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I wanted to ask you for help in better understanding two concepts that leave me a little perplexed regarding the first fundamental form:

1) The first fundamental form was defined to me as follows: given a parameterization $ϕ(u,v)$, the metric $A$ is found with the dot products like $ϕ_u​⋅ϕ_v$​ (through the 4 permutations of $u,v$), where the dot products are restricted to the tangent spaces induced by $R^3$. It was then explained that quantities dependent on $E,F,G$ (such as the Christoffel symbols, and therefore the curvature via Gauss's Theorema Egregium since it shows that they depend only on the Christoffel symbols) are intrinsic quantities, meaning they do not depend on how the surface is immersed in $R^3$, but are intrinsic to the object itself.

My confusion revolves around this: if I define the first fundamental form in this way, I note that $ϕ(u,v)$ is the map $ϕ:U→R^3$, so what comes out of this map is precisely the figure I have as a surface in R3R3, that is, the shape it takes. In fact, the map gives me the coordinates $(x,y,z)$. Now, $ϕ$​ and $ϕ_v$​ are the tangent vectors to that figure, so they indeed depend on the shape realized in $R^3$. Then I define the matrix $A$ through the dot products of $ϕ_u$​ and $ϕ_v$​, so what is intrinsic here? I am using tangent vectors to a figure that has a shape given by $ϕ:U→R^3$, so I would say it indeed depends on the immersion and how the figure is geometrically realized in it.

What seems to be suggested by the explanation is this: if I immerse the "abstract concept" of a sphere in $R^3$, I have different realizable figures. I do not understand this concept given the considerations above: curvature depends only on $E,F,G$ through the Christoffel symbols, but $E,F,G$ depend on the dot products of tangent vectors to a shape of the surface, so on an immersion of it in $R^3$.

2) The second question is this: if I change parameterization, I will have new $ϕ_i′​, i∈{u,v}$ which differ from the initial ones, so I will find $E′,F′,G′$. However, for an object (let's take the usual sphere), the curvature is fixed and depends only on $E,F,G$ right? Well, if I have changed parameterization and have $E′,F′,G′$, why wouldn't the curvature change? They are different values. It seems to me that by changing parameterization, the first fundamental form changes and therefore the curvature should change as well (but it shouldn't obviously be so).

Could you help me with these two questions? Thank you .


@Mikhail Katz

Thank you for your response and help, first of all.

So, if I understand correctly, the fact that the first fundamental form is intrinsic in the way it was explained to me (thus as a dot product of tangent vectors to a parameterized surface $phi$) cannot be understood.

I can see the point with your response: if I give an arbitrary first fundamental form (i.e., an abstract metric), then obviously by proving that the Gaussian curvature depends only on this metric, I detach it from the immersion. But as long as we talk about parameterization $phi$, it seems to me it must depend on the figure that is created in $R^3$ since the tangent vectors are tangent to that given immersion.

Is what I'm saying correct? Thank you.

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  • $\begingroup$ Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. $\endgroup$
    – Community Bot
    Commented Jun 20 at 14:37

2 Answers 2

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I prefer to think of it this way: The intrinsic geometry of a space is represented by the distance function, i.e., the distance between any two points on the surface. If there is a map between two surfaces in $\mathbb{R}$ such that the distances functions match up. Such surfaces are called isometric and the map an isometry. We know, by looking at flat surfaces, that it is possible for two different surfaces to be isometric. By differentiating the distance function, we discover the first fundamental form and that a map is an isometry if and only if the first fundamental form of one surface is mapped to the first fundamental form of the other.

Later, in Riemannian geometry, we can define the concept of the first fundamental form without using any embedding of the surface at all. That justifies the term "intrinsic". The idea is that an embedding uniquely determines a first fundamental form, but there are other first fundamental forms, with all of the same properties as the first fundamental form of an embedded surface, that do not arise from any embedding.

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In modern Riemannian geometry, as opposed to the classical differential geometry of curves and surfaces embedded in 3-space, the quantities $E,F,G$ are given abstractly without any Euclidean embedding. This is what is called a Riemannian metric on a manifold. As you mentioned, both the $\Gamma$ symbols and the Gaussian curvature can be expressed in terms of the metric, and in this sense are intrinsic invariants.

There are many different ways of isometrically embedding a given metric in Euclidean space. In addition, different parametrisations will give you different values of $E,F,G$. To add another level of abstraction, the metric viewed as a section of the symmetric square of the tangent bundle is the coordinate-invariant object, which will have different numerical representations depending on which coordinate chart you work in.

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  • $\begingroup$ Thank you for your answer. I tried to reply by modifying the original message; I'm not sure if it's the correct way to do it. $\endgroup$
    – user1339911
    Commented Jun 20 at 15:36
  • $\begingroup$ Well no, it does not really depends on the figure or shape of the embedding. What happens is that when you pass to a different parametrisation, the coefficients will transform accordingly. $\endgroup$ Commented Jun 20 at 15:40
  • $\begingroup$ Ok, what I wanted to say is that, if I define the first fundamental form with $ϕ_i*ϕ_j$ and $ϕ_i$ and $ϕ_j$ depend on the figure $ϕ$ (because are tangent on it). I have to demonstrate that the first fundamental form is independent of the figure. However, according to the professor's explanation, he says: since they depend on the first fundamental form => they are intrinsic. Is thath conclusion that i don't understand, beacuse $ϕ$ is the figure in $R^3$. $\endgroup$
    – user1339911
    Commented Jun 20 at 15:57
  • $\begingroup$ Right, a student cannot be expected to understand what this really means until she sees the intrinsic viewpoint on metrics. For the time being you should simply take "intrinsic" as defined by saying that it only depends on the first fundamental form (as opposed to, for example, the second fundamental form). $\endgroup$ Commented Jun 20 at 16:00
  • $\begingroup$ Thank you very much, this was what I was trying to understand because I felt stupid not seeing it :D $\endgroup$
    – user1339911
    Commented Jun 20 at 16:09

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