For L = the lie algebra of 2x2 upper triangular matrices over the C, is ad L = Der L?

Question

I am deeply confused about this. I have seen a proof of the fact: L semisimple over C implies ad L = Der L but i dont know if the converse is true or false.

Answer 1

The converse is already false for the $2$-dimensional non-abelian Lie algebra $L$. We may assume that the Lie brackets are given by $[x,y]=y$ for a basis $(x,y)$. A short computation shows that $Der(L)=ad(L)$, but of course $L$ is $2$-step solvable, because $[[L,L],[L,L]]=0$. In particular, $L$ is not semisimple.

One can view $L$ as the standard Borel subalgebra of $\mathfrak{sl}_2(\mathbb{C})$, i.e., as upper triangular matrices of size $2$ with trace zero. The $3$-dimensional Lie algebra $L$ of your title does not satisfy that $Der(L)=ad(L)$, because only $\mathfrak{sl}_2(\mathbb{C})$ satisfes this in dimension $3$ (this is well-known).