Let $H$ be a complex Hilbert space and let a group $G$ act on $H$ such that there are no invariant closed subspaces besides $H$ and $(0)$. Let $D$ be the ring of bounded operators which commute with the $G$ action. What can we say about $D$? What more can we say if

(1) $G$ is unitary or

(2) We assume the answer to the invariant subspace problem is "yes".

Some observations:

If $D$ is a division algebra, it is $\mathbb{C}$. Let $\theta \in D$.. By a standard lemma, the spectrum of $\theta$ is nonempty, so there is some $\lambda \in \mathbb{C}$ for which $\theta - \lambda \mathrm{Id}$ is not invertible. But every nonzero element of a division algebra is invertible, so $\theta - \lambda \mathrm{Id} = 0$ and $\theta = \lambda \mathrm{Id}$. We have shown that an arbitrary element of $D$ is a scalar. $\square$

One can modify this argument to show that any real division algebra of bounded operators is $\mathbb{R}$, $\mathbb{C}$ or $\mathbb{H}$. This argument shows that every element in $D$ is algebraic over $\mathbb{R}$, and a real division algebra in which every element is algebraic over $\mathbb{R}$ is one of these three. The proof of Frobenius' theorem in Wikipedia is easily modified to show this.

However, the usual proof that $D$ should a division algebra does not apply. The usual argument is that, if $\theta \in D$ were not injective, then $\mathrm{Ker}(\theta)$ would be an invariant subspace and, if $\theta$ were not surjective, then $\mathrm{Im}(\theta)$ would be an invariant subspace. But I am only requiring that there are no closed invariant subspaces, and there is no reason $\mathrm{Im}(\theta)$ has to be closed.

Indeed, if the invariant subspace problem is false then $D$ doesn't have to be a division algebra. Let $T:H \to H$ be an invertible bounded operator and let $\mathbb{Z}$ act on $H$ by $T^i$. Then there are no invariant subspaces and $T \in D$, so $D \supsetneq \mathbb{C}$.

Motivations: Thinking about this question and this one.