Bob Solovay posted the following question to the Foundations of Mathematics mailing list. I can't answer it, but it looked likely that members of this list would have useful comments, so with his permission I am posting it here. [There were two postings but I have just combined them.] -- Bob -- Robert L. Knighten Robert@Knighten.org Date: Mon, 3 Apr 2006 09:18:33 -0700 (PDT) From: solovay@Math.Berkeley.EDU Subject: [FOM] Fraenkel-Mostowski-Specker method and category theory To: "Foundations of Mathematics" <fom@cs.nyu.edu> I have come across a curious question related to the FMS method and category theory. Before stating the problem I need to recall some definitions. Let G be a group. A normal filter of subgroups of G, Gamma, is a non-empty collection of subgroups of G which has the following properties: 1) If H_1 and H_2 are members of Gamma then so is their intersection H_1 \cap H_2; 2) If H is in Gamma and K is a subgroup of G which is a supergroup of H, then K is in Gamma; 3) If H is in Gamma and x is an element of G then the conjugate subgroup xHx^{-1} is in Gamma. Now let G be a group and Gamma a normal filter of subgroups of G. To this data we associate a category C(G,Gamma) as follows: The objects of our category consist of sets X equipped with an action of H on X [ for some H in the filter Gamma] such that for every x in X the isotropy subgroup H_x [consisting of those elements of H which fix the element x] lies in Gamma. Now let X_1 and X_2 be objects of our category carrying actions of H_1 and H_2 respectively. The morphisms of our category from X_1 to X_2 are those maps from X_1 to X_2 [in the category of sets] which (for some subgroup K of H_1 \cap H_2) lying in Gamma are K-equivariant. The basic question is: when are C(G_1, Gamma_1) and C(G_2,Gamma_2) equivalent categories: Here are some obvious sufficient conditions: (a) If there is an isomorphism of G_1 with G_2 that carries Gamma_1 onto Gamma_2 then the two categories are equivalent. (b) Let G be a group and Gamma a normal filter of subgroups of G. Let H be a subgroup of G lying in Gamma, and let Gamma' be the collection of all subgroups of H lying in Gamma. Then C(G, Gamma) and C(H, Gamma') are equivalent. My question is this: Are (a) and (b) the only reasons that two such categories are equivalent. That is, if C(G_1, Gamma_1) and C(G_2, Gamma_2) are equivalent then are there subgroups H_1 of G_1 and H_2 of G_2 [lying in the respective filters] such that letting Gamma_1' and Gamma_2' be the evident restricted filters we have C(H_1, Gamma_1') equivalent to C(H_2, Gamma_2'). I suspect that the answer is no. Also welcome [for the undisclosed application I have in view] would be additional sufficient criteria other than (a) and (b). --Bob Solovay

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My question is this: Are (a) and (b) the only reasons that two such categories are equivalent. That is, if C(G_1, Gamma_1) and C(G_2, Gamma_2) are equivalent then are there subgroups H_1 of G_1 and H_2 of G_2 [lying in the respective filters] such that letting Gamma_1' and Gamma_2' be the evident restricted filters we have C(H_1, Gamma_1') equivalent to C(H_2, Gamma_2').

I must be missing something about this question. Isn't the answer automatically yes once we put H_1 = G_1 and H_2 = G_2 ? I feel tempted to replace the conclusion by (H_1, Gamma_1') isomorphic to (H_2, Gamma_2').

regards, Volodya Shavrukov

You are right. I stated my question wrongly and what I intended is your proposed repair. --Bob