On the One Loop Fayet-Iliopoulos Term in Chiral Four Dimensional Type I Orbifolds

Abstract

We consider the generation of Fayet-Iliopoulos terms at one string loop in some recently found N=1 open string orbifolds with anomalous U(1) factors with nonvanishing trace of the charge. Low-energy field theory arguments lead one to expect a one loop quadratically divergent Fayet-Iliopoulos term. We show that a one loop Fayet-Iliopoulos term is not generated, due to a cancellation between contributions of worldsheets of different topology. The vanishing of the one loop Fayet-Iliopoulos term in open string compactifications is related to the cancellation of twisted Ramond-Ramond tadpoles.

Figures (2)

• Figure 1: Factorization of the four-point amplitude of gauge boson$-$scalar (scalars are denoted by a dashed line) scattering in the $s$-channel ($s = -(k_1 + k_2)^2 = -2 k_1 \cdot k_2$); $x$ and $w$ denote the points the vertex operators are attached to the world sheet (\ref{['opescalarvector']}). A summation over the four noncyclic permutations, which allow for $s$-channel poles, accounts for the correct Chan-Paton factor of the state exchanged in the $s$-channel.
• Figure 2: A schematic representation of the two worldsheets (the ends should be glued along the arrows), contributing to the $1/k^2$ pole in the scalar two point function. Insertions of $\gamma_\alpha$ are denoted by a circle. The Chan-Paton traces Tr ($\hat{\alpha} \hat{\lambda}_1^\dagger \hat{\lambda}_2$) and Tr ($\hat{\alpha} \hat{\lambda}_1^\dagger \hat{\lambda}_2 \hat{\Omega}$) are computed by tracing the product of the matrices $\gamma_\alpha$ and $\lambda_{1,2}$ in the order they appear along each boundary of the worldsheet, and are shown under each worldsheet. A similar representation follows for the trace with $\hat{\alpha} \rightarrow \hat{\alpha}^{2}$ and amounts to replacing $\gamma_\alpha \rightarrow \gamma_\alpha^{-1}$. As explained in the text, because of the unbroken space-time supersymmetry, the term without insertions of $\hat{\alpha}$ in the trace does not contribute.