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For a displacement group representation, an atom contributes only if the operation leaves its site fixed. Its three Cartesian displacements then contribute the trace of the polar-vector transformation.

Take $z$ perpendicular to the ring and an in-plane $x$ axis through opposite carbon atoms. To avoid confusing the principal twofold rotation with the in-plane axes, write $C_2(z)$ for the former, $C_{2a}$ for an in-plane axis through atoms and $C_{2b}$ for one through opposite bond midpoints. The carbon displacement characters in the D6h molecular point group areA through-atom half-turn fixes two carbon sites and has Cartesian trace $-1$, giving $-2$. The molecular-plane reflection fixes all six carbons with trace $+1$, giving $6$. A through-atom vertical mirror fixes two sites with trace $+1$, giving $2$.

Using molecular character reduction, the 18-dimensional carbon-only result isThe six one-dimensional terms contribute six dimensions and the six copies of two-dimensional representations contribute twelve, totaling eighteen. The distinction between carbon-only and whole-molecule displacements is needed in part (ii).

For the character of a localized orbital basis, take the trace of the transformation matrix: both permutation of centres and signs of atomic orbitals matter. The two metal centres lie on the principal axis and remain fixed under rotation about $z$. At either metal the $(p_x,p_y)$ basis has the matrixThe full four-dimensional group representation is block diagonal with two such blocks. Consequently $\chi (R)=4\cos \theta$, and for the $60^{\circ }$ rotation of the D6h molecular point group, $\chi (C_6)=4\cos (\pi /3)=2$.

The planar benzene structure has a sixfold principal axis normal to its plane, six transverse twofold axes, the molecular mirror plane and an inversion centre. Its molecular point group is $D_{6h}$, the D6h molecular point group.