Example. An antiperiodic four-cycle has no parallel line section [fdip-006H]

Give the unit four-cycle total holonomy \(u=-1\), or \(\theta =\pi \). In the tree gauge, three links are one and the final link is minus one. The cycle modes have momenta \(-\pi /4,\pi /4,3\pi /4,5\pi /4\), so \[\operatorname {spec}L_\pi =\{2-\sqrt 2,2-\sqrt 2,2+\sqrt 2,2+\sqrt 2\}.\] All 4 eigenvalues are positive, and \(\det L_\pi =((2-\sqrt 2)(2+\sqrt 2))^2=4\).

The arrows show the chosen orientation; labels are connection values, not edge weights, which are all one. The thick edge carries the remaining holonomy after the other three form a tree. A parallel section would have \(f_0=f_1=f_2=f_3\) along those tree edges and \(f_0=-f_3\) along the last edge, forcing zero. Thus the massless complex Gaussian exists without a pin. With trivial holonomy the constant line is a kernel instead. This is a change of background connection, distinct from conditioning a scalar Gaussian on pinned values.