Timeline for Understanding the unreducedness of a subscheme supported on fixed points
Current License: CC BY-SA 2.5
22 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Oct 24, 2010 at 22:07 | history | edited | Ben Webster♦ | CC BY-SA 2.5 |
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| Oct 24, 2010 at 4:50 | answer | added | David Treumann | timeline score: 3 | |
| Oct 23, 2010 at 20:01 | history | edited | Ben Webster♦ | CC BY-SA 2.5 |
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| Oct 23, 2010 at 19:31 | comment | added | Sasha | Angelo, a simple example of a 3-dimensional affine quadratic cone shows that one can't expect to have equality for crepant resolutions. Indeed, the small resolution of this singularity is crepant, and has 2 torus fixed points on the exceptional locus, while the singular point has length 1. | |
| Oct 23, 2010 at 17:00 | comment | added | Ben Webster♦ | The examples I'm working with are symplectic, and thus crepant. | |
| Oct 23, 2010 at 15:05 | history | edited | BCnrd | CC BY-SA 2.5 |
Fixed up the description of the fixed-point scheme.
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| Oct 23, 2010 at 13:39 | comment | added | Angelo | This is a very good question. On the (rather feeble) basis of some examples, one could conjecture that the length of the fixed point scheme of the singular variety is at most equal to that of the desingularization, and that you have equality if the resolution is crepant (this would be somehow related to your conditions). I wonder if one can relate this with one of the many McKay type conjectures floating around. | |
| Oct 23, 2010 at 11:02 | comment | added | S. Carnahan♦ | Your definition fixed point subscheme is a little off, because it is missing a base change. See also BCnrd's answer to mathoverflow.net/questions/3190/… | |
| Oct 23, 2010 at 7:41 | history | edited | Ben Webster♦ | CC BY-SA 2.5 |
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| Oct 23, 2010 at 6:58 | comment | added | Ben Webster♦ | OK, question rewritten. Brian and Dave, I really appreciate the comments; it was quite valuable to realize I was thinking about things all wrong. As a note to future readers, all comments above this one are referring to a substantially different version of the question. | |
| Oct 23, 2010 at 6:53 | history | edited | Ben Webster♦ | CC BY-SA 2.5 |
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| Oct 23, 2010 at 6:00 | comment | added | BCnrd | Dave, very good. In an earlier version of the question, what is now called the "ring" $B$ was called an "ideal". I had no idea what was up with that. Now it is all clear, and I agree with your argument that the two notions coincide. | |
| Oct 23, 2010 at 5:51 | comment | added | Ben Webster♦ | I guess you're right...I think I was thrown off by the fact that I was coming from a non-commutative situation where it really is important to thing of this as a subquotient, not a quotient. I'm going to put off rewriting the question for a little while, though. | |
| Oct 23, 2010 at 5:45 | history | edited | Ben Webster♦ | CC BY-SA 2.5 |
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| Oct 23, 2010 at 5:29 | comment | added | Dave Anderson | BCrnd & Ben: Isn't this equal to the fixed point scheme? Could be wrong, but I think two things are true: (1) if $X$ is an affine $\Bbb{G}_m$-scheme (=$\Bbb{Z}$-graded affine $k$-algebra $R$), then the scheme $X^T$ is Spec of $R/I$, where $I = R^{>0}R+R^{<0}R$ is the ideal generated by nonzero-degree elements; and (2) the obvious map $R^0/(R^0 \cap R^{<0} R^{>0}) \to R/I$ is an isomorphism. (Please correct me!) | |
| Oct 23, 2010 at 4:22 | comment | added | Ben Webster♦ | Sasha- I added an example where it's non-reduced (actually of arbitary length). Obviously, you do need both positive and negative weights, but I think you'll agree that the example is not very complicated. I don't really know exactly what kind of geometric information is being stored in this subscheme; if I did, I probably wouldn't have needed to ask the question. | |
| Oct 23, 2010 at 4:19 | comment | added | Ben Webster♦ | BCnrd- I'm interested in this ad hoc construction (it shows up naturally in a certain representation theoretic problem, but explaining all of that just seemed going way too far afield). The title was just me having to run off to seminar, and having trouble coming up with a good title (the other contender was "A commutative algebra question I cannot concisely summarize"). I don't think this is the "fixed point scheme" though it would be really awesome if it were. | |
| Oct 23, 2010 at 4:16 | history | edited | Ben Webster♦ | CC BY-SA 2.5 |
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| Oct 23, 2010 at 3:25 | comment | added | Sasha | I don't quite understand. If $R$ is a nonnegatively graded with $R^0 = k$ (so $Spec R$ is a cone) and the action of $G_m$ corresponds to this grading, then $R^{<0} = 0$, so $R^0/(R^0 \cap R^{0}R^{>0}) = k$, while the vertex of the cone (which is the only invariant point) can have arbitrary bad singularity. Can you give an example, when $R^0/(R^0 \cap R^{0}R^{>0})$ is nontrivial? | |
| Oct 23, 2010 at 0:46 | comment | added | BCnrd | Dear Ben: Let $k$ be the (arbitrary) ground field, and $T$ a $k$-torus acting on sep'td $k$-scheme $X$ of finite type. Are you interested in your "ad hoc" construction (for affine $X$), or the closed subscheme $X^T$ that represents the functor associating to any $k$-algebra $A$ the set of points in $X(A)$ that are fixed by the $T_A$-action on the $A$-scheme $X_A$? (Not obvious to me that these agree.) See Prop. A.8.10 of "Pseudo-reductive groups" for existence of $X^T$, and proof that ${\rm{Tan}}_x(X^T) = {\rm{Tan}}_x(X)^T$ for any $x \in X^T(k)$. That may help to tell you about dimensions. | |
| Oct 23, 2010 at 0:21 | history | asked | Ben Webster♦ | CC BY-SA 2.5 |