= Solution
Let $F$ be an invertible, orientation-preserving <deformation gradient>, with $J=\det F>0$. To prove <Nanson's formula>, take two reference surface <tangent vectors> $a,b$. Their current images are $Fa,Fb$, and their oriented area vectors are proportional to $a\times b$ and $(Fa)\times(Fb)$. For every vector $c$, the scalar triple product gives
$$
(Fc)\cdot[(Fa)\times(Fb)]=Jc\cdot(a\times b).
$$
Thus $F^T[(Fa)\times(Fb)]=J(a\times b)$, and
$$
\boxed{dS=JF^{-T}dS_0}.
$$
This proof also identifies the area transformation as the cofactor of $F$, rather than a transformation by $F$ itself.
Use the paper's material-index-first <nominal stress tensor> $P_{Ii}$. Equality of force on the same material surface in its two configurations gives $P_{Ii}\,dS_{0I}=\sigma_{ji}\,dS_j$. Substitute the area transformation to obtain
$$
\boxed{P_{Ii}=J(F^{-1})_{Ij}\sigma_{ji}},\qquad \boxed{\tau_{ji}=J\sigma_{ji}=F_{jI}P_{Ii}}.
$$
Here $\tau$ is the <Kirchhoff stress tensor>. With symmetric <Cauchy stress tensor>, the matrix forms are $P=JF^{-1}\sigma$ and $\tau=FP$. This $P$ is the transpose of the <first Piola-Kirchhoff stress tensor>; keeping the index convention prevents a spurious transpose later.
For the <velocity gradient> $L=\dot FF^{-1}$, let $D=(L+L^T)/2$ be the <rate-of-strain tensor>. Then the reference-volume mechanical power is
$$
P_{Ii}\dot F_{iI}=\tau_{ji}(F^{-1})_{Ij}\dot F_{iI}=\tau_{ji}L_{ij}=\boxed{\tau_{ji}D_{ij}}.
$$
The last step uses stress symmetry, as supplied by angular-momentum balance in an ordinary continuum without couple stresses: a symmetric tensor has zero contraction with the skew part of $L$.
By <work-conjugate stress and strain>, a pair $(T,E)$ must satisfy $T:\dot E=\tau:D$ for every deformation rate, with both sides measured per reference volume. For the <Green-Lagrange strain tensor>, set $C=F^TF$ and calculate
$$
\dot E^{(2)}=\tfrac12(\dot F^TF+F^T\dot F)=F^TDF.
$$
Cyclically rearranging the contraction shows $T^{(2)}:\dot E^{(2)}=(FT^{(2)}F^T):D$. Hence the symmetric conjugate is
$$
\boxed{T^{(2)}=F^{-1}\tau F^{-T}},
$$
the <second Piola-Kirchhoff stress tensor>.
The <inverse right Cauchy-Green strain> is $E^{(-2)}=(I-C^{-1})/2$, because $C^{-1}=F^{-1}F^{-T}$. Differentiating $CC^{-1}=I$ gives
$$
\dot E^{(-2)}=\tfrac12C^{-1}\dot CC^{-1}=F^{-1}DF^{-T}.
$$
Consequently $T^{(-2)}:\dot E^{(-2)}=(F^{-T}T^{(-2)}F^{-1}):D$, and
$$
\boxed{T^{(-2)}=F^T\tau F}.
$$
The two stress measures have a geometric interpretation in the convected coordinate net. Its basis vectors are $g_I=Fe_I$ and its reciprocal basis is $g^I=F^{-T}e_I$. The entries of $T^{(2)}$ are the contravariant components in $\tau=T^{(2)}_{IJ}g_I\otimes g_J$, whereas $T^{(-2)}_{IJ}=g_I\cdot\tau g_J$ are the covariant components, equivalently $\tau=T^{(-2)}_{IJ}g^I\otimes g^J$. They describe the same spatial <Kirchhoff stress tensor> in a deforming, generally nonorthonormal coordinate net.
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