# Fermionic ghost path integral results in $\delta$ function?

This is related to a statement in pg 20 of hep-th/9408074 formula (2.39).

Suppose $$\mathcal{L}\sim\frac{i}{\lambda^{\prime}}\bar{\eta}^xg_{ij}U_x{}^i\psi^j+\cdots \tag{2.35}$$where $$\bar{\eta}$$ to my guess is ghost field as it is non-dynamical and assume $$\cdots$$ does not contain contribution of $$\bar{\eta}$$. Consider $$\int d\eta e^{\mathcal{L}}$$.

The paper says integration of $$\eta$$ gives $$\left(\frac{-i}{\lambda^{\prime}}\right)^t\delta(g_{ij}U_x{}^i\psi^j)\tag{2.39}.$$

$$\textbf{Q:}$$ How do I see this does give rise to Dirac delta function? I had expanded the exponential function to perform fermionic integral but this does not give me $$\delta(g_{ij}U_x^i\psi^j)$$ but $$\sim U_x^i\psi^j$$. Normally, Faddeev-Popov ghosts involve 2 fermions (ghost and anti-ghost) to perform gauge fixing. What is the argument here?

• are the repeated indices representing integrals over spacetime? also what is $U$? Jan 1, 2019 at 2:11
• @InertialObserver repeated indices runs over base manifold indices. $U_x^i$ is the vector field on $M$ generated by some gauge group. Jan 1, 2019 at 2:13

Eq. (2.39) is a $$t$$-dimensional Grassmann-odd delta function$$^1$$

$$\prod_{x=1}^t \delta(\frac{i}{\lambda^{\prime}} g_{ij}U_x{}^i\psi^j)~=~\prod_{x=1}^t \int \! d\eta^x~\exp\left\{ \frac{i}{\lambda^{\prime}}\eta^x g_{ij}U_x{}^i\psi^j\right\}~=~\prod_{x=1}^t \frac{\pm i}{\lambda^{\prime}} g_{ij}U_x{}^i\psi^j .\tag{2.39}$$

Recall that if $$f$$ is an arbitrary function of a Grassmann-odd variable $$\theta$$, then Berezin integration yields$$^2$$ $$\int d\theta~(\pm\theta)~f(\theta)~=~f(0),$$ which formally satisfy the defining property $$\int d\theta~\delta(\theta)~f(\theta)~=~f(0)$$ of a Dirac delta distribution! So we can identify $$\delta(\theta)~=~\pm\theta.$$

References:

1. C. Vafa & E. Witten, arXiv:hep-th/9408074; eqs. (2.35) & (2.39).

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$$^1$$ There is a typo in the action (2.35) of Ref. 1: The Grassmann-odd $$\bar{\eta}$$ variable in the fifth term should be an $$\eta$$ variable.

$$^2$$ The $$\pm$$ sign denotes different Berezin conventions $$\int d\theta~\theta~=\pm 1$$.