Is there a simple group that is torsion-free, type $\textrm{F}_\infty$, and infinite dimensional?

Question

Does there exist an example of a group that is:

1. Simple,
2. Torsion-free,
3. Of type $\textrm{F}_\infty$, and
4. Infinite dimensional (meaning of infinite cohomological dimension)?

Thompson's group $F$ has all these properties except not (1). Thompson's groups $T$ and $V$ have all these properties except not (2). The commutator subgroup $[F,F]$ has all these properties except not (3). My understanding is that things like Burger-Mozes groups have all these properties except not (4).

I'm curious whether there is an example with all four properties. Many "Thompson-like" constructions provide examples of groups with properties (3) and (4), but any of the examples I can think of only have one of property (1) or (2), not both. It's possible this question is just open, but outside the "Thompson world" I don't know much about infinite simple groups, so I may be missing something.

Answer 1

A while ago, James Hyde and Yash Lodha found a group with all four of these properties, I should mention this here. The paper is https://arxiv.org/abs/2302.04805 (to appear in Ann. Sci. Ecole Normale Sup.). The group is not only torsion-free, but left-orderable. They actually found an infinite family of examples, but let me explain the easiest one.

Take the group of all piecewise-linear, orientation-preserving homeomorphisms $f$ of $\mathbb{R}$ that have breakpoints in $\mathbb{Z}[1/6]$ and slopes of the form $2^i 3^j$, commute with translation by 1, and also satisfy that if the slope of $f$ at $x\in\mathbb{R}\setminus\mathbb{Z}[1/6]$ is $2^i 3^j$ then the difference $i-j$ equals the (signed) number of integers strictly in between $x$ and $x.f$.

They prove that the commutator subgroup of this group is simple and of type $\mathrm{F}_\infty$, hence satisfies properties (1) and (3), and obviously satisfies (2) and (4) since it's left-orderable and contains Thompson's group $F$.