Riemann Curvature and Riemannian Submanifolds — Examples

In Cartesian coordinates, $(g_{ij})=({\delta }_{ij})$ is a constant smooth symmetric matrix, and $v^T({\delta }_{ij})v={\sum }_i(v^i{)}^2>0$ for every nonzero $v$; hence [F1] makes $g_{\mathrm{E}}$ a Riemannian metric.

Every derivative ${\mathrm{\partial }}_i{g}_{j\mathrm{\ell }}={\mathrm{\partial }}_i{\delta }_{j\mathrm{\ell }}$ is zero, so [F2] gives ${\mathrm{\Gamma }}^k{}_{ij}=0$ identically for all indices; consequently every derivative ${\mathrm{\partial }}_i{\mathrm{\Gamma }}^{\mathrm{\ell }}{}_{jk}$ is also zero.

Substitution of step 2.1 into [F3] makes both derivative terms and both quadratic terms zero, so $R^{\mathrm{\ell }}{}_{kij}=0$ for every $i,j,k,\mathrm{\ell }$. Because the Cartesian coordinate vectors form a basis at every point, this is exactly $R^{g_{\mathrm{E}}}=0$ on all of ${\mathbb{R}}^n$.

For $n=0$, every index range is empty and the unique curvature field on the one-point manifold ${\mathbb{R}}^0$ is zero; for $n=1$, antisymmetry is not needed because steps 1.1–3.1 still give the sole coordinate component zero. The standard metric is nondegenerate by step 1.1, the global chart has neither a boundary nor a parameter endpoint, and all coordinates and tensors are explicit, so no choice principle is used. The claim is an equality, not a biconditional.

Put $Q(y)=\mathrm{\mid }y{\mathrm{\mid }}^2$. On $Q^{-1}(r^2)$, one has $d{Q}_y(w)=2⟨y,w⟩$ and $d{Q}_y(y)=2r^2\mathrm{\ne }0$, so $r^2$ is a regular value; the level is nonempty because $(r,0,\dots ,0)$ lies in it. By [F2], $S_r^n$ is an embedded hypersurface and $T_y{S}_r^n=\ker d{Q}_y=y^{\perp }$.

The field $\nu (y)=y\mathrm{/}r$ has unit length and is normal by step 1.1. In Cartesian coordinates the metric coefficients are ${\delta }_{ab}$, so [F3] gives ${\bar{\mathrm{\nabla }}}_{{\mathrm{\partial }}_a}{\mathrm{\partial }}_b=0$; applying the connection laws to $\nu ={\sum }_a(y^a\mathrm{/}r){\mathrm{\partial }}_a$ gives ${\bar{\mathrm{\nabla }}}_u\nu =u\mathrm{/}r$ for every tangent $u$. This derivative is tangent by step 1.1, and [F4] therefore gives $S_{\nu }u=-u\mathrm{/}r$. Since the normal space is spanned by $\nu$, [F4] then gives $\mathrm{I}\mathrm{I}(u,v)=-(1\mathrm{/}r)g(u,v)\nu$.

Let $(u,v)$ be any supplied basis of $\sigma$. Substitute $X=u$, $Y=Z=v$, $W=u$ and the formula from step 2.1 into [F5]. The ambient term is zero by [F6], while the two quadratic terms give ${\mathrm{Rm}}^{S_r^n}(u,v,v,u)=(1\mathrm{/}{r}^2)(g(u,u)g(v,v)-g(u,v{)}^2)$. The denominator is positive by [F7], so division yields $K(\sigma )=1\mathrm{/}{r}^2$, independently of $x$ and $\sigma$.

The explicit point in step 1.1 proves that every $S_r^n$ here is nonempty. When $n=0$ or $n=1$ there is no tangent two-plane, so the curvature function has empty domain rather than a numerical exception; for $n\ge 2$, [F7] excludes a degenerate Gram denominator. The required endpoint condition is $r>0$: at $r=0$ the level is not regular and $1\mathrm{/}{r}^2$ is undefined. Countable choice [F1] is assumed because [F4]–[F5] inherit it from their smooth orthogonal projection construction and [F7] inherits it through the Riemann-tensor symmetries; the point, normal, and the basis used above are explicit or supplied, so the calculation makes no additional choice. The result is a direct equality and asserts no biconditional.

Write $h=x^n>0$. The metric and inverse matrices are $g_{ij}=r^2{h}^{-2}{\delta }_{ij}$ and $g^{ij}=r^{-2}{h}^2{\delta }^{ij}$. The first is smooth and symmetric, and $g_{ij}{v}^i{v}^j=r^2{h}^{-2}{\sum }_i(v^i{)}^2>0$ for $v\mathrm{\ne }0$, so [F1] makes $g$ Riemannian.

Put ${ϵ}_i={\delta }_{in}$ and ${ϵ}^k={\delta }_n^k$. Since ${\mathrm{\partial }}_p{g}_{ij}=-2r^2{h}^{-3}{ϵ}_p{\delta }_{ij}$, substitution in [F2] gives ${\mathrm{\Gamma }}^k{}_{ij}=-h^{-1}({ϵ}_i{\delta }_j^k+{ϵ}_j{\delta }_i^k-{\delta }_{ij}{ϵ}^k)$.

Define $A^{\mathrm{\ell }}{}_{jk}={ϵ}_j{\delta }_k^{\mathrm{\ell }}+{ϵ}_k{\delta }_j^{\mathrm{\ell }}-{\delta }_{jk}{ϵ}^{\mathrm{\ell }}$, so step 2.1 says ${\mathrm{\Gamma }}^{\mathrm{\ell }}{}_{jk}=-h^{-1}{A}^{\mathrm{\ell }}{}_{jk}$ and ${\mathrm{\partial }}_i{\mathrm{\Gamma }}^{\mathrm{\ell }}{}_{jk}=h^{-2}{ϵ}_i{A}^{\mathrm{\ell }}{}_{jk}$. The derivative difference in [F3] expands to $h^{-2}({ϵ}_i{ϵ}_k{\delta }_j^{\mathrm{\ell }}-{ϵ}_i{\delta }_{jk}{ϵ}^{\mathrm{\ell }}-{ϵ}_j{ϵ}_k{\delta }_i^{\mathrm{\ell }}+{ϵ}_j{\delta }_{ik}{ϵ}^{\mathrm{\ell }})$, while the two contracted quadratic terms expand to $h^{-2}({\delta }_i^{\mathrm{\ell }}{ϵ}_j{ϵ}_k-{\delta }_j^{\mathrm{\ell }}{ϵ}_i{ϵ}_k-{\delta }_{ik}{ϵ}_j{ϵ}^{\mathrm{\ell }}+{\delta }_{jk}{ϵ}_i{ϵ}^{\mathrm{\ell }}-{\delta }_{jk}{\delta }_i^{\mathrm{\ell }}+{\delta }_{ik}{\delta }_j^{\mathrm{\ell }})$. The first four terms cancel pairwise, leaving $R^{\mathrm{\ell }}{}_{kij}=h^{-2}({\delta }_{ik}{\delta }_j^{\mathrm{\ell }}-{\delta }_{jk}{\delta }_i^{\mathrm{\ell }})$.

Since $g_{ik}=r^2{h}^{-2}{\delta }_{ik}$, step 3.1 is $R({\mathrm{\partial }}_i,{\mathrm{\partial }}_j){\mathrm{\partial }}_k=-(1\mathrm{/}{r}^2)(g_{jk}{\mathrm{\partial }}_i-g_{ik}{\mathrm{\partial }}_j)$. Tensoriality [F4] yields $R(X,Y)Z=-(1\mathrm{/}{r}^2)(g(Y,Z)X-g(X,Z)Y)$ for arbitrary tangent vectors. Pairing with $X=u$ after setting $Y=Z=v$, [F5] gives $\mathrm{Rm}(u,v,v,u)=-(1\mathrm{/}{r}^2)(g(u,u)g(v,v)-g(u,v{)}^2)$. For a basis $(u,v)$ of any tangent two-plane, the Gram determinant is positive, so [F5] yields $K=-1\mathrm{/}{r}^2$ at every point and on every plane.

The half-space is nonempty, for example at $(0,\dots ,0,1)$, and is open and boundaryless. Dimension zero is inapplicable because the defining coordinate $x^n$ requires $n\ge 1$; when $n=1$, step 3.1 gives zero curvature as it must, but there is no tangent two-plane, so the constant-sectional-curvature predicate is vacuous. For $n\ge 2$, [F5] excludes degenerate Gram denominators. The conditions $h>0$ and $r>0$ exclude the singular height endpoint and the degenerate scale $r=0$; no limiting assertion is made. Every coordinate, tensor, point, and plane basis is explicit or supplied, so no further family choice is made beyond the stated inherited assumption. The result assigns one value to every plane and states no biconditional.

Use [F2] to define, at $(p,q)$, $k_{(p,q)}((u_1,u_2),(v_1,v_2)):=g_p(u_1,v_1)+h_q(u_2,v_2)\mathrm{.}$ In the product chart supplied by [F1], its matrix is $\mathrm{diag}((g_{ij}(x)),(h_{\alpha \beta }(y)))$. Its entries are smooth and it is symmetric; moreover $k((u_1,u_2),(u_1,u_2))=g(u_1,u_1)+h(u_2,u_2)>0$ unless both components vanish. Thus [F3] proves that $k={\pi }_M^{\ast }g+{\pi }_N^{\ast }h$ is a Riemannian metric. Its inverse matrix is $\mathrm{diag}((g^{ij}(x)),(h^{\alpha \beta }(y)))$; all mixed entries vanish, the first block has no $y$-dependence, and the second has no $x$-dependence.

Applying [F4] to the blocks in step 1.1 gives ${\mathrm{\Gamma }}^k{}_{ij}={\mathrm{\Gamma }}^k{}_{ij}(g)$ and ${\mathrm{\Gamma }}^{\gamma }{}_{\alpha \beta }={\mathrm{\Gamma }}^{\gamma }{}_{\alpha \beta }(h)$, while every symbol whose indices meet both blocks is zero. For instance, ${\mathrm{\Gamma }}^k{}_{i\beta }=\frac{1}{2}{g}^{k\mathrm{\ell }}({\mathrm{\partial }}_i{G}_{\beta \mathrm{\ell }}+{\mathrm{\partial }}_{\beta }{g}_{i\mathrm{\ell }}-{\mathrm{\partial }}_{\mathrm{\ell }}{G}_{i\beta })=0$ and ${\mathrm{\Gamma }}^k{}_{\alpha \beta }=-\frac{1}{2}{g}^{k\mathrm{\ell }}{\mathrm{\partial }}_{\mathrm{\ell }}{h}_{\alpha \beta }=0$; exchanging the factors covers an upper $N$-index.

A factor lift has coefficients depending only on that factor. Expanding covariant derivatives with the connection laws in [F4] and the symbols from step 2.1 gives ${\mathrm{\nabla }}_{\tilde{X}}\tilde{Y}=\widetilde{{\mathrm{\nabla }}_X^{M}Y}$ and ${\mathrm{\nabla }}_{\tilde{U}}\tilde{V}=\widetilde{{\mathrm{\nabla }}_U^{N}V}$. In a cross derivative, the differentiated lift's coefficients are constant in the differentiating factor and all relevant mixed symbols vanish, so ${\mathrm{\nabla }}_{\tilde{X}}\tilde{V}={\mathrm{\nabla }}_{\tilde{V}}\tilde{X}=0$.

In [F5], if the output and all three lower indices lie in the $M$ block, step 2.1 reproduces exactly the coordinate formula for $R^M$ because the symbols and their $x$-derivatives agree; all-$N$ indices similarly reproduce $R^N$. If the indices meet both blocks, each derivative term is either the derivative of a zero mixed symbol or a cross derivative of a factor-only symbol, and every quadratic term contains a zero mixed symbol. Hence every mixed curvature component is zero. Tensoriality and [F2] now give $R^{M\times N}((u_1,u_2),(v_1,v_2))(w_1,w_2)=(R^M(u_1,v_1)w_1,R^N(u_2,v_2)w_2)$.

For $a=(u,0)$ and $b=(0,v)$, step 3.2 gives $R^{M\times N}(a,b)b=0$, so [F6] makes the sectional-curvature numerator zero. By step 1.1, $a$ and $b$ are orthogonal with squared norms $g(u,u)>0$ and $h(v,v)>0$, so their Gram determinant is the positive product $g(u,u)h(v,v)$; therefore their plane has sectional curvature zero.

If a factor is empty, the product and every assertion about its points or mixed planes are vacuous. A zero-dimensional factor contributes an empty coordinate block and no nonzero vector for a mixed plane; one-dimensional factors are fully covered by the same formulas. Positive definiteness in step 1.1 excludes degenerate product metrics and step 4.1 checks the only denominator. There is no interval, scale endpoint, or manifold-boundary claim in this example. All charts and vectors are supplied locally and the Levi–Civita connection is unique, so no further family choice is made beyond the stated inherited assumption. No biconditional is asserted.

Differentiation gives $X_u=(r^{\mathrm{\prime }}\cos v,r^{\mathrm{\prime }}\sin v,z^{\mathrm{\prime }})$ and $X_v=(-r\sin v,r\cos v,0)$. Their Euclidean inner products are $⟨X_u,X_u⟩=r^{\mathrm{\prime }2}+z^{\mathrm{\prime }2}=1$, $⟨X_u,X_v⟩=0$, and $⟨X_v,X_v⟩=r^2$. Thus $dX$ is injective because $r>0$, so [F1] gives the induced Riemannian metric $g=d{u}^2+r(u{)}^{2}d{v}^2$, with matrix $\mathrm{diag}(1,r^2)$ and inverse $\mathrm{diag}(1,r^{-2})$.

Write $x^1=u,x^2=v$. The only nonconstant metric entry is $g_{22}=r^2$, with ${\mathrm{\partial }}_1{g}_{22}=2r{r}^{\mathrm{\prime }}$ and ${\mathrm{\partial }}_2{g}_{22}=0$. Substitution in [F2] gives ${\mathrm{\Gamma }}^1{}_{22}=-r{r}^{\mathrm{\prime }}$ and ${\mathrm{\Gamma }}^2{}_{12}={\mathrm{\Gamma }}^2{}_{21}=r^{\mathrm{\prime }}\mathrm{/}r$; every other ${\mathrm{\Gamma }}^k{}_{ij}$ is zero.

In [F3], the component needed for the coordinate two-plane is $R^1{}_{212}={\mathrm{\partial }}_1{\mathrm{\Gamma }}^1{}_{22}-{\mathrm{\partial }}_2{\mathrm{\Gamma }}^1{}_{12}+{\mathrm{\Gamma }}^m{}_{22}{\mathrm{\Gamma }}^1{}_{1m}-{\mathrm{\Gamma }}^m{}_{12}{\mathrm{\Gamma }}^1{}_{2m}$. By step 2.1 the four terms are $-(r^{\mathrm{\prime }2}+r{r}^{\mathrm{\prime }\mathrm{\prime }})$, $0$, $0$, and $+r^{\mathrm{\prime }2}$, respectively; hence $R^1{}_{212}=-r{r}^{\mathrm{\prime }\mathrm{\prime }}$.

By [F4] and $g_{11}=1,g_{12}=0$, $\mathrm{Rm}({\mathrm{\partial }}_u,{\mathrm{\partial }}_v,{\mathrm{\partial }}_v,{\mathrm{\partial }}_u)=g(R({\mathrm{\partial }}_u,{\mathrm{\partial }}_v){\mathrm{\partial }}_v,{\mathrm{\partial }}_u)=R^1{}_{212}=-r{r}^{\mathrm{\prime }\mathrm{\prime }}$. The Gram determinant of $({\mathrm{\partial }}_u,{\mathrm{\partial }}_v)$ is $g_{11}{g}_{22}-g_{12}^2=r^2>0$, so the unique tangent two-plane has $K=(-r{r}^{\mathrm{\prime }\mathrm{\prime }})\mathrm{/}{r}^2=-r^{\mathrm{\prime }\mathrm{\prime }}\mathrm{/}r$.

If either parameter interval is empty, there are no points and the claim is vacuous; otherwise the chart is intrinsically two-dimensional, so zero- and one-dimensional curvature cases are inapplicable. The hypothesis $r>0$ makes the metric and Gram determinant nondegenerate; at $r=0$ this parametrization loses its angular direction, so the formula asserts neither a value nor a limit there. The intervals are open, so no parameter endpoint or manifold-boundary value is claimed. All functions, coordinates, and tangent vectors are explicitly supplied, and the computation makes no family selection, so no further family choice is made beyond the stated inherited assumption. The claim is an equality, not a biconditional.

For $x\in S_r^n$, $\parallel \nu (x)\parallel =\parallel x\parallel \mathrm{/}r=1$. If $u\in T_x{S}_r^n$ is represented by a smooth curve $c$ in the sphere with $c(0)=x$ and $c^{\mathrm{\prime }}(0)=u$, differentiating $\parallel c(t){\parallel }^2=r^2$ gives $2⟨x,u⟩=0$. Hence $\nu (x)$ is normal to the hypersphere; because it points in the radial direction away from the origin, it is the smooth outward unit normal.

In Cartesian coordinates the Euclidean metric coefficients are the constant matrix $({\delta }_{ab})$, so [F4] gives ${\bar{\mathrm{\Gamma }}}^a{}_{bc}=0$. Writing $\nu =(x^a\mathrm{/}r){\mathrm{\partial }}_a$, the connection Leibniz rule therefore yields ${\bar{\mathrm{\nabla }}}_u\nu =u(x^a\mathrm{/}r){\mathrm{\partial }}_a=u\mathrm{/}r$. This vector is tangent because it is a scalar multiple of $u$, and [F2] gives $S_{\nu }u=-u\mathrm{/}r$. Thus $S_{\nu }=-(1\mathrm{/}r)\mathrm{id}$ on every tangent space.

Every nonzero tangent vector is therefore an eigenvector with eigenvalue $-1\mathrm{/}r$, and the identity map on the $n$-dimensional tangent space has that eigenvalue with algebraic multiplicity $n$. By [F3] these are exactly all principal curvatures. Normal-linearity in [F2] gives $S_{-\nu }=-S_{\nu }=(1\mathrm{/}r)\mathrm{id}$, so [F3] gives $+1\mathrm{/}r$ for all principal curvatures with the inward normal.

The sphere is nonempty for every $n\ge 1$ and $r>0$; the one-dimensional case is the circle and steps 1.1–3.1 give its single curvature with the same sign. Dimension zero is excluded because [F3] defines the present principal-curvature package only in positive hypersurface dimension. The condition $r>0$ excludes the collapsed, non-hypersurface radius-zero case; there is no parameter endpoint or manifold boundary. The supplied global radial field fixes the orientation without a selection. The only choice assumption is precisely the stated ${\mathrm{A}\mathrm{C}}_{\omega }$ inherited through [F2] and [F3], and the explicit computation adds none. No biconditional is asserted.

Parametrize $C_r$ by $X(\theta ,z)=(r\cos \theta ,r\sin \theta ,z)$. The fields $e_{\theta }=(-\sin \theta ,\cos \theta ,0)$ and $e_z=(0,0,1)$ are an orthonormal tangent frame, and $\nu =(\cos \theta ,\sin \theta ,0)$ is the outward unit normal.

The Cartesian Euclidean metric has constant coefficients, so [F5] gives zero Christoffel symbols. Since $e_{\theta }=(1\mathrm{/}r)X_{\theta }$, differentiating the displayed normal gives ${\bar{\mathrm{\nabla }}}_{e_{\theta }}\nu =(1\mathrm{/}r){\mathrm{\partial }}_{\theta }\nu =e_{\theta }\mathrm{/}r$, whereas ${\bar{\mathrm{\nabla }}}_{e_z}\nu =0$. Both derivatives are tangent, so [F2] yields $S_{\nu }{e}_{\theta }=-e_{\theta }\mathrm{/}r$ and $S_{\nu }{e}_z=0$.

By [F3], the orthonormal frame from step 1.1 is a principal frame with principal curvatures $-1\mathrm{/}r$ and $0$. Their product is the extrinsic Gaussian curvature, so it is zero. By [F4], the sectional curvature of the unique tangent two-plane is also $(-1\mathrm{/}r)\cdot 0=0$.

Applying the scalar second-fundamental-form identity in [F2] to $e_{\theta }$ gives $⟨\mathrm{I}\mathrm{I}(e_{\theta },e_{\theta }),\nu ⟩=g(S_{\nu }{e}_{\theta },e_{\theta })=-1\mathrm{/}r\mathrm{\ne }0$. Therefore $\mathrm{I}\mathrm{I}(e_{\theta },e_{\theta })\mathrm{\ne }0$, so the second fundamental form is not the zero tensor despite both Gaussian curvatures vanishing.

For every $r>0$ the cylinder is nonempty and two-dimensional; zero- and one-dimensional cases are therefore inapplicable. The condition $r>0$ excludes the collapsed, non-hypersurface axis and makes the circumferential direction nonzero. The periodic angular coordinate and unbounded axial coordinate introduce no endpoint or manifold boundary. The displayed frame and normal are explicit. The only choice assumption is the stated ${\mathrm{A}\mathrm{C}}_{\omega }$ inherited through [F2]–[F4], and the calculation makes no further family choice. No biconditional is asserted.

Put $t=v\mathrm{/}a$, $c=\cosh t$, and $s=\sinh t$. Then $X_u=(-ac\sin u,ac\cos u,0)$ and $X_v=(s\cos u,s\sin u,1)$, so the first fundamental coefficients are $E=⟨X_u,X_u⟩=a^2{c}^2$, $F=⟨X_u,X_v⟩=0$, and $G=⟨X_v,X_v⟩=s^2+1=c^2$. Since $a,c>0$, these vectors are independent; [F1] therefore gives the induced Riemannian metric.

The image of $X$ is the level set $C_a=\{(x,y,z):x^2+y^2-a^2{\cosh }^2(z\mathrm{/}a)=0\}\mathrm{.}$ On this level set $(x,y)\mathrm{\ne }(0,0)$, so the differential of the defining function is nonzero; [F7] makes $C_a$ an embedded surface. The map $X:S^1\times \mathbb{R}\to C_a$ is bijective, with smooth inverse $(x,y,z)⟼(\frac{(x,y)}{a\cosh (z\mathrm{/}a)},z)\in S^1\times \mathbb{R}\mathrm{.}$ Thus $X$ is an embedding and the submanifold interfaces below apply to its image.

Their cross product is $X_u\times X_v=ac(\cos u,\sin u,-s)$ and has norm $a{c}^2$. Hence $N=c^{-1}(\cos u,\sin u,-s)$ is a smooth unit normal for the displayed orientation.

The second derivatives are $X_{uu}=(-ac\cos u,-ac\sin u,0)$, $X_{uv}=(-s\sin u,s\cos u,0)$, and $X_{vv}=(c\cos u\mathrm{/}a,c\sin u\mathrm{/}a,0)$. The Cartesian Euclidean symbols vanish by [F6], so the scalar second fundamental coefficients obtained from [F2] are $h_{uu}=⟨X_{uu},N⟩=-a$, $h_{uv}=0$, and $h_{vv}=1\mathrm{/}a$.

Because both $(g_{ij})=\mathrm{diag}(a^2{c}^2,c^2)$ and $(h_{ij})=\mathrm{diag}(-a,1\mathrm{/}a)$ are diagonal, the orthonormal fields $e_u=X_u\mathrm{/}(ac)$ and $e_v=X_v\mathrm{/}c$ are principal directions. The identity in [F2] gives $g(S_N{e}_u,e_u)=-1\mathrm{/}(a{c}^2)$, $g(S_N{e}_v,e_v)=1\mathrm{/}(a{c}^2)$, and the mixed entries zero; hence [F3] gives exactly the two displayed principal curvatures.

Their average is zero, so [F3] gives scalar mean curvature $H_N=0$. Since the normal space is spanned by $N$, step 3.1 gives $\mathrm{I}\mathrm{I}(e_u,e_u)={\kappa }_{u}N$ and $\mathrm{I}\mathrm{I}(e_v,e_v)={\kappa }_{v}N$; [F4] therefore gives $\mathbf{H}=\frac{1}{2}({\kappa }_u+{\kappa }_v)N=0$.

Because $a>0$ and $c=\cosh (v\mathrm{/}a)>0$, ${\kappa }_u$ and ${\kappa }_v$ are nonzero at every point. In particular, $\mathrm{I}\mathrm{I}(e_u,e_u)={\kappa }_{u}N\mathrm{\ne }0$, so $\mathrm{I}\mathrm{I}$ is not the zero tensor; [F5] says the catenoid is not totally geodesic.

The domain and embedded image from step 1.2 are nonempty fixed two-manifolds, so zero- and one-dimensional cases are inapplicable. The condition $a>0$ excludes the collapsed scale and steps 1.1–2.1 prove nondegeneracy; $\cosh t$ never vanishes. Neither factor has a boundary endpoint. The displayed normal and principal frame are explicit. The only choice assumption is the stated ${\mathrm{A}\mathrm{C}}_{\omega }$ inherited through [F2]–[F5], and the finite coordinate calculation adds none. Reversing $N$ reverses both principal curvatures but leaves both zero-mean conclusions and $\mathrm{I}\mathrm{I}\mathrm{\ne }0$ unchanged. No biconditional is asserted.

On ${\mathbb{R}}^{r+1}$ the function $q(z)=⟨z,z⟩$ has differential $d{q}_z(w)=2⟨z,w⟩$, which is nonzero on $q^{-1}(1)$. Thus [F2] makes each $S^r$ an embedded boundaryless hypersurface with $T_x{S}^r=x^{\perp }$. Consequently at $x\in S^k$ one has $T_x{S}^k=V\cap x^{\perp }\subseteq T_x{S}^n$, the equatorial inclusion is embedded with the usual induced round metric, and its normal space inside $T_x{S}^n$ is exactly $V^{\perp }$: this subspace lies in $x^{\perp }$, is orthogonal to $V\cap x^{\perp }$, and has the required dimension $n-k$.

For each standard basis vector $a\in \{e_{k+2},\dots ,e_{n+1}\}\subseteq V^{\perp }$, define the smooth tangent field $A(y)=a-⟨a,y⟩y$ on $S^n$. Along $S^k$ one has $⟨a,x⟩=0$, so $A(x)=a$ and this restriction is a normal field to $S^k$ inside $S^n$ by step 1.1. For $X\in T_x{S}^k\subseteq V$, Cartesian differentiation gives $D_{X}A=-⟨a,X⟩x-⟨a,x⟩X=0$. By [F3]–[F4], ${\bar{\mathrm{\nabla }}}_X^{S^n}A=(D_{X}A{)}^{\mathrm{\top }}=0$. Hence the shape operator for $S^k\subseteq S^n$ is $S_{a}X=-({\bar{\mathrm{\nabla }}}_X^{S^n}A{)}^{\mathrm{\top }}=0$.