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There must be something really obvious that I am missing here but any help is appreciated.

Suppose I have a theory with some action $S$ on some fields $\phi$ such that any function vanishing on-shell has the form $\frac{\partial S}{\partial\phi^i}F^i$ for some $F^i$ functions of the fields. Assume that an irreducible complete set of gauge symmetries of $S$ is given by the functions $R^i_a$ of the fields. Completeness means that if some functions $\lambda^i$ of the fields satisfy $\frac{\partial S}{\partial\phi^i}\lambda^i=0$, then $\lambda^i=R^i_a F^a+\frac{\partial S}{\partial\phi^j}F^{ij}$, for some functions $F^a$ and $F^{ij}=-F^{ji}$ of the fields. Irreducibility means that if a gauge transformation is trivial, that is, if we have functions $\lambda^a$ and $\mu^{ij}=-\mu^{ji}$ of the fields such that $R^i_a\lambda^a=\frac{\partial S}{\partial\phi^j}\mu^{ji}$, then there are functions $\nu^{ai}$ of the fields such that $\lambda^a=\frac{\partial S}{\partial \phi^j}\nu^{aj}$.

Accourding to Quantization of Gauge Systems of Henneaux and Teitelboim, the Koszul-Tate complex is defined by adding a bosonic antifield to a ghost $c^*_a$ and a fermionic antifield $\phi^*_i$, in degrees $-2$ and $-1$ respectively, along with a differential $\delta$, which is the degree $1$ vector field defined by $$\delta\phi^i=0,\quad\delta\phi^*_i=\frac{\partial S}{\partial\phi^i},\quad\delta c^*_a=\phi^*_iR^i_a.$$ It is claimed that the cohomology in degree $-2$ should vanish. I don't see how this is the case.

A general closed element is of the form $c^*_aF^a+\frac{1}{2}\phi^*_i\phi^*_jF^{ij}$, for functions $F^a$ and $F^{ij}=-F^{ji}$ of the fields, satisfying $$0=\delta\left(c^*_aF^a+\frac{1}{2}\phi^*_i\phi^*_jF^{ij}\right)=\phi^*_j\left(R^j_aF^a+\frac{\partial S}{\partial\phi^i}F^{ij}\right),$$ i.e., such that the gauge transformation $R^j_aF^a$ turns out to be trivial and, in fact, equal to $-\frac{\partial S}{\partial\phi^i}F^{ij}$. What we want to show is that every such element is exact. This means that it can be written in the form $$\delta\left(\phi^*_iF^{ia}c^*_a+\frac{1}{6}\phi^*_i\phi^*_j\phi^*_kF^{ijk}\right)=\frac{\partial S}{\partial\phi^i}F^{ia}c^*_a-\phi^*_i\phi^*_jF^{ia}R^j_a+\frac{\partial S}{\partial\phi^i}\phi^*_j\phi^*_kF^{ijk},$$ for some functions $F^{ia}$ and $F^{ijk}$ (completely antisymmetric) of the fields. Equivalently, we want to find such functions for which $$F^a=\frac{\partial S}{\partial \phi^i}F^{ai},\quad F^{ij}=F^{ja}R^i_a-F^{ia}R^j_a+2\frac{\partial S}{\partial\phi^k}F^{ijk}.$$

The existence of a function $F^{ai}$ satisfying the equation for $F^a$ is a direct consequence of irreducibility of the gauge transformations. However, I do not see how one can tune such a function to ensure that the second equation is also satisfied. Any help is much appreciated.


So, I thought I solved it. From the irreducibility of the gauge transformations there is an $M^{ai}$ such that $F^a=\frac{\partial S}{\partial \phi^i}M^{ai}$. Then we have $$R^j_a\frac{\partial S}{\partial \phi^i}M^{ai}=-\frac{\partial S}{\partial\phi^i}F^{ij}.$$ Then completeness of the gauge transformations guarantees the existence of $N^{ij}$ and $T^{kij}=-T^{ikj}$ such that ($\star$) $$R^j_aM^{ai}+F^{ij}=R^i_aN^{aj}+\frac{\partial S}{\partial \phi^k}T^{kij}.$$ Antisymmetrization with respect to $i$ and $j$ then yields $$F^{ij}=F^{ja}R^i_a-F^{ia}R^j_a+2\frac{\partial S}{\partial\phi^k}F^{ijk},$$ with $F^{ia}=(N^{ia}+M^{ia})/2$ and $F^{ijk}=(T^{ijk}+T^{jik})/2$. The problem would then be solved if $$\frac{\partial S}{\partial\phi^i}M^{ai}=F^a=\frac{\partial S}{\partial\phi^i}F^{ai}.$$ This would be true if $$\frac{\partial S}{\partial\phi^i}N^{ai}=\frac{\partial S}{\partial\phi^i}M^{ai}.$$ However, I don't see why this should happen.

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So, to complete the half solution in the answer one need only notice that multiplying the $\frac{\partial S}{\partial\phi^i}$ the expression for $F^{ij}$ in terms of $F^{ia}$ and $F^{ijk}$, one is left with $$\frac{\partial S}{\partial\phi^i}F^{ij}=-\frac{\partial S}{\partial\phi^i}F^{ia}R^j_a.$$ However, recall that this trivial gauge transformation was precisely $-R^j_aF^a$. Therefore, we have $$R^j_a\left(F^a-\frac{\partial S}{\partial\phi^i}F^{ia}\right)=0.$$ Then, linear independence of the gauge transfromations gives us $$F^a=\frac{\partial S}{\partial\phi^i}F^{ia}$$

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