如何求解高斯映射？解析参数化锥面的高斯映射及其导数与形态

Alright, let's break this down step by step to find the Gauss map and its derivative for the given cone parameterized by $\mathbf x(u,v)=(v\cos u, v\sin u, v)$. I'll keep things concrete and tie each step to the geometric intuition you mentioned.

Step 1: Calculate the tangent vectors to the cone #

First, we need the tangent vectors along the $u$ and $v$ parameter directions—these form the basis for the tangent plane at any non-singular point (so excluding the cone tip at $v=0$):

• $\mathbf x_u = \frac{\partial \mathbf x}{\partial u} = (-v\sin u, v\cos u, 0)$: This tangent vector circles around the cone's axis (the z-axis) at a radius $|v|$.
• $\mathbf x_v = \frac{\partial \mathbf x}{\partial v} = (\cos u, \sin u, 1)$: This tangent vector runs along the cone's straight generators, pointing away from the tip.

Step 2: Find the surface normal vector #

To get the normal vector, we take the cross product of the tangent vectors (this gives a vector perpendicular to the tangent plane):
$$
\mathbf x_u \times \mathbf x_v = \begin{vmatrix}
\mathbf i & \mathbf j & \mathbf k \
-v\sin u & v\cos u & 0 \
\cos u & \sin u & 1
\end{vmatrix}
$$
Expanding this determinant simplifies nicely:
$$
\mathbf x_u \times \mathbf x_v = (v\cos u, v\sin u, -v) = v(\cos u, \sin u, -1)
$$
We can ignore the scalar factor $v$ for now (we'll normalize it next), but note that $v \neq 0$—the cone tip has no well-defined normal.

Step 3: The Gauss map (unit normal vector) #

The Gauss map sends each surface point to its unit normal vector, which lives on the unit sphere $\mathbb{S}^2$. Let's normalize the cross product we found:

First, compute the magnitude of the normal vector:
$$
|\mathbf x_u \times \mathbf x_v| = |v| \sqrt{\cos^2 u + \sin^2 u + (-1)^2} = |v| \sqrt{2}
$$
Dividing the normal by its magnitude gives the unit normal $\mathbf N(u,v)$:
$$
\mathbf N(u,v) = \frac{v}{|v|} \cdot \frac{1}{\sqrt{2}} (\cos u, \sin u, -1)
$$

What does this Gauss map look like? #

Here's the key geometric insight you were curious about:

• For $v > 0$ (the "upper" half of the cone, if you imagine the tip at the origin pointing up), $\frac{v}{|v|}=1$, so $\mathbf N(u,v) = \frac{1}{\sqrt{2}}(\cos u, \sin u, -1)$. Notice this doesn't depend on $v$! Every point along a single generator (fixed $u$, varying $v$) maps to the same point on the unit sphere. The entire $v>0$ cone maps to a horizontal circle on $\mathbb{S}^2$: $z = -1/\sqrt{2}$, $x^2 + y^2 = 1/2$.
• For $v < 0$ (the "lower" half cone), $\frac{v}{|v|}=-1$, so $\mathbf N(u,v) = \frac{1}{\sqrt{2}}(-\cos u, -\sin u, 1)$. This maps to another horizontal circle on $\mathbb{S}^2$ at $z = 1/\sqrt{2}$, $x^2 + y^2 = 1/2$.

The cone tip ($v=0$) is excluded since it has no unique normal vector.

Step 4: Derivative of the Gauss map ($d\mathbf N$) #

The derivative of the Gauss map (called the differential $d\mathbf N$) is a linear map from the cone's tangent plane to the unit sphere's tangent plane. We calculate it by taking partial derivatives of $\mathbf N$ with respect to $u$ and $v$, then expressing those in terms of the original tangent vectors.

Let's focus on $v > 0$ first (the $v < 0$ case is analogous with sign changes):

• $\mathbf N_u = \frac{\partial \mathbf N}{\partial u} = \frac{1}{\sqrt{2}}(-\sin u, \cos u, 0)$
• $\mathbf N_v = \frac{\partial \mathbf N}{\partial v} = (0,0,0)$

Now, relate these to the tangent vectors:

• $\mathbf N_v = 0$: This means the derivative of the Gauss map along the generator direction ($v$-direction) is zero. Makes sense—all points along a generator have the same normal, so moving along the generator doesn't change the normal vector.
• For $\mathbf N_u$, notice that $\mathbf x_u = (-v\sin u, v\cos u, 0) = v\sqrt{2} \cdot \mathbf N_u$. Rearranging gives $\mathbf N_u = \frac{1}{v\sqrt{2}} \mathbf x_u$.

In matrix form (using ${\mathbf x_u, \mathbf x_v}$ as the tangent plane basis), the derivative $d\mathbf N$ is:
$$
d\mathbf N = \begin{pmatrix}
\frac{1}{v\sqrt{2}} & 0 \
0 & 0
\end{pmatrix}
$$

Geometric note #

The determinant of this matrix is 0, which tells us the cone has zero Gaussian curvature—a hallmark of developable surfaces (surfaces that can be flattened into a plane without stretching, like a paper cone).

内容的提问来源于stack exchange，提问作者Joojoos