Gaussian maximal graph equation on a compact spacelike slope cone

Work in the given Gaussian coordinate domain, with ambient indices $a,b,c,d$ from $0$ to $m$ and $x^0=z$. Define ${\mathrm{\Gamma }}^a{}_{bc}=\frac{1}{2}{g}^{ad}({\mathrm{\partial }}_b{g}_{dc}+{\mathrm{\partial }}_c{g}_{db}-{\mathrm{\partial }}_d{g}_{bc})$ and $({\mathrm{\nabla }}_{X}Y{)}^a=X^b{\mathrm{\partial }}_b{Y}^a+{\mathrm{\Gamma }}^a{}_{bc}{X}^b{Y}^c$. Smoothness follows from [F7] for the smooth invertible matrix $g$. The displayed formula satisfies the connection rules of [F5] by the product rule. Its symmetry in $b,c$ gives ${\mathrm{\nabla }}_{X}Y-{\mathrm{\nabla }}_{Y}X=[X,Y]$ by expanding the coordinate bracket. Lowering the first index gives ${\mathrm{\Gamma }}_{abc}=\frac{1}{2}({\mathrm{\partial }}_b{g}_{ac}+{\mathrm{\partial }}_c{g}_{ab}-{\mathrm{\partial }}_a{g}_{bc})$, so ${\mathrm{\Gamma }}_{cba}+{\mathrm{\Gamma }}_{abc}={\mathrm{\partial }}_b{g}_{ac}$; expanding a derivative of $g(Y,Z)$ therefore proves metric compatibility for all $X,Y,Z$. Conversely, compatibility and the symmetry in the two lower indices force this same formula by adding the compatibility equations in two index orders and subtracting the third. Hence it is the unique compatible torsion-free connection on this coordinate domain. This also makes the construction independent of a change of coordinates, since the connection rules, compatibility and torsion freedom are intrinsic. When $m=3$ it is precisely [F2].

By [F3], $x\mapsto (u(x),x)$ parametrizes a smooth embedded hypersurface; swapping the two product factors is a coordinate diffeomorphism. Its induced quadratic form is $h_{ij}(u(x),x){\xi }^i{\xi }^j-(u_i{\xi }^i{)}^2\ge W{h}_{ij}(u(x),x){\xi }^i{\xi }^j$, by [F8] for the positive-definite form $h$, so $W>0$ makes it spacelike. The conormal $\alpha =dz-u_{i}d{x}^i$ has squared norm $-W$. Its negative metric dual, normalized to unit length, is $n=-{\alpha }^{\mathrm{♯}}\mathrm{/}\sqrt{W}=({\mathrm{\partial }}_z+h^{ij}{u}_j{\mathrm{\partial }}_i)\mathrm{/}\sqrt{W}$. This vector is future-directed because $g(n,{\mathrm{\partial }}_z)=-W^{-1\mathrm{/}2}<0$ and ${\mathrm{\partial }}_z$ is future-directed. Hence the defined angle is $v=W^{-1\mathrm{/}2}$.

For the connection of step 1.1, cancellation of the first and third terms of the Christoffel formula after interchanging their summed indices gives ${\mathrm{\Gamma }}^a{}_{ab}=\frac{1}{2}{g}^{ac}{\mathrm{\partial }}_b{g}_{ac}$. Apply [F6] in each coordinate direction to $g$, which is invertible throughout the coordinate domain. Since $\det g=-\det h<0$ and $w=\sqrt{\mathrm{\mid }\det g\mathrm{\mid }}=J$, the scalar chain rule [F9] gives ${\mathrm{\partial }}_{b}w=\frac{1}{2}w{g}^{ac}{\mathrm{\partial }}_b{g}_{ac}$. Taking the trace of the coordinate formula for $\mathrm{\nabla }Y$ in step 1.1 yields ${\mathrm{div}}_{g}Y:=\mathrm{tr}(X\mapsto {\mathrm{\nabla }}_{X}Y)={\mathrm{\partial }}_a{Y}^a+{\mathrm{\Gamma }}^a{}_{ab}{Y}^b=w^{-1}{\mathrm{\partial }}_a(w{Y}^a)$. This proves the identity for every smooth $Y$ in every dimension under consideration; when $m=3$ it agrees with [F4].

Extend the normal from step 1.2 to nearby ambient points using the same spatial functions $u_i(x)$ and ambient coefficients $h^{ij}(z,x)$. Positivity of $W$ on the graph makes this a smooth unit timelike extension locally. At a graph point choose a $\gamma$-orthonormal tangent basis $e_1,\dots ,e_m$ by successive subtraction of projections and positive normalization; adjoining $n$ gives a $g$-orthonormal ambient basis. The trace formula and metric compatibility from step 1.1 give ${\mathrm{div}}_{g}n={\sum }_{i}g({\mathrm{\nabla }}_{e_i}n,e_i)-g({\mathrm{\nabla }}_{n}n,n)={\sum }_{i}g({\mathrm{\nabla }}_{e_i}n,e_i)$, since $g({\mathrm{\nabla }}_{n}n,n)=\frac{1}{2}n(g(n,n))=0$. The last sum is $-{\mathrm{tr}}_{\gamma }k=H$ by the Statement. Tangential derivatives of two extensions agreeing on the graph coincide, so the value is independent of the unit extension. Thus maximality is equivalent to ${\mathrm{div}}_{g}n=0$ in every dimension; for $m=3$ this recovers [F1].

By step 2.1, the divergence of the extension in step 2.2 is $J^{-1}\{{\mathrm{\partial }}_z(J\mathrm{/}\sqrt{W})+{\sum }_i{\mathrm{\partial }}_{x_i}(J{h}^{ij}{u}_j\mathrm{/}\sqrt{W})\}$, with spatial derivatives taken at fixed $z$. In each spatial derivative, the chain rule [F9] gives ${\mathrm{\partial }}_{x_i}{P}^i(z,x,Du(x))=P_{x_i}^i+P_{p_j}^i{u}_{ji}$; the $z$ derivative fixes $x$ and $p=Du(x)$. Evaluating at $z=u(x)$ therefore gives $J{\mathrm{div}}_{g}n=A^{ij}{u}_{ij}+b$. Since $J>0$, step 2.2 makes maximality equivalent to this expression vanishing.

Differentiate $P$ with respect to $p$ to obtain the displayed symmetric matrix. For any covector $\xi$, $A^{ij}{\xi }_i{\xi }_j\ge J{h}^{ij}{\xi }_i{\xi }_j\mathrm{/}\sqrt{W}$. On a compact background, [F10] gives $c_{h}I\le h\le C_{h}I$ and $C_h^{-1}I\le h^{-1}\le c_h^{-1}I$. Thus $v=W^{-1\mathrm{/}2}\le V$ gives $s\le 1-V^{-2}$ and $\mathrm{\mid }p{\mathrm{\mid }}^2\le C_{h}s\le C_h$. For a nonempty admissible set $V\ge 1$; choose ${\theta }_1^2=1-(2V^2{)}^{-1}$, strictly greater than $1-V^{-2}$. Over the compact background the closed jet set $s\le {\theta }_1^2$ is compact by [F12], since it is closed and $\mathrm{\mid }p{\mathrm{\mid }}^2\le C_h{\theta }_1^2$, and it contains the admissible jets in its relative interior in the slope variables. The compact background lies in the smooth coordinate domain; finitely many small closed coordinate balls inside that domain give, by [F12], a compact neighborhood of it, on which [F10] applies again. On the corresponding enlarged jet set, $W\ge (2V^2{)}^{-1}>0$. The coefficients are smooth on an open spacelike neighborhood, so [F11] bounds each of their continuous derivatives on this compact set, for every prescribed finite order. In particular the entries of $A$ have common bounds, giving a common upper quadratic-form bound by $\mathrm{\mid }{\sum }_{i,j}{A}^{ij}{\xi }_i{\xi }_j\mathrm{\mid }\le ({\max }_{i,j}\mathrm{\mid }{A}^{ij}\mathrm{\mid })({\sum }_i\mathrm{\mid }{\xi }_i\mathrm{\mid }{)}^2\le m({\max }_{i,j}\mathrm{\mid }{A}^{ij}\mathrm{\mid })\mathrm{\mid }\xi {\mathrm{\mid }}^2$. The lower bound follows from the first displayed inequality, $h^{-1}\ge C_h^{-1}I$, $W\le 1$ and the positive minimum of $J$ on the compact background supplied by [F11]. These bounds use only the compact background and the angle bound, with no initially uniform higher derivative bound for $u$.