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

Question #138980
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
1
Expert's answer
2020-10-19T17:56:58-0400

Jacobian= "\\begin{bmatrix}\n \\partial f_{1}\/\\partial u & \\partial f_{1}\/\\partial v \\\\\n \\partial f_{2}\/\\partial u & \\partial f_{2}\/\\partial v\\\\\n \\partial f_{3}\/\\partial u&\\partial f_{3}\/\\partial v \n\\end{bmatrix}" = "\\begin{bmatrix}\n -(sech \\ u \\ tanh \\ u)\\cos v & -sech \\ u \\ \\sin v\\\\ \n -(sech \\ u \\ tanh \\ u)\\sin v & sech \\ u \\ \\cos v\\\\ \nsech^2 \\ u &0 \n\\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."


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