Is there a commutative subalgebra $A\subset B(H)$ containing the 1 dimensional scalars with the following property:
The algebra $A$ has trivial intersection with the set of commutator elements $xy-yx$
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$\begingroup$ What is $B(H)$? $\endgroup$– Tom De MedtsCommented May 5, 2023 at 14:49
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3$\begingroup$ The algebra $A$ reduced to scalars works. $\endgroup$– YCorCommented May 5, 2023 at 14:58
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$\begingroup$ @YCor But I meant a proper inclusion. The motivation for question was the scalar case $\endgroup$– Ali TaghaviCommented May 29, 2023 at 17:42
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1 Answer
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From the results of
Brown, Arlen; Pearcy, Carl, Structure of commutators of operators, Ann. Math. (2) 82, 112-127 (1965). ZBL0131.12302.
an element of $B(H)$ is a commutator $xy-yx$ if and only if it is not of the form $\lambda I + C$ for some non-zero scalar $\lambda$ and compact $C$. So your algebra would have to contain an element of the form $\lambda I + C$ but not $C$ (unless $C=0$), and so cannot contain the scalars unless it consisted solely of the scalars, which are indeed not commutators by the classical results of Wintner and Wielandt.
answered May 6, 2023 at 0:32
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$\begingroup$ Thank you very much for your answer! $\endgroup$ Commented May 23, 2023 at 23:39