Adjoint representation of a Lie algebra of a same dimension that the basis representa

by yorik
Tags: adjoint, algebra, basis, dimension, representa, representation
yorik is offline
Feb17-09, 04:08 AM
P: 7

I hope it's not the wrong forum for my question which is the following:

Is there some list of Lie algebras, whose adjoint representations have the same dimension as their basic representation (like, e.g., this is the case for so(3))? How can one find such Lie algebras? Could you recommend some literature to me?

Thanks in advance!
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CompuChip is offline
Feb17-09, 04:21 AM
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I'm not very well-founded in Lie-algebra's, but the adjoint of an element x is the map
[tex]\operatorname{ad}_x: y \mapsto [x, y][/tex]
isn't it?
So the adjoint is linear, i.e.
[tex]\operatorname{ad}_x + \operatorname{ad}_y = \operatorname{ad}_{x + y}, \operatorname{ad}_x(y) + \operatorname{ad}_x(z) = \operatorname{ad}(y + z)[/tex]
etc. - then isn't the adjoint representation always of the same dimension.
I.e. the basic representation provides the generators [itex]g_i[/itex] of the Lie-algebra, and then [itex]\operatorname{ad}_{g_i}[/itex] generate the adjoint representation?
yorik is offline
Feb17-09, 06:22 AM
P: 7
Well, the basic idea of the Lie groups and lie algebras is the following. A Lie group G is a group, whose elements can be written as exp(i * ak Xk), where k runs from 1 to N and over k is summed (Einstein summation convention), ak are some real-valued parameters and Xk are some linearly independent hermitian operators (so called generators). They form an N-dimensional vector space (however, the dimension of the Hilbert space on which they act is not specified).

Now, the generators satisfy [Xk, Xl] = i fklmXm, where the f's are the so called structure constants. They define the Lie algebra of the Lie group G. It turns out, you can find infinitely many generator spaces (of different), which satisfy the algebra. The smallest irreducible generator space gives us the basic (fundamental) representation of the Lie Algebra.

We can also define the so called adjoint representation. We define
(Tk)lm = - i fklm. It has the same dimension as the Lie group. But, in generally, its dimension is not equal to the dimension of the fundamental representation. However, it's the case for so(3), the Lie Algebra of SO(3):

So my question is: Are there any other Lie groups / algebras for which the dimension of the two representations are equal like it is for SO(3) / so(3)?

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