# Are generic filters that produce the same forcing extension related by a ground-model automorphism?

Suppose $$M$$ is a countable transitive model of some fragment of $$\mathbf{ZFC}$$, $$\mathbb{P}\in M$$ is a forcing notion and $$G, H$$ are $$\mathbb{P}$$-generic such that $$M[G]=M[H]$$. Does it then follow that there is some automorphism $$\pi:\mathbb{P}\longrightarrow\mathbb{P}$$ such that $$\pi\in M$$ and $$\pi[G]=H$$?

If the answer is no, are there natural restrictions one could impose on $$\mathbb{P}$$ (maybe apart from $$\mathbb{P}$$ being finite) such that the above sentence holds?

• I have recently found Lemma 25.5 in an old version of "Set Theory" by Thomas Jech, which states the exact result im looking for, but only in a version for complete boolean algebras (i.e. when we take $\mathbb{P}$ to be a complete boolean algebra). However, i am unsure how to adapt it to my case, even when $\mathbb{P}$ is separative and we can use the representation theorem. – Hannes Jakob Jan 28 at 14:29
• The adaptation to posets other than complete Boolean algebras is, as I said in my answer, messy. One problem is that two posets can produce isomorphic complete Boolean algebras even when the posets are nowhere near isomorphic. Another is that the $b$ and $b'$ in my answer cannot generally be found in the posets (you can get one or the other in the poset you want, but not generally both). – Andreas Blass Jan 30 at 16:48

This works much better in terms of complete Boolean algebras. If $$\mathbb B$$ and $$\mathbb B'$$ are complete Boolean algebras and a $$\mathbb B$$-generic filter $$G$$ and a $$\mathbb B'$$-generic filter $$G'$$ generate the same forcing extension, then there are $$b\in G$$ and $$b'\in G'$$ such that the part of $$\mathbb B$$ below $$b$$ is isomorphic to the part of $$\mathbb B'$$ below $$b'$$ by an isomorphism in the ground model, that sends the restriction of $$G$$ to the restriction of $$G'$$.