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Three dimensional Lie algebra L with dim L' = 1

  1. Nov 3, 2012 #1
    Now suppose the derived algebra has dimension 1. Then there exits some non-zero [itex]X_{1} \in g[/itex] such that [itex]L' = span{X_{1}}[/itex]. Extend this to a basis [itex]{X_{1};X_{2};X_{3}}[/itex] for g. Then there exist scalars$\alpha, \beta , \gamma \in R (not all zero) such that
    [itex][X_{1},X_{2}] = \alpha X_{1}[/itex]
    [itex][X_{1},X_{3}] = \beta X_{1}[/itex]
    [itex][X_{2},X_{3}] = \gamma X_{1}[/itex]
    Suppose $\alpha \neq 0$. Then construct a new basis, as follows:
    [itex]e_{1} = X_{1}[/itex]
    [itex]e_{2} = \frac{1}{\alpha} X_{2}[/itex]
    [itex]e_{3} = \alpha X_{3} - \beta X_{2} + \gamma X_{1}[/itex]
    Since $\alpha \neq 0$, by assumption, this is a basis for the Lie algebra g. Let us
    calculate the Lie brackets for this basis:
    [itex][e_{1},e_{2}] = e_{1}[/itex]
    [itex][e_{1},e_{2}] = 0[/itex]
    [itex][e_{1},e_{2}] = 0[/itex]
    This Lie algebra is seen to be the direct sum of two Lie algebras, two dimensional non abelian lie algebra [itex]\oplus[/itex] 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 [itex]\oplus[/itex] 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

    source pdf where i get : http://math.ucsd.edu/~abowers/downloads/survey/3d_Lie_alg_classify.pdf
     
  2. jcsd
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