Fundamental form of explicit surface

Question

I'm trying to derive the formulae for the fundamental forms of an explicitly given surface $f(x,y)=z$ however I don't see how to set up my initial parametrisation. My intuition is that perhaps $\sigma (u,v)= (u,v, f(u,v))$ could be a natural starting point but I'm unsure if this is a correct assertion and whether this will induce the fundamental forms in terms of $f$ and it's derivatives.

Answer 1

Nick's answer is great, but in case you aren't comfortable with pull-backs and tensor products, you can use that the first fundamental form is expressed in coordinates by $${\rm d}s^2 = E\,{\rm d}u^2 + 2F\,{\rm d}u\,{\rm d}v + G\,{\rm d}v^2,$$where $E = \langle \sigma_u,\sigma_u\rangle$, $F=\langle \sigma_u,\sigma_v\rangle$ and $G = \langle \sigma_v,\sigma_v\rangle$. You can readily compute $E = 1+f_u^2$, $F = f_uf_v$ and $G = 1 + f_v^2$, so:

$${\rm d}s^2 = (1+f_u^2)\,{\rm d}u^2 + 2f_uf_v\,{\rm d}u\,{\rm d}v + (1+f_v^2)\,{\rm d}v^2.$$

Answer 2

Yes, this is the correct way to go. If the surface is the graph of a smooth function $z = f(x,y)$, then putting graph coordinates on it (as you suggest), via $\sigma(x,y) = (x,y,f(x,y))$ will give the formulas you want. The "First Fundamental Form" is just the Riemannian metric. You can get it, in coordinates, by "pulling back" the standard Euclidean metric of $\Bbb{R}^3$ by $\sigma$. If $g = dx \otimes dx + dy \otimes dy + dz \otimes dz$ is the standard Euclidean metric (standard dot product), then the metric (1st fund. form) $\tilde{g}$ on your surface is the pull back of $g$ by the embedding:

$$ \tilde{g} = \sigma^* g $$

To compute this, note that $\sigma^* dx = dx$, $\sigma^*dy = dy$, and $\sigma^*dz = df = f_x dx + f_y dy$, and substitute: $$ \begin {eqnarray} \tilde{g} &=& dx \otimes dx + dy \otimes dy + df \otimes df \\ &=& (1 + f_x^2) \, dx \otimes dx + (1+f_y^2) \, dy \otimes dy + f_x f_y \left( dx \otimes dy + dy \otimes dx \right) \end{eqnarray} $$

The Gram matrix for $\tilde{g}$ in these coordinates is $$ \left( \begin{array}{cc} 1+f_x^2 & f_x f_y \\ f_x f_y & 1+f_y^2 \end {array} \right) $$