Timeline for GIT and singularities
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 30, 2020 at 16:35 | comment | added | Jason Starr | @Angelo. That is precisely how I made the normal counterexample below. I started with a normal counterexample for the cyclic group of order 2 and induced an example for the multiplicative group. | |
| Aug 30, 2020 at 11:00 | vote | accept | Simon Parker | ||
| Aug 30, 2020 at 9:44 | answer | added | Jason Starr | timeline score: 6 | |
| Aug 30, 2020 at 8:08 | comment | added | Angelo | This is clearly false for finite groups (take the union of the two coordinate axes in $\mathbb A^2$, with the involution that switches the two axes. For a connected example, embed the cyclic group into $\mathbb G_\mathrm{m}$, and consider the induced action. | |
| Aug 29, 2020 at 19:03 | comment | added | Jason Starr | I was wrong! There are, indeed, nilpotent matrices that are regular. | |
| Aug 29, 2020 at 19:02 | comment | added | Simon Parker | @JasonStarr I'm trying to understand your counterexample in the 2 x 2 case, and it doesn't seem to work (I may be wrong). The non-invertible matrices are the locus of $ad - bc = 0$, so the only singular point is the zero matrix. Hence, any non-zero nilpotent matrix is a non-singular point of the fibre above zero. What am I missing? | |
| Aug 29, 2020 at 18:54 | comment | added | Jason Starr | That is still not true. The GIT quotient of the conjugation action on the locus of non-invertible matrices is the affine space given by the coefficients of the characteristic polynomial (in particular, the constant coefficient equals $0$). The fiber over the origin in this affine space is the nilpotent cone inside the locus of non-invertible matrices, and every nilpotent matrix gives a singular point of this locus. | |
| Aug 29, 2020 at 18:42 | history | edited | Simon Parker | CC BY-SA 4.0 |
added 76 characters in body
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| Aug 29, 2020 at 18:18 | history | edited | Simon Parker | CC BY-SA 4.0 |
added 317 characters in body
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| Aug 28, 2020 at 21:28 | comment | added | Spenser | I think that singular points can be mapped to smooth points. Take $xy = 0$ and let $\mathbb{C}^*$ act by $z\cdot(x, y) = (zx, y)$. The quotient is $\mathbb{C}$. | |
| Aug 28, 2020 at 21:18 | history | asked | Simon Parker | CC BY-SA 4.0 |