0
$\begingroup$

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?

$\endgroup$
2
1
$\begingroup$

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...

$\endgroup$
3
  • $\begingroup$ 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 '20 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 '20 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 '20 at 8:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.