# differential geometry about neither local isometry nor conformal map

hi I need help in geometry I try to get the the idea but I couldn't solve so. It becomes great if someone try solve with explain to me by using figures if possible. I see this question change a lot of geometry experts and thank you for trying to solve. the question

(Lambert’s cylindrical projection) Project the unit sphere (except for the north and south poles) radially outward to the cylinder of radius $1$ by sending $(x,y,z)$ to $(x/\sqrt{(x^2+y^2)},y/\sqrt{(x^2+y^2)},z)$

Check that this map preserves area, but is neither a local isometry nor conformal.

Hint: Let $x(u,v)$ be the spherical coordinates parametrization of the sphere, and consider $x^*(u,v)=(cos(v),sin(v),cos(u)$}

Thank you very much thanks

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 This goes back to Archimedes. What do you understand so far? What condition would tell you that a mapping from one surface to another preserved area? And, what book are you using? – Will Jagy Feb 23 at 6:53 Maybe this will work. This is the image Archimedes wanted on his tombstone: en.wikipedia.org/wiki/File:Archimedes_sphere_and_cylinder.svg – Will Jagy Feb 23 at 6:56 And believed discovery, later, by the Roman, Cicero: en.wikipedia.org/wiki/… – Will Jagy Feb 23 at 6:58 the book name elementary differential geometry 2nd edition by springer – leena adam Feb 23 at 7:20 I didnot get the idea also if anyone expert i hope help me to solve and explain – leena adam Feb 23 at 7:21

Use spherical coordinates $\phi$, $\theta$, where $\theta=0$ on the equator, and produce corresponding points on the sphere $S^2$ and the cylinder $C$ as follows: \eqalign{(\phi,\theta)\mapsto{\bf r}(\phi,\theta)&:=(\cos\phi\cos\theta,\sin\phi\cos\theta, \sin\theta)\ ,\cr (\phi,\theta)\mapsto{\bf f}(\phi,\theta)&:=(\cos\phi,\sin\phi, \sin\theta)\ .\cr} In this way Lambert's projection $L$ appears as $$L:\ S^2\to C\ ,\qquad {\bf r}(\phi,\theta)\to{\bf f}(\phi,\theta)\ .$$ Then compute \eqalign{{\bf r}_\phi&=(-\sin\phi\cos\theta,\cos\phi\cos\theta,0),\quad {\bf r}_\theta=(-\cos\phi\sin\theta,-\sin\phi\sin\theta,\cos\theta),\cr {\bf f}_\phi&=(-\sin\phi,\cos\phi,0),\quad {\bf f}_\theta=(0,0,\cos\theta)\ .\cr} The fact that $$|{\bf r}_\phi|=\cos\theta\ne 1=|{\bf f}_\phi|\qquad(\theta\ne0)\tag{1}$$ already proves that $L$ is not an isometry: Equation (1) says that a piece of parallel at latitude $\theta$ on the sphere has length $$\int_{\phi_0}^{\phi_1}|{\bf r}_\phi|\ d\phi=(\phi_1-\phi_0)\cos\theta\ ,$$ while the projected arc has length $(\phi_1-\phi_0)$, given that $\phi_1>\phi_0$. Second, from $${|{\bf r}_\phi|/|{\bf f}_\phi|\over |{\bf r}_\theta|/|{\bf f}_\theta|}=\cos^2\theta\ne1\qquad(\theta\ne0)\tag{2}$$ one then deduces that the projection cannot be conformal, even though the right angle between ${\bf r}_\phi$ and ${\bf r}_\theta$ is preserved. Equation $(2)$ says that at any point ${\bf p}:={\bf r}(\phi,\theta)\in S^2$ where $\theta\ne0$ the local dilatations of the parallel and the meridian through ${\bf p}$ under $L$ are different. For a conformal map the dilatations in all directions emanating from ${\bf p}$ would have to be equal.
Finally one computes $$|{\bf r}_\phi\times{\bf r}_\theta|=\cos\theta=|{\bf f}_\phi\times{\bf f}_\theta|\ .$$ This shows that to an infinitesimal rectangle in the parameter plane correspond isoareal infinitesimal rectangles on $S^2$, resp. $C$. Therefore the projection is area-preserving.