An autohomeomorphism of the Čech--Stone remainder $\beta N\setminus N$ is called trivial if it has a continuous extension to a map from $\beta N$ into itself. Such map is determined by an almost permutation, which is a bijection between cofinite subsets of $N$. By results of W. Rudin and S. Shelah, the question whether nontrivial autohomeomorphisms of $\beta N\setminus N$ exist is independent from ZFC. We will be considering the so-called rotary autohomeomorphisms. An autohomeomorphism is called rotary if it corresponds to a permutation of $N$ all of whose cycles are finite. If all autohomeomorphisms are trivial, then the problem of their conjugacy is also trivial (in the usual sense of the word). However the Continuum Hypothesis makes the conjugacy relation nontrivial. While our results are somewhat incomplete, they suffice to decide whether for example the rotary autohomeomorphisms whose cycles have lengths $2^{2n}$, for $n\in N$, and $2^{2n+1}$, for $n\in N$, are conjugate. This is a joint work with Will Brian.

7 years ago, also in Oxford, Sylvy Anscombe and I asked this question, which is part of the general effort to try and understand the model theory of henselian valued fields through dividing lines. In 2024, Sylvy Anscombe and Franziska Jahnke completely classified NIP henselian valued fields. Their methods can be extended, with the help of works of Chernikov, Kaplan and Simon and of Kuhlmann and Rzepka, to NTP2 henselian valued fields, obtaining the following:

• if a henselian valued field is NTP2, then it is semitame and its residue field is NTP2;
• if a henselian valued field is separably algebraically maximal Kaplansky and its residue field is NTP2, then it is NTP2.

This covers a large class of fields, but there is still a gap. Notably, Fp((Q)) is in the middle: it is semitame but not Kaplansky.

To answer this question, we studied so called tame henselian fields with finite residue field, and derived quantifier elimination results, namely, we prove that any formula in the language of valued fields reduces to a formula of the form (∃y f(x,y)=0) ∧ φ(v(x)) ∧ ψ(res(x)), where φ and ψ are formulas in the language of ordered groups and of rings, respectively.

In Fp((Q)) specifically, the valuation ring itself is definable with a diophantine formula (ie of the form ∃y f(x,y)=0), reducing further our quantifier elimination result.

Finally, a large chunk of these formulas are known to be NTP2: when f(x,y) is additive in y, the formula ∃y f(x,y)=z is NTP2 (with respect to x and z). Unfortunately, that does not cover all formulas, so the answer to the titular question is still unknown.

Non-trivial henselian valuations are often so closely related to the arithmetic of the underlying field that they are encoded in it, i.e., that their valuation ring is first-order definable in the language of rings. In this talk, I will survey and present old and new results around the definability of henselian valuations, also with a view towards parameters and uniformity of definitions.