Derived algebra of the Lie algebra of upper triangular matrices is the Lie algebra of strictly upper triangular matrices.

Question

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.

Answer 1

I think your method is fine! One needs to take extra care in examining the conditions.

From your expression for the commutator, it is clear that $[e_{ij}, e_{kl}]= 0$ unless $j = k$ or $i = l$, or both.

Let's deal with the case where both $j = k$ and $i = l$. Since $i \leq j$ and $k \leq l$, this case only arises when $i = j = k = l$. In this case, $$[e_{ij}, e_{kl}] = \delta_{jk}e_{il} - \delta_{il}e_{kj} = e_{ii} - e_{ii} = 0.$$

Next, let's deal with the case $j = k$ and $i \neq l$. Since $i \leq j$ and $k \leq l$, we have $i \leq j = k \leq l$ with $i$ strictly less than $l$. You can check that $$[e_{ij}, e_{kl}] = \delta_{jk}e_{il} - \delta_{il}e_{kj} = e_{il}.$$

The third case is really equivalent to the second one, but with $e_{ij}$ and $e_{kl}$ interchanged.

Moreover, for every $i < l$, the element $e_{il}$ can be realised as a commutator. Indeed, choosing any $j$ and $k$ such that $i \leq j = k \leq l$, we have $e_{il} = [e_{ij}, e_{kl}]$. So every strictly triangular matrix is in the derived subalgebra.

Answer 2

Here's more concise approach to that first part:

For all $a,b \in \mathfrak b$, $[a,b]$ is strictly upper triangular

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 $\mathfrak n$.

Answer 3

Is it not possible to show this directly?

Suppose $A$ and $B$ are upper triangular. Then \begin{eqnarray*} A \cdot B & : =& \begin{pmatrix} a_{11} & a_{12} & a_{13} & \cdots & a_{1n} \\ 0 & a_{22} & a_{23} & \cdots & a_{2n} \\ 0 & 0 & a_{33} & \cdots & a_{3n} \\ \vdots & \vdots & \ddots & \cdots & \vdots \\ 0 & 0 & \cdots & \cdots & a_{nn} \end{pmatrix} \cdot \begin{pmatrix} b_{11} & b_{12} & b_{13} & \cdots & b_{1n} \\ 0 & b_{22} & b_{23} & \cdots & b_{2n} \\ 0 & 0 & b_{33} & \cdots & b_{3n} \\ \vdots & \vdots & \ddots & \cdots & \vdots \\ 0 & 0 & \cdots & \cdots & b_{nn} \end{pmatrix} \\ &=& \begin{pmatrix} a_{11}b_{11} & \sum_{k=1}^{2} a_{1k}b_{k2} & \sum_{k=1}^3 a_{1k}b_{k3} & \cdots & \sum_{k=1}^n a_{1k}b_{kn} \\ 0 & a_{22}b_{22} & \cdots & \cdots & \sum_{k=2}^n a_{2k}b_{kn} \\ 0 & 0 & a_{33}b_{33} & \cdots & \sum_{k=3}^n a_{3k}b_{kn} \\ \vdots & \vdots & \ddots & \cdots & \vdots \\ 0 & 0 & \cdots & \cdots & a_{nn}b_{nn} \end{pmatrix}. \end{eqnarray*} Similarly, we have \begin{eqnarray*} B \cdot A &:= & \begin{pmatrix} b_{11} & b_{12} & b_{13} & \cdots & b_{1n} \\ 0 & b_{22} & b_{23} & \cdots & b_{2n} \\ 0 & 0 & b_{33} & \cdots & b_{3n} \\ \vdots & \vdots & \ddots & \cdots & \vdots \\ 0 & 0 & \cdots & \cdots & b_{nn} \end{pmatrix} \cdot \begin{pmatrix} a_{11} & a_{12} & a_{13} & \cdots & a_{1n} \\ 0 & a_{22} & a_{23} & \cdots & a_{2n} \\ 0 & 0 & a_{33} & \cdots & a_{3n} \\ \vdots & \vdots & \ddots & \cdots & \vdots \\ 0 & 0 & \cdots & \cdots & a_{nn} \end{pmatrix} \\ &=& \begin{pmatrix} b_{11}a_{11} & \sum_{k=1}^{2} b_{1k}a_{k2} & \sum_{k=1}^3 b_{1k}a_{k3} & \cdots & \sum_{k=1}^n b_{1k}a_{kn} \\ 0 & b_{22}a_{22} & \cdots & \cdots & \sum_{k=2}^n b_{2k}a_{kn} \\ 0 & 0 & b_{33}a_{33} & \cdots & \sum_{k=3}^n b_{3k}a_{kn} \\ \vdots & \vdots & \ddots & \cdots & \vdots \\ 0 & 0 & \cdots & \cdots & b_{nn}a_{nn} \end{pmatrix}. \end{eqnarray*} Hence, we see that \begin{eqnarray*} A \cdot B - B \cdot A &=& \begin{pmatrix} 0& \sum_{k=1}^2 (a_{1k}b_{k2} - b_{1k}b_{k2}) & \sum_{k=1}^3 (a_{1k}b_{k3} - b_{1k}a_{k3}) & \cdots & \sum_{k=1}^n (a_{1k}b_{kn}-b_{1k}a_{kn}) \\ 0 & 0 & \cdots & \cdots & \sum_{k=2}^n (a_{2k}b_{kn} - b_{2k}a_{kn}) \\ 0 & 0 & 0 & \cdots & \sum_{k=3}^n (a_{3k}b_{kn} - b_{3k}a_{kn}) \\ \vdots & \vdots & \ddots & \cdots & \vdots \\ 0 & 0 & \cdots & \cdots & 0 \end{pmatrix}, \end{eqnarray*} which is clearly strictly upper triangular.