The catenoid has zero mean curvature but is not totally geodesic

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.