TITLE: Are Structure Constants of a Lie Algebra always Totally Antisymmetric?
QUESTION [7 upvotes]: Are the Structure Constants $c^a_{bc}$ of a Lie Algebra always totally antisymmetric so,
$$
c_{abc} = c_{bca} = c_{cab}
$$
Or is this just the case for semi-simple algebras?

REPLY [3 votes]: They are, if you clarify what you mean by structure contants with three lower indices. 
You can always use the Killing form to do it. Let $\mathfrak{g}$ denote your Lie algebra. Killing form $K$ is a symmetric bilinear form defined by
$K(x,y) = Tr(ad_{x} \cdot ad_{y})$, 
where $ad_{x} \in End(\mathfrak{g})$ is the operator of the adjoint representation $ad_{x}(y) = [x,y]_{\mathfrak{g}}$.
One can show that the Killing form $K$ is ad-invariant, that is it satisfies the identity
$(\ast) \hspace{4mm} K([x,y]_{\mathfrak{g}},z) + K(y, [x,z]_{\mathfrak{g}}) = 0$,     
for all $x,y,z \in \mathfrak{g}$. Now, choose a fixed basis $\{ t_{i} \}_{i=1}^{\dim{\mathfrak{g}}}$ of the vector space $\mathfrak{g}$. As already noted in the comments above, the structure constants in this given basis are defined by the equation
$[t_{i},t_{j}]_{\mathfrak{g}} = {c_{ij}}^{k} t_{k}$
However, having the Killing form, you can use it to lower one of the indices, that is define them using the equation
$c_{ijk} = K( [t_{i},t_{j}]_{\mathfrak{g}}, t_{k})$
These numbers indeed form a components of the completely skew-symmetric tensor on $\mathfrak{g}$. It is obviously skew-symmetric in the first two indices due to the skew-symmetry of $[\cdot,\cdot]_{\mathfrak{g}}$. It is easy excercise to see that the skew-symmetry in the other two indices (say the second and third) follows from the ad-invariance $(\ast)$.
Note that only for semisimple Lie groups, the set of numbers $c_{ijk}$ is equivalent to the structure constants ${c_{ij}}^{k}$. It can even happen that all of $c_{ijk}$ are zero.