The cylinder has zero Gaussian curvature but nonzero second fundamental form

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.