Exploring the Trace of Lie Algebra Elements: A Proof and Reference Guide

[pellman]

Given a Lie algebra of elements G_i satisfying [G_i,G_j] = -if_ij^kG_k, it can be shown that

trace(G_iG_j) = -f_ik^lf_jl^k

where by "trace" we mean the trace in a particular matrix representation.

The text in which I encountered this does not have a proof and I'm having trouble convincing myself of it. Anyone have a reference or hint that could help?

 

[lethe]

that is just the definition of matrix multiplication, in the adjoint representation. this is the definition of the killing form, and i don t know if it s true in general, i think only in the adjoint representation.

do you know what representation you re working with?

 

[pellman]

I don't know. I'll have to look up "adjoint representation," which I suspect will alone be of big help. I'll come back if I need more after that. Thanks for your help, lethe!

 

[jeff]

Originally posted by pellman
Given a Lie algebra of elements G_i satisfying [G_i,G_j] = -if_ij^kG_k, it can be shown that

trace(G_iG_j) = -f_ik^lf_jl^k

where by "trace" we mean the trace in a particular matrix representation.

The text in which I encountered this does not have a proof and I'm having trouble convincing myself of it. Anyone have a reference or hint that could help?

The number n of lie algebra generators is representation independent, with the adjoint rep being the one that consists of nxn matrices. Developing what lethe said, the structure constants form the adjoint rep, i.e. f_i has matrix elements (f_i)_j^l so the formula may be rewritten Tr(G_iG_j) = Tr(f_if_j). Hint: This just expresses the fact that Tr(G_iG_j) is representation independent.

 

[lethe]

the adjoint representation is the mapping that takes elements of the lie algebra to the linear operator on the lie algebra (hence why it is a representation) which is given by taking the commutator with the element in question.

for example, we would map X to T_X, where X is an element of the lie algebra, and T_X is a linear transformation on the algebra given by T_X(Y)=[X,Y].

thus, since the commutators are completely specified by the structure constants (which depend on what basis you choose for your lie algebra), a matrix in the adjoint representation will be filled up with the structure constants for that element.

thus you can see, in the adjoint representation, the formula you have written is just the definition of matrix multiplication, and the definition of trace.

 

[lethe]

Originally posted by jeff
Hint: This just expresses the fact that Tr(G_iG_j) is representation independent.

ahh... so that is representation independent? i wasn t sure about that. it s not true for representations of groups in general, for example, the trace of the identity gives you the dimension of the representation, which certainly is not independent.

but we re not talking about groups in general, we re talking about lie algebras.

is there an easy way to see that this expression is representation independent?

 

[pellman]

Thanks for the further info. It's clear now. And the text I was looking at was indeed referring to the adjoint rep. But now I'm with lethe in not seeing that tr(G_iG_j) is representation independent.

 

[lethe]

in fact, i now think it is not true. for example, consider su(2) spanned by the pauli matrices: σ_x

i/2[0  1]
   [1  0]

square this matrix, take the trace, you get -1/2. now, the adjoint matrix is:

[0  0   0]
[0  0  -1]
[0  1   0]

square this, take the trace, you get -2.

 

[jeff]

Using the invariance of the trace under cyclic permutation of it's arguments, it's easy to see that Tr(G_iG_j) is invariant under a map r to another representation: Tr(rG_ir^-1rG_jr^-1) = Tr(G_iG_j).

 

[lethe]

Originally posted by jeff
Using the invariance of the trace under cyclic permutation of it's arguments, it's easy to see that Tr(G_iG_j) is invariant under a map r to another representation: Tr(rG_ir^-1rG_jr^-1) = Tr(G_iG_j).

such a representation is called an equivalent representation. in particular, the trace of every element remains invariant under such a similarity transformation, by the argument you just made, so the killing metric trivially remains invariant.

this amounts to a freedom in the choice of coordinate systems.

in particular, no such r can be found if your 2 representations have different dimension. in that case r^-1 is not defined in general.

see my counterexample above.

 

[jeff]

Originally posted by lethe
such a representation is called an equivalent representation. in particular, the trace of every element remains invariant under such a similarity transformation, by the argument you just made, so the killing metric trivially remains invariant.

this amounts to a freedom in the choice of coordinate systems.

in particular, no such r can be found if your 2 representations have different dimension. in that case r^-1 is not defined in general.

see my counterexample above.

Of course you're right, duh. Sorry about this.

Originally posted by lethe
...this is the definition of the killing form, and i don t know if it s true in general, i think only in the adjoint representation.

The killing form ( , ) is defined by (x,y) ≡ Tr(ad x)(ad y) in which x and y are elements of a lie algebra and ad is the operator you called T. In the case of matrix representations, ad G_i = f_i. The killing form satisfies an invariance property ([x,y],z)+(y,[x,z]) = 0. Now, a general compact lie algebra is a sum of simple lie algebras and U(1)s. From the definition of the killing form as a trace, TrG_iG_j obviously satisfies the invariance property. Then since for a simple lie algebra the killing form is unique up to normalization, we have in the given representation that TrG_iG_j = α(G_i,G_j) = αTrf_if_j where α is a constant whose value depends on the representation.