Let $L $ be a Lie algebra. A subalgebra $H$ of $L$ such that $ad_H:\frak g → g$ is diagonalizable for every $h ∈ H$ is called toral.
Now, every toral subalgebra is abelian. But, what is an example for an abelian subalgebra which is not toral?
Asked
Active
Viewed 107 times
2 Answers
2
Take any element $x\in L$ that is not semisimple and let $H$ be the span of $x$.
Eric Wofsey
- 330,363
-
2A larger expample (a maximal one, indeed!), the space of matrices described by David here. – Mariano Suárez-Álvarez Feb 29 '16 at 05:08
-
As a complementary of my question 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.. What about $\mathbb R^3 $,$^$? – Ronald Feb 29 '16 at 05:44
1
To give an explicit non-semisimple example, consider the Lie algebra $L=\mathfrak{r}_3(\mathbb{C})$, given by the Lie brackets $$ [e_1,e_2]=e_2,\; [e_1,e_3]=e_2+e_3. $$ We have $$ ad(e_1)=\begin{pmatrix} 0 & 0 & 0 \cr 0 & 1 & 1 \cr 0 & 0 & 1\end{pmatrix}. $$ This matrix is not diagonalizable, so that $\mathfrak{t}=\langle e_1\rangle$ is an abelian subalgebra, which is not toral.
Dietrich Burde
- 130,978