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I'm a bit confused about the phrase 'preserving the first fundamental form', or 'The Gaussian curvature is determined by the first fundamental form'.

For example, let's say I have two surfaces $M$ and $M'$. How can they possibly have 'the same fundamental form', when:

  1. The first fundamental form is a concept defined relative to some parametrisation
  2. The domains of the various fundamental forms will not even coincide: $T_pM\times T_pM$ vs $T_{p'}M'\times T_{p'}M'$.

I just don't see how to interpret the phrase 'having the same fundamental form' or something being 'determined by the fundamental form' given that it is such a 'relatively-defined' notion.

Any help in clearing this up would be much appreciated. Thanks


Oke here is a concrete example of something which confuses me:

This source defines the first fundamental form as follows:

$$I_P(U,V)=U\cdot V,\text{for } U,V\in T_PM\ \ (\subset \mathbb{R}^3)$$

From this definition it is clear that this is independent of parametrisation, since it is completely in terms of the inner product on $\mathbb{R}^3$. However, the confusion starts when they then give the following definition (These are notes by Theodore Shifrin):

Suppose $M$ and $M^*$ are surfaces. We say they are locally isometric if for each $P\in M$ there are a reg-param $x:U\to M$ with $x(u_0,v_0)=P$ and $x^*:U\to M^*$, with the property that $I_P=I_{P^*}^*$, whenever $P=x(u,v)$ and $P^*=x^*(u,v)$. That is, the function $f=x^*\circ x^{-1}$ is a one-to-one correspondence that preserves the first fundamental form.

Oke so I just don't see how this definition of a local isomertic surfaces is even well defined when working with the above definition of the first fundamental form. What does is possibly mean that $I_P=I_{P^*}^*$? Usually for a function equality they need to at least have the same domains, but that's not the case here.

The only thing I can imagine is that $f$ induces through it's derivative a map $T_PM\to T_{f(P)}M^*$, and so that we say that $$I_P=I_{P^*}^* \text{ iff } I_P(x,y)=I_{P^*}^*(f'(x),f'(y))$$

And so that in general for a map $f$ to preserve the first fundamental form, this above definition is what it means. At least this is independent of parametrisation, so that's nice.

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  • $\begingroup$ The first fundamental form is the inner product on the tangent space, not the matrix that represents it with respect to a basis, so that it is not «a concept defined relative to some parametrisation» $\endgroup$ Oct 11, 2015 at 20:13
  • $\begingroup$ @MarianoSuárez-Alvarez Yes I found that out, that seemed to have been my first source of confusion. There's more though, unfortunately. $\endgroup$ Oct 11, 2015 at 20:14

2 Answers 2

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First fundamental form is the traditional name of the Riemannian metric in a Riemannian manifold of dimension two, i.e. the scalar product on the tangent bundle. This is independent of a coordinate system but can be expressed (and is usually defined) using a coordindate system. A map preserving the first fundamental form is just a (local) isometry of Riemannian manifolds.

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  • $\begingroup$ Thanks for your answer. Unfortunately I do not really follow, since I have only studied the more 'concrete' case of surfaces in $\mathbb{R}^3$ so far. Would you say there is value in reading up on the more general theory to get a better view of what is going on here? $\endgroup$ Oct 11, 2015 at 18:18
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    $\begingroup$ There is always value in reading up more general theory :-) But it's reasonable to go for that value only if it fits into your plans. The important point here is that, while the first fundamental form is defined using a parametrization, it is a concept which does not depend on the parametriziation. Roughly speaking, if you have a 2 surface $M$ in $\mathbb{R}^3$ and look at the tangent space $T_pM$ for some $p\in M$, the first fundamental form ist the restriction of the ambient scalar product to $T_pM$. This does not depend on a parametrization, only it's local representation does. $\endgroup$
    – Thomas
    Oct 11, 2015 at 18:38
  • $\begingroup$ Oke, but if you don't mind me asking a more detailed question: I get that the function $I_p$ does not depend on the parametrisation, since this is, as you described, fairly easy to see. However, the proof of the Theorema Egregium expresses the curvature in terms of $E,F,G$ and their derivatives. These values DO depend on the parametrisation. So I find it confusing to speak as this depending only on $I_p$, while in fact these values also depend on the parametrisation... $\endgroup$ Oct 11, 2015 at 18:42
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    $\begingroup$ Ahh, the Theorema Egregium! Yes, it appears that the Gaussian curvature depends on the parametriziation, and if you can see it does not it still appears to depend on the way how the surface is embedded into the surrounding space. It's called 'egregium' (that means remarkable) because it states that this is not the case. That's the point of the theorem. You have a quantity which seems to depend on how the surface is embedded into the ambient space and you figure out it does not. There is no need to worry if this is confusing you, it means you got the point. $\endgroup$
    – Thomas
    Oct 11, 2015 at 19:01
  • $\begingroup$ Thanks for your answers. I would really appreciate it if you could comment on the example I added to my question, which illustrates my example more concretely. $\endgroup$ Oct 11, 2015 at 20:13
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This sort of question is very common in differential geometry. One does wonder if tangent plane at a point P of a surface M is dependent on the parametrization at that point. Such arguments are generally dealt with introduction of a function called change of parameters or change of coordinates.

You could look up section 2.3 of Curves and Surfaces, by Sebastian Montiel and Antonio Ros for a discussion on change of parameters. There is result in that section that proves that any change of parameters is a diffeomorphism. By having this result, we can see that the structures don't change irrespective of parametrization that we choose to work with.

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