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I'm trying to generalize a theorem on $\operatorname{SL}_n(\mathbb Z)$ to the Chevalley groups over $\mathbb Z$. In the theorem, there is a heavy use in the element $e_{1,n}(1)$ where $$e_{1,n}(m)= \begin{pmatrix} 1 & 0 & \cdots & m\\ & 1 & & \vdots\\ & & 1 & 0\\ & & & 1\\ \end{pmatrix}, $$
together with a general element from the Bruhat big cell.

I was wondering if there is a way to characterize $e_{1,n}(1)$ in the general Chevalley groups over $\mathbb Z$?

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    $\begingroup$ It is one of two generators of the highest-root subgroup. $\endgroup$
    – LSpice
    Commented Aug 6, 2019 at 13:35
  • $\begingroup$ @LSpice Can you please elaborate on the construction of this subgroup or direct me to a good source? $\endgroup$
    – Ami
    Commented Aug 7, 2019 at 12:15
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    $\begingroup$ 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. $\endgroup$
    – LSpice
    Commented Aug 7, 2019 at 13:43
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    $\begingroup$ 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. $\endgroup$
    – LSpice
    Commented Aug 7, 2019 at 14:17
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    $\begingroup$ 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? $\endgroup$
    – Ami
    Commented Aug 8, 2019 at 14:34

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