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I am looking for 3 dimensional non-nilpotent Lie algebra whose only toral subalgebra is $0$. In $sl_2$ the element $\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$ is diagonalizable so the space generated by it is toral.. In fact, if $L$ is a semisimple Lie algebra then $L$ has a nonzero toral subalgebra.

Ronald
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By definition, a toral Lie algebra is a Lie subalgebra of a general linear Lie algebra all of whose elements are semisimple. So, first we need a linear Lie algebra of dimension $3$. Here we take the $3$-dimensional solvable, non-nilpotent Lie algebra $L=\mathfrak{r}_3(\mathbb{C})$, given by the Lie brackets $$ [e_1,e_2]=e_2,\; [e_1,e_3]=e_2+e_3. $$ It has trivial center, so that the adjoint representation is faithful, and we can realize it as a subalgebra of $\mathfrak{gl}_3(\mathbb{C})$. Now let $\mathfrak{t}$ be a toral subalgebra of $ad(L)\cong L$. It must be abelian. However, $ad(e_1)$ is not diagonalisable, and also no nontrivial linear combination of $ad(e_2)$ and $ad(e_3)$. Hence we obtain $\mathfrak{t}=0$.

Dietrich Burde
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  • I actually have some questions in order to understand the whole situation: first, I want to make sure that algebra $\mathfrak r_3(\mathbb C)$ is the upper triangular $2\times 2$-matrices, if so why $[e_1,e_3]=e_2+e_3$. Second, why did you choose $ad(e_1)$ particularly? – Ronald Feb 29 '16 at 19:44
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    Because $\langle ad(e_1)\rangle$ is an abelian subalgebra (but not total), see your other question before. – Dietrich Burde Feb 29 '16 at 19:48
  • Alright, thank you very much! But sorry I am used to a different notations, what is $\mathfrak r_3(\mathbb C)$? – Ronald Feb 29 '16 at 19:51
  • $\mathfrak{r}_3(\mathbb{C})$ is the algebra given above. There are infinitely many different solvable non-nilpotent Lie algebras of dimension $3$, not only the $2\times 2$ upper-triangular matrices. – Dietrich Burde Feb 29 '16 at 20:53