# How "nondegenerate" are amalgamated free products of C*-algebras?

In the following, I assume all algebras are unital. Let $A$ and $B$ be C*-algebras that each contain (isomorphic copies of) a common C*-subalgebra $C$. Let $A *_C B$ denote the amalgamated free product of $A$ and $B$ over $C$, which is the pushout formed in the category of (unital) C*-algebras. In case $C = \mathbb{C}$ is the complex field, this yields the usual "maximal" free product of C*-algebras.

It is known that the naturally induced $*$-homomorphisms from $A$ and $B$ to $A *_C B$ are injective. For instance, see Theorem 3.1 of this paper by Blackadar or Theorem 4.2 of this paper by Pedersen.

On the other hand, we may form the "purely algebraic" pushout $A \circledast_C B$ in the category of (unital) rings. By its universal property, it is equipped with a canonical algebra homomorphism $$A \circledast_C B \to A *_C B.$$ The nondegeneracy result of Blackadar implies that the natural maps from $A$ and $B$ to $A \circledast_C B$ are also injective, which is already a nontrivial piece of information since there are many examples of (purely algebraic) amalgamated free products of complex algebras that trivialize.

But how much does $A *_C B$ "know" about the purely algebraic object $A \circledast_C B$?

Question: Is the natural map $A \circledast_C B \to A *_C B$ injective for all choices of $A$, $B$, and $C$?

I believe that $A *_C B$ may be constructed from $A \circledast_C B$ in a manner similar to the method for maximal free products described in the introduction of this paper by Avitzour, by defining $\|x\|$ for $x \in A \circledast_C B$ to be the supremum of the operator norm of $x$ over all Hilbert space $*$-representations and completing with respect to $\|\bullet\|$. If this defines an honest norm, then I can see that the answer to the question would be affirmative because the algebra will embed in its completion.

However, do we know that each $x \in A \circledast_C B$ truly acts nontrivially in some Hilbert space $*$-representation? It is not clear to me how to deduce this from the nondegeneracy proofs of Blackadar or Pedersen. I don't even see that this is the case for the free product when $C = \mathbb{C}$ as in Avitzour's paper, where he actually refers to $\|\bullet\|$ as a norm.

• Possibly relevant: arxiv.org/abs/1307.5609
– YCor
Feb 3, 2016 at 22:37
• In "The K-groups of free products of C$^*$-algebras" Cuntz says you must divide the algebraic free product by the null space of $\|\cdot\|$ but does not provide an example of when this is non-trivial. So either he knew of a degenerate example or your question was unknown in 1982. Feb 20, 2016 at 14:36
• @ChrisRamsey Thank you for that observation! Tracing back the references in Cuntz's paper, I see that the following, earlier paper of L.G. Brown refers to $\| \cdot \|$ a seminorm (but doesn't seem to discuss the question of whether it is truly a norm): theta.ro/jot/archive/1981-006-001/1981-006-001-012.pdf Feb 23, 2016 at 21:36
• The mystery deepens: Paschke in his review of Avitzour's paper (ams.org/mathscinet-getitem?mr=654842) declares that Blackadar's Theorem 3.1 proves non-degeneracy. However, I do not believe this statement. Mar 8, 2016 at 18:26
• @ChrisRamsey Your continued historical notes are most appreciated! Somehow I hadn't noticed Paschke's claim, but I agree with you that this does not seem to follow from Blackadar's proof (or at least not immediately). At this point I'm even curious to hear if anyone has suspicions about whether non-degeneracy should hold, both in case $C = \mathbb{C}$ and in general. Mar 8, 2016 at 19:57

If $\varphi$ and $\psi$ are faithful states on $A$ and $B$, respectively, then the proposition in subsection 2.3 in Avitzour proves that $A \circledast_\mathbb C B$ injects into the reduced free product of $(A,\varphi)$ and $(B,\psi)$. This is obtained by showing that the cyclic vector belonging to the state $\varphi * \psi$ is separating.
I can imagine that one can obtain a similar result when there are faithful expectations onto $C$ in the amalgamated case.