Proving Simplicity of Lie Algebra L: Basis Elements and Ideal Structure

[ElDavidas]

Take L to be a subspace of sl (2,R). (R is the real numbers)

$$L = \left( \begin{array}{ccc} 0 & -c & b\\ c & 0 & -a\\ -b & a & 0 \end{array}\right)$$

The basis elements of L are

$$e_1 = \left( \begin{array}{ccc} 0 & 0 & 0\\ 0 & 0 & -1\\ 0 & 1 & 0 \end{array}\right)$$

$$e_2 = \left( \begin{array}{ccc} 0 & 0 & 1\\ 0 & 0 & 0\\ -1 & 0 & 0 \end{array}\right)$$

$$e_3 = \left( \begin{array}{ccc} 0 & -1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 0 \end{array}\right)$$

What is the best way to show that this Lie Algebra is simple (i.e. the only ideals are {0} and L? I know it's non-abelian.

 

[matt grime]

By working out the [ ] of e_i and e_j, you should see how to show that the commutator spans the whole of the L again. That should do it.

Or you can just write down an isomorphism to sl_2(R). (Surely you meant to say L is a subspace of sl_3).

 

[tacman]

To show that a Lie Algebra is simple, we need to prove that the only ideals of the algebra are the trivial ideal {0} and the algebra itself. In other words, there are no non-trivial subspaces of the algebra that are closed under the Lie bracket operation.

One way to show this is to use the fact that L is a subspace of sl(2,R), which is a simple Lie algebra. This means that any ideal of sl(2,R) must also be an ideal of L. Since sl(2,R) only has the trivial ideal {0}, it follows that L also only has the trivial ideal {0}.

Another way to show this is to consider the basis elements of L, e1, e2, and e3. It can be shown that the only non-trivial commutators of these basis elements are [e1, e2] = e3 and [e2, e3] = e1. This means that any subspace of L that contains one of these basis elements must also contain the other two, making it the entire algebra L.

Therefore, we can conclude that L is a simple Lie algebra, as the only ideals are {0} and L.