A ternary $C^{\ast}$-ring is a complex Banach space $X$, equipped with a ternary product $[\cdot,\cdot,\cdot]:X^3 \to X$ which is linear in outer variables and conjugate linear in middle variable. Also $X$ is associative i.e. $$[[a,b,c],d,e]=[a,[d,c,b],e]=[a,b, [c,d,e]].$$ Moreover, $\lVert[a,a,a]\rVert= \lVert a\rVert^3$ and $\lVert[a,b,c]\rVert \leq \lVert a \rVert \lVert b\rVert\lVert c\rVert$.

Does there exist a ternary $C^{\ast}$-ring which is not an operator space?

One obvious class of examples of ternary $C^{\ast}$-rings is the class of ternary rings of operators but they all are operator spaces.

  • $\begingroup$ Is the middle condition in the definition of associativity really supposed to have $b$ and $d$ switched? $\endgroup$
    – LSpice
    Oct 28 '21 at 10:13
  • $\begingroup$ @LSpice: Yes. These are also known as ternary algebras of 2nd kind. In $1$st kind the associativity condition is the natural one. $\endgroup$
    – Math Lover
    Oct 28 '21 at 11:04
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    $\begingroup$ Your question is not well posed because any Banach space can be realized as an operator space, via the MIN or MAX constructions $\endgroup$
    – Yemon Choi
    Oct 28 '21 at 13:56
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    $\begingroup$ @LSpice I don't know off the top of my head, but that was not the question which was asked. As with some of the OP's history of questions, I start to find the lack of precision troubling. $\endgroup$
    – Yemon Choi
    Oct 28 '21 at 23:54
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    $\begingroup$ @YemonChoi, when the OP refers to an "operator space" in the context of ternary C*-rings, they mean a closed subspace $X\subseteq B(H)$ such that $XX^*X\subseteq X$, and equipped with the ternary product $$[x,y,z]=xy^*z.$$ $\endgroup$
    – Ruy
    Oct 29 '21 at 0:44

According to Zettl [1], a ternary ring of operators (TRO) is a ternary $C^*$-ring which is isomorphic to a closed subspace $X\subseteq B(H)$, such that $XX^*X\subseteq X$, equipped with the ternary multiplication $$ [x,y,z] := xy^*z. $$ On the other hand, an anti-TRO is a ternary $C^*$-ring defined as above, except that the multiplication operation is $$ [x,y,z] := -xy^*z. $$ It is a fundamental result of Zettl [1] that every ternary $C^*$-ring $X$ decomposes uniquely as $$ X=X_+\oplus X_-, $$ where $X_+$ is a TRO, and $X_-$ is an anti-TRO .

It seems to me that the reading of the question posed by the OP that makes the most sense is by taking the expression "operator space" to mean a TRO. In this case the answer is yes, there does exist a ternary $C^*$-ring which is not a TRO: just take any non-zero anti-TRO. For an even more concrete example, take $X=M_{n\times m}({\bf C})$, with ternary multiplication $[x,y,z] := -xy^*z$.

On the other hand, if one takes the expression "operator space" for its face value, Zettl's result implies that every ternary $C^*$-ring is an operator space in an even more canonical form than suggested by user @YemonChoi: write $X=X_+\oplus X_-$, embedd $X_+$ in $B(H_+)$, and $X_-$ in $B(H_-)$, whence $$ X\subseteq B(H_-\oplus H_+). $$ This embeding preserves the operator space structure (norms on matrix algebras) that a TRO canonical possesses.

It is interesting to remark that if you change the (binary) multiplication operation on a $C^*$-algebra by $$ x\circ y := -xy, $$ then the resulting object is strictly speaking a new C*-algebra, but it is isomorphic to the old one. The isomorphism is simply $a\mapsto -a$.

However, if you change the (ternary) multiplication on a ternary $C^*$-ring by inserting a minus sign as above, then the map $a\mapsto -a$ is no longer an isomorphism, essentially because 2 is even and 3 is odd! Indeed, Zettl's uniqueness result tells you that the new ternary $C^*$-ring might not be isomorphic to the old one at all!

EDIT: Here are some details of Zettl's proof which might shed some light into the reason an anti-TRO not isomorphic to a TRO.

Given a ternary $C^*$-ring $X$, let $A$ be the closed linear span within $B(X,X)$ (bounded operators on $X$) of the set of operators of the form $$ T_{y, z}:x\in X\mapsto [x,y,z]\in X, $$ as $y$ and $z$ range in $X$. It is easy to see that $A$ is a Banach algebra, and Zettl proves that $A$ is indeed a $C^*$-algebra for a unique involution operation "$^*$" satisfying $$ T_{y, z}^* = T_{z, y}. $$

Given this, it is clear that an operator of the form $T_{y,y}$ is self-adjoint but the key question is whether or not this is moreover positive.

If $X$ is a TRO, then $T_{y, y}\geq 0$, while in the anti-TRO case, one has that $T_{y, y}\leq 0$.

In other words, the positivity of $T_{y, y}$ is a signature of TRO's not shared by their anti-TRO cousins.

[1] Zettl, Heinrich, A characterization of ternary rings of operators, Adv. Math. 48, 117-143 (1983). ZBL0517.46049.

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    $\begingroup$ Is it clear that this does not embed in some (possibly different) operator algebra? Maybe this is what you mean by decomposing uniquely into a positive part and a negative part, but I don't know what that means. $\endgroup$
    – LSpice
    Oct 29 '21 at 2:26
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    $\begingroup$ @WlodAA, thanks! I have deleted my wrong comments about parity. $\endgroup$
    – LSpice
    Oct 29 '21 at 11:51
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    $\begingroup$ @LSpice, I've edited my answer hopping to clarify my point. Indeed my original answer was a bit terse! $\endgroup$
    – Ruy
    Oct 29 '21 at 14:31
  • $\begingroup$ Thanks! This definitely makes clear that you might obtain a new algebra from the old one in a particularly simple fashion. But is it clear that this new, different algebra does not embed in some new, different operator algebra? I think I must be misreading your description, because it seems to me that you are saying that every ternary $C^*$-ring does embed in an operator algebra in a particularly rigid way. $\endgroup$
    – LSpice
    Oct 29 '21 at 17:00
  • $\begingroup$ @LSpice, the last paragraph in my latest edit might help one to see what is wrong with anti-TRO's. $\endgroup$
    – Ruy
    Oct 29 '21 at 23:31

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