I may be conflating a number of different concepts, but I am confused about measurements using the first fundamental form.

Here is what I think I know:

  1. The first fundamental form can be used to compute distance on a parametric surface
  2. The coefficients of the first fundamental form can be re-parameterized through substitution, as with the relation given in Sec. 1.3 here.

Here is what confuses me:

  1. Re-parameterizing the sphere through a map projection (e.g. gnomonic projection, mercator projection, etc.) results in different distances on the mapped surface. That is, $ds_{map}^2 \ne ds_{sphere}^2$. I can walk through this derivation and it makes logical sense.
  2. Yet, "arclength is invariant to coordinate system transformations," at least according to the proof on the Metric Tensor Wikipedia page (screenshot pasted below) and a handful of other Google search results on the topic. But isn't a spherical projection a coordinate transformation $M : (\lambda, \phi) \rightarrow (x, y)$?

enter image description here

I believe that this text clears it up a bit (bottom of page 8):

A mapping of a portion of a manifold M to a portion of a manifold N is called isometric, if the length of any curve on N is the same as the length of its pre-image on M.

In other words, arclength is preserved by isometric transformations. Spherical map projections are provably not isometric, so, sure, that's why $ds_{map}^2 \ne ds_{sphere}^2$. I think I follow this.

Why I'm confused is because I thought I followed the logic of proof on the Wikipedia page, and I don't understand why it wouldn't generalize to non-isometric mappings. All the same derivatives can be calculated through a spherical projection, for example.

So does arclength hold or not?

[Note: when answering this question, feel free to use the example of spherical projections, because it's something I think I understand fairly well. Also, I'm not a mathematician, so formal mathematical language, while likely the most accurate way to answer this question, is probably going to be lost on me.]


Let $\dot S:=S^2\setminus\{(0,0,1)\}$ be the $2$-sphere with the north pole removed. There is the stereographic projection $$\sigma: \quad \dot S\to{\mathbb R}^2,\qquad (x,y,z)\mapsto(\xi,\eta)$$ according to well known formulas. This $\sigma$ is a map of one surface onto another one. It is not an isometry when we adopt on $\dot S$ and ${\mathbb R}^2$ their "usual" standard metrics, which come about as follows:

$\dot S$ acquires its standard metric from the embedding of $S^2$ into the euclidean ${\mathbb R}^3$. If $(\phi,\theta)$ are geographical coordinates on $\dot S$ you compute $ds_{\rm sphere}^2=E(\phi,\theta)d\phi^2+G(\phi, \theta)d\theta^2$, whereby $E$ and $G$ in fact depend only on the geographical latitude. On ${\mathbb R}^2$ you have the standard euclidean $ds^2_{\rm plane}=d\xi^2+d\eta^2$.

On $\dot S$ one might think of other coordinate systems, say $(x,y)$ in the vicinity of the south pole, or the $(\xi,\eta)$ furnished by $\sigma$. The latter is to be understood in the following way: Each point $P=(x,y,z)\in\dot S$ so far had geographical coordinates $(\phi,\theta)$, but $P$ obtains now new coordinates $(\xi,\eta):=\sigma(P)$. But we do not want to change the length measurement on $\dot S$. This means it is forbidden to write $ds^2_{\rm sphere}=d\xi^2+d\eta^2$. Instead we have to do some computation as quoted by you from Wikipedia, and we would then obtain $$ds^2_{\rm sphere}=G(\xi,\eta)(d\xi^2+d\eta^2)\ .$$ (In this example the resulting transformation formula is so simple because the stereographic projection $\sigma$ is conformal.) Note that in this second thought experiment we have not mapped one surface onto another, but we have (for whatever reasons) changed the local coordinate system on the given surface $\dot S$.

  • $\begingroup$ So in both coordinate systems, $(\phi, \theta)$ and $(\xi, \eta)$, the value of $d_{sphere}^2$ will be the same. However, the distance metric $ds_{plane}^2$ is different because it is not an isometric transformation, but rather a projection (a conformal one in this case). Is my understanding correct? $\endgroup$
    – marcman
    Apr 4 '20 at 17:05
  • $\begingroup$ Am I right to then understand that the "arclength being preserved" means that regardless of coordinate system the measure of distance on the parametric surface (e.g. the 2-sphere) will always be the same. Yet, the measure of distance in each coordinate space, however, will not always be (i.e. maybe $d_{plane}^2$ would be different in a different coordinate space?) $\endgroup$
    – marcman
    Apr 4 '20 at 17:07
  • $\begingroup$ Last followup for now: is a projection not a coordinate system transformation (or reparameterization)? $\endgroup$
    – marcman
    Apr 4 '20 at 17:11
  • 1
    $\begingroup$ @marcman: 1. comment: Basically yes. But $\sigma$ should not be called a "projection", since the word projection is nowadays reserved for projections in cartesian products. – 2. comment: Yes for the first part. "The measure of distance in each coordinate space" makes no sense, only the "measure of distance on a given surface". –3. comment: The formulas are the same, but the interpretation is different. Are we talking about a coordinate transformation in the space $X$ or about a map $f: \>X\to Y$? $\endgroup$ Apr 4 '20 at 18:37
  • $\begingroup$ Ahh. When I refer to a projection, I am talking about a map $f : X \rightarrow Y$. Perhaps therein lies the difference. If $f$ is not isometric, then arclength will not hold. If $f$ is isometric, is $f_{isometric} : X \rightarrow Y$ effectively a coordinate transform? $\endgroup$
    – marcman
    Apr 4 '20 at 19:07

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