Adjoint representation of a Lie algebra of a same dimension that the basis representaby yorik Tags: adjoint, algebra, basis, dimension, representa, representation 

#1
Feb1709, 04:08 AM

P: 7

Hello,
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! 



#2
Feb1709, 04:21 AM

Sci Advisor
HW Helper
P: 4,301

I'm not very wellfounded in Liealgebra'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 Liealgebra, and then [itex]\operatorname{ad}_{g_i}[/itex] generate the adjoint representation? 



#3
Feb1709, 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 * a_{k} X_{k}), where k runs from 1 to N and over k is summed (Einstein summation convention), a_{k} are some realvalued parameters and X_{k} are some linearly independent hermitian operators (so called generators). They form an Ndimensional vector space (however, the dimension of the Hilbert space on which they act is not specified).
Now, the generators satisfy [X_{k}, X_{l}] = i f_{klm}X_{m}, 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 (T_{k})_{lm} =  i f_{klm}. 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): http://math.ucr.edu/home/baez/lie/node5.html. 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)? 


Register to reply 
Related Discussions  
Linear Algebra: Basis and Dimension problem  Calculus & Beyond Homework  5  
adjoint representation/lie algebra question  General Math  21  
adjoint representation  Linear & Abstract Algebra  16  
adjoint representation  Differential Geometry  1  
Linear Algebra  Dimension, Basis  Introductory Physics Homework  3 