Smooth Euclidean hypersurface graphs and compact localization

Smooth inverse bootstrap. On coordinate lines the one-variable rules [F2] give ${\mathrm{\partial }}_j(uv)=({\mathrm{\partial }}_{j}u)v+u{\mathrm{\partial }}_{j}v$ and ${\mathrm{\partial }}_j(1\mathrm{/}u)=-({\mathrm{\partial }}_{j}u)\mathrm{/}{u}^2$ where $u\mathrm{\ne }0$. Induction on derivative order proves that products and nonvanishing quotients of $C^r$ functions are $C^r$. For compositions, [F2] gives ${\mathrm{\partial }}_j(f\circ g)={\sum }_k({\mathrm{\partial }}_{k}f\circ g){\mathrm{\partial }}_j{g}_k$; induction using the product rule and continuity proves $C^r$ closure under composition for each finite $r$. The earlier $C^1$ inverse theorem [F1] gives a $C^1$ inverse $g$ to a smooth map $f$ with invertible derivative, with $Dg=(Df\circ g{)}^{-1}$. For an invertible finite matrix $A$, cofactor expansion gives $A^{-1}=\mathrm{adj}(A)\mathrm{/}\det A$: multiplying either side by $A$ gives the identity by Laplace expansion, including the off-diagonal expansions with two equal rows. Its entries are polynomial quotients with nonzero denominator, hence smooth. If $g$ is $C^r$ and $f$ smooth, the repeated product and chain rules make $(Df\circ g{)}^{-1}$ $C^r$, so the displayed derivative makes $g$ $C^{r+1}$. Induction from $r=1$ proves $g$ smooth. The same bootstrap applies to the implicit theorem, whose solution is a component of the inverse of $(x,z)\mapsto (x,F(x,z))$.

Graphs from slices. In a slice chart $\theta$ near $p$, let $F$ be its last coordinate. Then $S\cap V=F^{-1}(0)$ and $DF$ has rank one, because $D\theta$ is invertible by differentiating the chart inverse identities. Apply the real spectral theorem in [F2] to the self-adjoint rank-one orthogonal projection $v\mapsto (v\cdot u)u$, where $u=\mathrm{\nabla }F(p)\mathrm{/}\mathrm{\mid }\mathrm{\nabla }F(p)\mathrm{\mid }$. Its eigenvalue-one space is $\mathbb{R}u$, so, after reordering and changing one sign, its orthonormal eigenbasis has last vector $u$; in these rigid coordinates ${\mathrm{\partial }}_{n}F(p)\mathrm{\ne }0$. The implicit theorem and step 1.1 solve $F(y,z)=0$ as $z=h(y)$ on a product neighbourhood, with $h$ smooth. Projection onto $y$ is the inverse of $X(y)=(y,h(y))$; its derivative has independent columns $(e_j,h_j)$, so $X$ is a smooth parametrization of the relatively open piece.

Coordinate independence and normal derivatives. If two parametrizations overlap, their transition is smooth and has invertible derivative, so the chain rule makes their derivative images identical and makes $D(\nu \circ X)(DX{)}^{-1}$ independent of the parametrization. The positive square root is smooth because it is the inverse of $t\mapsto t^2$ on $(0,\mathrm{\infty })$, to which step 1.1 applies. For a graph, ${\nu }_h=(-\mathrm{\nabla }h,1)\mathrm{/}\sqrt{1+\mathrm{\mid }\mathrm{\nabla }h{\mathrm{\mid }}^2}$ is therefore smooth, unit, and perpendicular to every $(e_j,h_j)$. Since the normal space has dimension one, any continuous unit normal is $ϵ{\nu }_h$ with $ϵ$ continuous and valued in $\{1,-1\}$, hence constant on a sufficiently small connected piece. It is therefore smooth. Differentiating $\mathrm{\mid }\nu {\mathrm{\mid }}^2=1$ gives $d\nu (v)\cdot \nu =0$, so $d\nu (v)\in T_{p}S$. Also differentiating $\nu \cdot X_k=0$ gives $-{\mathrm{\partial }}_j\nu \cdot X_k=\nu \cdot X_{jk}$, a symmetric expression by equality of mixed partials. Thus $S_{\nu }$ is self-adjoint.

Equivalence with the Euclidean Weingarten operator. Cartesian differentiation of an ambient field along a curve is its componentwise derivative; extending a field from a graph by holding its graph coordinates fixed in the last ambient coordinate gives exactly that derivative along tangent vectors, independent of the extension because two extensions agree on every curve in $S$. Orthogonal projection therefore gives the usual Euclidean second fundamental form $(D_{v}w{)}^{\perp }=(D_{v}w\cdot \nu )\nu$. The differentiated orthogonality in step 3.1 gives $⟨S_{\nu }v,w⟩=⟨(D_{v}w{)}^{\perp },\nu ⟩$ and $S_{\nu }=-(D_v\nu {)}^{\mathrm{\top }}=-d\nu (v)$. This is precisely the shape operator for the Euclidean ambient connection. Its self-adjoint eigenvalues are the usual principal curvatures, and their product is its determinant. Replacing $\nu$ by $-\nu$ replaces $S_{\nu }$ by $-S_{\nu }$, so $K_{-\nu }=(-1{)}^{n-1}{K}_{\nu }$ and nonvanishing is independent of local orientation. Rigid motions conjugate the shape operators by their orthogonal derivative and preserve the determinant.

Compact localization. For each point of $K$, choose a graph neighbourhood and an ambient ball whose closed, slightly larger ball meets $S$ only inside that graph neighbourhood. An ambient smooth bump supported in the larger ball and equal to one on the smaller ball exists by [F5]. Compactness gives finitely many smaller balls covering $K$, with bumps ${\beta }_j$. Their restrictions have supports compact in their graph pieces: inside the larger closed ball the hypersurface is a relatively closed zero set of the defining function from step 2.1. Put $B={\sum }_j{\beta }_j$. It has positive minimum on $K$. Choose a smooth real cutoff $\eta$ that is zero for $B\le c\mathrm{/}2$ and one for $B\ge c$, where $0<c<{\min }_{K}B$. Define ${\chi }_j=\eta (B){\beta }_j\mathrm{/}B$ where $B>0$, and zero where $B=0$. These functions are smooth, nonnegative, supported compactly in their graph pieces, and sum to $\eta (B)=1$ on a neighbourhood of $K$. If $K=S$, the sum is one on $S$. If $K$ is empty, take the empty family.