First fundamental form

Question

Wolfram MathWorld defines a paraboloid and its differential parameters as

\begin{align*} P&=\left(\frac{\partial x}{du}\right)^2+\left(\frac{\partial y}{du}\right)^2+\left(\frac{\partial z}{du}\right)^2= \\ &=1+\frac{1}{4u} \\ Q&=\frac{\partial x}{du}\frac{\partial x}{dv}+\frac{\partial y}{du}\frac{\partial y}{dv}+\frac{\partial z}{du}\frac{\partial z}{dv}= \\ &=\frac{1}{2\sqrt{u}}(\cos v - \sin v) \\ R&=\left(\frac{\partial x}{dv}\right)^2+\left(\frac{\partial y}{dv}\right)^2+\left(\frac{\partial z}{dv}\right)^2= \\ &=u \\ \end{align*}

Now, if these parameters correspond to the coefficients $E$, $F$ and $G$ described here, I don't understand how they arrived at the expression for $Q$.

I see... Nevermind. – Henrik Schumacher Aug 20 '20 at 20:43 — Henrik Schumacher, Aug 20 '20 at 20:43

score 3 · Accepted Answer · answered Aug 20 '20 at 20:35

3

Despite other comments/answers, these quantities are the usual first fundamental form. Note that the Wiki link defines $g_{ij} = X_i\cdot X_j$. These are the usual $E,F,G$, and they are the dot products of the derivatives of the parametrization with respect to the independent variables. In your case the first parameter is $u$ and the second parameter is $v$, and we do in fact have \begin{align*} P&=X_u\cdot X_u=E,\\ Q&=X_u\cdot X_v=F, \quad\text{and} \\ R&=X_v\cdot X_v=G. \end{align*} I'm not sure why Wolfram is using different letters.

If you want a further reference, check out my differential geometry text.

answered Aug 20 '20 at 20:35

Ted Shifrin

115,160

Thank you for clearing this up! When I plug in the parametrization given in the first link I don't know how they arrive at that non-zero expression for $Q$ (or $F$). I double-checked the partial derivatives but I still don't see it. – user2286339 Aug 20 '20 at 22:11
1

You're right. It's garbage. The parameter curves are orthogonal (as you can see pictorially) and $F=0$. – Ted Shifrin Aug 20 '20 at 22:16
FWIW, I figured it out where MathWorld went wrong: instead of calculating the product of the partial derivatives, e.g. $\frac{\partial x}{\partial u} \frac{\partial x}{\partial v}$, it is $\frac{\partial}{\partial u}\left(\frac{\partial x}{\partial v}\right)$, same for the other terms. Thanks again, Ted. – user2286339 Aug 20 '20 at 23:16

score 1 · Answer 2 · answered Aug 20 '20 at 19:25

The first fundamental form is the inner product of the tangent space in some point of the surface when you consider the surface contained in the ambient space $\mathbb{R}^3$. If you have a paraboloid $z=b(x^2+y^2)$, then the tangent vectors of the surface which generates the tangent space are

$v=[1,0, 2bx]$

and

$w=[0,1,2by]$

At this point the coefficients of the first fundamental form can be computed as follows

$E=\langle v, v \rangle=1+4b^2x^2$

$F=4b^2xy $

$G=1+4b^2y^2$

In your link about paraboloid, I guess the argument is a geodesic on the paraboloid.

First fundamental form

2 Answers2