The Turing equivalence relation on $\cal P(\mathbb{N})$ is defined by $A\equiv\_T B$ iff $A\leq\_T B$ and $B\leq\_T A$. This is a countable Borel equivalence relation on the polish space $\cal P(\mathbb{N})$. And as stated, the answer to my question is yes since a theorem by Feldman and Moore says: for any countable Borel equivalence relation $E$ on a standard Borel space $X$, there is a countable group $G$ and a Borel action of $G$ on $X$ such that $E$ is the orbit equivalence relation of the action.


But from where I have looked, the Turing equivalence isn't the orbit e.r. of any countable group that is "found in nature". I was wondering if anything else was known about this, as well as any other instances of equiv. relations that are provably orbit equivalences, but with no known "natural" group doing the acting. It seems like for many important equivalences (like $E\_0, E\_{\infty}$, hyperfinite e.r's) we do in fact have a natural group that is acting, but perhaps that is the reason why we know so much about them.


Any references or other information would be great, thanks.



Is the Turing equivalence relation the orbit equiv. relation of the action of a countable group?
These [slides of Simon Thomas](http://www.math.cmu.edu/~eschimme/Appalachian/athens2.pdf) for the Appalachian Set Theory Seminar 2007 are directly engaged with aspects of this question, particularly the question of which countable Borel equivalence relations arise as free Borel actions of countable groups or are Borel bireducidible with such an equivalence relation. These latter equivalence relations are called *essentially free*. (See also [part I](http://www.math.cmu.edu/~eschimme/Appalachian/athens1.pdf) of his talk series "Countable Borel Equivalence Relations", which contains a proof of Feldman-Moore; [part II](http://www.math.cmu.edu/~eschimme/Appalachian/athens2.pdf) was the link above.)


In particular, Simon Thomas proved that Turing equivalence is not essentially free. 


