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:


1. Is it true that $A\oplus B = \frac{A\hat\ast B}{(A,B)}$
where $(A,B)$ is the closed ideal generated by $\langle ab-ba: a\in A, b\in B\rangle$
2. 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.



Uniqueness of the direct sum of $C^*$ algebras as quotient of free products
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](https://en.wikipedia.org/wiki/Direct_sum#Direct_sum_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](https://www.ams.org/journals/tran/1982-271-02/S0002-9947-1982-0654842-1). 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?


