Proof. It is evident that $\widetilde{{\rm Lin}}_{\mathbb {C}}(G)$ is closed and normal.

The regular representation of the discrete group $G/G_0$ in the Hilbert space $\ell ^{2}(G/G_0)$ is faithful and automatically holomorphic, so $\widetilde{{\rm Lin}}_{\mathbb {C}}(G)\subset G_0$. On the other hand, the kernel of the adjoint representation coincides with the centralizer of $G_0$ in $G$ [Reference Hilgert and Neeb11, Lemma 9.2.21]. Thus, $\widetilde{{\rm Lin}}_{\mathbb {C}}(G)$ is contained in the centre of $G_0$.

Proof. By Corollary 2.2, it suffices to show that ${{\rm Lin}_\mathbb {C}}(G)\subset {{\rm Lin}_\mathbb {C}}(G_0)$. We need to prove that for any finite-dimensional holomorphic representation $\pi _0$ of $G_0$ there is a finite-dimensional holomorphic representation $\pi$ of $G$ such that $\pi (h)=1$ implies $\pi _0(h)=1$ when $h\in G_0$.

Fix $\pi _0$, denote by $X$ the representation space of $\pi _0$, and consider the induced representation $\pi$ of $G$, i.e., the representation associated with the ${\mathscr A}(G)$-module $Y:={\mathscr A}(G)\otimes _{{\mathscr A}(G_0)} X$. Here ${\mathscr A}(G)$ is considered as a right ${\mathscr A}(G_0)$-module with respect to an algebra homomorphism ${\mathscr A}(G_0)\to {\mathscr A}(G)$ induced by the group homomorphism $G_0\to G$.

We can consider each coset $\gamma \in G/G_0$ as a complex manifold. It is not hard to see that

\[ {\mathscr A}(G)\cong \bigoplus_{\gamma\in G/G_0}{\mathscr A}(\gamma) \]

as right ${\mathscr A}(G_0)$-modules. Since $X$ is finite-dimensional and $G/G_0$ is finite, the ${\mathscr A}(G)$-module $Y$ is a finite-dimensional linear space. Endowing $Y$ with the topology of the projective tensor product, we make it a Banach ${\mathscr A}(G)$-module. Hence, $\pi$ is a holomorphic representation of $G$ on $Y$.

Since the counit $\varepsilon :{\mathscr A}(G)\to \mathbb {C}$ is a quotient homomorphism of algebras, $\mathbb {C}$ is a right ${\mathscr A}(G)$-module with respect to the multiplication determined by $\lambda \cdot \delta _g:=\lambda$ (here $\delta _g$ is the delta-function at $g\in G$). Therefore, $\varepsilon$ is a right ${\mathscr A}(G)$-module morphism and so is a right ${\mathscr A}(G_0)$-module morphism. Hence, the linear map $\varepsilon \otimes 1:Y\to X:\nu \otimes x\mapsto \varepsilon (\nu )x$ is well defined.

Suppose that $h\in G_0$ and $x\in X$. Then $\delta _h\cdot (1\otimes x)=\delta _h\otimes x=1\otimes \delta _h\cdot x$. If, in addition, $\pi (h)=1$, then $\delta _h\cdot (1\otimes x)=1\otimes x$. Applying $\varepsilon \otimes 1$ to the equality $1\otimes \delta _h\cdot x= 1\otimes x$, we have $\delta _h\cdot x=x$. Since $x$ is arbitrary, we obtain $\pi _0(h)=1$. Thus, ${{\rm Lin}_\mathbb {C}}(G)\subset {{\rm Lin}_\mathbb {C}}(G_0)$.

Proof. First, note that it follows from [Reference Akbarov1, Theorem 5.10] that ${\mathscr A}_{\exp }(G)$ is a nuclear Fréchet space. Then $\mathcal {O}_{\exp }(G)$ is a nuclear (DF)-space and, by [Reference Aristov4, Proposition 5.5(B)], the natural map

\[ \mathcal{O}(G)\times\mathcal{O}(G)\to \mathcal{O}(G\times G) \]

induce a topological isomorphism

\[ \mathcal{O}_{\exp}(G)\mathbin{\widehat{\otimes}} \mathcal{O}_{\exp}(G)\cong \mathcal{O}_{\exp}(G\times G). \]

We claim that under this identification, the operations in (1.1) makes $\mathcal {O}_{\exp }(G)$ a Hopf $\mathbin {\widehat {\otimes }}$-algebra. The only non-trivial assertions are that $\mathcal {O}_{\exp }(G)$ is a $\mathbin {\widehat {\otimes }}$-algebra and $\Delta$ in (1.1) is a well-defined map $\mathcal {O}_{\exp }(G)\to \mathcal {O}_{\exp }(G\times G)$. The former is proved in [Reference Aristov4, Lemma 5.2]. To show the latter take for $f\in \mathcal {O}_{\exp }(G)$ a locally bounded submultiplicative weight $\omega$ such that $|f(g)|\leqslant \omega (g)$ for all $g\in G$. Then $\omega \times \omega :(g,\,h)\mapsto \omega (g)\omega (h)$ is a locally bounded submultiplicative weight on $G\times G$ such that

\[ |\Delta(f)(g,h)|=|f(gh)|\leqslant \omega(gh)\leqslant (\omega\times \omega)(g,h) \]

for all $g,\,h\in G$. Thus, $\Delta (f)\in \mathcal {O}_{\exp }(G\times G)$.

Proof. Since $\widetilde \pi _0'$ and $\widetilde \pi _0$ are Hopf $\mathbin {\widehat {\otimes }}$-algebra homomorphisms, it is sufficient to prove that they are topologically isomorphisms. But, by Theorem 3.2, ${\mathscr A}_{\exp }(G)\cong \widehat {\mathscr A}(G)$, so we can apply exactly the same argument as in the proof of [Reference Aristov4, Theorem 5.3] for ${{\rm Lin}_\mathbb {C}}(G)$ in the connected case.

Proof. The necessity follows from Part (C) of Theorem 3.4. To prove the sufficiency note that each function in $\mathcal {O}_{\exp }(G)$ is a coefficient of a holomorphic representation of $G$ in some Banach space (see, e.g., [Reference Aristov4, Proposition 5.1]). Since such functions are constant on $H$, the corresponding representation maps each element of $H$ to the identity operator.

Proof. By the definition of the holomorphic duality functor, $\mathcal {O}(G)^{\bullet }\cong \widehat {\mathscr A}(G)$, which in turn is topologically isomorphic to ${\mathscr A}_{\exp }(G)$ (Theorem 3.2).

Note that ${\mathscr A}_{\exp }(G)$ is a nuclear Fréchet space [Reference Akbarov1, Theorems 5.10]. So it is reflexive as a locally convex space [Reference Bogachev and Smolyanov7, Theorem 3.7.12], therefore ${\mathscr A}_{\exp }(G)'=\mathcal {O}_{\exp }(G)''\cong \mathcal {O}_{\exp }(G)$. Thus $\mathcal {O}(G)^{\bullet \bullet }\cong \widehat {\mathcal {O}}_{\exp }(G)$ and then $\mathcal {O}_{\exp }(G)\to \mathcal {O}(G)$ is an Arens-Michael envelope if and only if the canonical map $\mathcal {O}(G)^{\bullet \bullet }\to \mathcal {O}(G)$ is a topological isomorphism.

Proof. In fact, the result is proved in [Reference Akbarov1]; we give a sketch. Since $\mathcal {O}(\Gamma )$ is the algebra of all functions with the topology of uniform convergence on finite subsets, it suffices to show that for each continuous submultiplicative prenorm $\|\cdot \|$ on $\mathcal {O}_{\exp }(\Gamma )$, there are a finite subset $S$ in $\Gamma$ and $C>0$ such that $\|f\|\leqslant C \max _{\gamma \in S} |f(\gamma )|$.

For $\gamma \in \Gamma$ denote by $\chi _\gamma$ the characteristic function of $\{\gamma \}$. Since $\Gamma$ is discrete, $\chi _\gamma \in \mathcal {O}_{\exp }(\Gamma )$. By [Reference Akbarov1, Lemma 6.3], the set $S:=\{\gamma \in \Gamma :\|\chi _\gamma \|>0\}$ is finite. For any $f\in \mathcal {O}_{\exp }(\Gamma )$, the series $\sum \nolimits _{\gamma \in \Gamma } f(\gamma )\chi _\gamma$ converges to $f$ in the topology of $\mathcal {O}_{\exp }(\Gamma )$ [Reference Akbarov1, Lemma 6.1]. So we have

\[ \|f\|\leqslant \sum_{\gamma\in \Gamma} |f(\gamma)|\,\|\chi_\gamma\|=\sum_{\gamma\in S} |f(\gamma)|\|\chi_\gamma\|\leqslant C \max_{\gamma\in S} |f(\gamma)|, \]

where $C:=\sum \nolimits _{\gamma \in S} \|\chi _\gamma \|$.

Proof. Put $\Gamma := G/G_1$ and for any $\gamma \in \Gamma$ denote by $\chi _\gamma$ the characteristic function of $\gamma$ (now as a subset of $G$). Denote by ${}_gf$ the left shift a function $f$ on $g\in G$. Fix $g_\gamma \in \gamma$ for each $\gamma \in S$. Since every function in $\mathcal {O}_{\exp }(G)$ is a coefficient of a holomorphic representation of $G$ in some Banach space (see, e.g., [Reference Aristov4, Proposition 5.1]), $\mathcal {O}_{\exp }(G)$ is invariant under shifts. So for any finite subset $S$ of $G/G_1$, we have a continuous homomorphism

(3.2)\begin{equation} \psi:\mathcal{O}_{\exp}(G)\to\mathcal{O}_{\exp}(G_1)^{|S|}:f\mapsto({}_{g_\gamma}f\chi_{G_1}). \end{equation}

Note that $\psi _1$ is the restriction map, i.e., $\psi _1(f)= f\chi _{G_1}$. Hence, since the range of $\psi _1$ is dense, so is the range of $\psi$. Further, since $G_1$ is linear, [Reference Akbarov1, Theorem 5.9] implies that the range of $\mathcal {O}_{\exp }(G_1)\to \mathcal {O}(G_1)$ is dense; so is the range of $\mathcal {O}_{\exp }(G_1)^{|S|} \to \mathcal {O}(G_1)^{|S|}$.

Finally, note that any $f\in \mathcal {O}(G)$ can be approximated by $\sum \nolimits _{\gamma \in S}f \chi _\gamma$ for some finite subset $S$ of $G/G_1$. Thus, the range of $\iota$ is dense.

Proof. (The following argument is contained implicitly in [Reference Akbarov1].) We use the notation from the proof of Proposition 3.8. It is sufficient to show that, for a Banach algebra $B$ with a submultiplicative norm $\|\cdot \|$, any continuous homomorphism $\varphi :\mathcal {O}_{\exp }(G)\to B$ factors uniquely on $\iota :\mathcal {O}_{\exp }(G)\to \mathcal {O}(G)$.

Denote by $\sigma$ the continuous homomorphism $\mathcal {O}_{\exp }(\Gamma )\to \mathcal {O}_{\exp }(G)$ induced by the quotient map $G\to \Gamma$, where $\Gamma := G/G_1$. Note that $\Gamma$ is discrete and finitely generated; so Proposition 3.7 implies that $\varphi \sigma$ factors on $\mathcal {O}_{\exp }(\Gamma )\to \mathcal {O}(\Gamma )$. Since $\mathcal {O}(\Gamma )$ is just the algebra of all functions on $\Gamma$ endowed with the topology of uniform convergence on finite subsets, there are a finite subset $S$ of $\Gamma$ and $C>0$ such that

\[ \|\varphi\sigma(h)\|\leqslant C\max_{\gamma\in S}|h(\gamma)|\text{ for each }h\in \mathcal{O}_{\exp}(\Gamma). \]

In particular, since the characteristic function $\chi _\gamma$ equals the image of the delta-function at $\gamma$ under $\sigma$, we have $\|\varphi (\chi _\gamma )\|=0$ for $\gamma \notin S$. By [Reference Akbarov1, Lemma 6.2], the series $\sum \nolimits _{\gamma }\delta _\gamma$ converges to $1$ in $\mathcal {O}_{\exp }(\Gamma )$, hence $f=\sum \nolimits _{\gamma }f\chi _\gamma$ for any $f\in \mathcal {O}_{\exp }(G)$. Therefore,

(3.3)\begin{equation} \varphi(f)=\sum_{\gamma\in S }\varphi(f\chi_\gamma)\quad\text{for each }f\in \mathcal{O}_{\exp}(G). \end{equation}

Consider the homomorphism $\psi$ defined in (3.2) and note that the same formula determines a continuous homomorphism $\psi ':\mathcal {O}(G)\to \mathcal {O}(G_1)^{|S|}$. Put

\[ \upsilon: \mathcal{O}_{\exp}(G_1)^{|S|}\to B:(f_\gamma)\mapsto \sum_{\gamma\in S} \varphi({}_{g_\gamma^{{-}1}}f_\gamma \chi_\gamma). \]

It is clear that (3.3) implies $\varphi =\upsilon \psi$.

Note that the Arens-Michael envelope functor commutes with finite products. By the assumption, $\mathcal {O}_{\exp }(G_1)\to \mathcal {O}(G_1)$ is an Arens-Michael envelope, so is $\mathcal {O}_{\exp }(G_1)^{|S|}\to \mathcal {O}(G_1)^{|S|}$. Therefore $\upsilon$ factors on this map. Thus, we obtain the diagram

where $\beta$ is determined by the universal property and $\alpha :=\beta \psi '$. Since the square and the big triangle are commutative, we have $\varphi =\alpha \iota$.

By the hypothesis, the range of $\iota$ is dense in $\mathcal {O}(G)$, hence $\iota$ is an epimorphism. Thus, the homomorphism $\alpha$ satisfying $\varphi =\alpha \iota$ is unique.

Proof. Since $G_0$ is linear, it follows from Proposition 3.8 that $\iota$ has dense range. Moreover, [Reference Aristov4, Theorem 5.12(B)] implies that the natural embedding $\mathcal {O}_{\exp }(G_0)\to \mathcal {O}(G_0)$ is an Arens-Michael envelope because $G_0$ is linear and connected. Being compactly generated, $G$ satisfies to the hypothesis of Proposition 3.9, so $\iota$ is an Arens-Michael envelope.

Proof. Since $G$ is linear, so is $G_0$. By Corollary 3.10, it is sufficient to show that the restriction map $\psi _0:\mathcal {O}_{\exp }(G)\to \mathcal {O}_{\exp }(G_0)$ has dense range.

In fact, $\psi _0$ is surjective. Indeed, denote by $\mu (f)$ the extension of $f\in \mathcal {O}(G_0)$ such that $\mu (f)(g)=0$ when $g\notin G_0$. Since $G/G_0$ is finite, $G_0$ is obviously cocompact in $G$. Any cocompact closed subgroup of a compactly generated locally compact group is undistorted [Reference Varopoulos19, Lemma 1], that is, $\ell _0\lesssim \ell$ for any word length functions $\ell _0$ and $\ell$ on $G_0$ and $G$, respectively. Therefore, we have from (3.1) that $\mu (f)\in \mathcal {O}_{\exp }(G)$ for any $f\in \mathcal {O}_{\exp }(G_0)$. Note that $\psi _0\mu$ is the identity map on $\mathcal {O}_{\exp }(G_0)$, so $\psi _0$ is surjective.

Proof. The structure of the argument is the same as in [Reference Aristov4, Theorem 5.12]. First, we prove the necessity in Part (B), then Part (A) and, finally, the sufficiency in Part (B).

1. (1) If $G$ is linear, then so is $G_0$. Since any connected linear group is biholomorphically equivalent to an affine algebraic complex variety, we have that $G_0$ is a Stein group and so is $G$. On the other hand, Theorem 3.11 implies that $\iota$ is an Arens-Michael envelope. Thus, the condition in Part (B) is necessary.

2. (2) Write $L:={{\rm Lin}_\mathbb {C}}(G)$. It is obvious that $G/L$ is linear and have finitely many components. By Theorem 2.5, we have $L=\widetilde{{\rm Lin}}_{\mathbb {C}}(G)$. Since any Lie group with finitely many components is compactly generated, Theorem 3.4(B) implies that $\widetilde \pi$ is a topological isomorphism. The above argument shows that $j$ is an Arens-Michael envelope, so is $j\widetilde \pi ^{-1}$. Part (A) is proved.

3. (3) Suppose that $G$ is a Stein group and $\iota$ is an Arens-Michael envelope. Since, by Part (A), so is $j\widetilde \pi ^{-1}$, the universal property of the Arens-Michael enveloping functor implies that $\mathcal {O}(G/L)\to \mathcal {O}(G)$ is a topological isomorphism of Stein algebras. By Forster's Duality Theorem [Reference Forster10], the quotient map $G\to G/L$ is a biholomorphic equivalence, therefore $L$ is trivial. This completes the proof of the sufficiency in Part (B).

Proof. Since $\mathcal {O}_{\exp }(G)$ is dense in $\mathcal {O}(G)$, each function in $\mathcal {O}(G)$ is constant on $H$. Being a Stein manifold, $G$ is holomorphically separable; hence $H=\{1\}$.

Proof of Proposition 4.1. Since $\mathcal {O}(G)$ is holomorphically reflexive, it follows from Proposition 3.6 that the embedding $\mathcal {O}_{\exp }(G)\to \mathcal {O}(G)$ is an Arens-Michael envelope; in particular, it has dense range. On the other hand, by Theorem 3.4(C), each function in $\mathcal {O}_{\exp }(G)$ is constant on $\widetilde{{\rm Lin}}_{\mathbb {C}}(G)$. So Lemma 4.2 implies that $\widetilde{{\rm Lin}}_{\mathbb {C}}(G)=\{1\}$. Finally, ${{\rm Lin}_\mathbb {C}}(G_0)=\{1\}$ by Corollary 2.2.

Proof. Since $G$ is compactly generated, the implications $\mathrm {(i)}\Longrightarrow \mathrm {(ii)}$ and $\mathrm {(i)}\Longrightarrow \mathrm {(iii)}$ follow from Proposition 4.1. Theorem 2.5 implies that ${{\rm Lin}_\mathbb {C}}(G_0)={{\rm Lin}_\mathbb {C}}(G)=\widetilde{{\rm Lin}}_{\mathbb {C}}(G)$, therefore $\mathrm {(ii)}\Longleftrightarrow \mathrm {(iii)}$. Finally, the implication $\mathrm {(iii)}\Longrightarrow \mathrm {(i)}$ follows from Theorem 3.12(B) and Proposition 3.6.

Proof. We have to show that the map $\varkappa :{\mathscr A}_{\exp }(G)^{\bullet \bullet }\to {\mathscr A}_{\exp }(G)$ defined in (1.4) is an isomorphism.

Put $\widetilde L:=\widetilde{{\rm Lin}}_{\mathbb {C}}(G)$. Since ${\mathscr A}_{\exp }(G)'\cong \mathcal {O}_{\exp }(G)$, it follows from the definition of the holomorphic dual that ${\mathscr A}_{\exp }(G)^{\bullet }\cong \widehat {\mathcal {O}}_{\exp }(G)$. Moreover, $\mathcal {O}_{\exp }(G/\widetilde L)\cong \mathcal {O}_{\exp }(G)$ by Theorem 3.4. The assumption that $\mathcal {O}(G/\widetilde L)$ is holomorphically reflexive and Proposition 3.6 imply that $\widehat {\mathcal {O}}_{\exp }(G)\cong \mathcal {O}(G/\widetilde L)$. Hence ${\mathscr A}_{\exp }(G)^{\bullet }\cong \mathcal {O}(G/\widetilde L)$.

On the other hand, $\mathcal {O}(G/\widetilde L)'\cong {\mathscr A}(G/\widetilde L)$ and, by Akbarov's result (Theorem 3.2), ${\mathscr A}(G/\widetilde L)\to {\mathscr A}_{\exp }(G/\widetilde L)$ is an Arens-Michael envelope. It follows from Theorem 3.4 that the naturally defined map ${\mathscr A}_{\exp }(G)\to {\mathscr A}_{\exp }(G/\widetilde L)$ is an isomorphism. So $\mathcal {O}(G/\widetilde L)^{\bullet }\cong {\mathscr A}_{\exp }(G)$ and therefore $\varkappa$ is an isomorphism.

Proof. If follows from Theorems 2.5 and 3.12 that $j\widetilde \pi ^{-1}:\mathcal {O}_{\exp }(G)\to \mathcal {O}(G/\widetilde{{\rm Lin}}_{\mathbb {C}}(G))$ is an Arens-Michael envelope. So, applying Theorem 3.4 we get that $\mathcal {O}(G/\widetilde{{\rm Lin}}_{\mathbb {C}}(G))$ is holomorphically reflexive. Since $G$ is compactly generated, Theorem 4.5 implies that ${\mathscr A}_{\exp }(G)$ is also holomorphically reflexive.

Proof. Let $f\in \mathcal {O}_{\exp }(G)$. Consider the holomorphic homomorphism $\mu :\mathbb {C}^{\times }\to G$ that implements the isomorphism $\mathbb {C}^{\times }\to H$. Since $f\mu$ is holomorphic on $\mathbb {C}^{\times }$, we can write the Laurent series: $f\mu (z)=\sum \nolimits _{m\in \mathbb {Z}}c_m\ z^{m}$. To show that $f$ is constant on $H$ it suffices to prove that $c_m=0$ for each $m\ne 0$.

Let $U$ be a compact generating subset of $G$ and $\omega$ the corresponding word submultiplicative weight, i.e., the exponent of the word length function. We can assume that $\mu (\mathbb {T})$ (where $\mathbb {T}=\{z:|z|=1\}$) is contained in $U$. Then $\omega \mu (\lambda )=1$ for any $\lambda \in \mathbb {T}\setminus \{1\}$ and, by the submultiplicativity,

(5.1)\begin{equation} \omega\mu(z)= \omega\mu(|z|)\quad \text{for each }z\in \mathbb{C}. \end{equation}

Since $f$ is of exponential type and $\omega$ is a word weight, there are $C>0$ and $k\in \mathbb {N}$ such that

\[ |f(g)|\leqslant C\omega^{k}(g)\quad \text{for each }g\in G. \]

Write $M_R:=\max \{|f\mu (z)|:|z|=R\}$. Then, by (5.1), we have for every $R>0$ that

(5.2)\begin{equation} M_R\leqslant \max\{C\omega^{k}\mu(z): |z|=R\}= C \omega^{k}\mu(R). \end{equation}

Note that $z\mapsto \max \{|z|,\,|z|^{-1}\}$ is a submultiplicative weight on $\mathbb {C}^{\times }$ such that the corresponding length function is equivalent to a word length function. Since $H$ is distorted in $G$, there is a sequence $(z_n)$ in $\mathbb {C}^{\times }$ such that

(5.3)\begin{equation} \max\{|z_n|, |z_n|^{{-}1}\}>\omega^{n}\mu(z_n)\quad \text{for each }n\in\mathbb{N}. \end{equation}

Passing to a subsequence if necessary, we can assume that $(z_n)$ has a limit, finite or infinite, and (5.3) is still satisfied.

We claim first that $\lim _{n\to \infty } z_n$ does not belong to $\mathbb {C}^{\times }$. Indeed, assume the opposite. Then there is $K>0$ such that $\max \{|z_{n}|,\,|z_{n}|^{-1}\}\leqslant K$ and therefore, by (5.3), $\omega \mu (z_{n})< K^{1/n}$ for every $n$. Since $\omega$ takes values in $\{e^{p}:p\in \mathbb {Z}_+\}$, we have that $\omega \mu (z_{n})=1$ eventually. By the definition of a word weight, $\omega (g)=1$ if only if $g=1$. Since $\mu$ is injective, $z_{n}=1$ eventually. Applying again (5.3), we have that $\omega (1)<1$ and get a contradiction.

Thus, only two cases may occur: $\lim z_n=\infty$ and $\lim z_n=0$. Suppose that $\lim z_n=\infty$ and write $R_n:=|z_n|$. Applying consequently (5.2) and (5.3), we get

\[ M_{R_n}\leqslant C \omega^{k}\mu(R_n)< C \max\{R_n^{k/n},R_n^{{-}k/n}\}. \]

Now consider the coefficient $c_m$ of the Laurent series for $m\ne 0$. If $n\geqslant k(|m|-1/2)^{-1}$, then $|m|-k/n\geqslant 1/2$ and hence $R^{k/n-|m|}\leqslant R^{-1/2}$ when $R\geqslant 1$. Since $R_n\geqslant 1$ eventually,

\[ \frac{M_{R_n}}{R_n^{|m|}} \leqslant C R_n^{k/n-|m|}\leqslant C\ R_n^{{-}1/2} \]

also eventually. Finally, we have $\lim _{n\to \infty } M_{R_n}/R_n^{|m|}=0$ since $\lim R_n=\infty$. Cauchy's inequality for Laurent series asserts that $|c_m|\leqslant M_R/R^{m}$ for any $R>0$ and $m\in \mathbb {Z}$, therefore, putting $R=R_n$ for $m>0$ and $R=R_n^{-1}$ for $m<0$, we have $c_m=0$.

The proof in the case when $\lim z_n=0$ is similar.

Proof. The component $G_0$ of the identity is isomorphic to $\mathbb {C}^{\times }$ and distorted in $G$ (in fact, it is of strictly quadratic distortion). By Proposition 5.1, each holomorphic function of exponential type on $G$ is constant on $G_0$.

Suppose to the contrary that $\mathcal {O}(G)$ is holomorphically reflexive. Since $G$ is compactly generated, then, by Proposition 3.6, the embedding $\mathcal {O}_{\exp }(G)\to \mathcal {O}(G)$ is an Arens-Michael envelope. In particular, $\mathcal {O}_{\exp }(G)$ is dense in $\mathcal {O}(G)$. Lemma 4.2 implies that $G_0$ is trivial and we get a contradiction.

Since each function from $\mathcal {O}_{\exp }(G)$ is constant on $G_0$, it follows from Corollary 3.5 that $G_0\subset \widetilde{{\rm Lin}}_{\mathbb {C}}(G)$. So, by Lemma 2.3, $G_0= \widetilde{{\rm Lin}}_{\mathbb {C}}(G)$. Part (B) of Theorem 3.4 implies that $\mathcal {O}_{\exp }(G)\cong \mathcal {O}_{\exp }(\Gamma )$, where $\Gamma =G/G_0$. Since $\Gamma$ is discrete and finitely generated, $\mathcal {O}_{\exp }(\Gamma )\cong \mathcal {O}(\Gamma )$ by Proposition 3.7 and so $\mathcal {O}(\Gamma )$ is holomorphically reflexive. Finally, Theorem 4.5 implies that ${\mathscr A}_{\exp }(G)$ is holomorphically reflexive.

Proof. Let $K$ be a compact generating subset of $H$. Then

\[ S:=\{\exp \eta:\|\eta\|\leqslant 1\}\times K \]

is compact and generates $G$. We can assume that $\ell$ is the word length function associated with $S$.

Denote by $\alpha _h$ the action of $h\in H$ on $N$ and by $\mathbf {L}(\alpha _h)$ the corresponding automorphism of $\mathfrak {n}$. Since $K$ is compact, there is $B\geqslant 0$ such that $\|\mathbf {L}(\alpha _h)\|\leqslant B$ for all $h\in K$. We assume that $B>1$ for technical reasons.

Denote by $\ell _N$ a word length function on $N$. It is a standard result of the geometric theory of nilpotent Lie groups (see, e.g., [Reference Aristov5, Theorem 3.1]) that there are $A,\,A',\,k,\,k'\geqslant 1$ such that

(5.4)\begin{equation} \|\eta\|\leqslant A'(1+\ell_N(\exp\eta))^{k'}\quad\text{and}\quad \ell_N(\exp\eta)\leqslant A(1+\|\eta\|)^{k} \end{equation}

for every $\eta \in \mathfrak {n}$. The first inequality and $A'\geqslant 1$ imply that

\[ \log (1+\|\exp^{{-}1}(n)\|)\lesssim \log\ell_N(n) \]

for $n\in N\setminus \{1\}$. So it suffices to show that $\log \ell _N(n)\lesssim \ell (n)$.

Since $N$ is simply connected and nilpotent, the exponential map is bijective. Since the exponential map is natural, we have

(5.5)\begin{equation} \exp^{{-}1}\circ\alpha_h\circ\exp(\eta)=\mathbf{L}(\alpha_h)(\eta)\quad \text{for each }\eta\in\mathfrak{n} \text{ and }h\in H. \end{equation}

Suppose that $n\in N$ with $\ell (n)=m$, where $m>0$, and fix a representation $n=g_1\cdots g_m$, where $g_1,\,\ldots ,\, g_m\in S$, i.e., $g_j=(\exp \eta _j,\,\ h_j)$ for some $h_j\in K$ and $\eta _j\in \mathfrak {n}$ with $\|\eta _j\|\leqslant 1$. Then

\[ n=(\exp \eta_1)\alpha_{h_1}(\exp \eta_2)\cdots\alpha_{h_1\cdots h_{m-1}}(\exp \eta_m). \]

Put $h_0:=1$. Then $\|\mathbf {L}(\alpha _{h_0\cdots h_{j}}(\eta _j))\|\leqslant B^{j}$ for any $j\in \{0,\,\ldots ,\,m-1\}$. Applying (5.5) and the second inequality in (5.4), we get

\begin{align*} \ell_N(n)& \leqslant \sum_{j=0}^{m-1}\ell_N(\alpha_{h_0\cdots h_{j}}(\exp \eta_j))\\ & \leqslant\sum_{j=0}^{m-1} A(1+\|\mathbf{L}(\alpha_{h_0\cdots h_{j}}(\eta_j))\|)^{k}\leqslant A\sum_{j=0}^{m-1} (1+B^{j})^{k}\\ & \leqslant A\ 2^{k}\ \sum_{j=0}^{m-1} B^{jk}\leqslant \frac{A\ 2^{k}\ B^{mk}}{B^{k}-1}. \end{align*}

Since $m=\ell (n)$, we have $\log \ell _N(n)\lesssim \ell (n)$.