# Actions of finite permutation groups on hereditarily finite sets.

Model theorists have a lot to say about so-called definable imaginary elements of a structure. One way to formulate imaginaries is the following: Suppose $\mathcal{M}$ is a structure with universe $M$, and let $G$ be the automorphism group of $\mathcal{M}$. Define $$V_0(M) = M, \quad V_{\alpha+1}(M) = V_\alpha\cup P(V_\alpha(M)), \quad V_\lambda = \bigcup_{\alpha<\lambda}V_\alpha(M)$$ when $\lambda$ is a limit ordinal. Let $$V(M) = \bigcup_{\alpha<\infty}V_\alpha(M).$$ Then $G$ has a natural action on $V(M)$ defined inductively by $$g(x) = \{g(y):y\in x\}.$$ If $x\in V(M)$ and $S\subseteq M$, then $S$ supports $x$ if $G(S)\subseteq G(\{x\})$, where $G(S)$ is the pointwise stabilizer of $S$ and $G(\{x\})$ is the setwise stabilizer of $x$. The imaginaries are those $x\in V(M)$ which have finite support.

Now, for a finite structure $\mathcal{M}$, it makes sense to work with $HF(M)$ instead of $V(M)$ (stop the construction at the first countable infinite ordinal).

I would like to know if anyone has done any work regarding the action of a group of permutations of a finite set $X$ on the hereditarily finite sets above $X$. Ideally, I'd like to get results "off the shelf" if they're out there.

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I think you want V_{alpha+1}(M)=V_alpha(M)\cup P(V_alpha(M)). i.e. Missing M. – Joel David Hamkins Nov 23 '09 at 17:29
Also, your basic set-up presumes that the objects in M are not themselves sets, since otherwise your action is not well-defined. I guess you want to regard the objects of M as urlements for the purposes of this construction. – Joel David Hamkins Nov 23 '09 at 17:35

Well, since there have been no answers in a month, let me at least point out the easy fact that if M is finite, then every set in HF(M), and indeed, every set in V(M), is imaginary over M.

(I assume here that M is taken as urelements in the definition of V(M), as I mentioned in my comments to the question above, since otherwise there are problems with the action of G on V(M) and even HF(M) being well-defined.)

Theorem. If M is finite, then every object in HF(M), and indeed every object in V(M), is an imaginary element.

Proof: Since M is finite, we may take S=M. If pi fixes every element of M, then it is easy to see by transfinite induction that the action of pi on V(M) is the identity. Namely, if pi fixes every element of V_alpha(M), then it clearly also fixes every element of V_{alpha+1}(M). And so it fixes every element of V(M), including HF(M)=V_omega(M). QED

OK. What this answer really shows is that the question is not about the imaginaries over M, but rather, about gaining a greater understanding of the actiom of G on V(M). Perhaps it would be helpful to define the parameter-free version of imaiginary, where we might say that X in V(M) is pure imaginary over M if whenever pi is a permutation of M, then pi(X)=X, under the induced action of pi on V(M). For example, the set M itself has this property, as does the power set P(M), the set {M} and {emptyset,M}, and so on. In addition, any set whose transitive closure includes no urelements from M will be pure imaginary. The question would be to characterize the pure-imaginary sets over M.

This question shares many similarities with the various forcing arguments showing the consistency of the negation of the Axiom of Choice. Specifically, in the pre-forcing days, set theorists built what are called the symmetric models of set theory, by taking an infinite set of urelements M and restricting to the elements of V(M) having finite support. One can show that this is a model of ZF-with-urelements having no wellordering of M. The forcing proofs of the consistency of not-AC have exactly the same flavor, where one adds an infinite set of mutually generic Cohen reals, and then considers the sets that have names with finite support over this set. This is precisely how Cohen produced a model of ZF+not-AC, without urelements.

So one of the good reasons to study the imaginary elements over a set M is that they form a model of the set theory ZF-with-urelements. When M is infinite, however, then there can be no linear order of M in the pure imaginaries, since swapping elements outside the support of this set will not fix the order. In particular, M will not be well-orderable in this model of set theory, and so AC will not hold. For finite M, of course, there are linear orders of M having support M, and one can show that V(M) satisifes ZFC-with-urelements. But if one considers only the collection of pure-imaginaries, as I defined them above, then one will not even get ZF-with-urelements, unless M has only one element, since one will lose the Comprehensive (subset) axiom when there are parameters from M. For example, no proper nonempty subset of M can be pure imaginary. From this perspective, the pure imaginary sets are not so nice as the imaginary sets.

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