Timeline for Are linear algebraic groups rigid?
Current License: CC BY-SA 3.0
9 events
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Aug 28, 2014 at 10:29 | history | edited | Qfwfq | CC BY-SA 3.0 |
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Aug 28, 2014 at 3:54 | answer | added | user54268 | timeline score: 10 | |
Aug 28, 2014 at 3:43 | comment | added | user54268 | @StephenGriffeth: Yes, though I'd prefer to say "infinitesimally rigid" (and be explicit about the connectivity assumption; hmm, perhaps fibral connectivity can be dropped, but I haven't thought it through). The proof rests on quite a bit of the theory of reductive group schemes in SGA3. | |
Aug 28, 2014 at 2:04 | comment | added | Stephen | @user54268 So, you are saying that reductive groups are rigid, yes? | |
Aug 28, 2014 at 2:00 | comment | added | user54268 | If $G$ and $H$ are smooth affine group schemes with connected reductive fibers over an artin local ring $R$ (with arbitrary residue field) and their geometric special fibers are isomorphic then the functor ${\rm{Isom}}(G,H)$ is represented by a disjoint union of smooth affine $R$-schemes, as can be checked over a finite etale cover that splits $G$ and $H$ (and then using etale descent, which is avoids some effectivity problems since Spec($R$) is a fat point). Thus, an isomorphism of special fibers (if one exists) always lifts to an $R$-isomorphism. | |
Aug 28, 2014 at 1:57 | comment | added | Stephen | @Tony Pantev Huh. I guess I just wrote down this. For some reason your comment didn't show up while I was writing my answer... | |
Aug 28, 2014 at 1:56 | answer | added | Stephen | timeline score: 5 | |
Aug 28, 2014 at 1:48 | comment | added | Tony Pantev | The group law on a unipotent algebraic group deforms non-trivially in general. For instance by scaling the symplectic form you get a one parameter deformation of the three dimennsional Heisenberg group (upper triangular $3\times 3$ matrices with ones on the diagonal) to the additive group of a $3$ dimensional vector space. | |
Aug 27, 2014 at 23:35 | history | asked | Qfwfq | CC BY-SA 3.0 |