Equation involving structure constants of Lie algebra - is this proof correct?

Question

Let a Lie group $G$ with Lie algebra $\mathfrak{g}$ be given. Suppose further $G$ is compact and semi-simple. Let $X^a\in \mathfrak{g}$ be a basis of the Lie algebra, $a=1,\dots, \dim \mathfrak{g}$, and define the structure constants

$$[X^a,X^b]=f^{ab}_{\phantom{ab}c}X^c\tag{1}$$

where the summation convention is implied (repeated indices are summed over).

I wanted to prove that if $A,B,C\in \mathbb{C}$ the equation

\begin{eqnarray}Af^a_{\phantom{a}dc}f^{cb}_{\phantom{cb}e}+Bf^b_{\phantom{b}dc}f^{ac}_{\phantom{ac}e}= Cf^{ab}_{\phantom{c}}f^{c}_{\phantom{c}de}\tag{2}\end{eqnarray}

implies $A = C$. I have come up with a proof and wanted to know if it is correct.

I start by raising all the indices in (2), writing the equation in the equivalent form

\begin{eqnarray}Af^{adc}f^{cbe}+Bf^{bdc}f^{ace}= Cf^{abc}f^{cde},\tag{3}\end{eqnarray} then I take $b = d$ and sum over this index. Since $f^{bdc}$ is anti-symmetric in the first two indices, as we see easily from (1), it follows that the middle term vanishes when we do that. The equation becomes

$$Af^{abc}f^{cbe}=Cf^{abc}f^{cbe}\Longrightarrow (A-C)f^{abc}f^{cbe}=0\tag{4}.$$

To finish the proof we must argue that $f^{abc}f^{cbe}\neq 0$ at least for some pair of indices $a,e$. In that case, since $G$ is compact and semi-simple $f^{cbe}$ is totally anti-symmetric and we can write $f^{cbe}=-f^{ebc}$. But then we know that $f^{abc}f^{ebc}=\delta^{ae}$. It follows that for all $a = e$ the quantity $f^{abc}f^{cbe}$ is non-zero and therefore we must have $A =C $.

Now, is this proof correct? I have two very basic doubts about what I did:

1. I don't know if it is indeed legit to simply raise all the indices when going from (2) to (3).

2. I don't know whether the last step, where I invoked that $G$ is compact and semi-simple, is really necessary. I mean, can't we conclude $f^{abc}f^{cbe}\neq 0$ without these assumptions?

Answer 1

1.

Not in general. In index computations, raising and lowering indices usually requires a nondegenerate bilinear form $b$, so that one can define the raising and lowering $U_a:=b_{ab}U^b$, $V^a:=b^{ab}V_b$ where $b^{ab}$ are the coefficients of the inverse of $b$. In the case of a compact Lie algebra, such a form exists: the Killing form (or its negative). In the case of a positive definite form, a common choice is to work in $b$-orthonormal basis satisfying $b_{ab}=b^{ab}=\delta^a_b$, so indices can be raised and lowered "for free", and there's no need to distinguish between upper and lower indices. Your computation only works in such a basis.

2.

We certainly need some information about the structure of the algebra to conclude that the term is nonzero; otherwise, for example, $\mathfrak{g}$ could be Abelian, and all structure constant would vanish.