# The First Fundamental Form

In class we have been thinking about lengths and areas (in other words, some metric properties) of surfaces. An important tool in the study of these concepts is the first fundamental form, which will be the theme of this post. Not only is this the standard approach, I believe it better motivates further study; in particular the introduction of a Riemannian metric, and thus the field of Riemannian geometry, can be seen as an outgrowth of the first fundamental form. I will say a few words about this at the end of the post. Here’s the plan: introduce the first fundamental form, and show how it relates to some concepts you have covered so far. For instance,

• Length of curves
• Areas of regions on surfaces

To begin with I will be talking about surfaces embedded in $\mathbb{R}^3$. At the end we will see how the notion of first fundamental form relates to that of Riemannian manifolds when attempting to forget the embedding in Euclidean space.

### The Cross Product

We all recall our old friend the cross product. Let $u=(u_1, u_2, u_3), v=(v_1, v_2, v_3) \in \mathbb{R}^3$ and define the cross product, (or vector product) $u \wedge v$ by

$u\wedge v= \left|\begin{array}{cc}u_2 & u_3 \\v_2 & v_3\end{array}\right| e_1 -\left|\begin{array}{cc}u_1 & u_3 \\v_1 & v_3\end{array}\right|e_2+\left|\begin{array}{cc}u_1 & u_2 \\v_1 & v_2\end{array}\right|e_3$

Exercise. Do you think the wedge symbol $\wedge$ is justified in the notation for the cross product here? Why?

You hopefully remember that the cross product of two vectors produces a third vector, perpendicular to both. You may also recall that the length of this third vector is related to $A$, the area of the parallelogram defined by $u$ and $v$. In fact we have

Lemma. $|u\wedge v|^2=A^2$.

I’ll leave the proof as an easy exercise. This will be useful for us because it will relate the notion of area on a surface that Hubbard gave (in particular in terms of parallelograms) to the first fundamental form. But before we can do that, we’d better go ahead and actually define the first fundamental form.

### The First Fundamental Form

As I mentioned in the introduction, we are interested in surfaces $S$ sitting in $\mathbb{R}^3$. In particular, we want a way to study metric properties of the surface, for example, lengths of curves, areas of regions and angles between vectors. In $\mathbb{R}^3$ we already have a tool for doing this, the dot product. Now since our surface is sitting inside $\mathbb{R}^3$ we are allowed to  borrow its dot product. More precisely, there is an induced inner product on the tangent space $T_pS$ coming from the the standard inner product in $\mathbb{R}^3$ (the dot product!). That is, given $v_1, v_2 \in T_pS$ tangent vectors to a surface, then $\langle v_1, v_2\rangle_p$ is equal to the standard inner product of $v_1, v_2$ thought of as vectors in $\mathbb{R}^3$. We call this induced inner product the first fundamental form. In other words we have a quadratic form

$I_p: T_pS \rightarrow\mathbb{R}$

$v \mapsto \langle v, v\rangle_p$

$I_p$ is called the first fundamental form of $S$ at $p$. (I will stop writing the $p$ when there is no ambiguity). (Note, the first fundamental form is sometimes defined as the symmetric bilinear form $I_p: T_pS \times T_pS \rightarrow \mathbb{R}$ by $I_p(v_1, v_2) = \langle v_1, v_2 \rangle_p$).

Explicit description: There is a standard way to express quadratic forms in terms of a basis. Let $\phi: \mathbb{R}^2 \rightarrow S$ be a parametrisation (at $p$). We can think of tangent vectors as being tangent to a curve in our surface. That is, for a tangent vector $v \in T_pS$, there exists a parametrised curve $\alpha(t) = \phi(u(t), v(t))$ such that $v = \alpha'(0)$. We thus compute,

$I_p(v) = I_p(\alpha'(0)) = \langle\alpha'(0), \alpha'(0)\rangle_p = \langle\phi_u u' + \phi_v v', \phi_u u'+ \phi_v v'\rangle$

$= \langle\phi_u, \phi_u\rangle_p(u')^2+2\langle\phi_u, \phi_v\rangle_pu'v'+\langle\phi_v, \phi_v\rangle_p(v')^2$

We are motivated to define,

$E(u,v) = \langle \phi_u, \phi_u \rangle$

$F(u,v) = \langle \phi_u, \phi_v \rangle$

$G(u,v) = \langle \phi_v, \phi_v \rangle$

We have done nothing more then compute the entries in the matrix that determines the inner product, that is, writing tangent vectors in terms of the basis $\{\phi_u, \phi_v\}$, we have,

$I_p(v_1, v_2) = \langle v_1, v_2 \rangle_p = v_1^T\left(\begin{array}{cc}E & F \\F & G\end{array}\right)v_2$

### Examples

1. The Plane

Define a plane in $\mathbb{R}^3$ passing though a point $P \in \mathbb{R}^3$ containing orthonormal vectors $w_1, w_2$ (this uniquely determines the plane!). This plane $\pi$ is parametrised by

$\pi: \phi(u, v) = P + w_1 u + w_2 v$

We have that $\phi_u = w_1, \phi_v=w_2$, and since these were chosen to be orthonormal we get that $E=G=1, F=0$.

Exercise. Relate this result to Pythagoras’ theorem.

2. The Cylinder

Let $\phi(u, v) = (\cos u, \sin u, v)$ be a parametrisation of a cylinder (where the domain of $\phi$ is $\{ u \in (0, 2\pi), v \in \mathbb{R} \}$). We have,

$\phi_u = (-\sin u, \cos u, 0), \phi_v=(0, 0, 1)$

and hence we see that $E=G=1$ and $F=0$. NB. The first fundamental form for the cylinder and the plane were the same!

### Relating the First Fundamental Form to Some Things We Care About

We are about to give the arc length of a parametrised curve in a surface $S$, and to give the area of a region $R$ of $S$ in terms of the first fundamental form. The thing to notice is that, in principle, we know how to do these things in $\mathbb{R}^3$ in terms of the standard inner product there. Given that we didn’t do much of anything to get the first fundamental form from that standard inner product (what did we do? We just forgot all those vectors not coming from the tangent planes of our surface!), it  perhaps shouldn’t come as a surprise that we can get at these with our new gadget $I_p$. To illustrate what I mean, let’s consider a motivating example. We know very well how to use the inner product in $\mathbb{R}^3$ to compute the angle between two vectors, in particular, for vectors $w_1, w_2 \in \mathbb{R}^3$ we get

$\cos \theta = \displaystyle\frac{\langle w_1, w_2 \rangle}{|w_1||w_2|}$

where $\theta$ is the angle between them. It shouldn’t be too surprising then to believe that we can compute the angle between tangent vectors to our surface in terms of the first fundamental form. Indeed, let $\alpha, \beta$ be two parametrised curves in $S$ that intersect at $t=0$. We have absolutely no additional work to do in order to say that the angle between them is given by

$\cos \theta = \displaystyle\frac{\langle \alpha'(0), \beta'(0)\rangle}{|\alpha'(0)||\beta'(0)|}$

Now I claim that this is already an expression in terms of the first fundamental form (if we allow the second definition given as a symmetric bilinear form). Going further – let $\phi(u, v)$ be a parametrisation of $S$, then the coordinate curves (i.e., the images of the lines $u=0$ and $v=0$) meet at an angle given by,

$\cos \theta = \displaystyle\frac{\langle \phi_u, \phi_v\rangle}{|\phi_u||\phi_v|}=\frac{F}{\sqrt{EG}}$

Exercise. Give a necessary and sufficient condition for the coordinate curves to be orthogonal for a given parametrisation.

### Relation to Arc Length

Let $\alpha: [0, 1] \rightarrow S$ be a parametrised curve (e.g., $\alpha(t)=(\cos t, \sin t, 0)$ is a parametrised curve in the cylinder). The arc length of $\alpha|[0,t]$ (read, $\alpha$ restricted to $[0, t]$), which we will denote $s(t)$ is given in Hubbard’s book by the formula

$s(t)$= $\displaystyle\int_0^t |\alpha'(t)|dt$

But we can express $|w| = \sqrt{I(w)}$ with the first fundamental form, and so the formula above becomes,

$s(t) = \displaystyle\int_o^t \sqrt{I(\alpha'(t))}dt$

and we have an expression for arc length in terms of the first fundamental form. If our parametrised curve was given by $\alpha(t)=(u(t), v(t))$ we would have,

$s(t) = \displaystyle\int_0^t \sqrt{E(u')^2+2Fu'v'+G(v')^2}dt$

giving us an expression in terms of $E, F, G$.

Exercise. Compute the arc length of $\alpha(t)=(\cos t, \sin t, 0)$ for $t \in (0, 2\pi)$ in the cylinder using the first fundamental form.

### Relation to Area

Consider a parametrisation $\phi:U \rightarrow S$ and a region of $R \subseteq S$ realised as the image of a bounded region $Q \subseteq U$. The way we learnt to compute the area was with the integrand that took a parallelogram and retuned its area. This is where we can use the lemma in the section on the cross product. With that in hand, we can translate exactly the definition you have been given for surface area to read

surface area of $R = \displaystyle\iint_Q |\phi_u \wedge \phi_v|dudv$

Exercise. Make sure you see why these definitions are the same.

Now for the trick. We use the famous formula related to the cross product,

$|\phi_u \wedge \phi_v|^2 = |\phi_u|^2|\phi_v|^2 - \langle \phi_u, \phi_v\rangle^2$

which says exactly that

$|\phi_u \wedge \phi_v|^2 = EG-F^2$

and thus we have that

surface area of $R=\displaystyle\iint_Q\sqrt{EG-F^2}dudv$

### Riemannian Manifolds

As I mentioned in the intro, everything above was concerned with surfaces sitting in $\mathbb{R}^3$. This is a very rigid kind of imposition, and we would like remove it! How should we proceed? We used the standard inner product from $\mathbb{R}^3$ to induce an inner product on our tangent planes $T_pS$. To that end, in order to stop thinking of surfaces as subsets of $\mathbb{R}^3$, but still to have some nice structure on them to be able to talk about lengths, areas, angles etc., we make the following powerful definition,

Definition. A Riemann surface is a surface $S$ with the choice of an inner product $\langle-,-\rangle_p$ at each $T_pS$ ($p \in S$) which is suitably well behaved.

Let me clarify a little what I mean by suitable well behaved. When our surface was sitting in $\mathbb{R}^3$ we have the following upshot, that as we varied the points $p \in S$ a little, the functions $E, F, G$ were well behaved, in fact they were differentiable functions (convince yourselves of this). It is therefore imposed that the inner product on our Riemann surfaces  vary differentiably.

Example. Let $S = \mathbb{R}^2$ (with coordinates $u, v$), but impose a different inner product. Let $E=1, F=0$ but now let $G=e^{2u}$. This is the famous hyperbolic plane. Although the underlying surface was the same, the inner product we imposed (we refer to this as the Riemannian metric) has changed. Eulid famously (and rather impressively) had five axioms of geometry, known as Euclid’s postulates. His fifth postulate was that no two distinct parallel lines can meet. The hyperbolic plane provides an example of a geometry in which this fails!