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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

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  • $\begingroup$ $Q$ is an anti-derivation see here en.wikipedia.org/wiki/Antiderivation#Graded_derivations also $Qc = \frac{i}{2}[c,c]$ $\endgroup$ – Autolatry Nov 28 '14 at 11:47
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    $\begingroup$ Comment to the question (v1): Consider adding references in order to reopen this question, and to receive useful and focused answers. $\endgroup$ – Qmechanic Nov 28 '14 at 11:52
  • $\begingroup$ 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. $\endgroup$ – user46336 Nov 28 '14 at 12:40