Principal curvatures of a round sphere

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.