Timeline for Are the unipotent and nilpotent varieties isomorphic in bad characteristics?
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13 events
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| Oct 4, 2022 at 10:48 | comment | added | user148212 | @SeanC Thank you for pointing this out. You are right that there is an error in the counting (I don't remember how I got the number $2^{e+1}$, though..). | |
| Oct 3, 2022 at 8:38 | comment | added | SeanC | @user148212: if this point count were correct, the Riemann hypothesis would imply that one of the varieties was of dimension 1. In fact, the unipotent variety is isomorphic to $\mathbf{P}^2 - \{x^2 = yz\}$, and the nilpotent variety is isomorphic to $\mathbf{A}^2$. These have the same number of points over any finite field, and moreover they are equal in the Grothendieck ring of varieties. However, they are not isomorphic: for instance, the former has Picard group $\mathbf{Z}/2\mathbf{Z}$ and the latter has trivial Picard group. | |
| Feb 21, 2022 at 14:07 | history | edited | LSpice | CC BY-SA 4.0 |
Link and `\operatorname` while this is on the front page
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| Jul 19, 2021 at 7:51 | comment | added | user148212 | Maybe it is no more relevant, but here is a simple way to see why the above unipotent variety and the nilpotent variety are not isomorphic in char=2. Indeed, if they are isomorphic, then they are isomorphic over some finite field $F_{2^e}$, thus their numbers of $F_{2^e}$-points are the same, however, the latter is not true by a direct check (one is $2^{2e}$ and the other is $2^{e+1}$). Note that this argument applies to both the reduced varieties and the original varieties. | |
| May 5, 2017 at 15:22 | comment | added | Jay Taylor | I like this idea but I think there are some subtleties. I checked with Singular and adding that single term does not generate the radical. Also, how do you know the image of $X$ in the affine algebra is irreducible? It could happen that a polynomial generator becomes reducible in the quotient. For instance $X \equiv (X+1)(Y+1)$ in $K[X,Y]/\langle XY+Y+1\rangle$. | |
| May 5, 2017 at 7:14 | comment | added | Paul Levy | I think the reduced coordinate ring is given by the further condition $X(X+1) = ST$. Now consider the zero locus of $X$, which is given by $Y = S^2$, $Z = T^2$, $ST = 0$, i.e. is a union of two lines. So $X$ is an irreducible element which isn't prime, and therefore $\mathcal{U}$ isn't factorial, so can't be isomorphic to ${\mathbb A}^2$. | |
| May 5, 2017 at 2:22 | comment | added | Jay Taylor | @David. I don't think it's at all clear that the unipotent variety and nilpotent cone aren't isomorphic in the PGL_2 case in characteristic 2. Certainly they are not isomorphic as schemes because one is reduced and the other isn't. However why are their reduced schemes not isomorphic? In other words, why are the underlying varieties not isomorphic? As you say, the nilpotent cone is affine $2$-space $\mathbb{A}^2$ but you only know the unipotent variety is a smooth $2$-dimensional affine variety. I think it's non-trivial to show it's different from $\mathbb{A}^2$. | |
| May 4, 2017 at 21:30 | comment | added | Paul Levy | I am not very familiar with the classification of orbits in bad characteristic, but do Lusztig's nilpotent and unipotent pieces help you? My memory is sketchy but I thought the nilpotent pieces were maximal connected smooth locally closed subsets, and similarly for unipotent pieces. So, while the number of pieces is the same (if I remember correctly) one could rule out the possibility of an isomorphism from ${\mathcal N}$ to ${\mathcal U}$ by showing that one of the pieces in ${\mathcal N}$ is not isomorphic to the corresponding piece in ${\mathcal U}$. | |
| May 4, 2017 at 21:12 | comment | added | Jim Humphreys | @Paul: I should have specified $p$ very good for this (still hypothetical) comment about isomorphism of nilpotent varieties for all isogeny types of group, since there is clearly a problem otherwise. Anyway, my main concern here is bad primes, where it's hard to compute even low rank examples when $G$ is simply connected. | |
| May 4, 2017 at 20:40 | comment | added | Paul Levy | I'm not sure what you mean by "nilpotent varieties are isomorphic regardless of the isogeny type of $G$", but it is not true in general that if $G_1$ and $G_2$ are isogenous then ${\mathcal N}({\rm Lie}(G_1))$ is isomorphic to ${\mathcal N}({\rm Lie}(G_2))$. David's example of $\mathfrak{pgl}_2$ vs. $\mathfrak{sl}_2$ shows this. | |
| May 3, 2017 at 13:06 | comment | added | Jim Humphreys | P.S. Though all primes are technically good for type $A_\ell$, the treatment by Slodowy of type $A_1$ for $p=2$ does raise interesting questions. Presumably nilpotent varieties are isomorphic regardless of the isogeny type of $G$ in all Lie types, but for $G$ the unipotent variety can vary more (for example in type $A_\ell$ when $p|(\ell+1)$. This shows up in the revised hypothesis for Springer's theorem. But I'm still puzzled about what is happening for bad $p$. | |
| May 2, 2017 at 20:52 | comment | added | Jim Humphreys | This adjoint case did bother me, but I was inclined at first to exclude it. What you suggest does make a lot of sense, since there is also a problem here with Springer's original formulation for groups which are not simply connected (already for good $p$). | |
| May 2, 2017 at 20:42 | history | answered | David Stewart | CC BY-SA 3.0 |