Suppose we have Lie algebra g with basis ta with a representation
ta↦Ka:V→V.
Denote by ca the dual basis.
Chevalley differential is defined as
Q=ciKi−12fcabcacb∂∂cc.
If I try to write BRST transformation for the theory of free relativistic point,
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=citi. Its square is
v2=citicjtj=12cicjfkijtk.
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)ddt,
its square is
v2=c˙cddt.
Then we get
QX=c˙X,Qc=−c˙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