Is upper triangular matrices maximal parabolic subgroup

Question

Let $B_n(K)$ be an upper triangular matrices, which is a standard Borel subgroup of $GL_n(K)$. $K$ is an algebraically closed field. So this is a parabolic subgroup of $GL_n(K)$. I was wondering is $B_n(K)$ a maximal proper parabolic subgroup? If not what is the maximal proper parabolic subgroup containing $B_n(K)$? thank you.

Answer 1

The conjugacy classes of parabolic subgroups correspond to partitions of $n$, in the following way: For the partition $\lambda$ given by $n=n_1+\cdots+n_k$, one can attach the parabolic subgroup $$P_\lambda:=\begin{pmatrix}\mathrm{GL}_{n_1}(K)&*&*\\ &\mathrm{GL}_{n_2}(K)&\\ 0&0&&\ddots\\ 0&\dots&0&&\mathrm{GL}_{n_k}(K)\end{pmatrix}.$$ In particular, the trivial partition corresponds to $P_{(n)}=\mathrm{GL}_n(K)$ and $n=1+\cdots+1$ gives rise to the Borel subgroup: $P_{(n=1+\cdots+1)}=B$.

Now, refinements of partitions $\lambda\le\mu$ (e.g, $(2=2)<(2=1+1)$) exactly corresponds to inclusions $P_\lambda\supset P_\mu$. Thus, what you want are the nontrivial partitions of $n$ which is not the refinement of any other nontrivial partitions (i.e., the minimal elements in the lattice of partitions of $n$ with respect to refinement, minus the trivial partition $n=n$.) These are exactly the partitions $n=n_1+n_2$.

In particular, there is no unique proper maximal parabolic subgroup (for $n>3$)!