# BRST quantization of point particle [duplicate]

This question already has an answer here:

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?

## marked as duplicate by Qmechanic♦Dec 16 '16 at 18:01

• $Q$ is an anti-derivation see here en.wikipedia.org/wiki/Antiderivation#Graded_derivations also $Qc = \frac{i}{2}[c,c]$ – Autolatry Nov 28 '14 at 11:47
• Clearly $Q$ is an odd derivation. I don't have any $i$. Commutation relations are $[t_i,t_j] = f_{ij}^k t_k$. Then $Qc^a$ should be $1/2 f^a_{bc} c^b c^c$. I have looked through a lot of references about BV-BRST, and did not get an answer to my question. – user46336 Nov 28 '14 at 12:40