Commutators in factors – Hannes Thiel

Question: Is every element of trace zero in a $\mathrm{II}_1$ factor a commutator?

Update (August 2024): A positive answer to the questions was recently given In the following article:

S. Wen, J. Fang, Z. Yao. A stronger version of Dixmier’s averaging theorem and some applications. J. Funct. Anal. 287 (2024).

Background on commutators: It is a classical question to determine the elements in an algebra $A$ (over a field $k$ ) that are commutators, that is, of the form $[a,b] := ab-ba$ for some $a,b \in A$ . A related (and often easier) question is to determine the commutator subspace $\mathrm{span} [A,A]$ , which is defined as the linear subspace of $A$ generated by the set $[A,A] := \{[a,b] : a,b \in A \}$ of commutators. (Warning: In the literature, $[A,A]$ is often used to denote $\mathrm{span} [A,A]$ .)

A linear functional $\tau \colon A \to k$ is said to be tracial if $\tau(ab)=\tau(ba)$ for all $a,b \in A$ ; equivalently, $\tau$ vanishes on $\mathrm{span} [A,A]$ . It follows that an element in $A$ belongs to $\mathrm{span} [A,A]$ 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 [1]. Since commutators are mapped to commutators in quotients, we obtain according obstructions quotients by closed, two-sided ideals.

To summarize, if $A$ is a unital Banach algebra, and $a \in [A,A]$ , then:

$\tau(a) = 0$ for every tracial functional $\tau \colon A \to \mathbb{C}$ .
for every closed ideal $J \subseteq A$ (equivalently: every maximal ideal $J \subseteq A$ ), we do not have $a + J = \lambda 1 + J$ for some $\lambda \in \mathbb{C} \setminus \{0\}$ .

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 $\mathrm{I}$ factors.

Let us specialize to the case that $A$ is a unital C*-algebra. In this case, one can consider the space $T(A)$ of tracial states on $A$ . Since every tracial state is continuous, it vanishes on the closure of $\mathrm{span} [A,A]$ . Further, it follows from Theorem 5 in [2] that

$\overline{\mathrm{span} [A,A]} = \bigcap_{\tau\in T(A)}\ker(\tau).$

In many cases, $\mathrm{span} [A,A]$ 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 $\mathrm{I}_n$ . After presenting the results for these cases, we discuss some partial results for the case of a $\mathrm{II}_1$ factor.

Type $\mathrm{I}_n$ : The matrix algebra $M_n(\mathbb{C})$ is simple and has a unique tracial state $\tau$ . A matrix $a \in M_n(\mathbb{C})$ is a commutator if and only if $\tau(a)=0$ . In particular, the set $[M_n(\mathbb{C}, M_n(\mathbb{C}]$ of commutators agrees with the commutator subspace. More generally, this holds for every matrix algebra $M_n(k)$ 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 $\mathrm{I}_\infty$ , $\mathrm{II}_\infty$ and $\mathrm{III}$ : Let $M$ be a properly infinite factor. Then $M$ has no tracial states, and $M$ has a unique maximal ideal $J$ . (If $M=B(H)$ for some infinite-dimensional Hilbert space, then $J$ is the closure of the ideal of operators whose closed range have dimension strictly less than that of $H$ . In particular, if $H$ is separable, then $J$ is the ideal $K(H)$ of compact operators on $H$ .) Then

$[M,M] = \{ a \in M : a+J\neq \lambda 1 + J \text{ for every } \lambda\in\mathbb{C}\setminus\{0\}\},$

that is, an element $a\in M$

is a commutator if and only if its image in the simple C*-algebra $M/J$

(if $M=B(H)$

and $H$

is separable, then this is the Calkin algebra $B(H)/K(H)$

) is not a nonzero scalar multiple of the identity. This was shown by Halpern in [3]. In particular, $[M,M]$

is neither closed, nor a subspace, but every element in $M$

is a sum of two commutators and thus $\mathrm{span}[M,M]=M$

Type $\mathrm{II}_1$ : Let $M$ be a $\mathrm{II}_1$ factor. Then $M$ is simple and has a unique tracial state $\tau$ . It is expected that $[M,M]=\{a\in M : \tau(a)=0\}$ . Partial results in this direction have been obtained by Dykema-Skripka in [4]. In particular, every nilpotent element in $M$ is a commutator, and every normal element with vanishing trace and with atomic spectral measure is a commutator. [Update: As noted above, in a recent article by S. Wen, J. Fang, and Z. Yao it is shown that every trace-zero element in a $\mathrm{II}_1$ factor 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 $_{1}$ –factors, Proc. Amer. Math. Soc. (2012) 931–940. https://doi.org/10.1090/s0002-9939-2011-10953-5.

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