Timeline for Does every reductive group scheme admit a maximal torus?
Current License: CC BY-SA 3.0
20 events
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| Jun 30, 2023 at 22:22 | comment | added | LSpice | The statement that "any smooth reductive group over $k$ admits a maximal torus over $k$" is trivial, since there is an upper bound on dimensions of tori, and a torus that achieves that dimension is maximal. The magic of Grothendieck's theorem is that any smooth reductive group $G$ over $k$ admits a maximal torus $T$ such that $T_E$ is a maximal torus in $G_E$ for every field extension $E/F$. | |
| S May 14, 2014 at 9:13 | history | bounty ended | CommunityBot | ||
| S May 14, 2014 at 9:13 | history | notice removed | CommunityBot | ||
| May 10, 2014 at 15:40 | comment | added | Ben Wieland | I don't know this stuff and was just trying to reconcile contradictory claims. The abelianization of a Borel doesn't depend on the choice of Borel and thus descends everywhere. In the quasi-split case, the torus of the Borel is better than other tori, though not necessarily a subgroup over general bases. In the general case, I think this construction yields the torus of the quasi-split inner form, not remotely like a subgroup of the original group, even in the field case (contrary to my previous claim). I don't know the purpose. | |
| May 7, 2014 at 11:14 | vote | accept | Daniel Loughran | ||
| May 7, 2014 at 8:05 | comment | added | Daniel Loughran | @Ben: Could you explain what this "other way" is? | |
| May 7, 2014 at 5:12 | comment | added | Ben Wieland | There are several definitions of a torus of a group. The super-restrictive definition, usually called a split torus, is a subgroup isomorphic to $\mathbb G_m^n$. The common restrictive definition is a subgroup which is a form of that group. But there is another way to associate a form of a torus to a reductive group that is unique, hence descends to any setting. And its uniqueness makes it better than the subgroup torus, although they are isomorphic. | |
| May 6, 2014 at 13:56 | answer | added | Kestutis Cesnavicius | timeline score: 13 | |
| May 6, 2014 at 13:45 | answer | added | David E Speyer | timeline score: 10 | |
| May 6, 2014 at 13:13 | comment | added | Daniel Loughran | This is very useful to know, thanks Kestutis. | |
| May 6, 2014 at 13:00 | comment | added | Kestutis Cesnavicius | Working etale locally for a maximal torus is indeed not necessary: Zariski locally suffices, see [SGA3, XIV, 3.20] (and the (*) footnote there in case it is of interest). | |
| S May 6, 2014 at 7:28 | history | bounty started | Daniel Loughran | ||
| S May 6, 2014 at 7:28 | history | notice added | Daniel Loughran | Improve details | |
| May 6, 2014 at 7:24 | comment | added | Daniel Loughran | I hope this is ok, but since there is some contention about the answer provided, I decided to unaccept it and offer a bounty. If anybody is able to conclusively answer my question, I would be most obligued. | |
| May 5, 2014 at 11:20 | vote | accept | Daniel Loughran | ||
| May 6, 2014 at 7:24 | |||||
| May 5, 2014 at 8:20 | vote | accept | Daniel Loughran | ||
| May 5, 2014 at 11:20 | |||||
| May 2, 2014 at 16:10 | comment | added | Daniel Loughran | I have been mostly reading Brian Conrad's notes on reductive group schemes. Here he has a result which says that a maximal torus exists étale locally (this is also in SGA3 I believe). However I don't know of an example which illustrates that one really does need to work étale locally. | |
| May 2, 2014 at 15:58 | answer | added | Victor Petrov | timeline score: 6 | |
| May 2, 2014 at 15:50 | comment | added | Jim Humphreys | What sources have you consulted? The most standard treatment of reductive schemes seems to be Demazure's Expose XIX in SGA3 (see section 2): math.jussieu.fr/~polo/SGA3. Here the usual assumption is that $S$ is an arbitrary scheme. | |
| May 2, 2014 at 12:04 | history | asked | Daniel Loughran | CC BY-SA 3.0 |