Timeline for Characterize an element of $\operatorname{SL}_n(\mathbb Z)$
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Aug 8, 2019 at 18:01 | comment | added | LSpice | (Visually, they're the simple roots because they lie on the main super-diagonal; i.e., they're not in the derived subgroup $[U, U]$ of the unipotent radical $U$, which consists of the "upper uni-triangular" matrices (with 1's on the diagonal).) | |
| Aug 8, 2019 at 17:50 | comment | added | LSpice | In terms of the Borel corresponding to upper-triangular matrices, the simple roots are the $\alpha(i, i + 1)$, so the height of $\alpha(i, j)$ is $j - i$. | |
| Aug 8, 2019 at 14:34 | comment | added | Ami | OK, I think I got most of it, but how do I calculate the height of the roots $\alpha(i,j): diag(a_1,...,a_n)\to a_i-a_j$?what is the default simple system? | |
| Aug 7, 2019 at 14:17 | comment | added | LSpice | The root subgroup associated to $\alpha$ is loosely the 'exponentiation' of the subspace of the Lie algebra on which the adjoint action of the torus is through the character $\alpha$. Of course you pick the approach that works best for you, but I encourage you to try to think abstractly at least in parallel with the matricial approach; the longer you think just of matrices, the harder it is to switch perspectives later. | |
| Aug 7, 2019 at 14:11 | comment | added | Ami | Thanks, I do know about root systems and highest roots however it's a first I hear about root subgroups. I do not have a lot of experience in this subject so until the abstract part will sink in I use $SL_n$ to try and understand better. | |
| Aug 7, 2019 at 13:43 | comment | added | LSpice | Also, based on your one other question mathoverflow.net/questions/336671/… that I happened to see, it looks like you want to think of the Chevalley groups in a very "coördinatised" form. While it's handy to have this option, at some point it is usually necessary to get used to the idea that there is an abstract structure theory for these groups, and to work with that structure theory rather than with a specific faithful representation when possible. | |
| Aug 7, 2019 at 13:40 | comment | added | LSpice | It'd help to know a little bit more about your background. Do you know about root systems, highest roots, and root subgroups? (I assumed from your mention of the big cell that you did.) If you do, then there's nothing to it; I literally mean to refer to the root subgroup associated to the highest root (with the usual Borel–torus pair for $\operatorname{SL}_n$, the highest root is $\operatorname{diag}(a_1, \dotsc, a_n) \mapsto a_1 a_n^{-1}$, and the subgroup consists of all matrices $e_{1, n}(m)$, in your notation). If not, then maybe Carter? | |
| Aug 7, 2019 at 12:15 | comment | added | Ami | @LSpice Can you please elaborate on the construction of this subgroup or direct me to a good source? | |
| Aug 6, 2019 at 13:36 | history | edited | LSpice | CC BY-SA 4.0 |
Minor typos
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| Aug 6, 2019 at 13:35 | comment | added | LSpice | It is one of two generators of the highest-root subgroup. | |
| Aug 6, 2019 at 13:30 | history | asked | Ami | CC BY-SA 4.0 |