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Let $V$ be a TRO (closed subspace of $B(H,K)$, closed under the product $xyz\to xy^*z$). Let $C(V)$, $D(V)$ denote the $C^{\ast}$-algebra generated by $VV^{\ast}$ and $V^*V$ respectively. We define $A(V)$, the linking $C^*$-algebra of $V$ as follows:

$$A(V) = \begin{bmatrix} C(V) & V\\ V^* & D(V) \end{bmatrix}$$ Using this linking $C^*$-algebra one obtains a functor from category of TROs to the category of $C^*$-algebras using which one studies representation theory, nuclearity, exactness and ideal theory of TROs. Moreover, One can show that two TROs are isomorphic if and only if their corresponding linking $C^*$-algebras are isomorphic.

Recall that a ternary Banach algebra is a complex associative Banach space $A$, equipped with a ternary product $[.,.,.]:A^3 \to A$ which is linear in outer variables and conjugate linear in middle variable and $$\|[a,b,c]\| \leq \| a \| \| b\|\| c\|$$ Since TROs are obvious examples of ternary Banach algebra. This motivates me to ask following:

Does there exist $C^*$-algebra corresponding to each ternary Banach algebra "similar" to the one we have for TROs?

Edit: This paper (section $3$) discusses what i'm looking for but unfortunately I don't get any intuition for the given construction of $C^*$-algebra. I would be glad if someone can explain me the construction given in the paper.

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    $\begingroup$ How general do you want your ternary Banach algebras to be? There exist (commutative!) Banach algebras A such that the only homomorphism from A to B(H) is the zero map, so that even if one tries to construct some kind of functorial "enveloping Cstar algebra" for Banach algebras, the resulting object might not detect any of the original structure of A. I imagine that things will only be worse in the setting of general ternary Banach algebras $\endgroup$
    – Yemon Choi
    Commented Jun 20, 2021 at 14:10
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    $\begingroup$ Indeed, given your opening paragraph, which discusses how a TRO can be analyzed in terms of an associated Cstar algebra, surely the most natural parallel for ternary Banach algebras is to try and analyse a "TBA" in terms of some associated Banach algebra? $\endgroup$
    – Yemon Choi
    Commented Jun 20, 2021 at 14:12
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    $\begingroup$ @YemonChoi: I was thinking in the setting of general ternary Banach algebras. After some googling I found out this paper which discusses what I am looking for but unfortunately I don't get any intuition for the construction given in the paper. $\endgroup$
    – Math Lover
    Commented Jun 20, 2021 at 15:59

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This is far from an area I am an expert in, but I would keep in mind the following class of examples.

Let $A$ be any Banach $*$-algebra. So $A$ is a Banach algebra, and $A$ is a $*$-algebra, and the $*$-operation is continuous. Then we can renorm $A$ to make the $*$-operation isometric: $\|a^*\| = \|a\|$. For example, $A$ could be a $C^*$-algbera; or something well-behaved but non-$C^*$, like $L^1(G)$ for a locally compact group $G$; or something very degenerate, like a complex Banach space $E$ with an involution, given the zero product.

Turn $A$ into a ternary Banach algebra by defining the triple product as $[a,b,c] = ab^*c$. We now see that it's going to be hopeless to study $A$ using just $C^*$-algebras. For example, in the language of the paper you link to, in the highly degenerate case $A$ may not have any non-zero $C^*$-seminorms.

As Yemon says in the comments, in some cases it might be possible to build a "linking algebra" in the framework of Banach $*$-algebras: my guess is that maybe this would work for the $L^1(G)$ example. However, we should keep in mind that a general Banach algebra can have rather little structure, and the same will be true of ternary Banach algebras.

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    $\begingroup$ Thank you Matthew. I think thats why the author in the linked paper assumes $$\|[a,a,a]\| = \| a \| ^3$$ in the definition of Ternary Banach algebra to restrict to the appropriate class of Ternary Banach Algebras? $\endgroup$
    – Math Lover
    Commented Jun 22, 2021 at 9:19
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    $\begingroup$ Yes, the paper you link says that this is a "$C^*$-seminorm": I guess it's like the different between a general Banach $*$-algebra, and the very special $C^*$-norm condition $\|a^*a\| = \|a\|^2$. $\endgroup$ Commented Jun 22, 2021 at 10:26

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