Let $H$ be an infinite dimensional separable Hilbert space.

**Definition**: The **commutant** $\mathcal{S}'$ of a subset $\mathcal{S} \subset B(H)$ is $ \{A \in B(H) : AB=BA \ , \ \forall B \in \mathcal{S} \} $.

**Definitions** : An operator $A \in B(H)$ is :

**Irreducible**(Halmos 1968) if its commutant $\{ A\}'$ does not contain projections other than $0$ and $I$ ($A \ne A_{1} \oplus A_{2}$, $A$ generates $B(H)$ as von Neumann algebra : $\{A,A^{*}\}''=B(H)$).**Nonnormal**if $\{ A\}'$ does not contain $A^{*}$ (i.e. $AA^{*} \ne A^{*}A$).**Noncompact commuting**if $\{ A\}'$ does not contain a compact operator.

**Definition** : The **spectrum** $\sigma(A)$ of $A \in B(H)$ is $\{\lambda \in \mathbb{C} : A - \lambda I \text{ not bijective} \}$.

It decomposes as follows:

*Point spectrum*: $\sigma_{p}(A) = \{\lambda \in \mathbb{C} : A - \lambda I \text{ not injective} \}$*Continuous spectrum*: $\sigma_{c}(A) = \{\lambda \in \mathbb{C} : A - \lambda I \ \text{injective, dense nonclosed range} \}$*Residual spectrum*: $\sigma_{r}(A) = \{\lambda \in \mathbb{C} : A - \lambda I \ \text{ injective, nondense range} \}$

The spectrum of $A$ is **strictly continuous** if $\sigma(A) = \sigma_{c}(A)$.

**Examples**:

- Let $S$ be the
**bilateral shift**defined on $H = l^{2}(\mathbb{Z})$ by $S.e_{n} = e_{n+1} $.

Its spectrum is*strictly continuous*: $\sigma(S) = \sigma_{c}(S) = \mathbb{S}^{1}$.

It's also a unitary operator ($SS^{*} = S^{*}S = I$), so a fortiori a*normal*operator.

It is*noncompact commuting*and*reducible*. - Let $T$ be the
**unilateral shift**defined on $H = l^{2}(\mathbb{N})$ by $T.e_{n} = e_{n+1} $.

Its spectrum is not*strictly continuous*because $0 \in \sigma_{r}(T)$.

It's a*nonnormal*operator because $[T^{*},T].e_{0} = e_{0}$.

It is*noncompact commuting*and*irreducible*. - Let $V$ the
**Volterra operator**defined on $H= L^{2}[0,1]$ by $(V.f)(t)=\int_0^tf(x)dx$.

Its spectrum is*strictly continuous*$\sigma(V) = \sigma_{c}(V) = \{ 0\}$.

It is*compact*,*irreducible*and*nonnormal*(see here). - Let $p$ be a non-constant polynomial (see here).

Then $p(V)$ is*nonnormal*,*compact commuting*and*irreducible*.

Its spectrum is*strictly continuous*$\sigma(p(V)) = \sigma_{c}(p(V)) = \{ p(0)\}$.

It's*compact commuting*,*nonnormal*and*irreducible*. - Let $S \oplus V$ defined on $l^{2}(\mathbb{Z}) \oplus L^{2}[0,1]$.

It is*reducible*,*compact commuting*,*nonnormal*and with spectrum*strictly continuous*.

*If you find a mistake, thank you let me know in comment.*

The main question: Is there an irreducible, noncompact commuting and nonnormal operator, with spectrum strictly continuous ?

**Bonus questions** : How classify these operators ?

thinkyou get noncompact commuting and strictly continuous spectrum, and clearly you can arrange for non-normal. But irreducibility is not clear to me right now $\endgroup$ – Yemon Choi Jul 19 '13 at 18:562more comments