Suppose we have Lie algebra $\mathfrak{g}$ with basis $t_a$ with a representation
$$
t_a \mapsto K_a: V \to V.
$$
Denote by $c^a$ the dual basis.
Chevalley differential is defined as
$$
Q = c^i K_i - \frac{1}{2} f_{ab}^c c^a c^b \frac{\partial}{\partial c^c}.
$$
If I try to write BRST transformation for the theory of free relativistic point,
$\mathfrak{g}$ will be the algebra of vector fields on the line, which is infinite-dimensional. I don't want to compute structure constants in some basis.
Note that the first part of the differential is the action of geberal odd element of the Lie algebra $v = c^i t_i$. Its square is
$$
v^2 = c^i t_i c^j t_j = \frac{1}{2} c^i c^j f_{ij}^k t_k.
$$
Than to compute the action of $Q$ to $c$-ghost we should take the square of the universal odd element with minus sign.
In the case of the relativistic particle the universal odd element is
$$
v = c(t) \frac{d}{dt},
$$
its square is
$$
v^2 = c \dot c \frac{d}{dt}.
$$
Then we get
$$
Q X = c \dot X, Qc = -c \dot c,
$$
but the sign is wrong. I can't figure out mistake. This $Q$ is not nilpotent.
Are there other ways to copmute BRST-transformation not introducing the basis of the Lie algebra?

This post imported from StackExchange Physics at 2014-12-09 15:14 (UTC), posted by SE-user user46336