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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 $n=3$,

$L^1 = [L, L] = _{\mathbb{C}}$

$L^2 = [L, L^1] = <0>_{\mathbb{C}} = \left\{ 0 \right\}$

So, $L$ is nilpotent.

If $n=3$

$L^1 = [L,L] = _\mathbb{C}$

$L^2 = [L,L^1] = _{\mathbb{C}}$

$L^3 = [L, L^2] = <0>_{\mathbb{C}} = \left\{ 0 \right\}$

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.

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    Are you familiar with the bracket product formula for $[e_{ij},e_{kl}]$? – 2017-02-12
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    Yes. $[e_{ij},e_{kl}]=\delta_{jk}e_{il}-\delta{il}e_{kj}$ right? – 2017-02-12

2 Answers 2

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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.

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EDIT: My previous answer was wrong, so I deleted it.

The thing to prove is: $L^k$ has a basis given by $e_{ij}$ with $i < j - k$.

By definition, $L^k = [L,L^{k-1}]$, which is the span of $$\{[e_{ij},e_{ab}]:i < j, a+k< b \} $$ by using the induction hypothesis on $L^{k-1}$. After applying the formula, you get a few different cases.