Gaussian Curvatures
Gaussian and Mean Curvatures
Let \(W\) be the Weingarten maps of some oriented surface \(\Sigma\) at point \(p\in \Sigma\). The Gaussian curvature \(K\) and mean curvature \(H\) of \(\Sigma\) at \(p\) are defined as
Theorem 1
Let \(\sigma\) be a surface patch of \(\Sigma\). Then, \(W_{p,\Sigma}\) w.r.t. basis \(\{\sigma_u, \sigma_v\}\) of \(T_p\Sigma\) is
proof. Known that \(W(\sigma_u) = -\mathbf N_u, W(\sigma_v) = -\mathbf N_v\). Let the linear matrix with 4 unknowns \(W = \begin{bmatrix}a&b\\c&d\end{bmatrix}\) so that we have equations
We also have that
Thus, we can solve the 4 unknowns with 4 equations as
Corollary 2
Example: Surface z=f(x,y)
For surface \(\Sigma = \{(x,y,z): z=f(x,y)\}\), define \(\sigma(u,v) = (u,v,f(u,v))\). We have that
Principal Curvatures
Observe that \(W\) is self-adjoint \(2\times 2\) matrix, by eigen decomposition, we have that
Call the eigenvalues \(\kappa_1, \kappa_2\) the principal curvatures and eigen vectors \(\mathbf t_1,\mathbf t_2\) the principal vectors.
In addition, using linear algebra, determinant is the product of all eigen values, trace is the sum of all eigen values,
Also, by eigen decomposition, another corollary is that \(\mathbf t_1,\mathbf t_2\) forms a orthonormal basis of \(T_p\Sigma\).
Finding the principal values and principal vectors follow the standard approach for eigen decomposition, i.e. solve
and each of the corresponding null space on \(T_p\Sigma\).
Euler's Theorem
Theorem Let \(\Sigma\) be an oriented surface, \(\gamma:\mathbb R\rightarrow\Sigma\) be a curve on \(\Sigma\). Then, the normal curvature of \(\gamma\) is
proof. Note that \(\mathbf t_1,\mathbf t_2\) is a orthonormal basis, WLOG assume \(\mathbf t_1\cdot\mathbf t_2 = \pi/2\) (otherwise, flip one of them). Therefore, by the turning angle
and the normal curvature, by bilinearity, is
By orthonormal,
Corollary WLOG assume \(\kappa_1 \geq \kappa_2\), then
proof. By Eulers' Theorem,
Since \(\kappa_1\geq \kappa_2, \sin^2\theta \in [0, 1]\). The maximum is at \(\sin^2\theta = 0\implies \kappa_n = \kappa_1\) and minimum is at \(\sin^2\theta = 1\implies \kappa_n = \kappa_2\)
Example: Helicoid
\(\sigma(u,v) = (v\cos u, v\sin u, \lambda u)\).
Example: Catenoid
\(\sigma(u, v) = (\cosh u\cos v, \cosh u \sin v, u)\)
Umbilics
When \(\kappa_1 = \kappa_2=:\kappa\), we must have that \(W = \kappa I\), where \(\mathbf t\) can be any tangent vector on \(T_p\Sigma\). We define such point \(p\in\Sigma\) being umbilics IFF \(W_{p,\Sigma} = \kappa I\) IFF principal curvatures \(\kappa_1 = \kappa_2\).
Theorem If all points \(p\in \Sigma\) is an umbilic, then \(\Sigma\) is an open subset of a plane or a sphere.
proof. By the assumption, we have that \(W\mathbf t = \kappa\mathbf t\) for all \(p\) and all tangent vector \(\mathbf t(p)\). Parameterize \(\Sigma\) with \(\sigma\), then
Then, note that \(\mathbf N_{uv} = \mathbf N_{vu}\) where
Then, we have that \(\kappa_v \sigma_u = \kappa_u \sigma_v\). However, \(\sigma_u, \sigma_v\) is linearly independent, the equation holds IFF \(\kappa_u = \kappa_v = 0\implies \kappa = c\).
Suppose \(\kappa = 0\), then \(\mathbf N_u = \mathbf N_v = 0\implies \mathbf N\) is constant, hence \(\sigma\) is an open subset of plane.
Suppose \(\kappa \neq 0\), then \(\mathbf N = \kappa\sigma + \mathbf a\), hence \(\sigma\) is an open subset of sphere.
Characterize the Points on the Surface
Principal values provides local information at \(p\) and its neighborhood. Let \(p\in \Sigma\)
By applying suitable translations, Euler angle rotations, reflections around the standard planes, WLOG, assume that
- \(p = \mathbf 0\)
- \(T_p\Sigma = \Pi_{XY}\)
- \(\mathbf t_1 = \mathbf e_1, \mathbf t_2 = \mathbf e_2\)
- \(\mathbf N = \mathbf e_3\)
- \(\sigma_0 = \sigma(0, 0) = p = \mathbf 0\)
Then, the neighborhood \(\sigma(u,v)\) of \(p = \sigma_0\) can be approximated by
The first-orders are on the \(T_p\Sigma = \Pi_{XY}\) plane, and the second order terms are perpendicular to \(xy\) plane, hence
Change the basis from \(\sigma_u, \sigma_v\) to \(\mathbf e_1,\mathbf e_2\), i.e. \(u\sigma_u + v\sigma_v = x\mathbf e_1 + y\mathbf e_2\).
Therefore, we have 4 cases, the point \(p\) is 1. elliptic if \(\kappa_1,\kappa_2\) are both non-zero and have the same sign 2. hyperbolic if \(\kappa_1,\kappa_2\) are both non-zero and have different sign 3. parabolic if exactly one of \(\kappa_1,\kappa_2\) is 0. 4. planar if both of \(\kappa_1,\kappa_2\) are 0.
Line of Curvature
A curve \(\gamma: \mathbb R\rightarrow \Sigma\) is a line of curvature if the tangent vector of \(\gamma\) is a principal vector of \(\Sigma\) at all \(\gamma(t)\).
Claim This definition is equivalent to \(\mathbf N' = -\lambda \gamma'\) for some constant \(\lambda\in\mathbb R\)
proof. By definition of principal vector, \(\forall t. W(\gamma(t)) = \kappa \mathbf t(t)\) for some principal value \(\kappa\). Note that \(\mathbf N'= -W\) and \(\mathbf t = \gamma' / \|\gamma'\|\), take \(\lambda = \kappa / \|\gamma'\|\) we have the equality.
Claim This definition if equivalent to that for \(\gamma(t) = \sigma(u(t), v(t))\)
proof. Note that \(\gamma' = u'\sigma_u + v'\sigma_v\), we have that \(F_I^{-1}F_{II} = \lambda I\implies F_{II} = \lambda F_I\)