= Solution
The overall vibrational <ground state> is invariant under every <molecular symmetry operation>, so it transforms as $A_{1g}$. Combining it with the electronic <ground state> gives
$$
\boxed{\Gamma_{\rm initial}=A_{1g}\otimes A_{1g}=A_{1g}.}
$$
For a degenerate harmonic <normal mode>, the zero-quantum <wavefunction> depends on the invariant sum $Q_x^2+Q_y^2$, so the ground vibrational symmetry remains totally symmetric even when excited vibrational states form multiplets.
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