# Schur's Lemma for Hilbert spaces

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 ("Generalization of a theorem of Burnside to non-compact group") and this one ("Schur's lemma for antiunitary operators on complex Hilbert spaces").

• In the unitary case (or more generally, if the group $G$ is closed under taking adjoints), the ring $D$ is a von Neumann algebra and so there is a lot you can say. In particular, for instance, by spectral theory $D$ is generated by its projections. Commented Aug 10, 2015 at 1:24
• A belated addition to these comments: if the representation is a unitary one, then we are in the setting of topologically irreducible $*$-representations of ${\rm C}^*$-algebras on Hilbert spaces, and it turns out that these are all automatically algebraically irreducible: this is Kadison's transitivity theorem en.wikipedia.org/wiki/Kadison_transitivity_theorem Commented Jul 13, 2020 at 17:36

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.

• The spectral projections should also commute with $G$, and hence be either $0$ or $I$ by the assumption of topological irreducibility. Commented Aug 10, 2015 at 3:11
• Indeed: it is known that the analogue of Schur's theorem for unitary, topologically irreducible representations is true, i.e. the commutant of the representation is trivial. The proof is, I imagine, essentially the one you have just given, and my guess is that it should be somewhere in the second half of Dixmier's Cstar algebras book (among many other likely sources) Commented Aug 10, 2015 at 3:12

Adding one more observation: $D$ is an integral domain.

Proof The kernel of a bounded operator is always closed, so all the nonzero elements of $D$ have no kernel and are thus injective. The composition of two injective maps is injective, and hence has no kernel. $\square$

• (but of course "integral domain" is often only used for commutative rings and it is not clear where commutativity should come from)
– Tom
Commented Oct 31, 2016 at 9:29