 Solution

Outside the core, $f^{\prime \prime }=0$ and the metric is locally flat: set $\tilde{\varphi }=(1-\lambda )\varphi$. Static test particles have no Newtonian-like attraction because $g_{00}$ is constant and ${\Gamma }^r{}_{00}=0$. Local trajectories are straight lines in locally unwrapped coordinates, so a very thin dark core is hard to identify through ordinary orbital accelerations or local tidal forces.

The global angular range is nevertheless $2\pi (1-\lambda )$ rather than $2\pi$. The missing angle $\delta =2\pi \lambda$ can give gravitational lensing with two nearly identical, essentially undistorted images of a background galaxy, and multiple such pairs aligned along the projected object. For a small deficit, a straight string and favourable geometry, their angular separation is of order $\delta D_{ls}/D_s$, with an additional inclination factor for a tilted string. The matching spectra and common line of image pairs distinguish this from an ordinary compact lens.

A moving cosmic string can also induce a step in the cosmic microwave background temperature across its path. That signature needs motion in addition to the static geometry supplied here. These are consequences of the exterior conical geometry; interpreting the inferred deficit as a positive mass per length requires the regular-axis/core assumptions distinguished in part b.