 Solution

The independent head-tail reversal symmetry means that the squared dot product interaction energy is unchanged when any individual ${\mathbf{s}}_i$ is replaced by $-{\mathbf{s}}_i$. All configurations ${\mathbf{s}}_i={\sigma }_i\mathbf{n}$, with ${\sigma }_i=\pm 1$ independently, attain the aligned ground-state pair energy. There is consequently no energetic preference for either end of a molecular axis. Averaging over these equally weighted signs gives $⟨{\mathbf{s}}_i⟩=0$ and $⟨\mathbf{m}⟩=0$ even in the nematically ordered phase. The surviving order is the unoriented axis, represented by a nematic director with $\mathbf{n}\sim -\mathbf{n}$. A second-moment tensor can detect this order while a polar vector cannot.^[ref]

The word “always” needs the statistical ensemble or apolar interpretation: it is not an identity for each finite configuration. For example, choosing all ${\sigma }_i=+1$ is also a ground-state representative and gives $\mathbf{m}=\mathbf{n}$. In an unbiased sign statistical ensemble the typical residual polar vector has size of order $N^{-1/2}$, because $⟨|\mathbf{m}{|}^2⟩=1/N$ for these independent ground-state signs. This does not make polar order an equilibrium nematic order parameter. The tensor below is unchanged by each sign flip and thus removes the arbitrary labeling of rod ends.