Category theory question from Solovay

Category theory question from Solovay

[Robert Knighten]

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

>>      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