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Consider algebraic manifold M (e.g. vector space) and algebraic group G (e.g. unitriangular matrices UT(n)) acting on it.

Question: Is there some way to put algebraic structure on M/G i.e. on the SET of ALL orbits ("stack" "algebraic space" whatever) ?

More precisely I am interested in the following phenomenons: M1 and M2 are NOT isomorphic as G-manifolds, but M1/G and M2/G are "isomorphic" as these kind of "stacks-whatever" and isomorphism cannot be seen on the level of standard invariant-theory factors e.g. Fun(M1)^G and Fun(M2)^G are different.

I.e. geometrically I want such kind of notion of "isomorphism" which will not respect dimensions of orbits, however will respect moduli spaces of orbit. Something like this may happen: fixed POINTS in M1 could be mapped to orbits of highest dimension in M2.

Have you seen such examples ? Or in "stack" theory notion of "isomorphism" such things are impossible ?

(In my situation I see some examples of this type, but I do not know what kind structures to put on M/G and what should be corresponding notion of "isomorphism". I just see that some things are "similar" in some informal way and so hope that may be there is some theory which has been already developed).


Some background In Kirillov's "orbit method" it is important to study the set of ALL coadjoint orbits - actually it has natural factor topology. This is an example where whole set M/G appears in a natural way.
Moreover there is Brown's theorem which states that Kirillov's correspondence between coadjoint orbits and irreps is actually continuous with respect to factor topology on the set of coadjoint orbits and some natural topology on the set of irreps. (At least for nilpotent groups). There was some recent generalizations http://arxiv.org/abs/math/0608126

Question can one think about further generalization of Brown's theorem - define somehow "stack-or-whatever" structure on the set of all irreps and prove that it is "isomorphic" to "stack-or-whatever" structure on the set of coadjoint orbits ? (Where the last one comes from M/G , where $M=g^*$).

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    $\begingroup$ From just reading the title, is the answer Madam / Gentlemen? :-) $\endgroup$ – Seva Oct 24 '12 at 20:42
  • $\begingroup$ @Seva об этом я и не подумал ))))) прикольно )))) $\endgroup$ – Alexander Chervov Oct 24 '12 at 20:44
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    $\begingroup$ It's a dry cleaning service in Chicago: yelp.co.uk/biz/m-g-cleaners-chicago . Is there a prize for least informative title? $\endgroup$ – Tom Leinster Oct 24 '12 at 23:38
  • $\begingroup$ If only the questioner had framed it as let $M/G = D$. Then Miller Golden Draft would be my answer to win 'least informative' $\endgroup$ – aginensky Oct 25 '12 at 14:03

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