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 $\sigma(A) = \sigma_{c}(A)$.


  • 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 ?

  • $\begingroup$ 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 $\endgroup$
    – Yemon Choi
    Jul 19, 2013 at 18:56
  • 2
    $\begingroup$ @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 $B(H)$. $\endgroup$ Jul 19, 2013 at 20:12
  • 2
    $\begingroup$ Could the "downvoter" and "closer" indicate what's wrong ? Thank you. $\endgroup$ Jul 20, 2013 at 7:54
  • 2
    $\begingroup$ Cross-posted on MSE. Hi Sébastien, you are supposed to let people know when you ask the same question on both sites. $\endgroup$
    – Julien
    Jul 21, 2013 at 13:43
  • 1
    $\begingroup$ Definitely a question for MO. Don't cross post unless you say so and give the URL for the other post. $\endgroup$ Jul 23, 2013 at 19:12

1 Answer 1


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.

  • 5
    $\begingroup$ 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. $\endgroup$ Jul 23, 2013 at 18:20
  • $\begingroup$ Thanks, I think it would've taken me a while to think of something so slick! $\endgroup$
    – Ollie
    Jul 24, 2013 at 5:16

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