Answer to Question #138980 in Differential Geometry | Topology for shweta

Question

Show that the Mercator projection

σ(u, v) = (sech u cos v,sech u sin v,tanh u)
is a regular surface patch of the unit sphere

Answer 1

Jacobian= $\begin{bmatrix} \partial f_{1}/\partial u & \partial f_{1}/\partial v \\ \partial f_{2}/\partial u & \partial f_{2}/\partial v\\ \partial f_{3}/\partial u&\partial f_{3}/\partial v \end{bmatrix}$ = $\begin{bmatrix} -(sech \ u \ tanh \ u)\cos v & -sech \ u \ \sin v\\ -(sech \ u \ tanh \ u)\sin v & sech \ u \ \cos v\\ sech^2 \ u &0 \end{bmatrix}$ . Now if we look at the minor deleting the first row we get $sech^3 \ u \cos v.$ Now $sech \ u\neq 0$ , hence minor vanishes if $\cos v=0.$ Now taking the minor after deleting 2nd row gives, $-sech^3 \ u \sin v.$ This shows $\sin v=0.$ But both $\sin v , \cos v$ cannot be zero hence one of the two minors never vanish showing the Jacobian has rank two. Also, all points lie on the sphere since $(sech \ u)^2 \cos^2u+(sech \ u)^2\sin^2u+(tan \ h)^2u=1.$