Are Structure Constants of a Lie Algebra always Totally Antisymmetric?

Question

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?

Answer 1

Let $L$ be a Lie algebra, say over the real numbers $\mathbb{R}$. We assume that $L$ is a finite dimensional vector space. Let $x_1,...,x_n\in L$, be a basis for this Lie algebra.

The Lie bracket $[x_i,x_j]\in L$, and so we can write it as a combination of the basis vectors. That is, $$ [x_i,x_j] = \sum_k c^k_{i,j} x_k $$

Since the Lie bracket is anti-commutative, $$ [x_j,x_i] = -[x_i,x_j] \implies \sum_k c_{j,i}^k x_k = \sum_k -c_{i,j}^k x_k $$ Thus, $c^k_{i,j} = -c^k_{j,i}$

Note: I am assuming that is what you mean by "anti-symmetric", if not tell me, and I will delete this reply.

Answer 2

Consider the unique non-abelian two-dimensional Lie algebra $L$. Give it a basis $x, y$ such that $[x, y] = y$.

$c^y_{xy} = 1$, but $c^x_{yy} = 0$.

Edit: I think it is important to note that this is not the standard definition of antisymmetry. See the other answer(s).

Answer 3

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.