Timeline for Soft(?) algebraic groups question
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 11, 2012 at 4:33 | comment | added | Misha | I meant published, not available: For some reason, they were never officially published in English and have essentially the same status as, say, Thurston's Lecture Notes. One more remark: The point of my remark is that every complex semisimple Lie group (like $H$ in your question) can be realized as the set of complex points of a group scheme over the integers, i.e., Chevalley group. | |
| Mar 11, 2012 at 1:26 | comment | added | Igor Rivin | @Misha: thanks for the Steinberg reference. Certainly, the thing you pointed to is English, since these are lectures at Yale (a notoriously English-speaking university). | |
| Mar 10, 2012 at 20:14 | comment | added | Misha | In my comments above I only looked at the case when $G=SL(n, {\mathbb C})$ with the standard integer structure, so it is split. I do not know enough about non-split groups to say much in general. | |
| Mar 10, 2012 at 19:17 | comment | added | anon | Well first you need to do it over Q. If H and G are semisimple and G is split over Q, you can take the split form of H. If G is not split, it's a twist of the split form, so you need the cocycle to come from one on H. Looks dubious to me. | |
| Mar 10, 2012 at 17:25 | comment | added | Misha | It is not particularly easy but could be found for instance in Steinberg's "Lectures on Chevalley Groups" math.ucla.edu/~rst (published in Russian but not in English, I think). One can trace this result to rationality of characters of algebraic tori. | |
| Mar 10, 2012 at 16:45 | answer | added | Jim Humphreys | timeline score: 7 | |
| Mar 10, 2012 at 15:34 | comment | added | Igor Rivin | @Misha: I WAS thinking of semisimple when asking the question! Why is the answer positive then? Is this obvious? | |
| Mar 10, 2012 at 14:02 | comment | added | Vladimir Dotsenko | Along the lines of Misha's counterexample, is there a reason to expect the subgroup $\exp(t(E_{1,2}+\sqrt{2}E_{2,3}))$ of $SL_3(\mathbb{C})$ to be defined over $\mathbb{Z}$? | |
| Mar 10, 2012 at 11:57 | comment | added | Misha | Igor, another thought: If your groups $G$ is $SL(n, {\mathbb C})$ with the standard integer structure (which is all what you probably care about) and $H$ is semisimple, then the answer is positive. I do not know enough about integer structures to make a more general statement even in the $SL(n, {\mathbb C})$ case. | |
| Mar 10, 2012 at 11:20 | comment | added | Misha | Igor, take $G={\mathbb C}^2$ with obvious integral structure and the subgroup given by equation $z=\sqrt{2}w$. Since $G$ is an abelian group, no conjugation will help you. Thus, you probably want to assume that $G$ is semisimple and $H$ is too. Then the answer could be positive. | |
| Mar 10, 2012 at 6:29 | comment | added | naf | I think there exist continuous families of nilpotent complex Lie algebras of dimension $7$. The general such Lie algebra would give rise to a unipotent algebraic group which cannot be defined over $\mathbb{Z}$. If you embed such a group in $SL_n$ for some $n$ you would get an algebraic subgroup $H$ such that no conjugate is defined over $\mathbb{Z}$. | |
| Mar 10, 2012 at 3:17 | comment | added | Mariano Suárez-Álvarez | @Qiaochu, sure. | |
| Mar 10, 2012 at 2:44 | comment | added | Qiaochu Yuan | Is a finite cyclic subgroup of $\text{SL}_2(\mathbb{C})$ defined over $\mathbb{Z}$? | |
| Mar 10, 2012 at 1:13 | history | asked | Igor Rivin | CC BY-SA 3.0 |