Commutators in factors

Von | Oktober 8, 2022

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

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:

  1. \tau(a) = 0 for every tracial functional \tau \colon A \to \mathbb{C}.
  2. 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.

  1. [1]
    H. Wielandt, �ber die Unbeschr�nktheit der Operatoren der Quantenmechanik, Math. Ann. (1949) 21–21.
  2. [2]
    N. Ozawa, Dixmier approximation and symmetric amenability for C*-algebras, J. Math. Sci. Univ. Tokyo. 20 (2013) 349–374.
  3. [3]
    H. Halpern, Commutators in properly infinite von Neumann algebras, Trans. Amer. Math. Soc. (1969) 55–73.
  4. [4]
    K. Dykema, A. Skripka, On single commutators in II_{1}–factors, Proc. Amer. Math. Soc. (2012) 931–940.

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