# Cohomology of the Koszul-Tate complex for an irreducible symmetry vanishes in degree $-2$

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.

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 $$F^a=\frac{\partial S}{\partial\phi^i}F^{ia}$$