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The first fundamental form of a surface $S$ at a point $p$ is "the quadratic form on the tangent plane $S_p$ inherited from the inner product structure of $\mathbb R^3$".

At the same time, the first fundamental form is apparently supposed to describe the surface at $p$ in some way, for instance you can compute Gaussian curvature from it. In particular, the first fundamental form should be different for different surfaces $S$.

So what is the first fundamental form?

  1. It can't be the actual quadratic form $\langle x, x\rangle$ on $S_p$, that is, it can't simply be a function from $\mathbb R^3$ into $\mathbb R$, because that's the same for any surface $S$ that has the same tangent plane at $p$.

  2. It can't be the triplet of coefficients $(E, F, G)$ either, since they depend on the parameterization used, and in fact I think by using the right parameterization we can get any triplet $(E, F, G)$ we want.

Jack M
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    The point is it's a (smoothly varying) inner product on the tangent plane $T_pS$ for all $p$ in $S$, not just one. It does depend on how the surface $S$ is embedded in $\Bbb R^3$. Equivalently it's a triplet of coefficients $(E, F, G)$ at each point $p$ on $S$. The point is that triplet defines an inner product at each tangent plane. – Balarka Sen Jun 09 '17 at 20:34

1 Answers1

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$\newcommand{\Vec}[1]{\mathbf{#1}}\newcommand{\Brak}[1]{\left\langle #1\right\rangle}$There's a specific, formally simple relationship between a coordinate system on a regular surface and the components of the first fundamental form: In customary notation, if $\Vec{x}$ is a regular parametrization, then $$ E(u, v) = \Brak{\Vec{x}_{u}, \Vec{x}_{u}},\qquad F(u, v) = \Brak{\Vec{x}_{u}, \Vec{x}_{v}},\qquad G(u, v) = \Brak{\Vec{x}_{v}, \Vec{x}_{v}}. $$ The subtle points are:

  1. Although two surfaces sharing a tangent plane at a point $p$ have "the same first fundamental form at $p$", they do not generally have equivalent first fundamental forms in any open neighborhood of $p$;

  2. Perhaps surprisingly, it's not true that by using the right parameterization we can get any triplet $(E, F, G)$ we want. If it were, there would be no local invariants of a surface, because one could always pick, say, $E = G = 1$ and $F = 0$.

Globally, the first fundamental form of a surface is an inner-product-valued function on a surface.

Locally, the first fundamental form of a surface is represented by an association to each coordinate system of a $2 \times 2$ matrix-valued function whose value at each point is positive-definite, and which "transforms like an inner product" under change of coordinate. The first fundamental form itself may be viewed as the resulting equivalence class of all such pairs of "coordinate system and matrix-valued function".

Andrew D. Hwang
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    But we can always find a parameterization $\mathbb x$ such that at $p$, the vectors $\mathbb x_u$ and $\mathbb x_v$ are equal to any two desired vectors of the tangent plane, right? So (at a specific point) we can get any triplet of the form written in your question for some vectors $\mathbb x_u, \mathbb x_v$. Obviously there are some restrictions ($E$ and $G$ have to be positive, for one). – Jack M Jun 09 '17 at 20:44
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    You can get an arbitrary positive-definite symmetric matrix at one point, but not in an open neighborhood. By analogy, if you cut a piece of plastic from a toy ball (such as a ping-pong ball) and try to flatten it on a table top, you can get make the tangent planes match at any particular point of the ball, but not in any open neighborhood. – Andrew D. Hwang Jun 09 '17 at 20:55
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    So in what way are the FFFs on a cylinder and a sphere "different"? If I rotate/translate the cylinder to line up a tangent plane at $p$ on the cylinder with a tangent plane at $p$ on the sphere, the FFFs on the cylinder and the sphere become the same, don't they? So it seems that the FFF on different surfaces are only "different" in that they're defined on different planes of $\mathbb R^3$, but they're still isomorphic in some sense. – Jack M Jun 10 '17 at 15:54
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    On a cylinder and sphere (of the same radii) you can make the FFFs agree along a circle, but not on an open set. Another analogy: Declare that two real-valued functions $f$ and $F$ of one variable are "equivalent" if there exists an invertible function $g$ of one variable such that $F = f \circ g$. It can happen that for each real $p$, there exists an invertible $g$ such that $F(p) = (f \circ g)(p)$, yet $f$ and $F$ are not equivalent. [...] – Andrew D. Hwang Jun 10 '17 at 17:03
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    The way it seems you're thinking of it is, "Two surfaces $S$ and $S'$ have equivalent (i.e., isometric) FFFs at an arbitrary point, so why aren't the FFFs equivalent?" (This is a good question, and why I upvoted!) The reasons is, a FFF of a surface has an invariant depending not merely on the FFF at an arbitrary point, but on a certain differential expression, namely the Gaussian curvature. Just because two FFFs are equivalent at an arbitrary point does not mean they are equivalent on some open set: You'd have to be able to "choose the same isometry everywhere", which is not always possible. – Andrew D. Hwang Jun 10 '17 at 17:08
  • My problem is that I don't see what knowing the FFF at $p$ gives you beyond knowing the tangent plane at $p$, that is, it seems to me that surfaces $S, T$ have the same FFF at $p$ iff they have the same tangent plane at $p$. And of course, if two surfaces identically have the same tangent plane on an open set, they are the same on that open set. – Jack M Jun 10 '17 at 17:32
  • To be honest, the FFF at an arbitrary point of a smooth surface is just a Euclidean plane; any two Euclidean planes are isometric. Asking whether or not two surfaces are tangent at a point is a red herring. (And again, it's therefore all the more remarkable that there exists a local invariant of a Riemannian metric on a surface, since there are no pointwise invariants.) – Andrew D. Hwang Jun 11 '17 at 00:16