Equivalence of forcing automorphisms - MathOverflow

Equivalence of forcing automorphisms

2012-08-11T09:19:22Z

Suppose that $P$ is a forcing poset in $V$. If $\pi$ is an automorphism of $P$ then $\pi$ extends to automorphisms of the names by induction: $$\pi\dot x = \lbrace(\pi p,\pi\dot y)\mid (p,\dot y)\in\dot x\rbrace$$

I've been stuck on the following proposition for quite some time now and I don't see an argument for, nor an obvious counterexample again:

Let $\mathscr G$ be a group of automorphisms of $P$. Suppose that $\dot x$ is a $P$-name and $G$ is $P$-generic over $V$ then the equivalence relation over $\mathscr G$ defined as: $$\pi\sim_G\sigma\iff (\pi\dot x)^G=(\sigma\dot x)^G$$ is multiplicative? Namely, is the set $\lbrace\pi\in\mathscr G\mid\pi\sim_G\mathrm{id}_P\rbrace$ is a normal subgroup of $\mathscr G$?

I am particularly interested in the case where $P$ is a Cohen forcing, in case it isn't true in general.

Edit: To restrict the question even more (after Joel's answer), what if we assume that $\dot x$ is hereditarily symmetric with respect to a normal filter of subgroups over $\mathscr G$?

Answer by Joel David Hamkins for Equivalence of forcing automorphisms

2012-08-11T13:13:22Z

It depends on the forcing notion, on the generic filter, on the group of automorphisms and on the name $\dot x$.

First, there are some trivial cases where it is normal, such as the case of a check name $\check x$, which is fixed by every automorphism $\pi$, and so that subgroup is the whole group and hence normal. Similarly, if the forcing notion $P$ is rigid, then it has no nontrivial automorphisms at all, in which case the subgroup again is normal.

Meanwhile, in the case of Cohen forcing, your main case, it is sometimes not normal, depending on the name $\dot x$. Consider the case of the canonical name for the generic object $\dot x=\dot G$, and let $\pi$ be an automorphism that acts only below a condition that happens not to be in the filter $G$. For example, perhaps $\pi$ swaps the second bit of the Cohen real, but only if the first bit is $0$, whereas the first bit of $G$ happens to be $1$. Thus, $(\pi\dot G)^G=(\dot G)^G=G$, since this operation has no effect on the part of the name relevant for $G$. Meanwhile, let $\tau$ be an automorphism that moves that cone into $G$, such as flipping the first bit of the Cohen real. Observe now that $(\tau^{-1}\pi\tau\dot G)^G\neq G$, since the former will flip the second bit of $G$, and so the subgroup is not normal.

A similar argument will work in many other cases, with highly homogeneous forcing, and so I think we should think of the property usually as failing except in very special circumstances. The way I think about it is this: your equivalence relation is able to ignore huge parts of the automorphism, since it is interpreting the name by the filter $G$, and $G$ is very local to particular conditions in the poset, but the normality requirement on the automorphisms is much more global, and so they will sometimes conflict.

Answer by Andreas Blass for Equivalence of forcing automorphisms

2012-08-11T15:24:21Z

I think Joel's example can be easily modified to use a hereditarily symmetric name. Use Cohen's original model for the negation of choice, i.e., use Cohen forcing to add an $\omega$-sequence of generic reals, and let the filter be the finite-support filter on the group $\mathcal G$ of permutations of the positions in the $\omega$-sequence. Then use Joel's example working on the first real in the $\omega$-sequence. (If you don't like the fact that Joel's $\pi$ and $\tau$ aren't in $\mathcal G$, then adjoin them and extend the filter accordingly.)