= Solution
The initial <vibronic state> is $A_{1g}$ and the final one is $E_{1u}$, so an in-plane dipole component satisfies
$$
E_{1u}\otimes E_{1u}\otimes A_{1g}
=A_{1g}+A_{2g}+E_{2g}.
$$
The product contains the <totally symmetric representation>. \b[The combined electronic–vibrational transition is therefore symmetry-allowed], with an in-plane <electric dipole transition moment>.
A nonzero intensity still needs <vibronic coupling>. In a coordinate-independent separation, the forbidden electronic <matrix element> would remain zero. Its first derivative with respect to the $E_{2g}$ coordinate can instead be nonzero, and the vibrational factor $\langle1|Q|0\rangle$ supplies the required one-quantum transition. This is <vibrational activation of an electronic transition> and explains why the band can be weak.
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