# Why does elements in the parabolic subgroups of $SL(n,k)$ take the form of upper block triangular matrices?

Let $$G$$ be either $$SL(n,k)$$ (or I guess any linear subgroup of $$GL(n,k)$$) for a field $$k$$. And $$P$$ be a parabolic subgroup of $$G$$, I have seen the fact that any $$A\in P$$ looks like the block diagonal matrix:

$$A = \left( \begin{matrix} A_1 & * & * \\ 0 & \ddots & * \\ 0 & 0 & A_m \end{matrix} \right)$$

But how to prove this fact? Any solution or reference will be appreciated!

• An appropriate answer to this depends on your context. If nothing else, you'd probably want to say either that "every parabolic is conjugate to" a block upper-triangular subgroup of the sort you mention, or that these are the "standard parabolic subgroups". How do you define "parabolic..."? Commented Dec 14, 2020 at 22:33
• See 9. in the lecture here, together with some good references. Commented Dec 14, 2020 at 22:34

Do you know how to show that the set of upper triangular matrices $$B$$ of $$G= \operatorname{SL}_n(k)$$ forms a Borel subgroup of $$G$$? This follows from the Lie-Kolchin theorem.
A parabolic subgroup of $$G$$ is by definition a subgroup $$P$$ of $$G$$ such that $$gPg^{-1}$$ contains $$B$$ for some $$g \in G$$. So what you want to show is that all subgroups of $$G$$ containing $$B$$ are of the block diagonal form.
If you let $$T$$ be the group of diagonal matrices in $$G$$, and $$N$$ the normalizer of $$T$$ in $$G$$, this follows by showing that $$(G,B,N)$$ form a Tits system. In any Tits system, the subgroups of $$G$$ containing $$B$$ are completely classified in terms of the Weyl group $$N/(B \cap N)$$. If you write out what these are, you will see they all take the form you want.