Timeline for Do rational points in a split reductive group act transitively on the orbits of the Cartan subalgebra (w.r.t. automorphism group of Lie algebra)?
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Mar 12, 2020 at 16:52 | answer | added | Mikhail Borovoi | timeline score: 2 | |
| Mar 12, 2020 at 13:33 | comment | added | Mikhail Borovoi | OK, tonight I will write an answer containing all this. | |
| Mar 12, 2020 at 13:12 | comment | added | S. Carnahan♦ | @MikhailBorovoi Thank you for this argument. Could you explain why we may assume $G_R$ is semisimple and simply connected? | |
| Mar 11, 2020 at 18:29 | comment | added | Mikhail Borovoi | (cont.) I explain how we obtain $T'_R$. We may assume that ${\frak g}_R$ is semisimple and that $G_R$ is semisimple and simply connected. Then ${\rm Aut}\,{\frak g}_R={\rm Aut}\, G_R$, and we set $T'_R=a(T_R)$. | |
| Mar 11, 2020 at 17:54 | comment | added | Mikhail Borovoi | (cont.) (at least in char 0). Indeed, let $a$ be an automorphism of ${\frak g}_R$. Write ${\frak t}'_R=a({\frak t}_R)$. Then it is easy to see that ${\frak t}'_R={\rm Lie}(T'_R)$ for some split maximal torus $T'_R$ of $G_R$. Then there exists $g\in G_R(R)$ such that $T'_R=g\cdot T_R\cdot g^{-1}.$ Then ${\frak t}'_R={\rm Ad}(g)({\frak t}_R),$ as required. | |
| Mar 11, 2020 at 17:38 | comment | added | Mikhail Borovoi | When $R$ is a field, the answer to Question 1 is Yes, with the same proof as for Question 2. | |
| Mar 11, 2020 at 3:39 | history | edited | S. Carnahan♦ | CC BY-SA 4.0 |
Affirmative answer to question 2 in the field case, and possibly Zariski locally in R.
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| Mar 10, 2020 at 22:21 | comment | added | S. Carnahan♦ | @YCor Thank you for cleaning up the question text, and for taking the time to provide a very interesting example (even though it is not directly applicable to my question). | |
| Mar 10, 2020 at 22:20 | comment | added | S. Carnahan♦ | @LSpice Yes, $M$ is the character lattice. Sorry, I should have explained that in the text. | |
| Mar 10, 2020 at 21:43 | comment | added | YCor | @LSpice it's a typo: I just mean neither $3$ or $-3$ is a sum of 2 squares (in $\mathbf{Q}$. Yes I didn't guess whether OP's exclusively interested in split $T$ although it seems so; anyway I did the comment. | |
| Mar 10, 2020 at 21:32 | comment | added | LSpice | @YCor, I thought the question was asking specifically about the case where $T$ is split? Also, what do you mean by asking whether a diagonal matrix is or is not a sum of two scalar squares? | |
| Mar 10, 2020 at 21:00 | comment | added | YCor | (With $R$ field, $T$ nonsplit) For $G=\mathrm{SL}_2$, write $s=\mathrm{diag}(3,1)$ (so neither $s$ nor $-s$ is sum of two squares in $\mathbf{Q}$), $T$ the standard $\mathrm{SO}_2$. Then $sT_\mathbf{Q}s^{-1}$ is not in the $\mathrm{SL}_2(\mathbf{Q})$-orbit of $T_\mathbf{Q}$. Indeed, if it were equal to $s_1T_\mathbf{Q}s_1^{-1}$ with $s_1\in\mathrm{SL}_2(\mathbf{Q})$, then $t=s_1^{-1}s\in\mathrm{GL}_2(\mathbf{Q})$ normalizes $T_\mathbf{Q}$, so is in $\mathrm{O}(2)_\mathbf{Q}$ and has determinant $3$. But the determinant of a matrix in $\mathrm{O}(2)$ has determinant of the form $\pm (x^2+y^2)$. | |
| Mar 10, 2020 at 20:21 | comment | added | LSpice | (Based on my skimming of math.stanford.edu/~conrad/papers/luminysga3.pdf to try to answer your question, I guess that $M$ is the character lattice of $T$.) | |
| Mar 10, 2020 at 18:33 | comment | added | LSpice | In $(G, T, M)$, I guess $G$ is the ambient reductive group and $T$ is a split maximal torus, but what is $M$? Anyway, over $R$ a field, all split tori are $G(R)$-conjugate, so the answer to Question 2 in that case is 'yes'. | |
| Mar 10, 2020 at 18:22 | history | edited | YCor | CC BY-SA 4.0 |
minor formatting
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| Mar 10, 2020 at 18:10 | history | asked | S. Carnahan♦ | CC BY-SA 4.0 |