Lemma: What XIV-I's display presupposes

Let $a_2$ be a smooth map from symmetric tensors to symmetric tensors with $a_2[0]=0$ and ${\dot{a}}_2[G;h]:=\frac{d}{ds}{a}_2[G+sh]{|}_{s=0}$. Then ${\dot{a}}_2[G;h]$ is bilinear in $(G,h)$—as the display (eq:Dfrom) requires for a constant $\mathsf{D}$—if and only if $a_2$ is a homogeneous quadratic polynomial; in that case ${\dot{a}}_2[G;h]$ is symmetric in $(G,h)$, ${\dot{a}}_2[r;r]=2a_2[r]$, and $\mathsf{D}$ is twice the polarization of $a_2$, symmetric in its index pairs $(I,J)$.

Proof. ${\dot{a}}_2[G;h]=D{a}_2(G)h$ is linear in $G$ iff $D{a}_2$ is linear iff $a_2$ is a polynomial of degree at most two with $D{a}_2(0)=0$; with $a_2[0]=0$ this is a quadratic form. The rest is the polarization identity.