I am trying to show that the Lie algebra of strictly upper triangular matrices $\mathfrak{u}(n,\mathbb{C})$ is soluble for all $n\geq 2$. This is not an assessed question, but is an exercise in the notes. I am having real difficulty visualising the derived series. I know how to calculate it, and have done some test cases for $n=2$, $n=3$ and $n=4$. But I just can't quite visualise what is going on to generalise it. I know the formula for the bracket of the basis vectors $e_{ij}$ and $e_{jk}$: $$[e_{ij},e_{kl}] = \delta_{jk}e_{il}-\delta_{il}e_{kj}$$

and have obtained the following

for $L = \mathfrak{u}(2,\mathbb{C})$

$$ L^{(1)} = \left\{0\right\} $$

for $L = \mathfrak{u}(3,\mathbb{C})$

$$ L^{(1)} = \left<e_{13}\right>_\mathbb{C} $$

$$ L^{(2)} = \left\{0\right\} $$

and for $L = \mathfrak{u}(3,\mathbb{C})$

$$ L^{(1)} = \left<e_{13}, e_{14}, e_{24}\right>_\mathbb{C} $$

$$ L^{(2)} = \left\{0\right\} $$

but I can't visualise what is actually going on here and explain it, and that is preventing me from going any further.

How should I visualise this? I do notice that $L^{(1)}$ always removes the entries on the next diagonal, but I don't understand why this is the case.


You're exactly on the correct track when you say that $L^{(1)}$ removes the entries on the next diagonal (i.e., entries $e_{ij}$ with $j=i+1$). The idea is that $L^{(2)}$ will have zeros on the next diagonal after that, as well. In other words, $L^{(2)}$ will have $e_{ij} = 0$ for $j=i+1$ and $j=i+2$. Using induction and your formula $[e_{ij}, e_{kl}] = \delta_{jk}e_{il} - \delta_{il}e_{kj}$, it is not too hard to show that an element $(e_{ij})\in L^{(n)}$ satisfies $e_{ij} = 0$ if $j \leq i + n$.

So, intuitively, the diagonals of zeros keep "moving up" as you continue along the derived series.

  • $\begingroup$ indeed an easy insight $\endgroup$ – Dac0 Mar 17 '17 at 16:29

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