# Three dimensional Lie algebra L with dim L' = 1

1. Nov 3, 2012

### valtz

Now suppose the derived algebra has dimension 1. Then there exits some non-zero $X_{1} \in g$ such that $L' = span{X_{1}}$. Extend this to a basis ${X_{1};X_{2};X_{3}}$ for g. Then there exist scalars$\alpha, \beta , \gamma \in R (not all zero) such that $[X_{1},X_{2}] = \alpha X_{1}$ $[X_{1},X_{3}] = \beta X_{1}$ $[X_{2},X_{3}] = \gamma X_{1}$ Suppose$\alpha \neq 0$. Then construct a new basis, as follows: $e_{1} = X_{1}$ $e_{2} = \frac{1}{\alpha} X_{2}$ $e_{3} = \alpha X_{3} - \beta X_{2} + \gamma X_{1}$ Since$\alpha \neq 0\$, by assumption, this is a basis for the Lie algebra g. Let us
calculate the Lie brackets for this basis:
$[e_{1},e_{2}] = e_{1}$
$[e_{1},e_{2}] = 0$
$[e_{1},e_{2}] = 0$
This Lie algebra is seen to be the direct sum of two Lie algebras, two dimensional non abelian lie algebra $\oplus$ 1 dimensional lie algebra

i get this from pdf in internet , i want to ask, how can that be the direct sum of two lie algebra , two dimensional non abelian lie algebra $\oplus$ 1 dimensional lie algebra

what use from construct new basis here? what can we see from construct new basis here?
i'm sorry before, please give me a detail explanation, because my essay is about lie algebra. so i'm a little new about lie algebra