Is every bounded representation of Z unitarisable when all sets are measurable?

Question

For the purpose of this question, a group is amenable iff there exists a Følner sequence.

Dixmier unitarisability problem asks whether a (countable discrete) group G is amenable iff every bounded representation of G on a Hilbert space is unitarisable.

The problem is currently not solved, but at least argument "$\Rightarrow$" is not very difficult: let G be amenable and let $\rho\colon G \curvearrowright H$ be a bounded representation. We need to make a new scalar product on $H$ for which $\rho$ is unitary, and such that vectors of norm one in the old product have norms bounded from above in the new product (so that the identity mapping on $H$ is continuous) and bounded from below in the new product (so that $H$ with the new product is complete).

Let $F$ be a mean on $G$. For a pair of vectors $v,w$ consider a function $f_{vw}$ on $G$: $g \mapsto \langle gv, gw \rangle$, define the new scalar product as $\langle v, w \rangle_{\mathrm{new}} := F(f_{vw})$. It's easy to check that this new product has the desired properties.

But in the proof we used axiom of choice, because we use the mean on $G$. I tried to figure out a proof which uses less than a full or almost full axiom of choice, but so far I failed.

Question. Is the above implication true in a set of axioms in which also the statement "every subset of R is measurable" is true?

Maybe it's easier to answer this question specifically for the infinite cyclic group Z. Note that Z is amenable, according to our definition of amenability, in Zermelo-Fraenkel set theory.

Even if the anser is no, I'd be interested to learn about the proof of the above fact which uses less than Boolean prime ideal theorem (AFAIU using only BPIM is easy, because to prove existence of a mean one uses Banach-Aleoglu theorem, which uses Tychonoff theorem for Hasudorff space, which, according to wikipedia, uses only BPIM)

Answer 1

The answer is yes for separable Hilbert spaces.

If the Hilbert space is separable with basis $\lbrace e_n \mid n \in \mathbb N\rbrace$, you only have to fix countably many inner products and define $\langle e_n,e_m \rangle_{\rm new}$ for $n,m \in \mathbb N$. Assuming Folner's condition, you know that

$$\frac{1}{n} \sum_{i=1}^n \langle \pi(i)e_n,\pi(i)e_m \rangle$$

is a bounded sequence which defines an asymptotically $\mathbb Z$-invariant inner product. So, using the Axiom of Dependent Choice, which allows you to make a countable number of dependent choice (like -- for example - if you choose a convergent subsequences in a bounded sequence), you can now choose a sequence of integers $(n_k)_{k \in \mathbb N}$ such that

$$\frac{1}{n_k} \sum_{i=1}^{n_k} \langle \pi(i)e_n,\pi(i)e_m \rangle$$

will converge for all $n,m \in \mathbb N$ as $k \to \infty$. This of course needs the usual diagonalization procedure, but no futher set theoretic complications.

Solovay's model of ZF + Dependent Choice has the property that all subsets of $\mathbb R$ are Lebesgue measurable. Hence, in this world uniformly bounded representations of $\mathbb Z$ (and in fact any amenable group $G$) on separable Hilbert spaces are unitarisable and at the same time, all subsets of the real numbers are Lebesgue measurable.

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Now, one may ask whether separability is an issue or not, but that is surprisingly unclear. It (even with AC) is not clear if you can decompose every uniformly bounded representation into cyclic representations or even in somewhat smaller pieces. In particular, it is unclear whether the above result implies something for arbitrary uniformly bounded representations. In some sense, this is not surprising, since one is limiting oneself to situations with additional countability assumptions.

Answer 2

You might want to look also at the statement "every subset of the real line has the property of Baire". Here a set has the Baire property if it has a meager symmetric difference with some open set.

You can prove amenability of the group of integers from the existence of a free ultrafilter on the set of natural numbers. A free ultrafilter on the natural numbers gives you a non-measurable set of reals. However, for the amenability of $\mathbb Z$ you might actually get away with the existence of a finitely additive probability measure on $\mathbb N$.
(Needs to be checked.) As far as I know it is not known that the existence of such a measure implies the existence of a non-measurable set of reals. However, I believe that such a measure does imply the existence of a set of reals without the Baire property.

It is consistent with ZF that every set of reals has the Baire property (Solovay, same model as for all sets of reals are measurable). If it is true what I said above, $\mathbb Z$ is not amenable in the Solovay model (since amenability gives you a finitely additive measure). Actually, I am quite sure that Solovay explicitly mentions in his paper that there is no finitely additive measure on the natural numbers in his model.