Let $H$ be a Hilbert space. Let $A$ be a closed unbounded operator, and let $B\in B(H)$ be a bounded operator.

Definition:   $A$ and $B$ strong-commute if the partial isometry in the polar decomposition of $A$ commutes with $B$, and all the spectral projections of $|A|$ also commute with $B$.

In this question, I learned that, when $A$ is self-adjoint, we have $$ BA \subset AB \quad\Longrightarrow\quad \text{$A$ and $B$ strong-commute}. $$

When $A$ is not self-adjoint, then $BA \subset AB$ is not enough to ensure the strong-commuting of $A$ and $B$ (see below for a counterexample). One should think of strong-commuting as saying, roughly, that $B$ commutes with both $A$ and $A^*$.

Question:   Is it true that $$ (\,\,\,BA \subset AB \quad\text{and}\quad BA^* \subset A^*B\,\,\,) \quad\Longrightarrow\quad \text{$A$ and $B$ strong-commute}. $$

Ok, now the counterexample:
Let $D \subset E$ be two closed operators (both densely defined) on some Hilbert space $K$. Then I take $A := ...\oplus D\oplus D\oplus D\oplus E\oplus E\oplus E...$ on the Hilbert space $H:=\ell^2(\mathbb Z)\otimes K$, and I take $B$ to be the shift operator.

  • $\begingroup$ Strong-commuting is a very strong condition --- for "typical" non-normal $A$ it would imply that $B$ is a scalar multiple of the identity. Is that really what you want? $\endgroup$
    – Nik Weaver
    Apr 17, 2018 at 22:46
  • $\begingroup$ @Nik Weaver. Yes, that's what I mean. I agree with you that for "most" non-normal $A$ only scalar multiples of the identity will strong-commute with it. That's not too surprising: even when $A$ is bounded, commuting with both $A$ and $A^*$ is typically a very strong condition. $\endgroup$ Apr 17, 2018 at 23:01

1 Answer 1


Yes, your statement is true.

First observe that $$M := \{B \in \mathcal{B}(\mathcal{H}) : BA \subset AB \,\,\mbox{ and }\,\, BA^* \subset A^*B\}$$ is a von Neumann algebra (on the nose, without needing to take a strong closure). To see that M is closed under adjoints, it may be helpful to note that $BA^* \subset A^*B$ is equivalent to $B^*A \subset AB^*$ when $B$ is bounded. You wish to show that $M \subset N:=\{E_S , V\}'$ where $E_S$ and $V$ are the spectral projections and polar part of A, respectively. It suffices to show that $U \in N$ for all unitary $U \in M$.

If $U \in M$, then $U$ maps $D(A)$ onto itself, and similarly $D(A^*)$. One can check that $$A = UAU^* = (UVU^*)(U|A|U^*),$$ that $U|A|U^*$ is positive self-adjoint, and that $UVU^*$ is a partial isometry with the correct initial and final spaces, so by the uniqueness of the polar decomposition (see Reed and Simon, Functional Analysis, Thm. VIII.32) $U$ commutes with $V$ and $U$ commutes (strongly, by the spectral theorem/your earlier question) with $|A|$.


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