Pointwise Gauss–Codazzi identities for the normal angle

Along $\mathrm{\Sigma }$, $g(P(W),n)=0$ and $P$ is the identity on tangent vectors. The tangential derivative $D_{X}Y$ is independent of the ambient extension of $Y$: in an adapted chart the components of a field vanishing along $\mathrm{\Sigma }$ have zero tangential derivatives there, and its connection term also vanishes. Thus ${\mathrm{\nabla }}_{X}Y=P(D_{X}Y)$ is well defined and smooth. The affine-connection rules follow by applying $P$ to those of $D$. For tangent $X,Y,Z$, metric compatibility gives $Xh(Y,Z)=h({\mathrm{\nabla }}_{X}Y,Z)+h(Y,{\mathrm{\nabla }}_{X}Z)$, and torsion freedom gives ${\mathrm{\nabla }}_{X}Y-{\mathrm{\nabla }}_{Y}X=P([X,Y])=[X,Y]$. To see uniqueness, add the compatibility identities for $Xh(Y,Z)$ and $Yh(Z,X)$, subtract the one for $Zh(X,Y)$, and substitute torsion freedom; this gives $2h({\mathrm{\nabla }}_{X}Y,Z)=Xh(Y,Z)+Yh(Z,X)-Zh(X,Y)+h([X,Y],Z)+h([Z,X],Y)-h([Y,Z],X)$ for any such connection. Nondegeneracy of $h$ determines ${\mathrm{\nabla }}_{X}Y$ uniquely. Finally $g(D_{X}Y,n)=-g(Y,D_{X}n)=k(X,Y)$, so $D_{X}Y={\mathrm{\nabla }}_{X}Y-k(X,Y)n$. This constructs the induced connection and its Gauss decomposition without a positive-definite ambient hypothesis.

Fix a point $p$ and a boundaryless intrinsic neighborhood of it. Apply [F3] to its Riemannian metric $h$ and the connection identified in step 1.1. A finite basis of $T_p\mathrm{\Sigma }$ can be orthonormalized by successive subtraction of projections and division by positive lengths; this uses only finite choices. The associated normal coordinate fields $E_i$ satisfy $h(E_i,E_j)(p)={\delta }_{ij}$, $\mathrm{\nabla }{E}_i(p)=0$ and $d(h(E_i,E_j))(p)=0$. Apply the same Gram–Schmidt formulas smoothly to $E_i$ near $p$. Their denominators remain positive after shrinking the neighborhood. At $p$ every normalization factor is one and every off-diagonal projection coefficient is zero; differentiating these formulas, the vanishing first metric derivatives make all coefficient derivatives zero. Thus the resulting local $h$-orthonormal frame $e_a$ agrees with $E_a$ to first order at $p$ and satisfies ${\mathrm{\nabla }}_{e_a}{e}_b(p)=0$. Only the inherited ${\mathrm{A}\mathrm{C}}_{\omega }$ of the normal-coordinate and ambient geometric inputs is used, as supplied by ZFC.

At the chosen point metric compatibility gives $D_{e_a}n=-k_{ab}{e}_b$ and step 1.1 gives $D_{e_a}{e}_b=-k_{ab}n$ because its tangential covariant derivative vanishes by step 2.1. Differentiate $v=-g(n,U)$ and $Z_b=g(U,e_b)$ to obtain ${\mathrm{\nabla }}_{a}v=k_{ab}{Z}_b-g(n,D_{e_a}U)$ and ${\mathrm{\nabla }}_a{Z}_b=g(D_{e_a}U,e_b)+v{k}_{ab}$. The components $Z_b$ are those of the tangential covector $Z(X)=g(U,X)$. With its covariant derivative as defined in [F1], these are tensorial first-derivative formulas and hold in every induced frame.

Commuting the two ambient derivatives of $n$ and pairing with $e_c$ gives $({\mathrm{\nabla }}_{a}k{)}_{bc}-({\mathrm{\nabla }}_{b}k{)}_{ac}=-g(R(e_a,e_b)n,e_c)$. Indeed, differentiate $k(e_b,e_c)=-g(D_{e_b}n,e_c)$, subtract the expression with $a,b$ interchanged and the induced connection corrections, and use the Gauss decomposition from step 1.1; the normal components of $D{e}_c$ pair to zero with the tangent vector $Dn$. Torsion freedom accounts for the bracket term in [F2]. Contract $a=c$. Symmetry of $k$ and ${\mathrm{tr}}_{h}k=0$ make the left side ${\mathrm{\nabla }}^a{k}_{ab}$. In the orthonormal ambient frame $(n,e_1,e_2,e_3)$ the remaining normal term in the Ricci trace is $-g(R(n,e_b)n,n)=0$ by skew-adjointness. Therefore ${\mathrm{\nabla }}^a{k}_{ab}=-\mathrm{Ric}(n,e_b)$.

Differentiate the first angle identity in step 3.1 once more and trace in the normal frame of step 2.1. Differentiating $k_{ab}{Z}_b$ gives $({\mathrm{\nabla }}^a{k}_{ab})Z_b+k_{ab}g(D_{e_a}U,e_b)+v\mathrm{\mid }k{\mathrm{\mid }}^2$. Differentiating $-g(n,D_{e_a}U)$ gives a second copy of $k_{ab}g(e_b,D_{e_a}U)$ and $-g(n,{\sum }_a{D}_{e_a}{D}_{e_a}U)$. By the definition of $D^{2}U$ in the Statement and $D_{e_a}{e}_a=-k_{aa}n$, the difference between this last trace and $-g(n,{\sum }_a(D^{2}U)(e_a,e_a))$ is $({\mathrm{tr}}_{h}k)g(n,D_{n}U)=0$. The traced scalar Hessian is $\mathrm{\Delta }v$ because the induced frame is normal at the point. This proves exactly the stated Laplacian identity; the contracted term is identified in step 4.1. The point was arbitrary, and neither compactness nor division by $v^2-1$ was used.