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i think that universal algebraists miiiight be able to say more about this then logicians. i also think it might be more natural to ask this for a language containing all 16 binary connectives. (There is actually a psychologist Zellweger who created and advocated for such a system.)
The models of ZFC in continuous logic are the same as the models in usual first order logic, this should be true because the metric is built in to the syntax. It seems somewhat possible to me that there could be a connection between boolean valued models of ZFC and continuous logic, but I don't think anything is known along those lines.
Assuming that claim, it's not hard to show that $(C,R_{ij})$ is trace definable in the trivial structure. But I don't see how to generalize this to the general Hrushovski construction that $M$ is a special case.
Let $f_1 =f$ ($f$ as before) and $f_2$ be the function $C\to D$ where $f_2(a)=b$ if there is $c$ such that $aEc$ and $f_1(c)=b$. For all $(i,j) \in \{1,2\}^2$ let $R_{ij}$ be the binary relation on $C$ where $R_{ij}(a,a')$ iff $f_i(a) = f_j(a')$. So $(C,R_{ij})$ is bi-interpretable with $M$. I'm pretty sure that $(C,R_{ij})$ is homogeneous.
Now of course I want to know if this thing is trace definable in the trivial structure. I think that it should be, the main thing would be to figure out what the definable sets in this structure actually are.
Cool. If I recall correctly, there is a theorem that if $M$ and $N$ are $\omega$-categorical structures then $M$ interprets $N$ iff there is a surjective morphism $\mathrm{Aut}(M) \to \mathrm{Aut}(N)$. I think your argument basically rules out such a morphism directly.
As $V^\times$ is abelian any finite index subgroup contains the $m$th powers for some $m$. So we just need to know that the collection of finite index subgroups in $V^\times$ forms a nbhd basis at $1$. This follows as $V^\times$ is a profinite group.
