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Let $A$ be a bounded linear operator on a separable Hilbert space ${\cal H}$, and suppose that $A$ is distinct from its adjoint $A^*$.

Question: Can the double commutant of $A$ be distinct from the double commutant of $\{A,A^*\}$? If so, is there a simple example?

This question was inspired by the following statement from section 3.3 in Vaughan Jones (2009), Von Neumann Algebras (https://math.berkeley.edu/~vfr/VonNeumann2009.pdf):

If $S \subseteq {\cal B(H)}$, we call $(S \cup S^*)''$ the von Neumann algebra generated by $S$.

I don't know if this was meant to be the most efficient definition, and that's exactly what prompted my question. Maybe the Hilbert space wasn't assumed to be separable in that context, but I am interested in the separable case (if it matters).

Martin Argerami
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Chiral Anomaly
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2 Answers2

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No, this need not be the case. Actually, operators with the property you mention are said to belong to the class (dc).

Turner studied weighted (two-sided) shifts on the Hilbert space $\ell_2(\mathbb Z)$. He proved the following theorem (see Theorem 3.4 here)

Theorem. A two-sided shift on $\ell_2(\mathbb Z)$ is not invertible if and only if it is in the class (dc).

It may actually happen that an operator $T$ is in the class (dc) but $T\oplus 0$ is not (see Section 4 of the linked paper).

Tomasz Kania
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  • There is a confusion. Operators with the property as in the question are exactly normal operators, it follows from the Fuglede's theorem. The paper you cite defines (dc) class operators as operators which satisfy the non self-adjoint bicommutant theorem. See https://math.stackexchange.com/questions/445640/how-generalize-the-bicommutant-theorem – Boris Bilich Jan 06 '22 at 16:42
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Here is a simple example. Let $H=\mathbb C^2$ and $A=\begin{bmatrix}0&1\\0&0\end{bmatrix}$. Then $\{A,A^*\}''=M_2(\mathbb C)$, while $$ \{A\}''=\left\{\begin{bmatrix}a&b\\0&a\end{bmatrix}:\ a,b\in\mathbb C\right\}. $$

Martin Argerami
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