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

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?

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.

• Thanks @Kajelad ! So considering the two point I think that the proof I have provided is indeed valid in the compact semi-simple case, right?
– Gold
Commented Jul 18, 2021 at 0:18
• Also, regarding your point (1) I'm aware of Cartan's criterion that says that $\mathfrak{g}$ is semi-simple if and only if the Killing form is non-degenerate. Isn't that already enough for both points (1) and (2)?
– Gold
Commented Jul 18, 2021 at 0:22
• @Gold You can raise or lower indices in the semisimple case, but you won't necessarily have an orthonormal basis, so you have to be more careful with your index notation. For instance, your claim that $f^{abc}f^{ebc}=\delta^{ae}$ won't make sense there (and I'm not sure it's true even in the compact case). Instead, you can define $T_e{}^a=K_{bd}f^{ab}{}_cf^{cd}{}_e$ where $K_{bd}$ is the inverse Killing form, and compute its trace $T_a{}^a=-K_{ab}K^{ab}=-\operatorname{dim}(\mathfrak{g})$. That said, things do work out more or less the same in the semisimple case. Commented Jul 18, 2021 at 18:08