Why first fundamental form?

Question

Here is an excerpt from the notes we are using:

The first fundamental form dictates how one computes dot products of vectors tangent to the surface assuming they are expanded according to the basis $\frac{\partial q}{\partial u},\frac{\partial q}{\partial v}$.

In particular, we see that while the metric coefficients depend on our parametrization, the dot product $\text{I} (X, Y )$ of two tangent vectors remains the same if we change parameters.

I assume that first fundamental form is really a map from $T_pM \times T_pM\to \text{R}$, but I don't understand what the first part is talking about. Why would we need a matrix to tell us how to perform dot products? I mean, suppose we have vectors $<a,b,c>$ and $<d,e,f>$, we just multiply them term by term, isn't that correct? There is a calculation in the notes, showing that $\text{I}(X,Y)=X \cdot Y$, that even made me more confused.

Answer 1

$\newcommand{\Reals}{\mathbf{R}}\newcommand{\dd}{\partial}$If you have a surface embedded in $\Reals^{3}$, you can (as you note) use the "ambient" Euclidean inner product to take dot products of tangent vectors. However, that's mild overkill; in order to take dot products of tangent vectors to a surface, all you really need is the first fundamental form.

If $$ E = \left\langle\frac{\dd q}{\dd u}, \frac{\dd q}{\dd u}\right\rangle,\quad F = \left\langle\frac{\dd q}{\dd u}, \frac{\dd q}{\dd v}\right\rangle,\quad G = \left\langle\frac{\dd q}{\dd v}, \frac{\dd q}{\dd v}\right\rangle, $$ and if $$ v = a_{1}\, \frac{\dd q}{\dd u} + a_{2}\, \frac{\dd q}{\dd v},\qquad w = b_{1}\, \frac{\dd q}{\dd u} + b_{2}\, \frac{\dd q}{\dd v}, $$ i.e., if $v = (a_{1}, a_{2})$ and $w = (b_{1}, b_{2})$ with respect to the coordinate basis fields, then $$ \langle v, w\rangle = a_{1} b_{1} E + (a_{1} b_{2} + a_{2} b_{1}) F + a_{2} b_{2} G. $$ That's the content of the first paragraph of your excerpt. The second paragraph asserts that $\langle v, w\rangle$ is independent of the choice of local coordinates. This is in some sense "obvious" if you know you have a surface embedded in $\Reals^{3}$, but it's surprising if you think of the first fundamental form as "extra structure" imposed on an open set in $\Reals^{2}$, i.e., in a coordinate neighborhood.

Remarkably, there's non-trivial "intrinsic" geometry of a surface captured by its first fundamental form, even though the first fundamental form doesn't uniquely determine an embedding of the surface in $\Reals^{3}$.

Answer 2

Let $\langle \cdot,\cdot \rangle_{\Bbb R^3}$ denote the usual dot product in $\Bbb R^3$. Recall that one way of seeing $T_{\bf p}M$ is as pairs $({\bf p},{\bf v})$, where ${\bf v}$ is the velocity of some curve in $M$. Define: $${\rm I}_{\bf p} = \langle \cdot, \cdot\rangle_{\bf p}: T_{\bf p}M \times T_{\bf p}M \to \Bbb R, \qquad {\rm I}_{\bf p}({\bf v},{\bf w}) = \langle ({\bf p},{\bf v}),({\bf p},{\bf w})\rangle_{\bf p} := \langle {\bf v},{\bf w}\rangle_{\Bbb R^3}.$$

Once this is understood, we drop the pairs and look only at ${\bf v}$ and ${\bf w}$. The value of $\langle \cdot, \cdot \rangle_{\Bbb R^3}$ on any pair of vector does not depend on which base you are using, hence the same goes for ${\rm I}_{\bf p}$. So, we use the basis for $T_{\bf p}M$ associated to the parametrization $q: U \subset \Bbb R^2 \to M \subset \Bbb R^3$ we have in our hands, that is: $$\frac{\partial q}{\partial u}(q^{-1}({\bf p}))\quad \text{ and } \quad\frac{\partial q}{\partial v}(q^{-1}({\bf p})).$$ Since we're just too lazy, we'll drop the point of application $q^{-1}({\bf p})$, and while we're at it, might as well abbreviate the derivatives by $q_u$ and $q_v$. Call: $$E = \langle q_u,q_u\rangle, \quad F = \langle q_u,q_v\rangle, \quad G = \langle q_v, q_v \rangle.$$ Then if ${\bf v} = a q_u + bq_v$, we get: $$\langle {\bf v},{\bf v}\rangle = Ea^2+2Fab+Gb^2.$$ Take another parametrization $\overline{q} = \overline{q}(\overline{u},\overline{v})$, write ${\bf v} = \overline{a} \overline{q}_{\overline{u}} + \overline{b}\overline{q}_{\overline{v}}$ and try to check that: $$Ea^2+2Fab+Gb^2=\overline{E}\overline{a}^2+2\overline{F}\overline{a}\overline{b}+\overline{G}\overline{b}^2.$$ See how the barred and unbarred objects will be related, using the jacobian of $q \circ \overline{q}^{-1}$ at ${\bf p}$.