- Can someone provide two concrete examples the Lie algebra which is solvable, but not nilpotent?
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- And further explain the subtle differences between the solvable Lie algebra and the nilpotent Lie algebra?
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P.S. Any nilpotent Lie algebra is solvable.
Many thanks~~
Define $\mathfrak{t}_n$ to be the set of triangular $n \times n$ matrices with entries in some field $k$. To be clear, by triangular, I mean entries are allowed on the diagonal or anywhere above the diagonal, while everything below the diagonal is zero.
Now define $\mathfrak{u}_n$ to be the set of upper triangular $n \times n$ matrices with entries in some field $k$. Here I want all entries on or below the diagonal to be zero.
You can verify that $\left[ \mathfrak{t}_n,\mathfrak{t}_n \right] =\mathfrak{u}_n$ and that $\left[ \mathfrak{t}_n,\mathfrak{u}_n \right] =\mathfrak{u}_n$. In particular, this implies that $\mathfrak{t}_n$ is not nilpotent.
With a bit of effort, one can also show that $\mathfrak{t}_n$ is solvable, and that $\mathfrak{u}_n$ is nilpotent. Recall that all nilpotent algebras are solvable, since the $i$th term of the derived series is a subalgebra of the $i$th term of the lower central series.
Another example of a solvable Lie algebra is
\begin{equation*} \mathfrak{g} := \Bigg\{ \pmatrix{ 0&\theta &x \\ -\theta &0 &y \\ 0&0&0 } \Bigg\}_{\theta,x,y\in \mathbb{R}}, \end{equation*}
and this algebra is not nilpotent. You can verify these facts by direct computation of the brackets.
I hope this helps :)
Here is the easiest example in terms of dimension. The $2$-dimensional Lie algebra $L=\mathfrak{r}_2(K)$ over a field $K$ is defined by the basis $(x,y)$ with Lie bracket $[x,y]=y$. It is clear that $L$ is solvable because $$ [[L,L],[L,L]]=[Ky,Ky]=0. $$ On the other hand, $L$ is not nilpotent, because it has no center.
