1
$\begingroup$

Let $A$ be a $C^*$-algebra and $E$ be a (right) Hilbert $C^*$-module over $A$. Assume $F$ is a closed submodule of $E$ such that $F^\perp := \{x \in E: \langle x, F\rangle=0\}$ is orthogonally complemented, i.e. we have $F^\perp \oplus F^{\perp \perp} = E.$ Can we conclude that $F$ is orthogonally complemented, and if this is the case do we have $F= F^{\perp \perp}$?

I know that in general we don't need to have $F= F^{\perp \perp}$. However, when $F$ is orthogonally complemented, i.e. $F\oplus F^\perp = E$ this is obvious. For Hilbert spaces (i.e. $A = \mathbb{C}$), there are no counterexamples.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

If $F^\perp =\{0\}$ then $F^\perp \oplus F^{\perp\perp} = E$ is automatic, so all you need is an example of this where $F \neq E$. For instance, $C[0,1]$ as a Hilbert module over itself with $F =$ the functions which vanish at 0.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thanks! Nice example. I should keep this in mind. $\endgroup$
    – Andromeda
    Commented Aug 20, 2021 at 20:35
  • 3
    $\begingroup$ Yeah, I guess the first places I look for Hilbert module counterexamples are when $E$ is commutative, or when $E$ is a module over itself (or, in this case, both). $\endgroup$
    – Nik Weaver
    Commented Aug 20, 2021 at 20:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .