1
$\begingroup$

EDIT: There are serious problems with the definition below; see the comment thread below for those problems and some thoughts on addressing them. I'm leaving the question up for now since I think the idea is still interesting, but it is definitely broken at the moment.

Suppose $\mathbb{P}$ is a forcing notion and $\nu$ is a $\mathbb{P}$-name. Say that a $\mathbb{P}$-name $\mu$ is $\nu$-invariant if, whenever $G, H$ are $\mathbb{P}$-generic and $\nu[G]=\nu[H]$, then $\mu[G]=\mu[H]$. OK fine, that's nonsense, but it's easy to fix: formally, $\mu$ is $\nu$-invariant if whenever $\mathbb{Q}$ is a forcing notion and $\gamma_0,\gamma_1$ are $\mathbb{Q}$-names which are forced to be $\mathbb{P}$-generic and have $\nu[\gamma_0]=\nu[\gamma_1]$, then it is forced that $\mu[\gamma_0]=\mu[\gamma_1]$.

Say further that a name $\mu$ is hereditarily $\nu$-invariant if it and each name in its transitive closure is $\nu$-invariant, and let $HI_\nu$ denote the class of hereditarily $\nu$-invariant $\mathbb{P}$-names.

If $G$ is $\mathbb{P}$-generic, then it's not hard to show that $\{\mu[G]: \mu\in HI_\nu\}$ is a model of ZF; call it "$V[G]_\nu$". My question is this:

Does the model $V[G]_\nu$ have a snappy description as HOD of something, or as a symmetric submodel? By Grigorieff we know that it is such a model; but I don't immediately see how to describe it nicely as such.

(I'm particularly interested in the case when $\nu$ is a name for a set of reals, if that matters.)

I'm interested in the $HI_\nu$-construction because it gibes nicely with ideas from computability theory - specifically on a recasting of Steel forcing I've used, but also other things - and it makes a couple arguments I'm working on nicer. Of course, it has no broader mathematical value since there are provably no models of this form which aren't of better-known forms; but it's still something I find neat.

Improve this question
$\endgroup$
16
  • $\begingroup$ To be sure, a natural guess is $HOD(V\cup tc(\nu[G]))^{V[G]}$, but I don't see how to show that this is correct. $\endgroup$
    – Noah Schweber
    Commented Mar 11, 2017 at 23:32
  • $\begingroup$ Unfortunately for all of us, Grigorieff's work about finding the "generating set" (this $\nu[G]$ you seek) is negative in the sense that his proofs really just apply "almost abstract nonsense" in the form of the Reflection theorem. Almost good news, though, Yair Hayut and I have worked out a structure theorem for symmetric extensions as part of a recent work; not good news, though, we ditched it in favor of a far simpler notion that gave us cleaner results. It's half-written somewhere... $\endgroup$
    – Asaf Karagila ♦
    Commented Mar 12, 2017 at 11:24
  • $\begingroup$ So, are you interested in some specific $\Bbb P$, or at least some constraints on $\Bbb P$ (e.g. homogeneity assumptions, or something like that)? $\endgroup$
    – Asaf Karagila ♦
    Commented Mar 12, 2017 at 11:27
  • $\begingroup$ @AsafKaragila Re: Grigorieff's theorem, yeah :(. I'd be super interested in the structure theorem you and Yair wrote up - is that available anywhere? As to which $\mathbb{P}$ I'm interested in, there is a specific one (and a specific $\nu$), but I'm really interested in the general case: the invariant-names construction is in the spirit of stuff I've done before in higher reverse mathematics, and feels more computability-theoretic to me than HOD, and much more than symmetric submodels. In lieu of full generality, I'd be happy to understand the homogeneous (or weakly homogeneous) case, though. $\endgroup$
    – Noah Schweber
    Commented Mar 12, 2017 at 14:21
  • $\begingroup$ Well, we have a very... erm... basic write up. Probably riddled with inaccuracies, but I'll see him later this week, and I'll ask him if it's okay to send you a copy. As far as this go, the "invariant-names are more natural to me" sounds exactly like "Go symmetric extensions!", which is good. :-) $\endgroup$
    – Asaf Karagila ♦
    Commented Mar 12, 2017 at 14:24

1 Answer 1

Reset to default
1
$\begingroup$

The original definition has a problem with the hereditary requirement, unless your name is something particularly nice, e.g. a name for a subset of the ground model. But in the latter case, you just get a model of $\sf ZFC$, since you essentially get $V[A]$ for some $A\subseteq V$.

In the comments you suggest instead considering something which is invariant under interpreting some subset of $\nu$ correctly, but this turns out to be quite similar to just the usual symmetric extensions: you have a name, and you consider all the automorphisms which preserve it (not necessarily actual equality, generic equality is all the same here), and then you look at things which get preserved when some subset of $\nu$ is being interpreted "correctly" rather than the whole set.

This just translates to an automorphism group of $\Bbb P$, and some normal filter of subgroups. At least in the "relatively intuitive case".

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .