Timeline for units in distinct division algebras over number fields---are they definitely not isomorphic as abstract groups?
Current License: CC BY-SA 2.5
11 events
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| Mar 1, 2017 at 3:08 | comment | added | Venkataramana | @Victor: your guess is correct. The algebraicity of an abstract homomorphism follows from Borel Tits in the isotropic case for arbitrary fields but for global fields, in the anisotropic case, it is still true, and is a consequence of the Margulis super-rigidity (this consequence is deduced in Margulis' book, Chapter 8) | |
| Jun 17, 2010 at 8:26 | history | edited | Victor Protsak | CC BY-SA 2.5 |
updated in view of Kevin's and BCnrd's examples
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| Jun 16, 2010 at 22:39 | comment | added | Boyarsky | Small clarification: in the above argument with 14.26 of Borel's book, I should have said "nontrivial split smooth connected unipotent subgroup" (the nontriviality is where one uses that the nilpotent element in the Lie algebra is nonzero, and it is what forces rank to be nonzero). Actually, one can get the conclusion by more direct means without 14.26 of Borel (using Jacobson-Morozov and simple connectedness of ${\rm{SL}}_2$), but no need to get into that here. | |
| Jun 16, 2010 at 19:49 | comment | added | Boyarsky | Kevin & Victor: if $G$ and $G'$ are connected ss groups over a non-archimedean local field $K$ of char. 0, any $K$-analytic isomorphism $G(K) \simeq G'(K)$ (same as topological isomorphism if $K = \mathbb{Q}_p$ for some $p$) is algebraic if both are simply connected or both adjoint. Indeed, $G(K)$ as a $K$-analytic group recovers Lie algebra and adjoint representation, and $G^{\rm{ad}}$ is the id. component of Zariski closure of the image of the adjoint representation of $G(K)$, so $G(K)$ recovers $G^{\rm{ad}}$ functorially. The simply connected central cover is functorial in isomorphisms. | |
| Jun 16, 2010 at 19:41 | comment | added | Boyarsky | Kevin, $g \mapsto g^{-1}$ is an algebraic group isomorphism! (Doesn't extend to an algebra isomorphism, so no contradiction.) I was mistaken about the 3-dimensional Lie algebras: I had in mind Jacobson-Morozov, but can't use it unless there's a nonzero nilpotent in the Lie algebra, and in the anisotropic case there isn't any such element (since 14.26 of Borel's book + Galois descent with intersections in char. 0 would produce a split smooth connected unipotent subgroup inside, which forces rank $> 0$ in any connected reductive group). Good news in the local case in the next comment! | |
| Jun 16, 2010 at 19:04 | comment | added | Kevin Buzzard | If D'=D^{opp} then I think D_v^* and (D'_v)^* are isomorphic as topological groups but D_v and D_v' may not be isomorphic as F_v-algebras. So I think that your "I think" might be too optimistic---possibly this is the only thing that can go wrong though... | |
| Jun 16, 2010 at 17:03 | comment | added | Boyarsky | The Lie algebra of $D_v^{\times}$ as an $F_v$-analytic group is the same as for the analogous algebraic group: direct product of 1-dimensional central part and 3-dimensional semisimple part that must be $\mathfrak{sl}_ 2$ since we're in characteristic 0. The same goes with $D'_v$. Hence, $D_v^{\times}$ and ${D'}_v^{\times}$ have $F_v$-isomorphic Lie algebras, so they're $F_v$-analytically isomorphic near the identity. So it seems a fine line to infer anything about the algebraic group from the analytic group (equivalently from the topological group, when $F_v = \mathbb{Q}_p$!). | |
| Jun 16, 2010 at 13:32 | history | edited | Victor Protsak | CC BY-SA 2.5 |
added remark suggested by BCnrd
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| Jun 16, 2010 at 13:29 | comment | added | Victor Protsak | I agree. I started optimistically with "I think that this follows from BT", then replaced it with "the corresponding local fact follows from BT", and finally with conservative "this is close to... unfortunately...". Does $D^{\times}_v \simeq D'^{\times}_v$ as topological groups imply $D_v\simeq D'_v$, at least? | |
| Jun 16, 2010 at 13:07 | comment | added | BCnrd | Victor, perhaps it would be good to clarify that the abstract hom. may also "imply" a field hom. mixed up with the algebraic hom. (e.g., the abstract hom. could just be a field aut. applied to rational points of a group defined over a subfield, so no nontrivial alg. aut. in sight). Also, it is unclear (to me) how to promote an isomorphism $D^{\times} \simeq {D'}^{\times}$ of rational points over the number field to one of rational points over local fields, so I don't see how Kevin could access your suggestion. (I had considered a related idea using the BT paper, but gave up due to this.) | |
| Jun 16, 2010 at 12:54 | history | answered | Victor Protsak | CC BY-SA 2.5 |