Return to Answer

According to Zettl [1], a ternary ring of operators (TRO) is a ternary $C^*$-ring which is isomorphic to a closed subset $X\subseteq B(H)$, such that $XX^*X\subseteq X$, and is 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 an 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.

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.

According to Zettl [1], a ternary ring of operators (TRO) is a ternary $C^*$-ring which is isomorphic to a closed subset $X\subseteq B(H)$, such that $XX^*X\subseteq X$, and is 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 an 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!

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

According to Zettl [1], a ternary ring of operators (TRO) is a ternary $C^*$-ring which is isomorphic to a closed subset $X\subseteq B(H)$, such that $XX^*X\subseteq X$, and is 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 an 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.

For a counter-example let $X$ be a ternary $C^*$-ring which is an operator space (e.g. any closed subspace $X$ of $B(H)$ such that $XX^*X\subseteq X$) and define a new ternary operation by $$\{x,y,z\}=-[x,y,z].$$ It is a fundamental result of Zettl [1] that ternary $C^*$-rings decompose uniquely as a "positive" part and a "negative" part.

It is interesting to remark that you do not change the structure of an algebra by inserting a minus sign in front of the product because 2 is even, but you do change a ternary $C^*$-ring by doing so because 3 is odd!

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

According to Zettl [1], a ternary ring of operators (TRO) is a ternary $C^*$-ring which is isomorphic to a closed subset $X\subseteq B(H)$, such that $XX^*X\subseteq X$, and is 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 an 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!

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