For general odd $m_+$ and $m_-$, the same rectangular circuit gives
$$
\Delta\theta=\frac{(m_--m_+)\pi}{2},
\qquad
q_{\rm total}=\frac{m_--m_+}{4}.
$$
Every elementary two-dimensional nematic defect has $|q|=1/2$, so the minimum number is
$$
\boxed{N_{\rm 2D}=\frac{|m_+-m_-|}{2}.}
$$

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 $N_{\rm 2D}$ remains:
$$
\boxed{
N_{\rm 3D}=
\begin{cases}
0,&(m_+-m_-)/2\text{ even},\\
1,&(m_+-m_-)/2\text{ odd}.
\end{cases}}
$$
When one is required, it is a disclination line extending through the unbounded $z$ direction.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 344 / 2 / g / Solution

A two-dimensional director winding can sometimes unwind when the director may tilt out of the plane. For a three-dimensional nematic the fundamental group of the order-parameter space is $\pi_1(\mathbb{RP}^2)=\mathbb Z/2\mathbb Z$, so two disclination lines are topologically equivalent to none.


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 Incoming links: Escape into the third dimension