Timeline for $K$-ranks of some algebraic groups in Lubotzky's "Discrete groups, expanding graphs and invariant measures"
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Sep 21, 2022 at 23:31 | comment | added | F Zaldivar | In my first comment above, the even case is $\text{SO}_{2n}$ of course, and the odd case is $\text{SO}_{2n+1}$ | |
| Sep 21, 2022 at 18:00 | comment | added | F Zaldivar | For the odd case the torus is the same adjusting its action about a fixed axis. The hypothesis ensures that $\sqrt{-1}\in{\mathbb Q}_5$ and hence you work as you will do with ${\mathbb C}$ and the rotations are diagonalizable. | |
| Sep 21, 2022 at 17:54 | comment | added | F Zaldivar | By definition, a split torus $T$ over $K$ is diagonalizable over the given field $K$. For the special linear group, a torus $T\subseteq \text{SL}_n(K)$ has diagonal matrices of determinant $1$ and therefore its last entry is determined as you wrote (the inverse of the products of the $n-1$ previous ones). For the special orthogonal group ones has to consider the even case $\text{SO}_n$ and the odd case $\text{SO}_{n-1}$. In the even case the maximal torus is given as a block-diagonal matrices with blocks of size $2\times 2$ each one a rotation in dimension $2$. This gives the rank as $n$ . . | |
| Sep 21, 2022 at 11:39 | history | edited | user267839 | CC BY-SA 4.0 |
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| Sep 21, 2022 at 5:20 | history | edited | YCor | CC BY-SA 4.0 |
formatting, added tag
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| Sep 21, 2022 at 4:31 | comment | added | Asaf | 2. By this congruence you may generate isotropic vectors (and subspaces, essentially by choosing $(x_{1},\ldots, x_{n/2}, ix_{1},\ldots, ix_{n/2})$... Anyhow, this is not a research level question... | |
| Sep 21, 2022 at 4:22 | comment | added | user473423 | For 1: The set of diagonal elements is its own centraliser. Therefore, there is no bigger abelian subgroup. For 2: Find a torus, that does the job, I guess that the quadratic space decomposes into a sum of hyperbolic spaces (see Scharlau's book on quadratic forms). Then compare it to the case $K=\mathbb C$. If over the complex field, there is no bigger torus, you've got it. | |
| Sep 21, 2022 at 4:21 | review | Close votes | |||
| Oct 6, 2022 at 3:05 | |||||
| Sep 21, 2022 at 0:19 | history | edited | user267839 | CC BY-SA 4.0 |
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| Sep 20, 2022 at 23:43 | history | asked | user267839 | CC BY-SA 4.0 |