# Complemented submodules of a Hilbert C*-module

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.

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.
• 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). Aug 20, 2021 at 20:40