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Do we have some rudimentary understanding of some properties that a forcing can have in order to guarantee that it doesn't violate the axiom of determinacy?

To be more specific, in Which forcings preserve (some) determinacy? the question, and answer, talk about preserving things like projective determinacy, or determinacy in $L(\Bbb R)$. But much like many other discussions of similar flavor, we just re-compute the classes or models in question there. My question is different.

Suppose that $M\models\sf ZF+AD$. Let $\Bbb P\in M$ be a notion of forcing with property $\varphi$, then $M^\Bbb P\models\sf ZF+AD$.

What can be said about $\varphi$?

For example, we know that if $\Bbb P$ adds a well-ordering of $\Bbb R$, or adds an ultrafilter over $\omega$, or so on, then it must violate $\sf AD$. I suspect that trivially we can say that any forcing which doesn't add subsets to $V_{\Theta^+}$, or some other large enough $\alpha$, will not violate $\sf AD$ either.

So to be more concrete, let us narrow down $\Bbb P$ a little bit.

Question. Let $M$ be a model of $\sf ZF+AD$, and suppose $\Bbb P$ is a forcing in $M$ which adds a new subset of $\Bbb R$. What properties will ensure that $M^\Bbb P\models\sf ZF+AD$?

(For the sake of the question, assuming $M\models\sf DC$ works just fine.)

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  • $\begingroup$ Could you clarify: do you mean that $\mathbb{P}$ adds a subset of $\mathbb{R}$, or only that it adds a subset of ${\cal P}(\mathbb{R})$? (Your phrasing, "adds a subset to ${\cal P}(\mathbb{P})$" could be interpreted as adding an element to that power set, or as adding a subset to the power set.) $\endgroup$
    – Joel David Hamkins
    Commented Aug 25, 2014 at 17:19
  • $\begingroup$ @Joel: Subset of $\mathcal P(\Bbb R)$. Since we can think of $\Bbb R$ as $\mathcal P(\omega)$, adding a real, a set of reals, or a set of sets of reals are all included in this definition. But both interpretations are interesting anyway. :-) $\endgroup$
    – Asaf Karagila ♦
    Commented Aug 25, 2014 at 17:27
  • $\begingroup$ But if you add only a set of sets of reals, but no reals and no set of reals, then of course AD is preserved, since it has to do with sets of reals and reals. $\endgroup$
    – Joel David Hamkins
    Commented Aug 25, 2014 at 17:33
  • $\begingroup$ Alright then, your point is a good one. Let me rule that option out. $\endgroup$
    – Asaf Karagila ♦
    Commented Aug 25, 2014 at 17:34

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