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$\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}$Let $G$ be a semisimple algebraic group and $K$ any field. Then the $K$-rank of $G$ is the maximal rank $r$ of a $K$-splitting torus $T \cong (K^*)^r $ contained as a subgroup in $G(K)$, the group of $K$-valued points of $G$.

In Alexander Lubotzky's 'Discrete Groups, Expanding Graphs and Invariant Measures' is claimed without proof on

I) page 27, 2nd line: that for every field $K$, the $K$-rank of $\SL_n(K) $ is $n-1$

II) page 29, Example 3.2.4 (B): the $\mathbb{Q}_5$-rank of $ \SO(n)(\mathbb{Q}_5)$ is $[\frac{n}{2}]$ (due to the text that's since $x^2 \equiv -1$ has a solution in $\mathbb{Q}_5$)

How to see these two statements? For I) we can obviously embed for every field $K$ the torus $(K^*)^{(n-1)} $ into $\SL_n(K)$ via $(a_1,..., a_{n-1}) \mapsto \operatorname{diag}(a_1,..., a_{n-1}, \frac{1}{\prod_{i=1}^{n-1} a_i})$. How to show that it's not possible to embed in $\SL_n(K)$ somehow a torus of bigger rank?

On II) I have no clue how to exploit that the congruence relation $x^2 \equiv_5 -1$ has a solution in $\mathbb{Q}_5$, to conclude that the rank of $ \SO(n)(\mathbb{Q}_5)$ is $[\frac{n}{2}]$.

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    $\begingroup$ 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. $\endgroup$
    – user473423
    Commented Sep 21, 2022 at 4:22
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    $\begingroup$ 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... $\endgroup$
    – Asaf
    Commented Sep 21, 2022 at 4:31
  • $\begingroup$ 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 $n$ . . $\endgroup$
    – F Zaldivar
    Commented Sep 21, 2022 at 17:54
  • $\begingroup$ 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. $\endgroup$
    – F Zaldivar
    Commented Sep 21, 2022 at 18:00
  • $\begingroup$ In my first comment above, the even case is $\text{SO}_{2n}$ of course, and the odd case is $\text{SO}_{2n+1}$ $\endgroup$
    – F Zaldivar
    Commented Sep 21, 2022 at 23:31

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