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In section 2.5 of do Carmo, given an embedded regular surface $S\subset\mathbb R^3$, the author defines the first fundamental form $\mathrm I_p:T_pS\to \mathbb R$ as the quadratic form $\mathrm I_p(w)= \left\langle w,w \right\rangle _{\mathbb R^3}$.

In a first course on differential geometry, the lecturer gave the same definition. I really feel I'm missing the the idea here, because it seems to me $\mathrm I_p$ "does not depend on $p$" in the sense the inner product stays the same. The lecturer said the whole point of the first fundamental form is to capture the local geometry of a regular surface at a given point in an intrinsic fashion, which definitely seems like a great thing, but the very definition seems to be independent of $p$. The only dependence is through the domain, but not the geometry.

A fellow student told me the dependence is through $E,F,G$, but I don't understand: While $E,F,G$ are calculated via parametrizations of coordinate neighborhoods of $p$, they must be independent of parametrization... Their definition uses an inner product which is the same at every $p$...

What am I missing here? What's the picture I should have in mind? I think I understand this answer, which explains in what says the fundamental form allows computations without further reference to an embedding in Euclidean space, but fear I am missing something else. (My fellow student spoke with confidence.)

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While the inner product $\left< \cdot, \cdot \right>_{\mathbb{R}^3}$ doesn't change, if you think of the tangent space $T_pS$ as a subspace of $\mathbb{R}^3$, then $T_pS$ itself changes as $p$ moves around. The first fundamental form at two different points $p,q \in S$ tells you the restriction of the same object (the standard inner product of $\mathbb{R}^3$) to two different two-dimensional subspaces $T_pS, T_qS$ of $\mathbb{R}^3$.

How can you describe this dependence more concretely? By choosing a coordinate system $\varphi \colon V \rightarrow U \subseteq S$, we get two tangent vectors $\frac{\partial \varphi}{\partial x}(x,y), \frac{\partial \varphi}{\partial y}(x,y)$ at each point $\varphi(x,y) \in S$ that form a basis of $T_{\varphi(x,y)}S \subseteq \mathbb{R}^3$. This allows us to identify all the different subspaces $T_pS \subseteq \mathbb{R}^3$ with a single fixed subspace $\mathbb{R}^2$. Explicitly, we identify $(a,b) \in \mathbb{R}^2$ with $a \frac{\partial \varphi}{\partial x}(x,y) + b \frac{\partial \varphi}{\partial y}(x,y) \in T_{\varphi(x,y)} S$. Now, for each $(x,y)$ we can pull back the inner product from $T_{\varphi(x,y)} S$ to $\mathbb{R}^2$ using the identification and obtain an inner product $\left< \cdot, \cdot \right>_{(x,y)}$ on $\mathbb{R}^2$. Different points will give you different inner products that reflect both the fact that the tangent spaces in $\mathbb{R}^3$ have possibly changed and the specific way the coordinate system describes them. All this is encoded in a matrix

$$ \begin{pmatrix} E(x,y) & F(x,y) \\ F(x,y) & G(x,y) \end{pmatrix} $$

which describes the inner products $\left< \cdot, \cdot \right>_{(x,y)}$ in the sense that

$$ \left< (a,b), (c,d) \right>_{(x,y)} = (a,b) \begin{pmatrix} E(x,y) & F(x,y) \\ F(x,y) & G(x,y) \end{pmatrix} \begin{pmatrix} c \\ d \end{pmatrix} = \left< a \frac{\partial \varphi}{\partial x}(x,y) + b \frac{\partial \varphi}{\partial y}(x,y), c \frac{\partial \varphi}{\partial x}(x,y) + d \frac{\partial \varphi}{\partial y}(x,y) \right>_{\mathbb{R}^3}. $$

The quantities $E,F,G$ do depend on the parametrization (they aren't even defined without a parametrization). What doesn't depend on the parametrization is the first fundamental form.

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  • $\begingroup$ This is exactly what I was confused about - now I see how we might end up with different inner products at different tangent spaces. $\endgroup$ – Arrow Nov 23 '16 at 12:16
  • $\begingroup$ The idea of the first fundamental form is to abstract away the specific way in which the tangent spaces rotate in $\mathbb{R}^3$, forget it, and remember only the inner product that is induced on each tangent space. $\endgroup$ – levap Nov 23 '16 at 12:18
  • $\begingroup$ @Arrrow: I just want to warn you that "having different tangent spaces" is not the only issue. If you think of $\mathbb{R}^2 \setminus \{ 0 \}$ as a surface in $\mathbb{R}^3$, then the tangent plane doesn't move at all. However, if you use a polar coordinate system and describe the first fundamental form, you get non-constant $G$! This reflects the fact that the identification induced by the coordinate system between the different tangent spaces (which, in this case, are all the same) is non-trivial. Of course, if you use a cartesian coordinate system, this goes away but in general, if I give $\endgroup$ – levap Nov 23 '16 at 12:23
  • $\begingroup$ you a coordinate system with $E,F,G$ that describes a specific surface $S \subseteq \mathbb{R}^3$, you won't be able to tell if the surface it came from has "different tangent spaces" or not. This information is (partially) lost in the first fundamental form. $\endgroup$ – levap Nov 23 '16 at 12:26
  • $\begingroup$ I will think about this carefully. Thanks again :) $\endgroup$ – Arrow Nov 23 '16 at 12:37
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A fellow student told me the dependence is through $E,F,G$, but I don't understand: While $E,F,G$ are calculated via parametrizations of coordinate neighborhoods of $p$, they must be independent of parametrization... Their definition uses an inner product which is the same at every $p$...

Firstly, the bold statement is untrue. $E, F, G$ are not invariant but covariant under changes of parametrization. This means that the geometrical content of the form $I_p$ is invariant, but its components in a particular basis change to compensate for my change of basis. We might say "the first fundamental form is invariant, but its components are covariant".

To be concrete, suppose I live on the 1D line of the graph $y = f(x)$ in 2D, so we only have an $E$ in the metric (matrix) expression for $I_p(w) = \left<w,w\right>_{\mathbb{R}^2}$. If I use the $x$ coordinate as a parameter, then the tangent vector is $\frac{d}{dx} (x,f(x)) = (1,f'(x))$, and $E = 1+f'(x)^2$. If I use $u=2x$ as a parameter, then the tangent vector at $u=2x$ is $\frac{d}{du} (u/2,f(u/2)) = (1/2,1/2 f'(u/2))$ and $\tilde{E} = \frac{1}{4}(1+f'(u/2)^2)$.

Notice $\tilde{E} = E/4$. This change is the covariant transformation of $E$ to guarantee its physical content stays the same. For example, suppose I ask what the length of a path in the surface $x(t)$ is. It is given by $L = \int \sqrt{E \dot{x}^2} dt$ in the $x$ parametrization. But alternatively, we could use $u(t) = 2x(t)$ for the same path and calculate its length. We get $$\tilde{L} = \int \sqrt{\tilde{E} \dot{u}^2} dt = \int \sqrt{\frac{1}{4} E (2\dot{x})^2} dt = \int \sqrt{E \dot{x}^2} dt = L$$

Notice that the geometrical information in $E,\tilde{E}$ is invariant: we calculate the length of the curve as being the same whichever one we use. But the values of $E,\tilde{E}$ are necessarily different if this is to happen.

I really feel I'm missing the the idea here, because it seems to me $\mathrm I_p$ "does not depend on $p$" in the sense the inner product stays the same. The lecturer said the whole point of the first fundamental form is to capture the local geometry of a regular surface at a given point in an intrinsic fashion, which definitely seems like a great thing, but the very definition seems to be independent of $p$. The only dependence is through the domain, but not the geometry.

Notice also that in the above formula for $L$, say, $E = E(x(t))$ depends on where we are in the surface. This makes sense: if you just told me how far I went in $x$ units, I couldn't tell you how far we went in physical space (the line living in $\mathbb{R}^2$) without knowing where I was on the graph $f(x)$, and what shape the graph $f(x)$ was. The first fundamental form captures this.

The point is that because the domain is the tangent space, when I ask how far a small change in $x$ gets me, I'm obliged to also make a small change in $y$ (the size of that change is $y' = f'(x)$ of course). Hence when I evaluate $\left<w,w\right>_{\mathbb{R}^2}$ the information about the shape of the surface $f(x)$ comes along for the ride. Indeed, this is exactly where the $(1+f'(x)^2)$ is coming from.

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  • $\begingroup$ Do you mind clarifying that last paragraph.."The point is that because the domain is the tangent space" I dont understand in particular $\endgroup$ – user123124 Jan 16 '19 at 14:17
  • $\begingroup$ @Maxed Draw a graph of $f(x)$, and pick some point $x_0$. Now suppose I make a small change $x_0 + \delta x$ and move along the graph. The direction I move is along the tangent to $f(x)$ at $x_0$. The vector whose length I am measuring is really $(\delta x, \delta f(x)) \approx (\delta x, f'(x) \delta x)$, and it lives inside the (infinite) tangent line to the graph. The length of this vector, by Pythagoras, is $\delta x\sqrt{1 + f'(x)^2} \neq \delta x$. The reason is that my deviation was not really in the $x$ direction, but along the tangent. $\endgroup$ – not all wrong Jan 16 '19 at 14:47
  • $\begingroup$ right, are me measureing the lenght of the tanget then? and the first fundamental form is doing the same thing? Is it the length of tangets and not thier angles that we are intersted in? $\endgroup$ – user123124 Jan 16 '19 at 14:52
  • $\begingroup$ @Maxed You're measuring distance travelled in the surface using the form $E$; infinitesimally, that's approximated by moving along the tangent. In this example, there are no angles to measure inside the surface because the surface is only 1d (parametrized by e.g. $x$). In higher dimensions, you are also interested in angles. In the 2d example with $E, F, G$, we can compute dot products between 2 component vectors. $\endgroup$ – not all wrong Jan 16 '19 at 15:58
  • $\begingroup$ right, and is the intuition the same? moving from the formulas I got it looks like the norm of the normal or equivalently the area of the paralleogram span by the basis in the tanget spaces is used. I do not really see how the latter object is playing the same role as in the 1 d case. Should it? $\endgroup$ – user123124 Jan 17 '19 at 10:22
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I found this very confusing when I started learning DG too...

The confusion will be cleared up with you know a bit more about Riemannian-manifolds in general, since then you will really understand what people mean when they say that something is intrinsic to the surface.

The right picture to have in mind is that something intrinsic in this context means "A property of $S$ when it is considered as a Riemannian-manifold". So the 'trick' here is people define the first fundamental form to obtain a Riemannian structure on $S$.

I think you are rightfully confused because in the context of surfaces in $\mathbb{R}^3$ it is difficult to see the advantages of this approach. The best solution is to just go with it until you know a bit more Riemannian geometry and then read this material again, and it will make a lot more sense.

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  • $\begingroup$ I don't think I am confused at all by the idea of Riemannian geometry - instead of studying a manifold through a chosen embedding in Euclidean space, instead append to it an additional structure encoding the geometry of interest, and proceed "intrinsically". I am confused because my fellow student has said something which seems to me false, and I'm not sure who is right :) $\endgroup$ – Arrow Nov 23 '16 at 11:53
  • $\begingroup$ @Arrow What does "capturing the local geometry at a point" mean to you? $\endgroup$ – user2520938 Nov 23 '16 at 11:59
  • $\begingroup$ I'm not sure what the lecturer meant. What makes sense to me is that the first fundamental form lets us discuss metric questions using the tangent bundle alone, without reference to $\mathbb R^3$. $\endgroup$ – Arrow Nov 23 '16 at 12:01
  • $\begingroup$ @Arrow If you don't know what the lecturer meant by local geometry, then I think you should ask a question about that instead. Because right now it seems like youre confused based on your hand-wavey interpretation of the term "local geometry" $\endgroup$ – user2520938 Nov 23 '16 at 12:06

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