# Example of a ternary $C^{\ast}$-ring which is not an operator space

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.

• Is the middle condition in the definition of associativity really supposed to have $b$ and $d$ switched? Oct 28 '21 at 10:13
• @LSpice: Yes. These are also known as ternary algebras of 2nd kind. In $1$st kind the associativity condition is the natural one. Oct 28 '21 at 11:04
• Your question is not well posed because any Banach space can be realized as an operator space, via the MIN or MAX constructions Oct 28 '21 at 13:56
• @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. Oct 28 '21 at 23:54
• @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 $$[x,y,z]=xy^*z.$$
– 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.

• 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. Oct 29 '21 at 2:26
• @WlodAA, thanks! I have deleted my wrong comments about parity. Oct 29 '21 at 11:51
• @LSpice, I've edited my answer hopping to clarify my point. Indeed my original answer was a bit terse!
– Ruy
Oct 29 '21 at 14:31
• 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. Oct 29 '21 at 17:00
• @LSpice, the last paragraph in my latest edit might help one to see what is wrong with anti-TRO's.
– Ruy
Oct 29 '21 at 23:31