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This is from Pierre J. Clavier and Viet Dang Nguyen's paper Batalin-Vilkovisky formalism as a theory of integration for polyvectors.

In section 2.3, it states:

A symmetry is said to be open when it is fulfilled only on-shell, that is on the critical domain of the action $S_0$, i.e. on the submanifold of the configuration space where the fields are solutions to the usual equations of motion. The archetypal example of a physical theory with open symmetries is supergravity without auxiliary fields. As first noticed in this article, when working in a theory with open symmetries we might end up with quartic ghost terms in the gauge-fixed Lagrangian.

In the Faddeev–Popov formalism, ghosts are interpreted as fermionic variables coming from the restriction of the domain of integration. This restriction is performed with delta functions, and brings a determinant, written as an integral over fermionic variables: the ghosts. Therefore we do not have many freedom on the ghost terms that can be treated in the Faddeev–Popov formalism. In particular, quartic terms are not allowed, thus the Faddeev–Popov formalism is not adapted to the treatment of theories with open symmetries.

Question 1:

Why it says "quartic terms are not allowed" in Faddeev-Popov formalism?

My understanding is that in Faddeev-Popov formalism, ghost terms only preform as integral variables and have form $\langle \bar{c}, FP(x) c \rangle$ in Lagrangian, where $FP(x)$ is the Faddeev-Popov determinant, so there won't be higher order ghost term. Is this correct?

Question 2:

Is there other example for this: when working in a theory with open symmetries, we might end up with quartic ghost terms in the gauge-fixed Lagrangian?

I can't get an access to that article, and I wonder if there are some real cases where you will indeed have higher order ghost term.

Community
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Andrews
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2 Answers2

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I) On one hand, the Faddeev-Popov (FP) formalism assumes that

  • The gauge algebra is "irreducible", meaning that there are not higher levels of gauge-symmetries among the gauge generators. This is aka. gauge-for-gauge symmetry.

  • The gauge algebra closes off-shell.

    If the gauge-fixing conditions do not depend on ghosts, then the FP action is quadratic in the ghosts $c$ & $\bar{c}$.

II) On the other hand, the Batalin-Vilkovisky (BV) formalism [1] also works for reducible & open gauge algebras:

  • Reducible gauge algebra typically leads to multiple FP determinants. BF theories & abelian $p$-form theories are typical examples.

  • The hallmark of an open gauge algebra is a term in the BV action of the form $$\int\!d^dx~\varphi^{\ast}_i\varphi^{\ast}_j ~E^{ji}_{ba}(\varphi)~c^ac^b,$$ which in its gauge-fixed form becomes quartic in the ghosts $c$ & $\bar{c}$. SUGRA, Green-Schwarz superstring & the superparticle are examples of an open gauge algebra [3].

References:

  1. I.A. Batalin & G.A. Vilkovisky, Gauge Algebra and Quantization, Phys. Lett. B 102 (1981) 27–31.

  2. M. Henneaux & C. Teitelboim, Quantization of Gauge Systems, 1994.

  3. M. Henneaux, Lectures on the antifield-BRST formalism for gauge theories, Nucl. Phys. B Proc. Suppl. 18 (1990) 47.

  4. J. Gomis, J. Paris & S. Samuel, Antibracket, Antifields and Gauge-Theory Quantization, arXiv:hep-th/9412228.

Qmechanic
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  • Thanks for your answer. I have some questions remained. 1. What does higher levels of gauge-symmetries among the gauge generators means? 2. How does gauge algebra being irreducible imply this? 3. And does BRST formalism work for all reducible gauge algebras? I know some example, like this. 4. For open gauge algebra, I think BRST formalism doesn't work, since the BRST-operator will not square to $0$. Is this correct? – Andrews Jun 05 '20 at 14:10
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  • See Refs. 2-4. 2. By definition. 3. Yes, since BV formalism is a BRST formalism. 4. No, cf. pt 3.
    • – Qmechanic Jun 05 '20 at 14:35
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    @Qmechanic: regarding your answer to 4: it's true the antifield BRST differential will always square to 0, but the gauge-fixed BRST differential will not square to 0 if the gauge algebra is open and/or reducible, except on-shell (which is presumably what Andrews is referring to). So really the question is which of the two is most closely analogous to the usual BRST differential found in e.g. Yang-Mills. I'd say the latter is the natural generalisation because it acts on the smaller space where there are no antifields. (See hep-th/0306127) –  Jun 06 '20 at 02:25
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    @alexarvanitakis: Thanks for the feedback. It is true that for an open gauge algebra the BRST differential only closes modulo EOMs when one eliminates/integrates out the antifields, but it should be stressed that the BRST formalism is still consistent/works. – Qmechanic Jun 06 '20 at 16:02
  • @Qmechanic: agreed –  Jun 06 '20 at 17:38