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Apr 13, 2017 at 12:57 history edited CommunityBot
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Oct 25, 2010 at 21:22 vote accept Ben Webster
Oct 25, 2010 at 14:25 answer added Angelo timeline score: 8
Oct 25, 2010 at 12:26 comment added BCnrd Dear Angelo: Good point. So perhaps one should "remember" a shadow of connectedness of tori in the form that the resulting finite group action preserves generic points in fibers...though maybe there's another obstruction to flatness lurking around.
Oct 25, 2010 at 8:58 comment added Angelo This is false for finite groups, though. Consider two copies of the affine line glued at the origin, with an action of a cyclic group of order 2 switching the two copies, mapping to the affine line.
Oct 25, 2010 at 6:06 comment added Dave Anderson There's an old theorem of Iversen that, for any field $k$, for a smooth (& separated) $k$-scheme $X$ and a $k$-torus $T$, the fixed scheme $X^T$ is smooth. I'd wager that one could replace "smooth" with "Cohen-Macaulay" in this statement -- though the proof would have to be somewhat different from Iversen's. Anyone know a counterexample? (Since in your motivating situation, I think you know that the dimension of fibers of $X^T\to S$ is constant, knowing $X^T$ is C-M gives flatness over a smooth base $S$.)
Oct 25, 2010 at 0:11 comment added BCnrd Note also that by the "valuative criterion" for flatness, to handle the case when (noetherian) $S$ is reduced it suffices to handle the case when $S$ is the spectrum of a dvr.
Oct 25, 2010 at 0:09 comment added BCnrd Dear David: Good thought. Consider noetherian scheme $S$ and flat $S$-scheme $X$ of f. type equipped with action by $S$-torus $T$. Then $X^T$ exists as closed subscheme of $X$, & formation commutes with base change. To check flatness, WLOG $S$ is local. Pick prime $\ell$ invertible on $S$, and observe that the collection of finite etale $S$-subgroups $G_n=T[\ell^n]$ is relatively sch. dense in $T$. Thus, $X^T = X^{G_n}$ for large $n$ by noetherianness. By finite etale base change on $S$, $G_n$ constant. So problem reduces to analogue for action by finite gp of order invertible on $S$. Hmm...
Oct 25, 2010 at 0:01 comment added Ben Webster The behavior of reductive groups is quite different that of unipotent ones, so the counterexample doesn't worry me so much. After all, the degree 0 part of the ring has to vary flatly, and that's the coordinate ring of the categorical quotient, so nothing too horrible can happen.
Oct 24, 2010 at 23:36 comment added David Treumann Another thought: maybe one can reduce to the case of a finite group, thinking of a (char zero) torus as a limit of its finite subgroups.
Oct 24, 2010 at 23:35 comment added David Treumann This is false for general group actions. Think of the additive group acting on P^1 x C with x sending (z,t) to (z + tx, t); the dimension of the fixed points at t = 0 jumps. I haven't been able come up with an example like this for torus actions, so I'm starting to believe it.
Oct 24, 2010 at 22:06 history asked Ben Webster CC BY-SA 2.5