As a nit-picking question, I wanted to clarify a point of confusion. This arises from definitions found in a plethora of books, lectures notes and even the Wikipage on structure constants and Lie algebras. See this link to the Wiki page.

A generic Lie algebra that defines a group is written in terms of the group generators $X^a$. The Lie algebra is

$$[X^a,X^b] = i f_{abc}X^c.$$

I am happy that the structure constant $f_{abc}$ is a real number, and that $f_{abc} = - f_{bac} $ by commutator properties.

My issue is this; in the $SO(3)$ rotations, the Lie algebra is in terms of AM operators $J_i$. Hence, the Lie algebra is now

$$[J_i,J_j] = i \varepsilon_{ijk}J_k.$$ However, I am a bit uncomfortable with why all the indices are down. My only explanation for this is that since the structure constant $f_{ijk} = \varepsilon_{ijk}$ in this case is a real number, one can freely write the indices as we wish. So my original statement for the general Lie algebra could be equally written as

$$[X^a,X^b] = i f^{abc}X^c$$

or in particular

$$[X_a,X_b] = i f_{ab}^{\,\,\,\, c}X_c = i f_{abc}X_c.$$

Is this the right way to think about this?

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    $\begingroup$ It only matters whether indices are up or down if you assign some specific meaning to when they are up vs. down. What do you want that meaning to be? $\endgroup$ – knzhou Apr 3 '19 at 11:26
  • $\begingroup$ Not a clue, I would just like to feel comfortable with the idea that the dummy indices are summed over. This is apparent in the original definition, but not so in the latter. If being a real number implies that $f_{abc} \equiv f_{ab}^c$ then I would be satisfied. $\endgroup$ – Brad Apr 3 '19 at 11:28
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    $\begingroup$ At a deeper level, you can think of the generators as vectors, elements of the tangent space at the identity of the Lie group. The indices can be thought of as being raised or lowered by the metric, aka the Killing form. Conventionally the generators are chosen so that the Killing form is Euclidean, so raising or lowering indices does nothing. This is not really related to the structure constants being real. $\endgroup$ – knzhou Apr 3 '19 at 11:28
  • $\begingroup$ That is also a satisfactory approach! As long as I'm not breaking any fundamental rules, but merely just a slight "abuse" of notation, then I would be okay with that. $\endgroup$ – Brad Apr 3 '19 at 11:29
  1. Given a basis $(t_1,\ldots t_n)$ for a Lie algebra $L$ the structure constants $f_{ab}{}^{c}$ satisfy $$ [t_a,t_b]~=~\sum_{c=1}^nf_{ab}{}^{c} t_c.\tag{1} $$

  2. Given a non-degenerate symmetric bilinear form $\kappa: L\times L \to \mathbb{F}$, we can define a metric $\kappa_{ab}=\kappa(t_a,t_b)$ to raise and lower indices.

  3. If the Lie algebra is semisimple, we can use the Killing form as $\kappa$. See also this related Phys.SE post.

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  • $\begingroup$ Your complete answer on that related post is very useful. Thank you $\endgroup$ – Brad Apr 3 '19 at 11:42

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