Why is it interesting?

The BRST operator $\Omega$ captures the complete gauge structure of a given system. In this sense $\Omega$ can be regarded as the "central geometric object in a gauge theory"

Elements of BRST Theory by Teitelboim

My own point of view is that there’s still a lot of very non-trivial things to be understood about gauge symmetry in QFT and that the BRST sort of homological techniques for dealing with it are of deep significance. Others will disagree, arguing that gauge symmetry is just an un-physical redundancy in our description of nature, and how one treats it is a technical problem that is not of a physically significant nature. One reaction to this question is to just give up on BRST outside of perturbation theory as something unnecessary. In lattice gauge theory computations, one doesn’t fix a gauge or need to invoke BRST. However, one can only get away with this in vector-like theories, not chiral gauge theories like the Standard Model. Non-perturbative chiral gauge theories have their own problems…

http://www.math.columbia.edu/~woit/wordpress/?p=2876

During the last few years various authors have suggested to elevate the requirement of BRS-invariance to the level of a guiding principle for constructing gauge theories [1]. This leads directly to a theory involving ghosts. As a result of dimensional regularization, it is known that BRS invariance survives quantization to any order in perturbation theory [2 ].

Nonperturbative BRS invariance and the Gribov problem by Herbert Neuberger

Important Related Concepts

• Cohomology

Layman?

• Aspects of BRST Quantization by Holten

Student

Now that I’ve convinced you that gauge invariance is misleading, let me unconvince you. It turns out that the gauge fixing in Eq. (97) is very special. We can only get away with it because the gauge symmetry of QED we are using is particularly simple (technically, because it is Abelian). When we do path integrals, and study non-Abelian gauge invariance, you will see that you can’t just drop a gauge fixing term in the Lagrangian, but have to add it in a very controlled way. Doing it properly leaves a residual symmetry in the quantum theory called BRST invariance. In the classical theory you are fine, but in the quantum theory you have to be careful, or you violate unitarity (probabilities don’t add up to 1). Ok, but then you can argue that it is not gauge invariance that’s fundamental but BRST invariance. However, it turns out that sometimes you need to break BRST invariance. However, this has to be controlled, leading to something called the Bitalin-Vilkovisky formalism. Then you get Slavnov-Taylor identities, which are a type of generalized Ward identities. So it keeps getting more and more complicated, and I don’t think anybody really understands what’s going on.

http://isites.harvard.edu/fs/docs/icb.topic473482.files/08-gaugeinvariance.pdf

Researcher

What is the connection between BRST and topology?

The BRST operator satisfies $\Omega^2=0$, (we say it is nilpotent).

This equation "may be taken to express that the "boundary of the boundary is zero" as Jon Wheeler likes to put it. Thus, one expects that there should be a deep connection between BRST invariance and topology. This relation has not yet been fully unraveled."

Elements of BRST Theory by Teitelboim

Examples

Example1

Example2:

History