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)

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

The basis elements of L are

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

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

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

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).