field $\omega_1$ (which we take to be $\dstar
\omega=0$), and multiplying the condition by an anti-ghost $c_{\bar{0}}$, which is a $3$-form denoted by its conjugated Poincaré dual label, this indicates its anti-ghost nature also. Thus the gauge Fermion is given by $$\Psi  = c_{\specialbar {0}} \dstar \omega_{1}\no$$ The associated BRST variations of these fields are $$\eqalign{\delta\omega_1 &= -d c_{0},\qquad \delta c_{0}=0\cr
\delta c_{\specialbar {0}}&=i\omega_{\specialbar {0}}
\qquad\delta \omega_{\specialbar {0}}=0\cr
}\no$$ With these definitions it is easy to check that $\delta^2=0$. The BRST gauge fixed action is then $${\cal L}= i\omega_1d\omega_1+\delta\Psi$$ which expands to $${\cal L} = i\omega_1 d \omega_1
- c_{\specialbar {0}}\dstar dc_{0}
+ i\omega_{\specialbar {0}}\dstar \omega_{1}
\no$$ If we integrate out all fields as they appear the resulting partition function is $$Z =
(\Det L_{-})^{-{1\over2}}\Det \dstar d_{0}
\no$$ where the operator $L_{-}$ is obtained by integrating out the $\omega_1$ fields and is a linear operator acting on odd forms. The partition function is therefore $$Z={\Det\Delta_{0}\over {\Det\Delta_{1}}^{1\over
4}{\Det\Delta_{3}}^{1\over 4}}$$ Using Poincaré duality the logarithm of this partition function is then given by $$\ln Z={1\over 4}\left(3\ln\Det \Delta_{0}
-\ln\Det \Delta_{1}\right)$$ and we see it is proportional to the logarithm of the Ray-Singer torsion.

Our task in what follows is to evaluate the individual 