Math-6139 · mathematics · Analysis

Math-6139

Ultraproduct II_1 Factors — Problem 21

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Statement

Definitions and mathematical statement

Let $F$ be Thompson's group: the group, under composition, of increasing piecewise linear homeomorphisms of $[0,1]$ having finitely many breakpoints, all dyadic rational, and slopes in $\{2^k:k\in{\mathbb{Z}}\}$. Let $M=L(F)$ be its group von Neumann algebra, generated on $\ell^2(F)$ by the left regular representation $\lambda(g)\delta_h=\delta_{gh}$, with canonical trace $\tau(x)=\langle x\delta_e,\delta_e\rangle$.

For a nonprincipal ultrafilter $\mathcal U$ on ${\mathbb{N}}$, define the tracial ultrapower \[ M^{\mathcal U}=\ell^\infty({\mathbb{N}},M)\big/\{(x_n):\lim_{n\to\mathcal U}\|x_n\|_{2,\tau}=0\},\qquad \|x\|_{2,\tau}=\tau(x^*x)^{1/2}. \] Identify $M$ with its diagonal image in $M^{\mathcal U}$. Its central sequence algebra is \[ C_{\mathcal U}(M)=M'\cap M^{\mathcal U}=\{x\in M^{\mathcal U}:xa=ax\text{ for every }a\in M\}. \] A $\mathrm{II}_1$ factor $M$ is McDuff if $M\cong M\,\overline\otimes\,\mathcal R$, where $\mathcal R$ is the hyperfinite $\mathrm{II}_1$ factor. Using the standard convention for separable factors, call $M$ super McDuff if it is McDuff and $C_{\mathcal U}(M)$ is a factor for every nonprincipal $\mathcal U$ on ${\mathbb{N}}$; being a factor means that its center consists exactly of scalar multiples of its identity.

Question (Elayavalli, Problem 21, 2025). Is $L(F)$ super McDuff? In the convention just specified, does \[ L(F)\cong L(F)\,\overline\otimes\,\mathcal R \quad\text{and}\quad Z\!\left(L(F)'\cap L(F)^{\mathcal U}\right)={\mathbb{C}}1 \quad\text{for every nonprincipal }\mathcal U\text{ on }{\mathbb{N}} \] hold?

The supplied excerpt does not define "super McDuff" or specify an ultrafilter convention; the displayed formulation makes the standard interpretation explicit and requires confirmation against the complete source. Its introductory phrase "The above" refers to preceding material absent from this record and supplies no recoverable additional hypothesis.

Short English statement

The question asks whether the group von Neumann algebra of Thompson's group $F$ is super McDuff: it absorbs the hyperfinite $\mathrm{II}_1$ factor, and its central sequence algebras have trivial center. The supplied registry labels the problem open; that label is not an independently verified literature assessment.

Sources

  • Local UnsolvedMath registry, permanent ID $801120$, source code catalogue entry, title "Ultraproduct $\mathrm{II}_1$ Factors --- Problem 21." The supplied record attributes the problem to Elayavalli, with proposed year 2025 and status open.
  • Elayavalli, 50 Open Problems: Ultraproduct $\mathrm{II}_1$ Factors (2025), Problem 21. Supplied source URL: https://arxiv.org/abs/2511.20377. The record reports access on 2026-10-16 and extraction by arxiv_latex; these are supplied provenance statements, not a claim of independent source consultation.
  • The supplied introductory sentence cites the unresolved bibliography key boutonnetchifanioana and attributes to that reference the construction of many McDuff factors that are not elementarily equivalent. No bibliographic entry or link for this reference, nor the preceding material described as complementary to it, is supplied.

Catalogue entry

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