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Lie algebra

by bartadam
Tags: algebra
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bartadam
#1
Nov18-08, 02:09 PM
P: 41
I have five generators of a lie algebra, [tex]g_1,g_2,g_3,g_4,g_5[/tex] which at first glance I believe are independent, although I could be wrong.

I have calculated the structure constants, i.e.

[tex]\left[g_i,g_j\right]=f_{ij}^k g_k[/tex]

And from that I have calculated a matrix rep using [tex]\left(T_i\right)_j^k=f_{ij}^k[/tex]

I get [tex]T_1, T_2, T_3, T_4[/tex] all linearly independent.

However I get [tex]T_4=-T_5[/tex] which I do not understand. Does this mean there algebra is only 4D rather than 5D?
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bartadam
#2
Nov25-08, 04:14 AM
P: 41
Hi again, I could really use some help on this please.

I have also realised T1+T2+T3=0.

I do not understand. Does this mean g1+g2+g3=0. I do not believe this at all.
guedes
#3
Dec5-08, 07:13 AM
P: 3
The dimension of a lie algebra depends on the representation you choose for it.
Recall that a representation is a map [tex]d:L(G)\to M(V)[/tex], where [tex]L(G)[/tex] is the lie algebra of some lie group [tex]G[/tex] and [tex]M(V)[/tex] is the space of linear transformations [tex]V\to V[/tex] ([tex]V=\mathbb{R}, \mathbb{C}[/tex]).

So, for example, you could have what is called the trivial representation where [tex]d(g)=0[/tex] for all [tex]g\in L(G)[/tex], which is one dimensional, even though your algebra could has more than one element.

Another example is the fundamental representation, where [tex]d(g)=g[/tex] for all [tex]g\in L(G)[/tex]. Here the dimension of the lie algebra is equal to the number of elements in the algebra (called the rank of the algebra).

In the example that you state, where the matrix representation equals the structure constants, is called the adjoint representation. I would guess, from the information that you give, that that representation is 3 dimensional, as you have 2 constraints over 5 five "variables".


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