Three dimensional Lie algebra L with dim L' = 1

  • Thread starter valtz
  • Start date
  • #1
7
0

Main Question or Discussion Point

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
 
Last edited by a moderator:

Answers and Replies

  • #2
13,215
10,111
We have the Heisenberg algebra spanned by ##\{\,e_1,e_2,e_3\,\}## here. It is ##\mathfrak{g}= \mathfrak{Z(g)} \oplus [\mathfrak{g},\mathfrak{g}]##. Since both, center and derived Lie algebra are ideals, it is a direct sum of Lie algebras, not only vector spaces. The two dimensional part, the center ##\mathfrak{Z(g)}=\langle e_2,e_3 \rangle## is Abelian.

Another way to see it is to observe that ##\mathfrak{g}/[\mathfrak{g},\mathfrak{g}]## is always Abelian. Since ##[\mathfrak{g},\mathfrak{g}]## is one dimensional and ##\mathfrak{g}## three dimensional, ##\mathfrak{g}/[\mathfrak{g},\mathfrak{g}]## is two dimensional and Abelian.
 

Related Threads on Three dimensional Lie algebra L with dim L' = 1

  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
1K
Replies
4
Views
736
  • Last Post
Replies
1
Views
2K
Replies
20
Views
10K
  • Last Post
Replies
12
Views
5K
Replies
7
Views
885
Replies
1
Views
3K
  • Last Post
Replies
2
Views
711
Top