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](https://en.wikipedia.org/wiki/Invariant_subspace_problem) is "yes".




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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](https://en.wikipedia.org/wiki/Frobenius_theorem_%28real_division_algebras%29) 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 $D \supsetneq \mathbb{C}$.




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Motivations: Thinking about [this question](https://mathoverflow.net/questions/212950)
("Generalization of a theorem of Burnside to non-compact group")
and [this one](https://mathoverflow.net/questions/188022)
("Schur's lemma for antiunitary operators on complex Hilbert spaces").



Schur's Lemma for Hilbert spaces
To elaborate on my comment, let us suppose that $G$ is closed under taking adjoints (in particular, this holds if $G$ is unitary). Then it is easy to see $D$ is also closed under adjoints, so for any $A\in D$, the self-adjoint operators $\operatorname{Re} A=(A+A^\*)/2$ and $\operatorname{Im} A=(A-A^\*)/2i$ are also in $D$. It follows from the spectral theorem for self-adjoint operators that the spectral projections of $\operatorname{Re} A$ and $\operatorname{Im} A$ are also in $D$. By hypothesis, $D$ contains no nontrivial projections, so $\operatorname{Re} A$ and $\operatorname{Im} A$ must be scalar multiples of the identity. Since $A=\operatorname{Re} A+i\operatorname{Im} A$, the same is true of $A$. Thus $D$ consists only of $\mathbb{C}$.


More generally, even if you don't assume $D$ contains no nontrivial projections, it is a von Neumann algebra and the above argument shows that any von Neumann algebra is generated by its projections. There is quite a lot known about the structure of von Neumann algebras, but I'll leave it to others who know more than I do to elaborate on what can be said.


