# Lie Algebra's question

1. Apr 12, 2007

### 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.

2. Apr 12, 2007

### 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).