Uniqueness of the direct sum of $C^*$ algebras as quotient of free products Suppose that you have $A, B$ two unital $C^*$ algebras and let $A \ast B$ the reduced free product (I think that it is the reduced amalgamated product over the common $*$-subalgebra $\mathbb{C} 1$) and let $A\hat{*} B$ be the full free product.
The questions are:

*

*Is it true that $A\oplus B = \frac{A\hat\ast B}{(A,B)}$
where $(A,B)$ is the closed ideal generated by $\langle ab-ba: a\in A, b\in B\rangle$


*Is $\frac{A\hat\ast B}{(A,B)} \cong \frac{A \ast B}{(A,B)}$. If it is not, is it right to say that there are many direct sums? What defines a direct sum?
Thank you in advance.
 A: This question initially confused me, as "of course" the direct sum of $C^\ast$-algebras is simply the vector space direct sum, with the max norm.  However, this is not the direct product of rings exactly because of the unital issue: the natural maps $A\rightarrow A\oplus B$ and $B\rightarrow A\oplus B$ are not unital.
I must say that, in my humble opinion, while it can be very profitable to seek inspiration from ring theory, it is wrong just to think of $C^\ast$-algebras as just being special sorts of rings.  There is a lot of interesting, extra structure: for example, that $C^\ast$-algebras might naturally be non-unital, with many different, interesting notions of unitisation.
So, let me answer a slightly different question to the one you asked.  Initially follow Paulsen's presentation, which is to form the (perhaps amalgamated) free product of algebras, and then the take the supremum over all $C^\ast$-representations.  If we take the full free product, not amalgamated over anything, then we do not obtain a category-theoretic coproduct, exactly because the map $A\rightarrow A \ast B$ is not unital.
So, we need to amalgamated over units, then take the supremum over all $C^\ast$-representations, say obtaining $A \hat\ast_1 B$.  This has the universal property that given a unital $C^*$-algebra $C$ and unital $*$-homomorphisms $\phi_A:A\rightarrow C, \phi_B:B\rightarrow C$, there is a unique unital $*$-homomorphism $A \hat\ast_1 B \rightarrow C$ extending $\phi_A, \phi_B$.  If we now quotient by the ideal generated by $\{ ab-ba : a\in A,b\in B \}$ then we force commutativity of the factors.  This corresponds to the analogous universal property where the ranges of $\phi_A$ and $\phi_B$ commute.  This should remind us of tensor products: specifically, the maximal tensor product of $C^*$-algebras, which has the same universal property.  So we construct $A \otimes_\max B$.
The free product construction considered in the book of Brown and Ozawa is common in free-probability theory, and I think originates with the nice paper of Avitzour.  We need an extra ingredient: faithful states $\mu_A, \mu_B$ on $A,B$ respectively.  Then essentially we look at the GNS spaces for these states and take a free-product like construction of the Hilbert spaces to obtain a potentially smaller representation of $A \hat\ast_1 B$ say $A \star_1 B$.
If we now quotient by the ideal, examining the construction of Avitzour shows that we obtain $A\otimes B$,
represented on the tensor product of the GNS spaces.  Completing gives the minimal or spatial tensor product
$A\otimes_\min B$.
So, yes, there are multiple ways to carry out your construction, basically corresponding to the fact that there are multiple tensor norms on (non-nuclear) $C^\ast$-algebras.  Again, I view this as "a feature, not a bug".
Follow-up question: Avitzour says that "there seems no natural way to define a free product of representations of two $C^\ast$-algebras..." and then gives up on an analogue of the spatial tensor product when you don't have chosen faithful states.  I wonder if any progress has been made in this direction?
