Analytic flattening and the normal principal coefficient

After relabeling coordinates take ${\mathrm{\partial }}_{z_{d+1}}\varphi (a)\mathrm{\ne }0$. The map $\mathrm{\Psi }(z)=(z_1-a_1,\dots ,z_d-a_d,\varphi (z))$ has determinant ${\mathrm{\partial }}_{z_{d+1}}\varphi (a)$. F1 supplies an analytic inverse $z=\mathrm{\Phi }(x,t)$, and the initial surface is exactly t=0.

For any derivative $D_z^{\alpha }u$ of order m, repeated chain rule shows the coefficient of ${\mathrm{\partial }}_t^m(u\circ \mathrm{\Phi })$ is $(d\varphi {)}^{\alpha }$: obtaining m derivatives on u requires that each differentiation hit u, while differentiation of a coordinate coefficient leaves lower order on u. All transformed jet expressions are analytic and linear in the top-order jets. Differentiating the transformed nonlinear P in the pure t-jet therefore gives ${\sum }_{\mathrm{\mid }\alpha \mathrm{\mid }=m}{P}_{u_{\alpha }}(d\varphi {)}^{\alpha }$, the principal symbol of its linearization. This agrees with F2 applied to that linearized operator.

At the specified compatible jet P=0, and step 2.1 makes its derivative in the selected scalar slot nonzero. F1 gives a unique analytic branch expressing this slot in terms of the remaining jets and (x,t). The remaining derivatives all have total order at most m and normal order strictly less than m. This proves the claimed solved form only near the selected compatible jet.

For normal data, step 1.1 gives an analytic parametrization X of the surface. The nonvanishing analytic gradient has an analytic positive length: apply F1 to $s^2-\mathrm{\mid }\mathrm{\nabla }\varphi {\mathrm{\mid }}^2=0$ at its positive root. Thus $\nu =\mathrm{\nabla }\varphi \mathrm{/}\mathrm{\mid }\mathrm{\nabla }\varphi \mathrm{\mid }$ is analytic. The map $\mathrm{\Theta }(x,t)=X(x)+t\nu (x)$ has independent tangent columns and its unit normal column, hence invertible derivative, so F1 again gives an analytic inverse. By F3 and the fact that Theta is affine in t, ${\mathrm{\partial }}_t^j(u\circ \mathrm{\Theta })(x,0)=D^{j}u(X(x))[\nu (x),\dots ,\nu (x)]$. The conormal dt in these coordinates is proportional to dphi on the surface; homogeneity of the degree-m symbol preserves its nonvanishing. Thus this flattening handles exactly the stated Euclidean normal data.