# Definition of center of ternary ring of operators

Let $$H$$ and $$K$$ be Hilbert spaces and $$B(H,K)$$ denotes the space of bounded operators from $$H$$ to $$K$$. Recall that a ternary ring of operators (TRO) $$V$$ is a closed subspace of $$B(H,K)$$ which is closed under the operation $$(x,y,z) \to xy^*z$$. Moreover, $$V$$ is called commutative if $$ab^*c=cb^*a$$ for all $$a,b,c \in V$$.

Is there any definition of center of TRO In literature?

• Where is your definition of "commutative" for a TRO from? Jun 13, 2020 at 8:44
• @MatthewDaws: Page 340 of msp.org/pjm/2003/209-2/pjm-v209-n2-p10-p.pdf Jun 13, 2020 at 8:53

What about defining $$C = \{ v\in V : av^*c = cv^*a \ (a,c\in V) \}.$$ This is evidently a closed linear subspace of $$V$$. Then, given $$d,e,f\in C$$ and $$a,c\in V$$, $$a(de^*f)^*c = (af^*e)d^*c = cd^*(af^*e) = cd^*(ef^*a) = c(d^*ef^*)a = c(f^*ed^*)a = c(de^*f)^* a$$ using that $$d\in C$$, then $$f\in C$$, then $$e\in C$$. Thus $$de^*f\in C$$, so $$C$$ is a sub-TRO of $$V$$, and clearly $$C$$ is commutative. If $$V$$ were itself commutative, then $$C=V$$. This seems like a reasonable definition of "center" to me.
• If $V$ is $C^{\ast}-$Algebra, then it is clear that $C$ is contained in usual Center but it’s not clear whether usual Center is part of $C$ or not? Jun 13, 2020 at 8:10
• Yes, I think I agree. Given that every $C^*$-algebra is a TRO, but not conversely, why do you really expect a notion of "centre" for a TRO to agree with that for a $C^*$-algebra? As I said, I don't have a reference... Jun 13, 2020 at 8:43
• If they don’t agree then there would be two different notions of center for $C^{\ast}-$ Algebras. In that sense, definition of center like you defined won’t really be a good choice. Jun 13, 2020 at 8:51