Proof Let $B\in M_n(\mathbb {Q})$ be the matrix whose rows are $b^1,\dots ,b^n$ , where for $1\leq j \leq n$ , the vector $b^j\in (\mathbb {Q}^n)^\dagger $ , $b^j=(b^j_1,\dots , b^j_n)$ is as in (1.1), which can be written in the form (3.1) using the notation introduced above. Notice that if any one of the vectors $b^j$ is 0, then $\mathscr {U}$ is empty, since the inequality $\rho (z)^{b^j}<1$ becomes $1<1$ .

If $\delta _j>0$ is a common denominator for the rational numbers $b^j_1,\dots , b^j_n$ , then notice that for a $z\in {\mathbb {C}}^n$ , the quantity $\rho (z)^{b^j}$ is defined and less than 1 if and only if $\rho (z)^{\delta _j b^j}$ is defined and less than 1. Therefore, we can assume without loss of generality that the matrix B has integer entries. Note also that interchanging the rows of B simply corresponds renumbering the equations in (1.1), so we can further assume without loss of generality that $\det B\geq 0$ .

Now, assuming that $B\in M_n(\mathbb {Z})$ and $\det B\geq 0$ , suppose for a contradiction that $\det B=0$ . We will show that $\mathscr {U}$ is unbounded, which will contradict the hypothesis of Theorem 1.2. Since $\det B=0$ , there is a nonzero vector $x\in {\mathbb {R}}^n$ such that $Bx=0$ . Let $r\in \mathscr {U}$ be such that $r \succ 0$ (such an r exists since $\mathscr {U}$ is nonempty, open, and Reinhardt). Consider the curve $f : {\mathbb {R}} \to {\mathbb {C}}^n$ parametrized by

$$ \begin{align*} f(t)=r\odot \exp(tx),\end{align*} $$

where $\exp $ is as in (2.8). Observe that

$$ \begin{align*}f(t)^B=r^B\odot \exp(tx)^B=r^B\odot \exp(tBx)= r^B, \end{align*} $$

since $Bx=0$ . Therefore,

$$ \begin{align*}\rho(f(t))^B= \rho(f(t)^B)=r^B. \end{align*} $$

But since $r \in \mathscr {U}$ , it follows that , so in fact f is a curve in $\mathscr {U}$ . As $x \in {\mathbb {R}}^n $ is nonzero, some component $x_k$ of x is nonzero, so that the kth component of the curve f, given by $f_k(t)= r_k e^{tx_k}$ is unbounded as $t\to \pm \infty $ . Thus, $\mathscr {U}$ contains the unbounded image of the curve f, showing that $\mathscr {U}$ is unbounded if $\det B=0$ . Since by assumption $\mathscr {U}$ is bounded, we have $\det B\neq 0$ . So $\det B>0$ .

To show that $B^{-1}\succeq 0$ , we let A be the adjugate of B, i.e., $A = \operatorname {\mathrm {adj}} B$ , and we show that

$$ \begin{align*} \Phi_A((\mathbb{D}^*)^n) \subset \mathscr{U},\end{align*} $$

where $\Phi _A$ is the monomial map as in (2.6), and $(\mathbb {D}^*)^n$ is the product of n punctured unit disks. Since $\mathscr {U}$ is bounded this implies that $\Phi _A((\mathbb {D}^*)^n)$ is bounded. Let $w \in \Phi _A((\mathbb {D}^*)^n)$ and let $z \in (\mathbb {D}^*)^n$ be such that $z^A=w$ . Notice that $z \in (\mathbb {D}^*)^n$ is equivalent to , and we also have $z^A=w \in ({\mathbb {C}}^*)^n$ . By applying (2.11) and Cramer’s rule, we find

$$ \begin{align*}\rho(w)^{B} = \left(\rho(z)^A\right)^{B}= \rho(z)^{BA}=\rho(z)^{(\det B) I}=\left(\left\vert z_1\right\vert{{\hspace{-0.0001pt}}}^{\det B},\dots, \left\vert z_n\right\vert{{\hspace{-0.0001pt}}}^{\det B}\right)^T.\end{align*} $$

As $\det B> 0$ we find that , i.e., $w \in \mathscr {U}$ , yielding the desired inclusion $\Phi _A((\mathbb {D}^*)^n) \subset \mathscr {U}$ .

If $A \not \succeq 0$ then there exist some $1 \leq j,k \leq n$ such that $a^j_k < 0$ . Let $z\in (\mathbb {D}^*)^n$ be such that all its components, except the kth, are $1/2$ . Then the jth component of $\Phi _A(z)$ is given by

$$ \begin{align*} z^{a^j} = {z_k^{a^j_k}}\cdot \frac{1}{2^m},\quad \text{where }\quad m= \sum_{\substack{1 \leq \ell \leq n \\ \ell \neq k}} a^j_\ell. \end{align*} $$

As $a^j_k < 0$ we see that as $z_k\to 0$ , the monomial $z^{a^j}$ is unbounded, showing that $\Phi _A((\mathbb {D}^*)^n)$ is unbounded. Therefore, the boundedness of $\mathscr {U}$ guarantees that $A = \operatorname {\mathrm {adj}} B \succeq 0$ .

To complete the proof note that $A \succeq 0$ implies that $B^{-1}=(\det B)^{-1} \operatorname {\mathrm {adj}} B\succeq 0$ . ▪

Proof For a bounded monomial polyhedron $\mathscr {U}$ , let $z\in \mathscr {U}$ so that $\rho (z)^B\prec 1$ , where B is as in Proposition 3.2. Since $B^{-1}\succeq 0$ , we have for any $r\in {\mathbb {R}}^n$ with, $0\preceq r \prec 1$ that $r^{B^{-1}}\prec 1$ . Therefore, we have $(\rho (z)^{B})^{B^{-1}}=\rho (z)\prec 1$ , i.e., $z\in \mathbb {D}^n$ . The second assertion follows on noting that in the log absolute representation (1.7) of $\mathscr {U}$ , the origin (which corresponds to the unit torus of ${\mathbb {C}}^n$ ) is a boundary point.  ▪

Proof Notice that switching two rows of B does not change the complexity of $\mathscr {U}$ , we can assume that the conditions (3.3) hold for the representing matrix. In the log-absolute coordinates $\xi _k=\log \left \vert z_k\right \vert $ the domain $\mathscr {U}$ is represented by the equations (1.7), which define the open polyhedral cone $\mathscr {C}_{\mathscr {U}}=\{B\xi \prec 0\}\subset {\mathbb {R}}^n$ . We claim that for each $1\leq j\leq n$ , the hyperplane $H_j=\{ b^j \xi = 0 \}$ determined by the jth row of the matrix B is a face of $\mathscr {C}_{\mathscr {U}}$ , i.e., the intersection $\overline {\mathscr {C}_{\mathscr {U}}}\cap H_j\not =\emptyset $ . Indeed, since by Proposition 3.2, $B^{-1}\succ 0$ and B is a linear automorphism of $\mathbb {Q}^n$ , it is easy to verify that the image $B^{-1}(\mathscr {C}_{\mathscr {U}})$ is precisely the negative orthant $\{\eta \prec 0\}\subset {\mathbb {R}}^n$ . But the negative orthant clearly has faces $\{\eta _j=0\}, j=1,\dots , n$ , which corresponds to the faces $H_j$ of $\mathscr {C}_{\mathscr {U}}$ .

Now, if $C\in M_n(\mathbb {Q})$ is another matrix (satisfying (3.3)) such that

then the faces of $\mathscr {C}_{\mathscr {U}}$ are also given by $\{ c^j \xi = 0 \}, 1\leq j\leq n$ , so that there exists a permutation $\sigma \in S_n$ such that

$$ \begin{align*} \{ c^j \xi = 0 \} = H_{\sigma(j)}= \{b^{\sigma(j)} \xi = 0 \} \end{align*} $$

for each $1 \leq j \leq n$ . Then the transposes $(c^j)^T$ and $(b^{\sigma (j)})^T$ are both normal to $H_{\sigma (j)}$ , so that there is a $\delta _j \in \mathbb {Q}\setminus \{0\}$ such that $c^j = \delta _j b^{\sigma (j)}$ . Therefore, $C=DPB$ , where D is the diagonal matrix $\mathrm {diag}(\delta _1,\dots ,\delta _n)$ and P is the permutation matrix whose jth column is $e_{\sigma (j)}$ . Note that $P^{-1}$ is also a permutation matrix, and hence $P^{-1} \succ 0$ . Therefore, since $C^{-1}=B^{-1}P^{-1}D^{-1}$ and we have $C^{-1}\succ 0, B^{-1}\succ 0$ by Proposition 3.2, we have $D\succ 0$ since $B^{-1}P^{-1}\succ 0$ . Therefore,

$$ \begin{align*} \max_{1 \leq j \leq n} \mathsf{h}(C^{-1}e_j) &= \max_{1 \leq j \leq n} \mathsf{h}(B^{-1}P^{-1}D^{-1}e_j) = \max_{1 \leq j \leq n} \mathsf{h}(\frac{1}{\delta_j}B^{-1}e_{\sigma(j)}) \\ &=\max_{1 \leq j \leq n} \mathsf{h}(B^{-1}e_{\sigma(j)}) =\max_{1 \leq j \leq n} \mathsf{h}(B^{-1}e_j). \end{align*} $$

This shows that $\kappa (\mathscr {U})$ as in 1.5 is well-defined, being independent of the matrix B. ▪

Proof First assume that z is such that $z_k{\neq }0$ for each k. For $1\leq j,k \leq n$ , the entry in the jth row and kth column of the complex derivative matrix $\Phi _A'(z)$ is $\dfrac {\partial z^{a^j}}{\partial z_k}$ , where recall that $a^j$ denotes the multi-index in $(\mathbb {Z}^n)^\dagger $ whose entries constitute the jth row of A. If $a_k^j \neq 0$ then

$$ \begin{align*}\frac{\partial z^{a^j}}{\partial z_k}=a_k^j z_k^{a_k^j-1} \prod_{\substack{\ell=1 \\ \ell\neq k}}^n z_\ell^{a_\ell^j} = a_k^j\cdot \frac{z^{a^j}}{z_k}.\end{align*} $$

If $a_k^j=0$ , then $\dfrac {\partial z^{a^j}}{\partial z_k}=0$ , so in fact the above formula holds for all $j,k$ . Now, by the representation of a determinant as an explicit polynomial in the matrix entries, we find

where we have used the fact that

Also,

Therefore,

for z such that $z_k{\neq }0$ for each k. The result follows by analytic continuation.  ▪

Proof By the conditions (3.3), we have $A =(\det B)B^{-1} \succeq 0$ , so $\Phi _A$ is defined on all of $\mathbb {D}^n_{\mathsf {L}(B)}$ . We first claim that the image $\Phi _A(\mathbb {D}^n_{\mathsf {L}(B)})$ is contained in ${\mathbb {C}}^n_{\mathsf {K}(B)}$ . Indeed, if $z\in \Phi _A(\mathbb {D}^n_{\mathsf {L}(B)})$ , then $z_\ell {\neq }0$ if $a^k_\ell {\neq }0$ for some $k\in \mathsf {K}(B)$ . Therefore the kth element of the vector $z^A$ , i.e.,

(3.15) $$ \begin{align} z^{a^k}=z_1^{a^k_1}\dots z_n^{a^k_n} \end{align} $$

is nonzero, so $z^A\in {\mathbb {C}}^n_{\mathsf {K}(B)}.$ Now notice that

since $\rho (z)^B$ (and $z^B$ ) are defined if and only if $z\in {\mathbb {C}}^n_{\mathsf {K}(B)}.$ Therefore, as in the proof of Proposition 3.2, we have

for each point $z\in \mathbb {D}^n_{\mathsf {L}(B)} $ since $\det B>0$ . This shows that $\Phi _A : \mathbb {D}^n_{\mathsf {L}(B)} \to \mathscr {U}$ is a well defined holomorphic map.

We now show that, if we think of $\Phi _A$ as a map from ${\mathbb {C}}^n$ to itself (recall that $A\succeq 0$ ), then we have

(3.16) $$ \begin{align} \Phi_A^{-1}(\mathscr{U})\subset \mathbb{D}^n_{\mathsf{L}(B)}. \end{align} $$

If $z\in {\mathbb {C}}^n$ is such that $\Phi _A(z)=z^A\in \mathscr {U}$ , then $\rho (z^A)^B$ is well defined and

. The fact that $\rho (z^A)^B$ is well defined is equivalent to $z^A\in {\mathbb {C}}^n_{\mathsf {K}(B)}$ , i.e., for each $k\in \mathsf {K}(B)$ we have the entry $z^{a^k}{\neq }0.$ Now from (3.15), we see that whenever $a^k_\ell {\neq }0$ (so $a^k_\ell \geq 1$ since $A\succeq 0$ ), we must have $z_\ell {\neq }0$ , i.e., $z\in {\mathbb {C}}^n_{\mathsf {L}(B)}$ . The other condition

on the point z shows that

Since $\det B>0$ , this shows that

, i.e., $z\in \mathbb {D}^n$ . Together these conditions show that $z\in \mathbb {D}^n_{\mathsf {L}(B)}$ , proving the desired inclusion of (3.16).

We can now show that $\Phi _A:\mathbb {D}^n_{\mathsf {L}(B)}\to \mathscr {U}$ is a proper holomorphic map. Let $K \subset \mathscr {U}$ be compact. Since the topology induced on K from $\mathscr {U}$ is the same as that induced from ${\mathbb {C}}^n$ , it follows that K is closed in ${\mathbb {C}}^n$ (and bounded). As $\Phi _A$ is continuous, $\Phi _A^{-1} (K)$ is closed in ${\mathbb {C}}^n$ and in fact $\Phi _A^{-1} (K) \subseteq \mathbb {D}^n_{\mathsf {L}(B)}$ by (3.16), so that it is bounded as well. Thus, $\Phi _A^{-1} (K)$ is compact for every compact $K \subset \mathscr {U}$ , which is to say $\Phi _A$ is proper.

Now we want to verify that the proper holomorphic map $\Phi _A:\mathbb {D}^n_{\mathsf {L}(B)} \to \mathscr {U}$ is of quotient type in the sense of Definition 3.10 with group $\Gamma = \{\sigma _\nu : \nu \in \mathbb {Z}^n \},$ with $\sigma _\nu $ as in (3.14) above. Let

be the union of the coordinate hyperplanes, which is an analytic variety in ${\mathbb {C}}^n$ of codimension one. Notice that the formula (3.9) shows that the set of regular points of the map $\Phi _A$ (i.e., the set of points $\{z\in \mathbb {D}^n_{\mathsf {L}(B)}: \det \Phi _A'(z){\neq }0 \}$ ) contains $\mathbb {D}^n_{\mathsf {L}(B)}\setminus Z_1$ , and from the fact that $\Phi _A(z)=z^A$ , we see that the regular values of $\Phi _A$ contains the open set $\mathscr {U}\setminus Z_2$ . Notice that both $\mathbb {D}^n_{\mathsf {L}(B)}\setminus Z_1$ and $\mathscr {U}\setminus Z_2$ are open subsets of the “algebraic torus” $({\mathbb {C}}^*)^n$ , and it is clear that $\Phi _A$ maps $({\mathbb {C}}^*)^n$ into itself (and is even a group homomorphism, if $({\mathbb {C}}^*)^n$ is given the group structure from the elementwise multiplication operation $\odot $ ). Since a proper holomorphic map is a covering map restricted to its regular points, it now follows that $\Phi _A: \mathbb {D}^n_{\mathsf {L}(B)}\setminus Z_1\to \mathscr {U}\setminus Z_2 $ is a holomorphic covering map. We need to show that it is regular.

Let $w\in \mathscr {U}\setminus Z_2$ have polar representation $w=\rho (w)\odot \exp (i\theta )$ , where $\theta \in {\mathbb {R}}^n$ . Let $z\in \mathbb {D}^n_{\mathsf {L}(B)}$ be such that $\Phi _A(z)=z^A=w$ . Comparing the radial and angular parts, one such preimage is

$$ \begin{align*}z=\rho(w)^{A^{-1}}\odot \exp\left( i A^{-1}\theta\right). \end{align*} $$

Now notice that the angular vector $\theta $ is known only up to an additive ambiguity of $2\pi \mathbb {Z}^n$ , i.e., two values of $\theta $ corresponding to the same w differ by $2\pi \nu $ for some integer vector $\nu \in \mathbb {Z}^n$ . Therefore two different preimages $z,z'$ of the point w (corresponding to the choices $\theta $ and $\theta +2\pi \nu $ of the angular vector in the polar representation of w) are related by

$$ \begin{align*} z' &= \rho(w)^{A^{-1}}\odot \exp\left( i A^{-1}(\theta+2\pi\nu)\right) = \rho(w)^{A^{-1}}\odot \exp\left( i A^{-1}\theta\right)\odot \exp\left(2\pi i A^{-1}\nu\right)\\ &= \exp\left(2\pi i A^{-1}\nu\right)\odot z =\sigma_\nu(z), \end{align*} $$

which shows that the group $\Gamma $ acts transitively on the fiber $\Phi ^{-1}(w)$ .

Now we want to show that the group $\Gamma $ has exactly $\det A$ elements. Consider the map $\psi : \mathbb {Z}^n\to \Gamma $ given by $\psi (\nu )= \sigma _\nu . $ For any $\nu , \mu \in \mathbb {Z}^n$ and any $z \in \mathbb {D}^n_{\mathsf {L}(B)}$ we find that

$$ \begin{align*} \psi(\nu+\mu)(z)=\sigma_{\nu + \mu}(z) &=\exp \left(2 \pi i A^{-1}(\nu + \mu) \right) \odot z \\ &= \exp \left(2 \pi i A^{-1}\nu \right)\odot \left( \exp\left(2 \pi i A^{-1} \mu \right) \odot z \right) \\ &= \sigma_\nu \circ \sigma_\mu (z)\\ &=\psi(\nu) \circ \psi(\mu) (z). \end{align*} $$

Hence, $\psi : \mathbb {Z}^n \to \Gamma $ is a surjective group homomorphism, and so we have an isomorphism

$$ \begin{align*} \overline{\psi}: \mathbb{Z}^n/\ker\psi\to \Gamma. \end{align*} $$

given by $\psi ([\nu ])=\sigma _\nu $ where $[\nu ] \in \mathbb Z^n/ \ker \psi $ is the class of $\nu \in \mathbb Z^n$ . Notice that $\nu \in \ker \psi $ if and only if . This means that $A^{-1}\nu \in \mathbb {Z}^n$ , so it follows that $\ker \psi = A(\mathbb Z^n)$ . Therefore, the group $\Gamma $ is isomorphic to the quotient $\mathbb {Z}^n/A(\mathbb {Z}^n)$ by the isomorphism $ \overline {\psi }: \mathbb {Z}^n/A(\mathbb {Z}^n)\to \Gamma $

To complete the proof, it is sufficient to recall that for any matrix $A\in M_n(\mathbb {Z})$ with $\det A{\neq }0$ we have

(3.17) $$ \begin{align} \left\vert\mathbb{Z}^n/A(\mathbb{Z}^n)\right\vert=\left\vert\det A\right\vert. \end{align} $$

For completeness, we give two proofs of (3.17), neither of which unfortunately avoids high technology. The first (see [Reference Nagel and PramanikNP20] and [Reference ZwonekZwo99, Theorem 2.1]) is based on the Smith Canonical Form of an integer matrix (see [Reference MunkresMun84, p. 53 ff.]): for any $A \in M_n (\mathbb Z)$ there exist $P,Q \in GL_n(\mathbb Z)$ and a diagonal matrix $D \in M_n(\mathbb Z)$ , say $D = \mathrm {diag}(\delta _1 ,\dots ,\delta _n)$ , such that $A = PDQ$ . Consider the automorphism of $\mathbb {Z}^n$ given by $x\mapsto P^{-1}x$ . This maps $A(\mathbb {Z}^n)$ to $D(\mathbb {Z}^n)$ isomorphically, and therefore leads to an isomorphism $\mathbb {Z}^n/A(\mathbb {Z}^n)$ with $\mathbb {Z}^n/D(\mathbb {Z}^n)$ , but $\mathbb {Z}^n/D(\mathbb {Z}^n)$ is isomorphic to the product abelian group $\prod _{j=1}^n\mathbb {Z}/\delta _j\mathbb {Z}$ , which has exactly $\prod _{j=1}^n\left \vert \delta _j\right \vert $ elements. On the other hand note that

$$ \begin{align*}&\left\vert\det A\right\vert=\left\vert \det(PDQ)\right\vert= \left\vert\det P\det D\det Q\right\vert= \left\vert\pm 1\cdot \prod_{j=1}^n \delta_j \cdot \pm 1\right\vert,\\[-6pt]\end{align*} $$

which completes the first proof.

The second proof of (3.17) is geometric, and based on noticing that the inclusion of abelian groups $A(\mathbb {Z}^n)\hookrightarrow \mathbb {Z}^n$ gives rise to a covering map of tori

(3.18) $$ \begin{align} &p:{\mathbb{R}}^n/A(\mathbb{Z}^n)\to {\mathbb{R}}^n/\mathbb{Z}^n\\[-6pt]\nonumber\end{align} $$

given by $x+A(\mathbb {Z}^n)\mapsto x+\mathbb {Z}^n$ . If we endow ${\mathbb {R}}^n$ with the Euclidean metric (thought of as a Riemannian metric), we obtain quotient (flat) Riemannian structures on both ${\mathbb {R}}^n/A(\mathbb {Z}^n)$ and ${\mathbb {R}}^n/\mathbb {Z}^n$ , so that the covering map (3.18) is actually a local isometry. Therefore, the number of sheets of this covering map coincides with the ratio $\mathrm {vol}({\mathbb {R}}^n/A(\mathbb {Z}^n))/\mathrm {vol}( {\mathbb {R}}^n/\mathbb {Z}^n)$ . But we know that $\mathrm {vol}( {\mathbb {R}}^n/\mathbb {Z}^n)$ is 1, being also the volume of the unit parallelepiped, and similarly $\mathrm {vol}({\mathbb {R}}^n/A(\mathbb {Z}^n))$ is the volume of the image of the unit parallelepiped under the action of A, which is therefore $\left \vert \det A\right \vert $ . On the other hand, by covering space theory, we see that the number of sheets is equal to the order of the group of deck transformations of the covering p, which is in turn isomorphic to $\pi _1({\mathbb {R}}^n/\mathbb {Z}^n)/p_*\pi _1({\mathbb {R}}^n/A(\mathbb {Z}^n))=\mathbb {Z}^n/A(\mathbb {Z}^n)$ . ▪

Proof Denote by $w_j$ the jth component of the map $\Phi _A$ , where $1\leq j \leq n$ . Notice that $w_j$ can be identified with the function $\Phi _A^*(\pi _j)=\pi _j\circ \Phi _A$ where $\pi _j$ is the jth coordinate function on $\mathscr {U}$ . Since $\mathrm {Rat}(\mathscr {U})={\mathbb {C}}(\pi _1,\dots , \pi _n)$ , the field k is generated over ${\mathbb {C}}$ by the functions $w_1,\dots , w_n$ , i.e., $k={\mathbb {C}}(w_1,\dots , w_n)$ . Denoting the coordinate functions on $\mathbb {D}^n_{\mathsf {L}(B)}$ by $z_1,\dots , z_n$ , we see that $\mathrm {Rat}(\mathbb {D}^n_{\mathsf {L}(B)})= k(z_1,\dots , z_n)$ . To show that $\mathrm {Rat}(\mathbb {D}^n_{\mathsf {L}(B)})/k$ is Galois, we need to show that each $z_j$ is algebraic over k, and all the conjugates of $z_j$ over k are already present in $\mathrm {Rat}(\mathbb {D}^n_{\mathsf {L}(B)}).$

Notice that by definition for each $1\leq j \leq n$ , we have $w_j=z_1^{a^j_1}\dots z_n^{a^j_n}=z^{a^j}$ , which follows from the fact that $w=\Phi _A(z)=z^A$ . Therefore we have

$$ \begin{align*}z^{\det A\cdot I} = z^{ \operatorname{\mathrm{adj}} A\cdot A} =(z^A)^{ \operatorname{\mathrm{adj}} A} =w^{ \operatorname{\mathrm{adj}} A}. \end{align*} $$

Denoting the jth row of the integer matrix $ \operatorname {\mathrm {adj}} A$ by $c^j$ we see that $z_j^{\det A} = w^{c^j}\in k,$ so that $z_j$ is a root of the polynomial $t^{\det A}-w^{c^j}\in k[t]$ . Further, all the roots of this polynomial (which are of the form $\omega \cdot z_j $ for a $(\det A)$ th root of unity, $\omega \in {\mathbb {C}}$ ) are clearly in $\mathrm {Rat}(\mathbb {D}^n_{\mathsf {L}(B)})$ . It follows that $\mathrm {Rat}(\mathbb {D}^n_{\mathsf {L}(B)})$ is Galois over the subfield k.

For each $\sigma \in \Gamma $ , it is clear that $\Theta (\sigma )$ is an automorphism of the field $\mathrm {Rat}(\mathbb {D}^n_{\mathsf {L}(B)})$ . We need to show that it fixes k. But for $h=\Phi _A^*g\in k$ , where $g\in \mathrm {Rat}\left (\mathscr {U}\right )$ , we have

$$ \begin{align*} &\Theta(\sigma)h=h\circ\sigma^{-1}=g\circ \Phi_A\circ \sigma^{-1}=g\circ \Phi_A=h,\\[-4pt]\end{align*} $$

since $\sigma $ is a deck transformation of $\Phi _A$ . It follows that $\Theta (\sigma )\in \mathrm {Gal}(\mathrm {Rat}(\mathbb {D}^n_{\mathsf {L}(B)})/k)$ . It is easily verified by direct computation that $\Theta $ is an injective group homomorphism. To show that $\Theta $ is surjective, for $\gamma \in \mathrm {Gal}(\mathrm {Rat}(\mathbb {D}^n_{\mathsf {L}(B)})/k)$ , we find a deck transformation $\theta \in \Gamma $ such that $\gamma (f)=f\circ \theta $ , so that $\gamma =\Theta ( \theta ^{-1})$ . Let $\theta _j\in \mathrm {Rat}(\mathbb {D}^n_{\mathsf {L}(B)})$ be the function $\theta _j=\gamma (z_j)$ , where $z_j\in \mathrm {Rat}(\mathbb {D}^n_{\mathsf {L}(B)})$ is the jth coordinate function. As we saw above, we have $\theta _j=\omega _j\cdot z_j$ , where $\omega _j$ is a $(\det A)$ th root of unity. Then $\theta =(\theta _1,\dots ,\theta _n)$ defines a diagonal unitary mapping of ${\mathbb {C}}^n$ and therefore maps $\mathbb {D}^n_{\mathsf {L}(B)}$ (a product of discs and punctured discs) into itself. Also, the jth component of $\Phi _A\circ \theta $ is given by

$$ \begin{align*}&\theta^{a^j}=\theta_1^{a^j_1}\dots \theta_n^{a^j_n}= \gamma(z_1)^{a^j_1}\dots \gamma(z_n)^{a^j_n}=\gamma(z^{a^j})=\gamma(w_j)=w_j,\\[-4pt] \end{align*} $$

which is precisely the jth component of $\Phi _A$ . Therefore, $\Phi _A\circ \theta =\Phi _A$ , and $\theta \in \Gamma $ . Finally, for a rational function $f=f(z_1,\dots ,z_n)\in \mathrm {Rat}(\mathbb {D}^n_{\mathsf {L}(B)})$ , we have

$$ \begin{align*}\gamma(f)=f(\gamma(z_1),\dots, \gamma(z_n))=f(\theta_1,\dots,\theta_n)=f\circ\theta. \end{align*} $$

 ▪

Proof Choosing a local continuous branch $\theta _j$ of $\arg w_j$ for each coordinate of $w\in \mathscr {U}$ , we can write $w=\rho (w)\odot \exp (i\theta )$ for $\theta \in {\mathbb {R}}^n$ . Then $\psi (w)= w^{A^{-1}}=\rho (w)^{A^{-1}}\odot \exp (iA^{-1}\theta )$ is a local inverse to the proper holomorphic map $\Phi _A:\mathbb {D}^n_{\mathsf {L}(B)}\to \mathscr {U}$ . Since $\Phi _A$ is of the quotient type, it follows that the local branches of $\Phi ^{-1}_A$ are $ \{\sigma \circ \psi : \sigma \in \Gamma \}.$ By Theorem 3.12, the map $\sigma :\mathbb {D}^n_{\mathsf {L}(B)}\to \mathbb {D}^n_{\mathsf {L}(B)}$ is the restriction of a unitary linear map on ${\mathbb {C}}^n$ , so, noting that the argument behind the proof of (3.9) also applies for (locally defined) rational matrix powers, we have

Therefore, by the Bell transformation formula, for $z\in \mathbb {D}^n_{\mathsf {L}(B)}$ and $w\in \mathscr {U}$ , the Bergman kernels $K_{\mathscr {U}}$ and $K_{\mathbb {D}^n_{\mathsf {L}(B)}}=K_{\mathbb {D}^n}$ are related by

For a fixed $z\in \mathbb {D}^n_{\mathsf {L}(B)}$ , consider the function L on $\mathbb {D}^n_{\mathsf {L}(B)}$ given by

where $\overline {z}=(\overline {z_1},\dots , \overline {z_n})\in \mathbb {D}^n_{\mathsf {L}(B)} $ . Recalling that

(3.23) $$ \begin{align} K_{\mathbb{D}^n}(z,w)= \frac{1}{\pi^n}\prod_{j=1}^n \frac{1}{(1-z_j\overline{w_j})^2}, \end{align} $$

we see that $L\in \mathrm {Rat}(\mathbb {D}^n_{\mathsf {L}(B)})$ , and

We now claim that $L\in k$ , where $k\subset \mathrm {Rat}(\mathbb {D}^n_{\mathsf {L}(B)})$ is as in (3.19). Since the extension $\mathrm {Rat}(\mathbb {D}^n_{\mathsf {L}(B)})/k$ is Galois, it suffices to show that for each field-automorphism $\tau \in \mathrm {Gal}(\mathrm {Rat}(\mathbb {D}^n_{\mathsf {L}(B)})/k)$ , we have $\tau (L)=L$ . But by Lemma 3.20, for each $\tau \in \mathrm {Gal}(\mathrm {Rat}(\mathbb {D}^n_{\mathsf {L}(B)})/k)$ there is a $\theta \in \Gamma $ such that $\tau (L)=L\circ \theta $ . Notice that thanks to (3.14), we can think of the map $\theta :\mathbb {D}^n_{\mathsf {L}(B)}\to \mathbb {D}^n_{\mathsf {L}(B)}$ as a linear map of ${\mathbb {C}}^n$ represented by a diagonal matrix. Therefore, we have

Consequently

establishing the claim.

Now since $L\in k$ , there is a function $R(z,\cdot )\in \mathrm {Rat}(\mathscr {U})$ such that $L=\Phi _A^* R(z,\cdot )$ , i,e,

$$ \begin{align*} L(\zeta)= R(z, \Phi_A(\zeta))= R(z, \zeta^A).\end{align*} $$

Therefore, we have

which shows that for each fixed $z\in \mathbb {D}^n_{\mathsf {L}(B)}$ , the function $K_{\mathscr {U}}(z,\cdot )$ is rational. By the Reinhardt symmetry of $K_{\mathscr {U}}$ , there is a function $\widetilde {K}$ such that

$$ \begin{align*} K_{\mathscr{U}}\left(z,w\right)=\widetilde{K}(z_1\overline{w_1},\dots, z_n\overline{w_n})= \widetilde{K}(z\odot \overline{w}).\end{align*} $$

Since for a fixed z, the function $w\mapsto \widetilde {K}(z\odot \overline {w})$ is rational, it follows that the function $t\mapsto \widetilde {K}(t)=\widetilde {K}(t_1,\dots , t_n)$ is also rational. Therefore, the function $K_{\mathscr {U}}$ is rational on $\mathscr {U}\times \mathscr {U}$ . ▪

Proof Let f be a function on $\Omega _2$ , and let $g=\Phi ^\sharp f$ be its pullback to $\Omega _1$ , then we have for each $\sigma \in \Gamma $ that

$$ \begin{align*} \sigma^\sharp(g)=\sigma^\sharp (\Phi^\sharp f) = (\Phi\circ \sigma)^\sharp f = \Phi^\sharp f =g, \end{align*} $$

where we have used the contravariance of the pullback $(\Phi \circ \sigma )^\sharp =\sigma ^\sharp \circ \Phi ^\sharp $ and the fact that $\Phi \circ \sigma =\Phi $ which follows since the action of $\Gamma $ on $\Omega _2$ restricts to actions on each of the fibers. This shows that the range of $\Phi ^\sharp $ consists of $\Gamma $ -invariant functions. Special cases of this invariance were already noticed [Reference Misra, Roy and ZhangMSRZ13, Reference Chen, Krantz and YuanCKY20].

To complete the proof of (4.6) we must show that

1. (i) for each $f\in L^p(\Omega _2)$ ,

   (4.8) $$ \begin{align} \left\Vert\Phi^\sharp f\right\Vert{{\hspace{-0.0001pt}}}^p_{L^p\left(\Omega_1, \left\vert\det \Phi'\right\vert{{\hspace{-0.0001pt}}}^{2-p}\right)}= \left\vert\Gamma\right\vert\cdot \left\Vert f\right\Vert{{\hspace{-0.0001pt}}}^p_{L^p(\Omega_2)}, \end{align} $$

2. (ii) The image $\Phi ^\sharp (L^p(\Omega _2))$ is precisely $\left [L^p\left (\Omega _1, \left \vert \det \Phi '\right \vert {{\hspace{-0.0001pt}}}^{2-p}\right )\right ]^\Gamma $ .

Let $Z_1, Z_2$ be as in the Definition 3.10 of proper holomorphic maps of quotient type, i.e., $\Phi $ is a regular covering map from $\Omega _1\setminus Z_1$ to $\Omega _2\setminus Z_2$ . Let U be an open set in $\Omega _2\setminus Z_2$ which is evenly covered by $\Phi $ , and let V be an open set of $\Omega _1\setminus Z_1$ which is mapped biholomorphically by $\Phi $ onto U. Then the inverse image $\Phi ^{-1}(U)$ is the finite disjoint union $\bigcup _{\sigma \in \Gamma } \sigma V$ . Therefore, if $f\in L^p(\Omega _2)$ is supported in U, then $\Phi ^\sharp f$ is supported in $\bigcup _{\sigma \in \Gamma } \sigma V$ , and we have

$$ \begin{align*} &\left\Vert\Phi^\sharp f\right\Vert{{\hspace{-0.0001pt}}}^p_{L^p\left(\Omega_1, \left\vert\det \Phi'\right\vert{{\hspace{-0.0001pt}}}^{2-p}\right)}\\[10pt]&\quad =\int_{\Omega_1} \left\vert f\circ \Phi \cdot \det \Phi'\right\vert{{\hspace{-0.0001pt}}}^{p} \left\vert\det \Phi'\right\vert{{\hspace{-0.0001pt}}}^{2-p} dV \\[10pt]&\quad =\sum_{\sigma \in \Gamma}\int_{\sigma V} \left\vert f\circ \Phi\right\vert{{\hspace{-0.0001pt}}}^p \left\vert\det \Phi'\right\vert{{\hspace{-0.0001pt}}}^2 dV= \sum_{\sigma\in \Gamma }\int_{U} \left\vert f\right\vert{{\hspace{-0.0001pt}}}^p dV = \left\vert\Gamma\right\vert\cdot\left\Vert f\right\Vert{{\hspace{-0.0001pt}}}_{L^p(\Omega_2)}^p, \end{align*} $$

where we have used the change of variables formula applied to the biholomorphic map $\Phi |_{\sigma V}$ along with the fact that the real Jacobian determinant of the map $\Phi $ is equal to $\left \vert \det \Phi '\right \vert {{\hspace{-0.0001pt}}}^2$ .

For a general $f\in L^p(\Omega _2)$ , modify the proof as follows. There is clearly a collection of pairwise disjoint open sets $\{U_j\}_{j\in J}$ in $\Omega _2\setminus Z_2$ such that each $U_j$ is evenly covered by $\Phi $ and $\Omega _2 \setminus \bigcup _{j\in J} U_j$ has measure zero. Set $f_j = f\cdot \chi _j$ , where $\chi _j$ is the indicator function of $U_j$ , so that each $f_j\in L^p(\Omega _2)$ and $f=\sum _{j}f_j$ . Also, the functions $\Phi ^\sharp f_j$ have pairwise disjoint supports in $\Omega _1$ . Therefore we have

$$ \begin{align*} \left\Vert\Phi^\sharp f\right\Vert{{\hspace{-0.0001pt}}}^p_{L^p\left(\Omega_1, \left\vert\det \Phi'\right\vert{{\hspace{-0.0001pt}}}^{2-p}\right)}&= \sum_{j\in J} \left\Vert\Phi^\sharp f_j\right\Vert{{\hspace{-0.0001pt}}}^p_{L^p\left(\Omega_1, \left\vert\det \Phi'\right\vert{{\hspace{-0.0001pt}}}^{2-p}\right)} \\[10pt]&= \left\vert\Gamma\right\vert \sum_{j\in J} \left\Vert f_j\right\Vert{{\hspace{-0.0001pt}}}_{L^p(\Omega_2)}^p = \left\vert\Gamma\right\vert \left\Vert f\right\Vert{{\hspace{-0.0001pt}}}_{L^p(\Omega_2)}^p. \end{align*} $$

To complete the proof, we need to show that $\Phi ^\sharp $ is surjective in both (4.6) and (4.7). Let $g\in \left [L^p\left (\Omega _1, \left \vert \det \Phi '\right \vert {{\hspace{-0.0001pt}}}^{2-p}\right )\right ]^{\Gamma } $ . Let $\{U_j\}_{j\in J}$ be as in the previous paragraph, and set $g_j= g\cdot \chi _{\Phi ^{-1}(U_j)}$ , where $\chi _{\Phi ^{-1}(U_j)}$ is the indicator function of $\Phi ^{-1}(U_j)$ . Notice that $g_j\in \left [L^p\left (\Omega _1, \left \vert \det \Phi '\right \vert {{\hspace{-0.0001pt}}}^{2-p}\right )\right ]^{\Gamma }$ . Let $V_j\subset \Phi ^{-1}(U_j)$ be such that $\Phi $ maps $V_j$ biholomorphically to $U_j$ , and let $\Psi : U_j\to V_j$ be the local inverse of $\Phi $ onto $V_j$ . Define

(4.9) $$ \begin{align} f_j = \Psi^\sharp (g_j). \end{align} $$

We claim that $f_j$ is defined independently of the choice of $V_j$ . Indeed, any other choice is of the form $\sigma V_j$ for some $\sigma \in \Gamma $ and the corresponding local inverse is $\sigma \circ \Psi $ . But we have

$$ \begin{align*} (\sigma\circ \Psi)^\sharp g_j = \Psi^\sharp \circ \sigma^\sharp g_j = \Psi^\sharp g_j =f_j, \end{align*} $$

where we have used the fact that $ \sigma ^\sharp g_j=g_j$ since since $g_j\in [L^p(\Omega _1, \left \vert \det \Phi '\right \vert {{\hspace{-0.0001pt}}}^{2-p})]^{\Gamma }$ .

Now we define $f=\sum _{j}f_j$ . Notice that the $f_j$ have pairwise disjoint support, and it is easily checked that $\Phi ^\sharp f =g$ . This establishes that (4.6) is a homothetic isomorphism.

It is clear that if f is holomorphic on $\Omega _2$ , then $\Phi ^\sharp f$ is holomorphic on $\Omega _1$ , therefore, $\Phi ^\sharp $ maps $A^p(\Omega _2)$ into $[A^p(\Omega _1, \left \vert \det \Phi '\right \vert {{\hspace{-0.0001pt}}}^{2-p})]^{\Gamma }$ . Now, in the argument in the previous paragraph showing that the image $\Phi ^\sharp (L^p(\Omega _2))$ is $[L^p(\Omega _1, \left \vert \det \Phi '\right \vert {{\hspace{-0.0001pt}}}^{2-p})]^{\Gamma }$ , local definition of the inverse map (4.9) shows that if $g\in [A^p(\Omega _1, \left \vert \det \Phi '\right \vert {{\hspace{-0.0001pt}}}^{2-p})]^{\Gamma }$ , then the f constructed by this procedure is holomorphic, and therefore lies in $A^p(\Omega _2)$ . This completes the proof of the proposition.  ▪

Proof For (1), note that $\sigma ^\sharp $ is a unitary operator on $L^2(\Omega )$ and $\boldsymbol {B}_{\Omega }$ is an orthogonal projection on $L^2(\Omega )$ , therefore the unitarily similar operator $Q=\sigma ^\sharp \circ \boldsymbol {B}_{\Omega }\circ (\sigma ^\sharp )^{-1}$ is also an orthogonal projection. Since $ \sigma ^\sharp $ (and therefore its inverse) leaves $A^2(\Omega )$ invariant, it follows that the range of Q is $A^2(\Omega )$ . Therefore, $Q=\boldsymbol {B}_{\Omega }$ .

For (2), let $f\in [L^2(\Omega )]^G$ . Then using (4.13), we have for $\sigma \in G$

$$ \begin{align*} \sigma^\sharp (\boldsymbol{B}_{\Omega}f) = \boldsymbol{B}_{\Omega}(\sigma^\sharp (f)) = \boldsymbol{B}_{\Omega}f,\end{align*} $$

which shows that $\boldsymbol {B}_{\Omega }f\in [A^2(\Omega )]^G$ , so that $\boldsymbol {B}_{\Omega }$ maps the G-invariant functions $[L^2(\Omega )]^G$ into the G-invariant holomorphic functions $[A^2(\Omega )]^G$ . Since $\boldsymbol {B}_\Omega $ restricts to the identity on $[A^2(\Omega )]^G$ , it follows that the range of $\boldsymbol {B}_{\Omega }$ is $[A^2(\Omega )]^G$ . Observe that

$$ \begin{align*} \ker\left(\boldsymbol{B}_{\Omega}|_{[L^2(\Omega)]^G}\right)\subseteq \ker \boldsymbol{B}_{\Omega}= A^2(\Omega)^\perp \subseteq \left([A^2(\Omega)]^G\right)^\perp, \end{align*} $$

which shows that kernel of the restriction of $\boldsymbol {B}_{\Omega }$ to $[L^2(\Omega )]^G$ is orthogonal to its range, and therefore an orthogonal projection.  ▪

Proof of Proposition 4.10 By Proposition 4.5, the $\Phi ^\sharp $ represented by the top (resp. bottom) horizontal arrow is a homothetic isomorphism from the Hilbert space $ {L}^2(\Omega _2)$ (resp. $ {A}^2(\Omega _2)$ ) onto the Hilbert space $[{L}^2(\Omega _1)]^\Gamma $ (resp. $[{A}^2(\Omega _1)]^\Gamma $ ). Therefore, $\Phi ^\sharp $ preserves angles and in particular orthogonality. Now consider the map $ P: [{L}^2(\Omega _1)]^\Gamma \to [{A}^2(\Omega _1)]^\Gamma $ defined by

(4.14) $$ \begin{align} P= \Phi^\sharp \circ \boldsymbol{B}_{\Omega_2}\circ (\Phi^\sharp)^{-1}, \end{align} $$

which, being a composition of continuous linear maps, is a continuous linear mapping of Hilbert spaces. Notice that

$$ \begin{align*} P^2 = \Phi^\sharp \circ \boldsymbol{B}_{\Omega_2}\circ (\Phi^\sharp)^{-1}\circ \Phi^\sharp \circ \boldsymbol{B}_{\Omega_2}\circ (\Phi^\sharp)^{-1}= \Phi^\sharp \circ \boldsymbol{B}_{\Omega_2}\circ (\Phi^\sharp)^{-1}=P, \end{align*} $$

so P is a projection in $[{L}^2(\Omega _1)]^\Gamma $ , with range contained in $[{A}^2(\Omega _1)]^\Gamma $ . Since $(\Phi ^\sharp )^{-1}$ and $\Phi ^\sharp |_{A^2(\Omega )}$ are isomorphisms, and $\boldsymbol {B}_{\Omega _2}$ is surjective, it follows that P is a projection onto $[{A}^2(\Omega _1)]^\Gamma $ . We claim that P is in fact the orthogonal projection on to $[{A}^2(\Omega _1)]^\Gamma $ , i.e., the kernel of P is $\left ([{A}^2(\Omega _1)]^\Gamma \right )^\perp $ , the orthogonal complement of $[{A}^2(\Omega _1)]^\Gamma $ in $[{L}^2(\Omega _1)]^\Gamma $ . Since in formula (4.14), the maps $(\Phi ^\sharp )^{-1}$ and $\Phi ^\sharp $ are isomorphisms, it follows that $f\in \ker P$ if and only if $(\Phi ^\sharp )^{-1}f\in \ker \boldsymbol {B}_{\Omega _2}$ . But $\ker \boldsymbol {B}_{\Omega _2}= A^2(\Omega _2)^\perp $ , since the Bergman projection is orthogonal. It follows that $\ker P = \Phi ^\sharp (A^2(\Omega _2)^\perp )$ . Notice that $\Phi ^\sharp $ , being a homothetic isomorphism of Hilbert spaces, preserves orthogonality, and maps $A^2(\Omega _2)$ to $[{A}^2(\Omega _1)]^\Gamma $ isomorphically, therefore, $\Phi ^\sharp ( (A^2(\Omega _2))^\perp )= \left ([{A}^2(\Omega _1)]^\Gamma \right )^\perp $ , which establishes the claim.

Therefore, we have shown that $P=\Phi ^\sharp \circ \boldsymbol {B}_{\Omega _2}\circ (\Phi ^\sharp )^{-1}$ is the orthogonal projection from $ [{L}^2(\Omega _1)]^\Gamma $ to the subspace $[{A}^2(\Omega _1)]^\Gamma $ . To complete the proof, we only need to show that the restriction of the Bergman projection $\boldsymbol {B}_{\Omega _1}$ to the $\Gamma $ -invariant subspace $[{L}^2(\Omega _1)]^\Gamma $ is also the orthogonal projection from $[{L}^2(\Omega _1)]^\Gamma $ onto $[{A}^2(\Omega _1)]^\Gamma $ . But this follows from Lemma 4.12 above.

Thus $P=\boldsymbol {B}_{\Omega _1}|_{[L^2(\Omega _1)]^\Gamma }$ , and the commutativity of (4.11) follows.  ▪

Proof Proposition 4.5 says that $\Phi ^\sharp $ is a homothetic isomorphism, mapping

$$ \begin{align*}L^p(\Omega_2) \to \left[L^p(\Omega_1, \left\vert\det \Phi'\right\vert{{\hspace{-0.0001pt}}}^{2-p})\right]^\Gamma,\end{align*} $$

and that it restricts to a homothetic isomorphism on the holomorphic subspaces. Similarly, $(\Phi ^\sharp )^{-1}$ has the same properties with the domains and ranges switched.

First assume Statement (2). From the diagram (4.11), we write

(4.17) $$ \begin{align} \boldsymbol{B}_{\Omega_2}= (\Phi^\sharp)^{-1}\circ \boldsymbol{B}_{\Omega_1}\circ\Phi^\sharp. \end{align} $$

By hypothesis, $\boldsymbol {B}_{\Omega _1}$ is a bounded linear operator mapping

$$ \begin{align*}\left[L^p(\Omega_1, \left\vert\det \Phi'\right\vert{{\hspace{-0.0001pt}}}^{2-p})\right]^\Gamma \to \left[A^p(\Omega_1, \left\vert\det \Phi'\right\vert{{\hspace{-0.0001pt}}}^{2-p})\right]^\Gamma.\end{align*} $$

Consequently, this composition maps $L^p(\Omega _2)$ boundedly into $A^p(\Omega _2)$ , giving statement (1). A similar argument shows that (1) implies (2).

For the commutativity of the diagram, rewrite (4.17) and see that on the subspace $L^p(\Omega _2)\cap L^2(\Omega _2)$ of $L^p(\Omega _2)$ we have the relation

(4.18) $$ \begin{align} \Phi^\sharp\circ\boldsymbol{B}_{\Omega_2}= \boldsymbol{B}_{\Omega_1}\circ\Phi^\sharp. \end{align} $$

Using the hypothesis (for the $\boldsymbol {B}_{\Omega _j}$ ) and Proposition 4.5 (for $\Phi ^\sharp $ ), we see that each of the four maps in the diagram (4.16) extends to the respective domain in that diagram and is continuous. By continuity, (4.18) continues to hold for the extended maps. This shows that the diagram (4.16) is commutative.  ▪

Proof Define subsets $\mathcal {J}_\beta , \mathcal {K}_\beta \subset \{1,2,\cdots ,n \}$ with

$$ \begin{align*} \mathcal{J}_\beta = \{\,j : \beta_j \ge 0 \}, \quad \mathcal{K}_\beta = \{\,k : \beta_k < 0 \}, \end{align*} $$

and let

$$ \begin{align*} f(w)= \prod_{j\in \mathcal{J}_\beta} w_j^{\beta_j} \times \prod_{k\in \mathcal{K}_\beta } \left(\overline{w_k}\right)^{-\beta_k}. \end{align*} $$

Then f is a bounded function on $\Omega $ , and therefore $f \in L^p(\Omega )$ . We now show that $\boldsymbol {B}_\Omega f = C \varphi _\beta $ for some $C{\neq }0$ . Since $\varphi _\beta \not \in L^p(\Omega )$ , this will show that $\boldsymbol {B}_\Omega $ fails to map the element $f\in L^p(\Omega )$ to a function in $L^p(\Omega )$ . This will imply that $ \boldsymbol {B}_\Omega $ is not bounded in the $L^p$ -norm, since if it were so, it would extend to a map from the dense subspace $L^p(\Omega )\cap L^2(\Omega )$ to the whole of $L^p(\Omega )$ .

Let $\gamma =(\left \vert \beta _1\right \vert , \dots , \left \vert \beta _n\right \vert )\in \mathbb {N}^n$ be the multi-index obtained by replacing each entry of $\beta $ by its absolute value. Write the polar form of w as $w=\rho (w)\odot \exp (i\theta )$ for a $\theta \in {\mathbb {R}}^n$ . Then $ f(w) = \rho (w)^\gamma e^{i \beta \theta }, $ and for $\alpha \in (\mathbb {Z}^n)^\dagger $ , we have

$$ \begin{align*} \varphi_\alpha(w)=w^\alpha= \left(\rho(w)\odot \exp(i\theta) \right)^\alpha= \rho(w)^\alpha e^{i \alpha \theta}. \end{align*} $$

Further, denote by $\left \vert \Omega \right \vert =\{\rho (z):z\in \Omega \}\subset {\mathbb {R}}^n$ the Reinhardt shadow of $\Omega $ , and let $\mathbb {T}^n$ be the unit torus of n dimensions. Then, for each $\alpha \in \mathbb {Z}^n $ , we have

Since (5.4) is an orthonormal basis of $A^2(\Omega )$ it follows that all the Fourier coefficients of $\boldsymbol {B}_\Omega f$ with respect this basis vanish, except the $\beta $ th coefficient, which is nonzero. Therefore, $\boldsymbol {B}_\Omega f = C \varphi _\beta \notin L^p(\Omega ), $ for some constant $C{\neq }0$ . ▪

Proof Recall the definition of the projective height function as in (1.4). Then we have, by projective invariance, for each $1\leq j \leq n$ :

$$ \begin{align*}\mathsf{h}(B^{-1}e_j)=\mathsf{h}(\det B\cdot B^{-1}e_j)=\mathsf{h}_j( \operatorname{\mathrm{adj}} B\cdot e_j)= \mathsf{h}(A e_j)=\mathsf{h}(a_j). \end{align*} $$

Since $A\succ 0$ , it follows that $a_j$ is a vector of non-negative integers, and the vector $\frac {1}{\gcd (a_j)}a_j$ is such that its entries are coprime non-negative integers. Therefore,

So

Since the function $x\mapsto \dfrac {2x}{x-1}$ is strictly decreasing for $x>1$ , we have

 ▪

Proof By Theorem 3.12, $\Phi _A:\mathbb {D}^n_{\mathsf {L}(B)}\to \mathscr {U}$ is a proper holomorphic map of quotient type with $\det A$ sheets. Therefore, by Proposition 4.5 (in particular (4.8)) we see that

where $a_j \in \mathbb Z^n$ is the jth column of A. It is clear that $\int _{\mathscr {U}} \left \vert \varphi _\beta \right \vert {{\hspace{-0.0001pt}}}^p dV<\infty $ if and only if

for each $1\leq j \leq n$ . This completes the proof.  ▪

Proof Let $\mathcal {O}=\{x\in ({\mathbb {R}}^m)^\dagger :xP\succeq q \} .$ Then $\mathcal {O}$ is an unbounded convex set, so at least one of the coordinates $x_1,\dots , x_m$ is unbounded on $\mathcal {O}$ . Rename the coordinates so that $x_1$ is unbounded. It follows that the projection

$$ \begin{align*} \{x_1\in{\mathbb{R}}: (x_1,\dots, x_m)\in \mathcal{O}\} \end{align*} $$

of $\mathcal {O}$ on the coordinate axis $x_1$ is an unbounded convex set, and therefore a ray or the whole of ${\mathbb {R}}$ . For an integer N let $C_N=\{x_1=N\}\cap \mathcal {O}$ , which is naturally thought of as a subset of $({\mathbb {R}}^{m-1})^\dagger $ . Therefore, either there is an $N_1$ such that $C_N$ is nonempty if $N\geq N_1$ or there is an $N_2$ such that $C_N$ is nonempty if $N\leq N_2$ . Assuming the former, we see that the sets $C_N$ are convex subsets of $({\mathbb {R}}^{m-1})^\dagger $ and similar to each other, i.e., they are dilations of the same set. As the size of each $C_N$ becomes infinite as $N\to \infty $ , for large N, the set $C_N$ contains cubes of arbitrarily large size, where a cube is a product of intervals of the same size in each coordinate. As soon as $C_N$ contains a cube of side $(1+\epsilon )$ for some $\epsilon>0$ , we see that there is a point $M\in (\mathbb {Z}^{m-1})^\dagger $ that belongs to $C_N$ . It follows that the point $(N,M)\in \mathbb {Z}\times (\mathbb {Z}^{m-1})^\dagger $ belongs to $\mathcal {O}$ . ▪

Proof of Proposition 5.12 (1) If $p=2$ , the condition (5.10) becomes

, which is equivalent to

(5.16)

Now the jth entry of the row vector on the left of the above equation is given by

which is a positive integer divisible by $\gcd (a_j)=\gcd (a^1_j,\dots , a^n_j).$ It follows that (5.16) holds if and only if

which is precisely the content of (5.13).

(2) Fix $1\leq j \leq n$ . By the Euclidean algorithm, $\Pi _j {\neq }\emptyset $ . Choose $y \in \Pi _j$ . Define a $\mathbb {Z}$ -module homomorphism $\phi : (\mathbb {Z}^n)^\dagger \to \mathbb {Z}$ by setting $\phi (x)=xa_j$ , i.e.,

$$ \begin{align*} \phi(x_1, \ldots, x_n) = \sum_{k=1}^n x_k a^k_j.\end{align*} $$

We then see that $\Pi _j = y + \ker \phi $ . Since $\mathbb {Z}$ is a principal ideal domain, $\ker \phi $ is a free $\mathbb {Z}$ -submodule of $(\mathbb {Z}^n)^\dagger $ of rank $\leq n$ (see [Reference Dummit and FooteDF04, Theorem 4, Chapter 12 (p. 460)]). Moreover as $\phi $ is surjective, the quotient $\mathbb {Z}$ -module $(\mathbb {Z}^n)^\dagger /\ker \phi $ is isomorphic to $\mathbb {Z}$ . It can be seen, by tensoring with $\mathbb Q$ for example, that the rank of $\ker \phi $ is $n-1$ . Let D be an $(n-1)\times n$ integer matrix whose rows are a $\mathbb {Z}$ -basis of $\ker \phi $ . Then the map $f:(\mathbb {Z}^{n-1})^\dagger \to (\mathbb {Z}^n)^\dagger $ given by $ f(t)=y+tD$ is a parametrization of $\Pi _j$ , i.e., it is one-to-one and its range is precisely $\Pi _j$ . To complete the proof of the result, it is sufficient to show that $f^{-1}(\mathscr {S}_2(\mathscr {U}))$ is a nonempty subset of $(\mathbb {Z}^{n-1})^\dagger $ . Notice now that an integer vector $t\in f^{-1}(\mathscr {S}_2(\mathscr {U}))$ , i.e., $f(t)\in \mathscr {S}_2(\mathscr {U})\cap \Pi _j$ if and only if

By Lemma 5.15, there is an integer vector $t\in (\mathbb {Z}^{n-1})^\dagger $ that satisfies the above system of inequalities. This concludes the proof of part (2)  ▪

Proof Throughout this proof, C will denote a constant that depends only on p and $\gamma $ . The actual value of C may change from line to line.

Proceed by induction on the dimension n. First consider the base case $n=1$ . We have, by the Bergman inequality (cf. [Reference Duren and SchusterDS04, Theorem 1]) that there is a $C>0$ such that

$$ \begin{align*} \mathop{\mathrm{sup}}\limits_{\left\vert z\right\vert< \frac{1}{2}} \left\vert f(z)\right\vert\leq C \left\Vert f\right\Vert{{\hspace{-0.0001pt}}}_{L^p(\mathbb{D})} \end{align*} $$

for all $f\in A^p(\mathbb {D})$ . Therefore, for $f\in A^p(\mathbb {D})$ we have the estimate

(6.6) $$ \begin{align} \int_{\left\vert z\right\vert<\frac{1}{2}}\left\vert f(z)\right\vert{{\hspace{-0.0001pt}}}^p \left\vert z\right\vert{{\hspace{-0.0001pt}}}^\gamma dV(z)\leq \mathop{\mathrm{sup}}\limits_{\left\vert z\right\vert< \frac{1}{2}} \left\vert f(z)\right\vert{{\hspace{-0.0001pt}}}^p \cdot \int_{\left\vert z\right\vert<\frac{1}{2}} \left\vert z\right\vert{{\hspace{-0.0001pt}}}^\gamma dV(z) < C \cdot\left\Vert f\right\Vert{{\hspace{-0.0001pt}}}_{L^p(\mathbb{D})}^p, \end{align} $$

where we have used the fact that since $\gamma>-2$ we have

$$ \begin{align*}\int_{\left\vert z\right\vert<\frac{1}{2}} \left\vert z\right\vert{{\hspace{-0.0001pt}}}^\gamma dV(z)=2\pi\int_0^{\frac{1}{2}} r^{\gamma+1}dr <\infty. \end{align*} $$

On the other hand,

(6.7) $$ \begin{align} \int_{\frac{1}{2}\leq \left\vert z\right\vert<1} \left\vert f(z)\right\vert{{\hspace{-0.0001pt}}}^p \left\vert z\right\vert{{\hspace{-0.0001pt}}}^\gamma dV(z)\leq C \left\Vert f\right\Vert{{\hspace{-0.0001pt}}}_{L^p(\mathbb{D})}^p, \end{align} $$

where we have used the fact that $\displaystyle { \mathop {\mathrm {sup}}\limits _{\frac {1}{2}\leq \left \vert z\right \vert <1} \left \vert z\right \vert {{\hspace{-0.0001pt}}}^\gamma <\infty . }$

Adding (6.6) and (6.7), the estimate (6.5) follows in the case $n=1$ .

For the general case, assume the result established in $n-1$ dimensions. Write the coordinates of ${\mathbb {C}}^n$ as $z=(z',z_{n})\in {\mathbb {C}}^{n-1}\times {\mathbb {C}}$ , and $\gamma =(\gamma ',\gamma _n)\in {\mathbb {R}}^{n-1}\times {\mathbb {R}}$ . Then, using Fubini’s theorem

$$ \begin{align*} \int_{\mathbb{D}^n}\left\vert f\right\vert{{\hspace{-0.0001pt}}}^p \rho^{\gamma} dV& = \int_{\mathbb{D}^{n-1}}\rho(z')^{\gamma'}\left(\int_{\mathbb{D}} \left\vert f(z',z_n)\right\vert{{\hspace{-0.0001pt}}}^p\left\vert z_n\right\vert{{\hspace{-0.0001pt}}}^{\gamma_n}dV(z_n)\right)dV(z')\\ &\leq C \int_{\mathbb{D}^{n-1}}\rho(z')^{\gamma'}\left(\int_{\mathbb{D}} \left\vert f(z',z_n)\right\vert{{\hspace{-0.0001pt}}}^p dV(z_n)\right)dV(z')\\ &\leq C \int_{\mathbb{D}}\left(\int_{\mathbb{D}^{n-1}}\left\vert f(z',z_n)\right\vert{{\hspace{-0.0001pt}}}^p \rho(z')^{\gamma'} dV(z')\right)dV(z_n)\\ &\leq C \int_{\mathbb{D}}\left(\int_{\mathbb{D}^{n-1}}\left\vert f(z',z_n)\right\vert{{\hspace{-0.0001pt}}}^p dV(z')\right)dV(z_n)\\ &=C \int_{\mathbb{D}^n}\left\vert f(z',z_n)\right\vert{{\hspace{-0.0001pt}}}^p dV(z',z_n), \end{align*} $$

which proves the result.  ▪

Proof We need only to prove the case in which $\lambda =(1,0,\dots ,0)$ , so that $\varphi _\lambda (z)=z_1$ . Once this special case is established, the general result follows by repeatedly applying it and permuting the coordinates.

In what follows, C will denote some positive constant that depends only on p and $\lambda $ , where the actual value of C may change from line to line. First, consider the one dimensional case, so that we have to show that for a holomorphic function f on the disc we have

$$ \begin{align*} \int_{\mathbb{D}}\left\vert f(z)\right\vert{{\hspace{-0.0001pt}}}^p dV(z) \leq C \int_{\mathbb{D}}\left\vert z f(z)\right\vert{{\hspace{-0.0001pt}}}^p dV(z), \end{align*} $$

where the left hand side is assumed to be finite (and therefore the right hand side is finite.) First note that we obviously have

(6.10) $$ \begin{align} \int_{\frac{1}{2}\leq \left\vert z\right\vert<1}\left\vert f(z)\right\vert{{\hspace{-0.0001pt}}}^p dV(z) \leq 2^p \int_{\frac{1}{2}\leq \left\vert z\right\vert<1}\left\vert z f(z)\right\vert{{\hspace{-0.0001pt}}}^p dV(z). \end{align} $$

On the other hand, if $\left \vert z\right \vert =\frac {1}{2}$ , we have

$$ \begin{align*} \left\vert f(z)\right\vert{{\hspace{-0.0001pt}}}^p &= 2^p \left\vert z f(z)\right\vert{{\hspace{-0.0001pt}}}^p\\ &\leq 2^p \mathop{\mathrm{sup}}\limits_{\left\vert w\right\vert\leq \frac{1}{2}} \left\vert w f(w)\right\vert{{\hspace{-0.0001pt}}}^p& \text{(maximum principle)}\\ &\leq C \int_{\mathbb{D}}\left\vert w f(w)\right\vert{{\hspace{-0.0001pt}}}^p dV(w) &\text{(Bergman's inequality)} \end{align*} $$

The maximum principle now implies

$$ \begin{align*} \mathop{\mathrm{sup}}\limits_{\left\vert z\right\vert\leq \frac{1}{2}} \left\vert f(z)\right\vert{{\hspace{-0.0001pt}}}^p \leq C \int_{\mathbb{D}}\left\vert z f(z)\right\vert{{\hspace{-0.0001pt}}}^p dV(z), \end{align*} $$

so that we have

(6.11) $$ \begin{align} \int_{\left\vert z\right\vert\leq \frac{1}{2}} \left\vert f(z)\right\vert{{\hspace{-0.0001pt}}}^p dV(z) \leq C \int_{\mathbb{D}}\left\vert z f(z)\right\vert{{\hspace{-0.0001pt}}}^p dV(z). \end{align} $$

Combining (6.10) and (6.11) the result follows for $n=1$ .

For the higher-dimensional case, denote the coordinates of ${\mathbb {C}}^n$ as $(z_1, z')$ where $z'=(z_2,\dots , z_n)$ . Then for $f\in A^p(\mathbb {D}^n)$ we have

$$ \begin{align*} \int_{\mathbb{D}^n} \left\vert f(z_1, z')\right\vert{{\hspace{-0.0001pt}}}^pdV(z_1, z')& = \int_{\mathbb{D}^{n-1}} \left(\int_{\mathbb{D}} \left\vert f(z_1,z')\right\vert{{\hspace{-0.0001pt}}}^p dV(z_1)\right)dV(z')\\ &\leq C \int_{\mathbb{D}^{n-1}} \left(\int_{\mathbb{D}} \left\vert z_1f(z_1,z')\right\vert{{\hspace{-0.0001pt}}}^p dV(z_1)\right)dV(z')\\ &= C\int_{\mathbb{D}^{n}} \left\vert z_1f(z_1,z')\right\vert{{\hspace{-0.0001pt}}}^p dV(z_1,z'). \end{align*} $$

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Proof Recall that, by definition, the monomial $\varphi _\alpha $ is invariant under the group action of $\Gamma $ if and only if $\sigma _\nu ^\sharp (\varphi _\alpha )= (\varphi _\alpha \circ \sigma _\nu ) \det \sigma _\nu ' = \varphi _\alpha $ for all $\sigma _\nu \in \Gamma $ , where for $\nu \in \mathbb {Z}^n$ , the automorphism $\sigma _\nu \in \Gamma $ is as in (3.14). Denote the rows of $A^{-1}$ by $c^1, \ldots , c^n$ . By the linearity of $\sigma _\nu $

for $\nu \in \mathbb {Z}^n$ . Also, for $z \in (\mathbb {D}^*)^n$ and $ \nu \in \mathbb {Z}^n$ ,

$$ \begin{align*}\varphi_\alpha \circ \sigma_\nu (z) = \varphi_\alpha( \exp( 2 \pi i A^{-1} \nu) \odot z ) = (\exp( 2 \pi i A^{-1} \nu) \odot z )^\alpha = e^{2 \pi i \alpha A^{-1} \nu} z^\alpha. \end{align*} $$

Therefore,

(6.13)

Hence, $\varphi _\alpha \in [\mathcal {O} ({\mathbb {D}^n_{\mathsf {L}(B)}})]^\Gamma $ if and only if $\sigma _\nu ^\sharp (\varphi _\alpha )=\varphi _\alpha $ for all $\nu \in \mathbb {Z}^n$ , i.e.,

for all $z \in (\mathbb {D}^*)^n$ and $ \nu \in \mathbb {Z}^n, $ i.e., if and only if

Now if there is $\beta \in (\mathbb {Z}^n)^\dagger $ such that , then clearly, for all $\nu \in \mathbb {Z}^n$ . Conversely, assume that for all $\nu \in \mathbb {Z}^n$ and let $e_1,\dots , e_n$ be the standard basis of $\mathbb {Z}^n$ . Then the jth column of (where I is the $n\times n$ identity matrix) is which is therefore in $\mathbb {Z}$ . Therefore, we have . It follows that as desired.  ▪

Proof Let $\displaystyle { f(z)=\sum _{\alpha \succeq 0}a_\alpha z^\alpha }$ , be the Taylor expansion of f. If f is $\Gamma $ invariant, then we claim that for each $\alpha $ such that $a_\alpha {\neq }0$ , we have that $z^\alpha $ is $\Gamma $ -invariant. Indeed using (6.13) we have

comparing this with the Taylor expansion of f and equating coefficients the claim follows. Therefore, the Taylor expansion of f is of the form

(6.15)

Notice that

. The integer $\beta a_j=\sum _{k=1}^n\beta _k a_j^k$ which occurs in the exponent is divisible by $\gcd (a_j)$ . Further since $\sum _{k=1}^n\beta _k a_j^k\geq 1$ it follows that $\beta a_j\geq \gcd (a_j).$ It follows that the monomial

is of the form

, where $\gamma \succeq 0$ . The corollary follows from (6.15).  ▪