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 2x2 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?
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