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More Examples of Surface Regularity and Smoothness

Examples of Surface Patches

  • \(\sigma(u,v) = (u,v,uv)\) is a regular surface patch
    proof. \(\sigma\) is clearly injective, and \(\sigma_u = (1, 0, u), \sigma_v = (0, 1, v)\) is clearly linearly independent.
  • \(\sigma(u,v) = (u, v^2, v^3)\) is injective, but \(\sigma_u = (1, 0, 0), \sigma_v = (0, 2v, 3v^2)\) is not linearly independent when \(v= 0\), hence not regular
  • \(\sigma(u,v) = (u+u^2 , v, v^2)\) is not injective, for example \(\sigma(0, 0) = \sigma(-1, 0)\). Also, \(sigma_u = (1+2u, 0, 0), \sigma_v = (0, 1, 2v)\) is not linearly independent when \(u=-1/2\).

Example: Ellipsoid

A ellipsoid is defined as

\[\frac{x^2}{p^2} + \frac{y^2}{q^2} + \frac{z^2}{r^2} = 1\]

Similar to how we parameterize a sphere, using the long-lat coordinate, we have

\[\sigma\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix} p\cos\theta\cos\psi\\ q\cos\theta\sin\psi\\ r\sin\theta \end{pmatrix}\]

We will still use the same 2 surface patches to cover the surface.

Then, consider regularity, we have that

\[\sigma_\theta = \begin{pmatrix} -p\sin\theta\cos\psi\\ -q\sin\theta\sin\psi\\ r\cos\theta \end{pmatrix}, \sigma_\psi = \begin{pmatrix} -p\cos\theta\sin\psi\\ q\cos\theta\cos\psi\\ 0 \end{pmatrix}\]

The two vectors are linearly independent for

\[U_1 = \{(\theta, \psi): \theta\in(-\pi/2, \pi/2), \psi\in(0, 2\pi)\}\]

In addition, surface patches are still regular if taken rotations, hence the ellipsoid is smooth and regular.

Example: Torus

A torus is obtained by rotating a circle \(C\) in plane \(\Pi\) around a line \(l \subset \Pi\) and \(l\) does not intersect \(C\). Let \(\Pi = \Pi_{XZ}\), \(l(t) = (0, t, 0)\), take \(a > 0\) be the distance from the center of \(C\) to \(l\) and \(b\) be the radius of \(C\). Then, the torus is a smooth surface with parameterization

\[\sigma(\theta, \varphi) = \begin{pmatrix} (a+b\cos\theta)\cos\varphi\\ (a+b\cos\theta)\sin\varphi\\ b\sin\theta \end{pmatrix}\]

proof. WLOG assume that the center of the circle always reside on the \(XY\) plane \((z=0)\).

Consider the circle defined on the XZ plane as \(\gamma(\theta) = (a + b\cos\theta, 0, b\sin\theta)\).
Then, rotate \(\varphi\) degrees along the \(z\)-axis, the xy coordinate will be

\[\begin{bmatrix}\cos\varphi&-\sin\varphi\\\sin\varphi&\cos\varphi\end{bmatrix}\begin{bmatrix}a + b\cos\theta\\0\end{bmatrix}\]

Therefore we can obtain the parameterization.

Then, note that the torus can be cover by 4 patches. By taking each of \(\theta, \varphi\) in \((0, 2\pi), (-\pi, \pi)\), and we check the regularity conditions by

\[\sigma_\theta = (-b\sin\theta\cos\varphi, -b\sin\theta\sin\varphi, b\cos\theta)\]
\[\sigma_\varphi = (-(a+b\cos\theta)\sin\varphi, (a+b\cos\theta)\cos\varphi, 0)\]

We can easily verify that for each domain, the vectors are linearly independent.

Source code 
import plotly.graph_objects as go
import numpy as np

theta, phi = np.mgrid[0:2*np.pi:20j, 0:2*np.pi:20j]

fig = go.Figure(data=[go.Surface(
    x=(2 + np.cos(theta)) * np.cos(phi), 
    y=(2 + np.cos(theta)) * np.sin(phi), 
    z=np.sin(theta)
    )])
fig.update_traces(showscale=False)
fig.update_layout(margin=dict(l=0, r=0, b=0, t=0))
with open("../assets/torus.json", "w") as f:
    f.write(fig.to_json())

Example: Helicoid

A helicoid can be parameterized as

\[\sigma(u, v) = (v\cos u, v\sin u, \lambda u)\]

Then, we have that

\[\sigma_u = (-v\sin u, v\cos u, \lambda), \sigma_v = (\cos u, \sin u, 0)\]
\[N = \sigma_u \times \sigma_v = (-\lambda \sin u, \lambda \cos u, -v)\]
\[\hat N = \frac{N}{\|N\|} = \frac{1}{\sqrt{\lambda^2 + v^2}} (-\lambda \sin u, \lambda \cos u, -v)\]
Source code 
import plotly.graph_objects as go
import numpy as np

u, v = np.mgrid[0:4*np.pi:20j, 0:4*np.pi:20j]

fig = go.Figure(data=[go.Surface(x=v * np.cos(u), y=v * np.sin(u), z=u)])
fig.update_traces(showscale=False)
fig.update_layout(margin=dict(l=0, r=0, b=0, t=0))
with open("../assets/helicoid.json", "w") as f:
    f.write(fig.to_json())

Example: Tube along some curve

Let \(\gamma\) be a unit-speed space curve with \(\kappa \neq 0\) everywhere. The tube is parameterized by

\[\sigma(s, \theta) = \gamma(s) + a(\mathbf n(s)\cos\theta, +\mathbf b(s)\sin \theta)\]

The tube is hence obtained by a circle of radius \(a\), where the center of the circle moves through \(\gamma\) and the circle is always on the plane perpendicular to the curve.

Claim \(\sigma\) is regular if \(\kappa < a^{-1}\) everywhere

proof. For \(s\in\mathbb R, \theta \in (0, 2\pi)\) (or any open interval of period \(2\pi\))

\[\begin{align*} \sigma_s &= \gamma'(s) + a(\cos\theta \mathbf n'(s) + \sin\theta \mathbf b'(s))\\ &= \mathbf t + a(\cos\theta (-\kappa\mathbf t+ \tau \mathbf b) - \sin\theta \tau\mathbf n)\\ &= (1 - a\cos(\theta)\kappa)\mathbf t + a\cos\theta\tau\mathbf b - a\sin\theta\tau \mathbf n\\ \sigma_\theta &= a\cos\theta \mathbf b - a\sin \theta \mathbf n\\ \sigma_s \times \sigma_\theta &= (1 - a\cos(\theta)\kappa)(a\cos\theta)(\mathbf t\times \mathbf b) - (1 - a\cos(\theta)\kappa)(a\sin\theta)(\mathbf t\times \mathbf n)\\ &= -a(1-a\kappa\cos\theta)(\cos\theta\mathbf n + \sin\theta\mathbf b) \end{align*}\]

Then, \(\sigma\) is regular if \(1-a\kappa\cos\theta\neq 0\), note that if \(\cos\theta\in (-1, 1)\), so that \(a\kappa < 1 \implies 1-a\kappa\cos\theta > 0\)

Tangent and Derivatives

Find the tangent plane at given point

\(\sigma(u,v) = (u, v, u^2 - v^2), (1, 1, 0)\)

\[\sigma_u(1, 1)= (1, 0, 2u) = (1, 0, 2), \sigma_v(1, 1) = (0, 1, 2v) = (0, 1, 2)\]
\[\sigma_u\times \sigma_v(1, 1) = (-2, 2, 1)\implies -2x+2y+z =0\]

\(\sigma(r,\theta) = (r\cosh \theta, r\sinh\theta, r^2), (1, 0, 1)\)

\[\sigma_r(1, 0) = (\cosh \theta, \sinh\theta, 2r) = (1, 0, 2)\]
\[\sigma_\theta(1, 0) = (r\sinh\theta, r\cosh\theta, 2) = (0, 1, 2)\]
\[\sigma_u\times \sigma_v(1, 1) = (-2, -2, 1)\implies -2x-2y+z =0\]

Claim If \(f:\Sigma_1\rightarrow\Sigma_2\) is a local diffeomorphism and \(\gamma\) is a regular curve on \(\Sigma_1\). Then \(f\circ \gamma\) is regular on \(\Sigma_2\).

proof. Let \(\tilde \gamma := f\circ\gamma : \mathbb R\rightarrow \Sigma_2\)

\[\frac{d\tilde\gamma}{dt} = D_\gamma f\frac{d\gamma}{dt}\]

where \(D_\gamma f\) is locally intertible and \(\gamma' \neq 0\) by regularity assumption. Therefore, \(\tilde \gamma\) is also regular.

Surface of Revolution and Ruled Surfaces

Example: Catenoid

The surface is obtained by rotating \(x=\cosh z\) in xz-plane around z-axis.

Take \(\gamma(t) = (\cosh t, 0, t)\) so that

\[\sigma(t,\theta) = (\cosh t\cos \theta, \cosh t\sin\theta, t)\]

Note that we need at least 2 patches \(t\in\mathbb R, \theta\in (0,2\pi)\) or \(\theta\in(-\pi,\pi)\)

Source code 
import plotly.graph_objects as go
import numpy as np

t, theta = np.mgrid[-1:1:10j, 0:1.5 * np.pi:50j]

fig = go.Figure(data=[go.Surface(
    x=np.cosh(t) * np.cos(theta), 
    y=np.cosh(t) * np.sin(theta), 
    z=t)])
fig.update_traces(showscale=False)
fig.update_layout(margin=dict(l=0, r=0, b=0, t=0))
with open("../assets/catenoid.json", "w") as f:
    f.write(fig.to_json())

Example: Mercator's Projection

Show that \(\sigma(u,v) = (\text{sech} u\cos v, \text{sech} u\sin v, \text{tanh} u)\) is a regular surface patch for \(\Sigma = \{x^2 + y^2 + z^2 = 1\}\).

\[\begin{align*} \|\sigma\|^2 &= \text{sech}^2 u\cos^2 v + \text{sech}^2 u\sin^2 v + \text{tanh}^2 u\\ &= \text{sech}^2 u + \text{tanh}^2 u\\ &= 1\\ \sigma_u &= (-\text{sech}(u)\text{tanh}(u)\cos v, -\text{sech}(u)\text{tanh}(u)\sin v, \text{sech}^2 u)\\ \sigma_v &= (-\text{sech} u\sin v, \text{sech} u\cos v, \text{tanh} u)\\ \sigma_u\times \sigma_v &= -\text{sech}^2 u (\text{sech} u\cos v, \text{sech} u\sin v, \text{tanh} u) \neq 0 \end{align*}\]