Solution
= Solution
The relevant decomposition is unchanged:
$$
\operatorname{Sym}^2T_2=A_1+E+T_2.
$$
Thus \b[the same $E$ and $T_2$ vibrational symmetries are linearly Jahn-Teller active]. Their coupling constants and the resulting <energy> surface can differ from those of a $T_1$ state. The ground-geometry <methane normal-mode representation> and the symmetry-allowed harmonic invariants are also unchanged; actual force constants in the excited state need not have the same values.