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?


1 Answer 1


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.

I don't know of a reference...

  • $\begingroup$ 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? $\endgroup$
    – Math Lover
    Jun 13, 2020 at 8:10
  • $\begingroup$ 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... $\endgroup$ Jun 13, 2020 at 8:43
  • $\begingroup$ 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. $\endgroup$
    – Math Lover
    Jun 13, 2020 at 8:51

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