2 The main result

Suppose $\mathcal A$ is a unital C $^*$ -algebra and $\alpha _i:\mathcal A\rightarrow \mathcal A, 1\leq i\leq n$ , are unital $*$ -endomorphisms. Recall that the tensor algebra ${\mathcal {T}}^+({\mathcal {A}}, \alpha )$ of the C $^*$ -dynamical system $(\mathcal A,\alpha )$ is generated by $v_1, \dots , v_n$ , which form a row isometry $v = [v_1\ v_2\ \dots \ v_n]$ and a faithful copy of $\mathcal A$ . The row isometry v encodes the dynamics of the dynamical system $(\mathcal A,\alpha )$ in the sense that $av_i=v_i\alpha _i(a)$ , for all $a \in {\mathcal {A}}$ and $1\leq i \leq n$ . By definition, the tensor algebra ${\mathcal {T}}^+({\mathcal {A}}, \alpha )$ is universal over all representations encoding the dynamics of $(\mathcal A,\alpha )$ , so that the generating isometries $v_1, v_2, \dots , v_n$ are mapped to a row isometry.

Due to its universality, the tensor algebra ${\mathcal {T}}^+({\mathcal {A}}, \alpha ) $ admits a gauge action

$$ \begin{align*} \zeta:{\mathbb{T}} \longrightarrow \operatorname{Aut} ({\mathcal{T}}^+({\mathcal{A}}, \alpha)); \ \ {\mathbb{T}}\ni \lambda \longmapsto \zeta_{\lambda}, \end{align*} $$

so that $\zeta _{\lambda } (a) = a $ , for all $a \in {\mathcal {A}}$ , and $\zeta _{\lambda } (v_p)=\lambda ^{|p|} v_p$ , where $p=i_1i_2\dots i_k \in {\mathbb {F}}^+_{n}$ is an element of the free semigroup with n generators, $|p| :=k$ denotes the length of p and $v_p:=v_{i_1}v_{i_2}\dots v_{i_k}$ (we write $p=0$ is the empty word, with the understanding that $|0| =0$ and $v_{0}:=I$ ). From this gauge action, we deduce that every element $x \in {\mathcal {T}}^+({\mathcal {A}}, \alpha ) $ admits a formal Fourier series development $x \sim \sum _{k =0}^{\infty } E_k(x)$ . Each $E_k: {\mathcal {T}}^+({\mathcal {A}}, \alpha ) \rightarrow {\mathcal {T}}^+({\mathcal {A}}, \alpha )$ is a completely contractive, ${\mathcal {A}}$ -module projection on the subspace of ${\mathcal {T}}^+({\mathcal {A}}, \alpha ) $ generated by elements of the form $v_pa_p$ , with $p \in {\mathbb {F}}^+_{n}$ , $|p|=k$ and $a_p \in {\mathcal {A}}$ . Furthermore, $E_0$ is a multiplicative expectation onto ${\mathcal {A}} \subseteq {\mathcal {T}}^+({\mathcal {A}}, \alpha )$ . Finally, the formal series $x \sim \sum _{k =0}^{\infty } E_k(x)$ is Cesaro-convergent to $x $ , that is,

$$\begin{align*}x = \lim_{n\rightarrow \infty} \sum_{k=0}^n \left(1 - \frac{k}{n+1}\right)E_k(x) \end{align*}$$

(see [Reference Davidson and Katsoulis8] for a comprehensive development of this theory).

The following is a key result in our investigation.

Theorem 2.1. If $b = [b_1 \ \dots \ b_n]$ is a strict row contraction in $\mathcal A$ , such that $ab_i = b_i\alpha _i(a)$ , for all $a\in \mathcal A$ and $1\leq i\leq n$ , then there is a completely isometric automorphism $\rho $ of ${\mathcal {T}}^+({\mathcal {A}}, \alpha )$ , such that $\rho (a) = a$ for $a\in \mathcal A$ ,

$$\begin{align*}\rho(v) = (I - bb^*)^{1/2}(I-vb^*)^{-1}(b-v)(I_n-b^*b)^{-1/2} \end{align*}$$

and $\rho \circ \rho = {\operatorname {id}}$ . Furthermore, $E_0(\rho (v_i)) = b_i, 1\leq i\leq n$ .

Proof. By hypothesis, b is a strict contraction and so $\|vb^*\| < 1$ , which implies that $I-vb^*$ and $D_{b*} := (I-bb^*)^{1/2}$ are indeed invertible in $\mathcal A$ . Similarly, $D_{b} := (I_n - b^*b)^{1/2}$ is invertible in $M_n(\mathcal A)$ . One calculates

$$ \begin{align*} (I-bv^*)^{-1}&(I-bb^*)(I-vb^*)^{-1} \\ &= (I-bv^*)^{-1}(I-bv^*vb^*)(I-vb^*)^{-1} \\ &= (I-bv^*)^{-1}(I - vb^* + (I -bv^*)vb^*)(I-vb^*)^{-1} \\ &= (I-bv^*)^{-1} + vb^*(I-vb^*)^{-1} \\ &= (I-bv^*)^{-1} + (I-vb^*)^{-1} - I, \end{align*} $$

which gives for $w(v) := (I - bb^*)^{1/2}(I-vb^*)^{-1}(b-v)(I-b^*b)^{-1/2}$ that

$$ \begin{align*} w(v)^*w(v) & = D_b^{-1}(b^*-v^*)(I-bv^*)^{-1}(I-bb^*)(I-vb^*)^{-1}(b-v)D_b^{-1} \\ & = D_b^{-1}v^*(vb^*- I)\Big((I-bv^*)^{-1} + (I-vb^*)^{-1} - I\Big)(bv^*- I)vD_b^{-1} \\ & = D_b^{-1}v^*\Big((I-vb^*) + (I-bv^*) - (I-vb^*)(I-bv^*)\Big)vD_b^{-1} \\ & = D_b^{-1}v^*(I - vb^*bv^*)vD_b^{-1} \\ & = D_b^{-1}(I_n-b^*b)D_b^{-1} \\ & = I_n. \end{align*} $$

Hence, $w(v)$ is a row isometry. Now by the conjugation relation in the hypothesis, we have $bb^*$ , $bv^*$ and $vb^*$ commute with $\mathcal A$ and $b^*b$ commutes with $\mathop{\mathrm {diag}}(\alpha _1(a),\dots , \alpha _n(a)), a\in \mathcal A$ . Using this one gets for $a\in \mathcal A$ that

$$ \begin{align*} aw(v) & = a(I-bb^*)^{-1}(I-vb^*)^{-1}(b-v)(I_n-b^*b)^{-1/2} \\ & = (I - bb^*)^{1/2}(I-vb^*)^{-1}a(b-v)(I_n-b^*b)^{-1/2} \\ & = (I - bb^*)^{1/2}(I-vb^*)^{-1}(b-v)\mathop{\mathrm{diag}}(\alpha_1(a), \dots, \alpha_n(a))(I-b^*b)^{-1/2} \\ & = (I - bb^*)^{1/2}(I-vb^*)^{-1}(b-v)(I-b^*b)^{-1/2}\mathop{\mathrm{diag}}(\alpha_1(a), \dots, \alpha_n(a)) \\ & = w(v)\mathop{\mathrm{diag}}(\alpha_1(a), \dots, \alpha_n(a)). \end{align*} $$

Thus, by the universal property, there exists a unique completely contractive homomorphism $\rho $ of ${\mathcal {T}}^+({\mathcal {A}}, \alpha )$ to itself, such that

$$\begin{align*}\rho(a) = a, \forall a\in \mathcal A \quad \textrm{and} \quad [\rho(v_1) \dots \rho(v_n)] = w(v). \end{align*}$$

Lastly, recall the classical Halmos functional calculus trick: since

$$\begin{align*}b(I_n-b^*b) = (I-bb^*)b, \end{align*}$$

then

$$\begin{align*}bD_b^{-1} = b(I_n-b^*b)^{-1/2} = (I-bb^*)^{-1/2}b = D_{b^*}^{-1}b. \end{align*}$$

This allows us to compute

$$ \begin{align*} [\rho&\circ\rho(v_1), \dots ,\rho\circ\rho(v_n)] = w(w(v)) \\ & = D_{b^*}\Big(I - D_{b^*}(I-vb^*)^{-1}(b-v)D_b^{-1}b^*\Big)^{-1}\Big( b - D_{b^*}(I-vb^*)^{-1}(b-v)D_b^{-1} \Big)D_b^{-1} \\ &= D_{b^*}\Big(I - D_{b^*}(I-vb^*)^{-1}(b-v)b^*D_{b^*}^{-1}\Big)^{-1}\Big( b - D_{b^*}(I-vb^*)^{-1}(b-v)D_b^{-1} \Big)D_b^{-1} \\ &= D_{b^*}^2\Big(I - (I-vb^*)^{-1}(b-v)b^*\Big)^{-1}D_{b^*}^{-1}\Big( b - D_{b^*}(I-vb^*)^{-1}(b-v)D_b^{-1} \Big)D_b^{-1} \\ &= D_{b^*}^2\Big(I - (I-vb^*)^{-1}(b-v)b^*\Big)^{-1}\Big( D_{b^*}^{-1}b - (I-vb^*)^{-1}(b-v)D_b^{-1} \Big)D_b^{-1} \\ &= D_{b^*}^2\Big(I - (I-vb^*)^{-1}(b-v)b^*\Big)^{-1}\Big( b - (I-vb^*)^{-1}(b-v)\Big)D_b^{-2} \\ &= D_{b^*}^2\Big((I-vb^*) - (b-v)b^*\Big)^{-1}(I-vb^*)\Big( b - (I-vb^*)^{-1}(b-v)\Big)D_b^{-2} \\ &= D_{b^*}^2(I-bb^*)\Big((I-vb^*)b - (b-v)\Big)D_b^{-2} \\ &= (v-vb^*b)D_b^{-2} \\ &= v. \end{align*} $$

Therefore, by the universal property, $\rho \circ \rho = {\operatorname {id}}$ and $\rho $ is a completely isometric automorphism of ${\mathcal {T}}^+({\mathcal {A}}, \alpha )$ .

To find the first Fourier coefficients of the $\rho (v_i)$ , one needs to recall that $E_0:{\mathcal {T}}^+({\mathcal {A}}, \alpha ) \rightarrow \mathcal A$ is a completely contractive homomorphism given by sending $\mathcal A$ to itself and sending $v_i$ to 0, $1\leq i\leq n$ . Thus,

$$ \begin{align*} E_0([\rho(v_1) , \dots , \rho(v_n)]) & = E_0(w(v)) \\ & = E_0\big((I - bb^*)^{1/2}(I-vb^*)^{-1}(b-v)(I_n-b^*b)^{-1/2}\big) \\ & = (I - bb^*)^{1/2}((I-0b^*))^{-1}(b-0)(I_n-b^*b)^{-1/2} \\ & = (I - bb^*)^{1/2}b(I_n-b^*b)^{-1/2} \\ & = b, \end{align*} $$

with the last equality arising from a familiar functional calculus argument.

In the proof of our next result, we make use of the orbit representations for a tensor algebra ${\mathcal {T}}^+({\mathcal {A}}, \alpha )$ . Let $\sigma : {\mathcal {A}} \rightarrow B({\mathcal {H}})$ be a $*$ -representation. We define a representation

$$\begin{align*}\pi_{\sigma}:{\mathcal{A}} \longrightarrow B({\mathcal{H}} \otimes \ell^2({\mathbb{F}}^+_n)) \end{align*}$$

by

$$\begin{align*}\pi_{\sigma}(a)(x\otimes e_p )= \pi(\alpha_p(a))x\otimes e_p, \quad a \in {\mathcal{A}}, \end{align*}$$

where $\{e_p\}_{p \in {\mathbb {F}}^+_n}$ is the canonical orthonormal basis of the Fock space $\ell ^2({\mathbb {F}}^+_n)$ . Let $L_1, L_2, \dots , L_n$ be the Cuntz-Toeplitz isometries on $\ell ^2({\mathbb {F}}^+_n)$ . Then the representation $\pi _{\sigma }$ , together with the isometries $I\otimes L_1, I\otimes L_2, \dots , I\otimes L_n$ , form a covariant representation of $({\mathcal {A}}, \alpha )$ and determine the orbit representation of ${\mathcal {T}}^+({\mathcal {A}}, \alpha )$ associated with $\rho $ , which we also denote as $\pi _{\sigma }$ .

Proof. Let $\sigma : {\mathcal {T}}^+({\mathcal {A}}, \beta ) \longrightarrow B(\mathcal H)$ be a completely isometric representation of $T^+({\mathcal {A}}, \beta )$ . Then $\sigma \circ \psi $ is a completely isometric representation of $T^+({\mathcal {A}}, \alpha )$ , which coincides with $\sigma $ on ${\mathcal {A}}$ . This allows us to view both tensor algebras as being embedded in $B({\mathcal {H}})$ via maps that coincide on ${\mathcal {A}}$ .

Let $w = [ w_1, w_2, \dots , w_{n_b}]$ be the generating row isometryFootnote ² for ${\mathcal {T}}^+({\mathcal {A}}, \beta )$ . By the Fourier analysis discussed earlier,

(2.1) $$ \begin{align} \begin{aligned} \psi(v)&= \lim_{n\rightarrow \infty} \sum_{k=0}^n \left(1 - \frac{k}{n+1}\right)E_k(\psi(v))\\ &= \lim_{n\rightarrow \infty} \sum_{k=0}^n \sum_{|p| = k }\left(1 - \frac{k}{n+1}\right)w_pb_p, \end{aligned} \end{align} $$

with each $b_p=[b_{p,1},b_{p,2},\dots ,b_{p,n_{a}}]$ being a row contraction.

Assume by contradiction that $\|b_0\| = \|E_0(\psi (v))\| = 1$ . Then there exists a sequence of unit vectors $\xi _j \in \mathcal H^{(n_b)}$ , such that $\lim _{j\rightarrow \infty }\|b_0\xi _j\| =1$ . Using these contractive Cesaro sums in conjunction with the orbit representation $\pi := \pi _{\sigma _{\mid {\mathcal {A}}}}$ applied to the vectors $\vec {\xi }_j=\xi _j\otimes e_0$ , we get

$$ \begin{align*} 1 & \geq \lim_{j\rightarrow \infty} \left\| \pi\left( \sum_{k=0}^n \sum_{|p| = k }\left(1 - \frac{k}{n+1}\right)w_pb_p \right) \vec{\xi}_j \right\|^2 \\ & = \lim_{j \rightarrow \infty} \sum_{k=0}^n \sum_{|p| = k } \left(1 - \frac{k}{n+1}\right)^2 \|b_p\xi_j\|^2 \\ & \geq \lim_{j\rightarrow \infty} \|b_0\xi_j\|^2= 1. \end{align*} $$

This implies that $\lim _{j\rightarrow \infty } \|b_p\xi _j\| = 0$ , for all nonempty words $p \in {\mathbb {F}}^+_{n_a}$ .

Pick an $n\geq 0$ , such that

$$\begin{align*}\left\| \sum_{k=0}^n \left(1 - \frac{k}{n+1}\right)E_k(\psi(v)) - \psi(v) \right\| \leq \frac{1}{2}. \end{align*}$$

Then for every $j\geq 1$ , we have

$$ \begin{align*} \left\| E_0\circ\psi^{-1}\left( \sum_{k=0}^n \left(1 - \frac{k}{n+1}\right)E_k(\psi(v))\right)\xi_j\right\| \\ & \hskip -130 pt \leq \left\| E_0\circ\psi^{-1}\left(\sum_{k=0}^n \left(1 - \frac{k}{n+1}\right)E_k(\psi(v))- \psi(v)\right)\xi_j\right\| + \|E_0\circ\psi^{-1}\circ\psi(v)\xi_j\| \\ & \hskip -130 pt \leq \left\| \sum_{k=0}^n \left(1 - \frac{k}{n+1}\right)E_k(\psi(v))- \psi(v)\right\|\|\xi_j\| + \|E_0(v)\xi_j\| \\ & \hskip -130 pt \leq \frac{1}{2} + 0 = \frac{1}{2}. \end{align*} $$

Using the fact that $\lim _{j\rightarrow \infty } b_p\xi _j = 0$ , for all nonempty words $p \in {\mathbb {F}}^+_{n_a}$ , we obtain that

$$ \begin{align*} \lim_{j\rightarrow \infty} \left\| E_0\circ\psi^{-1}\left( \sum_{k=0}^n \left(1 - \frac{k}{n+1}\right)E_k(\psi(v))\right)\xi_j \right\| \\ & \hskip -80 pt = \lim_{j\rightarrow \infty} \left\|E_0\left( \sum_{k=0}^n \sum_{|p| = k }\left(1 - \frac{k}{n+1}\right)\psi^{-1}(w_p) b_p \right)\xi_j\right\| \\ & \hskip -80 pt = \lim_{j\rightarrow \infty} \left\| \sum_{k=0}^n \sum_{|p| = k }\left(1 - \frac{k}{n+1}\right)E_0\big(\psi^{-1}(w_p)\big) b_p \xi_j\right\| \\ & \hskip -80 pt = \lim_{j\rightarrow \infty} \left\| E_0 \big(\psi^{-1}(w_0)\big) b_0\xi_j \right\| \\ & \hskip -80 pt = 1, \end{align*} $$

which is a contradiction. Therefore, $b_0 = E_0(\psi (v))$ must be a strict contraction.

Continuing with the assumptions and notation of the previous proposition and its proof, the matrix $[b_{i j }] \in M_{n_b, n_a}({\mathcal {A}})$ , with entries $b_{i j}$ as appearing in (2.1) with $|p|=1$ , is called the matrix associated with $\psi $ . It has the property of intertwining the endomorphisms $\alpha _1, \alpha _2, \dots , \alpha _{n_a}$ and $\beta _1, \beta _2, \dots , \beta _{n_b}$ in the sense,

(2.2) $$ \begin{align} \beta_i(a)b_{ij}=b_{ij}\alpha_j(a), \end{align} $$

for all $a\in {\mathcal {A}}$ , $1\leq i \leq n_b$ and $1\leq j \leq n_a$ . A matrix $[a_{i j }] \in M_{n_a, n_b}({\mathcal {A}})$ with similar properties is associated with $\psi ^{-1}$ . Let us observe this more closely.

Indeed, assume that $\psi (v)$ admits a Fourier development

(2.3) $$ \begin{align} \begin{aligned} \psi(v)&= \lim_{n\rightarrow \infty} \sum_{k=0}^n \sum_{|p| = k }\left(1 - \frac{k}{n+1}\right)w_pb_p \\ &= b_0+w[b_{ij}]+\lim_{n\rightarrow \infty} \sum_{k=2}^n \sum_{|p| = k }\left(1 - \frac{k}{n+1}\right)w_pb_p \end{aligned} \end{align} $$

as in (2.1). For any $a \in {\mathcal {A}}$ and $1\leq j\leq n_a$ , we have

$$ \begin{align*}a\psi(v_j) = \psi(av_j)=\psi(v_j\alpha_j(a)) = \psi(v_j) \alpha_j(a),\end{align*} $$

and so

(2.4) $$ \begin{align} \begin{aligned} a \Big( b_{0j}+\sum_{i=1}^{n_b}w_ib_{ij}+ \lim_{n\rightarrow \infty} \sum_{k=2}^n & \sum_{|p| = k }\left(1 - \frac{k}{n+1}\right)w_pb_{p j} \Big) =\\ &=\Big(b_{0j}+\sum_{i=1}^{n_b}w_ib_{ij}+\sum_{k=2}^n \sum_{|p| = k }\left(1 - \frac{k}{n+1}\right)w_pb_{pj} \Big) \alpha_j(a). \end{aligned} \end{align} $$

Apply $E_0$ to (2.4) to obtain

(2.5) $$ \begin{align} ab_{0j}=b_{0j} \alpha_j(a), \end{align} $$

for all $a \in {\mathcal {A}}$ and $1\leq j \leq n_a$ . The relations (2.2) are obtained by applying $E_1$ to (2.4).

Proof. Assume that $\psi (v)$ admits a Fourier development as in (2.3) and similarly

$$ \begin{align*} \psi^{-1}(w)= a_0+v[a_{ij}]+\lim_{n\rightarrow \infty} \sum_{k=2}^n \sum_{|q| = k }\left(1 - \frac{k}{n+1}\right)v_qa_q. \end{align*} $$

Assume that $E_0(\psi ^{-1}(w)) = a_0=0$ .

Claim. $E_0(\psi ^{-1}(w_p)) = E_1(\psi ^{-1}(w_p))=0$ , for all $p \in {\mathbb {F}}^+_{n_b}$ with $|p|\geq 2$ .

Indeed, if $p=p_1p_2\dots p_l$ , $l \geq 2$ , then

(2.6) $$ \begin{align} \begin{aligned} \psi^{-1}(w_p)&= \prod_{i=1}^{l}\psi^{-1}(w_{p_i}) \\ &=\lim_{n\rightarrow \infty} \prod_{i=1}^{l} \left(\sum_{k=1}^n \sum_{|q| = k }\left(1 - \frac{k}{n+1}\right)v_qa_{q, p_i}\right). \end{aligned} \end{align} $$

Since in the limit above we have $l\geq 2 $ and $k\geq 1$ , a development of the product involved will reveal only terms of the form $v_u a_u$ , with $u \in {\mathbb {F}}^+_{n_a}$ , $|u|\geq 2$ and $a_u\in {\mathcal {A}}$ . Since both $E_0$ and $E_1$ are continuous, this suffices to prove the claim.

For the proof, consider $i=1,2$ . Apply $\psi ^{-1}$ to (2.3), and use the Claim to obtain

$$ \begin{align*} \begin{aligned} E_i(v)&=E_i(\psi^{-1}(\psi(v)))\\ &= E_i((b_0)+E_i(\psi^{-1}(w)) [b_{ij}]+\lim_{n\rightarrow \infty} \sum_{k=2}^n \sum_{|p| = k }\left(1 - \frac{k}{n+1}\right) E_i(\psi^{-1}(w_p)) b_p \\ &= E_i(b_0) +E_i(\psi^{-1}(w))[b_{ij}] \\ &=E_i(b_0) +E_i\Big(v[a_{ij}]+\lim_{n\rightarrow \infty} \sum_{k=2}^n \sum_{|q| = k }\left(1 - \frac{k}{n+1}\right)v_qa_q \Big) [b_{ij}] \\ & = E_i(b_0) +E_i(v) [a_{ij}][b_{ij}]. \end{aligned} \end{align*} $$

For $i=0$ , we obtain $0=b_0=E_0(\psi (v))$ . For $i=1$ , we obtain $v =v [a_{ij}][b_{ij}]$ , and so

$$\begin{align*}I_{n_a} =v^*v=v^*v [a_{ij}][b_{ij}]= [a_{ij}][b_{ij}]. \end{align*}$$

By reversing the roles of $\psi $ and $\psi ^{-1}$ and using what has been proven so far, we obtain $[b_{ij}].[a_{ij}] =I_{n_b}$ . This completes the proof.

Proof. According to (2.2), the matrix $b:=[b_{i j}] \in M_{n_b, n_a}({\mathcal {A}})$ associated with $\psi $ intertwines the endomorphisms $\alpha _1, \alpha _2, \dots , \alpha _{n_a}$ and $\beta _1, \beta _2, \dots , \beta _{n_b}$ , that is,

(2.7) $$ \begin{align} \mathop{\mathrm{diag}}(\beta_1(a), \dots, \beta_{n_b}(a))b= b\mathop{\mathrm{diag}}(\alpha_1(a),\dots,\alpha_{n_a}(a)), \end{align} $$

for all $a \in {\mathcal {A}}$ . Hence, $b^*b$ (and, therefore, $|b|$ ) commutes with $\mathop{ \mathrm {diag}}(\alpha _1(a),\dots ,\alpha _{n_a}(a))$ , for all $a\in {\mathcal {A}}$ . By Lemma 2.3, b is invertible, and so it admits a polar decomposition $b = u|b|$ , with $u=M_{n_b, n_a}({\mathcal {A}})$ a unitary matrix. Furthermore, (2.7) implies that

$$ \begin{align*} \mathop{\mathrm{diag}}(\beta_1(a), \dots, \beta_{n_b}(a)) u|b| &= u|b| \mathop{\mathrm{diag}}(\alpha_1(a),\dots,\alpha_{n_a}(a))\\ &= u\mathop{\mathrm{diag}}(\alpha_1(a),\dots,\alpha_{n_a}(a)) |b|, \end{align*} $$

for all $a \in {\mathcal {A}}$ . Since $|b|$ is invertible, $\mathop{ \mathrm {diag}}(\beta _1(a), \dots , \beta _{n_b}(a)) u= u \mathop{ \mathrm {diag}}(\alpha _1(a),\dots ,\alpha _{n_a}(a))$ , and the conclusion follows.

Motivated by the statement of the previous result, we introduce the following.

Note that the above definition does not require that the multivariable systems $({\mathcal {A}}, \alpha )$ and $({\mathcal {A}}, \beta )$ should have the same number of maps, that is, $n_{\alpha }=n_{\beta }$ , where $\alpha = (\alpha _1, \alpha _2, \dots , \alpha _{n_{\alpha }})$ and $\beta =(\beta _1, \beta _2, \dots , \beta _{n_{\beta }})$ . On the contrary, it is possible for two dynamical systems with a different number of maps to be unitarily equivalent (see [Reference Kakariadis and Katsoulis11, Example 5.1]). Nevertheless, in that case, at least one of the systems will fail to be automorphic, as [Reference Kakariadis and Katsoulis11, Theorem 4.4] clearly indicates.

Recall that two (single variable) dynamical systems are said to be outer conjugate if there is a $*$ -isomorphism $\gamma : \mathcal A \rightarrow \mathcal B$ and a unitary $u\in \mathcal A$ , such that

$$\begin{align*}\alpha(c) = u(\gamma^{-1}\circ\beta\circ\gamma(c))u^*. \end{align*}$$

Therefore, in that case, the concept of unitary equivalence after a conjugation coincides with that of outer conjugacy. Davidson and Kakariadis [Reference Davidson and Kakariadis5] established that outer conjugacy of the systems implies that the associated tensor algebras are completely isometrically isomorphic. They showed that the converse is true in several broad cases (injective, surjective, etc.) and, specifically, when $\|E(\psi (v))\| < \frac {2}{3}\sqrt 3 - 1 \approx 0.1547$ [Reference Davidson and Kakariadis5, Remark 3.6]. For multivariable dynamical systems consisting of automorphisms, Kakariadis and Katsoulis have shown [Reference Kakariadis and Katsoulis11, Theorem 4.5] that the isomorphism of the tensor algebras is equivalent to unitary equivalence after a conjugation for the associated dynamical systems. A similar result was shown for arbitrary dynamical systems, provided that the pertinent $\mathrm {C}^*$ -algebras are stably finite [Reference Kakariadis and Katsoulis11, Theorem 5.2]. Our next result removes all these conditions and establishes a complete result for the unital case.

Proof. Without loss of generality, we may assume that ${\mathcal {A}} = {\mathcal {B}}$ and $\psi |_{\mathcal A} = {\operatorname {id}}$ . Indeed, it is well known that the restriction of the isomorphism $\psi $ on ${\mathcal {A}} \subseteq {\mathcal {T}}^+({\mathcal {A}}, \alpha )$ induces a $*$ -isomorphism $\gamma : \mathcal A \rightarrow \mathcal B$ of the diagonals of the two tensor algebras (see, for instance, [Reference Davidson and Katsoulis7, Proposition 3.1]). The dynamical systems $({\mathcal {B}} , \beta )$ and $({\mathcal {A}} , \gamma ^{-1}\circ \beta \circ \gamma )$ are conjugate via $\gamma $ , and so there exists a completely isometric isomorphism $\varphi : {\mathcal {T}}^+({\mathcal {B}} , \beta ) \rightarrow {\mathcal {T}}^+({\mathcal {A}} , \gamma ^{-1}\circ \beta \circ \gamma )$ , so that $\varphi |_{{\mathcal {B}}} = \gamma ^{-1}$ . Hence, $\varphi \circ \psi $ establishes a completely isometric isomorphism between ${\mathcal {T}}^+({\mathcal {A}}, \alpha )$ and ${\mathcal {T}}^+({\mathcal {A}} , \gamma ^{-1}\circ \beta \circ \gamma )$ , whose restriction on ${\mathcal {A}}$ is the identity map. Therefore, if $\gamma $ is not the identity map to begin with, then replace $\psi $ with $\varphi \circ \psi $ and establish the conclusion for the dynamical systems $({\mathcal {A}} , \alpha )$ and $({\mathcal {A}} , \gamma ^{-1}\circ \beta \circ \gamma )$ .

For the proof, if $v = [v_1\ v_2\ \dots \ v_n]$ is the generating row isometry in ${\mathcal {T}}^+({\mathcal {A}}, \alpha ) $ , then Proposition 2.2 gives that $b_0 = E_0(\psi (v))$ is a strict row contraction. Combined with (2.5), this implies that $b_0$ satisfies the conditions of Theorem 2.1, and so there exists a completely isometric automorphism $\rho $ of ${\mathcal {T}}^+({\mathcal {A}}, \alpha )$ , such that

$$\begin{align*}\rho(v) = (I - b_0b_0^*)^{1/2}(I-vb_0^*)^{-1}(b_0-v)(I_n-b_0^*b_0)^{-1/2}. \end{align*}$$

Since $E_0$ is multiplicative, we obtain that

$$ \begin{align*} E_0(\psi\circ \rho(v) ) &= E_0\circ\psi\left( D_{b_0^*}(I-vb_0^*)^{-1}(b_0-v)D_{b_0}\right) \\ &=D_{b_0^*}E_0((I-\psi(vb_0^*))^{-1}\left(b_0-E_0(\psi(v))\right)D_{b_0} \\ &=D_{b_0^*}E_0((I-\psi(vb_0^*))^{-1}\left(b_0-b_0\right)D_{b_0} \\ & = 0. \end{align*} $$

Therefore, $\psi \circ \rho $ satisfies the requirements of Corollary 2.4 and the conclusion follows.

Of course, the converse of Theorem 2.6 is also true. If two multivariable systems $(\mathcal A,\alpha )$ and $(\mathcal B,\beta )$ are unitarily equivalent after a conjugation, then the associated $\mathrm {C}^*$ -correspondences are unitarily equivalent, and so the tensor algebras ${\mathcal {T}}^+({\mathcal {A}}, \alpha )$ and ${\mathcal {T}}^+({\mathcal {B}} , \beta )$ are completely isometrically isomorphic (see [Reference Kakariadis and Katsoulis12, Theorem 4.5] and the discussion preceding it). Hence, Theorem 2.6 provides a complete classification of tensor algebras up to complete isomorphism. For semicrossed products, we can say something more.

3 Concluding remarks and open problems

(i) As we mentioned in the Introduction, Davidson and Kakariadis [Reference Davidson and Kakariadis5, Theorem 1.1] provide six different properties for a dynamical system, with each one of them guaranteeing that the isomorphism problem has a positive resolution. It is instructive to observe that there are dynamical systems that do not satisfy any of their properties. Therefore, our Corollary 2.7 verifies the conjecture of Davidson and Kakariadis by going beyond the realm of [Reference Davidson and Kakariadis5, Theorem 1.1].

Each one of the following conditions on a $\mathrm {C}^*$ -dynamical system $({\mathcal {A}}, \alpha )$ is sufficient for [Reference Davidson and Kakariadis5, Theorem 1.1] to apply:

1. (1) ${\mathcal {A}}$ has trivial centre.

2. (2) ${\mathcal {A}}$ is abelian.

3. (3) ${\mathcal {A}}$ is finite, that is, no proper isometries.

4. (4) $\alpha ({\mathcal {A}})'$ is finite.

5. (5) $\alpha (R_{\alpha })=R_{\alpha }$ , where $R_{\alpha }=\overline {\cup _{k \geq 1} \ker (\alpha ^k )}$ .

6. (6) $\alpha (R_{\alpha }^{\perp }) \subseteq R_{\alpha }^{\perp }$ .

We claim that for each $i=1,2,\dots , 6$ , there exists a dynamical system $({\mathcal {A}}_i, \alpha _i)$ that fails the corresponding condition $(i)$ from the above list. It is easy to see then that the dynamical system $(\oplus _{i=1}^6 {\mathcal {A}}_i, \oplus _{i=1}^6 \alpha _i)$ will fail all six conditions.

The existence of dynamical systems that do not satisfy any one of the conditions (1), (2) or (3) is a trivial matter. For condition $(4)$ , let ${\mathcal {H}}$ be a separable Hilbert space and let $B({\mathcal {H}}), K({\mathcal {H}})$ denote the bounded and compact operators, respectively, acting on ${\mathcal {H}}$ . Let

$$\begin{align*}{\mathcal{A}}_4 = B({\mathcal{H}}) \otimes K({\mathcal{H}}) + {\mathbb{C}}(I\otimes I) \end{align*}$$

acting on ${\mathcal {H}} \otimes {\mathcal {H}}$ , and let $\alpha _4$ be the unital endomorphism of ${\mathcal {A}}$ defined as

$$\begin{align*}\alpha_4 (S +\lambda I\otimes I)= \lambda I\otimes I, \quad S \in B({\mathcal{H}}) \otimes K({\mathcal{H}}), \,\, \lambda \in {\mathbb{C}}. \end{align*}$$

The dynamical system $({\mathcal {A}}_4, \alpha _4)$ fails property (4) from the above list.

For condition (6), let ${\mathcal {H}}$ and $K({\mathcal {H}})$ be as in the previous paragraph and consider

$$\begin{align*}{\mathcal{A}}_6=( K({\mathcal{H}})+{\mathbb{C}} I )\oplus {\mathbb{C}} I \end{align*}$$

acting on ${\mathcal {H}}\oplus {\mathcal {H}}$ and let $\alpha _6$ be the unital endomorphism of ${\mathcal {A}}_6$ defined as

$$\begin{align*}\alpha_6(K + \lambda I, \mu I) = (\mu I , \lambda I), \quad K \in K({\mathcal{H}}), \, \, \lambda, \mu \in {\mathbb{C}}. \end{align*}$$

It is clear that $\ker (\alpha _6^k) = K({\mathcal {H}})\oplus 0$ , for all $k\geq 1$ , and so $R_{\alpha } = K({\mathcal {H}})\oplus 0$ . Therefore, $R_{\alpha _6}^{\perp } = 0\oplus {\mathbb {C}} I$ , and so

$$\begin{align*}\alpha_6 (R_{\alpha_6}^{\perp} )= {\mathbb{C}} I \oplus 0 \not\subseteq 0\oplus {\mathbb{C}} I = R_{\alpha_6}^{\perp}, \end{align*}$$

that is, $({\mathcal {A}}_6, \alpha _6)$ does not satisfy (6). Note also that

$$\begin{align*}\alpha_6(R_{\alpha_6})= 0 \neq R_{\alpha_6},\\[-30pt] \end{align*}$$

and so $({\mathcal {A}}_6, \alpha _6)$ does not satisfy (5) as well.

(ii) Could one obtain Corollary 2.7 by just using the earlier techniques of Davidson and Kakariadis [Reference Davidson and Kakariadis5]? Notice that their [Reference Davidson and Kakariadis5, Remark 3.6] would imply Corollary 2.7, provided that one could establish the existence of an isomorphism $\psi $ that satisfies the technical requirement $\|E(\psi (v))\| < \frac {2}{3}\sqrt 3 - 1 \approx 0.1547$ . However, the existence of such an isomorphism was not established in [Reference Davidson and Kakariadis5] and it does not seem likely that the techniques of [Reference Davidson and Kakariadis5] alone could do that. Our Theorem 2.1 and Proposition 2.2 show now that such an isomorphism $\psi $ exists and satisfies the much stronger condition $\|E(\psi (v))\| =0$ (this actually provides a slightly different proof of Corollary 2.7). Our approach in Theorem 2.1 is much different from that of [Reference Davidson and Kakariadis5], and it is inspired by the works of Muhly and Solel [Reference Muhly and Solel17] and Davidson, Ramsey and Shalit [Reference Davidson, Ramsey and Shalit4], which were actually available during the writing of [Reference Davidson and Kakariadis5].

Note that a multivariable analogue of [Reference Davidson and Kakariadis5, Remark 3.6] does not exist in the literature, thus, necessitating the approach that we followed in the proof of Theorem 2.6.

(iii) Even though we provide a complete invariant for completely isometric isomorphisms, Theorem 2.6 is not the end of the story for the isomorphism problem for tensor algebras of multivariable dynamical systems. There are still questions that need to be addressed, and the following two seem to be the most important.

Indeed, a great deal of work on isomorphisms between tensor algebras of abelian multivariable systems addresses algebraic isomorphisms. In general, it is possible for nonselfadjoint operator algebras, even tensor algebras, to be algebraically or bicontinuously isomorphic without being isometrically isomorphic. So Question 3.1 above whether the various concepts of isomorphism coincide for tensor algebras of multivariable dynamical systems. Remarkably, to this date, there has been no work addressing algebraic isomorphisms between tensor algebras of nonabelian multivariable systems. One reason for that is perhaps the fact that algebraic isomorphisms between nonselfadjoint operator algebras do not preserve the diagonal, and so the techniques developed for isometric isomorphisms are not applicable here. Hopefully, the progress achieved in this paper will finally facilitate the study of algebraic isomorphisms beyond the abelian case.

This important question goes back to the memoirs of Davidson and Katsoulis [Reference Davidson and Katsoulis8, Conjecture 3.26], and the second half of [Reference Davidson and Katsoulis8] was actually occupied with partial solutions to this question. In light of Theorem 2.6, this question essentially asks whether piecewise conjugacy and unitary equivalence after a conjugation coincide as invariants for multivariable dynamical systems over abelian $\mathrm {C}^*$ -algebras; in this form, the question was raised in [Reference Kakariadis and Katsoulis11, Question 2] . As we mentioned in the Introduction of this paper, this question does not make sense in the generality of multivariable dynamical systems over-nonabelian $\mathrm {C}^*$ -algebras. Hence, the invariant we offer here is the only one available beyond the abelian case.