0
$\begingroup$

Hello, I have recently started reading about Lie algebras. However all the examples I have encountered so far are simple and semisimple Lie algebras. Thus I would love to see an example of a real or complex finite dimensional Lie algebra $A$ with the following property :

$A$ is non abelian and it contains non trivial ideals.

Improve this question
$\endgroup$
2
  • 3
    $\begingroup$ Usually, the difficulty is in finding the simple ones :) $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Oct 14, 2011 at 17:13
  • 1
    $\begingroup$ @ Srifo B: Here is an exercise for you: given simple Lie algebras $s_1,...,s_n$, construct a Lie algebra with abelian radical $r$ and semi-simple part $s_1\times...\times s_n$, such that $[r,s_i]\neq 0$ for $i=1,...,n$. (Hint: think of the adjoint representation). $\endgroup$
    – Alain Valette
    Commented Oct 15, 2011 at 8:00

5 Answers 5

Reset to default
7
$\begingroup$

A nice example to play around with is the Lie algebra of upper triangular matrices. It is solvable, so has plenty of ideals and things like that.

Improve this answer
$\endgroup$
1
  • $\begingroup$ And if you take the strictly upper triabgular, then it is nilpotent. $\endgroup$
    – Yiftach Barnea
    Commented Oct 14, 2011 at 19:01
4
$\begingroup$

I think it is also a good idea to take a look at the following papers by Willem de Graaf et al. about nilpotent and solvable Lie algebras of small dimension over arbitrary fields:

http://arxiv.org/abs/math/0404071

http://arxiv.org/abs/math/0511668

http://arxiv.org/abs/1011.0361

Improve this answer
$\endgroup$
2
$\begingroup$

The 3-dimensional Heisenberg Lie algebra can be described by the presentation:

$$\mathcal{H}=\big\langle x, y, z\,\big\vert\,[x,y] = z, [x,z]=[y,z]=0\big\rangle$$

The derived subalgebra $[\mathcal{H},\mathcal{H}]$ is a central ideal spanned by $z$, and the whole Lie algebra is a nilpotent Lie algebra (thus not simple or semi-simple).

Improve this answer
$\endgroup$
0
$\begingroup$

Take any non-abelian Lie algebra $L$ and consider $L\oplus {\mathbb C}^n$. This Lie algebra is non-abelian, and non-semisimple because it has a non-trivial radical.

Improve this answer
$\endgroup$
0
$\begingroup$

To get a good idea about the relative paucity of simple Lie algebras (as Mariano says in his comment above), you could take a look at a list of low-dimensional Lie algebras. For instance, here: J. Patera,R.T. Sharp,P. Winternitz, and H. Zassenhaus, Invariants of real low dimension Lie algebras, J. Mathematical Phys. 17 (1976), no. 6, 986–994.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .