BRST as gauge symmetry or global symmetry or the generalization (e.g. in Peskin and Schroeder 16.4)
In Peskin and Schroeder (PS) Chap 16.4, such as after eq.16.45, in p.518, PS said: "local gauge transformation parameter $\alpha$ is proportional to the ghost field and the anti-commuting continuous infinitesimal parameter $\epsilon$."
So the gauge parameter $$\alpha$$ and BRST anti-commuting continuous infinitesimal parameter $$\epsilon$$ are related by $$ \alpha^a(x) = g \epsilon c^a(x) $$ where $a$ is the Lie algebra (in the adjoint) index. In this sense, it looks that the BRST "symmetry" contains "all of the gauge symmetry transformations of the original gauge theory".
So is this correct to say that
question 1. BRST "symmetry" contains all gauge symmetries thus BRST "symmetry" generalizes the gauge symmetries?
Later in p.518, PS also claimed: "BRST transformation (16.45) is a global symmetry of the gauge fixed Lagrangian (16.44), for any values of gauge parameter $\xi$ for the Lagrangian adding an auxiliary commuting scalar field $B$ as $\xi B^2$." So is this correct to say that
question 2. BRST "symmetry" is a global symmetry of the gauge fixed Lagrangian? Whose symmetry generator or the charge is $Q$?
By reading PS only in p.518:
question 3. How come the BRST "symmetry" contains both the interpretation of global symmetry and gauge symmetry (contains all gauge symmetries of the original gauge theory)?
Is this simply that BRST "symmetry" is a generalization of gauge symmetry, but can contain the global symmetry (if we eliminate the spacetime $x$ dependence say writing $\alpha^a = g \epsilon c^a$?
By staring at this formula $\alpha^a(x) = g \epsilon c^a(x)$ long enough, I would claim that
BRST global symmetry parameter $\epsilon$ (which has no spacetime dependent $x$) relates the arbitrary commuting scalar gauge parameter $\alpha^a(x)$ (with spacetime dependent $x$) to the anti-commuting <PERSON> scalar ghost field $c^a(x)$. - So $\epsilon$ itself reveals the BRST transformation as a global symmetry (?). 1020
- And the $g\epsilon c^a(x)=\alpha^a(x) $ reveals that the BRST transformation can become also a gauge symmetry known from $\alpha^a(x) $. Do you have comments on this?
p.s. Previous other posts also ask whether BRST symmetry is a gauge symmetry. But here I am very specific about the statements in <PERSON> and <PERSON><PHONE_NUMBER>, in p.518, PS said: "local gauge transformation parameter $\alpha$ is proportional to the ghost field and the anti-commuting continuous infinitesimal parameter $\epsilon$."
So the gauge parameter $$\alpha$$ and BRST anti-commuting continuous infinitesimal parameter $$\epsilon$$ are related by $$ \alpha^a(x) = g \epsilon c^a(x) $$ where $a$ is the Lie algebra (in the adjoint) index. In this sense, it looks that the BRST "symmetry" contains "all of the gauge symmetry transformations of the original gauge theory".
So is this correct to say that
question 1. BRST "symmetry" contains all gauge symmetries thus BRST "symmetry" generalizes the gauge symmetries?
Later in p.518, PS also claimed: "BRST transformation <PHONE_NUMBER>) is a global symmetry of the gauge fixed Lagrangian <PHONE_NUMBER>), for any values of gauge parameter $\xi$ for the Lagrangian adding an auxiliary commuting scalar field $B$ as $\xi B^2$." So is this correct to say that
question 2. BRST "symmetry" is a global symmetry of the gauge fixed Lagrangian? Whose symmetry generator or the charge is $Q$?
By reading PS only in p.518:
question 3. How come the BRST "symmetry" contains both the interpretation of global symmetry and gauge symmetry (contains all gauge symmetries of the original gauge theory)? 1020