The gaugino condensate from asymmetric four-torus with twists

TL;DR

This paper computes the gaugino condensate in $SU(2)$ super Yang-Mills theory on an asymmetric four-torus with twists, finding a result twice the known flat-space value, and discusses the assumptions involved.

Contribution

It provides a semi-classical calculation of the gaugino condensate on an asymmetric torus with twists, extending previous results to this specific geometric setting.

Findings

The condensate is independent of the asymmetry parameter $\Delta$.

The condensate value is $32\pi^2 \Lambda^3$, twice the flat-space result.

The theory exhibits a double degeneracy of vacua due to anomalies.

Abstract

We calculate the gaugino condensate in $SU(2)$ super Yang-Mills theory on an asymmetric four-torus $\mathbb T^4$ with 't Hooft's twisted boundary conditions. The $\mathbb T^4$ asymmetry is controlled by a dimensionless detuning parameter $\Delta$, proportional to $L_3 L_4 - L_1 L_2$, with $L_i$ denoting the $\mathbb T^4$ periods. We perform our calculations via a path integral on a $\mathbb T^4$. Its size is taken much smaller than the inverse strong scale $\Lambda$ and the theory is well inside the semi-classical weak-coupling regime. The instanton background, constructed for $\Delta\ll 1$ in arXiv:hep-th/0007113, has fractional topological charge $Q=\frac{1}{2}$ and supports two gaugino zero modes, yielding a non-vanishing bilinear condensate, which we find to be $\Delta$-independent. Further, the theory has a mixed discrete chiral/$1$-form center anomaly leading to double degeneracy of the energy eigenstates on any size torus with 't Hooft twists. In particular, there are two vacua, $|0\rangle$ and $|1\rangle$, that are exchanged under chiral transformation. Using this information, the $\Delta$-independence of the condensate, and assuming further that the semi-classical theory is continuously connected to the strongly-coupled large-$\mathbb T^4$ regime, we determine the numerical coefficient of the gaugino condensate: $\langle 0| \mbox{tr}\lambda\lambda|0\rangle=|\langle 1| \mbox{tr}\lambda\lambda|1\rangle|=32\pi^2 \Lambda^3$, a result equal to twice the known $\mathbb R^4$ value. We discuss possible loopholes in the continuity approach that may lead to this discrepancy.