Check that the parametrization x(u,v)is conformal if and only if E=G and F=0.

Question

Check that the parametrization x(u,v)is conformal if and only if E=G and F=0. I am slightly confused with what this question is asking me. Could someone please walk me through this question. I believe that for --> we need to choose two convenient pairs of orthogonal directions. However, I am unsure of where to start and how to proceed with this question. Thank yoU!

Answer 1

I am simplifying notation and being slightly informal. If anything is unclear, I will be happy to clarify.

I will use the notation $$\mathrm{g}(a,b)=a^T\begin{bmatrix}E & F \\ F & G\end{bmatrix}b.$$

Note that the definition of conformal is precisely that $a\cdot b=\lambda\mathrm{g}(a,b)$, for some positive function $\lambda: (u,v)\mapsto \lambda(u,v)$. Thus, if $E=G$ and $F=0$, then $$\mathrm{g}(a,b)=a^T\begin{bmatrix}E & 0 \\ 0 & E\end{bmatrix}b=E^2a^Tb=E^2a\cdot b.$$ Conversely, if $x$ is conformal, then $a\cdot b=\lambda\mathrm{g}(a,b)=\lambda~a^T\begin{bmatrix}E & F \\ F & G\end{bmatrix}b$, so $\lambda\begin{bmatrix}E & F \\ F & G\end{bmatrix}=\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$, so $F=0$ and $E=G=\frac{1}{\lambda}$.

Answer 2

This question seems to be the same as exercise $2.1.6$ on page $42$ in these nice notes by Theodore Shifrin on a differential geometry course. I will walk through my solution with that in mind.

The material defines conformal on page $40$ as a parametrization that preserves angles: if angles measured in the $uv-$plane agree with corresponding angles in the tangent space $T_PS$ for all points $P$.

We have to fill in with definitions of angles and inner products.

We have a parametrization of the surface $x : U \to S$ with $x = x(u, v)$. Where $U \subset \mathbb{R}^2$ and $S \subset \mathbb{R}^3$.

Define angle $\theta$ between two vectors x, y by

$x \cdot y = |x||y| \cos \theta$

In $U$ use the usual inner product which means that $x \cdot y = x_1 y_1 + x_2 y_2$

In surface $S$ (for vectors of the tangent spaces) use the inner product produced by the first fundamental form (or metric tensor), which means $w \cdot z = \sum_{i, j} g_{ij} w_i z_j$.

Pick a point $Q \in U$ which maps to a point $P \in S$. Pick two vectors x, y based at $Q$. The vectors map componentwise to $T_PS$ like this: $(x_1, x_2) \mapsto x_1 \mathbf{x}_u + x_2 \mathbf{x}_v$, $(y_1, y_2) \mapsto y_1 \mathbf{x}_u + y_2 \mathbf{x}_v$. Where $\mathbf{x}_u$ is the partial derivative $$\left.{\partial \mathbf{x}(u,v) \over \partial u }\right|_Q $$, and so on. (Again on page $40$ of the notes.)

Now we can write out the angle in $U$:

$$ \begin{aligned} \cos \theta_U &= { x \cdot y \over |x| |y| } \\ &= { x_1 y_1 + x_2 y_2 \over \sqrt{(x_1^2 + x_2^2) (y_1^2 + y_2^2) }} \end{aligned} $$

Now we can write out the angle in $S$:

$$\begin{aligned} \cos \theta_S &= { w \cdot z \over |w| |z| } \\ &= { Ex_1 y_1 + F x_1 y_2 + Fx_2 y_1 + Gx_2 y_2 \over \sqrt{(Ex_1^2 + 2Fx_1x_2 + Gx_2^2)(Ey_1^2 + 2Fy_1y_2 + Gy_2^2)} } \\\end{aligned} $$

Where we are using the $E, F, G$ parts of the first fundamental form.

Part 1: Show that if angles are preserved, then $E = G$ and $F = 0$.

Pick these vectors $x = (a, b)$ and $y = (-b, a)$ in $U$ with $a, b \geq 0$, not both zero. These are by design orthogonal in $U$.

$ \cos \theta_U = \dfrac{ x \cdot y}{ \cdots } = 0$ by orthogonality.

We require that $ Ex_1 y_1 + F x_1 y_2 + Fx_2 y_1 + Gx_2 y_2 = 0$ so that the angle in $S$ is correspondingly orthogonal.

We have $Ex_1 y_1 + F x_1 y_2 + Fx_2 y_1 + Gx_2 y_2 = -E ab + Fa^2 - Fb^2 + Gab = 0$.

• Examine the case $a = 1$, $b = 0$ which gives $F = 0$
• Examine the case $a > 0$, $b > 0$ which leads to $(G - E) ab = 0$ and we conclude that $E = G$ (because $ab \neq 0$) to preserve orthogonality.

No further argument is needed for this part. I think this is in the spirit of the hint “pick two orthogonal pairs of vectors” from the question, except slightly more general.

Part 2: Show that if $E = G$ and $F = 0$, angles are preserved.

Use previous expressions for the the angles in $U, S$ and simplify using $E = G$ and $F = 0$. It produces $\cos \theta_U = \cos \theta_S$ because $E$ can be factored from both numerator and denominator and cancels out.