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 $$\|[a,b,c]\| \leq \| a \| \| b\|\| c\|$$



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> Does there exist a unique $C^\*$-algebra corresponding to each ternary $C^\*$-ring?
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$3.2$ Proposition of [this](https://reader.elsevier.com/reader/sd/pii/000187088390083X?token=01B8E5AC9D0B68D9B1C3816A5724216F3593567C5DB7B706F1E7859093E031641858701A9AA2ECD58D273F5FFED0E1D4&originRegion=eu-west-1&originCreation=20210901064606) 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.



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> Can someone please explain me why do we need to consider opposite algebra and the motivation behind this proof?
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Trying to understand construction of $C^*$-algebra corresponding to a ternary $C^*$-ring from a paper
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.)


