# Lie Algebra isomorphism

1. May 2, 2007

### ElDavidas

1. The problem statement, all variables and given/known data

Take

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

where a,b,c are complex numbers.

2. Relevant equations

I find that a basis for the above Lie Algebra is

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

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

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

I then calculate all the products $[e_i,e_j]$ and see that L is non-abelian and simple

3. The attempt at a solution

The question then asks show L is isomorphic to sl(2,C). I have found $e,f,h \in L$ such that $[h,e] = 2e, [h,f] = -2f, [e,f] = h$
where,

$$h = \left(\begin{array}{ccc}0 & 0 & 0 \\0 & 2 & 0 \\0 & 0 & -2 \end{array}\right)$$

$$e = \left(\begin{array}{ccc}0 & 0 & -\sqrt{2} \\ \sqrt{2} & 0 & 0 \\0 & 0 & 0 \end{array}\right)$$

$$f = \left(\begin{array}{ccc}0 & -\sqrt{2} & 0 \\0 & 0 & 0 \\ \sqrt{2} & 0 & 0 \end{array}\right)$$

if I haven't made any mistakes. I see L is homomorphic to sl(2,C) but how do I show it's an isomorphism? (i.e show the injection and surjection). I know the definitions for an injection map and a surjection map but don't know how to apply it in this case.

Thanks in advance for any help.

Last edited: May 2, 2007
2. May 2, 2007

### matt grime

You've just written down a sub-algebra isomorphic to sl_2, and clearly it is all of the space (just by dimension arguments). There is nothing more to show.

You haven't actually written down a map so you can't apply the notion of injection or surjection. If you want to put in a map - there is an obvious one - then it is trivially an injection (and a surjection).