 Solution

As a function of the full deformation gradient, use $e=\mathrm{sym}(F-I)$. Let $Q=\mathrm{diag}(1,-1,-1)$, a proper rotation. Assumed full frame indifference would give $\tilde{W}(Q)=\tilde{W}(I)=0$. For $F_s=I+s(Q-I)$, its infinitesimal strain is $s\mathrm{sym}(Q-I)$. Quadratic homogeneity therefore givesChoose $s=1/4$. ThenThe homogeneous deformation $y(X)=F_{s}X$ changes volume but has zero energy. This illustrates why a nondegenerate small-strain quadratic energy cannot also be imposed as a fully frame-indifferent finite-deformation energy.