2 Invariant subspaces of shift operators

Let E and F be two complex separable Hilbert spaces. Let $k_{1}$ and $k_{2}$ be two reproducing kernels on $\Omega .$ If an operator-valued function $\Phi :\Omega \rightarrow B(F,E)$ is a multiplier from $H_{k_{1}}\otimes F$ into $H_{k_{2}}\otimes E,$ then for $y\in E$

(2.1) $$ \begin{align} M_{\Phi}^{\ast}\left[ k_{2}(z,\overline{\lambda})\otimes y\right] =k_{1}(z,\overline{\lambda})\otimes\Phi(\lambda)^{\ast}y. \end{align} $$

In particular, for $x\in E$ , $k(z,\overline {\lambda })\otimes x$ is an eigenvector of $M_{z_{i}}^{\ast }$ on $H_{k}\otimes E$ such that

(2.2) $$ \begin{align} M_{z_{i}}^{\ast}\left[ k(z,\overline{\lambda})\otimes x\right] =\overline{\lambda_{i}}\left[ k(z,\overline{\lambda})\otimes x\right]. \end{align} $$

We will also use operator-valued reproducing kernel $K(\zeta ,\overline {\lambda })$ where for each $(\zeta ,\overline {\lambda }),$ $K(\zeta ,\overline {\lambda })\in B(E).$ For example, $H_{K}=H_{k}\otimes E$ is the reproducing kernel space with operator-valued kernel $K(\zeta ,\overline {\lambda })=k(\zeta ,\overline {\lambda })I_{E}.$ The following result is useful to check (1.3) by using reproducing kernel.

Proof Set $R(\zeta ,\overline {\lambda })=AK(\zeta ,\overline {\lambda }).$ Since the set

$$\begin{align*}\left\{ \sum_{i=1}^{n}K(\zeta,\overline{\lambda_{i}})e_{i}:\{\lambda _{i}\}_{i=1}^{n}\subset\Omega,e_{i}\in E\right\} \end{align*}$$

is dense in $H_{K}$ , $A\geq 0$ if and only if

$$ \begin{align*} 0 & \leq\left\langle A\sum_{i=1}^{n}K(\zeta,\overline{\lambda_{i}} )e_{i},\sum_{i=1}^{n}K(\zeta,\overline{\lambda_{i}})e_{i}\right\rangle _{H_{K}}\\ & =\left\langle \sum_{i=1}^{n}AK(\zeta,\overline{\lambda_{i}})e_{i} ,\sum_{j=1}^{n}K(\zeta,\overline{\lambda_{j}})e_{j}\right\rangle _{H_{K}}\\ & =\sum_{i,j=1}^{n}\left\langle R(\lambda_{j},\overline{\lambda_{i}} )e_{i},e_{j}\right\rangle _{E}\geq0, \end{align*} $$

if and only if $R(\zeta ,\overline {\lambda })$ is a positive kernel.

We recall that in this paper, when we write $\frac {1}{k}(M_{z},M_{z}^{\ast })$ , we always assume that $\frac {1}{k}$ has the property that $\frac {1}{k}(M_{z},M_{z}^{\ast })$ exists as the limit of finite sums in the weak operator topology and it is a bounded operator. The usefulness of Lemma 2.1 lies in the fact that for the operator $A=f(T|M,\left ( T|M\right ) ^{\ast })$ in this note, $AK(\zeta ,\overline {\lambda })$ can often be computed.

Now we can prove the main result Theorem 1.1.

Proof of Theorem 1.1

Let P denote the orthogonal projection from $H_{k}\otimes E$ onto $L.$ The invariance of L can be represented as $PM_{z_{i}}P=M_{z_{i}}P.$ Taking adjoint, we have

(2.3) $$ \begin{align} PM_{z_{i}}^{\ast}P=PM_{z_{i}}^{\ast},1\leq i\leq d. \end{align} $$

It follows from $\frac {1}{k_{1}}(M_{z}|L,\left ( M_{z}|L\right ) ^{\ast })\geq 0$ that for any $x,y\in E,$

(2.4) $$ \begin{align} \left\langle \frac{1}{k_{1}}(M_{z}|L,\left( M_{z}|L\right) ^{\ast })Pk(z,\overline{\lambda})\otimes x,k(z,\overline{\zeta})\otimes y\right\rangle =\left\langle H(\zeta,\overline{\lambda})x,y\right\rangle , \end{align} $$

where $H(\zeta ,\overline {\lambda })$ is a positive $B(E)$ -valued (reproducing) kernel. By [Reference Agler and McCarthy2, Theorem 2.62], there exists a Hilbert space F and an operator-valued function $\Phi :\Omega \rightarrow B(F,E)$ such that

(2.5) $$ \begin{align} H(\zeta,\overline{\lambda})=\Phi(\zeta)\Phi(\lambda)^{\ast}. \end{align} $$

Suppose,

$$\begin{align*}\frac{1}{k_{1}(\zeta,\overline{\lambda})}=\sum_{\alpha,\beta\geq0} c_{\alpha,\beta}\zeta^{\alpha}\overline{\lambda}^{\beta}. \end{align*}$$

Since $\left ( M_{z}|L\right ) ^{\ast }=PM_{z}^{\ast }|L,$ for $\alpha ,\beta \geq 0,$

(2.6) $$ \begin{align} \left( M_{z}|L\right) ^{\ast\beta}P=PM_{z}^{\ast\beta}P=PM_{z}^{\ast\beta}, \end{align} $$

where the second equality follows from (2.3). Therefore,

(2.7) $$ \begin{align} & \left\langle \frac{1}{k_{1}}(M_{z}|L,\left( M_{z}|L\right) ^{\ast })Pk(z,\overline{\lambda})\otimes x,k(z,\overline{\zeta})\otimes y\right\rangle \\ & =\sum_{\alpha,\beta\geq0}c_{\alpha,\beta}\left\langle \left( M_{z}|L\right) ^{\alpha}\left( M_{z}|L\right) ^{\ast\beta}Pk(z,\overline {\lambda})\otimes x,k(z,\overline{\zeta})\otimes y\right\rangle \nonumber\\ & =\sum_{\alpha,\beta\geq0}c_{\alpha,\beta}\left\langle \left( M_{z}|L\right) ^{\ast\beta}Pk(z,\overline{\lambda})\otimes x,\left( M_{z}|L\right) ^{\ast\alpha}Pk(z,\overline{\zeta})\otimes y\right\rangle \nonumber\\ & =\sum_{\alpha,\beta\geq0}c_{\alpha,\beta}\left\langle PM_{z}^{\ast\beta }k(z,\overline{\lambda})\otimes x,PM_{z}^{\ast\alpha}k(z,\overline{\zeta })\otimes y\right\rangle \nonumber\\ & =\sum_{\alpha,\beta\geq0}c_{\alpha,\beta}\zeta^{\alpha}\overline{\lambda }^{\beta}\left\langle Pk(z,\overline{\lambda})\otimes x,k(z,\overline{\zeta })\otimes y\right\rangle \nonumber\\ & =\frac{1}{k_{1}(\zeta,\overline{\lambda})}\left\langle Pk(z,\overline {\lambda})\otimes x,k(z,\overline{\zeta})\otimes y\right\rangle ,\nonumber \end{align} $$

where the equality before the last one follows from (2.2). Then, by (2.4), (2.5), and (2.7) we have

(2.8) $$ \begin{align} & \left\langle Pk(z,\overline {\lambda})\otimes x,k(z,\overline{\zeta})\otimes y\right\rangle\\ &=k_{1}(\zeta,\overline{\lambda}) \left\langle \frac{1}{k_{1}}(M_{z}|L,\left( M_{z}|L\right) ^{\ast })Pk(z,\overline{\lambda})\otimes x,k(z,\overline{\zeta})\otimes y\right\rangle \nonumber\\ & = \langle k_{1}(\zeta,\overline{\lambda}) \Phi(\zeta)\Phi(\lambda)^{\ast} x, y\rangle. \nonumber \end{align} $$

Thus, $\Phi $ is a contractive multiplier from $H_{k_{1}}\otimes F$ to $H_{k}\otimes E$ , see [Reference Paulsen and Raghupathi27, Theorem 6.28]. By (2.1) with $k_{2}=k,$

$$ \begin{align*} & \left\langle M_{\Phi}M_{\Phi}^{\ast}k(z,\overline{\lambda})\otimes x,k(z,\overline{\zeta})\otimes y\right\rangle \\ & =\left\langle M_{\Phi}^{\ast}k(z,\overline{\lambda})\otimes x,M_{\Phi }^{\ast}k(z,\overline{\zeta})\otimes y\right\rangle \\ & =\left\langle k_{1}(z,\overline{\lambda})\otimes\Phi(\lambda)^{\ast} x,k_{1}(z,\overline{\zeta})\otimes\Phi(\zeta)^{\ast}y\right\rangle \\ & =\left\langle k_{1}(z,\overline{\lambda}),k_{1}(z,\overline{\zeta })\right\rangle \left\langle \Phi(\lambda)^{\ast}x,\Phi(\zeta)^{\ast }y\right\rangle \\ & =k_{1}(\zeta,\overline{\lambda})\left\langle \Phi(\zeta)\Phi(\lambda)^{\ast}x, y\right\rangle. \end{align*} $$

It thus follows from (2.8) that $M_{\Phi }M_{\Phi }^{\ast }=P,$ and $M_{\Phi }$ is a partial isometry with range $L.$

In the case, k and $k_{1}$ are holomorphic kernels, $\Phi $ is an analytic operator-valued function. In this note, we will assume all kernels are holomorphic. Since L is the range of the partial isometry $M_{\Phi },$ both L are $L^{\perp }$ are reproducing kernel spaces. In particular, $L^{\perp }$ is a special case of de Branges–Rovnyak spaces, see [Reference Fricain and Mashreghi11].

Corollary 2.1 Let L be a closed subspace of $H_{k}\otimes E$ that is $\frac {1}{k_{1}}$ -invariant for $M_{z}$ , where $\frac {1}{k_{1}}$ is a polynomial. Then, L are $L^{\perp }$ are reproducing kernel spaces with kernels

$$\begin{align*}\Phi(z)\Phi(\lambda)^{\ast}k_{1}(z,\overline{\lambda})\text{ and }k(z,\overline{\lambda})I_{E}-\Phi(z)\Phi(\lambda)^{\ast}k_{1}(z,\overline {\lambda}), \end{align*}$$

respectively, where $\Phi $ is as in Theorem 1.1.

Note that if $L=\Phi \cdot \left [ H_{k_{1}}\otimes F\right ] $ , where $M_{\Phi }$ is a partial isometry from $H_{k_{1}}\otimes F$ to $H_{k}\otimes E,$ then L is an invariant subspace for $M_{zI_{E}}.$ The fact that such an L is in fact $\frac {1}{k_{1}}$ -invariant for $M_{zI_{E}}$ is contained implicitly in the proof of Theorem 1.1. We state a slightly more general result and also include a direct proof.

Proof The proof is contained implicitly in the proof Theorem 1.1, we will use the notation there. Write

$$\begin{align*}\frac{1}{k_{2}(\zeta,\overline{\lambda})}=\sum_{\alpha,\beta\geq0} a_{\alpha,\beta}\zeta^{\alpha}\overline{\lambda}^{\beta}. \end{align*}$$

Since L has reproducing kernel $\Phi (z)\Phi (\lambda )^{\ast }k_{1}(z,\overline {\lambda })$ and $Pk(z,\overline {\lambda })\otimes x = \Phi (z)\Phi (\lambda )^{\ast } k_{1}(z,\overline {\lambda })x, x \in E$ , by Lemma 2.1, $\frac {1}{k_{2}}(M_{z}|L,\left ( M_{z}|L\right ) ^{\ast })\geq 0$ if and only if

$$ \begin{align*} & \frac{1}{k_{2}}(M_{z}|L,\left( M_{z}|L\right) ^{\ast})\Phi(z)\Phi(\lambda)^{\ast}k_{1}(z,\overline{\lambda})\\ & =\sum_{\alpha,\beta\geq0}a_{\alpha,\beta}M_{z}^{\alpha}PM_{z}^{\ast\beta }\Phi(z)\Phi(\lambda)^{\ast}k_{1}(z,\overline{\lambda})\\ & =\sum_{\alpha,\beta\geq0}a_{\alpha,\beta}M_{z}^{\alpha}\overline{\lambda }^{\beta}\Phi(z)\Phi(\lambda)^{\ast}k_{1}(z,\overline{\lambda})\\ & =\sum_{\alpha,\beta\geq0}a_{\alpha,\beta}z^{\alpha}\overline{\lambda }^{\beta}\Phi(z)\Phi(\lambda)^{\ast}k_{1}(z,\overline{\lambda})\\ & =\frac{1}{k_{2}(z,\overline{\lambda})}\Phi(z)\Phi(\lambda)^{\ast} k_{1}(z,\overline{\lambda})=\Phi(z)\Phi(\lambda)^{\ast}\frac{k_{1} (z,\overline{\lambda})}{k_{2}(z,\overline{\lambda})}\geq0, \end{align*} $$

where the first equality follows from (2.6).

3 Invariant subspaces on the unit ball

Let $\mathbb {B}^{m}$ be the unit ball of $\mathbb {C}^{m},$

$$\begin{align*}\mathbb{B}^{m}=\left\{ z=(z_{1},\dots,z_{m})\in\mathbb{C}^{m}:\left\vert z_{1}\right\vert ^{2}+\cdots+\left\vert z_{m}\right\vert ^{2}<1\right\}. \end{align*}$$

Let $H(\mathbb {B}^{m}\mathbb {)}$ be the space of all holomorphic functions on $\mathbb {B}^{m}$ . For $f\in H(\mathbb {B}^{m}\mathbb {)}$ , we always assume $f(z)$ has the following power series expansion

$$\begin{align*}f(z)= {\textstyle\sum_{\alpha\geq0}} f_{\alpha}z^{\alpha}. \end{align*}$$

Let $\lambda =(\lambda _{1},\dots ,\lambda _{m})\in \mathbb {B}^{m}$ and $\left \langle z,\lambda \right \rangle =\left \langle z,\lambda \right \rangle _{\mathbb {C}^{m}}=\sum \nolimits _{i=1}^{m}z_{i}\overline {\lambda _{i}}$ .

Let $\mathcal {A}_{\rho }(\mathbb {B}^{m})$ (for $\rho>0$ ) be the Hilbert space of holomorphic functions $\mathbb {B}^{m}$ with reproducing kernel

(3.1) $$ \begin{align} k_{\rho,\mathbb{B}^{m}}(z,\overline{\lambda})=\frac{1}{\left( 1-\left\langle z,\lambda\right\rangle \right) ^{\rho}}=\sum_{\alpha\geq0}\frac{\Gamma (\rho+\left\vert \alpha\right\vert )}{\alpha!\Gamma(\rho)}z^{\alpha} \overline{\lambda}^{\alpha}. \end{align} $$

That is,

$$\begin{align*}\mathcal{A}_{\rho}(\mathbb{B}^{m}){\mathbb{=}}\left\{ f\in H(\mathbb{B} ^{d}{\mathbb{)}}:\left\Vert f\right\Vert _{\mathcal{A}_{\rho}}^{2} ={\sum_{\alpha\geq0}}\frac{\alpha!\Gamma(\rho)}{\Gamma(\rho+\left\vert \alpha\right\vert )}\left\vert f_{\alpha}\right\vert ^{2}<\infty\right\}. \end{align*}$$

This scale of spaces contains the Drury–Arveson space $H_{m}^{2}$ ( $\rho =1$ ), the Hardy space $H^{2}(\mathbb {B}^{m})$ ( $\rho =m$ ), the Bergman space $L_{a}^{2}(\mathbb {B}^{m})$ ( $\rho =m+1$ ), and the weighted Bergman spaces ( $\rho>m$ ). In the Drury–Arveson space $H_{m}^{2}$ , m could also be countable infinity [Reference Agler and McCarthy2].

Theorem 1.1 generalizes a theorem of McCullough and Trent on the Drury–Arveson space $H_{m}^{2}$ in the way that an invariant subspace of $M_{z}$ is automatically $\frac {1}{k_{\rho ,\mathbb {B}^{m}}}$ -invariant ( $\rho =1)$ for $M_{z}.$ A special case of the theorem of McCullough and Trent was proved by Arveson [Reference Arveson5].

Proof Let P denote the orthogonal projection from $H_{m}^{2}\otimes E$ onto $L.$ Then,

$$\begin{align*}M_{z_{i}}PM_{z_{i}}^{\ast}\leq M_{z_{i}}M_{z_{i}}^{\ast}. \end{align*}$$

Since $1/k_{1,\mathbb {B}^{m}}=1-\sum \nolimits _{i=1}^{m}z_{i}\overline {\lambda _{i}},$

$$ \begin{align*} \frac{1}{k_{1,\mathbb{B}^{m}}}(M_{z}|L,\left( M_{z}|L\right) ^{\ast}) & =P\left( I-\sum\nolimits_{i=1}^{m}M_{z_{i}}PM_{z_{i}}^{\ast}\right) P\\ & \geq P\left( I-\sum\nolimits_{i=1}^{m}M_{z_{i}}M_{z_{i}}^{\ast}\right) P\geq0. \end{align*} $$

The proof is complete.

The invariant subspace theorem of McCullough and Trent [Reference McCullough and Trent24] actually is on $H_{k}\otimes E,$ where k is any irreducible complete pick kernel on $\Omega $ , see [Reference McCullough and Trent24, Theorem 0.7] and [Reference Agler and McCarthy2, Theorem 8.67] for precise statements. See also more refined results in this direction [Reference Clouâtre, Hartz and Schillo10, Reference Greene, Richter and Sundberg13, Reference McCullough and Richter23]. We use the argument in Theorem 1.1 to prove the following Theorem 1.1 obtained in [Reference Clouâtre, Hartz and Schillo10].

Proof The kernel s on $\Omega $ can be represented as

$$\begin{align*}s(z,\overline{\lambda})=\frac{1}{1-\left\langle b(z),b(\lambda)\right\rangle _{H}}, \end{align*}$$

where H is a Hilbert space and $b_{i}$ are the coordinate functions [Reference Agler and McCarthy2]. Then the assumption $k/s$ being a positive kernel implies that b is a contractive multiplier from $H_{k} \otimes H$ to $H_{k}$ . If L is a subspace of $H_{k}\otimes E$ invariant under $\text {Mult}(H_{s})$ , then on $H_{k}\otimes E$ (P is the projection onto L)

$$ \begin{align*} P\left( I-\sum\nolimits_{i}M_{b_{i}}PM_{b_{i}}^{\ast}\right) P \geq P\left( I-\sum\nolimits_{i}M_{b_{i}}M_{b_{i}}^{\ast}\right) P\geq0. \end{align*} $$

So for any $x,y\in E,$

$$ \begin{align*} \left\langle P\left( I-\sum\nolimits_{i}M_{b_{i}}PM_{b_{i}}^{\ast}\right) Pk(z,\overline{\lambda})\otimes x,k(z,\overline{\zeta})\otimes y\right\rangle =\left\langle H(\zeta,\overline{\lambda})x,y\right\rangle , \end{align*} $$

where $H(\zeta ,\overline {\lambda })$ is a positive $B(E)$ -valued (reproducing) kernel. It is clear that the left hand side is equal to

$$ \begin{align*} \left( 1-\sum\nolimits_{i} b_{i}(\zeta) \overline{b_{i}(\lambda)}\right) \left\langle Pk(z,\overline{\lambda})\otimes x,k(z,\overline{\zeta})\otimes y\right\rangle = \frac{1}{s(\zeta,\overline{\lambda})}\left\langle Pk(z,\overline {\lambda})\otimes x,k(z,\overline{\zeta})\otimes y\right\rangle. \end{align*} $$

If we assume $H(\zeta ,\overline {\lambda })=\Phi (\zeta )\Phi (\lambda )^{\ast }$ for some operator-valued function $\Phi :\Omega \rightarrow B(F,E)$ , then by the same argument in Theorem 1.1, we have $P = M_{\Phi }M_{\Phi }^{\ast }$ and $L=\Phi \cdot \left [ H_{s}\otimes F\right ] $ .

Similarly, we have the following result of Sarkar [Reference Sarkar34] which further generalizes Theorem of McCullough and Trent.

Proof Let $A= I-\sum \nolimits _{i=1}^{m}M_{z_{i}}M_{z_{i}}^{\ast }$ . Then,

$$ \begin{align*} Ak_{\rho,\mathbb{B}^{m}}(z,\overline{\lambda})I_{E} & =\left( 1-\left\langle z,\lambda\right\rangle \right) \frac{1}{\left( 1-\left\langle z,\lambda \right\rangle \right) ^{\rho}}I_{E}\\ & =\frac{1}{\left( 1-\left\langle z,\lambda\right\rangle \right) ^{\rho-1} }I_{E} \end{align*} $$

is a positive kernel. So by Lemma 2.1 $A\geq 0$ on $\mathcal {A}_{\rho }(\mathbb {B}^{m})\otimes E$ . It then follows that

(3.2) $$ \begin{align} \frac{1}{k_{1,\mathbb{B}^{m}}}(M_{z}|L,\left( M_{z}|L\right) ^{\ast}) & =P\left( I-\sum\nolimits_{i=1}^{m}M_{z_{i}}PM_{z_{i}}^{\ast}\right) P\\ & \geq P\left( I-\sum\nolimits_{i=1}^{m}M_{z_{i}}M_{z_{i}}^{\ast}\right) P\geq0.\nonumber \end{align} $$

The proof is complete.

The following relation between $\frac {1}{k_{l,\mathbb {B}^{m}}}$ -invariant subspaces and $\frac {1}{k_{n,\mathbb {B}^{m}}}$ for two natural numbers l and n is a consequence of Proposition 2.1, we give a different proof.

If $L = \mathcal {A}_{\rho }(\mathbb {B}^{m})\otimes E$ , then L is $\frac {1}{k_{n,\mathbb {B}^{m}}}$ -invariant, so L is $\frac {1}{k_{l,\mathbb {B}^{m}} }$ -invariant, $\rho \geq n>l$ . The above proposition leads to n different representations of L if L is $\frac {1}{k_{n,\mathbb {B}^{m}}}$ -invariant. Namely

$$\begin{align*}L=\Phi_{l}\cdot\left[ \mathcal{A}_{l}(\mathbb{B}^{m})\otimes F_{l}\right] , \text{ }1\leq l\leq n, \end{align*}$$

where $M_{\Phi _{l}}$ is a partial isometry from $\left [ \mathcal {A} _{l}(\mathbb {B}^{m})\otimes F_{l}\right ] $ to $\mathcal {A}_{\rho } (\mathbb {B}^{m})\otimes E.$ It is interesting to study how these $M_{\Phi _{l} }$ are related.

4 Invariant subspaces on the polydisk

Let $\mathbb {D}^{m}$ be the polydisk and $\mathcal {A}_{\rho }(\mathbb {D}^{m})$ (for $\rho>0$ ) be the Hilbert space of holomorphic functions on $\mathbb {D} ^{m}$ with reproducing kernel

(4.1) $$ \begin{align} k_{\rho,\mathbb{D}^{m}}(z,\overline{\lambda})=\prod_{i=1}^{m}\frac{1}{\left( 1-z_{i}\overline{\lambda_{i}}\right) ^{\rho}}. \end{align} $$

This scale of spaces contains the Hardy space $H^{2}(\mathbb {D}^{m})$ ( $\rho =1$ ), the Bergman space $L_{a}^{2}(\mathbb {D}^{m})$ ( $\rho =2$ ) and the weighted Bergman spaces ( $\rho>1$ ).

The study of invariant subspaces on $H^{2}(\mathbb {D}^{m})$ is extensive, see, for example, [Reference Agrawal, Clark and Douglas3, Reference Izuchi18, Reference Izuchi, Izuchi and Izuchi19, Reference Izuchi, Nakazi and Seto20, Reference Mandrekar22, Reference Radlow28, Reference Sarkar32, Reference Sarkar, Sasane and Wick35]. In fact, this study is a central theme in operator theory on $H^{2}(\mathbb {D}^{m}),$ see the survey paper [Reference Yang38] (mostly on $H^{2}(\mathbb {D}^{2})$ ) for more references. One of the elegant results on invariant subspaces of $H^{2}(\mathbb {D}^{m})$ is the following theorem. Recall an operator-valued inner function $\Theta $ is a bounded analytic operator-valued function $\Theta :\mathbb {D}^{m}\rightarrow B(F,E)$ such that

$$\begin{align*}\Theta(z)^{\ast}\Theta(z)=I_{F},\text{a.e. }z\in\mathbb{T}^{m} \end{align*}$$

for some subspace $F\subset E.$ A commuting tuple of operators $T=(T_{1} ,\dots ,T_{m})$ is doubly commuting if $T_{i}T_{j}^{\ast }=T_{j}^{\ast }T_{i}$ for all $i\neq j.$

Theorem 4.1 [Reference Kumari, Sarkar, Sarkar and Timotin21, Reference Mandrekar22, Reference Radlow28, Reference Sarkar, Sasane and Wick35]

Let $\ L\subset H^{2} (\mathbb {D}^{m})\otimes E$ be a closed subspace invariant for $M_{z}.$ Set $T=(T_{1},\dots ,T_{m})$ where

$$\begin{align*}T_{i}=M_{z_{i}}|L,i=1,\dots,m. \end{align*}$$

Then, T is doubly commuting if and only if there exist a subspace $F\subset E$ and an operator-valued inner function $\Theta :\mathbb {D}^{m}\rightarrow B(F,E)$ such that $L=\Theta \cdot \left [ H^{2}(\mathbb {D}^{m})\otimes F\right ] .$

Proof The “if” direction is by a direct verification, see also (4.4) and (4.5). We now prove the “only if” direction. Assume T is doubly commuting. Let P be the projection from $H^{2}(\mathbb {D}^{m})\otimes E$ to $L.$ Then, for $i\neq j,$

$$ \begin{align*} (I-T_{i}T_{i}^{\ast})(I-T_{j}T_{j}^{\ast}) & =I-T_{i}T_{i}^{\ast}-T_{j} T_{j}^{\ast}+T_{i}T_{i}^{\ast}T_{j}T_{j}^{\ast}\\ & =I-T_{i}T_{i}^{\ast}-T_{j}T_{j}^{\ast}+T_{i}T_{j}T_{i}^{\ast}T_{j}^{\ast.} \end{align*} $$

Therefore,

$$\begin{align*}\frac{1}{k_{1,\mathbb{D}^{m}}}(T,T^{\ast})=\prod_{i=1}^{m}(I-T_{i}T_{i}^{\ast })\geq0, \end{align*}$$

since for all $i,$

(4.2) $$ \begin{align} I-T_{i}T_{i}^{\ast}=I-M_{z_{i}}PM_{z_{i}}^{\ast}\geq I-M_{z_{i}}M_{z_{i} }^{\ast}\geq0. \end{align} $$

By Theorem 1.1, there exist an analytic operator-valued function $\Phi :\mathbb {D}^{m}\rightarrow B(F,E)$ for some Hilbert space F such that $M_{\Phi }$ is a partial isometry and $L=\Phi \cdot \left [ H^{2} (\mathbb {D}^{m})\otimes F\right ] .$ We show F can be chosen such that $\dim F\leq \dim E$ and $\Phi $ is inner by explicitly identifying F and $\Phi $ as W and $\Theta $ below following the proof of Theorem 1.3 in [Reference Sarkar, Sasane and Wick35]. Since T is a tuple of doubly commuting isometries, by the generalization of classical Wold decomposition theorem [Reference Gaşpar and Suciu12, Reference Sarkar31, Reference Timotin37],

(4.3) $$ \begin{align} L=\bigoplus_{\alpha\geq0}z^{\alpha}W, \end{align} $$

where

$$\begin{align*}W=\bigcap_{i=1}^{m}(L\ominus T_{i}L)=\bigcap_{i=1}^{m}\mathrm{ran}(I-T_{i}T_{i}^{\ast })=\mathrm{ran}\left( \prod_{i=1}^{m}(I-T_{i}T_{i}^{\ast})\right). \end{align*}$$

Let $V:H^{2}(\mathbb {D}^{m})\otimes W\rightarrow H^{2}(\mathbb {D}^{m})\otimes E$ be defined by

$$\begin{align*}V\left( {\textstyle\sum_{\alpha\geq0}} w_{\alpha}z^{\alpha}\right) = {\textstyle\sum_{\alpha\geq0}} M_{z}^{\alpha}w_{\alpha}, w_{\alpha} \in W. \end{align*}$$

Then, by (4.3), V is an isometry. It is also clear that $ran(V)=L.$ Furthermore, since $M_{z_{i}}V=VM_{z_{i}}$ for all $i,$ so $V=M_{\Theta }.$ $M_{\Theta }^{\ast }M_{\Theta }=T_{\Theta ^{\ast }\Theta }=I$ ( $T_{\Theta ^{\ast }\Theta }$ is the Toeplitz operator) implies that $\Theta $ is inner. By (4.2),

$$ \begin{align*} \dim W & =\mathrm{rank}\left( \prod_{i=1}^{m}(I-T_{i}T_{i}^{\ast})\right) \\ & \leq \mathrm{rank}\left( \prod_{i=1}^{m}(I-M_{z_{i}I_{E}}M_{z_{i}I_{E}}^{\ast })\right) =\dim E. \end{align*} $$

The proof is complete.

It is interesting to remark that the above result when $m=2$ and in the scalar case ( $\dim E=1$ ) is attributed to the highly cited paper by Mandrekar [Reference Mandrekar22] whose proof uses the generalization of classical Wold decomposition theorem for two doubly commuting isometries [Reference Słociński36]. But this result for any m in the scalar case is in fact proved earlier by Radlow [Reference Radlow28] in a slightly different formulation. The above theorem is proved by Sarkar, Sasane, and Wick [Reference Sarkar, Sasane and Wick35].

Theorem 4.2 Let $\ L\subset \mathcal {A}_{\rho }(\mathbb {D}^{m})\otimes E$ (for $\rho>1$ ) be a closed subspace invariant for $M_{z}.$ Set $T=(T_{1},\dots ,T_{m})$ where

$$\begin{align*}T_{i}=M_{z_{i}}|L,i=1,\dots,m. \end{align*}$$

If T is doubly commuting, then there exist a Hilbert space F and an analytic operator-valued function $\Phi :\mathbb {D}^{m}\rightarrow B(F,E)$ such that $M_{\Phi }$ is a partial isometry and $L=\Phi \cdot \left [ H^{2} (\mathbb {D}^{m})\otimes F\right ] .$

Proof Let $M_{z}=(M_{z_{1}}\cdots ,M_{z_{m}})=M_{z}|\mathcal {A}_{\rho }(\mathbb {D} ^{m})\otimes E$ for $\rho>1$ . Since $M_{z}$ is doubly commuting and each $M_{z_{i}I_{E}}$ is a contraction,

$$\begin{align*}\frac{1}{k_{1,\mathbb{D}^{m}}}(M_{z},M_{z}^{\ast})=\prod_{i=1}^{m} (I-M_{z_{i}I_{E}}M_{z_{i}I_{E}}^{\ast})\geq0. \end{align*}$$

Assume T is doubly commuting. Let P be the projection from $\mathcal {A} _{\rho }(\mathbb {D}^{m})\otimes E$ to $L.$ Similarly,

$$\begin{align*}\frac{1}{k_{1,\mathbb{D}^{m}}}(T,T^{\ast})=\prod_{i=1}^{m}(I-T_{i}T_{i}^{\ast })\geq0 \end{align*}$$

since by (4.2) as in the proof of the above theorem, each $T_{i}$ is a contraction. Now the result follows from Theorem 1.1 with $k_{1}=k_{1,\mathbb {D}^{m}}$ and $k=k_{\rho ,\mathbb {D}^{m}}$ (for $\rho>1$ ).

When $m=1,$ there is no need of doubly commuting condition and the above theorem in the case $m=1$ and $\rho $ is a natural number is proved by Ball and Bolotnikov [Reference Ball and Bolotnikov6] and [Reference Ball and Bolotnikov7]. The more general case $m=1$ and $\rho>1$ is proved by Sarkar [Reference Sarkar33].

In the case $M_{\Phi }$ is an isometry, the converse of the above theorem holds. That is, if $L=\Phi \cdot \left [ H^{2}(\mathbb {D}^{m})\otimes F\right ] $ , where $M_{\Phi }$ is an isometry, then T is doubly commuting. Since $P=M_{\Phi }M_{\Phi }^{\ast },$

(4.4) $$ \begin{align} T_{i}T_{j}^{\ast}P & =M_{z_{i}}PM_{z_{j}}^{\ast}P=M_{z_{i}}M_{\Phi}M_{\Phi }^{\ast}M_{z_{j}}^{\ast}P=M_{\Phi}M_{z_{i}}M_{z_{j}}^{\ast}M_{\Phi}^{\ast }P\nonumber\\ & =M_{\Phi}M_{z_{j}}^{\ast}M_{z_{i}}M_{\Phi}^{\ast}M_{\Phi}M_{\Phi}^{\ast }=M_{\Phi}M_{z_{j}}^{\ast}M_{z_{i}}M_{\Phi}^{\ast} \end{align} $$

(4.5) $$ \begin{align} T_{j}^{\ast}T_{i}P & =PM_{z_{j}}^{\ast}M_{z_{i}}P=M_{\Phi}M_{\Phi}^{\ast }M_{z_{j}}^{\ast}M_{z_{i}}M_{\Phi}M_{\Phi}^{\ast}\nonumber\\ & =M_{\Phi}M_{z_{j}}^{\ast}M_{\Phi}^{\ast}M_{\Phi}M_{z_{i}}M_{\Phi}^{\ast }=M_{\Phi}M_{z_{j}}^{\ast}M_{z_{i}}M_{\Phi}^{\ast}, \end{align} $$

thus $T_{i}T_{j}^{\ast }=T_{j}^{\ast }T_{i}$ for $i\neq j.$ It is an interesting question to characterize $\Phi $ such that T is doubly commuting when $M_{\Phi }$ is only a partial isometry. We mention that there are nontrivial $M_z$ -invariant subspaces of $\mathcal {A}_{\rho }(\mathbb {D}^{m})$ such that T is doubling commuting. For example, if $L = z_{1}^{n}z_{2}^{m} \mathcal {A}_{2}(\mathbb {D}^{2}), n, m \in \mathbb {N}$ , then T is doubling commuting, see [Reference Chattopadhyay, Das, Sarkar and Sarkar9, Reference Redett and Tung29]. We also mention the wandering subspace property is proved for an invariant subspace L such that T is doubly commuting on Bergman space of polydisk ( $\mathcal {A}_{2}(\mathbb {D}^{m})$ ) and Dirichlet space of polydisk in [Reference Chattopadhyay, Das, Sarkar and Sarkar9]. We note the following result.

Proof We claim $W=\Phi \cdot F$ . If the claim holds, then

$$\begin{align*}\bigvee_{\alpha\geq0}M_{z}^{\alpha}W=\bigvee_{\alpha\geq0}z^{\alpha}\Phi\cdot F=\Phi\cdot\left[ H^{2}(\mathbb{D}^{m})\otimes F\right] =L \end{align*}$$

and L has the wandering subspace property. To prove the claim, first note that $\Phi \cdot F\subset W:=\cap _{i-1}^{m}(L\ominus z_{i}L)$ since for any $f\in F$ and $h\in H^{2}(\mathbb {D}^{m})\otimes F,$

$$ \begin{align*} \left\langle \Phi\cdot f,z_{i}\Phi h\right\rangle & =\left\langle \Phi\cdot f,\Phi z_{i}h\right\rangle \\ & =\left\langle M_{\Phi}^{\ast}M_{\Phi}f,z_{i}h\right\rangle =\left\langle f,z_{i}h\right\rangle =0. \end{align*} $$

On the other hand, if $g=\Phi f\in W$ where $f\in H^{2}(\mathbb {D}^{m})\otimes F,$ then $g\perp z_{i}L$ implies that for all $h\in H^{2}(\mathbb {D} ^{m})\otimes F,$

$$\begin{align*}0=\left\langle \Phi\cdot f,z_{i}\Phi h\right\rangle =\left\langle f,z_{i}h\right\rangle. \end{align*}$$

Thus, $f\in F$ and $g\in \Phi \cdot F.$ So $W\subset \Phi \cdot F$ and the claim $W=\Phi \cdot F$ is established.

When $m=1,$ the wandering subspace property for invariant subspaces of the Bergman space and Dirichlet space are deep results by Aleman, Richter, and Sundberg [Reference Aleman, Richter and Sundberg4] and Richter [Reference Richter30]. The wandering subspace property does not hold in general on weighted Bergman spaces of the unit disk even for finite zero based invariant subspaces [Reference Hedenmalm and Zhu17]. It has been shown that on a class of holomorphic reproducing kernel Hilbert spaces of the unit disk, including all weighted Bergman spaces, the wandering subspace property does hold for finite zero based invariant subspaces if those zeros lie in a small disk centered at zero [Reference Gu and Park16].

5 Invariant subspaces on the polyball

The general nature of Theorem 1.1 applies to invariant subspaces of a multi-shift on polyball $\mathbb {B}^{m_{1}}\times \mathbb {B}^{m_{2}} \times \cdots \times \mathbb {B}^{m_{d}}.$ For notational simplicity, we state a result for $\mathbb {B}^{m}\times \mathbb {B}^{n}.$ For $\rho _{1},\rho _{2}>0,$ let

$$\begin{align*}\mathcal{A}_{\rho_{1},\rho_{2}}:=\mathcal{A}_{\rho_{1}}(\mathbb{B}^{m} )\otimes\mathcal{A}_{\rho_{2}}(\mathbb{B}^{n}). \end{align*}$$

Then, $\mathcal {A}_{\rho _{1},\rho _{2}}$ is a Hilbert space of holomorphic functions on $\mathbb{B}^{m}\times \mathbb{B}^{n}$ with reproducing kernel

$$\begin{align*}k_{\rho_{1},\rho_{2}}(z,w,\lambda,\zeta)=\frac{1}{(1-\left\langle z,\lambda\right\rangle )^{\rho_{1}}}\frac{1}{(1-\left\langle w,\zeta \right\rangle )^{\rho_{2}}}, \end{align*}$$

where $z,\lambda \in \mathbb {B}^{m}$ and $w,\zeta \in \mathbb {B}^{n}.$

Theorem 5.1 Let $L\subset \mathcal {A}_{\rho _{1},\rho _{2}}\otimes E$ (for $\rho _{1} ,\rho _{2}\geq 1$ ) be a closed subspace invariant for $(M_{z},M_{w}).$ Set $T=(T_{1},\dots ,T_{m})$ , $U=(U_{1},\dots ,U_{n})$ where

$$ \begin{align*} T_{i} & =M_{z_{i}}|L,i=1,\dots,m,\\ U_{j} & =M_{w_{j}}|L,j=1,\dots,n. \end{align*} $$

If T and U are doubly commuting, that is, for all $i,j$ , $T_{i}U_{j} =U_{j}T_{i}$ which is automatic and

$$\begin{align*}T_{i}U_{j}^{\ast}=U_{j}^{\ast}T_{i},\text{ } \end{align*}$$

then there exist a Hilbert space F and an analytic operator-valued function $\Phi :\mathbb {B}^{m}\times \mathbb {B}^{n}\rightarrow B(F,E)$ such that $M_{\Phi }$ is a partial isometry from $\mathcal {A}_{1,1}\otimes F$ to $\mathcal {A}_{\rho _{1},\rho _{2}}\otimes E$ and $L=\Phi \cdot \left [ \mathcal {A}_{1,1}\otimes F\right ] .$

Proof Since T and U are doubly commuting,

$$\begin{align*}\frac{1}{k_{1,1}}(T,U,T^{\ast},U^{\ast})=\left( I- {\textstyle\sum_{i=1}^{m}} T_{i}T_{i}^{\ast}\right) \left( I- {\textstyle\sum_{i=1}^{n}} U_{i}U_{i}^{\ast}\right). \end{align*}$$

As in (3.2), for $\rho _{1},\rho _{2}\geq 1$ , $I- {\textstyle \sum _{i=1}^{m}} T_{i}T_{i}^{\ast }\geq 0$ and $I- {\textstyle \sum _{i=1}^{n}} U_{i}U_{i}^{\ast }\geq 0.$ Thus

$$\begin{align*}\frac{1}{k_{\rho_{1},\rho_{2}}}(T,U,T^{\ast},U^{\ast})\geq0. \end{align*}$$

Hence, L is $\frac {1}{k_{1,1}}$ -invariant. The result now follows from Theorem 1.1.

There are nontrivial $(M_{z},M_{w})$ -invariant subspaces of $\mathcal {A}_{\rho _{1},\rho _{2}}$ such that T and U are doubly commuting. For example, if $L = z^{n}w^{m} \mathcal {A}_{1}(\mathbb {B}^{1} )\otimes \mathcal {A}_{2}(\mathbb {B}^{1}), n, m \in \mathbb {N}$ , then T and U are doubling commuting.