Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 3 / 25I / c

Use angular coordinate $v$. The two surfaces have parametrizationsandThe supplied surface-of-revolution formula givesSince $s>0$, the change of coordinateis a diffeomorphism from $(0,\infty )$ to $\mathbb{R}$ and satisfies $1+e^{2u}={\cosh }^{2}s$. Thereforeis a diffeomorphism and obeys $K_R\circ \varphi =K_S$. This is the curvature-matching diffeomorphism between two surfaces of revolution.

No such local isometry exists. The curvatures above are strictly monotone functions of $u$ and $s$, so curvature preservation would force the $R$-coordinate of any local isometry to beThe first fundamental forms areWrite the putative map as $(s,v)\mapsto (u(s),\theta (s,v))$. The mixed coefficient in its pullback isA local isometry is locally nonsingular, so ${\theta }_v\ne 0$; matching the zero mixed coefficient of $I_S$ therefore gives ${\theta }_s=0$. The $d{v}^2$ coefficients would then requireThe left side depends only on $v$, while the right side varies with $s$, an impossibility. Equal pointwise Gaussian curvature under a diffeomorphism therefore does not suffice for local isometry.