Outward graph stability has the negative-Laplacian principal symbol

By the exact decomposition in [F2], $\mathrm{\Theta }(u)=-q_u^{ab}h({\nu }_u,T_r){\mathrm{\partial }}_a{\mathrm{\partial }}_{b}u+R(x,u,\mathrm{\partial }u)$, with all coefficients and $R$ smooth and real. At the zero graph the same supplier gives $h({\nu }_0,T_r)=1$ and $q_0=q$. Along $u=ϵ\psi$, the derivative of the coefficient multiplying ${\mathrm{\partial }}_a{\mathrm{\partial }}_{b}u$ contributes zero at $ϵ=0$, since the zero graph has zero second partials. The second-order part of $\mathcal{L}$ is consequently $-q^{ab}{\mathrm{\partial }}_a{\mathrm{\partial }}_b$, and the remaining terms are smooth real first- and zeroth-order terms by the supplied linearization. The intrinsic Laplacian from [F1] has second-order part $q^{ab}{\mathrm{\partial }}_a{\mathrm{\partial }}_b$, so $P=\mathcal{L}+{\mathrm{\Delta }}_q$ has order at most one. Set $c=P(1)$. In coordinates $P-c$ has the form $b^a{\mathrm{\partial }}_a$ with smooth real coefficients; coordinate independence of $P-c$ and the chain rule give the vector-field transformation law for these coefficients. They therefore define a global smooth real vector field $b$, and $\mathcal{L}=-{\mathrm{\Delta }}_q+b\cdot \mathrm{\nabla }+c=-{\mathrm{\Delta }}_q+2t\cdot \mathrm{\nabla }+c$ for $t=b\mathrm{/}2$. In the quadratic-symbol convention of [F4], the principal symbol is $p_2(x,\xi )=-q^{ab}(x){\xi }_a{\xi }_b$. For every nonzero covector $\xi$, positive definiteness of $q$ gives $q^{ab}{\xi }_a{\xi }_b>0$, so $p_2(x,\xi )<0$. Thus the principal symbol is nonzero on every nonzero covector at every point, and [F4] gives the asserted ellipticity.

The null pairings follow from $g(n,n)=-1$, $g(\nu ,\nu )=1$ and $g(n,\nu )=0$: $g(n+\nu ,n-\nu )=-2$ and $g(n+\nu ,\nu )=1$. Direct substitution yields $\nu =-k\mathrm{/}2+l\mathrm{/}2$. The supplier defines ${\zeta }_A=-g(\underset{‾}{L},{\mathrm{\nabla }}_{A}l)$; because $k=2\underset{‾}{L}$, the source convention $s_A=-g(k,{\mathrm{\nabla }}_{A}l)\mathrm{/}2$ is exactly ${\zeta }_A$. For each smooth test function the compact-surface tube supplies a common parameter interval for the graph path, and its velocity is $\psi \nu$. Thus ${\delta }_{\psi \nu }{\theta }_l=\mathcal{L}\psi$ in this same fixed outward direction. This proves all assertions, independently of any specialization of the zeroth-order curvature coefficient.