# 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. – Eric Wofsey Aug 10 '15 at 1:24

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. – Yemon Choi Aug 10 '15 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) – Yemon Choi Aug 10 '15 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 Oct 31 '16 at 9:29