# (Levi) Decomposition of group of upper triangular matrices with block matrices on its diagonal into two subgroups

Our teacher told us something about the decomposition of the subgroup $G$ of $GL_n(K)$ (invertible matrices with coefficients in a field $K$) of matrices of the form :

$$\{ \begin{pmatrix} \begin{bmatrix} & & \\ & & \end{bmatrix} & * & * \\ 0 & ... & * \\ 0 & 0 & \begin{bmatrix} & & & \\ & & & \\ & & & \end{bmatrix} \\ \end{pmatrix} \}$$ (i.e. the subgroup that consists of all the invertible matrices with a sequence of block matrices on its diagonal (where the size of each block is set) and other values in the upper part) into two subgroups $H$ and $N$, respectively of the form :

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

with the second one being normal in $G$, so that we have $NH = G$ (and $N \cap H = \{e\}$, which would lead us to an isomorphism of $G$ with the semidirect product of $N$ and $H$)

Also, I don't know if the block matrices must all be square matrices. There could be some mistakes on what I said though. Feel free to correct me.

I'm looking for a proof of this statement (so any reference would be okay).

Our teacher told us it was linked to the group $GL(V)$ (where $V$ is a finite dimensional vector space over $K$) and the two following subgroups :

If we denote $0 = V_0 \subseteq V_1 \subseteq V_2 \subseteq ... \subseteq V_r \subseteq V$ a flag of subspaces of $V$, the subgroups are $P = \{g \in GL(V) \mid g(V_i)\subseteq V_i \ \forall i \}$ and $U = \{g \in GL(V) \mid gv - v \in V_{i-1} \ \forall v \in V_i \ \forall i \geq 1 \}$. Some basic results we have is that $P \leq N_{GL(V)}(U)$ and $U \leq P$, where $N_{GL(V)}$ denotes the normalizer and $\leq$ the relation of being a subgroup.

Thank you.

• You are scratching the surface of the fascinating subject of linear algebraic groups. The group $G$ is called a parabolic subgroup of $\textrm{GL}_n$, and the decomposition $G = NH$ is called a Levi decomposition. – D_S Jan 12 '17 at 0:04
• These notes (people.brandeis.edu/~igusa/Math101b/parabolic.pdf) give a proof of the result along the lines of what you mentioned. – D_S Jan 12 '17 at 4:18
• Thank you, I'm now able to make further research with more accuracy. (I've looked at Alperin's book "Groups and representations" and there is this proposition, but it is given as an exercise for the reader.. and other web sources don't seem to prove it either.) – Desura Jan 12 '17 at 10:44

I was going to write an answer explaining the deep ideas behind this decomposition, but I realized it would end up way too long. It's not difficult to check that $H \cap N = 1$ and $N$ is normal in $G$; the trouble is showing rigorously that the product set $HN$ is equal to all of $G$. I know a way to do this using algebraic geometry when the field is algebraically closed, but hopefully someone else will just post a more elementary proof.

I'll post what I have written, which is an explanation of how the subgroup $H$ arises naturally from combinatorial data associated with $\textrm{GL}_n$. Instead of an arbitrary field $K$, I'm working with an algebraically closed field $\Omega$, such as $\mathbb{C}$. Although what I ended up saying seems to work for any field.

Let $B$ denote the subgroup of $\textrm{GL}_n(\Omega)$ consisting of upper triangular matrices. Let $T$ denote the subgroup of $\textrm{GL}_n(\Omega)$ consisting of diagonal matrices.

Define homomorphisms $e_1, ... , e_n: T \rightarrow \textrm{GL}_1(\Omega)$ by the formula

$$e_i (\begin{pmatrix} x_1 & & \\\ & x_2 & \\ & & \ddots \\ & & & x_n\end{pmatrix} )= x_i$$

Define $a \cdot e_i(x) = e_i(x^a)$ (for $x \in T$ and $a$ an integer), and $(e_i + e_j)(t) = e_i(t)e_j(t)$. Then we can talk about the abelian group $X(T)$ consisting of all homomorphisms of the form $a_1 e_1 + \cdots + a_ne_n: T \rightarrow \textrm{GL}_1(\Omega)$ ($a_1, ... , a_n \in \mathbb{Z}$). It is a free abelian group with basis $e_1, ... , e_n$.

Consider the finite subset

$$\Delta = \{e_1 - e_2, ... , e_{n-1} - e_n \}$$

To each subset $\Theta \subseteq \Delta$, let

$$S_{\Theta} = \bigcap\limits_{\alpha \in \Theta} (\textrm{Ker } \alpha)$$

This is a subgroup of $T$. For example, if $\Theta = \{e_1 - e_2, e_{n-1}-e_n\}$, then $S$ consists of all invertible diagonal matrices of the form

$$S_{\Theta} = \begin{pmatrix} x \\ & x \\ & & x_3 \\ &&&\ddots\\ & && & x_{n-2} \\ & & & & & z \\ & & && & & z \end{pmatrix}$$

for $x, x_3, ... , x_{n-2}, z \in \Omega$.

Let $\alpha_i = e_i - e_{i+1}$ for $1 \leq i \leq n -1$. Think of $\alpha_1, ... , \alpha_{n-1}$ as distinguished points on a line segment starting at $\alpha_1$ and ending at $\alpha_{n-1}$. To give a subset of $\Delta$ is to break up this line segment at some of the points $\alpha_i$, leaving unbroken segments of lengths $m_1, ... , m_t$ with $m_1 + \cdots + m_t = n$. Then the centralizer $M_{\Theta}$ of $S_{\Theta}$ in $\textrm{GL}_n$, that is

$$M_{\Theta} = \{ x \in \textrm{GL}_n(\Omega) : xs = sx \textrm{ for all } s \in S_{\Theta} \}$$

will consist of block diagonal invertible matrices of sizes $m_1, ... , m_t$, that is

$$M_{\Theta} = \{ \begin{pmatrix} A_1 \\ & A_2 \\ && \ddots \\&&& A_t \end{pmatrix} : A_i \in \textrm{GL}_{m_i}(\Omega) \}$$

This is very easy to check! For example, if $\Theta = \Delta - \{e_{n-2} - e_{n-1} \}$, then

$$M_{\Theta} = \{ \begin{pmatrix} A_1 \\ & A_2 \end{pmatrix} : A_1 \in \textrm{GL}_{n-2}(\Omega), A_2 \in \textrm{GL}_2(\Omega) \}$$

These subgroups $M_{\Theta}$ are exactly the block diagonal groups you described! And they are in one to one order preserving correspondence with subsets of $\Delta$. For example, $M_{\Delta} = \textrm{GL}_n(\Omega)$, and $M_{\emptyset} = T$.

This is half of the story, the other half are those subgroups with $1$s on the diagonal.