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Example: upper triangular matrices (a solvable Lie algebra)

Upper triangular matrices form a Lie algebra whose derived subalgebra is strictly upper triangular, giving an explicit derived series.
Example: upper triangular matrices (a solvable Lie algebra)

Let B⊂HAHAHUGOSHORTCODE1380s0HBHBB\subset be the subgroup of invertible upper triangular matrices:

B={(ab0d):a,d≠0}. B=\left\{\begin{pmatrix}a&b\\0&d\end{pmatrix}: a,d\ne 0\right\}.

It is a Lie subgroup, and its Lie algebra is the upper triangular matrices

b={(xy0z):x,y,z∈R}⊂HAHAHUGOSHORTCODE1380s1HBHB. \mathfrak b=\left\{\begin{pmatrix}x&y\\0&z\end{pmatrix}: x,y,z\in\mathbb R\right\}\subset .

Commutator calculation

Take A=(xy0z),A′=(x′y′0z′). A=\begin{pmatrix}x&y\\0&z\end{pmatrix},\quad A'=\begin{pmatrix}x'&y'\\0&z'\end{pmatrix}. Then [ [A,A’]=AA’-A’A

\begin{pmatrix} 0 & (x-z)y’-(x’-z’)y\ 0 & 0 \end{pmatrix}, ] which is strictly upper triangular.

Therefore the is [ [\mathfrak b,\mathfrak b]

\left{\begin{pmatrix}0&u\0&0\end{pmatrix}: u\in\mathbb R\right} \cong \mathbb R, ] matching the pattern.

Derived series (explicit)

Since strictly upper triangular 2×22\times 2 matrices commute with each other, we get

b(1)=[b,b]={(0u00)},b(2)=[b(1),b(1)]=0. \mathfrak b^{(1)}=[\mathfrak b,\mathfrak b]=\left\{\begin{pmatrix}0&u\\0&0\end{pmatrix}\right\}, \qquad \mathfrak b^{(2)}=[\mathfrak b^{(1)},\mathfrak b^{(1)}]=0.

Hence b\mathfrak b is solvable (see and ).

Context. Upper triangular (Borel) subalgebras are the archetypal solvable subalgebras inside semisimple Lie algebras, and computations like the one above are the linear-algebraic shadow of triangularization phenomena.