8
$\begingroup$

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 $\phi = \langle x, - \rangle$.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Jochen Wengenroth Jan 25 '17 at 8:03
  • 2
    $\begingroup$ 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. $\endgroup$ – InfiniteLooper Jan 25 '17 at 8:20
5
$\begingroup$

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.

$\endgroup$

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.