Proof We define $t : E \rightarrow {\mathcal O}_{E} \otimes \mathrm {C}(G)$ and $\pi : A \rightarrow {\mathcal O}_{E}\otimes \mathrm {C}(G)$ by the following:

$$\begin{align*}t=(k_E \otimes \mathrm{id}_{\mathrm{C}(G)}) \circ \lambda, \text{ and, } \pi=(k_A \otimes \mathrm{id}_{\mathrm{C}(G)})\circ \alpha.\end{align*}$$

We claim that $(t,\pi )$ is a covariant representation of the $\mathrm {C}^*$ -correspondence $(E,\phi )$ over A. Now, for $\xi ,\eta \in E$ , we have

$$ \begin{align*} \begin{aligned} t(\xi)^*t(\eta)={}&((k_E \otimes \mathrm{id}_{\mathrm{C}(G)}) \lambda(\xi))^*((k_E \otimes \mathrm{id}_{\mathrm{C}(G)}) \lambda(\eta))\\ ={}&(k_A \otimes \mathrm{id}_{\mathrm{C}(G)})(\langle \lambda(\xi),\lambda(\eta) \rangle)\\ ={}&(k_A \otimes \mathrm{id}_{\mathrm{C}(G)})(\alpha(\langle \xi,\eta \rangle))\\ ={}&\pi(\langle \xi,\eta \rangle), \end{aligned} \end{align*} $$

where the second equality is by Definition 3.3; and the third equality is by Definition 2.6. Again, for $a \in A$ and $\xi \in E$ , we have

$$ \begin{align*} \begin{aligned} t(\phi(a)\xi)={}&(k_E \otimes \mathrm{id}_{\mathrm{C}(G)}) \lambda(\phi(a).\xi)\\ ={}&(k_E \otimes \mathrm{id}_{\mathrm{C}(G)})(\phi \otimes \mathrm{id}_{\mathrm{C}(G)})(\alpha(a))\lambda(\xi)\\ ={}&(k_A \otimes \mathrm{id}_{\mathrm{C}(G)})(\alpha(a))(k_E \otimes \mathrm{id}_{\mathrm{C}(G)})\lambda(\xi)\\ ={}&\pi(a)t(\xi), \end{aligned} \end{align*} $$

where the second equality is by Definition 2.6; and the third equality is by Definition 3.6. Thus, $(t,\pi )$ is a representation of the $\mathrm {C}^*$ -correspondence $(E,\phi )$ over A. We recall that there is a unital $*$ -homomorphism $\psi _t : {\mathcal K}(E) \rightarrow {\mathcal O}_{E} \otimes \mathrm {C}(G)$ given by $\psi _t(\theta _{\xi ,\eta })=t(\xi )t(\eta )^{\ast }$ such that $\psi _t(k)t(\xi )=t(k\xi )$ for all $\xi \in E$ and $k \in \mathcal {K}(E)$ ; in particular, we have $\psi _t(\phi (a))t(\xi )=t(\phi (a)\xi )=\pi (a)t(\xi )$ for all $\xi \in E$ and all $a \in A$ . Since $[\lambda (E)(1_A \otimes \mathrm {C}(G))]=E \otimes \mathrm {C}(G)$ , using the linearity and the continuity of the maps involved, we get for all $\xi \in E$ and $a \in A$ ,

$$\begin{align*}\psi_t(\phi(a))(k_E(\xi) \otimes 1_{\mathrm{C}(G)})=\pi(a)(k_E(\xi)\otimes 1_{\mathrm{C}(G)}). \end{align*}$$

Consequently, for any $\xi ,\eta \in E$ , and $a \in A$ , we have

$$\begin{align*}\psi_t(\phi(a))(k_E(\xi)k_E(\eta)^*\otimes 1_{\mathrm{C}(G)})=\pi(a)(k_E(\xi)k_E(\eta)^*\otimes 1_{\mathrm{C}(G)}), \end{align*}$$

i.e.,

$$\begin{align*}\psi_t(\phi(a))(\psi_{k_E}(\theta_{\xi,\eta})\otimes 1_{\mathrm{C}(G)})=\pi(a)(\psi_{k_E}(\theta_{\xi,\eta})\otimes 1_{\mathrm{C}(G)}).\end{align*}$$

So again by linearity and continuity of the maps involved, we get

$$\begin{align*}\psi_t(\phi(a))(\psi_{k_E}(k)\otimes 1_{\mathrm{C}(G)})=\pi(a)(\psi_{k_E}(k)\otimes 1_{\mathrm{C}(G)}), \end{align*}$$

for all $k \in \mathcal {K}(E)$ . As the $\mathrm {C}^*$ -algebra A is unital and E is finitely generated, ${\mathcal K}(E)={\mathcal L}(E)$ and since $\psi _{t}$ is a unital $*$ -homomorphism, we obtain, by plugging $k=\mathrm {id}_E \in {\mathcal L}(E)$ in the identity above,

$$\begin{align*}\psi_t(\phi(a))=\pi(a), \end{align*}$$

for all $a \in A$ , which in particular, shows that $(t,\pi )$ is a covariant representation of the $\mathrm {C}^*$ -correspondence $(E,\phi )$ over A on the unital $\mathrm {C}^*$ -algebra ${\mathcal O}_{E} \otimes \mathrm {C}(G)$ . Therefore, by the universality of ${\mathcal O}_{E}$ , we get a necessarily unique $*$ -homomorphism

$$\begin{align*}t \times \pi=\omega : {\mathcal O}_{E} \rightarrow {\mathcal O}_{E} \otimes \mathrm{C}(G)\end{align*}$$

such that

$$\begin{align*}\omega \circ k_E=(k_E \otimes \mathrm{id}_{\mathrm{C}(G)}) \circ \lambda, \text{ and, } \omega \circ k_A=(k_A \otimes \mathrm{id}_{\mathrm{C}(G)})\circ \alpha,\end{align*}$$

completing the first part of the theorem. To see that $(\mathcal {O}_E,\omega )$ is indeed a G- $\mathrm {C}^*$ -algebra, we start with the coassociativity of $\omega $ , i.e., $(\omega \otimes \mathrm {id}_{\mathrm {C}(G)})\circ \omega =(\mathrm {id}_{\mathcal {O}_E} \otimes \Delta _G)\circ \omega $ . Again, by universality (and uniqueness), it suffices to show that the two $*$ -homomorphisms $(\omega \otimes \mathrm {id}_{\mathrm {C}(G)})\circ \omega $ and $(\mathrm {id}_{\mathcal {O}_E} \otimes \Delta _G)\circ \omega $ agree on the images $k_E(E)$ and $k_A(A)$ , i.e.,

$$\begin{align*}(\omega \otimes \mathrm{id}_{\mathrm{C}(G)})\circ \omega \circ k_E=(\mathrm{id}_{\mathcal{O}_E} \otimes \Delta_G)\circ \omega \circ k_E, \end{align*}$$

and

$$\begin{align*}(\omega \otimes \mathrm{id}_{\mathrm{C}(G)})\circ \omega \circ k_A=(\mathrm{id}_{\mathcal{O}_E} \otimes \Delta_G)\circ \omega \circ k_A. \end{align*}$$

However, using

$$\begin{align*}\omega \circ k_E=(k_E \otimes \mathrm{id}_{\mathrm{C}(G)}) \circ \lambda, \text{ and, } \omega \circ k_A=(k_A \otimes \mathrm{id}_{\mathrm{C}(G)})\circ \alpha,\end{align*}$$

respectively, we see that we are reduced to the coassociativity of $\lambda $ and $\alpha $ , respectively. Thus $\omega $ is indeed coassociative. For the Podleś condition, we consider the set

$$\begin{align*}S=\{x \in \mathcal{O}_E \mid x \otimes 1_{\mathrm{C}(G)} \in [\omega(\mathcal{O}_E)(1_{\mathcal{O}_E} \otimes \mathrm{C}(G))]\}. \end{align*}$$

Since $[\lambda (E)(1_A \otimes \mathrm {C}(G))]=E \otimes \mathrm {C}(G)$ and $[\alpha (A)(1_A \otimes \mathrm {C}(G))]=A \otimes \mathrm {C}(G)$ , we see that for each $\xi \in E$ , $k_E(\xi ) \in S$ and for each $a \in A$ , $k_A(a) \in S$ . Let x and y be in S. Then we see that

$$ \begin{align*} \begin{aligned} xy \otimes 1_{\mathrm{C}(G)}=(x \otimes 1_{\mathrm{C}(G)})(y \otimes 1_{\mathrm{C}(G)}) \in{}& [\omega(\mathcal{O}_E)(1_{\mathcal{O}_E} \otimes \mathrm{C}(G))(y \otimes 1_{\mathrm{C}(G)})]\\ ={}&[\omega(\mathcal{O}_E)(y \otimes 1_{\mathrm{C}(G)})(1_{\mathcal{O}_E} \otimes \mathrm{C}(G))]\\ \subseteq{}&[\omega(\mathcal{O}_E)\omega(\mathcal{O}_E)(1_{\mathcal{O}_E} \otimes \mathrm{C}(G))(1_{\mathcal{O}_E} \otimes \mathrm{C}(G))]\\ ={}&[\omega(\mathcal{O}_E)(1_{\mathcal{O}_E} \otimes \mathrm{C}(G))], \end{aligned} \end{align*} $$

i.e., S is closed under multiplication. But S contains, as shown above, $k_E(E)$ and $k_A(A)$ ; thus again by universality, S equals $\mathcal {O}_E.$ Therefore, $(\mathcal {O}_E,\omega )$ is indeed a G- $\mathrm {C}^*$ -algebra and this completes the proof.

Proof We begin by observing that it suffices to show, by universality (and uniqueness), that the two $*$ -homomorphisms $(\gamma _z \otimes \mathrm {id}_{\mathrm {C}(G)})\circ \omega $ and $\omega \circ \gamma _z$ agree on $k_E(E)$ and $k_A(A)$ , i.e.,

$$\begin{align*}(\gamma_z \otimes \mathrm{id}_{\mathrm{C}(G)})\circ \omega \circ k_E=\omega \circ \gamma_z \circ k_E, \end{align*}$$

and

$$\begin{align*}(\gamma_z \otimes \mathrm{id}_{\mathrm{C}(G)})\circ \omega \circ k_A=\omega \circ \gamma_z \circ k_A. \end{align*}$$

However, using

$$\begin{align*}\omega \circ k_E=(k_E \otimes \mathrm{id}_{\mathrm{C}(G)}) \circ \lambda, \text{ and, } \omega \circ k_A=(k_A \otimes \mathrm{id}_{\mathrm{C}(G)})\circ \alpha,\end{align*}$$

respectively, and the explicit form of $\gamma $ provided after Definition 3.6, we see that the above two identities are indeed satisfied. This completes the proof.

Proof We begin by observing that $\rho $ so defined is clearly gauge-equivariant and “non-linear.” We have to show that it indeed defines an action of $\mathbb {T}^n$ . First, let us check that $\rho $ is a well-defined $*$ -homomorphism.

To begin with, using the unitarity of $z_{i}$ , for $i=1,\dots ,n$ , it is easy to see that u is a unitary element of ${\mathcal O}_{n}\otimes \mathrm {C}(\mathbb {T}^{n})$ , and so for all $1 \leq i,j \leq n$ ,

$$\begin{align*}\rho(S^*_i)\rho(S_j)=\delta_{ij}(1\otimes 1).\end{align*}$$

Moreover,

$$\begin{align*}\sum_{k=1}^{n}\rho(S_k)\rho(S^*_k)=1 \otimes 1,\end{align*}$$

so that by the universality of ${\mathcal O}_{n}$ , $\rho $ is well-defined. It is also easy to observe that for all $1 \leq i,j \leq n$ ,

$$\begin{align*}\rho(S_iS^*_j)=S_iS^*_j \otimes 1\end{align*}$$

and coassociativity of $\rho $ follows. Again using

$$\begin{align*}\rho(S_{i}S_{i}^*)=S_{i}S_{i}^* \otimes 1\end{align*}$$

for all $i=1,\dots ,n$ , one can see that

$$\begin{align*}\sum_{k=1}^{n}\rho(S_{k}S_{k}^*)(1 \otimes z_{k}^*)=u^*\end{align*}$$

and consequently,

$$\begin{align*}\sum_{k=1}^{n}\rho(S_{i}S_{k}S_{k}^*)(1 \otimes z_{k}^*)=(S_{i} \otimes 1_{\mathrm{C}(\mathbb{T}^n)})uu^*=S_{i} \otimes 1_{\mathrm{C}(\mathbb{T}^n)}\end{align*}$$

for all $i=1,\dots ,n$ . Thus $S_{i} \otimes 1_{\mathrm {C}(\mathbb {T}^n)}$ , for all $i=1,\dots ,n$ , belongs to $[\rho ({\mathcal O}_{n})(1_{\mathcal {O}_n} \otimes \mathrm {C}(\mathbb {T}^{n}))]$ and arguing as in the last part of the proof of Theorem 3.9, the Podleś condition follows. This completes the proof.

Proof We begin by observing that the gauge-equivariance of $\rho $ yields

$$\begin{align*}\rho(A(0)) \subseteq A(0) \otimes \mathrm{C}(G), \text{ and } \rho(A(1)) \subseteq A(1) \otimes \mathrm{C}(G), \end{align*}$$

i.e., $\rho $ preserves the $\mathrm {C}^*$ -algebra and the module $A(1)$ . Let us denote the restricted actions $\rho |_{A(0)}$ and $\rho |_{A(1)}$ on $A(0)$ and $A(1)$ by $\alpha $ and $\lambda $ , respectively. We will show that $(A(0),\alpha )$ is a G- $\mathrm {C}^*$ -algebra, $(A(1),\phi ,\lambda )$ is a G-equivariant $\mathrm {C}^*$ -correspondence over the G- $\mathrm {C}^*$ -algebra $(A,\alpha )$ and that $\rho $ coincides with $\omega $ , where $\omega $ as in Theorem 3.9.

To that end, we first remark that $\alpha $ is indeed a unital $*$ -homomorphism and $\lambda $ is indeed a linear map; these follow from the corresponding property of $\rho $ . Next, we observe that coassociativity of both $\alpha $ and $\lambda $ follow from that of $\rho $ . Now, we fix $\xi , \eta \in A(1)$ and $a \in A(0)$ . Then

$$\begin{align*}\lambda(\xi a)=\rho(\xi a)=\rho(\xi)\rho(a)=\lambda(\xi)\alpha(a), \end{align*}$$

$$\begin{align*}\langle \lambda(\xi),\lambda(\eta) \rangle=\rho(\xi)^*\rho(\eta)=\rho(\xi^*\eta)=\rho(\langle \xi,\eta \rangle)=\alpha(\langle \xi,\eta \rangle), \end{align*}$$

and

$$\begin{align*}\lambda(\phi(a)\xi)=\lambda(a\xi)=\rho(a\xi)=\rho(a)\rho(\xi)=(\phi \otimes \mathrm{id}_{\mathrm{C}(G)})(\alpha(a))\lambda(\xi), \end{align*}$$

all the equalities being clear from the facts that $\rho $ is a $*$ -homomorphism and that $\lambda =\rho |_{A(1)}$ and $\alpha =\rho |_{A(0)}$ . Thus, we see that indeed condition (1) of Definition 2.2, conditions (1) and (2) of Definition 2.6 and conditions (1) and (3 $^{\prime }$ ) of Definition 3.2 are satisfied by $\alpha $ and $\lambda $ . The proof will be complete, by uniqueness of $\omega $ in Theorem 3.9, provided we could show the Podleś conditions for $\alpha $ and $\lambda $ . We now proceed to do so.

We recall that there is a conditional expectation $\operatorname {E} :A \rightarrow A$ , given by

$$\begin{align*}\operatorname{E}(a)=\int_{\mathbb{T}}\gamma_{z}(a) \ dz, \end{align*}$$

and that $A(0)$ coincides with $\operatorname {E}(A)$ . Since $\rho $ is gauge-equivariant, we have

$$\begin{align*}\rho \circ \operatorname{E}=(\operatorname{E} \otimes \mathrm{id}_{\mathrm{C}(G)})\circ \rho. \end{align*}$$

Then

$$ \begin{align*} \begin{aligned} A(0) \otimes \mathrm{C}(G)={}&\operatorname{E}(A) \otimes \mathrm{C}(G)\\ ={}&(\operatorname{E} \otimes \mathrm{id}_{\mathrm{C}(G)})(A \otimes \mathrm{C}(G))\\ ={}&(\operatorname{E} \otimes \mathrm{id}_{\mathrm{C}(G)})[\rho(A)(1_A \otimes \mathrm{C}(G))]\\ ={}&[\rho(\operatorname{E}(A))(1_A \otimes \mathrm{C}(G))]\\ ={}&[\alpha(A(0))(1_{A(0)} \otimes \mathrm{C}(G))], \end{aligned} \end{align*} $$

where the first and fifth equalities use the fact that $A(0)=\operatorname {E}(A)$ ; the third equality uses the Podleś condition for $\rho $ ; the fourth equality uses the identity just above this computation; and finally, the fifth equality uses the fact that $\alpha =\rho |_{A(0)}$ . Therefore, $(A(0),\alpha )$ is a G- $\mathrm {C}^*$ -algebra.

Now, let $\operatorname {P} : A \rightarrow A$ denote the projection onto $A(1)$ , obtained from the Banach space decomposition of A,

$$\begin{align*}A=A(1) \oplus \overline{\bigoplus_{n\neq 1}A(n)}. \end{align*}$$

Again, since $\rho $ is gauge-equivariant, we have

$$\begin{align*}\rho \circ \operatorname{P}=(\operatorname{P} \otimes \mathrm{id}_{\mathrm{C}(G)})\circ \rho. \end{align*}$$

Then

$$ \begin{align*} \begin{aligned} A(1) \otimes \mathrm{C}(G)={}&\operatorname{P}(A) \otimes \mathrm{C}(G)\\ ={}&(\operatorname{P} \otimes \mathrm{id}_{\mathrm{C}(G)})(A \otimes \mathrm{C}(G))\\ ={}&(\operatorname{P} \otimes \mathrm{id}_{\mathrm{C}(G)})[\rho(A)(1_A \otimes \mathrm{C}(G))]\\ ={}&[\rho(\operatorname{P}(A))(1_A \otimes \mathrm{C}(G))]\\ ={}&[\lambda(A(1))(1_{A(0)} \otimes \mathrm{C}(G))], \end{aligned} \end{align*} $$

where the first and fifth equalities use the fact that $A(1)=\operatorname {P}(A)$ ; the third equality uses the Podleś condition for $\rho $ ; the fourth equality uses the identity just above this computation; and finally, the fifth equality uses the fact that $\lambda =\rho |_{A(1)}$ . Therefore, $(A(1),\phi ,\lambda )$ is a G-equivariant $\mathrm {C}^*$ -correspondence over the G- $\mathrm {C}^*$ -algebra $(A(0),\alpha )$ . Thus as observed above, by the uniqueness of $\omega $ in Theorem 3.9, the proof is now complete.

Proof We assume first that $\varphi $ is G-equivariant and show that $\tau =\varphi \circ k_A$ is G-equivariant too. To that end, we fix $a \in A$ . Then we have

$$ \begin{align*} \begin{aligned} (\tau \otimes \mathrm{id}_{\mathrm{C}(G)})\alpha(a)={}&(\varphi \otimes \mathrm{id}_{\mathrm{C}(G)})(k_A \otimes \mathrm{id}_{\mathrm{C}(G)})\alpha(a)\\ ={}&(\varphi \otimes \mathrm{id}_{\mathrm{C}(G)})\omega(k_A(a))\\ ={}&\varphi(k_A(a)) 1_{\mathrm{C}(G)}\\ ={}&\tau(a) 1_{\mathrm{C}(G)}, \end{aligned} \end{align*} $$

where the first equality uses the fact that $\tau =\varphi \circ k_A$ ; the second equality follows from Theorem 3.9; the third equality is from our assumption that $\varphi $ is G-equivariant.

We now assume that $\tau $ is G-equivariant and show that $\varphi $ too is G-equivariant. We will show that for any $x \in \mathcal {O}_E$ ,

$$\begin{align*}(\varphi \otimes h)\omega(x)=\varphi(x), \end{align*}$$

where h is the Haar state of the CQG G. Before proving this, let us see how we can conclude the proof from this identity. We observe first that the identity can be written, using the standard convolution notation, as

$$\begin{align*}\varphi \ast h=\varphi. \end{align*}$$

Now, for any $\psi \in \mathrm {C}(G)^*$ ,

$$\begin{align*}\varphi \ast \psi=(\varphi \ast h) \ast \psi=\varphi \ast (h \ast \psi)=(\varphi \ast h)=\varphi, \end{align*}$$

where the first equality uses the identity just above the computation; the second uses associativity of convolution; the third equality uses invariance of the Haar state h. Therefore, for all $\psi \in \mathrm {C}(G)^*$ and all $x \in {\mathcal O}_{E}$ ,

$$\begin{align*}(\varphi \ast \psi)(x)=\varphi(x), \end{align*}$$

i.e.,

$$\begin{align*}\psi((\varphi \otimes \mathrm{id}_{\mathrm{C}(G)})\omega(x))=\psi(\varphi(x)1_{\mathrm{C}(G)}). \end{align*}$$

Therefore, we indeed have for all $x \in {\mathcal O}_{E}$ ,

$$\begin{align*}(\varphi \otimes \mathrm{id}_{\mathrm{C}(G)})\omega(x)=\varphi(x)1_{\mathrm{C}(G)},\end{align*}$$

which is what we wanted. So we can now proceed to prove that for any $x \in \mathcal {O}_E$ ,

(4.2) $$ \begin{align} \begin{aligned} (\varphi \otimes h)\omega(x)=\varphi(x), \end{aligned} \end{align} $$

holds. As the Hilbert A-module is assumed to be full and finitely generated, the linear span of the elements of the form $k_E(\xi _{1})\dots k_E(\xi _{m})k_E(\eta _{n})^*\dots k_E(\eta _{1})^* \in {\mathcal O}_{E}$ is dense in ${\mathcal O}_{E}$ . Therefore, by linearity and continuity of the maps involved, it suffices to prove (4.2) only for $k_E(\xi _{1})\dots k_E(\xi _{m})k_E(\eta _{n})^*\dots k_E(\eta _{1})^* \in {\mathcal O}_{E}$ . Moreover, $\xi _1,\dots ,\xi _m,\eta _1,\dots ,\eta _n$ further can be chosen from the spectral submodule $\mathcal {S}(E)$ , where we have an algebraic coaction of $\mathbb {C}[G]$ , hence allowing us to leverage Sweedler notation. However, we shall refrain from saying so and use Sweedler notation freely in what follows. We also remind the reader that $\varphi $ is determined through $\tau $ via the formula (4.1), as in Theorem 4.7. We therefore begin by fixing $\xi _1,\dots ,\xi _m,\eta _1,\dots ,\eta _n \in E$ and assume first that $m \neq n$ . Then we have

$$ \begin{align*} \begin{aligned} {}&(\varphi \otimes h)\omega(k_E(\xi_{1})\dots k_E(\xi_{m})k_E(\eta_{n})^*\dots k_E(\eta_{1})^*)\\ ={}&(\varphi \otimes h)\Big(\omega(k_E(\xi_{1}))\dots \omega(k_E(\xi_{m}))\omega(k_E(\eta_{n}))^*\dots \omega(k_E(\eta_{1}))^\ast\Big)\\ ={}&(\varphi \otimes h)\bigg((k_E \otimes \mathrm{id}_{\mathrm{C}(G)})\lambda(\xi_1)\dots(k_E \otimes \mathrm{id}_{\mathrm{C}(G)})\lambda(\xi_m)\\ &{}((k_E \otimes \mathrm{id}_{\mathrm{C}(G)})\lambda(\eta_n))^*\dots ((k_E \otimes \mathrm{id}_{\mathrm{C}(G)})\lambda(\eta_1))^*\bigg)\\ ={}&\varphi\Big(k_E(\xi_{1(0)})\dots k_E(\xi_{m(0)})k_E(\eta_{n(0)})^*\dots k_E(\eta_{1(0)})^*\Big)h\Big(\xi_{1(1)}\dots\xi_{m(1)}\eta^*_{n(1)}\dots\eta^*_{1(1)}\Big)\\ ={}&0\\ ={}&\varphi(k_E(\xi_{1})\dots k_E(\xi_{m})k_E(\eta_{n})^*\dots k_E(\eta_{1})^*), \end{aligned} \end{align*} $$

where the second equality is by Theorem 3.9; the fourth equality is because of $m\neq n$ ; the fifth equality is again because of $m\neq n$ . So (4.2) holds in this case. Now, let us consider the case where $m=n$ . We have

(4.3) $$ \begin{align} \begin{aligned} {}&(\varphi \otimes h)\omega(k_E(\xi_{1})\dots k_E(\xi_{m})k_E(\eta_{m})^*\dots k_E(\eta_{1})^*)\\ ={}&(\varphi \otimes h)\Big(\omega(k_E(\xi_{1}))\dots \omega(k_E(\xi_{m}))\omega(k_E(\eta_{m}))^*\dots \omega(k_E(\eta_{1}))^*\Big)\\ ={}&(\varphi \otimes h)\bigg((k_E \otimes \mathrm{id}_{\mathrm{C}(G)})\lambda(\xi_1)\dots(k_E \otimes \mathrm{id}_{\mathrm{C}(G)})\lambda(\xi_m)\\ &{}((k_E \otimes \mathrm{id}_{\mathrm{C}(G)})\lambda(\eta_m))^*\dots ((k_E \otimes \mathrm{id}_{\mathrm{C}(G)})\lambda(\eta_1))^*\bigg)\\ ={}&\varphi\Big(k_E(\xi_{1(0)})\dots k_E(\xi_{m(0)})k_E(\eta_{m(0)})^*\dots k_E(\eta_{1(0)})^*\Big)\\ &h\Big(\xi_{1(1)}\dots\xi_{m(1)}\eta^*_{m(1)}\dots\eta^*_{1(1)}\Big), \end{aligned} \end{align} $$

where we have used nothing but Theorem 3.9. Let us consider the expression

(4.4) $$ \begin{align} \begin{aligned} {}&(\tau \otimes h)(\langle\lambda_m(\eta_{1}\otimes \dots \otimes \eta_{m}),\lambda_m(e^{-\beta D}\xi_{1} \otimes \dots e^{-\beta D}\xi_{m})\rangle)\\ ={}&(\tau \otimes h)\bigg(\langle\eta_{1(0)} \otimes \dots \otimes \eta_{m(0)},e^{-\beta D}\xi_{1(0)} \otimes \dots \otimes e^{-\beta D}\xi_{m(0)}\rangle \otimes\\ &\eta_{m(1)}^*\dots\eta_{1(1)}^*\xi_{1(1)}\dots\xi_{m(1)}\bigg)\\ & ={}\tau\, \bigg(\langle\eta_{1(0)} \otimes \dots \otimes \eta_{m(0)},e^{-\beta D}\xi_{1(0)} \otimes \dots \otimes e^{-\beta D}\xi_{m(0)}\rangle\bigg)\\ &h(\eta_{m(1)}^*\dots\eta_{1(1)}^*\xi_{1(1)}\dots\xi_{m(1)})\\ ={}&\varphi(k_E(\xi_{1(0)}) \dots k_E(\xi_{m(0)})k_E(\eta_{m(0)})^*\dots k_E(\eta_{1(0)})^*)\\ &h(\xi_{1(1)}\dots\xi_{m(1)}\eta_{m(1)}^*\dots\eta_{1(1)}^*), \end{aligned} \end{align} $$

where the first equality is by the expression of $\lambda _m$ as in Corollary 4.5 and the assumption that U is G-equivariant; the third equality uses the traciality of h and the expression of $\varphi $ in formula (4.1). Now, by combining (4.3) and (4.4), we obtain

(4.5) $$ \begin{align} \begin{aligned} &{}(\varphi \otimes h)\omega(k_E(\xi_{1}) \dots k_E(\xi_{m})k_E(\eta_{m})^* \dots k_E(\eta_{1})^*)\\ ={}&(\tau \otimes h)(\langle\lambda_m(\eta_{1} \otimes \dots \otimes \eta_{m}),\lambda_m(e^{-\beta D}\xi_{1} \otimes \dots \otimes e^{-\beta D}\xi_{m})\rangle). \end{aligned} \end{align} $$

But the last expression of (4.5) can be further simplified as follows:

(4.6) $$ \begin{align} \begin{aligned} {}&(\tau \otimes h)(\langle\lambda_m(\eta_{1} \otimes \dots \otimes \eta_{m}),\lambda_m(e^{-\beta D}\xi_{1} \otimes \dots \otimes e^{-\beta D}\xi_{m})\rangle)\\ ={}&h((\tau \otimes\mathrm{id}_{\mathrm{C}(G)})\alpha(\langle\eta_{1}\otimes \dots \otimes \eta_{m},e^{-\beta D}\xi_{1} \otimes \dots \otimes e^{-\beta D}\xi_{m}\rangle))\\ ={}&h(\tau(\langle\eta_{1} \otimes \dots \otimes\eta_{m},e^{-\beta D}\xi_{1}\otimes \dots \otimes e^{-\beta D}\xi_{m}\rangle)1_{\mathrm{C}(G)})\\ ={}&\tau(\langle\eta_{1} \otimes \dots \otimes\eta_{m},e^{-\beta D}\xi_{1}\otimes \dots \otimes e^{-\beta D}\xi_{m}\rangle)\\ ={}&\varphi(k_E(\xi_{1}) \dots k_E(\xi_{m})k_E(\eta_{m})^*\dots k_E(\eta_{1})^*), \end{aligned} \end{align} $$

where the first equality is by Corollary 4.5; the second is by our assumption that $\tau $ is G-equivariant; the third is because h is a state; and finally the fourth is by formula (4.1). Now, we combine (4.5) and (4.6), and obtain

$$ \begin{align*} \begin{aligned} &{}(\varphi \otimes h)\omega(k_E(\xi_{1}) \dots k_E(\xi_{m})k_E(\eta_{m})^* \dots k_E(\eta_{1})^*)\\ ={}&\varphi(k_E(\xi_{1}) \dots k_E(\xi_{m})k_E(\eta_{m})^*\dots k_E(\eta_{1})^*), \end{aligned} \end{align*} $$

which shows that (4.2) holds, settling this case as well for good. This completes the proof.

Proof Let G be compact quantum group of Kac type. Then by (2) and (3) of Remark 5.3, all the conditions of Theorem 4.10 are satisfied and the KMS state $\varphi $ on $\mathrm {C}^*({\mathcal G})$ is G-equivariant.

Conversely, we assume that the KMS state $\varphi $ is G-equivariant. Then it follows from the proof of [Reference Joardar and MandalJM18, Proposition 3.8] that G is a quantum subgroup of $\text {U}_{n}^{+}$ (observe that the matrix $F^{{\mathcal G}}$ in [Reference Joardar and MandalJM18, Proposition 3.8] is the identity matrix). Hence, G is a compact quantum group of Kac type.

Proof We consider the $\mathrm {C}^*$ -correspondence coming from the graph of the Cuntz algebra with n-generators. Then since the graph has only one vertex, $\mathrm {C}(\mathcal {G}^0)=\mathbb {C}$ and hence the Hilbert $\mathrm {C}^*$ -module $\mathrm {C}(\mathcal {G}^{1})$ is an n-dimensional Hilbert space. So if $(u_{ij})_{i,j=1,\dots ,n}$ is the unitary matrix corresponding to the fundamental unitary corepresentation u, denoting an orthonormal basis of $\mathrm {C}(\mathcal {G}^{1})$ by $e_{1},\dots ,e_{n}$ , we get a G-action on $\mathrm {C}(\mathcal {G}^1)$ by

$$\begin{align*}\lambda(e_{j})=\sum_{i=1}^{n}e_{i}\otimes u_{ij}, \quad j=1,\dots,n. \end{align*}$$

The G-action $\alpha $ on $\mathrm {C}(\mathcal {G}^{0})=\mathbb {C}$ is the trivial one and it is easy to see that the above $\alpha $ , and $\lambda $ make the $\mathrm {C}^*$ -correspondence $(\mathrm {C}(\mathcal {G}^1),\phi )$ a G-equivariant $\mathrm {C}^*$ -correspondence. Thus, there is a G-action $\omega $ on ${\mathcal O}_{n}$ given on the generators by

$$\begin{align*}\omega(S_{j})=\sum_{i=1}^{n}S_{i} \otimes u_{ij}, \quad j=1,\dots,n. \end{align*}$$

But the unique KMS state on ${\mathcal O}_{n}$ is G-equivariant if and only if $u^{t}=((u_{ji}))$ is unitary (see [Reference Joardar and MandalJM21a]). Now, the unique KMS state is a distinguished KMS state in the above sense. So applying Proposition 5.4, we obtain the desired conclusion.

Proof of Theorem 6.7

We have to show first that $\mathrm {Aut}^{+}_{\text {Ban}}({\mathcal G})$ can be made into an object of the category ${\mathcal C}_{\text {Ban}}({\mathcal G})$ . We proceed to do so now. As in Theorem 6.2, we denote the generators of the underlying $\mathrm {C}^*$ -algebra $\mathrm {C}(\mathrm {Aut}^{+}_{\text {Ban}}({\mathcal G}))$ of $\mathrm {Aut}^{+}_{\text {Ban}}({\mathcal G})$ by $\{q_{wv}\}_{v,w \in \mathcal {G}^0}$ . Now, we define $\lambda _{\text {Ban}}$ and $\alpha _{\text {Ban}}$ as

(6.7) $$ \begin{align} \lambda_{\text{Ban}}(\delta_{e})=\sum_{f\in \mathcal{G}^1}\delta_{f}\otimes q_{s(f)s(e)}q_{r(f)r(e)}, \quad e \in \mathcal{G}^1, \end{align} $$

(6.8) $$ \begin{align} \alpha_{\text{Ban}}(\delta_{v})=\sum_{w\in \mathcal{G}^0}\delta_{w}\otimes q_{wv}, \quad v \in \mathcal{G}^0. \end{align} $$

The relations (6.1),(6.2), and (6.3) imply that $\alpha _{\text {Ban}}$ is a unital $*$ -homomorphism. The formula (6.4) for the comultiplication shows coassociativity of both $\alpha _{\text {Ban}}$ and $\lambda _{\text {Ban}}$ . The Podleś conditions for $\alpha _{\text {Ban}}$ and $\lambda _{\text {Ban}}$ follow from the facts that $\mathrm {C}(\mathcal {G}^0)$ and $\mathrm {C}(\mathcal {G}^1)$ are finite dimensional, respectively. Furthermore, the fact that $\{q_{vw}\}_{v,w\in \mathcal {G}^0}$ are generators of $\mathrm {C}(\mathrm {Aut}^{+}_{\text {Ban}}({\mathcal G}))$ implies that $\alpha _{\text {Ban}}$ is a faithful G-action on $\mathrm {C}(\mathcal {G}^0)$ . Therefore, to show that $(\mathrm {Aut}^{+}_{\text {Ban}}({\mathcal G}),\alpha _{\text {Ban}},\lambda _{\text {Ban}})$ is an object of the category ${\mathcal C}_{\text {Ban}}({\mathcal G})$ , it suffices to show

(6.9) $$ \begin{align} \lambda_{\text{Ban}}(\phi(\delta_{v})\delta_{e})=(\phi \otimes \mathrm{id}_{\mathrm{C}(G)})\alpha_{\text{Ban}}(\delta_{v})\lambda_{\text{Ban}}(\delta_{e}), \quad v \in \mathcal{G}^0, e \in \mathcal{G}^1, \end{align} $$

(6.10) $$ \begin{align} \lambda_{\text{Ban}}(\delta_{e}\delta_{v})=\lambda_{\text{Ban}}(\delta_{e})\alpha(\delta_{v}), \quad v \in \mathcal{G}^0, e \in \mathcal{G}^1, \end{align} $$

(6.11) $$ \begin{align} \langle\lambda_{\text{Ban}}(\delta_{e}),\lambda_{\text{Ban}}(\delta_{f})\rangle=\alpha_{\text{Ban}}(\langle\delta_{e},\delta_{f}\rangle), \quad e, f \in \mathcal{G}^1, \end{align} $$

(6.12) $$ \begin{align} (r_{\ast} \otimes \mathrm{id})\alpha_{\text{Ban}}(f)=\lambda_{\text{Ban}}(r_{\ast}(f)), \quad f \in \mathrm{C}(\mathcal{G}^0). \end{align} $$

We remind the reader that $\phi $ is given as in Example 5.1. Now to check identity (6.9), we note that $\phi (\delta _{v})\delta _{e}=\delta _{s(e),v}\delta _{e}$ ; hence the left-hand side of (6.9) reduces to

(6.13) $$ \begin{align} \lambda_{\text{Ban}}(\phi(\delta_{v})\delta_{e})=\delta_{s(e),v}\sum_{f\in \mathcal{G}^1}\delta_{f}\otimes q_{s(f)s(e)}q_{r(f)r(e)}. \end{align} $$

For the right-hand side of (6.9), we have the following expression:

(6.14) $$ \begin{align} \begin{aligned} &{}(\phi \otimes \mathrm{id}_{\mathrm{C}(G)})\alpha_{\text{Ban}}(\delta_{v})\lambda_{\text{Ban}}(\delta_{e})\\ ={}&(\sum_{w\in \mathcal{G}^0}\phi(\delta_{w}) \otimes q_{wv})(\sum_{f\in \mathcal{G}^1}\delta_{f}\otimes q_{s(f)s(e)}q_{r(f)r(e)})\\ ={}&\sum_{w\in \mathcal{G}^0}\sum_{w=s(f)}\delta_{f}\otimes q_{wv}q_{s(f)s(e)}q_{r(f)r(e)}. \end{aligned} \end{align} $$

But for $v \neq s(e)$ , $q_{wv}q_{s(f)s(e)}=0$ for all $w=s(f)$ , from the relation (6.1); in that case, the final expression of (6.14) is zero, coinciding with (6.13) and so we have (6.9). For $v=s(e)$ , the final expression of (6.14) reduces to

(6.15) $$ \begin{align} \sum_{w\in \mathcal{G}^0}\sum_{w=s(f)}\delta_{f}\otimes q_{ws(e)}q_{s(f)s(e)}q_{r(f)r(e).} \end{align} $$

On the other hand, for $v=s(e)$ , we now consider (6.13).

(6.16) $$ \begin{align} \begin{aligned} \lambda_{\text{Ban}}(\phi(\delta_{v})\delta_{e})={}&\delta_{s(e),v}\sum_{f\in \mathcal{G}^1}\delta_{f}\otimes q_{s(f)s(e)}q_{r(f)r(e)}\\ ={}&\sum_{f \in \mathcal{G}^1}\delta_{f}\otimes q_{s(f)s(e)}q_{r(f)r(e)}\\ ={}&\sum_{f\in \mathcal{G}^1}\delta_{f}\otimes \sum_{w \in \mathcal{G}^0}q_{ws(e)}q_{s(f)s(e)}q_{r(f)r(e)}\\ ={}&\sum_{w\in \mathcal{G}^0}\sum_{w=s(f)}\delta_{f}\otimes q_{ws(e)}q_{s(f)s(e)}q_{r(f)r(e)}, \end{aligned} \end{align} $$

where the third equality is by (6.2); and the fourth equality is by (6.1). However, the final expression in (6.16) is nothing but (6.15). Therefore, we indeed have (6.9). We leave the checking of (6.10) to the reader, which is similar to what has just been done for (6.9). To check the identity (6.11), we note that

$$\begin{align*}\langle\delta_{e},\delta_{f}\rangle=\delta_{e,f}\delta_{r(e)}.\end{align*}$$

Then it reduces to the calculations already done in [Reference SchafhauserSW18, Section 4.1.2]. So we are left with the checking of (6.12). To that end, we note that

$$\begin{align*}r_{\ast}(\delta_{v})=\sum_{e \in r^{-1}(v)}\delta_{e}. \end{align*}$$

Therefore, the right-hand side of (6.12) becomes

(6.17) $$ \begin{align} \begin{aligned} \lambda_{\text{Ban}}(r_{\ast}(\delta_{v}))={}&\sum_{e \in r^{-1}(v)}\sum_{f \in \mathcal{G}^1}\delta_{f}\otimes q_{s(f)s(e)}q_{r(f)r(e)}\\ ={}& \sum_{f \in \mathcal{G}^1}\delta_{f}\otimes \Big(\sum_{e \in r^{-1}(v)}q_{s(f)s(e)}q_{r(f)r(e)}\Big). \end{aligned} \end{align} $$

On the other hand, the left-hand side of (6.12) reduces to

(6.18) $$ \begin{align} \begin{aligned} (r_{\ast}\otimes \mathrm{id}_{\mathrm{C}(G)})\alpha_{\text{Ban}}(\delta_{v})={}& (r_{\ast}\otimes\mathrm{id})(\sum_{w\in \mathcal{G}^0}\delta_{w}\otimes q_{wv})\\ ={}&\sum_{w\in \mathcal{G}^0}\Big(\sum_{f \in r^{-1}(w)}\delta_{f}\otimes q_{wv}\Big). \end{aligned} \end{align} $$

Now, we observe that the set $\{f \in \mathcal {G}^1 \mid r(f)=w, w \in \mathcal {G}^0\}$ is the whole of $\mathcal {G}^1;$ and therefore, the last expression in (6.18) reduces to

(6.19) $$ \begin{align} \sum_{f \in \mathcal{G}^1}\delta_{f}\otimes q_{r(f)v}. \end{align} $$

We also have, for each $f \in \mathcal {G}^1$ ,

(6.20) $$ \begin{align} q_{r(f)v}=\sum_{w \in \mathcal{G}^0}q_{s(f)w}q_{r(f)v}; \end{align} $$

moreover, since $q_{s(f)w}q_{r(f)v}=0$ whenever $(v,w)$ is not an edge,

(6.21) $$ \begin{align} \sum_{w \in \mathcal{G}^0}q_{s(f)w}q_{r(f)v}=\sum_{e \in r^{-1}(v)}q_{s(f)s(e)}q_{r(f)r(e)}. \end{align} $$

Combining (6.18), (6.19), (6.20), and (6.21), we obtain

$$\begin{align*}(r_{\ast}\otimes \mathrm{id}_{\mathrm{C}(G)})\alpha_{\text{Ban}}(\delta_{v})=\sum_{f \in \mathcal{G}^1}\delta_{f}\otimes \Big(\sum_{e \in r^{-1}(v)}q_{s(f)s(e)}q_{r(f)r(e)}\Big), \end{align*}$$

which is exactly the last expression in (6.17). Therefore (6.12) holds. This proves that indeed $(\mathrm {Aut}^{+}_{\text {Ban}}({\mathcal G}),\alpha _{\text {Ban}},\lambda _{\text {Ban}})$ is an object in the category ${\mathcal C}_{\text {Ban}}({\mathcal G})$ .

Now, we turn to the proof of the fact that $(\mathrm {Aut}^{+}_{\text {Ban}}({\mathcal G}),\alpha _{\text {Ban}},\lambda _{\text {Ban}})$ is indeed the universal object in the category ${\mathcal C}_{\text {Ban}}({\mathcal G})$ . To that end, let $(G,\alpha ,\lambda )$ be an object of the category ${\mathcal C}_{\text {Ban}}({\mathcal G})$ .

$$\begin{align*}\lambda(\delta_{e})=\sum_{f \in \mathcal{G}^1}\delta_{f}\otimes a_{fe}, \quad a_{fe} \in \mathrm{C}(G), \quad e\in \mathcal{G}^1, \end{align*}$$

$$\begin{align*}\alpha(\delta_{v})=\sum_{w\in \mathcal{G}^0}\delta_{w}\otimes b_{wv}, \quad b_{vw} \in \mathrm{C}(G), \quad v \in \mathcal{G}^0. \end{align*}$$

Now, as already observed, for any $v \in \mathcal {G}^0$ ,

$$\begin{align*}r_{\ast}(\delta_{v})=\sum_{e \in r^{-1}(v)}\delta_{e}.\end{align*}$$

As the graph does not have any multiple edges or sources, for $w \in \mathcal {G}^0$ with $s(e)=w, r(e)=v$ ,

$$\begin{align*}\phi(\delta_{w})r_{\ast}(\delta_{v})=\delta_{e}.\end{align*}$$

Hence, we have

(6.22) $$ \begin{align} \lambda(\phi(\delta_{w})r_{\ast}(\delta_{v}))=\sum_{f\in \mathcal{G}^1}\delta_{f}\otimes a_{fe}, \text{ provided } s(e)=w,r(e)=v. \end{align} $$

On the other hand, since $(r_{\ast }\otimes \mathrm {id}_{\mathrm {C}(G)})\circ \alpha =\lambda \circ r_{\ast }$ and $\lambda (\phi (f)\xi )=(\phi \otimes \mathrm {id}_{\mathrm {C}(G)})\alpha (f)\lambda (\xi )$ for $f \in \mathrm {C}(\mathcal {G}^0)$ and $\xi \in \mathrm {C}(\mathcal {G}^1)$ ,

(6.23) $$ \begin{align} \begin{aligned} \lambda(\phi(\delta_{w})r_{\ast}(\delta_{v}))={}&(\phi \otimes \mathrm{id}_{\mathrm{C}(G)})\alpha(\delta_{w})(r_{\ast}\otimes\mathrm{id}_{\mathrm{C}(G)})\alpha(\delta_{v})\\ ={}&\sum_{f : u \rightarrow p}\delta_{f}\otimes b_{uw}b_{pv}. \end{aligned} \end{align} $$

By comparing coefficients, we get

(6.24) $$ \begin{align} \lambda(\delta_{e})=\sum_{f \in \mathcal{G}^1}\delta_{f}\otimes b_{s(f)s(e)}b_{r(f)r(e)}. \end{align} $$

Then the fact that $(b_{vw})_{v,w \in \mathcal {G}^0}$ satisfy the relations of $\mathrm {C}(\mathrm {Aut}^{+}_{\text {Ban}}({\mathcal G}))$ can be checked, along the lines of [Reference SchafhauserSW18, Section 4.3.3], using

$$\begin{align*}\langle\lambda(\delta_{e}),\lambda(\delta_{e})\rangle=\alpha(\langle\delta_{e},\delta_{e}\rangle),\end{align*}$$

together with the fact that

$$\begin{align*}\langle\delta_{e},\delta_{e}\rangle=\delta_{r(e)};\end{align*}$$

we leave this to the reader. We therefore have shown the universality of the triple $(\mathrm {Aut}^{+}_{\text {Ban}}({\mathcal G}),\alpha _{\text {Ban}},\lambda _{\text {Ban}})$ , thus completing the proof of the theorem.

Proof We begin by fixing a vertex $v \in \mathcal {G}^0$ . Then, if $s^{-1}(v)$ is nonempty, then the arguments used in the proof of the fact $(r_{\ast }\otimes \mathrm {id}_{\mathrm {C}(G)})\alpha _{\text {Ban}}(f)=\lambda _{\text {Ban}}(r_{\ast }(f))$ can be repeated with obvious modifications. So we are left with the case when $s^{-1}(v)$ is empty. Then $s_{\ast }(\delta _{v})=0$ so that $\lambda _{\text {Ban}}(s_{\ast }(\delta _{v}))=0$ . On the other hand, denoting the generators of the $\mathrm {C}^*$ -algebra $\mathrm {C}(\mathrm {Aut}^{+}_{\text {Ban}}({\mathcal G}))$ by $\{q_{vw}\}_{v,w\in \mathcal {G}^0}$ , as in Theorem 6.2, we obtain

$$\begin{align*}(s_{\ast}\otimes\mathrm{id}_{\mathrm{C}(G)})\alpha_{\text{Ban}}(\delta_{v})=\sum_{w \in \mathcal{G}^0}s_{\ast}(\delta_{w})\otimes q_{wv}=\sum_{f \in \mathcal{G}^1}(\sum_{s(f)=w}\delta_{f}\otimes q_{s(f)v}). \end{align*}$$

But, for a fixed edge $f \in \mathcal {G}^1$ , $q_{s(f)v}=\sum _{w \in \mathcal {G}^0}q_{s(f)v}q_{r(f)w}$ . Since $s^{-1}(v)$ is empty, $q_{s(f)v}q_{r(f)w}=0$ , which implies that $(s_{\ast }\otimes \mathrm {id}_{\mathrm {C}(G)})\alpha _{\text {Ban}}(\delta _{v})=0$ as well.