I'm currently studying Lie algebras and trying to prove a result from a textbook. Namely if we let $ \mathfrak{b}(n,F) $ be the Lie algebra of upper triangular $ n \times n $ matrices over a field $F$. That the derived algebra $ \mathfrak{b}(n,F) ' = \mathfrak{n}(n,F)$, the Lie algebra of strictly upper $ n \times n $ matrices over the field $F$.
I've attempted a proof but it doesn't look very thorough.
Since any element of a Lie algebra can be expressed as a linear combination of its basis elements – we need only look at the commutators of the basis elements.
We take the standard basis.
So we know that $ \mathfrak{b}(n,F) $ has basis elements $ e_{ij} $, for $ j \geq i $, that is $ e_{ij} = 0 $ if $ i > j $ for $ i,j \in \{ 1,…,n \}.$
Note we can make use of the formula $ [ e_{ij}, e_{kl} ] = \delta _{jk} e_{il} – \delta_{li} e_{kj} $ where $ \delta $ denotes the Kronecker delta.
I've tried to split it into cases and work like that. However all cases result in the commutators equal to $0 $ – except this one.
$ i= j = k, $ with $l > i $. We have $ [ e_{ii}, e_{il} ] = \delta _{ii} e_{il} – \delta_{li} e_{ii} = e_{il} $. Where $ l > i $ and hence represents only strictly upper triangular matrices.
Now the only elements of this derived algebra are contained in the span of $ e_{il} $ where $ e_{il} $ are all strictly upper triangular matrices.
My question is – does this provide us with a basis for $\mathfrak{n}(n,F) $ , I don't know if the proof is complete enough to say yes this is the derived algebra.
Intuitively I know it is , and I've double checked that if I use the standard basis for both I get the respective Lie algebras. It's also worth noting in these bases that the number of elements for each is as I double checked for a few cases.
$$ \mathfrak{b}(n,F) : \dfrac{n(n+1)}{2} $$
and $$ \mathfrak{n}(n,F): \dfrac{n(n-1)}{2} $$
I'm not sure if I've gone wildly off track here or not. Any help would be appreciated, thank you in advance.
Best Answer
Here's more concise approach to that first part:
To prove that this is the case, it suffices to note that $ab$ and $ba$ have the same diagonal entries, namely $a_{ii}b_{ii}$. With that, we have $\mathfrak b' \subset \mathfrak n$.
For the reverse inclusion, it suffices to note that for $i < j$, we have $$ [e_{ii},e_{ij}] = e_{ij} $$ and these form a basis of $\mathfrak n$.