I read a paper in which the author uses the propetry that if $A$ is a submodule of a

Hilbert $C^\ast$ module ($C$ is a $C^*$ algebra) such that

$A^\bot=0$ then $A$ is dense. I don't know how to prove it since in a gneral Hilbert module over a $C^*$ algebra $(A^\bot)^\bot$ is greater than $Closure(A)$ and the standard Hilbert space proof does not work.

The paper in question is: UNBOUNDED OPERATORS ON HILBERT C∗-MODULES OVER C∗-ALGEBRAS OF COMPACT OPERATORS by Guljas you can find on the web.

  • 2
    $\begingroup$ In general, I agree with the answer by MTS. However, if you look at the paper you mention, then it only considers modules over the algebra of compact operators, and such things are rather special. See the 2nd paragraph of the paper: "These modules are characterized by the property that each closed submodule is orthogonally complemented or orthogonally closed" $\endgroup$ Oct 2 '11 at 20:23

Unless I am misunderstanding your question, I don't believe the statement is true in general. For example, take $C = C(X)$ for some $X$. Take your $C^*$-module to be $C$ itself with the standard inner product $\langle f, g \rangle = f^*g$. Let $x_0 \in X$ and define $$ A = \{ f \in C \mid f(x_0) =0 \}. $$ Then $A^\perp = 0$ but $A$ is not dense in $C$.


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.