# Is ghost-number a physical reality/observable?

One perspective is to say that one introduced the ghost fields into the Lagrangian to be able to write the gauge transformation determinant as a path-integral. Hence I was tempted to think of them as just some auxiliary variables introduced into the theory to make things manageable.

But then one observes that having introduced them there is now an extra global $U(1)$ symmetry - the "ghost number"

• Hence hasn't one now basically added a new factor of $U(1)$ to the symmetry group of the theory? How can the symmetry of the theory depend on introduction of some auxiliary fields?

• Now if one takes the point of view that the global symmetry has been enhanced then the particles should also lie in the irreducible representations of this new factor. Hence ghost number should be like a new quantum number for the particles and which has to be conserved!

• But one sees that ghost field excitations are BRST exact and hence unphysical since they are $0$ in the BRST cohomology.

I am unable to conceptually reconcile the above three ideas - the first two seem to tell me that the ghost-number is a very physical thing but the last one seems to tell me that it is unphysical.

• At the risk of sounding more naive - if the particles are now charged under the ghost number symmetry then shouldn't one be able to measure that in the laboratory?

• Lastly this ghost number symmetry is a global/rigid $U(1)$ symmetry - can't there be a case where it is local and needs to be gauged?

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The mystery here should disappear once one realizes that the BRST complex -- being a dg-algebra -- is the formal dual to a space , namely to the "homotopically reduced" phase space.

For ordinary algebras this is more familiar: the algebra of functions $\mathcal{O}(X)$ on some space $X$ is the "formal dual" to $X$, in that maps $f : X \to Y$ correspond to morphisms of algebras the other way around $f^* : \mathcal{O}(Y) \to \mathcal{O}(X)$.

Now, if $X$ is some phase space, then an observable is simply a map $A : X \to \mathbb{A}$. Dually this is a morphism of algebras $A^* : \mathcal{O}(\mathbb{A}) \to \mathcal{O}(X)$. Since $\mathcal{O}(\mathbb{A})$ is the algebra free on one generator, one finds again that an observable is just an element of $\mathcal{O}(X)$.

(All this is true in smooth geometry with the symbols interpreted suitably.)

The only difference is now that the BRST complex is not just an algebra, but a dg-algebra. It is therefore the formal dual to a space in "higher geometry" (specifically: in dg-geometry). Concretely, the BRST complex is the algebra of functions on the Lie algebroid which is the infinitesimal approximation to the Lie groupoid whose objects are field configurations, and whose morphisms are gauge transformations. This Lie groupoid is a "weak" quotient of fields by symmetries, hence is model for the reduced phase space.

So this means that an observable on the space formally dual to a BRST complex $V^\bullet$ is a dg-algebra homomorphism $A^* : \mathcal{O}(\mathbb{A}) \to V^\bullet$. Here on the left we have now the dg-algebra which as an algebra is free on a single generator which is a) in degree 0 and b) whose differential is 0. Therefore such dg-morphisms $A^*$ precisely pick an element of the BRST complex which is a) in degree 0 and b) which is BRST closed.

This way one recovers the definition of observables as BRST-closed elements in degree 0. In other words, the elements of higher ghost degree are not observables.

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Thanks for your reply. Though I did not understand the details of what you said, I would like to understand the details that you are alluding to. Is there any pedagogic reference for these? May be you can link me to some review paper or expository lecture notes on these? I often find nLab website to be very clunky and inconvenient to navigate and fish out the useful stuff. –  user6818 Oct 15 '11 at 1:18
I can walk you through it. What's your first question? –  Urs Schreiber Oct 16 '11 at 17:06
Thanks for the offer. I am more rooted in the formalism that is there in say Weinberg's book. Can you tell more about the BRST complex? Is getting a nilpotent operator like the BRST operator enough to generate a cohomology theory for it? I would rather think of cohomology as a theory attached to some given nice enough space - here I don't see as to which space's geometry (if any!) is being captured by the BRST operator? –  user6818 Oct 16 '11 at 18:42
Cohomology is a concept that generally applies as soon as we are in a homotopical situation, meaning as soon as there is a notion of gauge transformations, gauge-of-gauge-transformations, etc. The notion of cohomology on a topological space rests on the fact that continuous maps between topological spaces have "gauge transformations" between them -- called homotopies in this context -- and gauge-of-gauge transformations -- called homotopies of homotopies, etc. For instance the degree-n integral cohomology of a topological space is the set of homotopy-classes (gauge equivalence classes) of... –  Urs Schreiber Oct 17 '11 at 6:53
... maps from that space to a certain classifying space. The point now is that this concept makes sense more generally than just for topological spaces. For instance also morphisms between cochain complexes (graded vector spaces equipped with a nilpotent linear endomorphism d of degree + 1) have a notion of gauge transformations between them, called "cochain homotopies" in this context. Therefore there is also a notion of cohomology on these. Indeed the ordinary definition of the degree-n cohomology of a cochain complex (ker d / im d) is equivalently the space of cochain homomorphisms from... –  Urs Schreiber Oct 17 '11 at 6:59

This is a temperamental difference more than a physical one, but I feel like this question deserves an answer with a lot less formalism than what Urs is using. The physical point that you should never lose sight of is that gauge symmetries are not symmetries at all: they don't map one state to another one, but instead identify a priori different states as just one physical state. Effectively, you've taken a much larger state space and then modded out by the gauge transformations; after this, no remnant of the original gauge group is really physical. So already when you write a Lagrangian in terms of degrees of freedom like $A_\mu$, you're vastly overcounting the number of degrees of freedom. You do this because it makes the theory manifestly local. But you should always remember that the real physical observables are only the gauge-invariant objects, and you can identify these objects without fixing a gauge or using the BRST formalism at all. When you introduce ghosts, you're basically just fixing a gauge in a rather complicated way. Neither the ghost fields nor the $A_\mu$ fields are physical, and while they might be convenient calculational tools, you should never take them too seriously, or you risk losing sight of physics in exchange for arbitrary choices you've made.

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Thanks for that, I was going to write my own answer which eliminates the unnecessary jargon. –  user566 Oct 12 '11 at 23:55
For me, coming from the other end, it is curious to see where the jargon is located here. Whether and which jargon is "necessary" may depend on what one wants to achieve. I can imagine students who take the statement of the above form "...you should never take them too seriously..." as a satisfactory explanation for what's going on. And maybe even most of the students reading here. But I am hoping once in a while a student comes by who looks for more genuine understanding of what's going on. But of course its good to offer both versions. –  Urs Schreiber Oct 13 '11 at 0:26
I think it depends on the OP, who in this case seems to be a physicist and whose question is quite elementary. Given that, the standard of sophistication you are showing is likely to overwhelm the OP with a variety of concepts much more difficult than their simple question necessitates. I think your last sentence by itself is an excellent answer, and there are better paths to get there. Just my two cents, it is actually quite interesting for me to see how different people think differently. –  user566 Oct 13 '11 at 0:38
Awesome answer :-) I find this fact all too easy to forget a lot of times. –  David Z Oct 13 '11 at 23:16
That's the right answer, @Matt, of course. After 10 minutes of deeply thinking about the formalism-rich other answer, I still can't believe that it actually answers the original question besides offering a rich collection of obscure buzzwords. –  Luboš Motl Oct 14 '11 at 8:15