What is the unipotent radical of a parabolic subgroup? For example, let
$$
P=\{ \left( \begin{matrix} a & b & k & c \\ d & e & f & g \\ 0 & 0 & h & i \\ 0 & 0 & 0 & j \end{matrix} \right) \mid a, b, c, d, e, f, g, h, i, j, k \in \mathbb{C} \}.
$$
What is the unipotent radical of $P$? Thank you very much.
[Math] the unipotent radical of a parabolic subgroup
algebraic-groupsgroup-theory
Best Answer
It is defined as the largest normal subgroup consisting entirely of unipotent elements (elements satisfying the equation $(x-1)^n=0$ for some $n$).
In the case of a parabolic subgroup of $GL_n(k)$, the parabolic can always be written as block upper triangular, and the unipotent radical is the subgroup where the block diagonal elements are the identity.
In your example, corresponding to the partition [2,1,1], the requirements on the variables to be in $P$ are not specified, but they are $(ae-bd)hj \neq 0$. To be in the unipotent radical the requirements are $a=e=h=j=1$, $b=d=0$, and $k,c,f,g,i$ are free.