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Given a Hilbert space $\mathcal{H}$.
Regard the bounded operators $\mathcal{B}:=\mathcal{B}(\mathcal{H})$ on $\mathcal{H}$.
Consider a sub-C*-algebra $A\subseteq\mathcal{B}$.

Then I guess it holds: $$\overline{\mathcal{A}\mathcal{B}}=\mathcal{B}\iff A'=\mathbb{C}1\iff\overline{\mathcal{A}\mathcal{H}}=\mathcal{H}$$ How could I check this?

I was thinking about applying Hilbert-C*-module theory:
Any C*-algebra is prototypically itself a Hilbert-C*-mooule.
In particular, the bounded operators become a Hilbert-C*-module.
However there may be obstacles in Hilbert-C*-modules.

Similar statement: Wegge-Olsen, Lemma 2.2.1*

C-star-W-star
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1 Answers1

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For a counter-example to $AB = B \implies A' = \mathbb{C}$, take $A$ to be $\mathbb{C}1$. For a counter-example to the converse, take $A = K(H)$, the compact operators, which form a norm-closed ideal in $B(H)$.

So the first two properties you listed aren't related by any implications. The scalars ($\mathbb{C}1$) also give a counter-example for the implication that $AH = H \implies A' = \mathbb{C}$. So the only implication that might hold is $A' = \mathbb{C} \implies AH$ is dense in $H$.

The property that $A' = \mathbb{C}$ is equivalent to saying that the identity representation of $A$ on $H$ is irreducible (see here), so we must have $\overline{AH} = H$, since $\overline{AH}$ is a nonzero $A$-invariant subspace of $H$.

More generally, a similar argument shows that if $A$ is irreducible, then we even have that $A\xi$ is dense in $H$ for any $\xi \in H$, $\xi \neq 0$.

Josh Keneda
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