# Levi decomposition for the parabolic subgroups

This question is for the algebraic groups. I find I cannot understand Levi decomposition for the parabolic subgroups well.

Denote the parabolic subgroup is P=LV, L is Levi subgroup. I guess that for the classical group, L is the diagonal element and the left part of it and V is right part of it with all the diagonal elements are 1.

Am I right? If yes, how to show it; if no, please give other interested examples.

No, $L$ could be a block $L = GL(2) \times GL(1) \subset GL(3)$,i.e, matrix of type $$\begin{pmatrix} * & * & 0 \\ * & * &0 \\ 0 &0 & *\end{pmatrix}$$ and $V$ being matrices of the form

$$\begin{pmatrix} 1 & 0 & * \\ 0 & 1 &* \\ 0 &0 & 1\end{pmatrix}.$$

Being Borel subgroup in GL(n) is being a minimal parabolic subgroup. There, you are right.

For general reductive $G$, this is more difficult.

First I suppose by "classical group" you mean linear groups ? If not things are more complicated.

Then in this situation, your decomposition is correct if you work with P=B the standard Borel subgroup (of diagonal matrices).

More precisely, for GLn, you can show that parabolic subgroups are exactly upper triangular matrices, except that you allow to add blocks on the diagonal (for P=B, the blocks on the diagonal are all of size =1). Then, the corresponding Levi decomposition is L=blocks of the diagonal and 0 elsewhere, and V = diagonal of 1's and the other terms elsewhere.