# Is there an irreducible, noncompact commuting, nonnormal operator, with spectrum strictly continuous?

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 ?

• One more possible example (though not an answer to your question): I think that if you take any non-zero element of the von Neumann algebra of a discrete group $\Gamma$, thought of as an operator on $\ell^2(\Gamma)$, then I think you get noncompact commuting and strictly continuous spectrum, and clearly you can arrange for non-normal. But irreducibility is not clear to me right now Jul 19, 2013 at 18:56
• @YemonChoi : Thank you for this comment. What's not clear with irreducibility ? If the group $\Gamma$ is ICC (infinite conjugacy classes), it generates a $II_{1}$ factor, and there are plenty of projections in its commutant, so there is no irreducibility for a single element. Next, for any discrete group $\Gamma$, if we take an irreducible representation $H$, the von Neumann algebra it generates is the whole $B(H)$. Jul 19, 2013 at 20:12
• Could the "downvoter" and "closer" indicate what's wrong ? Thank you. Jul 20, 2013 at 7:54
• Cross-posted on MSE. Hi Sébastien, you are supposed to let people know when you ask the same question on both sites. Jul 21, 2013 at 13:43
• Definitely a question for MO. Don't cross post unless you say so and give the URL for the other post. Jul 23, 2013 at 19:12

Essentially I think weighted shifts should be a sufficiently rich class of operators. Consider, for instance, the following example.

Take the doubly infinite sequence $$w_k=\left\{\begin{array}{ll} 2 & \text{if } k<0 \\ 1 & \text{if } k\geq 0 \end{array}\right.$$ and let $W$ be the weighted shift on $\ell^2(\mathbb{Z})$ defined by $$We_j=w_je_{j+1} \qquad \forall j\in\mathbb{Z}\qquad (\text{here }(e_j)\text{ is the usual basis}).$$ Then

1. W has strictly continuous spectrum. This is an almost identical argument the usual bilateral shift.

2. W is not normal. The adjoint is given by $W^\ast e_j=w_{j-1}e_{j-1}$ and so $$W^\ast We_j=w_j^2e_{j},\quad \text{but}\quad WW^\ast e_j = w_{j-1}^2e_j.$$

3. $W$ is irreducible. This follows from Corollary 2 of R. Gellar, Operators commuting with a weighted shift, Proc. Amer. Math. Soc. 23 (1969), 538-545.

4. I do not seem to be able to show $W$ is non-compact-commuting. See Bill's comment below.

• It is non-compact-commuting. First, $\|Wx\| \ge \|x\|$, so if $S$ is compact and $S e_n \not= 0$ then $\|W^k S e_n\|$ is bounded away from zero. However, $(W_k e_n)_k$ tends weakly to zero, so $(SW_k e_n)_k$ tends in norm to zero. Jul 23, 2013 at 18:20
• Thanks, I think it would've taken me a while to think of something so slick! Jul 24, 2013 at 5:16