Past exam of the geometry course of the University of Oxford / 2011 / 3

For a regular parametrization $\mathbf{r}(u,v)$, the first fundamental form is the induced Riemannian metricIt measures squared lengths of tangent vectors and therefore lengths of curves. In the plane put $x=\rho \cos \theta$, $y=\rho \sin \theta$. The derivatives are orthogonal, of lengths one and $\rho$, soThese polar coordinates are regular for $\rho >0$.

Two regular coordinate patches are locally isometric under a coordinate correspondence exactly when it preserves their first fundamental forms. More explicitly, a diffeomorphism $\Phi :U\to \tilde{U}$ of parameter domains induces a local isometry if and only ifIn common coordinates this means equality of the three coefficient functions $E,F,G$. The criterion follows because the first fundamental form gives the inner product of every pair of tangent vectors. Equality therefore preserves lengths of all curves; conversely a smooth isometry preserves tangent-vector lengths and their inner products by polarization.

Using the two first fundamental forms calculated below, chooseThe surface of revolution is then a catenoid, with metric ${\cosh }^{2}s(d{s}^2+d{v}^2)$. On the helicoid, take $u=\sinh s/a$, assuming $a\ne 0$ so that its parametrization is regular. Substitution givesThus the coordinate mapis a local isometry on sufficiently small patches. The constant $a$ only changes the radial parametrization of the given helicoid, whose pitch is fixed at one. For $a=0$ the displayed helicoid parametrization is not a regular surface.

Differentiate to getTheir squared lengths are $a^2$ and $a^2{u}^2+1$, and their inner product is zero. Hence the helicoid has first fundamental form

For the surface of revolution,The mixed inner product vanishes, givingTaking $f(u)=\cosh u$ makes both coefficients ${\cosh }^{2}u$, as used in the parent calculation.