= Solution
A defect of charge $q$ has far-field elastic energy proportional to $q^2\log(L/r_0)$. Splitting $q$ into allowed charges $q_i$ with $\sum q_i=q$ lowers $\sum q_i^2$ unless the pieces already have the smallest permitted magnitude. Polar order distinguishes $\mathbf p$ from $-\mathbf p$, so its smallest nonzero charge is $|q|=1$. A <nematic liquid crystal> identifies $\widehat{\mathbf n}\sim-\widehat{\mathbf n}$; a rotation by only $\pi$ closes a loop in its order-parameter space, permitting $q=\pm1/2$.
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