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
\[K = \det (W), H = \frac{1}{2}Tr(W)\]
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
\[W = \begin{bmatrix}E&F\\F&G\end{bmatrix}^{-1}\begin{bmatrix}L&M\\M&N\end{bmatrix} = F_I^{-1}F_{II}\]
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
\[-\mathbf N_u = a\sigma_u + b\sigma_v, -\mathbf N_v = c\sigma_u + d\sigma_v\]
We also have that
\[\begin{align*} L &= -\sigma_{u}\mathbf N_u = \sigma_u (a\sigma_u + b\sigma_v) = aE + bF\\ M &= -\sigma_u\mathbf N_v = \sigma_u (c\sigma_u + d\sigma_v) = cE+dF\\ &= -\sigma_v\mathbf N_u = aF+bG\\ N &= -\sigma_v\mathbf N_v = \sigma_v(c\sigma_u + d\sigma_v) = cF + dG \end{align*}\]
Thus, we can solve the 4 unknowns with 4 equations as
\[\begin{bmatrix}L&M\\M&N\end{bmatrix} = \begin{bmatrix}E&F\\F&G\end{bmatrix}\begin{bmatrix}a&b\\c&d\end{bmatrix}\]
Corollary 2
\[\begin{align*} K &= \det(W) = \det(F_I)^{-1}\det(F_{II}) = \frac{LN-M^2}{EG-F^2}\\ H&= \frac{1}{2}tr(F_I^{-1}F_{II}) \\ &= \frac{1}{2(EG-F^2)}tr(\begin{bmatrix}G&-F\\-F&E\end{bmatrix}\begin{bmatrix}L&M\\M&N\end{bmatrix})\\ &= \frac{1}{2(EG-F^2)}tr\begin{pmatrix}LG-MF&MG-NF\\ME-LF&NE-MF\end{pmatrix}\\ &= \frac{LG-2MF+NE}{2(EG-F^2)} \end{align*}\]
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
\[\begin{align*} \sigma_u &= (1, 0, f_x), \sigma_v = (0, 1, f_y)\\ \mathbf N &= \frac{(-f_x,-f_y,1)}{\sqrt{1+f_x^2+f_y^2} }\\ E &= 1+f_x^2, F = f_xfy, G = 1+f_y^2\\ \sigma_{uu} &= (0, 0, f_{xx}), \sigma_{uv} = (0, 0, f_{xy}), \sigma_{vv} = (0, 0, f_{yy})\\ L &= \frac{f_{xx} }{\sqrt{1+f_x^2+f_y^2} }, M = \frac{f_{xy} }{\sqrt{1+f_x^2+f_y^2} }, N = \frac{f_{yy} }{\sqrt{1+f_x^2+f_y^2} }\\ K &= \frac{LN-M^2}{EG-F^2} \\ &= \frac{f_{xx}f_{yy} - f_{xy}^2}{1+f_x^2+f_y^2}\frac{1}{(1+f_x^2)(1+f_y^2) - f_x^2f_y^2}\\ &= \frac{f_{xx}f_{yy} - f_{xy}^2}{(1+f_x^2+f_y^2)^2}\\ H &= \frac{1}{2(1+x^2+y^2)}\frac{(1+f_y^2)f_{xx} - 2f_xf_yf_{xy} + (1+f_x^2)f_{yy} }{\sqrt{1+f_x^2+f_y^2} }\\ &= \frac{(1+f_y^2)f_{xx} - 2f_xf_yf_{xy} + (1+f_x^2)f_{yy} }{2(1+f_x^2+f_y^2)^{3/2} } \end{align*}\]
Principal Curvatures
Observe that \(W\) is self-adjoint \(2\times 2\) matrix, by eigen decomposition, we have that
\[W(\mathbf t_1) = \kappa_1\mathbf t_1, W(\mathbf t_2) = \kappa_2\mathbf t_2\]
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,
\[K = \det(W) = \kappa_1\kappa_2, H = \frac{1}{2}Tr(W) = \frac{\kappa_1+\kappa_2}{2}\]
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
\[\begin{align*}\det(F_I^{-1}F_{II} - \kappa I) &= \det(F_I^{-1}(F_{II} - \kappa F_I))\\ &= \det(F_{II} - \kappa F_I)\\ &= (L-\kappa E)(N-\kappa G) - (M-\kappa F)^2\\ &= 0\end{align*}\]
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
\[\kappa_n = \kappa_1\cos^2\theta + \kappa_2\sin^2\theta, \theta = \arccos(\mathbf t_1\cdot \gamma')\]
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
\[\gamma' = \cos\theta\mathbf t_1 + \sin\theta \mathbf t_2\]
and the normal curvature, by bilinearity, is
\[\kappa_n = II(\gamma', \gamma') = \cos^2\theta II(\mathbf t_1,\mathbf t_1) + 2\sin\theta\cos\theta II(\mathbf t_1,\mathbf t_2) +\sin^2\theta II(\mathbf t_2,\mathbf t_2 )\]
By orthonormal,
\[II(\mathbf t_1, \mathbf t_1) = \mathbf t_1 \kappa_1 \mathbf t_1 = \kappa_1, II(\mathbf t_1, \mathbf t_2) = 0, II(\mathbf t_2, \mathbf t_2) \mathbf t_2 \kappa_2 \mathbf t_2 =\kappa_2\]
Corollary WLOG assume \(\kappa_1 \geq \kappa_2\), then
\[\kappa_1 = \arg\max_{\gamma:\mathbb R\rightarrow \Sigma}\kappa_{n,\gamma}, \kappa_2 = \arg\min_{\gamma:\mathbb R\rightarrow \Sigma}\kappa_{n, \gamma}\]
proof. By Eulers' Theorem,
\[\kappa_n = \kappa_1 - (\kappa_1 - \kappa_2) \sin^2\theta\]
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
\(\sigma(u,v) = (v\cos u, v\sin u, \lambda u)\).
\[\sigma_u = (-v\sin u, v\cos u, \lambda), \sigma_v = (\cos u, \sin u, 0), \mathbf N = \frac{(-\lambda \sin u, \lambda \cos u, -v)}{\sqrt{\lambda^2+v^2} }\]
\[E = v^2 + \lambda^2, F = 0, G = 1\]
\[\sigma_{uu} = (-v\cos u, -v\sin u, 0), \sigma_{uv} = (-\sin u, \cos u, 0), \sigma_{vv} = 0\]
\[L = (1-1)(v\lambda \cos u\sin u) + 0 = 0, M = \frac{\lambda}{\sqrt{\lambda^2+v^2} }, N = 0\]
\[K = \frac{1}{\lambda^2+v^2} \frac{-\lambda^2}{\lambda^2+v^2}= \frac{-\lambda^2}{(\lambda^2+v^2)^2}\]
\[(-\kappa (v^2 + \lambda^2))(-\kappa) - (\frac{\lambda}{\sqrt{\lambda^2+v^2} })^2 = \kappa^2 (v^2 + \lambda^2) - \lambda^2 (v^2 + \lambda^2)^{-1} = 0\]
\[\implies \kappa = \pm\lambda (v^2 + \lambda^2)^{-1}\]
Example: Catenoid
\(\sigma(u, v) = (\cosh u\cos v, \cosh u \sin v, u)\)
\[\sigma_u = (\sinh u\cos v, \sinh u\sin v, 1), \sigma_v = (-\cosh u\sin v, \cosh u\cos v, 0)\]
\[E = \sinh^2 u + 1 = \cosh^2 u, F = (-1 + 1)(\cdots) + 0 = 0, G = \cosh ^2 u\]
\[\mathbf N = \cosh^{-1} u (-cos v, -\sin v, \sinh u)\]
\[\begin{align*} \sigma_{uu} &= (\cosh u\cos v, \cosh u\sin v, 0)\\ \sigma_{uv} &= (-\sinh u \sin v, \sinh u\cos v, 0)\\ \sigma_{vv} &= (-\cosh u \cos v, -\cosh u \sin v, 0) \end{align*}\]
\[\begin{align*} L &= -\cosh^{-1}u(\cosh u) = -1\\ M &= (-1+1)(\cdots) + 0 = 0\\ N &= -\cosh^{-1}u(-\cosh u) = 1 \end{align*}\]
\[K = \frac{-1}{\cosh^4 u}\]
\[(-1 -\kappa \cosh^2 u)(1 - \kappa \cosh ^2 u) = 0\implies \kappa = \pm \cosh^{-2} u\]
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
\[W(\sigma_u) = -\mathbf N_u = \kappa \sigma_u, W(\sigma_v) = -\mathbf N_v = \kappa\sigma_v\]
Then, note that \(\mathbf N_{uv} = \mathbf N_{vu}\) where
\[\mathbf N_{uv} = -\frac{\partial N_u}{\partial v} = -(\kappa_v \sigma_u + \kappa \sigma_{uv}) = -\frac{\partial N_v}{\partial u} = -(\kappa_u \sigma_v + \kappa \sigma_{uv})\]
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
\[\sigma(u,v) = u\sigma_u + v\sigma_v + \frac{1}{2}(u^2 \sigma_{uu} + 2uv\sigma_{uv} + v^2\sigma_{vv}) + rem.\]
The first-orders are on the \(T_p\Sigma = \Pi_{XY}\) plane, and the second order terms are perpendicular to \(xy\) plane, hence
\[\begin{align*} z &= \frac{1}{2}(u^2 \sigma_{uu} + 2uv\sigma_{uv} + v^2\sigma_{vv})\cdot \mathbf N \\ &= \frac{1}{2}(Lu^2 + 2M uv + Nv^2)\\ &= \frac{1}{2}II(u\sigma_u +v\sigma_v, u\sigma_u +v\sigma_v) \end{align*}\]
Change the basis from \(\sigma_u, \sigma_v\) to \(\mathbf e_1,\mathbf e_2\), i.e. \(u\sigma_u + v\sigma_v = x\mathbf e_1 + y\mathbf e_2\).
\[\begin{align*} z &= \frac{1}{2}II(u\sigma_u +v\sigma_v, u\sigma_u +v\sigma_v) \\ &= \frac{1}{2}W( x\mathbf e_1 + y\mathbf e_2)\cdot ( x\mathbf e_1 + y\mathbf e_2) \\ &= \frac{1}{2}(xW(\mathbf t_1) + yW(\mathbf t_2)) \cdot ( x\mathbf e_1 + y\mathbf e_2)\\ &= \frac{1}{2}(\kappa_1 x,\kappa_2 y, 0)\cdot (x, y, 0)\\ &= \frac{\kappa_1 x^2 + \kappa_2x^2}{2} \end{align*}\]
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 \(\gamma(t) = \sigma(u(t), v(t))\)
\[(EM-FL)u'^2 + (EN-GL)u'v' + (FN-GM)v'^2 = 0\]
proof. Note that \(\gamma' = u'\sigma_u + v'\sigma_v\), we have that \(F_I^{-1}F_{II} = \lambda I\implies F_{II} = \lambda F_I\)