# Counterexample to Riesz representation for Hilbert modules

For a Hilbert space $H$, the Riesz representation theorem states that $H$ is isomorphic to its dual $H^*$ via $x \mapsto \langle x, -\rangle$.

It is often stated in the literature that this does not work in full generality for a Hilbert module over a $C^*$ algebra. For example, attempts to define the adjoint of a morphism between Hilbert modules run into the lack of such a representation theorem. To avoid this, we insist that a morphism between Hilbert modules admit an adjoint.

Is there a simple counterexample to Riesz representability for Hilbert modules?

To be more precise, take $E$ an Hilbert module over $A$ and $\phi : E \to A$ being $A$ linear. The question is on the existence of an $x$ in $E$ such that $\phi = \langle x, - \rangle$.

• What do you mean by a representation of $f\in H^\ast$? For a Hilbert module the "scalar product" maps $H\times H$ into the $C^*$-algebra. – Jochen Wengenroth Jan 25 '17 at 8:03
• edited. Hope it is clear now. I also found a counterexemple : Take a unital C∗ algebra A an a closed ideal I in A. The inclusion I⊂A, which is A linear, can be written as ⟨x,−⟩ only if x∈I is the unit of I. This is equivalent to the splitting A≃I⊕A/I and is not true in full generality. – InfiniteLooper Jan 25 '17 at 8:20

Take $A= \mathcal{C}([0,1])$ and $H$ the ideal of $A$ of functions that vanish at $0$.
$H$ is a Hilbert $A$ module (as any ideal, with the natural multiplication of $A$ and the scalar product $(x,y)=x^*y$ of $xy^*$ depending on if you are talking of right or left modules) the inclusion of $H$ into $A$ is a continuous $A$ linear map and it has no adjoint.
Indeed an adjoint would be a map $p$ from $A$ to $H$ such that for all $y$ in $H$ and $x$ in $A$, $y^* p(x) = y^*x$ hence $p(1)$ would be an element of $H$ which is a unit for the ideal $H$, that does not exists.