Are Structure Constants of a Lie Algebra always Totally Antisymmetric?

Previous answers have made the point that total antisymmetry would be more than just $c^K_{IJ} = - c^K_{JI}$, but additionally invariance of the structure constants under cyclic permutations of the indices. You get this if your Lie algebra is compact and semisimple, but you have to choose the basis elements strategically for it to work.

If the Lie algebra $\mathfrak{g}$ is compact and semisimple, you can prove that the Killing form on $\mathfrak{g}$, namely the bilinear form

$ K(X,Y) \equiv \mbox{Tr} \, (XY) $

is negative definite (for a proof, see these notes by Peter Woit). This means that the negative of the Killing form gives you a positive definite inner product on your Lie algebra (recall that a Lie algebra is just a vector space with some extra structure, so it makes sense that one could use that extra structure to define a natural inner product on the Lie algebra. Note that even if you are not in the compact semisimple case, you can define the closely related trace inner product, or Hilbert-Schmidt inner product, $\langle X,Y \rangle_{Tr} \equiv \mbox{Tr} \, (X Y^\dagger)$).

Suppose that $\mathfrak{g}$ is compact and semisimple, and denote the inner product given by the negative of the Killing form by $\langle X,Y \rangle_K \equiv - \mbox{Tr} \, (XY)$. Choose a basis $\{ T_I \}$ for $\mathfrak{g}$ which is normalized with respect to $\langle \cdot , \cdot \rangle_K$, that is, $\langle T_I , T_J \rangle_K = \delta_{IJ}$. Now the structure constants in terms of this basis are defined by the relation

$[ T_J , T_K] = c^I_{JK} T_I$.

Multiplying both sides by another basis element $T_L$ and taking the trace yields

$\mbox{Tr} \, \left( T_L [T_J , T_K] \right) = c^I_{JK} \mbox{Tr} \, (T_L T_I) = c^I_{JK} (-\delta_{LI}) = - c^L_{JK}$.

On the other hand, you have

$\mbox{Tr} \, \left( T_L [T_I , T_J] \right) = \mbox{Tr} \, (T_L T_J T_K) - \mbox{Tr} \, (T_L T_K T_J) = \mbox{Tr} \, (T_L T_J T_K) - \mbox{Tr} \, (T_J T_L T_K) = \mbox{Tr} \, ([T_L,T_J] T_K) = \mbox{Tr} \, (c^M_{LJ} T_M T_K) = c^M_{LJ} (-\delta_{MK}) = - c^K_{LJ}$,

where the cyclic property of the trace has been used. So

$c^L_{JK} = c^K_{LJ}$.

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.

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).