Measuring the conformality or equal-areaness of a projection from the sphere to some surface embedded in $R^3$

Question

Suppose I want to map from the sphere to some other surface that is embedded into $\Bbb R^3$, which we will treat as an embedding into $\Bbb R^3$. In this situation, suppose I already have some formula to map from latitudes and longitudes, denoted $(\lambda, \phi)$ to $(x, y, z)$ coordinates in $\Bbb R^3$. I would like to see if this map is equal area, conformal, etc, and in general just evaluate what the metric looks like at some arbitrary point.

I haven't studied this stuff formally, but I would guess that the method would be something like this. Given that I have some function $f: (\lambda, \phi) \mapsto (x, y, z)$, I can get the Jacobian matrix of partial derivatives. This will give two vectors which span a plane tangent to the surface at that point. Then I would guess that, for instance, we can:

• See if these vectors are orthogonal and have the same norm. If this is true everywhere, then the map is conformal.
• Compute the area of the parallelogram formed by the two vectors, and see if it has the same area everywhere. If so, the map is equal area.

However I don't think this is exactly correct, because if we differentiate with respect to the coordinate basis of latitudes and longitudes on the sphere, what we're really doing is mapping from the equirectangular projection to $\Bbb R^3$, and the equirectangular projection is neither a conformal nor equal-area map from the sphere to begin with. So some kind of adjustment term will surely be necessary, but I am hoping I have the correct basic idea with the Jacobian. How does this work?

Answer 1

tl; dr: In fact, your instincts are precisely correct.

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In differential geometry, we associate to a parametrization $(x, y, z) = f(\lambda, \phi)$ the three "components of the first fundamental form," \begin{align*} E &= f_{\lambda} \cdot f_{\lambda} = \|f_{\lambda}\|^{2}, \\ F &= f_{\lambda} \cdot f_{\phi}, \\ G &= f_{\phi} \cdot f_{\phi} = \|f_{\phi}\|^{2}. \end{align*}

• We say $f$ is conformal if $E = G$ (coordinate tangent vectors have the same length) and $F = 0$ (orthogonality). (Geometers also speak of orthogonal coordinates when $F = 0$.)
• We say $f$ is area preserving if $EG - F^{2} = 1$ (the parallelogram spanned by the coordinate tangents has squared area $1$). (In geography, it sounds that the condition is instead that $EG - F^{2}$ is constant.)

For the equirectangular projection (a.k.a., spherical coordinates rolled out flat), we have (up to translation and relabeling of coordinates) $$ f(\lambda, \phi) = (R\cos\lambda\cos\phi, R\sin\lambda\cos\phi, R\sin\phi). $$ A short calculation gives \begin{align*} f_{\lambda} &= (-R\sin\lambda\cos\phi, R\cos\lambda\cos\phi, 0), \\ f_{\phi} &= (-R\cos\lambda\sin\phi, -R\sin\lambda\sin\phi, R\cos\phi); \end{align*} thus $$ E = R^{2}\cos^{2}\phi,\qquad F = 0,\qquad G = R^{2}. $$ This confirms the parametrization is orthogonal, but neither conformal nor area-preserving.