I have been attempting this problem for a while, it is an assignment problem so I don't want somebody to just post the answer, I'm just looking for hints.
Let $\mathfrak{u}(n,\mathbb{C})$ be the Lie algebra of strictly upper triangular $n\times n$ matrices over $\mathbb{C}$.
I am trying to show that $\mathfrak{u}(n,\mathbb{C})$ is a nilpotent Lie algebra for $n\geq 2$.
This is what I have looked at so far.
I have done some explicit calculations (I shall omit the bulk of them and just summarise below) for $n=2,3,4$ and have come up with some ideas.
Let $L=\mathfrak{u}(n, \mathbb{C})$.
If $n=2$,
$L^1 = [L, L] = <0>_{\mathbb{C}}$
So $L$ is nilpotent.
If $n=3$,
$L^1 = [L, L] = <e_{13}>_{\mathbb{C}}$
$L^2 = [L, L^1] = <0>_{\mathbb{C}} = \left\{ 0 \right\}$
So, $L$ is nilpotent.
If $n=3$
$L^1 = [L,L] = <e_{13},e_{14},e_{24}>_\mathbb{C}$
$L^2 = [L,L^1] = <e_{14}>_{\mathbb{C}}$
$L^3 = [L, L^2] = <0>_{\mathbb{C}} = \left\{ 0 \right\}$
So, $L$ is nilpotent.
So my hypothesis so far is that for an arbitrary $m\in\mathbb{N}$, $L^{m-1}=\{0\}$
My instinct is to do this by induction. My base case, $n=2$ has already been shown. Now I assume that $\mathfrak{u}(n,\mathbb{C})$ is nilpotent for $n=k$ ($k\in\mathbb{N}$) and consider the case for n=k+1.
However, I have no idea how to link these two cases together. I have read in Introduction to Lie Algebras and Representation Theory – J.E. Humphreys that $L^k$ should be spanned by basis vectors $e_{ij}$ where $j-i=k+1$ – however in my calculations for $L^1$, I obtain an $e_{14}$ which does not satisfy 4-1=2.
Am I missing something completely obvious? Or have I made a mistake somewhere? (I have checked my calculations numerous times, but cannot find any errors!)
Thanks,
Andy.
Best Answer
Your computations easily generalize to arbitrary $n\ge 1$. We have to compute $L^1=[L,L]$, $L^2=[L,[L,L]]$, $L^3=[L,[L,[L,L]]$ and so on. In each step the strictly upper triangular matrices of $L$ obtain a new side diagonal consisiting of zeros. In other words, the zero diagonal is expanding to the right upper corner each time we we pass from $L^k$ to $L^{k+1}$. The reson is, that $L^k=[L,L^{k-1}]$ and the commutator is given by $[A,B]=AB-BA$. Consequently we arrive at $L^m=0$ for some $m\le n$. Hence one can even argue without induction. But of course, also an induction will work.