= Solution
For small <displacement> <gradients> write $F=I+H$. <Taylor expansion> around the stress-free reference state gives
$$
W(I+H)=W(I)+\frac12C_{i\alpha j\beta}H_{i\alpha}H_{j\beta}+o(|H|^2),
\qquad C_{i\alpha j\beta}=D^2W(I)_{i\alpha j\beta}.
$$
The linear term vanishes because $DW(I)=0$. Drop the irrelevant constant and retain the quadratic term:
$$
\boxed{W^{\mathrm{lin}}(H)=\frac12C_{i\alpha j\beta}H_{i\alpha}H_{j\beta}.}
$$
The major symmetry $\boxed{C_{i\alpha j\beta}=C_{j\beta i\alpha}}$ follows from equality of mixed second <derivatives>. For a skew matrix $K$, stress covariance along $e^{tK}$ gives
$$
\widehat T(e^{tK})=e^{tK}\widehat T(I)=0.
$$
Differentiation at zero yields $D\widehat T(I)[K]=C[K]=0$. Thus $C_{i\alpha j\beta}K_{j\beta}=0$ for every skew matrix, which forces the input minor symmetry $C_{i\alpha j\beta}=C_{i\alpha\beta j}$. Combining it with the major symmetry gives the output minor symmetry as well:
$$
\boxed{C_{i\alpha j\beta}=C_{\alpha i j\beta}=C_{i\alpha\beta j}.}
$$
These identify reference and spatial indices at $F=I$. Decompose $H=e+K$ into symmetric and skew parts. Since $C[K]=0$ and the major symmetry also eliminates the other cross term,
$$
\boxed{W^{\mathrm{lin}}(H)=\tfrac12C[e,e],\qquad e=(H+H^T)/2.}
$$
Thus infinitesimal rotations cost no energy in the quadratic approximation.
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