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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.

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  • $\begingroup$ For those reading along, these definitions appear on p. 9 of the paper. $\endgroup$ – Dylan Thurston May 9 '15 at 16:27
  • $\begingroup$ Why is setting $C$ to be $(A \times B)/\langle R \rangle$ not right? $\endgroup$ – Dylan Thurston May 9 '15 at 16:44
  • $\begingroup$ @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. $\endgroup$ – Transcendental May 9 '15 at 16:58
  • $\begingroup$ @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 $. $\endgroup$ – Transcendental May 9 '15 at 18:44
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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 $
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