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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