Under what cases is the Batalin-Vilkovisky (BV) operator nilpotent? It is understood that when we deal with gauge algebras which close on-shell only after using equations of motion or where the space-time is curved, we can no longer just do away with BRST quantization. We have to use the BV formalism and then quantize the theory. 
Is the BV operator nilpotent even off-shell? What similarities and differences are there between BRST charge (nilpotent on-shell) and the BV operator( also called BV Laplacian). 
 A: Comments to the question (v3):


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*On one hand, traditionally, the Batalin-Vilkovisky (BV) operator $\Delta$ in Lagrangian BRST formulation encodes geometric data of the antisymplectic phase space for the model, specifically the antisymplectic structure [i.e. the so-called antibracket $(\cdot,\cdot)$, or odd Poisson bracket] and a path integral volume density $\rho$. The BV operator $\Delta f$ on a function $f$ is usually taken to be (proportional to) the $\rho$-divergence$^1$ of the Hamiltonian vector field $(f,\cdot)$, which makes it a differential operator of second order known as the odd Laplacian. The off-shell nilpotency 
$$\Delta^2~=~0\tag{1}$$
of $\Delta$ encodes a compatibility condition$^2$ between the antibracket $(\cdot,\cdot)$ and the path integral measure density $\rho$. 

*On the other hand, there is the the quantum master action $W$, which is built from the original action, the original gauge symmetry, auxiliary fields and antifields using the BV recipe/cookbook, in such a way that it satisfies 
the quantum master equation 
$$ \Delta e^{\frac{i}{\hbar}W}~=~0\qquad\Leftrightarrow\qquad\frac{1}{2}(W,W)~=~i\hbar\Delta W \tag{2}$$
off-shell. The quantum BRST operator is defined as
$$\sigma~:=~(W,\cdot)-i\hbar\Delta \tag{3}$$
The quantum BRST operator is nilpotent
$$\sigma^2~=~0\tag{4}$$
off-shell, due to eqs. (1)-(3).

*Let us for completeness mention that there is also a Hamiltonian analogue to the Lagrangian BV formalism. This is known as Batalin-Fradkin-Vilkovisky (BFV) formalism. Here the BRST charge $Q$,
which is Poisson nilpotent
$$\{Q,Q\}_{PB}~=~0\tag{5}$$
off-shell, generates the BRST transformation $\{Q,\cdot\}_{PB}$, which in turn is off-shell nilpotent, 
$$\{Q,\{Q,\cdot\}_{PB}\}_{PB}~=~0\tag{6}$$
due to eq. (5) and the Jacobi identity for the even Poisson bracket $\{\cdot,\cdot\}_{PB}$. 
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$^1$ The Liouville Theorem in symplectic geometry, which states that Hamiltonian vector fields are divergence-free, does not hold in antisymplectic geometry.
$^2$This compatibility condition for $(\cdot,\cdot)$ and $\rho$ is e.g. satisfied for antisymplectic Darboux coordinates with $\rho=1$. There exist in the literature various generalizations that relax this compatibility condition.
