 Solution

For general odd $m_{+}$ and $m_{-}$, the same rectangular circuit givesEvery elementary two-dimensional nematic defect has $|q|=1/2$, so the minimum number is

In three dimensions the director takes values in the real projective plane, whose fundamental group is $\mathbb{Z}/2\mathbb{Z}$. The signs of half-charge line defects are no longer distinct topological classes, and two such lines can annihilate by escape into the third dimension. Hence only the parity of $N_{2D}$ remains:When one is required, it is a disclination line extending through the unbounded $z$ direction.