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Let the cocone of a Turing degree ${\bf d}$ be the set $cc({\bf d}):\{{\bf c}: {\bf c}\not\ge_T {\bf d}\}$. I'm curious what's known about the various partial orders (isomorphic to ones) of the form $cc({\bf d})$ for a nonzero Turing degree ${\bf d}$. Upwards cones have been extensively studied of course, and it turns out that in general the cones above distinct degrees could look very different - see e.g. this paper by Shore - but I can't find much information about cocones, and they seem more finicky objects.

For example, it's not immediately clear to me how to construct a pair of degrees yielding isomorphic cocones or how to construct a pair of degrees yielding nonisomorphic cocones! One natural guess is that since $0'$ is definable in the Turing degrees, the cocone of $0'$ should not be isomorphic to the cocone of $0''$, since the latter has $0'$ while the former doesn't. But the definition of $0'$ I'm aware of isn't $\Sigma_1$, so conceivably $cc(0')$ could have something that looks like what $0'$ looks like in $cc(0'')$, and so on.

Basically, I'm trying to get a picture of whether, and if so to what extent, removing an upper cone from the Turing degrees could change their global structure. For example, it's generally conjectured these days that the Turing degrees are rigid (that is, the partial order of Turing degrees has no nontrivial automorphisms), but I don't see an obvious reason why this would mean that cocones can't have nontrivial automorphisms. My only relevant vague recollection is that the Slaman-Woodin machinery should imply that, assuming rigidity, every cocone $cc({\bf d})$ with base ${\bf d}$ sufficiently high will also be rigid. (In fairness, I also have a couple sillier interests - e.g. what happens when we force with $cc({\bf d})$ (ordered by "stronger = higher"), and to what extent it depend on what ${\bf d}$ we pick?)

To give this question focus, I'll ask specifically:

Are there nonzero Turing degrees with nonisomorphic cocones?

EDIT: Note that ${\bf d}\not\in cc({\bf d})$. Were we to include it - that is, replace "$\not\ge_T$" with "$\not>_T$" in the definition of the cocone - this question would have an easy affirmative answer, since ${\bf d}$ would be definable (= the unique maximal element) and so for example the cocone of a minimal degree would be non-elementarily-equivalent to the cocone of a non-minimal degree. It's plausible to me that omitting ${\bf d}$ ultimately doesn't change much, but my current suspicion is that it changes a lot in general.


As a side note, there's some flexibility here with respect to how we view cocones as structures. Turing reducibility is insanely expressive: from $\le_T$ alone we can define all sorts of more complicated degree operations/relations, most importantly the jump; by contrast, in a cocone $cc({\bf d})$ it's not even clear to me that the relation "${\bf a}\vee{\bf b}\ge_T{\bf a}\vee{\bf c}$" is definable from $\le_T$ alone (where "$\vee$" in the preceding is the join in the Turing degrees; joins in cocones won't exist in general). So we could conceivably get very different answers depending on what structure we provide the cocone

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    $\begingroup$ I'm glad there's no such thing as the dual object to a Jones. $\endgroup$ Commented Jul 22, 2018 at 3:59
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    $\begingroup$ @MortezaAzad One quick example is showing there are incomparable Turing degrees using forcing. $\endgroup$
    – Wojowu
    Commented Jul 22, 2018 at 8:20
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    $\begingroup$ @MortezaAzad See the Slaman-Woodin manuscript linked in my answer; forcing/absoluteness arguments play a crucial role in showing e.g. that the automorphism group of the Turing degrees is at most countable. $\endgroup$ Commented Jul 22, 2018 at 14:03
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    $\begingroup$ @GerryMyerson: I'd like there to be a dual object to a nant. $\endgroup$
    – Jim Conant
    Commented Jul 22, 2018 at 19:45
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    $\begingroup$ @MortezaAzad Also, determinacy hypotheses can play a role, via Martin's cone theorem. One might be tempted to use Martin's theorem here to give a negative answer on a cone, but it doesn't seem applicable since for the obvious attack we would need to first know that there are only countably many possible isomorphism types of cocones. You can use the cone theorem to show (in ZFC alone, in fact) that the cocones of all sufficiently large ${\bf d}$ are elementarily equivalent, but that's a very weak result. $\endgroup$ Commented Jul 23, 2018 at 17:51

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