Equivalence of forcing automorphisms - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T05:12:42Z http://mathoverflow.net/feeds/question/104465 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/104465/equivalence-of-forcing-automorphisms Equivalence of forcing automorphisms Asaf Karagila 2012-08-11T09:19:22Z 2012-08-11T15:24:21Z <p>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: $$\pi\dot x = \lbrace(\pi p,\pi\dot y)\mid (p,\dot y)\in\dot x\rbrace$$</p> <p>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:</p> <blockquote> <p>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 $\mathscr G$?</p> </blockquote> <p>I am particularly interested in the case where $P$ is a Cohen forcing, in case it isn't true in general.</p> <p><strong>Edit:</strong> 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 $\mathscr G$?</p> http://mathoverflow.net/questions/104465/equivalence-of-forcing-automorphisms/104478#104478 Answer by Joel David Hamkins for Equivalence of forcing automorphisms Joel David Hamkins 2012-08-11T13:13:22Z 2012-08-11T13:18:56Z <p>It depends on the forcing notion, on the generic filter, on the group of automorphisms and on the name $\dot x$.</p> <p>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.</p> <p>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. </p> <p>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.</p> http://mathoverflow.net/questions/104465/equivalence-of-forcing-automorphisms/104496#104496 Answer by Andreas Blass for Equivalence of forcing automorphisms Andreas Blass 2012-08-11T15:24:21Z 2012-08-11T15:24:21Z <p>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.)</p>