In most textbooks (Georgi, for example) a scalar product on the generators of a Lie Algebra is introduced (the Cartan-Killing form) as
$$tr[T^{a}T^{b}]$$
which is promptly diagonalised (for compact algebras) and the generators scaled such that
$$tr[T^{a}T^{b}] = \delta^{ab}.$$
In this basis we get that, for example,
$$f_{abc} = -i\, tr ([T^{a}, T^{b}]T^{c})$$
that are fully antisymmetric.
Yet I have seen the these relations used for arbitrary (it particular the fundamental) representation as matter of course (maybe up to some normalisation). Is this because $tr[T^{a}T^{b}]$ defines a symmetric matrix in any rep that can thus be diagonalised? Is it a general truth? Or does the diagonalisation in the adjoint imply a diagonal for in any other rep?
I know that the structure constants are essentially fixed for all reps by smoothness and the group product — is this why fixing the form in one basis for one rep fixes it for that basis in all reps?
For a concrete example, let's suppose I look at SU(2). The adjoint rep is 3 dimensional and I can linearly transform and scale my generators (i.e. the structure constants) so that I get the trace to be diagonal and normalised. This fixes once and for all that the structure constants of SU(2) are $f_{ijk} = \epsilon_{ijk}$, say.
Now I ask someone to construct the fundamental rep; they look for 2×2 matrices satisfying the Lie algebra with these structure constants. They find the Pauli matrices.
Why do these come out such that the trace $tr [\sigma^{a} \sigma^{b}] \propto \delta_{ab}$ automatically? It's a different rep…why is it guaranteed?
Best Answer
More generally, let there be given an $n$-dimensional complex Lie algebra $(L, [\cdot,\cdot])$.
The adjoint representation ${\rm ad}:L\to {\rm End}(L)$ is defined as $$({\rm ad}x)y~:=~[x,y], \qquad x,y~\in~L.\tag{1}$$
The Killing form $\kappa:L\times L\to \mathbb{C}$ is defined as $$ \kappa(x,y)~:={\rm tr}({\rm ad}x \circ {\rm ad}y), \qquad x,y~\in~L, \tag{2}$$ is bilinear, symmetric, associative. It is non-degenerate iff $L$ is semisimple.
One may show that any bilinear, symmetric, associative form $L\times L\to \mathbb{C}$ is proportional to the Killing form $\kappa$ if $L$ is simple.
Given an arbitrary basis $(t_a)_{a=1,\ldots, n}$ for $L$, define the structure constants $f_{ab}{}^c\in\mathbb{C}$ via $$ [t_a,t_b]~=~\sum_{c=1}^nf_{ab}{}^ct_c , \qquad a,b~\in~\{1,\ldots, n\}. \tag{3}$$
Define Killing metric $$ \kappa_{ab}~:=~\kappa(t_a,t_b) , \qquad a,b~\in~\{1,\ldots, n\}.\tag{4}$$
One may show that the lowered structure constants $$ f_{abc}~:=~\sum_{d=1}^nf_{ab}{}^d\kappa_{dc}\, \qquad a,b,c~\in~\{1,\ldots, n\},\tag{5}$$ are always totally antisymmetric in the indices $abc$.
It is possible to choose the basis $(t_a)_{a=1,\ldots, n}$ such that the Killing metric $\kappa_{ab}$ is diagonal; and in the semisimple case, such that $\kappa_{ab}$ is proportional to the identity $\delta_{ab}$.