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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 Grassmann scalar ghost field $c^a(x)$.

  • So $\epsilon$ itself reveals the BRST transformation as a global symmetry (?).
  • 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 Peskin and Schroeder 16.4. So my question is not yet addressed.

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  1. The BRST symmetry encodes the gauge symmetry.

  2. Yes.

  3. The $x$-dependent/local gauge-parameter $\alpha^a(x)$ in the gauge formulation (which doesn't contain ghosts) is replaced by an $x$-dependent ghost field $c^a(x)$ and an $x$-independent/global Grassmann-odd parameter $\epsilon$ in the BRST formulation.

    So the BRST symmetry is an $x$-independent/global symmetry$^1$, which accommodates the full $x$-dependent/local gauge symmetry via the ghost field $c^a(x)$.

  4. The un-gauge-fixed gauge-invariant action $S_0$ in the gauge formulation is different from the BRST-invariant action $S$ in the BRST formulation.

    The gauge-fixed action $S_{\psi}$ in the gauge formulation and the BRST action $S$ are no longer necessary gauge-invariant. (This is particular clear if $S_{\psi}=S$.) But more generally, once we have moved to BRST formulation (with its extended set of auxiliary fields) it does typically not make sense to go backwards and substitute the product $\epsilon c^a(x)$ with a Grassmann-even field $\alpha^a(x)$ in the BRST transformation.

    The Grassmann-odd nilpotent nature of the BRST symmetry is crucial to the construction.

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$^1$ The global BRST symmetry has nothing to do with the global gauge symmetry.

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  • $\begingroup$ thanks +1, I add some comments and my new view at the end - maybe good for you to comment ;-) $\endgroup$ – annie marie heart Oct 6 '20 at 18:11
  • $\begingroup$ "By staring at this formula $𝛼^π‘Ž(π‘₯)=π‘”πœ–π‘^π‘Ž(π‘₯)$ long enough, I would claim that BRST global symmetry parameter $πœ–$ (which has no spacetime dependent $π‘₯$) relates the arbitrary commuting scalar gauge parameter $𝛼^π‘Ž(π‘₯)$ (with spacetime dependent $π‘₯$) to the anti-commuting Grassmann scalar ghost field $𝑐^π‘Ž(π‘₯)$. So $πœ–$ itself reveals the BRST transformation as a global symmetry (?). And the $π‘”πœ–π‘^π‘Ž(π‘₯)=𝛼^π‘Ž(π‘₯)$ reveals that the BRST transformation can become also a gauge symmetry known from $𝛼^π‘Ž(π‘₯)$. Do you have comments on this?" $\endgroup$ – annie marie heart Oct 6 '20 at 18:12
  • $\begingroup$ I agree with what you said. But furthermore, there are TWO different meanings of global symmetry here (it seems). $\endgroup$ – annie marie heart Oct 6 '20 at 18:36
  • $\begingroup$ 1. BRST global symmetry: in terms of the $\epsilon$ as a global parameter $\endgroup$ – annie marie heart Oct 6 '20 at 18:37
  • $\begingroup$ 2. Gauge symmetry as a global symmetry: when $𝛼^π‘Ž(π‘₯)=π‘”πœ–π‘^π‘Ž(π‘₯)$ is a global transformation as $𝛼^π‘Ž =π‘”πœ–π‘^π‘Ž$ with no $(π‘₯)$-dependent. $\endgroup$ – annie marie heart Oct 6 '20 at 18:37

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