Timeline for Comparing algebraic group orbits over big and small algebraically closed fields
Current License: CC BY-SA 2.5
6 events
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Dec 20, 2010 at 16:10 | comment | added | Jim Humphreys | @Thomas This way of looking at the question is intriguing, though from a pedagogical viewpoint it adds extra prerequisites to the papers I cited. Something like a Lefschetz principle did seem to me to be lurking here. Certainly not all the specifics of the group actions in these papers can be needed for a comparison principle. At the same time, I wonder whether one can build into your approach the refined version in my added paragraph? In the applications, one wants orbit representatives to be compatible over the two fields beyond just counting numbers of orbits. | |
Dec 20, 2010 at 15:57 | comment | added | BCnrd | Thomas, thanks for the clarifications. | |
Dec 20, 2010 at 6:55 | comment | added | Thomas Scanlon | For a fixed G and fixed action of G on a variety X, it is fairly routine to formalize the assertion that G has n obits on X. For more sophisticated assertions, the coding can be more complicated. For instance, in the problem under consideration, we would apply completeness to the assertions that for every semisimple group G and action of G on a variety X for which G, X and the action are described by polynomials in at most n variables of degree at most d there are finitely many unipotent orbits. Finiteness is usually not a first-order condition but is for algebraically closed fields. | |
Dec 20, 2010 at 6:52 | comment | added | Thomas Scanlon | The Encyclopedia of Mathematics entry (eom.springer.de/T/t110050.htm ) on this transfer principle, a weak formalized version of the Lefshetz Principle, explains what I mean by complete. | |
Dec 20, 2010 at 6:35 | comment | added | BCnrd | Dear Thomas: The theorem is for connected semisimple groups, not all linear algebraic groups. Is that still first-order? Or if we fix a specific $G$ over some $k$ and only consider that single $G$ over all algebraically closed extensions of $k$ then is one in a setting where "completeness" applies (whatever it means; sorry, it is unfamiliar material to me)? That aside, "simple" is probably in the eye of the beholder. :) | |
Dec 20, 2010 at 6:24 | history | answered | Thomas Scanlon | CC BY-SA 2.5 |