{1/2}X_{-\gamma}^{\prime}=\\\
\\\
&=c(\gamma)|c(\gamma)|^{-1}X_{-\gamma}^{\prime}=\mathrm{sgn}c(\gamma)X_{-\gamma}^{\prime}\end{array}$

This proves the first assertion. The second is immediate from
$X_{\gamma}^{\mathrm{conj}}=-X_{\gamma}^{*}$, by the very definitions. ∎

Let $\beta:\mathcal{U}(\mathfrak{g})\longrightarrow\mathcal{U}(\mathfrak{h})$
be the Harish-Chandra homomorphism as in Sec. 2.5.

######  Theorem 2.21.

Let $\lambda\in\mathfrak{h}^{*}$ and let $\pi_{\lambda}$ be the irreducible
highest weight module of highest weight $\lambda$. The $\pi_{\lambda}$ is
unitary if and only if $\beta(a^{*}a)(\lambda)\geq 0$ for all
$a\in\mathcal{U}(\mathfrak{g})$. In particular it is necessary that:

$\lambda(H_{\gamma})\geq 0,\quad\hbox{for
compact}\quad\gamma,\qquad\lambda(H_{\gamma})\leq 0\quad\hbox{ for non
compact}\quad\gamma.$

###### Proof.

Assume that $\pi_{\lambda}$ acting on $V$ is unitary and write $v$ for the
highest weight vector. Since, for $H\in\mathfrak{h}_{0}$

$\lambda(H)(v,v)=(Hv,v)=-(v,Hv)=-\lambda(H)^{\mathrm{conj}}(v,v),$

it follows that $\lambda$ is pure imaginary on $\mathfrak{h}_{0}$. Hence all
weights are pure imaginary on $\mathfrak{h}_{0}$ (as the roots are so). A
similar argument shows that the weight spaces are mutually orthogonal. By
definition of $\beta$ it follows that for $c\in\mathcal{U}(\mathfrak{g})$, we
have $(cv,v)=\beta(c)(\lambda)(v,v)$. In particular,

$0\leq(av,av)=\beta(a^{*}a)(\lambda)(v,v)$

from which we get the necessary part.

With $X_{\gamma}$ as in Lemma 2.20, we have $X_{-\gamma}^{*}X_{-\gamma}=\pm
X_{\gamma}X_{-\gamma}$ according as $\gamma$ is compact or not. Hence

$\beta(X_{-\gamma}X_{-\gamma})=\beta(\pm X_{\gamma}X_{-\gamma})=\pm
H_{\gamma}$

so that $\lambda(H_{\gamma})$ is $\geq 0$ or $\leq 0$ according as $\gamma$ is
compact or not. For the sufficiency of the condition for unitarity we define
$(a,b)=\beta(b^{*}a)(\lambda)$ for $a,b\in\mathcal{U}(\mathfrak{g})$. Then
$(a,b)$ is sesqui-linear and $(a,a)$ is real and $\geq 0$ for all $a$. Hence
$(a,b)$ is Hermitian symmetric and $(a,a)\geq 0$ for all $a$. Let $R$ be the
radical of this Hermitian form: $a\in R$ if and only if $(a,a)=0$ or
equivalently $(a,b)=0$ for all $b\in\mathcal{U}(\mathfrak{g})$. Moreover
$(ca,b)=\beta(b^{*}ca)=(a,c^{*}b)$. We claim that $R$ has the following
properties.

  1. (i)

$R$ is a left ideal.

  2. (ii)

$X_{\gamma}$ $(\gamma>0)$, $H-\lambda(H)$ $(H\in\mathfrak{h}_{0})$ are in $R$.

Let us prove these assertions. Since $(1,1)=1$ we have $1\not\in R$. If $a\in
R$ and $c\in\mathcal{U}(\mathfrak{g})$, $(ca,b)=(a,c^{*}b)=0$ for all $b$ and
so $ca\in R$, proving (i). For $\gamma>0$,
$(X_{\gamma},b)=\beta(b^{*}X_{\gamma})=0$ for all $b$ and so $X_{\gamma}\in
R$.

$(H-\lambda(H),H-\lambda(H))=(-H+\lambda(H))(H-\lambda(H))(\lambda)=0,$

proving (ii). At this stage we know that $W=\mathcal{U}(\mathfrak{g})/R$ is a
$\mathcal{U}(\mathfrak{g})$-module of highest weight $\lambda$ and $1$ as the
corresponding weight vector, and that $W$ is unitary. We claim that $W$ is
irreducible, hence isomorphic to $\pi_{\lambda}$. Suppose $W^{\prime}\neq W$
is a submodule. Then $W^{\prime\perp}$ is also a submodule. Since the weight
spaces are mutually orthogonal and finite dimensional it follows that
$W=W^{\prime}\oplus(W^{\prime})^{\perp}$. Since $W^{\prime}\neq W$, we must
have $1\not\in W^{\prime}$. As

$W[\lambda]=W^{\prime}[\lambda]\oplus(W^{\prime})^{\perp}[\lambda],$

we must have $1\in W^{\prime\perp}$. Hence $(W^{\prime})^{\perp}=W$, showing
that $W^{\prime}=0$. ∎

######  Observation 2.22.

These remarks imply that the only irreducible finite dimensional unitary
module is the trivial representation. Suppose $\pi_{\lambda}$ is finite
dimensional and unitary. Then $\lambda(H_{\gamma})\geq 0$ for all roots
$\gamma>0$. By Theorem 2.21, we then conclude that $\lambda(H_{\gamma})=0$ for
all non compact roots $\gamma$. By Lemma 2.15 $\lambda=0$, hence
$\pi_{\lambda}$ is the trivial one dimensional representation.

######  Remark 2.23.

By using global methods Harish-Chandra proved that when
dim($\mathfrak{c}_{0}$) = 1 and $(\lambda+\delta)(H_{\gamma})\leq 0$ for all
non compact positive roots $\gamma$, then the module $U^{\lambda}$ of Theorem
2.8 is unitary. The full set of unitary highest weight modules was later on
determined by Enright, Howe, and Wallach [14], (see also [31], [26] and refs.
therein). A generalization to the super setting is due to Jakobsen [22] (see
also [4], [5] and refs. therein).

##  3\. Representations of the group

###  3.1. Geometry

The objective now is to construct the representations of the group $G_{0}$
that correspond to the highest weight HC modules constructed in Sec. 2. We
shall eventually assume that rk($\mathfrak{g}_{0}$)=rk($\mathfrak{k}_{0}$) and
that dim($\mathfrak{c}_{0}$)=1. But initially we drop the condition on the
center of $\mathfrak{k}_{0}$.

######  Lemma 3.1.

Let $M$ be a Lie group (real or complex), $A_{1}$, $A_{2}$ Lie subgroups with
$\mathrm{Lie}(A_{1})$ \+ $\mathrm{Lie}(A_{2})=\mathrm{Lie}(M)$. Then
$A_{1}A_{2}$ is open in $M$. If $A_{1}\cap A_{2}=\\{1\\}$, then $A_{i}$ are
closed and $a_{1},a_{2}\mapsto a_{1}a_{2}$ is an analytic diffeomorphism of
$A_{1}\times A_{2}$ with $A_{1}A_{2}$.

###### Proof.

Let $\mathfrak{a}_{i}$ = Lie$(A_{i})$, $\mathfrak{m}$ = Lie($M$). If $f$ is
the map $a_{1}$, $a_{2}\mapsto a_{1}a_{2}$ of $A_{1}\times A_{2}$ into $M$,
then $df$ is submersive at $(1,1)$ and hence everywhere since $f$ intertwines
the actions $(b_{1},b_{2})$: $(a_{1},a_{2})\mapsto(b_{1}a_{1},a_{2}b_{2})$ and
$(b_{1},b_{2}):x\mapsto b_{1}xb_{2}$. Thus $A_{1}A_{2}=f(A_{1}\times A_{2})$
is open in $M$. For the second part, note that when $A_{1}\cap A_{2}=\\{1\\}$,
$f$ is bijective and $\mathfrak{m}$ is the direct sum
$\mathfrak{a}_{1}\oplus\mathfrak{a}_{2}$ so that $df$ is also bijective; $f$
is thus a diffeomorphism. So the $A_{i}$ are closed in $A_{1}A_{2}$, i.e.,
locally closed. This means that they are open in their closures, and being
subgroups, are therefore closed. ∎

Let $G$ be a complex semisimple Lie group, $G_{0}\subset G$ a connected real
form. Let $R_{0}\subset G_{0}$ a closed subgroup and let

$\mathrm{Lie}(G_{0})=\mathfrak{g}_{0},\qquad\mathrm{Lie}(G)=\mathfrak{g}\cong\mathbb{C}\otimes_{\mathbb{R}}\mathfrak{g}_{0},\qquad\mathrm{Lie}(R_{0})=\mathfrak{r}_{0}.$

######  Lemma 3.2.

Suppose there exists a complex Lie subalgebra
$\mathfrak{q}\subset\mathfrak{g}$ such that
$\mathfrak{g}_{0}+\mathfrak{q}=\mathfrak{g}$,
$\mathfrak{g}_{0}\cap\mathfrak{q}=\mathfrak{r}_{0}$. Assume that the subgroup
$Q\subset G$ defined by $\mathfrak{q}$ is closed. Then

  * (a)

$G_{0}Q$ is open in $G$.

  * (b)

If $R_{1}=Q\cap G_{0}$, then $G_{0}/R_{1}\cong G_{0}Q/Q$.

  * (c)

$G_{0}/R_{0}$ has a $G_{0}$-invariant complex structure.

###### Proof.

(a) follows from Lemma 3.1. The map $g,q\mapsto gq$ of $G_{0}\times
Q\longrightarrow G$ induces a $G_{0}$-equivariant map of $G_{0}/R_{1}\cong
G_{0}Q/Q$ with bijective differential. The complex structure on $G_{0}/R_{1}$
is the pull back from $G_{0}Q/Q$. But $R_{0}$ and $R_{1}$ have the same Lie
algebra, namely $\mathfrak{r}_{0}$, and so $G_{0}/R_{0}\longrightarrow
G_{0}/R_{1}$ is a covering map. So the complex structure on $G_{0}/R_{1}$ can
be pulled back to $G_{0}/R_{0}$. ∎

From now on we assume that $\mathfrak{g}_{0}$ is semisimple and use notation
of Sec. 2. We take $G$ to be simply connected. At first we assume only that
$\mathfrak{g}_{0}$ and $\mathfrak{k}_{0}$ have the same rank and that
$\mathfrak{h}_{0}\subset\mathfrak{k}_{0}$ is a CSA for both $\mathfrak{g}_{0}$
and $\mathfrak{k}_{0}$. Fix a positive system $P$ of roots for
$(\mathfrak{g},\mathfrak{k})$. Define

$\mathfrak{b}^{\pm}=\mathfrak{h}\oplus\sum_{\alpha\in\pm
P}\mathfrak{g}_{\alpha},\qquad\mathfrak{n}^{\pm}=\oplus_{\alpha\in\pm
P}\mathfrak{g}_{\alpha}.$

Let $B^{\pm}$, $N^{\pm}$ be the subgroups of $G$ defined by
$\mathfrak{b}^{\pm}$, $\mathfrak{n}^{\pm}$ We will drop the suffix “+” when
clear from the context. Let $A$ (resp. $A_{0}$) be the subgroup of $G$ (resp.
$G_{0}$) defined by $\mathfrak{h}$ (resp. $\mathfrak{h}_{0}$). It follows from
Lemma 3.2 that $\Gamma=N^{-}B^{+}=N^{-}AN^{+}$ is open in $G$ (the big Bruhat
cell). Let $\mathfrak{u}=\mathfrak{k}_{0}\oplus i\mathfrak{p}_{0}$. Then
$\mathfrak{u}$ is a compact form of $\mathfrak{g}$ and the corresponding
subgroup $U$ of $G$ is compact and simply connected. The subgroup $A_{0}$
defined by $\mathfrak{h}_{0}$ is a torus and is a maximal torus of $U$ as well
as $G_{0}$.

######  Lemma 3.3.

If $\mathfrak{m}_{0}$ is a real form of $\mathfrak{g}$ containing
$\mathfrak{h}_{0}$, we have $\mathfrak{m}_{0}+\mathfrak{b}^{+}=\mathfrak{g}$.
In particular $M_{0}B^{\pm}$ is open in $G$, $M_{0}$ being the group defined
by $\mathfrak{m}_{0}$.

###### Proof.

If we denote the conjugation of $\mathfrak{g}$ with respect to
$\mathfrak{m}_{0}$ by $X\mapsto X\widetilde{\,\,}$, then

$\mathfrak{g}\widetilde{\,\,\,}_{\beta}=\mathfrak{g}_{-\beta}\quad\hbox{and}\quad
X_{-\beta}=X_{\beta}\widetilde{\,\,}=(X_{\beta}+X_{\beta}\widetilde{\,\,})-X_{\beta}\in\mathfrak{m}_{0}+\mathfrak{n},$

showing that $\mathfrak{n}^{-}\subset\mathfrak{m}_{0}+\mathfrak{n}^{+}$. So
$\mathfrak{m}_{0}+\mathfrak{b}^{+}$ contains
$\mathfrak{g}=\mathfrak{n}^{+}+\mathfrak{h}+\mathfrak{n}^{-}$. The second
assertion now follow from Lemma 3.2. ∎

######  Corollary 3.4.

The conjugation of $\mathfrak{g}$ with respect to $\mathfrak{m}_{0}$ takes
$\mathfrak{g}_{\beta}$ to $\mathfrak{g}_{-\beta}$ for all roots $\beta$.

###### Proof.

Follows from the previous proof. ∎

If we regard $\mathfrak{g}$ as a Lie algebra over the reals, it is semisimple
and $\mathfrak{g}=\mathfrak{u}+i\mathfrak{u}$ is its Cartan decomposition.
Then $\mathfrak{g}=\mathfrak{u}+i\mathfrak{h}_{0}+\mathfrak{n}^{\pm}$ are its
Iwasawa decompositions. Thus $G=UA_{\mathbb{R}}N^{\pm}$ are the global Iwasawa
decompositions of $G$, where we write $A_{\mathbb{R}}$ for the subgroup
defined by $i\mathfrak{h}_{0}$. The exponential maps:

$\exp:\mathfrak{n}^{\pm}\longrightarrow N^{\pm},\qquad
i\mathfrak{h}\longrightarrow A_{\mathbb{R}}$

are analytic diffeomorphisms (resp. complex and real). Also $A$ normalizes
$N^{\pm}$. We write $\mathrm{conj}$ for the conjugation of $\mathfrak{g}$ and
$G$ with respect to $G_{0}$ and $\mathfrak{g}_{0}$ respectively.

######  Lemma 3.5.

We have

$N^{-}\cap B^{+}=\\{1\\},\qquad A\cap N^{+}=\\{1\\}.$

In particular the map

$\psi:N^{-}\times A\times N^{+}\longrightarrow
G,\qquad\psi(n^{-},h,n^{+})=n^{-}hn^{+}$

is an analytic diffeomorphism onto the big Bruhat cell $\Gamma$, and $A$,
$N^{\pm}$ are closed.

###### Proof.

If $\exp Z\in B^{+}$ for some $Z\in\mathfrak{n}^{-}$, then
$e^{\mathrm{ad}(Z)}$ normalizes $\mathfrak{b}^{+}$, hence
$[Z,\mathfrak{b}^{+}]\subset\mathfrak{b}^{+}$ which is impossible unless
$Z=0$. If $\exp Z\in A$ for some $Z\in\mathfrak{n}^{+}$, $e^{\mathrm{ad}(Z)}$
is semisimple and unipotent, hence ad$(Z)=0$, or $Z=0$. ∎

######  Lemma 3.6.

We have

$G_{0}\cap B^{\pm}=A_{0}$

and $G_{0}B^{\pm}$ are open in $G$ while $G=UB^{+}$. Moreover
$G_{0}/A_{0}\cong G_{0}B^{\pm}/B^{\pm}$ acquires a $G_{0}$ -invariant complex
structure. Finally, $N^{-}$ may be viewed as a section for $\Gamma/B^{+}$, and
the left action of $A$ on $\Gamma/B^{+}$ is given by
$h,nB^{+}\mapsto(hnh^{-1})B^{+}$.

###### Proof.

Let $a=hn^{+}\in G_{0}$ where $h\in A$, $n^{+}\in B^{+}$. Then
$a=h^{\mathrm{conj}}n^{-}$ where $n^{-}=n^{\mathrm{conj}}\in N^{-}$. So
$hn^{+}=h^{\mathrm{conj}}n^{-}$ giving $(h^{\mathrm{conj}})^{-1}hn^{+}=n^{-}$.
Thus $n^{-}=n^{+}=1$, $h=h^{\mathrm{conj}}$ which give $a\in A_{0}$. The rest
follow from Lemmas 3.2 and 3.3. Since $B^{+}$ is closed and $U$ is compact,
$UB^{+}$ is closed, and as it is also open, it is $G$. The last statement is
trivial. ∎

###  3.2.  Holomorphically induced sheaves and group representations

The characters of $A_{0}$ extend uniquely to holomorphic characters of $A$.
Since $G$ is simply connected, these are precisely of the form

$\chi_{\lambda}:\exp H\mapsto e^{\lambda(H)}$

where $\lambda$ is integral, i.e., $\lambda(H_{\gamma})$ is an integer for all
roots $\gamma$. We write $\chi_{\lambda}$ again for the character
$hn^{+}\mapsto\chi_{\lambda}(h)$ of $B^{+}$. Throughout what follows,
$\lambda$ will be a fixed holomorphic character of $A$ and we shall suppress
mentioning $\lambda$ whenever there is no confusion. Let $\pi$ be the natural
map $G\longrightarrow G/B^{+}=:X$. For fixed $\chi=\chi_{\lambda}$ and any
open set $E\subset X$ we define $L(E)$ to be the linear space of all functions
$f$ on $\pi^{-1}(E)$ that are holomorphic and satisfy:

(4) $f(ub)=f(u)\chi(b)\quad\hbox{for all}\quad b\in B,\,u\in\pi^{-1}(E)$

By abuse of notation we also write sometimes $L(\pi^{-1}(E))$ for $L(E)$. Then
$L:E\longrightarrow L(E)$ is a sheaf on $X$. For any $E$, $L(E)$ is a Fréchet
space in the topology of uniform convergence on compact sets. This sheaf may
be naturally interpreted as the sheaf of sections of a holomorphic line bundle
on $X$ canonically associated to $\chi$. Let $G_{0}$ be a connected real form
of $G$. Write $S=G_{0}B^{+}$. The restriction of $L$ to $S/B^{+}$ (which is
open in $X$) is a seaf on $S/B^{+}$. The group $G_{0}$ acts naturally from the
left on the space $L(S/B^{+})$ of global sections of this sheaf. It is easy to
verify that this action gives a representation of $G_{0}$ on $L(S/B^{+})$. Our
goal is to study this representation by analyzing the action of $A_{0}$. Note
that at this stage we are not asserting that this space is even $\neq 0$. For
the definition of a representation of a locally compact group on a Fréchet
space, or more generally, a complete locally convex space, see [18] pp. 5-14
(for basic facts on Fréchet spaces see [33]). In our case the holomorphy
implies that all vectors are $C^{\infty}$, that is the action of the group on
them is smooth; we do not use this fact however.

Let $F$ be a Fréchet space, $H$ a Lie group, and $K\subset H$ a compact
subgroup of $H$. In our applications we will have $H=G_{0}$ and $K=A_{0}$ or
$K_{0}$. Let $R$ a representation of $H$ in $F$. For any irreducible character
$\tau$ of $K$ of degree $d(\tau)$, we define the operator

$P(\tau)=d(\tau)\int_{K}\tau(k)^{\mathrm{conj}}R(k)dk\qquad\int_{K}dk=1.$

Then $P(\tau)$ is a continuous projection $F\longrightarrow F(\tau)$ where
$F(\tau)=P(\tau)F$. $F(\tau)$ is a closed subspace of $F$. We have
$P(\tau)P(\tau^{\prime})=0$, $(\tau\neq\tau^{\prime})$. The $P(\tau)$ commute
with the $K$-action, and further, commute with any continuous endomorphism $E$
of $F$ that commutes with $K$. It is easy to show that $F(\tau)$ is the
algebraic linear span of all finite dimensional subspaces of $F$ which are
stable under $K$ and on which $K$ acts irreducibly according to a
representation with character $\tau$. The $F(\tau)$ are linearly independent
(see [35]).

We say that the representation $R$ is $K$-finite, if each $F(\tau)$ is finite
dimensional. In this case we write $F^{0}=F_{K}^{0}=\sum_{\tau}F(\tau)$. Let
$F^{\infty}$ be the subspace of $C^{\infty}$ vectors for $R$. Then,
$F^{\infty}$ is dense in $F$. Thus, $F^{\infty}(\tau):=P(\tau)F^{\infty}$ is
dense in $F(\tau)$. For any $f\in F$ let $f_{\tau}=P(\tau)f$. Then,
$\sum_{\tau}f_{\tau}$ is called the Fourier series of $f$. One knows ([18],
[35]) that, when $f\in F^{\infty}$,

$f=\sum_{\tau}f_{\tau},\qquad(f\in F^{\infty})$

where the series converges absolutely.

######  Lemma 3.7.

Suppose $F_{0}\subset F$ is a dense subspace, with $F_{0}=\sum_{\tau}L_{\tau}$
where the sum is algebraic and all $L_{\tau}$ are finite dimensional with
$L_{\tau}\subset F(\tau)$ Then $L_{\tau}=F(\tau)$ for all $\tau$,
$F^{0}=\sum_{\tau}F(\tau)$, and $F^{0}\subset F^{\infty}$. Suppose $L\subset
K$ is a compact subgroup and $F$ is $L$-finite; then $F$ is $K$-finite and
$F_{K}^{0}=F_{L}^{0}$.

###### Proof.

Clearly $L_{\tau}=P(\tau)F_{0}$ is dense in $P(\tau)F=F(\tau)$. But $L_{\tau}$
is closed in $F$ because it is finite dimensional, so that $F(\tau)=L_{\tau}$.
Since $F^{\infty}$ is dense in $F$, the same argument shows that
$F^{\infty}(\tau)$ is dense in $F(\tau)$, hence $F^{\infty}(\tau)=F(\tau)$. In
particular $F_{0}=\sum_{\tau}F(\tau)\subset F^{\infty}$. For the last
assertion let s be an irreducible representation of $K$ with character $\tau$.
The restriction of $s$ to $L$ will contain irreducible representation
(inequivalent) $t_{1},\dots t_{r}$ of $K$ with characters
$\sigma_{1},\dots,\sigma_{r}$ respectively. It is then clear that
$F_{K}(\tau)\subset\sum_{i}F_{L}(\sigma_{i})$. This proves that $F_{K}(\tau)$
is finite dimensional, hence that $F$ is $K$-finite, and that
$F_{K}^{0}\subset F_{L}^{0}$. To prove equality here, let us consider
$F_{L}(\sigma)$ where $\sigma$ is the character of an irreducible
representation $t$ of $L$. We know that $F_{L}(\sigma)\subset F_{K}^{\infty}$
and so each $f\in F_{L}(\sigma)$ has an absolutely convergent expansion
$f=\sum_{\tau}f_{\tau}$ where the $\tau$ vary over the irreducible characters
of $K$. It is enough to show that $P(\tau)F_{L}(\sigma)=0$ for almost all
$\tau$; for then by the Fourier expansion, $F_{L}(\sigma)$ is contained in the
sum of finitely many $F(\tau)$ and so is contained in $F_{K}^{0}$. Indeed,
suppose for some $\tau$ we have $P(\tau)F_{L}(\sigma)\neq 0$. Since $P(\tau)$
commutes with $K$, hence with $L$, this means that
$P(\tau)F_{L}(\sigma)\subset F(\tau)\cap F_{L}(\sigma)$. Actually we have
equality here; for, if $v\in F(\tau)\cap F_{L}(\sigma)$, then $P(\tau)v=v$.
Let $S(\tau)=F(\tau)\cap F_{L}(\sigma)$. Then $S(\tau)\neq 0$, and, when
$\tau$ varies over the set of characters for which $P(\tau)F_{L}(\sigma)\neq
0$, the $S(\tau)$ are non-zero linear subspaces of $F_{L}(\sigma)$ which are
linearly independent (as the $F(\tau)$ are linearly independent). Hence the
number of such $\tau$ cannot exceed dim $F_{L}(\sigma)$. ∎

###  3.3. Action of $\mathcal{U}(\mathfrak{g})$ on holomorphic sections

For any (second countable) complex manifold $M$, write $\mathrm{Hol}(M)$ for
the algebra of holomorphic functions on $M$ equipped with the topology of
uniform convergence on compact sets. Then $\mathrm{Hol}(M)$ is a Fréchet space
[33]. If $H$ is a locally compact group acting on $M$ via holomorphic
diffeomorphisms of $M$, then the corresponding natural action of $H$ on
$\mathrm{Hol}(M)$ is a representation (see [18] Lemma 1). Let

$\Gamma=N^{-}B^{+}$

be the big Bruhat cell. We begin by studying $F=L(\Gamma)$. Although $G$ does
not act on $F$, $\mathcal{U}(\mathfrak{g})$ does, from the left, as explained
below. For any $Z\in\mathfrak{g}$ we have the right invariant vector field
$\partial_{r}(Z)$ on $G$. Then for any open $W\subset G$ and any
$f\in\mathrm{Hol}(W)$,

$-(\partial_{r}(Z))f(u)=\left(\frac{d}{dt}\right)_{t=0}f(\exp(-tZ)u)\qquad(u\in
W).$

The map $Z\longrightarrow-\partial_{r}(Z)$ is a Lie homomorphism of
$\mathfrak{g}$ into the Lie algera of holomorphic vector fields on $W$. So it
extends to a representation $\ell$ of $\mathcal{U}(\mathfrak{g})$ on
$\mathrm{Hol}(W)$. We refer to this as the left action of
$\mathcal{U}(\mathfrak{g})$ on $\mathrm{Hol}(W)$. Similarly there is an action
from the right using the elements of $\mathfrak{g}$ viewed as left invariant
vector fields:

$(\partial(Z))f(u)=\left(\frac{d}{dt}\right)_{t=0}f(u\exp(tZ))\qquad(u\in W).$

For $u=X_{1}X_{2}\dots X_{r}\in\mathcal{U}(\mathfrak{g})$,
$X_{i}\in\mathfrak{g}$, we have

$(\partial(u)f)(g)=\Big{(}(\partial^{r}f/\partial t_{1}\dots\partial
t_{r})(g\exp(t_{1}X_{1})\dots\exp(t_{r}X_{r}))\Big{)}_{0}$

where $(\dots)_{0}$ means the derivative is calculated at
$t_{1}=\dots=t_{r}=0$. In contrast to this is

$(\ell(u)f)(g)=\Big{(}(\partial^{r}f/\partial t_{1}\dots\partial
t_{r})(\exp(-t_{r}X_{r})\dots\exp(-t_{1}X_{1})g)\Big{)}_{0}.$

Clearly $\partial$ and $\ell$ commute with each other. We shall determine
$\ell$ explicitly on $\mathrm{Hol}(N^{-}AN^{+})$ (Theorem 3.12). For
$g\in\mathrm{Hol}(N^{-})$ we write $g\widetilde{\,\,\,}$ for the unique
element of $F$ whose restriction to $N^{-}$ is $g$, so that
$g\widetilde{\,\,\,}(nb)=g(n)\chi(b)$ for $n\in N^{-}$, $b\in B^{+}$. The
correspondence $g\leftrightarrow g\widetilde{\,\,\,}$ is a linear topological
isomorphism between $\mathrm{Hol}(N^{-})$ and $F$, the inverse being the
restriction map from $F$ to $\mathrm{Hol}(N^{-})$. Since
$\exp:\mathfrak{n}^{-}\longrightarrow N^{-}$ is an analytic diffeomorphism, we
can introduce global coordinates $(t_{\alpha})_{\alpha\in P}$ on $N^{-}$ by

$t_{\alpha}(n)=y_{\alpha},\qquad n=\exp\left(\sum_{\alpha\in
P}y_{\alpha}X_{-\alpha}\right)$

We have

$t_{\alpha}(a^{-1}na)=t_{\alpha}(n)\chi_{\alpha}(a),\qquad(a\in A_{0},n\in
N^{-}).$

The polynomials in the $t_{\alpha}$ form a dense subalgebra $\mathcal{P}$ of
$\mathrm{Hol}(N^{-})$ and $\mathcal{P}\widetilde{\,\,\,}$ is the corresponding
dense subalgebra of $F$. If $i_{a}$ $(a\in A_{0})$ is the map $n\mapsto
ana^{-1}$ of $N^{-}$, it is immediate that

$\ell_{a}g\widetilde{\,\,\,}=\chi(a)^{-1}(i_{a}g)\widetilde{\,},\qquad(\ell_{a}g\widetilde{\,\,\,}=g\widetilde{\,\,\,}\circ\ell_{a}^{-1},\,\,i_{a}g=g\circ
i_{a}^{-1}).$

For $r=(r_{\alpha})$ let $t^{r}=\prod t_{\alpha}^{r_{\alpha}}$. We have

$i_{\alpha}t^{r}=\prod(t_{\alpha}\circ
i_{a}^{-1})^{r_{\alpha}}=\chi_{r^{*}}t_{r}\qquad\left(r^{*}=\sum_{\alpha}r_{\alpha}\alpha\right).$

Then, we obtain the following explicit decompositions of $\mathcal{P}$ and
$\mathcal{P}\widetilde{\,\,\,}$ under $A_{0}$:

$\mathcal{P}=\oplus_{r}\mathbb{C}.t^{r},\qquad
i_{a}t^{r}=\chi_{r^{*}}t_{r},\qquad\mathcal{P}_{\lambda}\widetilde{\,\,\,}=\oplus_{r}\mathbb{C}.(t^{r})\widetilde{\,},\qquad\ell_{a}(t_{r})\widetilde{\,}=\chi_{-\lambda+r^{*}}(t_{r})\widetilde{\,\,\,}.$

The elements of $\mathcal{P}$ are polynomials in the coordinates $t_{\alpha}$.
One can describe $\mathcal{P}$ also intrinsically. If we identify $N^{-}$ with
$\mathfrak{n}^{-}$ via the exponential map, then $\mathcal{P}$ is simply the
space of polynomials on the vector space $\mathfrak{n}^{-}$. By the Baker-
Campbell-Hausdorff (BCH) formula the multiplication in $\mathfrak{n}^{-}\cong
N^{-}$ is given by

(5) $X\cdot Y=p(X,Y),\qquad X^{-1}=-X$

where $p$ is a polynomial map
$\mathfrak{n}^{-}\times\mathfrak{n}^{-}\longrightarrow\mathfrak{n}^{-}$. We
can also write this as

(6) $n_{1}n_{2}=m(n_{1},n_{2}),\qquad n^{-1}=i(n)$

where $m$, $i$ are polynomial maps $N^{-}\times N^{-}\longrightarrow N^{-}$
and $N^{-}\longrightarrow N^{-}$ respectively. There is also another way of
thinking about this. $G$ may be viewed as a complex affine algebraic group in
a unique manner (the regular functions on $G$ form the algebra generated by
the matrix elements of holomorphic finite dimensional representations of $G$);
$N^{-}$ is an algebraic subgroup of $G$ whose underlying variety is the vector
space $\mathfrak{n}^{-}$. The polynomials on $N^{-}$ are thus the regular
functions on $N^{-}$, denoted as $\mathcal{O}_{N^{-}}(N^{-})$. Using either
point of view, one can conclude that the adjoint representation Ad of $N^{-}$
on $\mathfrak{g}$ is rational and so its matrix elements are polynomials on
$N^{-}$. Let $D^{+}$ be the semi group in $\mathfrak{h}^{*}$ generated by the
positive roots, namely, the set of elements of the form
$\sum_{i}m_{i}\alpha_{i}$ where the $m_{i}$ are integers $\geq 0$ and the
$\alpha_{i}$ are the simple roots in $P$. Then

$\mathcal{P}=\oplus_{d\in
D^{+}}\mathcal{P}_{d},\qquad\mathcal{P}_{d}=\oplus_{r^{*}=d}\mathbb{C}.t^{r}$

while

$\mathcal{P}=\oplus_{d\in
D^{+}}\mathcal{P}_{d}\widetilde{\,\,\,},\qquad\mathcal{P}_{d}\widetilde{\,\,\,}=\oplus_{r^{*}=d}\mathbb{C}.(t^{r})\widetilde{\,\,\,}.$

######  Lemma 3.8.

We have $F(\tau)\neq 0$ if and only if $\tau=\chi_{-\lambda}+d$ for some $d\in
D^{+}$. Moreover

(7) $\hbox{dim}F(\lambda+d)=\\#\left\\{r=(r_{\alpha})\,\big{|}\,r_{\alpha}\geq
0,\,\sum_{\alpha}r_{\alpha}\alpha=d\right\\}$

###### Proof.

Since $\mathcal{P}$ is dense in $\mathrm{Hol}(N^{-})$, we see that
$\mathcal{P}_{\lambda}\widetilde{\,\,\,}$ is dense in $F_{\lambda}$
($F=L(\Gamma)$, in (4) with $E=\Gamma$). The first statement now follows from
Lemma 3.7. The second statement is obvious. ∎

We have used the fact that the algebra $\mathrm{Pol}(V)$ polynomials on a
finite dimensional complex vector space $V$ form a dense subspace of
$\mathrm{Hol}(V)$. This is in general not true for arbitrary open subdomains
of $V$; such domains where this is true are called Runge domains.

For our purpose it is enough to know the following.

######  Lemma 3.9.

Let $V$ be a vector space with a compact torus $T$ acting on it. Suppose that,
for any character $\tau$ of $T$, the subspace of $\mathrm{Pol}(V)$ of all $p$
such that $p(t^{-1}v)=\tau(t)p(v)$ for all $v\in V$, is finite dimensional.
Then, any open connected subset of $V$, which is $T$-invariant and contains
the origin, is a Runge domain.

###### Proof.

We may assume that $V=\mathbb{C}^{N}$ with $T$-action

$t,(z_{1},\dots,z_{N})\mapsto(f_{1}(t)z_{1},\dots,f_{N}(t)z_{N})$

where the $f_{j}$ are characters of $T$. Let $U$ be an open connected subset
of $\mathbb{C}^{N}$ containing the origin and stable under $T$. The action of
$T$ induces an action on $\mathrm{Hol}(U)$ which is a representation. It is
enough to prove that the closure of $\mathrm{Pol}(\mathbb{C}^{N})$ contains
$\mathrm{Hol}(U)^{\infty}$. Since the Fourier series of any $f$ in
$\mathrm{Hol}(U)^{\infty}$ converges to $f$, it is enough to show that any
eigenfunction of $T$ in $\mathrm{Hol}(U)$ is a polynomial. Suppose $g\neq 0$
is in $\mathrm{Hol}(U)$ such that $g(t^{-1}(u))=f(t)g(u)$ for all $t\in T$ and
$u\in U$, $f$ being a character of $T$. Since $0\in U$, we can expand $g$ as a
power series $g(u)=\sum_{r}c_{r}u^{r}$ where we write $r=(r_{i})$ $1\leq i\leq
N$, $u=u_{1}\dots u_{N}$, valid in a polydisk. Then $c_{r}f^{r}=c_{r}f$
whenever $c_{r}\neq 0$ So only the $r$ with $f=f^{r}$ appear in the expansion
of $g$. We claim that there are only finitely many such $r$; once this claim
is proven we are done, because $g$ is a linear combination of the monomials
$u_{r}$ with $f^{r}=f$, hence $g$ is a polynomial. To prove the claim, note
that all such $u_{r}$ are eigenfunctions for $T$ for the eigencharacter $f$,
and by assumption, there are only finitely many of these. ∎

Recall that $\Gamma=N^{-}B^{+}$ and that

$anb=(n^{a})ab,\qquad(n^{a}=ana^{-1},\,a\in A_{0})$

Thus, $A_{0}$-invariant open connected subsets $U$ of $N^{-}$ correspond via
$U\leftrightarrow U\widetilde{\,\,\,}=UB^{+}$ to connected open sets
$U\widetilde{\,\,\,}=A_{0}U\widetilde{\,\,\,}B^{+}$. From the remark above we
know that any such $U$ is a Runge domain. Let $\Gamma^{1}=A_{0}\Gamma B^{+}$
be connected and open in $G$, $\Gamma^{2}=(\Gamma^{1}\cap\Gamma)^{0}$ where
the superscript $0$ refers to the connected component containing $1$. Clearly
$\Gamma^{2}=U_{2}B^{+}$ where $U_{2}$ is a connected $A_{0}$-invariant open
subset of $N^{-}$ containing 1. By what we saw above, $U_{2}$ is a Runge
domain in $N^{-}$. Let

$F^{1}=L(\Gamma^{1}),\qquad F^{2}=L(\Gamma^{2}),\qquad(\Gamma^{1}=A_{0}\Gamma
B^{+},\,\,\Gamma^{2}=(\Gamma^{1}\cap\Gamma)^{0})$

######  Lemma 3.10.

Let notation be as above. Then the restriction map $F^{1}\longrightarrow
F^{2}$ is a continuous injection and $F^{1}(\tau)$ maps into $F^{2}(\tau)$ for
all characters $\tau$ of $A_{0}$. Moreover $F^{2}(\tau)=F(\tau)$.

###### Proof.

The first statement is just the principle of analytic continuation. The second
follows from the fact that the polynomials are dense in $\mathrm{Hol}(U_{2})$.
∎

######  Lemma 3.11.

The left action of $\mathcal{U}(\mathfrak{g})$ on $L(W)$ for any open
$W\subset\Gamma$ leaves $\mathcal{P}\widetilde{\,\,\,}$ invariant.

###### Proof.

Let $Z\in\mathfrak{g}$. Then, for fixed $n\in N^{-}$ and $t\in\mathbb{C}$
sufficiently small, we have $\exp(-tZ^{n^{-1}})n\in N^{-}B^{+}$ and so
$\exp(-tZ^{n^{-1}})nb\in N^{-}B^{+}$ for all $b\in B^{+}$. Write

(8) $\exp(-tZ)nb=n\exp(-tZ^{n^{-1}})b,\qquad\exp(-tZn)=\nu(n,t)\beta(n,t)$

with

$\nu(n,t)\in N^{-},\quad\nu(n,0)=1,\quad\beta(n,t)\in
B^{+},\quad\beta(n,0)=1.$

Hence, for any $f\in\mathcal{P}$,

$(\exp(tZ)\cdot
f\widetilde{\,\,\,})(nb)=f\widetilde{\,\,\,}(\exp(-tZ)nb)=f\widetilde{\,\,\,}(n\nu(n,t)\beta(n,t)b)$

resulting in

$(\exp(tZ)\cdot f\widetilde{\,\,\,})(nb)=f(n\nu(n,t))\chi(\beta(n,t))\chi(b).$

We now differentiate with respect to $t$ at $t=0$. Let

$V(n)=(d/dt)_{t=0}\nu(n,t)\in\mathfrak{n}^{-},\qquad
W(n)=(d/dt)_{t=0}\beta(n,t)\in\mathfrak{b}^{+}.$

Then

$(Z\cdot f\widetilde{\,\,\,})(nb)=(V(n)f)(n)\chi(b)+f(n)d\chi(W(n))\chi(b).$

To determine $V(n)$ and $W(n)$, we differentiate (8) at $t=0$ to get

$-Z^{n^{-1}}=V(n)+W(n)$

so that $V(n)$ (resp. $W(n)$) is the projection of $-Z^{n^{-1}}$ on
$\mathfrak{n}^{-}$ (resp. $\mathfrak{b}^{+}$) corresponding to the direct sum
$\mathfrak{g}=\mathfrak{n}^{-}\oplus\mathfrak{b}^{+}$. We have seen already
that the adjoint action of $N^{-}$ on $\mathfrak{g}$ is rational. Hence if
$(Y_{j})$, $(H_{i})$, $(X_{k})$ are bases of $\mathfrak{n}^{-}$,
$\mathfrak{h}$, $\mathfrak{n}^{+}$ respectively, then

$V(n)=\sum_{j}f_{j}(n)Y_{j},\qquad
W(n)=\sum_{i}h_{i}(n)H_{i}+\sum_{k}g_{k}(n)X_{k}$

where $f_{j}$, $h_{i}$, $g_{k}$ are polynomial functions on $N^{-}$. Hence

$(Z\cdot f\widetilde{\,\,\,})(nb)=g\widetilde{\,\,\,}(nb)$

where

$g(n)=\sum_{j}f_{j}(n)(Y_{j}f)(n)+\sum_{i}\lambda(H_{i})h_{i}(n).$

From (5) and (6) we know that the action of $Y_{j}$ on $f$ is by a polynomial
differential operator of degree 1. Hence we get a polynomial differential
operator of degree 1, say $D_{Z}$, such that

$Z\cdot f\widetilde{\,\,\,}=g\widetilde{\,\,\,},\qquad g=D_{Z}f.$

But $D_{Z}f\in\mathcal{P}$ for all $f\in\mathcal{P}$. This proves that for any
$f\in\mathcal{P}$, $Z\cdot
f\widetilde{\,\,\,}\in\mathcal{P}\widetilde{\,\,\,}$. ∎

###  3.4. Pairings of $U(\mathfrak{g})$ modules of holomorphic sections

Let $M_{1}$ , $M_{2}$ be two modules for $\mathfrak{g}$. By a
$\mathfrak{g}$-pairing between them we mean a bilinear form
$\langle\cdot,\cdot\rangle$ on $M_{1}\times M_{2}$ with the property that

$\langle Xm_{1},m_{2}\rangle=\langle m_{1},-Xm_{2}\rangle\qquad(m_{i}\in
M_{i},X\in\mathfrak{g}).$

Since the $M_{i}$ are modules for $\mathcal{U}(\mathfrak{g})$ this implies
that

$\langle X_{1}\dots X_{r}m_{1},m_{2}\rangle=\langle m_{1},(-1)^{r}X_{r}\dots
X_{1}m_{2}\rangle\qquad(m_{i}\in M_{i},X_{j}\in\mathfrak{g}).$

The map $X\longrightarrow-X$ of $\mathfrak{g}$ is an involutive anti-
automorphism of $\mathfrak{g}$. It extends uniquely to an involutive anti-
automorphism $u\longrightarrow u^{T}$ of $\mathcal{U}(\mathfrak{g})$. The
$\mathfrak{g}$-pairing requirement is equivalent to

$\langle um_{1},m_{2}\rangle=\langle m_{1},u^{T}m_{2}\rangle(m_{i}\in
M_{i},u\in\mathcal{U}(\mathfrak{g})).$

We refer to this as a $\mathcal{U}(\mathfrak{g})$-pairing also. The pairing is
said to be non-singular if $\langle m_{1},m_{2}\rangle=0$ for all $m_{2}$
(resp. for all $m_{1}$) implies that $m_{1}=0$ (resp. $m_{2}=0$).

######  Theorem 3.12.

There is a non-singular $\mathcal{U}(\mathfrak{g})$-pairing between
$\mathcal{P}\widetilde{\,\,\,}$ and the Verma module $V_{\lambda}$. Moreover
every non-zero submodule of $\mathcal{P}\widetilde{\,\,\,}$ contains the
element $1\widetilde{\,\,\,}$ corresponding to the constant function
$1\in\mathcal{P}$. In particular, the submodule
$\mathcal{I}\widetilde{\,\,\,}$ of $\mathcal{P}\widetilde{\,\,\,}$ generated
by $1\widetilde{\,\,\,}$ is irreducible and is the unique irreducible
submodule of $\mathcal{P}\widetilde{\,\,\,}$. Finally,
$\mathcal{I}\widetilde{\,\,\,}$ is the unique irreducible module of lowest
weight $-\lambda$.

###### Proof.

It is clear that

$(\ell(u)f)(1)=(\partial(u^{T})f)(1);$

this is seen by taking $u=X_{1}\dots X_{r}$, $X_{i}\in\mathfrak{g}$. We now
define, for $u\in\mathcal{U}(\mathfrak{g})$,
$f\in\mathcal{P}\widetilde{\,\,\,}$,

(9) $\langle f,u\rangle=(\partial(u)f)(1)=(\ell(u^{T})f)(1)$

Then, for $c\in\mathcal{U}(\mathfrak{g})$,

$\begin{array}[]{rl}\langle\ell(c)f,u\rangle&=(\partial(u)\ell(c)f)(1)=(\ell(c)\partial(u)f)(1)=(\partial(c^{T})\partial(u)f)(1)=\\\
\\\ &=(\partial(c^{T}u)f)(1)=\langle f,c^{T}u\rangle.\end{array}$

We thus have a $\mathfrak{g}$-pairing between $\mathcal{P}\widetilde{\,\,\,}$
and $\mathcal{U}(\mathfrak{g})$ where the latter is regarded as a
$\mathcal{U}(\mathfrak{g})$-module under left multiplication. For
$f\in\mathcal{P}\widetilde{\,\,\,}$ we have, for all $w\in\Gamma$,

$f(w\exp tX_{\alpha})=f(w)\quad(\alpha>0),\qquad f(w\exp
tH)=e^{t\lambda(H)}f(w)$

from which we get

$\partial(Z)f=0\qquad(Z=X_{\alpha},H-\lambda(H)).$

If $M_{\lambda}$ is the left ideal generated by the $X_{\alpha}$ $(\alpha>0)$
and $H-\lambda(H)$ for $H\in\mathfrak{h}$, then we have

$\langle f,u\rangle=0\qquad(u\in M_{\lambda}).$

Hence $\langle\cdot,\cdot\rangle$ defines a $\mathfrak{g}$-pairing between
$\mathcal{P}\widetilde{\,\,\,}$ and
$\mathcal{U}(\mathfrak{g})/M_{\lambda}=V_{\lambda}$, the Verma module. We
shall now show that this is a non-singular pairing. If $\langle f,u\rangle=0$
for all $u\in\mathcal{U}(\mathfrak{g})$, then $(\partial(u)f)(1)=0$ for all
$u\in\mathcal{U}(\mathfrak{g})$ which implies that $f=0$. Conversely, suppose
that $\langle f,u\rangle=0$ for all $f\in\mathcal{P}\widetilde{\,\,\,}$. We
wish to prove that $u\in M_{\lambda}$. Now we can write $u$ as

$u=v+\mu$

where $\mu$ is in the enveloping algebra of $N^{-}$ and $v\in M_{\lambda}$.
Since $(\partial(v)f)(1)=0$, we have $(\partial(\mu)f)(1)=0$ for all
$f\in\mathcal{P}\widetilde{\,\,\,}$. This means that $(\partial(\mu)f)(1)=0$
for all polynomials $g$ on $N^{-}$. It is elementary to show that this implies
that $\mu=0$, proving that $u\in M_{\lambda}$.

The remainder of the proof is a formal consequence of the existence of the
non-singular pairing. The functions $(t^{r})\widetilde{\,\,\,}$ corresponding
to the coordinate polynomials $t^{r}$ on $N^{-}$ are weight vectors for the
action of $\mathfrak{h}$ for the weight $r-\lambda$. Hence
$\mathcal{P}\widetilde{\,\,\,}$ is a weight module with the multiplicities
defined by Lemma 3.8. We shall prove that every non-zero $\ell$-invariant
subspace $W$ of $\mathcal{P}\widetilde{\,\,\,}$ contains the vector
$1\widetilde{\,\,\,}$ defined by the constant function 1 on $N^{-}$. Now $W$
is a sum of weight spaces and if it does not contain $1\widetilde{\,\,\,}$,
then $W$ is contained in the sum of all weight spaces corresponding to the
weights $\lambda-r$ where $r=(r_{\alpha})$ with some $r_{\alpha}>0$. Now

$\langle Hm_{1},m_{2}\rangle=-\langle m_{1},Hm_{2}\rangle$

for all $H\in\mathfrak{h}$, $m_{1}\in\mathcal{P}\widetilde{\,\,\,}$, $m_{2}\in
V_{\lambda}$. This shows that the weight space of
$\mathcal{P}\widetilde{\,\,\,}$ for the weight $\theta$ is orthogonal to the
weight space of $V_{\lambda}$ for the weight $\phi$ unless $\theta=-\phi$. Let
$v$ be a non-zero vector of highest weight $\lambda$ in $V_{\lambda}$. Since
$W\subset\mathcal{P}$ is contained in the span of weights other than
-$\lambda$, we have $\langle W,v\rangle=0$. Hence, for all
$g\in\mathcal{U}(\mathfrak{g})$, $w\in W$ we have
$\langle\ell(g)w,v\rangle=0$. So $\langle w,g^{T}v\rangle=0$ for all
$g\in\mathcal{U}(\mathfrak{g})$. But $v$ is cyclic for $V_{\lambda}$ and so we
have $\langle w,V_{\lambda}\rangle=0$ for all $w\in W$. This means that $W=0$,
contradicting the hypothesis that $W\neq 0$. Thus every non-zero submodule of
$\mathcal{P}\widetilde{\,\,\,}$ contains the submodule
$\mathcal{I}\widetilde{\,\,\,}$ generated by $1\widetilde{\,\,\,}$. This
submodule is then the unique irreducible submodule of
$\mathcal{P}\widetilde{\,\,\,}$. The weights of
$\mathcal{I}\widetilde{\,\,\,}$ are of the form $-\lambda+d$, where $d$ is a
positive integral linear combination of the simple roots, and
$1\widetilde{\,\,\,}$ has weight -$\lambda$. It is then clear that
$1\widetilde{\,\,\,}$ is the lowest weight of $\mathcal{I}\widetilde{\,\,\,}$.
This fact, together with its irreducibility, characterizes it uniquely.
Theorem 3.12 is thus fully proved. ∎

###  3.5. Representations of the real group

We now come to the real group $G_{0}$.

######  Theorem 3.13.

Let $S=G_{0}B^{+}$ and $F^{1}=L(S)$. Then:

  1. (1)

$F^{1}\neq 0$ if and only if $F^{1}$ contains an element $\psi$ which is an
analytic continuation of $1\widetilde{\,\,\,}$ to $S$, i.e., which coincides
with $1\widetilde{\,\,\,}$ on $S\cap\Gamma$. In this case
$\psi\in(F^{1})^{\infty}$.

  2. (2)

If $F^{{11}}=\mathrm{Cl}(\mathcal{U}(\mathfrak{g})\psi)$ (Fréchet closure),
then $F^{11}$ is a Fréchet module for $G_{0}$, is $K_{0}$-finite, and its
$K_{0}$-finite part, which is also its $A_{0}$-finite part, is the irreducible
lowest weight module of lowest weight -$\lambda$.

  3. (3)

In particular $\lambda(H_{\alpha})$ is an integer $\geq 0$ for all compact
positive roots.

###### Proof.

(1) $F^{1}$ is a Fréchet module for $G_{0}$. We use Lemma 3.10 with
$\Gamma^{1}=S$ and $\Gamma^{2}=(S\cap\Gamma)^{0}$ where 0 refers to the
connected component containing the unit element of $G$. The restriction map
$F^{1}\longrightarrow F^{2}$ is a continuous injection which is
$A_{0}$-equivariant. so that $F^{1}(\tau)\subset F^{2}(\tau)=F(\tau)$ for all
characters $\tau$ of $A_{0}$. Thus $F^{1}$ is $A_{0}$-finite, hence also
$K_{0}$-finite, with $(F^{1})^{0}=\sum_{\tau}F^{1}(\tau)$ as the space of its
$K_{0}$-finite vectors (cf. Lemma 3.7). Suppose first that $F^{1}\neq 0$. The
weights corresponding to the characters $\tau$ are $-\lambda+r^{*}$ (earlier
notation) with $(t_{r})\widetilde{\,\,\,}$ as the corresponding eigenfunction,
the weight $-\lambda$ corresponding to $1\widetilde{\,\,\,}$. Suppose now that
the weights in $(F^{1})^{0}$ do not include $-\lambda$. Then the corresponding
eigenfunctions are the $(t_{r})\widetilde{\,\,\,}$ where $r=(r_{\alpha})=0$,
and so vanish at 1. Thus all elements of $(F^{1})^{0}$ vanish at 1, hence all
elements of $F^{1}$ vanish at 1 as $(F^{1})^{0}$ is dense in $F^{1}$. By left
translation by elements of $G_{0}$ all elements of $F^{1}$ vanish everywhere
on $S$, i.e., $F^{1}=0$. This contradicts the assumption that $L(S)\neq 0$.
Hence $F^{1}(\tau_{0})\neq 0$ for the trivial character $\tau_{0}$ of $A_{0}$.
Since

$F^{1}(\tau_{0})\subset F^{2}(\tau_{0})=\mathbb{C}\,1\widetilde{\,\,\,},$

we see that $F^{1}(\tau_{0})$ contains an element $\psi$ which restricts to
$1\widetilde{\,\,\,}$ on $(S\cap\Gamma)^{0}$. The converse is obvious; if
$\psi$ extends $1\widetilde{\,\,\,}$, then $F^{1}(\tau)^{0}\neq 0$, hence
$F^{1}\neq 0$.

(2) The algebra $\mathcal{U}(\mathfrak{g})$ acts on $(F^{1})^{0}$ and the
restriction map commutes with the actions of $\mathcal{U}(\mathfrak{g})$. Let
$J=\mathcal{U}(\mathfrak{g})\psi$. Then $J$ injects onto a non-zero submodule
of $\mathcal{I}\widetilde{\,\,\,}$, hence $J$ maps onto
$\mathcal{I}\widetilde{\,\,\,}$ isomorphically. Thus $J$ is the lowest weight
module for the weight $-\lambda$. Now $\psi$ is both $K_{0}$-finite and
$\zeta$-finite, where $\zeta$ is the center of $\mathcal{U}(\mathfrak{g})$,
the latter because $\mathcal{I}\widetilde{\,\,\,}\cong J$ has an infinitesimal
character (as all lowest (or highest) weight modules have). Hence Theorems 11
and 12 of [35] pg. 312 apply to tell us that $\mathrm{Cl}(J)$ is invariant
under $G$, is a Fréchet module whose $K_{0}$-finite part is precisely $J$.

(3) The last statement is clear because $F^{11}$ is a Harish-Chandra module
and $\psi$ is killed certainly by the compact negative roots, showing that
$-\lambda(-H_{\alpha})=\lambda(H_{\alpha})\geq 0$ for all compact positive
roots. This finishes the proof of the theorem. ∎

The shortcoming of Theorem 3.13 is that it does not tell us when $L(S)\neq 0$.
For instance, suppose that the real form $G_{0}$ is actually the maximal
compact subgroup $U_{0}$, i.e., $G_{0}=U_{0}$.Then $S=U_{0}B^{+}=G$, so that
$L=L(S)$ is the space of global sections of the holomorphic bundle. Then the
result above is essentially the Borel-Weil-Bott theorem in degree 0. Indeed,
the space of global sections, say $H$, is finite dimensional and carries an
action of $\mathcal{U}(\mathfrak{g})$ from the left. If $\lambda$ is a
dominant integral linear function, then for the corresponding representation
with highest weight $\lambda$ the matrix element $a_{11}(g)$ in a weight basis
$(v_{i})$, $0\leq i\leq N$ with $v_{0}$ as the highest weight vector is a
global section of the sheaf so that $L\neq 0$. Then we know that the module
$J$ generated by $a_{11}$ is irreducible and has lowest weight $-\lambda$. We
claim that $J=L$; otherwise we can find a complementary module $R$ such that
$L=J\oplus R$, contradicting the fact that $R$ must contain $a_{11}$.

###  3.6. Analytic continuation of $1\widetilde{\,\,\,}$ when the positive
system of roots is admissible

We shall now assume that the center $\mathfrak{c}_{0}$ of $\mathfrak{k}_{0}$
has dimension 1, and that $P$ is an admissible system of positive roots. From
the theory on the Lie algebra this means the following. We have

$\mathfrak{p}=\mathfrak{p}^{+}\oplus\mathfrak{p}^{-};$

$\mathfrak{p}^{\pm}$ are stable under $\mathfrak{k}$ and are irreducible, or
what comes to the same thing, stable and irreducible under $K_{0}$ (see Sec
2). Since $\mathfrak{h}_{0}$ is a CSA of $\mathfrak{g}_{0}$ also, it follows
that $\mathfrak{p}^{\pm}$ are sums of root spaces. The positive system of
roots is $P=P_{k}\cup P_{n}$ where $P_{k}$ is a positive system of compact
roots, and $P_{n}$ is the set of roots whose root spaces are contained in
$\mathfrak{p}^{+}$. Thus

$\mathfrak{p}^{\pm}=\sum_{\beta\in P_{n}}\mathfrak{g}^{\pm}_{\beta}.$

We know that the simple system for $P$ is of the form

$\\{\alpha_{1},\dots,\alpha_{r},\beta\\}$

where $\\{\alpha_{1},\dots,\alpha_{r}\\}$ form the simple system for the
compact roots, and $\beta$ is non-compact. Moreover the positive non-compact
roots are of the form

$\gamma=m_{1}\alpha_{1}+\dots+m_{r}\alpha_{r}+\beta$

where the $m_{i}$ are integers $\geq 0$. In particular
$[\mathfrak{p}^{\pm},\mathfrak{p}^{\pm}]=0$ (same sign in both). Obviously
$\mathfrak{p}^{\pm}$ are ideals in $\mathfrak{n}^{\pm}$. Let $P^{\pm}$ be the
complex analytic subgroups defined by $\mathfrak{p}^{\pm}$. Let

$\mathfrak{n}_{k}^{\pm}=\sum_{\beta\in P_{k}}\mathfrak{g}^{\pm}_{\beta}.$

Then

$\mathfrak{n}^{\pm}_{k}=\mathfrak{n}^{\pm}\cap\mathfrak{k}.$

The subgroups of $G$ defined by $\mathfrak{n}^{\pm}$ are closed and
$N^{\pm}=N_{k}^{\pm}P^{\pm}$. We can then set up the map

$f:P^{-}\times K\times P^{+}\longrightarrow G\qquad
f(p^{-},k,p^{+})=p^{-}kp^{+}.$

######  Lemma 3.14.

The set $\Omega=P^{-}KP^{+}$ is open in $G$ and the map $f$ is a complex
analytic diffeomorphism onto $\Omega$.

###### Proof.

This is standard and is valid under much greater generality. First we compute
the differential of $f$. We have, after a standard calculation,

$df_{(p_{1},k,p_{2})}(X_{1},Z,X_{2})=(X_{1}+Z^{k}+X_{2}^{kp_{2}})^{(kp_{2})^{-1}}$

from which it is clear $df$ is bijective everywhere. We must show that $f$ is
one-one. This reduces to showing that $P^{-}\cap KP^{+}=\\{1\\}$. Let
$p_{1}\in KP^{+}$. Since $p_{1}=kp^{+}$ where $k\in K$, $p^{+}\in P^{+}$, and
Ad($K$) leaves $\mathfrak{p}^{+}$ invariant, it follows that Ad($p_{1}$)
leaves $\mathfrak{p}^{+}$ invariant. Now $P^{-}$ is nilpotent, even abelian,
and so we can write $p_{1}=\exp X_{1}$ where $0\neq X_{1}\in\mathfrak{p}^{-}$.
Then $e^{\mathrm{ad}(X_{1})}=\mathrm{Ad}(p_{1})$ leaves $\mathfrak{p}^{+}$
invariant. Because ad($X_{1}$) is nilpotent, it is a polynomial in
$e^{\mathrm{ad}(X_{1})}$. Hence ad($X_{1}$) leaves $\mathfrak{p}^{+}$
invariant. Using a suitable lexicographic ordering of the roots we can write
$X_{1}=c_{1}X_{-\beta_{1}}+X_{-\beta_{2}}+\dots$ where
$\beta_{1}<\beta_{2}<\dots$ are non-compact roots and $c_{1}\neq 0$. Clearly

$[X_{1},X_{\beta_{1}}]=-c_{1}H_{\beta_{1}}+Y\qquad\hbox{where}\,\,Y\in\mathfrak{n}^{-}.$

Hence

$\mathrm{Ad}(p_{1})(X_{\beta_{1}})=e^{\mathrm{ad}(X_{1})}(X_{\beta_{1}})=X_{\beta_{1}}-c_{1}H_{\beta_{1}}+Y^{\prime}\qquad\hbox{where}\,\,Y^{\prime}\in\mathfrak{n}^{-}.$

But this must be in $\mathfrak{p}^{+}$. Thus
$H_{\beta_{1}}\in\mathfrak{n}^{-}+\mathfrak{n}^{+}$ which is impossible. ∎

$\Omega$ is the big cell corresponding to the parabolic subgroups $P^{\pm}$,
known also as the Harish-Chandra open cell. In particular, $K$, $P^{\pm}$ are
closed in $\Omega$, hence locally closed in $G$, hence closed in $G$. We write
$\theta$ for the automorphism of $\mathfrak{g}$ whose fixed point set is
$\mathfrak{k}$, namely the Cartan involution. Let

$\mathfrak{u}=\mathfrak{k}_{0}+i\mathfrak{p}_{0}.$

Then $\mathfrak{u}$ is a compact form of $\mathfrak{g}$. One knows that

$\mathfrak{g}=\mathfrak{u}\oplus i\mathfrak{h}_{0}\oplus\mathfrak{n}^{+}$

is an Iwasawa decomposition of $\mathfrak{g}$ (viewed as a real Lie algebra),
see [25] ch. VI. Let $U$ and $A^{+}$ be the real analytic subgroups of $G$
defined by $\mathfrak{u}$ and i$\mathfrak{h}_{0}$ respectively. Then $U$ is a
maximal compact subgroup of $G$ and simply connected, $A^{+}$ is a vector
space, and $G=UA^{+}N^{+}$ is the Iwasawa decomposition of $G$ (viewed as a
real Lie group). We write $\theta\widetilde{\,\,\,}$ and $\eta$ for the
conjugations of $\mathfrak{g}$ with respect to $\mathfrak{u}$ and
$\mathfrak{g}_{0}$ respectively. Since $G$ is simply connected it follows that
$\theta$, $\theta\widetilde{\,\,\,}$, and $\eta$ all lift to $G$; we denote
them by the same symbol. The actions of these on the Lie algebra are given by
the following table:

| $\mathfrak{k}_{0}$ | $i\mathfrak{k}_{0}$ | $\mathfrak{p}_{0}$ | $i\mathfrak{p}_{0}$
---|---|---|---|---
$\theta$ | $\mathrm{id}$ | $\mathrm{id}$ | $-\mathrm{id}$ | $-\mathrm{id}$
$\theta\widetilde{\,\,\,}$ | $\mathrm{id}$ | $-\mathrm{id}$ | $-\mathrm{id}$ | $\mathrm{id}$
$\eta$ | $\mathrm{id}$ | $-\mathrm{id}$ | $\mathrm{id}$ | $-\mathrm{id}$

It follows from this that

$\theta\widetilde{\,\,\,}=\eta\theta$

both on $\mathfrak{g}$ and $G$. Finally, we have

$\gamma(\mathfrak{n}^{+})=\mathfrak{n}^{-},\qquad\gamma(N^{+})=N^{-}\qquad(\gamma=\eta,\theta\widetilde{\,\,\,}).$

This follows from Lemma 3.3 and Lemma 2.20. We are thus able to formulate the
crucial lemma.

######  Lemma 3.15.

We have

$G_{0}B^{+}\subset P^{-}KP^{+}.$

###### Proof.

Let us denote:

$\mathfrak{n}_{k}^{+}=\mathrm{Lie}(N^{+}_{k}),\quad\mathfrak{n}_{n}^{+}=\mathrm{Lie}(N_{n}^{+}),\qquad(\mathfrak{n}_{k}^{+}=\sum_{\alpha>0\,{\mathrm{compact}}}\mathfrak{g}_{\alpha},\,\mathfrak{n}_{n}^{+}=\sum_{\alpha>0\,{\mathrm{noncompact}}}\mathfrak{g}_{\alpha})$

Since $KP^{+}$ contains $AN_{k}^{+}P^{+}=B^{+}$, we have $KP^{+}B^{+}=KP^{+}$
and so it is enough to prove that $G_{0}\subset P^{-}KP^{+}$. Now
$G_{0}=K_{0}\exp\mathfrak{p}_{0}$, while the left side is invariant by left
multiplication by $K_{0}$; indeed,

$K_{0}P^{-}KP^{+}\subset KP^{-}KP^{+}=P^{-}KKP^{+}=P^{-}KP^{+}.$

Hence it suffices to prove that $\exp\mathfrak{p}_{0}\subset P^{-}KP^{+}$. Let
$q=\exp X$ where $X\in\mathfrak{p}_{0}$. Write $p=\exp(X/2)$ so that
$q=p^{2}$. Then by the Iwasawa decomposition for $G$ we have $p=uan^{+}$ where
$u\in U,a\in A^{+},n^{+}\in N^{+}$. We apply $\theta\widetilde{\,\,\,}$ to
this relation. We have $\theta\widetilde{\,\,\,}(u)=u$ while
$\theta\widetilde{\,\,\,}=-\mathrm{id}$ on $i\mathfrak{k}_{0}$, so that
$\theta\widetilde{\,\,\,}(a)=a^{-1}$. Further we have observed that
$\theta\widetilde{\,\,\,}$ takes $N^{+}$ to $N^{-}$. On the other hand,

$p=\eta(p)=\theta\widetilde{\,\,\,}(\theta(p))=\theta\widetilde{\,\,\,}(p^{-1})$

so that $\theta\widetilde{\,\,\,}(p)=p^{-1}$. Hence we have

$p^{-1}=ua^{-1}n_{1},\qquad n_{1}\in N^{-}$

giving

$p=n_{1}^{-1}au^{-1}$

Multiplying this by $p=uan^{+}$ we get

$\exp X=q=p^{2}=n^{-1}_{1}a^{2}n^{+}$

showing that $q\in N^{-}AN^{+}$. But

$N^{-}AN+=P^{-}N_{n}^{-}AN_{k}^{+}P^{+}\subset P^{-}KP^{+}$

proving what we want. ∎

To get our result on analytic continuation of $1\widetilde{\,\,\,}$ we need
some preparation. Let $\lambda$ be an integral linear function on
$\mathfrak{h}$ which is dominant for compact positive roots, namely
$\lambda(H_{\alpha})\geq 0$ for all compact positive roots $\alpha$. We wish
to examine when there is an irreducible holomorphic representation
$\sigma_{\lambda}$ of $K$ with a highest weight $\chi_{\lambda}$, i.e.,
$\sigma_{\lambda}$ admits a non-zero vector $v$ such that

$\sigma_{\lambda}(h)v=\chi_{\lambda}(h)v\quad(h\in
A),\qquad\sigma_{\lambda}(X_{\alpha})v=0\quad(\alpha>0\quad\hbox{and
compact}).$

(Here we use the same symbol for a representation on the group and the
differentiated representation on the Lie algebra). We are interested in the
case when $\mathfrak{k}_{0}$ has a non-zero center $\mathfrak{c}_{0}$. Let
$\mathfrak{k}^{\prime}_{0}$ be the derived algebra
$[\mathfrak{k}_{0},\mathfrak{k}_{0}]$; then $\mathfrak{k}^{\prime}_{0}$ is
semisimple and
$\mathfrak{k}_{0}=\mathfrak{k}_{0}^{\prime}\oplus\mathfrak{c}_{0}$. As usual,
we drop the suffix 0 when we complexify. Thus $\mathfrak{c}$ is the center of
$\mathfrak{k}$, $\mathfrak{k}^{\prime}:=[\mathfrak{k},\mathfrak{k}]$ is
semisimple, $\mathfrak{k}=\mathfrak{k}^{\prime}\oplus\mathfrak{c}$, and
$\mathfrak{h}^{\prime}:=\mathfrak{h}\cap\mathfrak{k}^{\prime}$ is a CSA of
$\mathfrak{k}^{\prime}$. Also,
$\mathfrak{h}=\mathfrak{h}\cap\mathfrak{k}^{\prime}\oplus\mathfrak{c}$. Since
$K$ is the fixed point set of $\theta$, we know that $K$ is an algebraic group
and so is $C$, the connected component of the center of $K$. We have
$K=K^{\prime}C$; moreover $Z=K^{\prime}\cap C$ is finite since it is contained
in the center of $K^{\prime}$. Note that $C\subset A$. We then have the
following lemma.

######  Lemma 3.16.

Let $\chi_{\lambda}$ be a holomorphic character of $A$ such that $\lambda$ is
dominant for compact positive roots. For the existence of an holomorphic
finite dimensional representation of $K$ with highest weight $\lambda$ it is
necessary and sufficient that the representation of $\mathfrak{k}^{\prime}$
whose highest weight is the restriction of $\lambda$ to
$\mathfrak{h}^{\prime}$ lifts to a holomorphic representation of $K^{\prime}$.

###### Proof.

The condition is obviously necessary. We now prove its sufficiency. Let
$K_{1}$ be the universal covering group of $K^{\prime}$. Clearly the
representation of $\mathfrak{k}^{\prime}$ with highest weight $\lambda$ lifts
to a holomorphic representation $\tau^{\prime}$ of $K_{1}$. We thus have a
representation $\tau:=\tau^{\prime}\otimes\chi_{\lambda}$ of $K_{1}\times C$.
The map $k,c\mapsto kc$ from $K_{1}\times C$ to $K$ is onto $K$ and we shall
show that $\tau$ is trivial on the kernel of this map. Indeed, if $(k,c)$ is
in the kernel, we have $kc=1$ and so $k=c^{-1}\in\mathbb{Z}$. The
representation $\tau^{\prime}$ is a scalar on the center of $K^{\prime}$ and
so $\tau^{\prime}(t)=a(t)1$ for $t\in\mathbb{Z}$. But $\mathbb{Z}\subset A\cap
K^{\prime}$ since $A\cap K^{\prime}$ is a maximal torus of $K^{\prime}$ and
$A\cap K^{\prime}$ acts as the character $\chi_{\lambda}$ on the highest
weight vector. Hence $a(t)=\chi_{\lambda}(t)$ for $t\in Z$. But then

$\tau((k,c))=\tau^{\prime}(k)\chi_{\lambda}(c)=\chi_{\lambda}(c)^{-1}\chi_{\lambda}(c)=1.$

∎

Let $\sigma_{\lambda}$ be the representation of $K$ thus defined. It is
obvious that it is the required one. We say that $\lambda$ is of $K$-type if
the condition of the lemma is satisfied. In the notation of the lemma,
$\lambda$ is of $K$-type if and only if the representation $\tau_{1}$ is
trivial on the kernel of the covering map $K_{1}\longrightarrow K^{\prime}$.

######  Theorem 3.17.

Suppose that dim($\mathfrak{c}_{0}$) = 1. Let $\lambda$ be integral and
$\lambda(H_{\alpha})$ be an integer $\geq 0$ for all compact positive roots.
Assume that $\lambda$ is of $K$-type. Then $1\widetilde{\,\,\,}$ extends
analytically to an element $\psi\in L(S)$. In particular $L(S)\neq 0$ and
$(F_{\lambda})^{11}$ carries the representation whose $K_{0}$ -finite part is
the irreducible lowest weight representation of lowest weight $-\lambda$.
Finally and conversely, if $\lambda$ is not of $K$-type, then $L(S)=0$.

###### Proof.

We have a representation $\sigma=\sigma_{\lambda}$ of $K$ with highest weight
vector $v$ of weight $\chi_{\lambda}$. Since $P^{+}$ is normal in $KP^{+}$ and
$KP^{+}/P^{+}\cong K$ , we can view $\sigma$ as a representation of
$Q^{+}=KP^{+}$ trivial on $P^{+}$. We define the holomorphic function $f$ on
$K$ by

$R\sigma(k)v=f(k)v$

where $R$ is the unique projection on $\mathbb{C}v$ modulo the sum of the
remaining weight spaces. Extend $f$ to a holomorphic function $g$ on
$\Omega=P^{-}KP^{+}$ by

$g(p^{-}kp^{+})=f(k)\qquad(k\in K,p^{\pm}\in P^{\pm}).$

We now claim that the restriction of $g$ to $S=G_{0}B^{+}$ is the analytic
continuation of $1\widetilde{\,\,\,}$. Since $\Omega$ contains the big cell
$\Gamma=N^{-}AN^{+}$ it s enough to show that the restriction of $g$ to
$\Gamma$ is just $1\widetilde{\,\,\,}$. Fix $n^{\pm}\in N^{\pm}$, $h\in A$. As

$N^{\pm}=P^{\pm}N_{k}^{\pm}=N_{k}^{\pm}P^{\pm}$

we can write

$n^{-}=p^{-}n_{k}^{-},\quad n^{+}=n^{+}_{k}p^{+}\quad(n_{k}^{\pm}\in
N_{k}^{\pm},p^{\pm}\in P^{\pm})$

and so

$g(n^{-}hn^{+})=g(p^{-}n^{-}_{k}hn_{k}^{+}p^{+})=f(n_{k}^{-}hn_{k}^{+})=\chi_{\lambda}(h)f(n^{-}_{k}).$

It is thus enough to verify that $f(n^{-}_{k})=1$. But it is clear that
$\sigma(nk)v\cong v$ modulo the sum of weight spaces of weight less than
$\lambda$. Hence $f(n^{-}_{k})=1$ as we wanted to check. This proves the
theorem. ∎

######  Corollary 3.18.

Let $K_{0}^{\prime}$ be the analytic subgroup of $K_{0}$ defined by
$\mathfrak{k}^{\prime}_{0}:=[\mathfrak{k}_{0},\mathfrak{k}_{0}]$. If
$K_{0}^{\prime}$ is simply connected, Theorem 3.17 is valid for all integral
$\lambda$ whih are dominant with respect to the compact positive roots.

###### Proof.

Since $K_{0}^{\prime}$ is simply connected, and is a compact form of
$K^{\prime}$, it follows that $K^{\prime}$ is simply connected. So, for any
integral $\lambda$ dominant with respect to the positive compact roots, the
corresponding irreducible representation lifts to $K^{\prime}$. Lemma 3.16 now
shows that this representation extends to $K$. To prove the last statement,
assume that $L(S)\neq 0$. It is clear that the $K_{0}^{\prime}$-module
generated by $\psi$ has the lowest weight representation of
$\mathfrak{k}_{0}^{\prime}$ as its infinitesimal representation (same argument
as for $G_{0}$). Call this $\pi_{\lambda}$. So $\pi_{-\lambda}$ lifts to a
representation of $K_{0}^{\prime}$. Now $K_{0}^{\prime}$ is semisimple and is
the compact form of $K^{\prime}$, and so $\pi_{-\lambda}$ extends to a
holomorphic representation of $K^{\prime}$. This is contragredient to the
representation with highest weight $\lambda$ which then lifts to $K$. By Lemma
3.16 we conclude that $\lambda$ is of $K$-type. ∎

######  Example 3.19.

If $G=\mathrm{SU}(n,1)$, $K_{0}^{\prime}$ is $\mathrm{SU}(n)$ and so the above
Corollary applies. If $G={\mathrm{SO}}(2,2k)$. then $K_{0}^{\prime}$ is
${\mathrm{SO}}(2k)$ and the condition of $\lambda$ being of $K$-type is non-
trivial.

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