Pseudoconvexity and Steinness of Connected Complex Lie Groups: A Concise Lie-Theoretic Approach

Abstract

We give new concise Lie-theoretic proofs of basic analytic–geometric properties of connected complex Lie groups. Using Matsushima’s biholomorphic splitting $G\simeq {\mathbb{C}}^n\times \tilde{K}$ together with a refined analysis of the center via its Cousin factor, we show that every connected complex Lie group is pseudoconvex. Our approach is structural: it reduces to the reductive factor, separates the semisimple and central parts, and concludes using permanence of pseudoconvexity under products and finite quotients, together with standard triviality results for holomorphic principal bundles over Stein bases.

1. Introduction

In real differential geometry, Whitney’s embedding theorem asserts that any smooth real m-manifold admits a smooth embedding into some ${\mathbb{R}}^N$. The complex–analytic picture is subtler: not every complex manifold embeds holomorphically into ${\mathbb{C}}^N$. For instance, by Liouville’s theorem, every holomorphic function on a compact complex manifold is constant; hence a compact complex manifold cannot embed into any ${\mathbb{C}}^N$. Those complex manifolds that do embed are precisely the Stein manifolds, introduced by Karl Stein in the early 1950s. Identifying and characterizing Stein manifolds remains a central theme of several complex variables.

A classical route to Steinness goes through pseudoconvexity and holomorphic convexity. Following Levi’s program [1] and Cartan–Thullen [2], the Levi problem relates domains of holomorphy, holomorphic convexity, and (strong) pseudoconvexity. Modern accounts show that a complex manifold is Stein if and only if it admits a smooth strictly plurisubharmonic exhaustion function (Grauert, Narasimhan, Hörmander; see [3]). On the flexibility side, recent developments in Oka theory further illuminate the landscape of Stein (and Oka) manifolds and holomorphic maps between them [4].

For complex Lie groups, Matsushima–Morimoto [5] obtained a Lie-theoretic criterion for Steinness: a connected complex Lie group G is Stein if and only if the maximal complex Lie subalgebra of the Lie algebra of a maximal compact subgroup is trivial (condition $\left(P\right)$). Their argument combines complex geometry, Lie theory, holomorphic principal bundles, and stability properties of Stein manifolds.

A parallel line of work due to Kazama established pseudoconvexity for complex Lie groups in two papers—first for abelian complex Lie groups and then in full generality—via long and technical arguments [6,7]. (For the abelian case, see also Morimoto [8].)

Contributions: We give a short Lie-theoretic proof that every connected complex Lie group is pseudoconvex (Theorem 8). The argument uses Matsushima’s biholomorphic splitting $G\simeq {\mathbb{C}}^n\times \tilde{K}$ together with a refined analysis of the abelian center via its Cousin factor. We then revisit the Matsushima–Morimoto characterization and show that for connected complex Lie groups Steinness is equivalent to holomorphic separability; in particular, the Cousin factor (equivalently, the Steinizer) is the only obstruction (Theorems 4 and 5). Along the way we record auxiliary results showing that pseudoconvexity (and, when applicable, the Stein property) is preserved under finite products, finite quotients, and unramified coverings.

2. Preliminaries

A domain $\Omega \subset {\mathbb{C}}^n$ is a domain of holomorphy if, for every boundary point $p\in \partial \Omega$, there exists $f\in \mathcal{O}(\Omega )$ that does not admit a holomorphic extension to any neighborhood of p. In one complex variable every domain is a domain of holomorphy, but in higher dimensions the situation is different due to Hartogs’ phenomenon: if $U\subset {\mathbb{C}}^n$ is open with $n\ge 2$, $K\subset U$ is compact, and f is holomorphic on $U\setminus K$, then f extends holomorphically to all of U. Levi asked whether domains of holomorphy admit a geometric characterization [1]. Cartan and Thullen [2] showed that domains of holomorphy are exactly the holomorphically convex domains.

A domain $\Omega \subset {\mathbb{C}}^n$ is holomorphically convex if for every compact $K\subset \Omega$ its holomorphic hull

$${\widehat{K}}_{\mathcal{O}(\Omega )}:=\{z\in \Omega :|f\left(z\right)|\le \underset{K}{max}\left|f\right|\mathrm{for}\mathrm{all}f\in \mathcal{O}(\Omega )\}$$

is relatively compact in $\Omega$. For domains in ${\mathbb{C}}^n$, holomorphic convexity is closely related to pseudoconvexity (see, e.g., [3]).

Let $f:\Omega \to \mathbb{R}$ be a ${\mathcal{C}}^2$ function on a domain $\Omega \subset {\mathbb{C}}^n$. The Hermitian matrix $\left({\partial }^{2}f/\partial z_{\alpha }\partial {\overline{z}}_{\beta }\right)$ defines the Levi form $\mathcal{L}\left(f\right)$. The function f is plurisubharmonic (psh) if $\mathcal{L}\left(f\right)$ is positive semidefinite, and strictly plurisubharmonic if $\mathcal{L}\left(f\right)$ is positive definite. A complex manifold X is called pseudoconvex if it admits a continuous plurisubharmonic exhaustion function. For general complex manifolds, pseudoconvexity need not imply holomorphic convexity; see Grauert’s examples [9] and the survey by Siu [10].

A complex manifold X is Stein if it is holomorphically convex and holomorphically separable: for any distinct points $x\ne y$ in X there exists $f\in \mathcal{O}\left(X\right)$ with $f\left(x\right)\ne f\left(y\right)$. Equivalently, X is Stein if and only if it admits a smooth strictly plurisubharmonic exhaustion function (see [3]). By Remmert [11], Stein manifolds embed properly as closed complex submanifolds of some ${\mathbb{C}}^N$.

Standard examples of Stein manifolds include products of Stein manifolds and closed complex submanifolds of Stein manifolds. Moreover, $\mathrm{GL}(n,\mathbb{C})$ is Stein; for instance one may use the strongly plurisubharmonic exhaustion

$$\phi \left(A\right):=\mathrm{trace}\left(A{A}^{*}\right)+{\displaystyle \frac{1}{det\left(A{A}^{*}\right)}},$$

where $A^{*}$ denotes the conjugate transpose (see, e.g., [12]). Since every connected semisimple complex Lie group admits a faithful finite-dimensional holomorphic representation, it embeds as a closed complex subgroup of $\mathrm{GL}(n,\mathbb{C})$; hence any connected semisimple complex Lie group is Stein.

Definition 1.

A connected complex Lie group G is called a Cousin group if it has no nonconstant holomorphic functions, i.e., $\mathcal{O}\left(G\right)=\mathbb{C}$.

Cousin groups were introduced by Cousin [13] and later studied systematically by Morimoto [14]. They are also called toroidal groups; see [15].

Proposition 1.

Every Cousin group is abelian.

Proof. 

Consider the adjoint representation $Ad:G\to \mathrm{GL}\left(\mathfrak{g}\right)$. Since $\mathcal{O}\left(G\right)=\mathbb{C}$, every holomorphic map from G into a holomorphically separable complex manifold is constant; in particular, Ad is constant, and hence $Ad\left(G\right)=\left\{e\right\}$. Therefore $G\subset ker(Ad)$. For any connected complex Lie group, $ker(Ad)$ is the identity component of the center, so G is abelian. □

An important example of Cousin groups is given by compact complex Lie groups: by Liouville’s theorem they admit no nonconstant holomorphic functions, and hence are abelian by Proposition 1.

3. Stein Complex Lie Groups

The main goal of this section is to revisit the characterization of Stein complex Lie groups due to Matsushima and Morimoto [5]. They show that a connected complex Lie group G is Stein if and only if a certain Lie-algebraic obstruction (their condition $\left(P\right)$) vanishes. Equivalently, no positive-dimensional complex Lie subgroup can lie inside a maximal compact subgroup of G.

For the reader’s convenience, we summarize the notation used throughout this section in Table 1. These conventions remain in force for the rest of the paper.

Table 1. Notation used in Section 3.

Symbol	Meaning
G	a connected complex Lie group
$K\subset G$	a maximal compact connected Lie subgroup
$\mathfrak{k}$	the Lie algebra of K
${\mathfrak{k}}_{ss}$	a Levi (maximal semisimple) subalgebra of $\mathfrak{k}$
$\tilde{\mathfrak{k}}$	the complex Lie subalgebra $\mathfrak{k}+i\mathfrak{k}\subset \mathfrak{g}$
$\tilde{K}$	the connected complex Lie subgroup with Lie algebra $\tilde{\mathfrak{k}}$
$\mathfrak{m}$	the maximal complex subalgebra of $\mathfrak{k}$, i.e., $\mathfrak{m}:=\mathfrak{k}\cap i\mathfrak{k}$
$\mathcal{Z}\left(K\right)$	the center of K
Z	the identity component of $\mathcal{Z}\left(K\right)$
$\mathfrak{z}$	the center of $\mathfrak{k}$
$\tilde{\mathfrak{z}}$	the complexified center $\mathfrak{z}+i\mathfrak{z}\subset \tilde{\mathfrak{k}}$
$\mathcal{Z}\left(\tilde{K}\right)$	the center of $\tilde{K}$
$\tilde{Z}$	the identity component of $\mathcal{Z}\left(\tilde{K}\right)$
${\tilde{\mathfrak{k}}}_{ss}$	the complex semisimple Lie algebra ${\mathfrak{k}}_{ss}+i{\mathfrak{k}}_{ss}$
${\tilde{K}}_{ss}$	the connected complex Lie subgroup with Lie algebra ${\tilde{\mathfrak{k}}}_{ss}$
$\mathfrak{k}={\mathfrak{k}}_{ss}⋉\mathfrak{z}$	Levi–Malcev decomposition of $\mathfrak{k}$
$\tilde{\mathfrak{k}}={\tilde{\mathfrak{k}}}_{ss}⋉\tilde{\mathfrak{z}}$	Levi–Malcev decomposition of $\tilde{\mathfrak{k}}$
$\tilde{K}={\tilde{K}}_{ss}·\tilde{Z}$	the corresponding decomposition at the group level

Theorem 1

(Matsushima–Morimoto [5]). A connected complex Lie group G is Stein if and only if

$$\mathfrak{m}=\mathfrak{k}\cap i\mathfrak{k}=\left(0\right),$$

where $\mathfrak{k}$ is the Lie algebra of a maximal compact subgroup $K\subset G$.

In this article we provide a new proof of Theorem 1, based on Matsushima’s splitting theorem [16] together with a Lie-theoretic analysis of the center via its Cousin factor. Our approach is designed to be short and conceptual, and it also leads to a streamlined proof of pseudoconvexity for connected complex Lie groups.

To see that the condition $\mathfrak{m}=\left(0\right)$ is intrinsic (i.e., independent of the choice of maximal compact subgroup), recall Iwasawa’s theorem [17]: every connected Lie group G is diffeomorphic to $K\times {\mathbb{R}}^N$, where K is a maximal compact subgroup, which is connected and unique up to conjugation. In particular, $dim\left(\mathfrak{m}\right)$ is independent of the chosen maximal compact subgroup K.

• Outline of the proof of Theorem 1.

• We show that G is Stein if and only if the complexification $\tilde{K}$ of a maximal compact subgroup K is Stein.
• We then reduce Steinness of $\tilde{K}$ to Steinness of the identity component $\tilde{Z}$ of its center $\mathcal{Z}\left(\tilde{K}\right)$.
• Finally, we analyze the abelian complex Lie group $\tilde{Z}$ using the structure theorem for connected abelian complex Lie groups: it splits as a product of a Stein abelian factor and a Cousin factor C. We show that $\tilde{Z}$ is Stein if and only if the Cousin factor C is trivial, and we identify this obstruction with the Lie algebra $\mathfrak{m}=\mathfrak{k}\cap i\mathfrak{k}$.

3.1. Complexification of Maximal Compact Subgroups

Let $K\subset G$ be a maximal compact subgroup. Since compact Lie groups are reductive, K admits a Levi decomposition

$$K=K_{ss}·Z,$$

where $K_{ss}$ is a maximal connected semisimple subgroup of K and Z is the identity component of the center $\mathcal{Z}\left(K\right)$. Moreover, $K_{ss}\cap Z$ is finite (it is a compact abelian semisimple subgroup).

A similar structure appears on the complex side. We recall the following results from Matsushima [16] (see Lemma 3 and §2).

Lemma 1

(Matsushima, [16], Lemma 3 and §2). Let G be a connected complex Lie group and $K\subset G$ a maximal compact subgroup with Lie algebra $\mathfrak{k}$. Set

$$\tilde{K}:=exp(\mathfrak{k}+i\mathfrak{k})\subset G.$$

Then $\tilde{K}$ is a closed complex Lie subgroup of G, and the analytic quotient $G/\tilde{K}$ is biholomorphic to ${\mathbb{C}}^n$, where $n={dim}_{\mathbb{C}}G-{dim}_{\mathbb{C}}\tilde{K}$.

Lemma 2

(Oka–Grauert principle for principal bundles). Let H be a complex Lie group and $\pi :P\to B$ a holomorphic principal H-bundle over a Stein base B. If the underlying topological principal bundle is trivial, then π is holomorphically trivial; equivalently, $P\simeq B\times H$ as holomorphic principal H-bundles.

Proof. 

This is the Oka–Grauert principle for principal bundles with complex Lie structure group over Stein bases. See Grauert (1958) and, for example, Forstnerič [3] (Theorem 5.4.4 and Corollary 5.5.2). □

Proposition 2

(Holomorphic splitting). With G and $\tilde{K}$ as in Lemma 1, the quotient map

$$\pi :G\to G/\tilde{K}$$

is a holomorphic principal $\tilde{K}$-bundle with Stein base $G/\tilde{K}\simeq {\mathbb{C}}^n$. Since ${\mathbb{C}}^n$ is contractible, the underlying topological principal bundle is trivial; hence Lemma 2 implies that π is holomorphically trivial. In particular,

$$G\simeq (G/\tilde{K})\times \tilde{K}\simeq {\mathbb{C}}^n\times \tilde{K}\mathit{biholomorphically}.$$

Proof. 

By Lemma 1, $\tilde{K}:=exp(\mathfrak{k}+i\mathfrak{k})$ is a closed complex Lie subgroup of G and the quotient $G/\tilde{K}$ is biholomorphic to ${\mathbb{C}}^n$. Since $\tilde{K}$ is a closed complex subgroup, the quotient map

$$\pi :G\to G/\tilde{K}$$

is a holomorphic submersion and makes G into a holomorphic principal $\tilde{K}$–bundle over $G/\tilde{K}$ (local holomorphic trivializations follow from the holomorphic slice/local product structure for complex Lie group quotients; see, e.g., [18] (Ch. VII) or [19]). Topologically, every principal $\tilde{K}$-bundle over the contractible base ${\mathbb{C}}^n$ is trivial (classification by homotopy classes into $B\tilde{K}$, or equivalently because ${\mathbb{C}}^n$ is paracompact and has vanishing higher homotopy; see [20] (§ principal bundles)). Hence the underlying topological principal bundle of $\pi$ is trivial. Therefore, by Lemma 2, $\pi$ is holomorphically trivial: $G\cong (G/\tilde{K})\times \tilde{K}\cong {\mathbb{C}}^n\times \tilde{K}$ biholomorphically. □

Corollary 1.

The group G is Stein if and only if $\tilde{K}$ is Stein.

Proof. 

By Proposition 2, $G\simeq {\mathbb{C}}^n\times \tilde{K}$. Since ${\mathbb{C}}^n$ is Stein and the product of Stein manifolds is Stein, G is Stein if and only if $\tilde{K}$ is Stein. □

Next, consider the Levi–Malcev decomposition of the Lie algebra $\tilde{\mathfrak{k}}$:

$$\tilde{\mathfrak{k}}={\tilde{\mathfrak{k}}}_{ss}⋉\tilde{\mathfrak{z}},$$

where ${\tilde{\mathfrak{k}}}_{ss}:={\mathfrak{k}}_{ss}+i{\mathfrak{k}}_{ss}$ is semisimple and $\tilde{\mathfrak{z}}:=\mathfrak{z}+i\mathfrak{z}$ is the center ideal. Let ${\tilde{K}}_{ss}$ and $\tilde{Z}$ denote the connected complex Lie subgroups of $\tilde{K}$ with Lie algebras ${\tilde{\mathfrak{k}}}_{ss}$ and $\tilde{\mathfrak{z}}$, respectively. Then

$$\tilde{K}={\tilde{K}}_{ss}·\tilde{Z}.$$

We now record the following structural facts (cf. [16] (§2)), together with a self-contained proof of the Lie-algebra identity in (3), which we will use repeatedly.

Proposition 3.

With the terms and definitions of groups and algebras discussed above:

1. 

$\tilde{K}\cong {\tilde{K}}_{ss}·\tilde{Z}$ with ${\tilde{K}}_{ss}\cap \tilde{Z}$ being finite.

2. 

$\tilde{Z}$ is a closed subgroup of $\tilde{K}$.

3. 

$\mathfrak{k}\cap i\mathfrak{k}=\mathfrak{z}\cap i\mathfrak{z}$.

Proof. 

• Since the center of a semisimple Lie algebra is trivial we have ${\tilde{\mathfrak{k}}}_{ss}\cap \tilde{\mathfrak{z}}=\left(0\right)$. Thus, ${\tilde{K}}_{ss}\cap \tilde{Z}$ is finite since the center of a semisimple algebraic group is finite.
• The kernel of the adjoint representation $\mathrm{Ad}:\tilde{K}\to GL(n,\mathbb{C})$ is equal to the center $\mathcal{Z}\left(\tilde{K}\right)$ of $\tilde{K}$. Thus, $\mathcal{Z}\left(\tilde{K}\right)$ and its connected component $\tilde{Z}$ are closed subgroups of $\tilde{K}$.
• Consider the adjoint representation:

  $$\mathrm{Ad}:\tilde{K}\to \mathrm{GL}(n,\mathbb{C}).$$
  Since K is compact, we have $\mathrm{Ad}\left(K\right)\subset U\left(n\right)$, which implies that, at the Lie algebra level:

  $$\mathrm{ad}\left(\mathfrak{k}\right)\subset \mathfrak{u}\left(n\right).$$
  Now observe that

  $$\mathrm{ad}(\mathfrak{k}\cap i\mathfrak{k})\subset \mathrm{ad}\left(\mathfrak{k}\right)\cap i\mathrm{ad}\left(\mathfrak{k}\right)\subset \mathrm{ad}\left(\mathfrak{k}\right)+i\mathrm{ad}\left(\mathfrak{k}\right).$$
  By the linearity of the adjoint representation, it follows that

  $$\mathrm{ad}\left(\mathfrak{k}\right)+i\mathrm{ad}\left(\mathfrak{k}\right)=\mathrm{ad}(\mathfrak{k}+i\mathfrak{k})\subset \mathfrak{u}\left(n\right)+i\mathfrak{u}\left(n\right).$$
  Hence, we conclude that

  $$\mathrm{ad}(\mathfrak{k}\cap i\mathfrak{k})\subset \mathfrak{u}\left(n\right)\cap i\mathfrak{u}\left(n\right).$$
  Since the eigenvalues of a skew-Hermitian matrix are purely imaginary, the only eigenvalue of any matrix in $\mathfrak{u}\left(n\right)\cap i\mathfrak{u}\left(n\right)$ is 0. Therefore:

  $$\mathfrak{u}\left(n\right)\cap i\mathfrak{u}\left(n\right)=\left(0\right).$$
  Since $\tilde{\mathfrak{z}}$ is the kernel of $\mathrm{ad}$, we have

  $$\mathfrak{k}\cap i\mathfrak{k}\subset \tilde{\mathfrak{z}}.$$

  (1)
  To proceed with the proof, we need to establish that

  $$\mathfrak{z}=\tilde{\mathfrak{z}}\cap \mathfrak{k}.$$
  Now, consider the Levi–Malcev decomposition of $\mathfrak{k}$:

  $$\mathfrak{k}={\mathfrak{k}}_{ss}⋉\mathfrak{z},$$

  where ${\mathfrak{k}}_{ss}$ is the semisimple part and $\mathfrak{z}$ is the center of $\mathfrak{k}$. Let $X\in \tilde{\mathfrak{z}}\cap \mathfrak{k}$. Since $X\in \mathfrak{k}$, X can be expressed as

  $$X=Y+Z,$$

  where $Y\in {\mathfrak{k}}_{ss}$ and $Z\in \mathfrak{z}$. Moreover, since $X\in \tilde{\mathfrak{z}}$, we can write

  $$X=X_1+i{X}_2,$$

  with $X_1,X_2\in \mathfrak{z}$. Combining both decompositions, we have

  $$Y=(X_1-Z)+i{X}_2\in \tilde{\mathfrak{z}}.$$
  Thus, $Y\in {\mathfrak{k}}_{ss}\cap \tilde{\mathfrak{z}}$. Since ${\mathfrak{k}}_{ss}\subset \widetilde{{\mathfrak{k}}_{ss}}$, it follows that

  $$Y\in \widetilde{{\mathfrak{k}}_{ss}}\cap \tilde{\mathfrak{z}}=\left(0\right).$$
  This implies $Y=0$, and consequently, $X=Z\in \mathfrak{z}$. Therefore,

  $$\mathfrak{k}\cap \tilde{\mathfrak{z}}=\mathfrak{z}.$$

  (2)
  Finally, by Equation (1)

  $$\mathfrak{k}\cap i\mathfrak{k}=(\mathfrak{k}\cap \tilde{\mathfrak{z}})\cap i(\mathfrak{k}\cap \tilde{\mathfrak{z}}),$$

  and by Equation (2), one has

  $$\mathfrak{k}\cap i\mathfrak{k}=(\mathfrak{k}\cap \tilde{\mathfrak{z}})\cap i(\mathfrak{k}\cap \tilde{\mathfrak{z}})=\mathfrak{z}\cap i\mathfrak{z}.$$

This completes the proof. □

We will also use the following classical result of Stein; see [21] and also [19].

Theorem 2

(Stein). Any unramified covering of a Stein manifold is Stein.

For our purposes we also need a standard converse for finite quotients, which can be proved using Remmert’s theory of finite maps (equivalently, using direct images of coherent sheaves and Cartan’s theorems).

Lemma 3

(Finite maps preserve Steinness). Let $\pi :X\to Y$ be a finite surjective holomorphic map between reduced complex spaces. If X is Stein, then Y is Stein.

Background used below. If a finite group $\Lambda$ acts holomorphically on a complex space X, then the analytic quotient $X/\Lambda$ exists and the quotient map $\pi :X\to X/\Lambda$ is finite (hence proper); see [19] (Ch. V, §2) or [22] (Ch. I, §5). We will also use Grauert’s direct image theorem: for a proper holomorphic map $\pi$, the direct image of a coherent sheaf is coherent [23] (Thm. (1)). Finally, “Leray yields” refers to the Leray spectral sequence for $\pi$; for finite maps one has $R^q{\pi }_{*}\mathcal{E}=0$ for all $q>0$ and every coherent $\mathcal{E}$, and hence $H^q(Y,{\pi }_{*}\mathcal{E})\cong H^q(X,\mathcal{E})$. Cartan’s Theorems A and B mean that, on a Stein space, every coherent sheaf is globally generated (A) and has vanishing higher cohomology (B); see [19] (Ch. III).

Proof. 

Let $\mathcal{F}$ be a coherent sheaf on Y. Since $\pi$ is finite, $\pi$ is proper and ${\pi }_{*}{\mathcal{O}}_X$ is a coherent ${\mathcal{O}}_Y$-algebra; moreover, ${\pi }_{*}$ is exact on coherent sheaves and $R^q{\pi }_{*}\mathcal{G}=0$ for $q>0$ and any coherent $\mathcal{G}$. Hence, by Leray,

$$H^q\left(Y,{\pi }_{*}{\pi }^{*}\mathcal{F}\right)\cong H^q\left(X,{\pi }^{*}\mathcal{F}\right)(q\ge 0).$$

Since X is Stein, Cartan B gives $H^q(X,{\pi }^{*}\mathcal{F})=0$ for $q\ge 1$, and therefore $H^q(Y,{\pi }_{*}{\pi }^{*}\mathcal{F})=0$ for $q\ge 1$.

It remains to pass from ${\pi }_{*}{\pi }^{*}\mathcal{F}$ to $\mathcal{F}$. Because $\pi$ is finite and surjective, there exists a trace (Reynolds) morphism ${tr}_{\mathcal{F}}:{\pi }_{*}{\pi }^{*}\mathcal{F}\to \mathcal{F}$ which splits the natural map ${\eta }_{\mathcal{F}}:\mathcal{F}\to {\pi }_{*}{\pi }^{*}\mathcal{F}$ (locally this is the usual trace on finite ${\mathcal{O}}_Y$-algebras). Hence $\mathcal{F}$ is a direct summand of ${\pi }_{*}{\pi }^{*}\mathcal{F}$. It follows that $H^q(Y,\mathcal{F})=0$ for $q\ge 1$.

Cartan A for Y follows similarly from Cartan A for X applied to ${\pi }^{*}\mathcal{F}$ and the same splitting argument. Therefore Y is Stein. □

Lemma 4.

If a finite group Λ acts holomorphically on a Stein manifold X, then the analytic quotient $Y:=X/\Lambda$ is Stein.

Proof. 

Since $\Lambda$ is finite, the action is proper and the analytic quotient $Y=X/\Lambda$ exists; moreover, the quotient map $\pi :X\to Y$ is finite and surjective. Hence Y is Stein by Lemma 3. □

Corollary 2.

The group $\tilde{K}$ is Stein if and only if $\tilde{Z}$ is Stein.

Proof. 

By Proposition 3(1), the finite group $A:={\tilde{K}}_{ss}\cap \tilde{Z}$ satisfies

$$\tilde{K}\cong ({\tilde{K}}_{ss}\times \tilde{Z})/A,a·(x,z)=(xa,a^{-1}z).$$

Equivalently, the multiplication map

$$m:{\tilde{K}}_{ss}\times \tilde{Z}⟶\tilde{K},(x,z)⟼xz,$$

is a holomorphic homomorphism with finite kernel $\{(a,a^{-1}):a\in A\}$; hence m is a finite unramified covering.

Assume first that $\tilde{K}$ is Stein. Then ${\tilde{K}}_{ss}\times \tilde{Z}$ is Stein by Theorem 2. Since $\left\{e\right\}\times \tilde{Z}$ is a closed complex submanifold of a Stein manifold, it is Stein; hence $\tilde{Z}$ is Stein.

Conversely, suppose $\tilde{Z}$ is Stein. The semisimple group ${\tilde{K}}_{ss}$ is Stein (it embeds as a closed complex subgroup of some $\mathrm{GL}(N,\mathbb{C})$), so the product ${\tilde{K}}_{ss}\times \tilde{Z}$ is Stein. The finite group A acts holomorphically on ${\tilde{K}}_{ss}\times \tilde{Z}$; therefore the quotient

$$({\tilde{K}}_{ss}\times \tilde{Z})/A\cong \tilde{K}$$

is Stein by Lemma 4. □

This reduces the study of Steinness for a connected complex Lie group G to the structure of the identity component $\tilde{Z}$ of the center of $\tilde{K}$.

3.2. The Structure of the Center $\tilde{Z}$

We recall the standard structure theory for connected abelian complex Lie groups. Let A be a connected abelian complex Lie group with Lie algebra $\mathfrak{a}\simeq {\mathbb{C}}^n$. The exponential map $exp:\mathfrak{a}\to A$ is surjective and its kernel is a discrete subgroup $\Gamma \subset {\mathbb{C}}^n$, hence

$$A\cong {\mathbb{C}}^n/\Gamma$$

as a complex Lie group (see [8,24]).

Let ${\Gamma }_{\mathbb{R}}:={\mathrm{span}}_{\mathbb{R}}(\Gamma )$ and set

$$J:={\Gamma }_{\mathbb{R}}/\Gamma ,$$

the maximal compact subgroup of A (a real torus). Let

$$M_{\Gamma }:={\Gamma }_{\mathbb{R}}\cap i{\Gamma }_{\mathbb{R}}$$

be the maximal complex subspace contained in ${\Gamma }_{\mathbb{R}}$. The additive complex vector group $M_{\Gamma }$ acts on J by translations; we study the orbits $M_{\Gamma }·x\subset J$. Equivalently,

$$\pi \left(M_{\Gamma }\right)=M_{\Gamma }/(M_{\Gamma }\cap \Gamma )\subset J$$

is the connected complex subgroup of J generated by $M_{\Gamma }$.

The statements below about $\mathrm{rank}(\Gamma )$ concern the underlying real Lie group structure; in particular, when $\mathrm{rank}(\Gamma )=n$ the group is (isomorphic as a real Lie group to) ${\left({\mathbb{C}}^{*}\right)}^n$, but not canonically biholomorphic to it.

The following structural theorem (Morimoto–Remmert) will be used repeatedly; see, e.g., [8,14,24].

Theorem 3

(Structure of abelian complex Lie groups). Any connected abelian complex Lie group is isomorphic (as a complex Lie group) to

$${\mathbb{C}}^b\times {\left({\mathbb{C}}^{*}\right)}^d\times C,$$

where C is a Cousin group (equivalently, $\mathcal{O}\left(C\right)=\mathbb{C}$).

In this decomposition, the factor C measures the failure of holomorphic separability and is the unique obstruction to Steinness.

Proposition 4.

The abelian group $\tilde{Z}$ is Stein if and only if its Cousin factor is trivial. Equivalently,

$$\tilde{Z}\mathit{is}\mathit{Stein}\iff M_{\Gamma }=\left(0\right),$$

where $\tilde{Z}\simeq {\mathbb{C}}^n/\Gamma$ and $M_{\Gamma }={\Gamma }_{\mathbb{R}}\cap i{\Gamma }_{\mathbb{R}}$.

Proof. 

By Theorem 3, we may write $\tilde{Z}\simeq {\mathbb{C}}^b\times {\left({\mathbb{C}}^{*}\right)}^d\times C$ with C a Cousin group. The factors ${\mathbb{C}}^b$ and ${\left({\mathbb{C}}^{*}\right)}^d$ are Stein; hence $\tilde{Z}$ is Stein if and only if C is trivial. Moreover, the Cousin factor is precisely the complex subtorus generated by $M_{\Gamma }$ inside the maximal compact torus $J={\Gamma }_{\mathbb{R}}/\Gamma$; thus C is trivial if and only if $M_{\Gamma }=\left(0\right)$. □

Now return to the reductive group $\tilde{K}$. Let $J:=\tilde{Z}\cap K$ be the maximal compact torus of $\tilde{Z}$ (connected). Its Lie algebra is $\mathfrak{j}=\tilde{\mathfrak{z}}\cap \mathfrak{k}$. The maximal complex subalgebra of $\mathfrak{j}$ is $\mathfrak{j}\cap i\mathfrak{j}$. Using Proposition 3(3) and $\mathfrak{k}\cap i\mathfrak{k}\subset \tilde{\mathfrak{z}}$ (see (1)), we obtain

$$\mathfrak{j}\cap i\mathfrak{j}=\mathfrak{k}\cap i\mathfrak{k}.$$

Proposition 5.

The abelian subgroup $\tilde{Z}$ is Stein if and only if

$$\mathfrak{k}\cap i\mathfrak{k}=\left(0\right).$$

Proof. 

By Proposition 4, $\tilde{Z}$ is Stein if and only if the Cousin factor of $\tilde{Z}$ is trivial, equivalently if and only if the maximal complex subgroup of its maximal compact torus J is trivial. At the Lie algebra level this is $\mathfrak{j}\cap i\mathfrak{j}=\left(0\right)$, and as noted above $\mathfrak{j}\cap i\mathfrak{j}=\mathfrak{k}\cap i\mathfrak{k}$. □

Combining Corollaries 1 and 2, and Proposition 5, we recover Theorem 1 and also obtain the following formulations.

Theorem 4.

A connected complex Lie group G is Stein if and only if the maximal compact torus in the identity component of the center of the reductive factor $\tilde{K}$ is totally real (equivalently, has no positive-dimensional complex subtorus).

Theorem 5.

A connected complex Lie group G is Stein if and only if it is holomorphically separable.

Proof. 

In the splitting $G\simeq {\mathbb{C}}^n\times \tilde{K}$, holomorphic separability reduces to holomorphic separability of $\tilde{K}$. By Corollary 2 and the abelian structure theorem, the only obstruction to holomorphic separability is the presence of a nontrivial Cousin factor in $\tilde{Z}$. Thus holomorphic separability holds if and only if the Cousin factor is trivial, which is equivalent to Steinness. □

4. The Steinizer of a Complex Lie Group

For completeness, we briefly recall Morimoto’s notion of the Steinizer [14] and explain how it fits into the Lie-theoretic framework developed above. For a connected complex Lie group G, the Steinizer $G_0$ is the subgroup on which all holomorphic functions are constant. The corresponding quotient $G/G_0$ is therefore holomorphically separable and hence Stein by Theorem 5. Thus, the Steinizer captures precisely the obstruction to Steinness. Moreover, every connected complex Lie group admits a holomorphic principal bundle

$$G⟶G/G_0,$$

whose base $G/G_0$ is a connected Stein complex Lie group and whose fiber $G_0$ is a Cousin group. In particular, G is Stein if and only if its Steinizer $G_0$ is trivial.

Definition 2

(Steinizer of Complex Lie Groups). Let G be a connected complex Lie group. The Steinizer $G_0$ of G is defined as

$$G_0:=\left\{g\in G\mid f\left(g\right)=f\left(e\right)\mathit{for}\mathit{all}f\in \mathcal{O}\left(G\right)\right\}.$$

An important property of $G_0$ is that it forms a closed normal subgroup of G. Moreover, the coset space $G/G_0$ is holomorphically separable. To see this, we define the holomorphic function $\widehat{f}:G/G_0\to \mathbb{C}$ by $\widehat{f}\left(g{G}_0\right):=f\left(g\right)$, for all $f\in \mathcal{O}\left(G\right)$, and suppose $\widehat{f}\left(g_1{G}_0\right)=\widehat{f}\left(g_2{G}_0\right)$ for all $f\in \mathcal{O}\left(G\right)$. Now define the holomorphic function ${\widehat{f}}_{g_2}:G/G_0\to \mathbb{C}$ by

$${\widehat{f}}_{g_2}\left(g{G}_0\right):=\widehat{f}\left(g{g}_2^{-1}{G}_0\right).$$

Since ${\widehat{f}}_{g_2}\left(g_1{G}_0\right)={\widehat{f}}_{g_2}\left(g_2{G}_0\right)$, it follows that

$$\widehat{f}\left(g_1{g}_2^{-1}{G}_0\right)=\widehat{f}\left(e{G}_0\right).$$

This implies that $f\left(g_1{g}_2^{-1}\right)=f\left(e\right)$ for all $f\in \mathcal{O}\left(G\right)$. Consequently, we deduce that $g_1{g}_2^{-1}\in G_0$, or equivalently, $g_1{G}_0=g_2{G}_0$. Thus, the quotient space $G/G_0$ is holomorphically separable. By Theorem 5, this implies that $G/G_0$ is itself a Stein complex Lie group, thus justifying the terminology “Steinizer.”

We will need the following fundamental theorem in the proof of Theorem 7.

Theorem 6

([5]). Let $(P,\pi ,B,G)$ be a holomorphic principal bundle, where B is a Stein manifold and G is a Stein Lie group. Then, the total space P is also Stein.

Theorem 7

([14]). Let G be a connected complex Lie group, and let $G_0$ be its Steinizer. Then $G_0$ is a connected, central, closed complex subgroup of G, and $\mathcal{O}\left(G_0\right)=\mathbb{C}$. Furthermore, $G_0$ is the minimal closed normal complex subgroup of G such that $G/G_0$ is Stein.

Proof. 

Let $x\in G$ and $f\in \mathcal{O}\left(G\right)$. Define the function $f_x\in \mathcal{O}\left(G\right)$ by $f_x\left(g\right):=f\left(xg\right).$ For $x,y\in G_0$, then $f\left(e\right)=f\left(x\right)=f_x\left(e\right)=f_x\left(y\right)=f\left(xy\right),$ for all $f\in \mathcal{O}\left(G\right)$. Hence, $xy\in G_0$. Next, consider the function $f^{\prime }\in \mathcal{O}\left(G\right)$ defined by

$$f^{\prime }\left(g\right):=f\left(g^{-1}\right).$$

For $x\in G_0$, it follows that $f\left(e\right)=f^{\prime }\left(e\right)=f^{\prime }\left(x\right)=f\left(x^{-1}\right)$, which implies $x^{-1}\in G_0$. This proves that $G_0$ is a subgroup of G. A standard argument also shows that $G_0$ is both a closed and complex subgroup. To show that $G_0$ is normal, consider any holomorphic automorphism $\sigma :G\to G$. For $f\in \mathcal{O}\left(G\right)$, define $f^{\sigma }\in \mathcal{O}\left(G\right)$ by

$$f^{\sigma }\left(x\right):=f\left(\sigma \left(x\right)\right).$$

Then, for $x\in G_0$, $f\left(x\right)=f\left(e\right)=f^{\sigma }\left(e\right)=f^{\sigma }\left(x\right)=f\left(\sigma \left(x\right)\right).$ This implies $\sigma \left(G_0\right)\subseteq G_0$, so $G_0$ is a characteristic subgroup of G, and hence normal. We now verify that $G_0$ is connected and central and does not admit non-constant holomorphic functions. Before that, note that if N is any closed normal complex subgroup of G such that $G/N$ is Stein, then N must contain all points where holomorphic functions fail to separate, i.e., $G_0\subseteq N$. Thus, $G_0$ is the minimal closed normal complex subgroup of G such that $G/G_0$ is Stein. To verify that $G_0$ is connected, let $G_0^{\prime }$ denote the connected component of the identity in $G_0$, and consider the covering map $G/G_0^{\prime }\to G/G_0$. By Theorem 2, the quotient $G/G_0^{\prime }$ is Stein. However, by the minimality of $G_0$, it follows that $G_0^{\prime }=G_0$. Therefore, $G_0$ is connected. Next, to show that $G_0$ is central, consider the adjoint representation of G. This representation maps $G_0$ to the identity. If $G_0$ were not mapped to the identity, it would be possible to construct a holomorphic function on G that is non-constant on $G_0$, which would contradict the definition of $G_0$. Hence, $G_0$ must be contained in the center of G. Finally, to see that $\mathcal{O}\left(G_0\right)=\mathbb{C}$, consider the Steinizer $G_{00}$ of $G_0$. Then the principal holomorphic bundle $G/G_{00}\to G/G_0$ has a Stein base and a Stein connected Lie group as fiber. By Theorem 6 and the minimality of $G_0$, it follows that

$$G_{00}=G_0\mathrm{and}\mathcal{O}\left(G_0\right)\cong \mathbb{C}.$$

This completes the proof. □

It is evident that when G is a connected reductive complex Lie group, the Cousin group C constructed in Section 3 coincides with the Steinizer $G_0$. In fact, this holds for any connected complex Lie group. Indeed, if G is any connected complex Lie group, the Cousin group C, as constructed, is a central connected subgroup of $\tilde{K}$, and hence C is a subgroup of G. This subgroup C does not separate points, which implies $C\subseteq G_0$. Conversely, the only subgroup of G that does not separate points is C. Therefore, $G_0\subseteq C$. It follows that $G_0=C$.

5. Pseudoconvexity of G

We now prove one of the fundamental results in this paper that every connected complex Lie group is pseudoconvex.

Lemma 5.

Every Cousin group $C={\mathbb{C}}^n/\Gamma$ is pseudoconvex.

Proof. 

Let ${\Gamma }_{\mathbb{R}}={\mathrm{span}}_{\mathbb{R}}\Gamma$, and decompose ${\Gamma }_{\mathbb{R}}=M_{\Gamma }\oplus W$, where $M_{\Gamma }={\Gamma }_{\mathbb{R}}\cap i{\Gamma }_{\mathbb{R}}$ is the maximal complex subspace with ${dim}_{\mathbb{C}}{M}_{\Gamma }=q$, and W is a complementary totally real subspace. In suitable coordinates, $M_{\Gamma }$ corresponds to the first q complex directions, and W corresponds to the remaining $n-q$ real part directions. Thus:

$${\Gamma }_{\mathbb{R}}=\{\mathbf{z}=\mathbf{x}+i\mathbf{y}\in {\mathbb{C}}^n:y_{q+1}=\cdots =y_n=0\}.$$

Define the $\Gamma$-periodic ${\mathcal{C}}^{\infty }$-function on ${\mathbb{C}}^n$:

$$\widehat{\varphi }\left(\mathbf{z}\right):=\sum _{k=q+1}^n{y}_k^2,\mathbf{z}=\mathbf{x}+i\mathbf{y}\in {\mathbb{C}}^n.$$

This induces a non-negative ${\mathcal{C}}^{\infty }$-function $\varphi$ on ${\mathbb{C}}^n/\Gamma$ defined by

$$\varphi (z+\Gamma ):=\widehat{\varphi }\left(z\right).$$

Since $\widehat{\varphi }$ is $\Gamma$-periodic, for any $w\in \Gamma$,

$$\varphi \left((z+w)+\Gamma \right)=\widehat{\varphi }(z+w)=\widehat{\varphi }\left(z\right).$$

Thus, $\varphi$ is well-defined. Moreover, $\varphi$ is an exhaustion function. To see this, let $c>0$. Then,

$${\widehat{\varphi }}^{-1}\left([0,c)\right)=\left\{\mathbf{z}\in {\mathbb{C}}^n:\widehat{\varphi }\left(\mathbf{z}\right)<c\right\}={\Gamma }_{\mathbb{R}}\times B_c,$$

where

$$B_c=\{\mathbf{0}+i(0,\dots ,0,y_{q+1},\dots ,y_n)\in {\mathbb{C}}^n:y_{q+1}^2+\cdots +y_n^2<c\}.$$

Since, $\Gamma \cap B_c=\left(\mathbf{0}\right)$, then

$${\varphi }^{-1}\left([0,c)\right)={\Gamma }_{\mathbb{R}}/\Gamma \times B_c,$$

which is relatively compact in ${\mathbb{C}}^n/\Gamma$. Therefore, $\varphi$ is an exhaustion function.

Next, we prove that $\varphi$ is plurisubharmonic. The Levi form of $\widehat{\varphi }$ is computed as, for $u=({\zeta }_1,\dots ,{\zeta }_n)\in {\mathbb{C}}^n$,

$$\begin{array}{cc}\hfill \mathcal{L}\left(\widehat{\varphi }\right)(u,u)& =\sum _{j,k=1}^n{\displaystyle \frac{{\partial }^2\widehat{\varphi }}{\partial z_j\partial {\overline{z}}_k}}{\zeta }_j{\overline{\zeta }}_k\hfill \\ \hfill & ={\displaystyle \frac{1}{4}}\sum _{j,k=1}^n\left({\displaystyle \frac{{\partial }^2\widehat{\varphi }}{\partial x_j\partial x_k}}+{\displaystyle \frac{{\partial }^2\widehat{\varphi }}{\partial y_j\partial y_k}}\right){\zeta }_j{\overline{\zeta }}_k\hfill \\ \hfill & ={\displaystyle \frac{1}{4}}\sum _{j,k=1}^n{\displaystyle \frac{{\partial }^2\widehat{\varphi }}{\partial y_j\partial y_k}}{\zeta }_j{\overline{\zeta }}_k\hfill \\ \hfill & ={\displaystyle \frac{1}{4}}\sum _{j,k=1}^n{\displaystyle \frac{{\partial }^2}{\partial y_j\partial y_k}}\left(\sum _{k=q+1}^n{y}_k^2\right){\zeta }_j{\overline{\zeta }}_k\hfill \\ \hfill & ={\displaystyle \frac{1}{2}}\sum _{k=q+1}^n{\zeta }_k{\overline{\zeta }}_k={\displaystyle \frac{1}{2}}\sum _{k=q+1}^n{\left|{\zeta }_k\right|}^2.\hfill \end{array}$$

Let $\pi :{\mathbb{C}}^n\to C={\mathbb{C}}^n/\Gamma$ be the quotient map. Since $\Gamma$ is discrete, $\pi$ is a holomorphic covering map and hence a local biholomorphism. In particular, for every $z\in {\mathbb{C}}^n$ the differential

$$d{\pi }_z:T_z{\mathbb{C}}^n⟶T_{\pi \left(z\right)}C$$

is a complex-linear isomorphism.

By construction $\varphi \circ \pi =\widehat{\varphi }$. Therefore, for $u\in T_z{\mathbb{C}}^n$ we have the identity of Levi forms

$${\mathcal{L}}_{\varphi }\left(\pi \left(z\right)\right)\left(d{\pi }_z\left(u\right),d{\pi }_z\left(u\right)\right)={\mathcal{L}}_{\widehat{\varphi }}\left(z\right)(u,u).$$

Writing $u=({\zeta }_1,\dots ,{\zeta }_n)\in {\mathbb{C}}^n\cong T_z{\mathbb{C}}^n$, the above computation gives

$${\mathcal{L}}_{\widehat{\varphi }}\left(z\right)(u,u)={\displaystyle \frac{1}{2}}\sum _{k=q+1}^n{\left|{\zeta }_k\right|}^2\ge 0.$$

Hence ${\mathcal{L}}_{\varphi }\ge 0$ on C, so $\varphi$ is plurisubharmonic. Moreover, the Levi form vanishes precisely on those tangent vectors whose components in the last $n-q$ directions are zero (equivalently, on the complex directions corresponding to $M_{\Gamma }$). This completes the proof. □

Lemma 6.

Finite products of pseudoconvex complex manifolds are pseudoconvex.

Proof. 

If ${\varphi }_j$ is a psh exhaustion on $X_j$, then $\varphi :={\sum }_{j=1}^m{\varphi }_j\circ {\mathrm{pr}}_j$ is psh and proper on $X_1\times \cdots \times X_m$. □

Lemma 7.

If a finite group Λ acts holomorphically on a pseudoconvex complex manifold X, then the quotient $Y:=X/\Lambda$ is pseudoconvex.

Proof. 

Let $\pi :X\to Y$ be the analytic quotient map. Since $\Lambda$ is finite, $\pi$ is finite and hence proper. Let $\varphi$ be a plurisubharmonic exhaustion on X. For each $g\in \Lambda$, the function $\varphi \circ g$ is psh, and a finite average of psh functions is psh. Define

$$\tilde{\varphi }\left(x\right):={\displaystyle \frac{1}{|\Lambda |}}\sum _{g\in \Lambda }\varphi (g·x).$$

Then $\tilde{\varphi }$ is $\Lambda$–invariant, hence constant on the $\pi$–fibres, so it descends to a unique upper semicontinuous function $\psi :Y\to \mathbb{R}$ with $\tilde{\varphi }=\psi \circ \pi$.

To see that $\psi$ is plurisubharmonic, let $h:\Delta \to Y$ be a holomorphic disc. Let $B\subset Y$ be the set where $\pi$ fails to be a local biholomorphism. Then $h^{-1}\left(B\right)$ is a discrete subset of $\Delta$. On ${\Delta }^{\prime }:=\Delta \setminus h^{-1}\left(B\right)$ the map $\pi$ is a covering, so ${h|}_{{\Delta }^{\prime }}$ admits local holomorphic lifts $s:{\Delta }^{\prime }\to X$ with ${\pi \circ s=h|}_{{\Delta }^{\prime }}$. Hence

$${(\psi \circ h)|}_{{\Delta }^{\prime }}=\tilde{\varphi }\circ s$$

is subharmonic on ${\Delta }^{\prime }$. Since $\psi \circ h$ is upper semicontinuous on $\Delta$ (in particular locally bounded above near the punctures), it extends subharmonically across the discrete set $h^{-1}\left(B\right)$. Therefore $\psi$ is psh on Y.

Finally, $\psi$ is an exhaustion: for $c\in \mathbb{R}$,

$${\psi }^{-1}\left((-\infty ,c]\right)=\pi ({\tilde{\varphi }}^{-1}\left((-\infty ,c]\right)),$$

and the right-hand side is relatively compact in Y because $\tilde{\varphi }$ is proper and $\pi$ is proper. Thus Y is pseudoconvex. □

The following theorem is due to Kazama. In the abelian case, Kazama proves that a complex abelian Lie group admits a smooth plurisubharmonic exhaustion whose Levi form has controlled rank, yielding q-completeness; the proof is constructive, reducing to a quotient ${\mathbb{C}}^n/\Gamma$ and verifying the Levi-form positivity by explicit computation [6]. For general connected complex Lie groups, Kazama obtains q-completeness (and hence pseudoconvexity) within the q-complete framework, relying on results developed in his earlier work [7].

Our contribution is an alternative short Lie-theoretic proof: we avoid the q-complete/cohomological machinery and coordinate Levi-form computations, and instead use structural properties of complex Lie groups. Using Matsushima’s splitting $G\simeq {\mathbb{C}}^n\times \tilde{K}$, we reduce to the reductive factor, separate the semisimple and central (Cousin) contributions, and conclude by permanence of pseudoconvexity under products and finite quotients.

Theorem 8.

Every connected complex Lie group is pseudoconvex.

Proof. 

Let G be connected and let $K\subset G$ be a maximal compact subgroup. By Lemma 1 and Proposition 2,

$$G\cong {\mathbb{C}}^n\times \tilde{K}$$

biholomorphically. By Lemma 6, it suffices to prove that $\tilde{K}$ is pseudoconvex.

The group $\tilde{K}$ is connected and reductive. Let ${\tilde{K}}_{ss}$ be its connected semisimple part and let $\tilde{Z}$ be the identity component of its center. By Proposition 3(1),

$$\tilde{K}\cong ({\tilde{K}}_{ss}\times \tilde{Z})/A,A:={\tilde{K}}_{ss}\cap \tilde{Z}\mathrm{finite}.$$

Now ${\tilde{K}}_{ss}$ is Stein (hence pseudoconvex), and

$$\tilde{Z}\cong {\mathbb{C}}^{\ell }\times {\left({\mathbb{C}}^{*}\right)}^k\times C$$

with C a Cousin group (Theorem 3), hence pseudoconvex by Lemma 5. Therefore ${\tilde{K}}_{ss}\times \tilde{Z}$ is pseudoconvex by Lemma 6, and the finite quotient by A is pseudoconvex by Lemma 7. Hence $\tilde{K}$ is pseudoconvex and therefore so is G. □