# A question about correlations between $C^{*}$-algebras

I was studying J. M. G. Fell’s paper The Structure of Algebras of Operator Fields when I encountered the concept of a correlation between two $C^{*}$-algebras.

Definition. Let $A$ and $B$ be $C^{*}$-algebras. Then an $(A,B)$-correlation is a relation $R \subseteq A \times B$ such that there exist another $C^{*}$-algebra $C$ and surjective $*$-homomorphisms, $f: A \to C$ and $g: B \to C$, satisfying $$\forall a \in A, ~ \forall b \in B: \quad (a,b) \in R \iff f(a) = g(b).$$

In the same paper, another definition is given.

Definition. Let $A$ and $B$ be $C^{*}$-algebras. Then an $(A,B)$-correlation is a closed $C^{*}$-subalgebra of the direct product $C^{*}$-algebra $A \times B$ satisfying $$\{ a \in A \mid \exists b \in B: ~ (a,b) \in R \} = A \quad \text{and} \quad \{ b \in B \mid \exists a \in A: ~ (a,b) \in R \} = B.$$

It is very easy to see that the first definition implies the second one. However, I have difficulty in showing that the second implies the first. My difficulty lies chiefly in constructing the $C^{*}$-algebra $C$ required by the first definition. At first, I thought that $C$ would be $(A \times B) / \langle R \rangle$, where $\langle R \rangle$ denotes the closed two-sided ideal generated by $R$, but this does not appear right.

I suspect that to construct $C$, one must take the free $*$-algebra over the elements of $A$ and $B$, modulo the relations in $A$ and in $B$ and also the relations between elements of $A$ and elements of $B$ implemented by $R$. However, I am not sure if such an object has a faithful $C^{*}$-algebraic representation.

• For those reading along, these definitions appear on p. 9 of the paper. – Dylan Thurston May 9 '15 at 16:27
• Why is setting $C$ to be $(A \times B)/\langle R \rangle$ not right? – Dylan Thurston May 9 '15 at 16:44
• @Dylan: Hi Dylan. I’ve just worked out a solution (it’s been a rough night), and I’ll try to address your question in an answer that I’m formulating into a post. – Transcendental May 9 '15 at 16:58
• @Dylan: Let $C \stackrel{\text{df}}{=} (A \times B) / \langle R \rangle$, and suppose that we have surjective $*$-homomorphisms $f: A \to C$ and $g: B \to C$. Then the most natural definitions for $f$ and $g$ appear to be \begin{align} \forall a \in A: \quad f(a) & \stackrel{\text{df}}{=} (a,0_{B}) + \langle R \rangle, \\ \forall b \in B: \quad g(b) & \stackrel{\text{df}}{=} (0_{A},b) + \langle R \rangle. \end{align} However, this would mean that $f(a) g(b) = 0_{C} = g(b) f(a)$ for all $(a,b) \in A \times B$, which is an unrealistic condition on $f$ and $g$. – Transcendental May 9 '15 at 18:44

It turns out that I have figured it out. Here is a full solution.

Suppose that the hypotheses on $R$ in the second definition hold. Let $$I_{A} \stackrel{\text{df}}{=} \{ a \in A \mid (a,0_{B}) \in R \} \quad \text{and} \quad I_{B} \stackrel{\text{df}}{=} \{ b \in B \mid (0_{A},b) \in R \}.$$

Claim 1: $I_{A}$ and $I_{B}$ are closed two-sided ideals of $A$ and $B$ respectively.

Proof of Claim 1

As $R$ and $A \times \{ 0_{B} \}$ are $C^{*}$-subalgebras of $A \times B$, we see that $R \cap (A \times \{ 0_{B} \})$ is a $C^{*}$-subalgebra of $A \times B$ also. Then as $I_{A}$ is the image of $R \cap (A \times \{ 0_{B} \})$ under the projection $*$-homomorphism $$\pi: A \times B \to A,$$ it follows that $I_{A}$ is a $*$-subalgebra of $A$. To conclude that $I_{A}$ is a closed $*$-subalgebra of $A$, which would make it a $C^{*}$-subalgebra of $A$, use the well-known fact that any $*$-homomorphism from one $C^{*}$-algebra to another has a closed range.

It remains to show that $I_{A}$ is a two-sided ideal of $A$. To begin, let $a \in I_{A}$ and $x \in A$. By the hypotheses on $R$, there exists a $b \in B$ such that $(x,b) \in R$. Then as $R$ is a subalgebra of $A \times B$, we obtain $$(a x,0_{B}) = (a,0_{B}) (x,b) \in R \quad \text{and} \quad (x a,0_{B}) = (x,b) (a,0_{B}) \in R.$$ Therefore, $a x,x a \in I_{A}$, and as $x$ is arbitrary in $A$, we deduce that $I_{A}$ is indeed a closed two-sided ideal of $A$.

A similar argument works to show that $I_{B}$ is a closed two-sided ideal of $B$. $\quad \blacksquare$

Claim 2: $A / I_{A}$ and $B / I_{B}$ are isomorphic $C^{*}$-algebras.

Proof of Claim 2

Define a map $\Phi: A / I_{A} \to B / I_{B}$ by $$\forall a \in A: \quad \Phi(a + I_{A}) \stackrel{\text{df}}{=} b + I_{B},$$ where $b$ is any element of $B$ such that $(a,b) \in R$ (such a $b$ must exist). We now establish the following:

• $\Phi$ is well-defined. Let $(a,b),(a',b') \in R$, so that $(a - a',b - b') \in R$. Furthermore, suppose that $a - a' \in I_{A}$, i.e., $(a - a',0_{B}) \in R$. Then $$(0_{A},b - b') = (a - a',b - b') - (a - a',0_{B}) \in R,$$ so $b - b' \in I_{B}$.
• $\Phi$ is injective. Let $(a,b) \in R$, and suppose that $b \in I_{B}$, i.e., $(0_{A},b) \in R$. Then $$(a,0_{B}) = (a,b) - (0_{A},b) \in R,$$ so $a \in I_{A}$.
• $\Phi$ is surjective. Given any $b \in B$, the requirements on $R$ guarantee that there exists an $a \in A$ such that $(a,b) \in R$. Hence, $\Phi(a + I_{A}) = b + I_{B}$.

This concludes the proof of Claim 2. $\quad \blacksquare$

Claim 3: The required $C^{*}$-algebra $C$ is $B / I_{B}$.

Proof of Claim 3

Define a surjective $*$-homomorphism $f: A \to B / I_{B}$ by $f \stackrel{\text{df}}{=} \Phi \circ q_{A}$, where $q_{A}$ is the quotient map from $A$ to $A / I_{A}$. Similarly, define a surjective $*$-homomorphism $g: B \to B / I_{B}$ simply by $g \stackrel{\text{df}}{=} q_{B}$, where $q_{B}$ is the quotient map from $B$ to $B / I_{B}$.

Let $(a,b) \in A \times B$.

• If $(a,b) \in R$, then $$f(a) = \Phi({q_{A}}(a)) = \Phi(a + I_{A}) = b + I_{B} = g(b).$$
• Conversely, if $f(a) = g(b)$, then by the same relations above, we have $\Phi(a + I_{A}) = b + I_{B}$. From the definition of $\Phi$, there exists a $b' \in B$ satisfying $(a,b') \in R$ such that $b + I_{B} = b' + I_{B}$. Hence, $$(a,b) = (a,b') + \underbrace{(0_{A},b - b')}_{\text{ \in R , as  b - b' \in I_{B} }} \in R,$$ which concludes the proof of Claim 3. $\quad \blacksquare$