quantum field theory - 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"

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  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?

  
  


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  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!

  
  
  


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  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.

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  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?

  
  



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  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?

  
  

Answer

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.