Faithfulness of bi-free product states

TL;DR

This paper investigates conditions under which bi-free product states are faithful, showing that faithfulness implies the pairs of faces are minimal tensor products, with some partial converse results.

Contribution

It establishes a link between faithfulness of bi-free product states and the minimal tensor product structure of pairs of faces in unital C*-algebras.

Findings

Faithfulness of bi-free product states implies minimal tensor product structure.

Partial converse results connecting faithfulness and tensor products.

Conditions for faithfulness in bi-free product states.

Abstract

Given a non-trivial family of pairs of faces of unital C*-algebras where each pair has a faithful state, it is proved that if the bi-free product state is faithful on the reduced bi-free product of this family of pairs of faces then each pair of faces arises as a minimal tensor product. A partial converse is also obtained.

Equations

$$C^{\ast }(\{{\mathcal{A}}_i{\}}_{i\in \mathcal{I}})\simeq {\ast }_{i\in \mathcal{I}}({\mathcal{A}}_i,\phi {|}_{{\mathcal{A}}_i}),$$

$$(\mathcal{H},\xi )={\ast }_{i\in \mathcal{I}}({\mathcal{H}}_i,{\xi }_i),$$

$$\mathcal{H}:=\mathbb{C}\xi \oplus \underset{\begin{array}{c}n\in \mathbb{N}\\ i_1,\cdots ,i_n\in \mathcal{I}\\ i_1\ne \cdots \ne i_n\end{array}}{⨁}{\mathring{\mathcal{H}}}_{i_1}\otimes \cdots \otimes {\mathring{\mathcal{H}}}_{i_n}.$$

$$\mathcal{H}(l,i)$$

$$\mathcal{H}(r,i)$$

$${\lambda }_i(T)=V_i(T\otimes I_{\mathcal{H}(l,i)})V_i^{\ast }$$

$${\rho }_i(T)=W_i(I_{\mathcal{H}(r,i)}\otimes T)W_i^{\ast }.$$

$$\begin{CD}\check{*}_{i\in{\mathcal{I}}}({\mathcal{A}}_{l}^{(i)}\check{*}{\mathcal{A}}_{r}^{(i)})@>{\iota}>{}>{\mathcal{A}}@>{\varphi}>{}>{\mathbb{C}}\\ @V{*_{i\in{\mathcal{I}}}(\pi_{i}*\pi_{i})}V{}V\Big{\|}\\ \check{*}_{i\in{\mathcal{I}}}(B({\mathcal{H}}_{i})\check{*}B({\mathcal{H}}_{i}))@>{*_{i\in I}(\lambda_{i}*\rho_{i})}>{}>B({\mathcal{H}})@>{\langle\cdot\xi,\xi\rangle}>{}>{\mathbb{C}}\end{CD}$$

$$({\ast \ast }_{i\in \mathcal{I}}({\mathcal{A}}_l^{(i)},{\mathcal{A}}_r^{(i)}),{\ast \ast }_{i\in \mathcal{I}}{\phi }_i)={\ast \ast }_{i\in \mathcal{I}}(({\mathcal{A}}_l^{(i)},{\mathcal{A}}_r^{(i)}),{\phi }_i)$$

$${\ast \ast }_{i\in \mathcal{I}}({\mathcal{A}}_l^{(i)},{\mathcal{A}}_r^{(i)}):=C^{\ast }(({\lambda }_i\circ {\pi }_i({\mathcal{A}}_l^{(i)}),{\rho }_i\circ {\pi }_i({\mathcal{A}}_r^{(i)}){)}_{i\in \mathcal{I}})\subset B(\mathcal{H})$$

$${\ast \ast }_{i\in \mathcal{I}}{\phi }_i(\cdot ):=⟨\cdot \xi ,\xi ⟩.$$

$$C^{\ast }({\mathcal{A}}_l^{(i)},{\mathcal{A}}_r^{(i)})\simeq {\mathcal{A}}_l^{(i)}{\otimes }_{min}{\mathcal{A}}_r^{(i)}.$$

$$({\lambda }_i(a_l){\rho }_i(a_r){\lambda }_j\ast {\rho }_j(b)-{\rho }_i(a_r){\lambda }_i(a_l){\lambda }_j\ast {\rho }_j(b))\xi =0$$

$${\lambda }_i(a_l){\rho }_i(a_r){\lambda }_j\ast {\rho }_j(b)-{\rho }_i(a_r){\lambda }_i(a_l){\lambda }_j\ast {\rho }_j(b)=0$$

$$0$$

$$=({\lambda }_i(a_l){\rho }_i(a_r)-{\rho }_i(a_r){\lambda }_i(a_l))⟨b{b}^{\ast }{\xi }_j,{\xi }_j⟩\xi$$

$$=⟨b{b}^{\ast }{\xi }_j,{\xi }_j⟩(a_l{a}_r-a_r{a}_l){\xi }_i.$$

$$\sum _{k=1}^m{a}_{k,l}\odot a_{k,r}\mapsto \sum _{k=1}^m{a}_{k,l}{a}_{k,r}.$$

$$V_h:{\mathcal{H}}_i\otimes {\mathcal{H}}_i\to {\mathcal{H}}_i\otimes h\otimes {\mathcal{H}}_i$$

$${\mathcal{H}}_i\otimes h\otimes {\mathcal{H}}_i=\mathbb{C}h\oplus ({\mathring{\mathcal{H}}}_i\otimes h)\oplus (h\otimes {\mathring{\mathcal{H}}}_i)\oplus ({\mathring{\mathcal{H}}}_i\otimes h\otimes {\mathring{\mathcal{H}}}_i)$$

$$V_h^{\ast }{\lambda }_i(a){\rho }_i(b)V_h({\eta }_1\otimes {\eta }_2)$$

$$=a{\eta }_1\otimes b{\eta }_2.$$

$$V_h^{\ast }{C}^{\ast }({\lambda }_i({\mathcal{A}}_l^{(i)}),{\rho }_i({\mathcal{A}}_r^{(i)}))V_h={\mathcal{A}}_l^{(i)}{\otimes }_{min}{\mathcal{A}}_r^{(i)}.$$

$$0$$

$$=V_h^{\ast }\sum _{k=1}^m{\lambda }_i(a_{k,l}){\rho }_i(a_{k,r})V_h({\xi }_i\otimes {\xi }_i)$$

$$=\sum _{k=1}^m{\lambda }_i(a_{k,l}){\rho }_i(a_{k,r})h$$

$$0$$

$$=⟨{\left(\sum _{k=1}^m{\lambda }_i(a_{k,l}){\rho }_i(a_{k,r})\right)}^{\ast }\left(\sum _{k=1}^m{\lambda }_i(a_{k,l}){\rho }_i(a_{k,r})\right)\xi ,\xi ⟩$$

$$={\phi }_i\left({\left(\sum _{k=1}^m{a}_{k,l}{a}_{k,r}\right)}^{\ast }\left(\sum _{k=1}^m{a}_{k,l}{a}_{k,r}\right)\right)$$

$$q:C^{\ast }({\mathcal{A}}_l^{(i)},{\mathcal{A}}_r^{(i)})\to {\mathcal{A}}_l^{(i)}{\otimes }_{min}{\mathcal{A}}_r^{(i)}.$$

$${\iota }_i:{\mathcal{A}}_l^{(i)}\check{\ast }{\mathcal{A}}_r^{(i)}\to C^{\ast }({\mathcal{A}}_l^{(i)},{\mathcal{A}}_r^{(i)})(\subseteq B({\mathcal{H}}_i)\text{by assumption})$$

$${\lambda }_i\ast {\rho }_i(\cdot ){|}_{{\mathcal{H}}_i}={\iota }_i(\cdot ),$$

$$V_{b{\xi }_j}^{\ast }{\lambda }_i\ast {\rho }_i(\tilde{a})V_{b{\xi }_j}=q(a)=0,$$

$$0$$

Full text

2010 Mathematics Subject Classification. 46L30, 46L54, 46L09.

Key words and phrases: Free probability, operator algebras, bi-free

1. introduction

The reduced free product was given independently by Avitzour [1] and Voiculescu [7] and it has been foundational in the development of free probability. Dykema proved in [2] that the free product state on the reduced free product of unital C∗-algebras with faithful states is faithful. In consequence of this, if $\{{\mathcal{A}}_{i}\}_{i\in{\mathcal{I}}}$ is a free family of unital C∗-algebras in the non-commutative C∗-probability space $({\mathcal{A}},\varphi)$ and if $\varphi$ is faithful on $C^{*}(\{{\mathcal{A}}_{i}\}_{i\in{\mathcal{I}}})$ then

[TABLE]

the reduced free product of the ${\mathcal{A}}_{i}$’s with respect to the given states. This can be deduced from a paper of Dykema and Rørdam, namely [3, Lemma 1.3].

The present paper is the result of the author’s attempt to prove the same result in the new context of bi-free probability introduced by Voiculescu [8]. To this end, suppose $({\mathcal{A}}_{l}^{(i)},{\mathcal{A}}_{r}^{(i)})_{i\in{\mathcal{I}}}$ is a non-trivial family of pairs of faces in the non-commutative C∗-probability space $({\mathcal{A}},\varphi)$. If $\varphi_{i}=\varphi|_{C^{*}({\mathcal{A}}_{l}^{(i)},{\mathcal{A}}_{r}^{(i)})}$ is faithful on $C^{*}({\mathcal{A}}_{l}^{(i)},{\mathcal{A}}_{r}^{(i)})$, for all $i\in{\mathcal{I}}$, then it will be proven that if the bi-free product state ${**}_{i\in{\mathcal{I}}}\varphi_{i}$ is faithful on the reduced bi-free product ${**}_{i\in{\mathcal{I}}}({\mathcal{A}}_{l}^{(i)},{\mathcal{A}}_{r}^{(i)})$ then $C^{*}({\mathcal{A}}_{l}^{(i)},{\mathcal{A}}_{r}^{(i)})\simeq{\mathcal{A}}_{l}^{(i)}\otimes_{min}{\mathcal{A}}_{r}^{(i)},i\in{\mathcal{I}}$. A converse is shown with the added asumption that each $\varphi_{i}$ is a product state. Moreover, in this case there is a commensurate result to that which follows from Dykema and Rørdam, mentioned above.

It should be mentioned that the failure in general of the faithfulness of the bi-free product state has been pointed out in [4] and this failure has been the cause of the introduction of weaker versions of faithfulness in the bi-free context [4, 5]. Acknowledgements: The author would like to thank Scott Atkinson for sparking my interest into bi-free independence and for suggesting the reduced bi-free product, Paul Skoufranis for pointing out an error in a previous version of this paper, and the referee for their help in improving several difficult passages.

2. Bi-free independence and the reduced bi-free product

We will first take some time to recall the definition of bi-free independence from [8] and then define the reduced bi-free product of C∗-algebras and the bi-free product state.

Fix a non-commutative C∗-probability space $({\mathcal{A}},\varphi)$, that is a unital C∗-algebra and a state. Given a set ${\mathcal{I}}$, suppose that for each $i\in{\mathcal{I}}$ there is a pair of unital C∗-subalgebras ${\mathcal{A}}_{l}^{(i)}$ and ${\mathcal{A}}_{r}^{(i)}$ of ${\mathcal{A}}$, a “left” algebra and a “right” algebra. We call the set $({\mathcal{A}}_{l}^{(i)},{\mathcal{A}}_{r}^{(i)})_{i\in{\mathcal{I}}}$ a family of pairs of faces in ${\mathcal{A}}$. Such a family will be called non-trivial if $|{\mathcal{I}}|\geq 2$ and $C^{*}({\mathcal{A}}_{l}^{(i)},{\mathcal{A}}_{r}^{(i)})\neq{\mathbb{C}}$ for all $i\in{\mathcal{I}}$. That is, there are at least two pairs of faces and there are no trivial pairs of faces.

Let $(\pi_{i},{\mathcal{H}}_{i},\xi_{i})$ be the GNS construction for $(C^{*}({\mathcal{A}}_{l}^{(i)},{\mathcal{A}}_{r}^{(i)}),\varphi_{i})$ where $\varphi_{i}=\varphi|_{C^{*}({\mathcal{A}}_{l}^{(i)},{\mathcal{A}}_{r}^{(i)})}$. Voiculescu [8] (and even way back in [7]) observed that there are two natural representations of $B({\mathcal{H}}_{i})$ on the free product Hilbert space, which we will now introduce. The free product Hilbert space,

[TABLE]

is given by associating all of the distinguished vectors and then forming a Fock space like structure. Namely, if $\mathring{{\mathcal{H}}}_{j}={\mathcal{H}}_{j}\ominus{\mathbb{C}}\xi_{j}$, then

[TABLE]

To define these representations we need to first build some Hilbert spaces and some unitaries. To this end, define

[TABLE]

Then there are unitaries $V_{i}:{\mathcal{H}}_{i}\otimes{\mathcal{H}}(l,i)\rightarrow{\mathcal{H}}$ and $W_{i}:{\mathcal{H}}(r,i)\otimes{\mathcal{H}}_{i}$ given by concatenation (with appropriate handling of $\xi_{i}$ and $\xi$). Finally, the two natural representations are the left representation $\lambda_{i}:B(\mathcal{H}_{i})\rightarrow B({\mathcal{H}})$ which is defined as

[TABLE]

and the right representation $\rho_{i}:B(\mathcal{H}_{i})\rightarrow B({\mathcal{H}})$ which is defined as

[TABLE]

With all of this groundwork established we can finally define bi-free independence. Note that $\check{*}$ below refers to the full (or universal) free product of C∗-algebras.

Definition 2.1** (Voiculescu [8]).**

The family of pairs of faces $({\mathcal{A}}_{l}^{(i)},{\mathcal{A}}_{r}^{(i)})_{i\in{\mathcal{I}}}$ in the non-commutative probability space $({\mathcal{A}},\varphi)$ is said to be bi-freely independent with respect to $\varphi$ if the following diagram commutes

[TABLE]

where $\iota$ is the unique $*$-homomorphism extending the identity on each ${\mathcal{A}}_{\chi}^{(i)}$, for all $\chi\in\{l,r\}$ and $i\in{\mathcal{I}}$.

From this we can now define the main objects of this paper.

Definition 2.2**.**

Let $({\mathcal{A}}_{l}^{(i)},{\mathcal{A}}_{r}^{(i)})_{i\in{\mathcal{I}}}$ be a family of pairs of faces in the non-commutative C∗-probability space $({\mathcal{A}},\varphi)$. As before, denote $\varphi_{i}$ to be the restriction of $\varphi$ to $C^{*}({\mathcal{A}}_{l}^{(i)},{\mathcal{A}}_{r}^{(i)})$ and let $(\pi_{i},{\mathcal{H}}_{i},\xi_{i})$ be the GNS construction of $(C^{*}({\mathcal{A}}_{l}^{(i)},{\mathcal{A}}_{r}^{(i)}),\varphi_{i})$.

The reduced bi-free product of $({\mathcal{A}}_{l}^{(i)},{\mathcal{A}}_{r}^{(i)})_{i\in{\mathcal{I}}}$ with respect to the states $\varphi_{i}$ is

[TABLE]

which is made up of the unital C∗-subalgebra of $B({\mathcal{H}})$, called the reduced bi-free product of C∗-algebras,

[TABLE]

and the bi-free product state

[TABLE]

It is an immediate fact that the family of pairs of faces $(\lambda_{i}\circ\pi_{i}({\mathcal{A}}_{l}^{(i)}),\rho_{i}\circ\pi_{i}({\mathcal{A}}_{r}^{(i)}))_{i\in{\mathcal{I}}}$ is bi-freely independent with respect to the bi-free product state.

It should be noted that we are working within the framework of the original non-commutative C∗-probability space $({\mathcal{A}},\varphi)$. This means that the reduced bi-free product is taking into account the behaviour of $\varphi$ not just on the left and right faces but on the C∗-algebra they generate, $C^{*}({\mathcal{A}}_{l}^{(i)},{\mathcal{A}}_{r}^{(i)})$. Since bi-free independence is a statement about the behaviour in the original C∗-probability space this definition makes sense.

That being said, one can create the reduced bi-free product as an external product. Start with pairs of faces in different C∗-probability spaces and simply create a new C∗-probability space by taking the full free product of the C∗-algebras and their associated states and then proceed with the above reduced bi-free product construction.

3. Faithfulness of bi-free product states

We first establish what happens when the bi-free product state is faithful.

Theorem 3.1**.**

Let $({\mathcal{A}}_{l}^{(i)},{\mathcal{A}}_{r}^{(i)})_{i\in{\mathcal{I}}}$ be a non-trivial family of pairs of faces in the non-commutative $C^{*}$-probability space $({\mathcal{A}},\varphi)$ such that $\varphi_{i}=\varphi|_{C^{*}({\mathcal{A}}_{l}^{(i)},{\mathcal{A}}_{r}^{(i)})}$ is faithful on $C^{*}({\mathcal{A}}_{l}^{(i)},{\mathcal{A}}_{r}^{(i)})$ for each $i\in{\mathcal{I}}$. If ${**}_{i\in{\mathcal{I}}}\varphi_{i}$ is faithful on the reduced bi-free product ${**}_{i\in{\mathcal{I}}}({\mathcal{A}}_{l}^{(i)},{\mathcal{A}}_{r}^{(i)})$ then

[TABLE]

Proof.

First we will establish that ${\mathcal{A}}_{l}^{(i)}$ and ${\mathcal{A}}_{r}^{(i)}$ commute in ${\mathcal{A}}$, then we will show that they induce a C∗-norm on the algebraic tensor product ${\mathcal{A}}_{l}^{(i)}\odot{\mathcal{A}}_{r}^{(i)}$ and finally that this is in fact the minimal tensor norm.

We will be using the notation from Section 2. To simplify things a little bit, because the $\varphi_{i}$ are assumed to be faithful, consider $C^{*}({\mathcal{A}}_{l}^{(i)},{\mathcal{A}}_{r}^{(i)})$ as already a subalgebra of $B({\mathcal{H}}_{i})$ and so $\varphi_{i}(\cdot)=\langle\cdot\xi_{i},\xi_{i}\rangle$. That is, we are suppressing the $\pi_{i}$ notation from the GNS construction. Moreover, we will be using the convention that $\lambda_{i}(x),\rho_{i}(x),\lambda_{i}*\rho_{i}(x)$ all are living in $B({\mathcal{H}})$.

Suppose $a_{\chi}\in{\mathcal{A}}_{\chi}^{(i)}$ such that $\varphi_{i}(a_{\chi})=0$, $\chi\in\{l,r\}$ and $0\neq b\in{\mathcal{A}}_{l}^{(j)}\cup{\mathcal{A}}_{r}^{(j)}$ for $j\neq i$ such that $\varphi_{j}(b)=0$. Such a $b$ exists by the non-triviality of the family of pairs of faces. This gives that $\langle b^{*}\xi_{j},\xi_{j}\rangle=\varphi_{j}(b)=0$ and so $b^{*}\xi_{j}\in\mathring{{\mathcal{H}}}_{j}$ while $\langle b(b^{*}\xi_{j}),\xi_{j}\rangle=\varphi_{j}(bb^{*})\neq 0$ by the faithfulness of $\varphi_{j}$.

Now, [8, Section 1.5] establishes that $[\lambda_{i}({\mathcal{A}}_{l}^{(i)}),\rho_{i}({\mathcal{A}}_{r}^{(i)})]({\mathcal{H}}\ominus{\mathcal{H}}_{i})=0$ which gives that

[TABLE]

since $b\xi\in\mathring{{\mathcal{H}}}_{j}\subset{\mathcal{H}}$. The faithfulness of ${**}_{i\in{\mathcal{I}}}\varphi_{i}$ implies that $\xi$ is a separating vector for the reduced bi-free product and thus

[TABLE]

which gives that

[TABLE]

Since $\xi_{i}$ is separating for $C^{*}({\mathcal{A}}_{l}^{(i)},{\mathcal{A}}_{r}^{(i)})$ this implies that $a_{l}$ and $a_{r}$ commute. Thus, ${\mathcal{A}}_{l}^{(i)}$ and ${\mathcal{A}}_{r}^{(i)}$ commute in ${\mathcal{A}}$ for every $i\in{\mathcal{I}}$.

Claim: The canonical map from ${\mathcal{A}}_{l}^{(i)}\odot{\mathcal{A}}_{r}^{(i)}$ to $C^{*}({\mathcal{A}}_{l}^{(i)},{\mathcal{A}}_{r}^{(i)})$ is injective. Since ${\mathcal{A}}_{l}^{(i)}$ and ${\mathcal{A}}_{r}^{(i)}$ commute, the universal property of ${\mathcal{A}}_{l}^{(i)}\odot{\mathcal{A}}_{r}^{(i)}$ gives that there exists a $*$-homomorphism

[TABLE]

We need to establish its injectivity. To this end, consider $h\in\mathring{{\mathcal{H}}}_{j}$, $\|h\|=1$ where $j\neq i$ and the isometric map

[TABLE]

defined by $V_{h}(h_{l}\otimes h_{r})=h_{l}\otimes h\otimes h_{r}$ for $h_{l},h_{r}\in{\mathcal{H}}_{i}$. This map is inspired by Dykema’s proof of the faithfulness of the free product state [2, Theorem 1.1]. Note that in ${\mathcal{H}}$ we really have that

[TABLE]

but hopefully the reader will pardon the simplified notation.

Now ${\mathcal{H}}_{i}\otimes h\otimes{\mathcal{H}}_{i}$ is a reducing subspace of $C^{*}(\lambda_{i}({\mathcal{A}}_{l}^{(i)}),\rho_{i}({\mathcal{A}}_{r}^{(i)}))$ since for all $a\in{\mathcal{A}}_{l}^{(i)},b\in{\mathcal{A}}_{r}^{(i)}$ and $\eta_{1},\eta_{2}\in{\mathcal{H}}_{i}$ we have that

[TABLE]

Thus, compressing to ${\mathcal{H}}_{i}\otimes h\otimes{\mathcal{H}}_{i}$ gives

[TABLE]

So, if $\sum_{k=1}^{m}a_{k,l}\odot a_{k,r}\neq 0\in{\mathcal{A}}_{l}^{(i)}\odot{\mathcal{A}}_{r}^{(i)}$ then $\sum_{k=1}^{m}a_{k,l}\otimes a_{k,r}\neq 0\in{\mathcal{A}}_{l}^{(i)}\otimes_{min}{\mathcal{A}}_{r}^{(i)}$ which implies that

[TABLE]

since the state $\langle\cdot\xi_{i}\otimes\xi_{i},\xi_{i}\otimes\xi_{i}\rangle$ is faithful on the min tensor product. But then $\sum_{k=1}^{m}\lambda_{i}(a_{k,l})\rho_{i}(a_{k,r})\neq 0\in C^{*}(\lambda_{i}({\mathcal{A}}_{l}^{(i)}),\rho_{i}({\mathcal{A}}_{r}^{(i)}))$ which by the faithfulness of ${**}_{i\in{\mathcal{I}}}\varphi_{i}$ gives that $\sum_{k=1}^{m}\lambda_{i}(a_{k,l})\rho_{i}(a_{k,r})\xi\neq 0$. Finally,

[TABLE]

which gives by the faithfulness of $\varphi_{i}$ that $\sum_{k=1}^{m}a_{k,l}a_{k,r}\neq 0$. Therefore, the claim is verified.

Now, this implies that $C^{*}({\mathcal{A}}_{l}^{(i)},{\mathcal{A}}_{r}^{(i)})\simeq{\mathcal{A}}_{l}^{(i)}\otimes_{\alpha}{\mathcal{A}}_{r}^{(i)}$ where $\|\cdot\|_{\alpha}$ is a C∗-norm on ${\mathcal{A}}_{l}^{(i)}\odot{\mathcal{A}}_{r}^{(i)}$. So by Takesaki’s Theorem [6] we have that there exists a surjective $*$-homomorphism

[TABLE]

To finish the proof all we need to do is show that $q$ is injective.

To this end, let $a\in C^{*}({\mathcal{A}}_{l}^{(i)},{\mathcal{A}}_{r}^{(i)})$ such that $q(a)=0$. Again as in the first part of this proof, find $0\neq b\in{\mathcal{A}}_{l}^{(j)}\cup{\mathcal{A}}_{r}^{(j)}$ for $j\neq i$ such that $\varphi_{j}(b)=0$ and $h\in\mathring{{\mathcal{H}}}_{j}$ such that $\langle bh,\xi_{j}\rangle\neq 0$. Additionally, assume that $\|b\xi_{j}\|=1$.

In the second part of this proof we saw that compressing to ${\mathcal{H}}_{i}\otimes b\xi_{j}\otimes{\mathcal{H}}_{i}$ is tantamount to this quotient homomorphism $q$. Namely, suppose

[TABLE]

is the unique $*$-homomorphism extending the identity in each component. There then exists $\tilde{a}\in{\mathcal{A}}_{l}^{(i)}\check{*}{\mathcal{A}}_{r}^{(i)}$ such that $\iota_{i}(\tilde{a})=a$. An important fact to record is that, by uniqueness,

[TABLE]

remembering that we have that $\lambda_{i}*\rho_{i}(\cdot)\in B({\mathcal{H}})$. Thus,

[TABLE]

which implies, by the fact that $V_{b\xi_{j}}({\mathcal{H}}_{i}\otimes{\mathcal{H}}_{i})$ is reducing for $\lambda_{i}*\rho_{i}({\mathcal{A}}_{l}^{(i)}\check{*}{\mathcal{A}}_{r}^{(i)})$, that

[TABLE]

By the faithfulness of the bi-free product state $\lambda_{i}*\rho_{i}(\tilde{a})\lambda_{j}*\rho_{j}(b)=0$ and so

[TABLE]

Hence, by the faithfulness of $\varphi_{i}$ we have that $a=0$. Therefore, for all $i\in{\mathcal{I}}$, $C^{*}({\mathcal{A}}_{l}^{(i)},{\mathcal{A}}_{r}^{(i)})\simeq{\mathcal{A}}_{l}^{(i)}\otimes_{min}{\mathcal{A}}_{r}^{(i)}$. ∎

We turn now to a partial converse of the previous theorem. This is probably known among the experts in bi-free probability but we could not find a published proof. The following proof may be a tad clunky but we find it the clearest from a non-expert perspective.

Theorem 3.2**.**

Let $({\mathcal{A}}_{l}^{(i)},{\mathcal{A}}_{r}^{(i)})_{i\in{\mathcal{I}}}$ be a family of pairs of faces in the non-commutative C∗-probability space $({\mathcal{A}},\varphi)$. If $C^{*}({\mathcal{A}}_{l}^{(i)},{\mathcal{A}}_{r}^{(i)})\simeq{\mathcal{A}}_{l}^{(i)}\otimes_{min}{\mathcal{A}}_{r}^{(i)}$ and $\varphi_{i}=\varphi_{i}|_{{\mathcal{A}}_{l}^{(i)}}\otimes\varphi_{i}|_{{\mathcal{A}}_{r}^{(i)}}$ is a faithful product state on $C^{*}({\mathcal{A}}_{l}^{(i)},{\mathcal{A}}_{r}^{(i)})$, for all $i\in{\mathcal{I}},$ then ${**}_{i\in{\mathcal{I}}}\varphi_{i}$ is faithful on the reduced bi-free product and

[TABLE]

Proof.

As before, we will be using the notation of Section 2.

For each $i\in{\mathcal{I}}$, since $C^{*}({\mathcal{A}}_{l}^{(i)},{\mathcal{A}}_{r}^{(i)})\simeq{\mathcal{A}}_{l}^{(i)}\otimes_{min}{\mathcal{A}}_{r}^{(i)}$ and $\varphi_{i}$ is a product state we can a priori choose ${\mathcal{H}}_{i}={\mathcal{H}}_{i,l}\otimes{\mathcal{H}}_{i,r}$, unit vectors $\xi_{i,l}\in{\mathcal{H}}_{i,l},\xi_{i,r}\in{\mathcal{H}}_{i,r}$ such that $\xi_{i}=\xi_{i,l}\otimes\xi_{i,r}$ and $*$-homomorphisms $\pi_{i,\chi}:{\mathcal{A}}_{\chi}^{(i)}\rightarrow B({\mathcal{H}}_{i,\chi})$ such that $\pi_{i}=\pi_{i,l}\otimes\pi_{i,r}$. This will give for $a_{\chi}\in{\mathcal{A}}_{\chi}^{(i)}$, $\chi\in\{l,r\}$, that

[TABLE]

Along with the free product Hilbert space

[TABLE]

we need to also define, for $\chi\in\{l,r\}$, the free product Hilbert spaces

[TABLE]

Since there are multiple free product Hilbert spaces we will use subscripts to denote the different left and right representations, namely,

[TABLE]

for the left representations and

[TABLE]

for the right representations.

Dykema’s original result [2] proves that $\langle\cdot\xi_{\chi},\xi_{\chi}\rangle$ is faithful on $*_{i\in{\mathcal{I}}}({\mathcal{A}}_{\chi}^{(i)},\varphi)$ for $\chi\in\{l,r\}$ and it is a folklore result that the minimal tensor product of faithful states is faithful. Thus, $\langle\cdot\>\xi_{l}\otimes\xi_{r},\xi_{l}\otimes\xi_{r}\rangle$ is faithful on $*_{i\in{\mathcal{I}}}({\mathcal{A}}_{l}^{(i)},\varphi)\ \otimes_{min}\;*_{i\in{\mathcal{I}}}({\mathcal{A}}_{r}^{(i)},\varphi)$.

Fix $k\geq 1$ and $j_{1},\cdots,j_{k}\in{\mathcal{I}}$ such that $j_{i}\neq j_{i+1},1\leq i\leq k-1$. Now fix a unit vector

[TABLE]

If $k=1$ the only possible $h$ is $\xi=\xi_{j_{1}}=\xi_{j_{1},l}\otimes\xi_{j_{1},r}$. Call the collection of such $h$, as $k$ and the indices vary, ${\mathcal{S}}\subset{\mathcal{H}}$.

As will be shown below, this set of unit vectors ${\mathcal{S}}$ plays an important role in decomposing simple tensors in ${\mathcal{H}}$, in particular for every simple tensor $\eta\in{\mathcal{H}}$ that is also a simple tensor in each component there exists a unique $h\in{\mathcal{S}}$ such that $\eta\in{\mathcal{H}}_{l}\otimes h\otimes{\mathcal{H}}_{r}$. By abuse of tensor notation this is not very hard to see in one’s mind but the reality of proving this carefully needs plenty of indices.

To this end, for $m\geq 1$ suppose $s_{1},\dots,s_{m}\in{\mathcal{I}}$ such that $s_{t}\neq s_{t+1}$ for $1\leq t\leq m-1$, and $\eta_{t,l}\in{\mathcal{H}}_{s_{t},l}$, $\eta_{t,r}\in{\mathcal{H}}_{s_{t},r}$ such that $\eta_{t,l}\otimes\eta_{t,r}\in\mathring{{\mathcal{H}}}_{s_{t}}$ for $1\leq t\leq m$. This last condition implies that $\eta_{t,\chi}=\|\eta_{t,\chi}\|\xi_{t,\chi}$ cannot hold for both $\chi=l$ and $\chi=r$. In summary,

[TABLE]

Note that the conditions imposed on the $\eta_{t,\chi}$ in the above paragraph imply that the form of $\eta$ above is as reduced as it can be.

As mentioned above, it will be established that there exists $h\in{\mathcal{S}}$ such that

[TABLE]

To prove the required decomposition, let

[TABLE]

and

[TABLE]

This gives that $v$ is the number of terms in a row from the left with trivial right tensor components and $m+1-w$ is the number of terms in a row from the right with trivial left tensor components.

By the fact that $\eta_{t,l}\otimes\eta_{t,r}\in\mathring{{\mathcal{H}}}_{s_{t}}$, that is $\eta_{t,\chi}=\|\eta_{t,\chi}\|\xi_{t,\chi}$ cannot hold for both $\chi=l$ and $\chi=r$, we have that $v<w$. If $v=m$ then $w=m+1$ and $\eta\in{\mathcal{H}}_{l}$, and if $w=1$ then $v=0$ and $\eta\in{\mathcal{H}}_{r}$. Otherwise, when $0\leq v\leq m-1$ and $2\leq w\leq m$, define

[TABLE]

with $\eta_{\mathcal{S}}=\xi$ if $v+1=w$. Hence, by the usual slight abuse of the tensor notation, $\eta=\eta_{l}\otimes\eta_{\mathcal{S}}\otimes\eta_{r}\in{\mathcal{H}}_{l}\otimes\eta_{\mathcal{S}}\otimes{\mathcal{H}}_{r}$ with $\frac{1}{\|\eta_{\mathcal{S}}\|}\eta_{\mathcal{S}}\in{\mathcal{S}}$. Therefore,

[TABLE]

For any $h\in{\mathcal{S}}$, which is a unit vector, there is a natural isometric map $S_{h}:{\mathcal{H}}_{l}\otimes{\mathcal{H}}_{r}\rightarrow{\mathcal{H}}$ given by the concatenation ${\mathcal{H}}_{l}\otimes{\mathcal{H}}_{r}\mapsto{\mathcal{H}}_{l}\otimes h\otimes{\mathcal{H}}_{r}$ with the appropriate simplification of tensors when needed. In particular, there exist $k\geq 1$ and $j_{1},\cdots,j_{k}\in{\mathcal{I}}$ such that $j_{i}\neq j_{i+1},1\leq i\leq k-1$ and then

[TABLE]

We can now carefully specify that the isometric map is given by

[TABLE]

[TABLE]

[TABLE]

and

[TABLE]

if $i_{m}\neq j_{1}$ and $j_{k}\neq t_{1}$ with similar statements as the cases above when $i_{m}=j_{1}$ or $j_{k}=t_{1}$ or both happen. Perhaps the most natural case of $S_{h}$ is when $h=\xi$. It certainly minimizes, but doesn’t remove, the need for all of the cases above.

A careful examination of the $S_{h}$ isometric map implies that for $a\in{\mathcal{A}}_{l}^{(i_{1})}$, $b\in{\mathcal{A}}_{r}^{(i_{2})}$ and $\eta_{\chi}\in{\mathcal{H}}_{\chi}$ for $\chi\in\{l,r\}$ we have that, by abuse of the tensor notation,

[TABLE]

Hence, $S_{h}({\mathcal{H}}_{l}\otimes{\mathcal{H}}_{r})$ is a reducing subspace of the reduced bi-free product. Moreover,

[TABLE]

and

[TABLE]

Therefore, for any $h\in{\mathcal{S}}$,

[TABLE]

and furthermore, by the identities involving $S_{h},\lambda$ and $\rho$, $S_{h}^{*}aS_{h}=S_{\xi}^{*}(a)S_{\xi}$ for all $h\in{\mathcal{S}}$ and $a\in{**}_{i\in{\mathcal{I}}}({\mathcal{A}}_{l}^{(i)},{\mathcal{A}}_{r}^{(i)})$.

Finally, we want to show that compression to $S_{\xi}({\mathcal{H}}_{l}\otimes{\mathcal{H}}_{r})$ is a $*$-isomorphism. Note that this is the same as compression to $S_{h}({\mathcal{H}}_{l}\otimes{\mathcal{H}}_{r})$ being injective for any $h\in{\mathcal{S}}$. This gives us a way forward. Suppose that $a\in{**}_{i\in{\mathcal{I}}}({\mathcal{A}}_{l}^{(i)},{\mathcal{A}}_{r}^{(i)})$ such that ${**}_{i\in{\mathcal{I}}}\varphi_{i}(a^{*}a)=0$. This implies that

[TABLE]

By the faithfulness of $\langle\cdot\>\xi_{l}\otimes\xi_{r},\xi_{l}\otimes\xi_{r}\rangle$ this gives that $S_{\xi}^{*}aS_{\xi}=0$ or rather $a$ is 0 on the reducing subspace $S_{\xi}({\mathcal{H}}_{l}\otimes{\mathcal{H}}_{r})$. But then for all $h\in{\mathcal{S}}$ we have that

[TABLE]

and $a$ is 0 on the reducing subspace $S_{h}({\mathcal{H}}_{l}\otimes{\mathcal{H}}_{r})$. By what we proved about the set ${\mathcal{S}}$, we have that $a$ is 0 on

[TABLE]

Therefore, $a=0$ and thus ${**}_{i\in{\mathcal{I}}}\varphi_{i}$ is faithful. ∎

There may exist a full converse to Theorem 3.1 but the previous proof highly depends on the state $\varphi_{i}$ arising as a tensor product of states. In general, $\varphi_{i}$ need not be of this form. We should note here that if $\varphi_{i}|_{{\mathcal{A}}_{l}^{(i)}}$ or $\varphi_{i}|_{{\mathcal{A}}_{l}^{(i)}}$ is a pure state then $\varphi_{i}$ will be a tensor product of states.

To end this paper, we summarize with the following corollary.

Corollary 3.3**.**

Let $({\mathcal{A}}_{l}^{(i)},{\mathcal{A}}_{r}^{(i)})_{i\in{\mathcal{I}}}$ be a non-trivial family of pairs of faces in the non-commutative C∗-probability space $({\mathcal{A}},\varphi)$. If $\varphi$ is faithful on $C^{*}(({\mathcal{A}}_{l}^{(i)},{\mathcal{A}}_{r}^{(i)})_{i\in{\mathcal{I}}})$, $C^{*}({\mathcal{A}}_{l}^{(i)},{\mathcal{A}}_{r}^{(i)})\simeq{\mathcal{A}}_{l}^{(i)}\otimes_{min}{\mathcal{A}}_{r}^{(i)}$, $\varphi_{i}=\varphi_{i}|_{{\mathcal{A}}_{l}^{(i)}}\otimes\varphi_{i}|_{{\mathcal{A}}_{r}^{(i)}}$ and $({\mathcal{A}}_{l}^{(i)},{\mathcal{A}}_{r}^{(i)})_{i\in{\mathcal{I}}}$ is bi-freely independent with respect to $\varphi$, then

[TABLE]

Proof.

Recall, that by bi-free independence we know that the following diagram commutes

[TABLE]

Because both of the states are faithful on their algebras then for any $a^{*}a\in\check{*}_{i\in{\mathcal{I}}}({\mathcal{A}}_{l}^{(i)}\check{*}{\mathcal{A}}_{r}^{(i)})$, $a^{*}a$ is in the kernel of $\iota$ if and only if $a^{*}a$ is in the kernel of $*_{i\in I}(\lambda_{i}*\rho_{i})\circ*_{i\in{\mathcal{I}}}(\pi_{i}*\pi_{i})$. Therefore, both quotients are $*$-isomorphic and Theorem 3.2 gives the final $*$-isomorphism. ∎