# Commutators in factors

Von | Oktober 8, 2022

Question: Is every element of trace zero in a factor a commutator?

Background on commutators: It is a classical question to determine the elements in an algebra (over a field ) that are commutators, that is, of the form for some . A related (and often easier) question is to determine the commutator subspace , which is defined as the linear subspace of generated by the set of commutators. (Warning: In the literature, is often used to denote .)

A linear functional is said to be tracial if for all ; equivalently, vanishes on . It follows that an element in belongs to if and only if it vanishes under every tracial functional.

Thus, a commutator has to vanish under every tracial functional. Another obstruction occurs in the normed setting, since the unit of a normed algebra is not a commutator (although it may be a sum of two commutators). It follows that nonzero scalar multiples of the unit are not commutators either. A simple proof was given by Wielandt in ​​. Since commutators are mapped to commutators in quotients, we obtain according obstructions quotients by closed, two-sided ideals.

To summarize, if is a unital Banach algebra, and , then:

1. for every tracial functional .
2. for every closed ideal (equivalently: every maximal ideal ), we do not have for some .

We will see below that these are the only obstructions for an element to be a commutator in the case of properly infinite factors, as well as type factors.

Let us specialize to the case that is a unital C*-algebra. In this case, one can consider the space of tracial states on . Since every tracial state is continuous, it vanishes on the closure of . Further, it follows from Theorem 5 in ​​ that In many cases, is a closed subspace. If it is not closed, then the C*-algebra admits non-continuous tracial functionals.

Commutators in von Neumann factors. The set of commutators have been completely characterized in properly infinite factors, and in factors of type . After presenting the results for these cases, we discuss some partial results for the case of a factor.

Type : The matrix algebra is simple and has a unique tracial state . A matrix is a commutator if and only if . In particular, the set of commutators agrees with the commutator subspace. More generally, this holds for every matrix algebra over a field, which was shown by Shoda in 1936 for the case of characteristic zero, and by Albert-Muckenhoupt in 1957 for arbitrary characteristic.

Types , and : Let be a properly infinite factor. Then has no tracial states, and has a unique maximal ideal . (If for some infinite-dimensional Hilbert space, then is the closure of the ideal of operators whose closed range have dimension strictly less than that of . In particular, if is separable, then is the ideal of compact operators on .) Then that is, an element is a commutator if and only if its image in the simple C*-algebra (if and is separable, then this is the Calkin algebra ) is not a nonzero scalar multiple of the identity. This was shown by Halpern in ​​. In particular, is neither closed, nor a subspace, but every element in is a sum of two commutators and thus .

Type : Let be a factor. Then is simple and has a unique tracial state . It is expected that . Partial results in this direction have been obtained by Dykema-Skripka in ​​. In particular, every nilpotent element in is a commutator, and every normal element with vanishing trace and with atomic spectral measure is a commutator.

1. 
H. Wielandt, �ber die Unbeschr�nktheit der Operatoren der Quantenmechanik, Math. Ann. (1949) 21–21. https://doi.org/10.1007/bf01329611.
2. 
N. Ozawa, Dixmier approximation and symmetric amenability for C*-algebras, J. Math. Sci. Univ. Tokyo. 20 (2013) 349–374.
3. 
H. Halpern, Commutators in properly infinite von Neumann algebras, Trans. Amer. Math. Soc. (1969) 55–73. https://doi.org/10.1090/s0002-9947-1969-0251546-8.
4. 
K. Dykema, A. Skripka, On single commutators in II –factors, Proc. Amer. Math. Soc. (2012) 931–940. https://doi.org/10.1090/s0002-9939-2011-10953-5.