 Solution

For small displacement gradients write $F=I+H$. Taylor expansion around the stress-free reference state givesThe linear term vanishes because $DW(I)=0$. Drop the irrelevant constant and retain the quadratic term: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 $e^{tK}$ givesDifferentiation at zero yields $D\hat{T}(I)[K]=C[K]=0$. 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: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,Thus infinitesimal rotations cost no energy in the quadratic approximation.