# Trying to understand construction of $C^*$-algebra corresponding to a ternary $C^*$-ring from a paper

Recall that a ternary $$C^*$$-ring is a complex Banach space $$X$$, equipped with a associative ternary product $$[.,.,.]:X^3 \to X$$ which is linear in outer variables and conjugate linear in middle variable, $$\|[a,a,a]\|= \|a\|^3$$ and $$\|[a,b,c]\| \leq \| a \| \| b\|\| c\|$$

Does there exist a unique $$C^*$$-algebra corresponding to each ternary $$C^*$$-ring?

$$3.2$$ Proposition of this paper seems to answer what i am looking for. The construction given in paper goes as follows: For $$y,z \in X$$, consider the bounded linear map $$D_{y,z}: X \to X$$ defined as $$D_{y,z}(x)= [x,y,z]$$. Let $$V=$$ span $$\{D_{y,z}: y,z \in X \} \subset L(X)$$. After defining like this, the author shows that $$V$$ is pre $$C^*$$-algebra. Finally, author considers the opposite algebra of the norm closure of $$V$$ to construct the required $$C^*$$-algebra.

Can someone please explain me why do we need to consider opposite algebra and the motivation behind this proof?

• The title could perhaps be a bit more explicit: It suggests to me more a general question on how to read a difficult proof, a question of Shakespearean nature in my opinion. Sep 1 at 7:10
• @RolandBacher: Edited, thanks! Sep 1 at 7:18
• You can also complete $V$ and obtain a $C^*$-algebra. The opposite $C^*$-algebra of this $C^*$-algebra is then simply the $C^*$-algebra constructed in the paper. Sep 1 at 8:44
• Your terminology is not correct, if you look closely at the Zettl paper that you link to. What you call a "ternary Banach algebra" is what Zettl calls a "ternary Cstar ring", and the key point is the axiom $\Vert [x,x,x,] \Vert= \Vert x\Vert^3$ which is an analogue of the Cstar condition that is one of the axioms for a Cstar algebra. Calling these objects "ternary Banach algebras", as you have done, is misleading since that name should be reserved for an object which does not have the "3-variable Cstar condition" Sep 2 at 2:29
• It would be good if you could fix the wording of your title and your question. Note also that you have written that X is a "complex associative Banach space" which does not make sense. Sep 2 at 2:30

Yes, Proposition 3.2 from the article [MR0700979] that you link to gives you the answer for ternary $$\rm C^*$$-rings. The reason for taking the opposite algebra is that the algebra $$V$$ acts on the right of $$X$$ rather than on the left. A right action is the same thing as a left action by the opposite algebra. You can see this right action directly in (3) of this proposition.

As a special case, suppose $$X$$ is a right Hilbert $$\rm C^*$$-module over a $$\rm C^*$$-algebra $$A$$. Denoting the $$A$$-valued inner product on $$X$$ by $$\langle x|y\rangle_A$$ (which is linear on the right), we get a ternary $$\rm C^*$$-ring by $$[x,y,z] := x\langle y|z\rangle_A$$. In this setting, $$D_{y,z}$$ is right multiplication by $$\langle y|z\rangle_A$$. All ternary $$\rm C^*$$-rings are essentially like this one, except that the $$A$$-valued form need not be positive. (See the intro of the cited article, bottom of p118 after the displayed equation.)