 Solution

The transition strength depends on $|R_{ji}{|}^2$, so a necessary condition is that at least one Cartesian electric dipole transition moment is nonzero. An integral can be nonzero only if its integrand contains the totally symmetric representation:The harmonic vibrational ground state is totally symmetric, and a one-quantum state of normal mode $k$ transforms as ${\Gamma }^{(k)}$. For the real irreducible representations here, the fundamental is therefore symmetry-allowed precisely whenEquivalently, expand the electric dipole moment near equilibrium:The fundamental has $⟨1_k|Q_k|0_k⟩\ne 0$, so vibrational infrared activity requires a nonzero dipole derivative of the matching symmetry. In $D_{6h}$, $z\sim A_{2u}$ and $(x,y)\sim E_{1u}$, giving $A_{2u}$ and $E_{1u}$ as the IR-active species. Symmetry permission does not guarantee a large band intensity.