Surface Curvatures

Second Fundamental Form

To define the curvature of a surface, the first attempt is to imitate the curvature of a curve. In other words, how quick the surface moves away from the plane parallel to the tangent plane and intersecting \(p\).

For a surface \(\Sigma\) parameterized by \(\sigma\), Consider a point \(p = \sigma(u,v)\) and the tangent plane at \(p\) with unit normal \(\mathbf N\). The intuition above leads to

\[(\sigma(u+\Delta u, v+\Delta v) - \sigma(u,v))\cdot \mathbf N\]

Using Taylor expansion, \((\sigma(u+\Delta u, v+\Delta v) - \sigma(u,v))\) is linearly approximated (first two orders) as

\[(\sigma_u u' + \sigma_v v') + \frac{1}{2}(\sigma_{uu}u'^2 + 2\sigma_{uv}u'v' + \sigma_{vv}v'^2)+ rem.\]

Then, note that \(N \parallel (\sigma_u\times\sigma_v)\) and \(rem.\) approches 0, so that the equation becomes

\[\frac{1}{2}(\sigma_{uu}\cdot \mathbf N u'^2 + 2\sigma_{uv} \cdot \mathbf N u'v' + \sigma_{vv}\cdot \mathbf N v'^2)\]

Let \(L = \sigma_{uu}\cdot \mathbf N , M = \sigma_{uv}\cdot \mathbf N, N = \sigma_{vv}\cdot \mathbf N\) so that we can describe curvature by

\[Lu'^2 + 2M u'v' + N v'^2\]

which is called the second fundamental form of the surface patch \(\sigma\).

Example: Surface of Revolution

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

with assumption that \(f>0\) and \(\gamma(u) = (f(u), 0, g(u))\) is unit-speed.

Known that \(\sigma_u = (f'\cos v, f'\sin v, g'), \sigma_v = (-f\sin v, f\cos v, 0)\).

\[\begin{align*} \mathbf N &= \sigma_u\times \sigma_v = (-fg'cos v, -f g'\sin v, ff')\\ \|\mathbf N\| &= \sqrt{f^2g'^2(\cos^2 v + \sin^v) + f^2f'^2} = f\sqrt{f'^2+g'^2} = f\\ \hat{\mathbf N} &= (-g'\cos v, -g'\sin v, f')\\ L &= \sigma_{uu}\cdot \hat{\mathbf N} \\ &= (f''\cos v, f''\sin v, g'')\cdot (-g'\cos v, -g'\sin v, f')\\ &= f'g'' - f''g'\\ M &= \sigma_{uv}\cdot \hat{\mathbf N} \\ &= (-f'\sin v, f'\cos v, 0)\cdot (-g'\cos v, -g'\sin v, f')\\ &= 0\\ N &= \sigma_{vv}\cdot \hat{\mathbf N} \\ &= (-f\cos v, -f\sin v, 0)\cdot (-g'\cos v, -g'\sin v, f')\\ &= fg' \end{align*}\]

The second fundamental form is

\[(f'g''-f''g')u'^2 + fg' v'^2\]

For sphere using lat-long coordinate parameterization, inserting \(f = \cos u, g = \sin u\) so that the second fundamental form is

\[u'^2 + \cos^2(u)v'^2\]

which is the same as the first fundamental form.

Example: elliptic paraboloid

\[\sigma(u, v) = (u,v,u^2+v^2)\]

\[\begin{align*} \mathbf N &= \frac{\sigma_u\times \sigma_v}{\|\sigma_u\times \sigma_v\|}\\ &= \frac{(1, 0, 2u)\times (0, 1, 2v)}{\|(1, 0, 2u)\times (0, 1, 2v)\|}\\ &= \frac{1}{\sqrt{4u^2+4v^2+1} }(-2u, -2v, 1)\\ L &= \sigma_{uu} \cdot\mathbf N \\ &= \frac{1}{\sqrt{4u^2+4v^2+1} }(0, 0, 2)\\ N &= \sigma_{vv} \cdot\mathbf N \\ &= \frac{1}{\sqrt{4u^2+4v^2+1} }(0, 0, 2)\\ M &= \sigma_{uv}\cdot {\mathbf N } = 0 \end{align*}\]

Reparameterization

Claim If \(\tilde \sigma\) is a reparameterization of a surface patch \(\sigma\) with map \((u,v) = \Phi(\tilde u, \tilde v)\), then

\[\begin{bmatrix}\tilde L &\tilde M\\\tilde M&\tilde N\end{bmatrix} = \pm J(\Phi)^T \begin{bmatrix}L&M\\M&N\end{bmatrix}J(\Phi)\]

First, by reparameterization we have that \(\tilde{\mathbf N} = \pm \mathbf N\), then given that

\[\tilde\sigma(\tilde u, \tilde v) = \sigma(\Phi(\tilde u, \tilde v))\]

Differentiate both sides we have that

\[\frac{d\tilde \sigma}{d\tilde u} = \frac{d\sigma}{d u}\frac{d u}{d\tilde u} + \frac{d\sigma}{d v}\frac{d v}{d\tilde u}, \frac{d\tilde \sigma}{d\tilde v} = \frac{d\sigma}{d u}\frac{d u}{d\tilde v} + \frac{d\sigma}{d v}\frac{d v}{d\tilde v}\]

Differentiating twice, we have that

\[\begin{align*}\frac{d^2\tilde \sigma}{d \tilde u^2} &= \frac{d}{d\tilde u}(\frac{d\sigma}{d u}\frac{d u}{d\tilde u} + \frac{d\sigma}{d v}\frac{d v}{d\tilde u})\\ &= \frac{d\sigma}{d u}\frac{d^2 u}{d\tilde u^2} + \frac{d}{d\tilde u}(\frac{d\sigma}{d u})\frac{d u}{d\tilde u} + \frac{d\sigma}{d v}\frac{d^2 v}{d\tilde u^2} + \frac{d}{d\tilde u}(\frac{d\sigma}{d v})\frac{d v}{d\tilde u}\\ \frac{d}{d\tilde u}(\frac{d\sigma}{d u})&= \frac{d^2\sigma}{d u^2}\frac{d u}{d \tilde u} + \frac{d^2\sigma}{d udv}\frac{d v}{d \tilde u}\\ \frac{d}{d\tilde u}(\frac{d\sigma}{d v})&= \frac{d^2\sigma}{d v^2}\frac{d v}{d \tilde u} + \frac{d^2\sigma}{d udv}\frac{d v}{d \tilde u} \end{align*}\]

Inserting all differentials back with simplified notations

\[\tilde{\sigma}_{\tilde u\tilde u} = \sigma_u \frac{d^2u}{d\tilde u^2} + \sigma_v \frac{d^2u}{d\tilde u^2} + \sigma_{uu}(\frac{d u}{d\tilde u})^2 + 2\sigma_{uv}\frac{d u}{d\tilde u}\frac{d v}{d\tilde u} + \sigma_{vv}(\frac{d v}{d\tilde u})^2\]

Note that \(\tilde{\mathbf N} = \pm \mathbf N, \sigma_u\cdot \mathbf N = \sigma_v\cdot \mathbf N = 0\), we have that

\[\tilde L = \tilde{\sigma}_{\tilde u\tilde u} \cdot \tilde{\mathbf N} = L(\frac{d u}{d\tilde u})^2 + 2M\frac{d u}{d\tilde u}\frac{d v}{d\tilde u} +N(\frac{d v}{d\tilde u})^2\]

Similarly, we can obtain \(\tilde M,\tilde N\)

Example: Plane

Claim \(\sigma\) parameterizes an open subset of a plane IFF the second fundamental form is zero everywhere.

proof. \(\Rightarrow\) Note that if one parameterization of the plane has second fundamental form being 0, then all of its reparameterizations have 0 second fundamental form by the reparameterization theorem (\(\pm J^T \mathbf 0 J = \mathbf 0\) for any \(J\)).
We simply take \(\sigma(u,v) = u\mathbf a + v\mathbf b + \mathbf c\), so that \(\sigma_u = \mathbf a, \sigma_v = \mathbf b, \sigma_{uu} = \sigma_{uv} = \sigma_{vv} = 0\), hence its second fundemental form is \(0\).

\(\Leftarrow\) Assume that \(\sigma_{uu}\cdot \mathbf N = \sigma_{uv}\cdot \mathbf N = \sigma_{vv}\cdot \mathbf N = 0\). Note that by product rule

\[d_u (\sigma_u \cdot\mathbf N) = \sigma_u \cdot \mathbf N_u + \sigma_{uu}\cdot\mathbf N\]

while we also have that \(\sigma_u \cdot\mathbf N = 0\) since they are perpendicular, hence

\[\sigma_u \cdot \mathbf N_u = -\sigma_{uu}\cdot\mathbf N = 0\]

Similarly, we can obtain that

\[\sigma_u \cdot \mathbf N_u = \sigma_u \cdot \mathbf N_v= \sigma_v \cdot \mathbf N_u= \sigma_v \cdot \mathbf N_v = 0\]

Also, since \(\mathbf N\) is a unit vector, its partial derivatives

\[\mathbf N \cdot \mathbf N_u = \mathbf N \cdot \mathbf N_v = 0\]

However, \(N\) is perpendicular to \(\sigma_u, \sigma_v\) so that we must have that \(\mathbf N_u = \mathbf N_v = 0\), hence \(\mathbf N(u,v)\) is a constant.

Gauss and Weingarten Maps

Another approach for defining curvature is the speed of change in the unit normal \(\mathbf N\). Note that the values of \(\mathbf N\) on a surface belongs to a unit sphere \(S^2\). Therefore, we define the Gauss map

\[G: \Sigma\rightarrow S^2, G(p) = \mathbf N(p)\]

The rate of change of normal thus becomes

\[D_pG : T_p \Sigma\rightarrow T_{N(p)}S^2\]

Note that the tangent plane \(T_{N(p)}S^2\) is the tangent plane perpendicular to \(N(p)\), which is the same plane as \(T_p\Sigma\).

Define the Weingarten map

\[W_p = -D_pG\]

and define the blinear form of second fundamental form of \(\Sigma\) at \(p\) by

\[\langle\langle v, w\rangle\rangle = \langle W_p(v), w\rangle\]

We will then show how Weingarten map is related to the second fundamental form.

Lemma \(\mathbf N_u\cdot \sigma_u = -L, \mathbf N_u\cdot \sigma_v = \mathbf N_v\cdot \sigma_u = -M, \mathbf N_v\cdot \sigma_v = -N\).

proof. Note that \(\mathbf N = \frac{\sigma_u\times \sigma_v}{\|\sigma_u\times \sigma_v\|}\) is perpendicular to both \(\sigma_u, \sigma_v\), we have that

\[N\cdot \sigma_u = N\cdot\sigma_v = 0\]

Differentiating \(N\cdot \sigma_u\), we have that

\[\begin{align*} \frac{\partial}{\partial u}(\mathbf N\cdot\sigma_u) &= \mathbf N_u\cdot \sigma_u + \mathbf N\cdot\sigma_{uu} = 0\\ \mathbf N_u\cdot \sigma_u &= - \mathbf N\cdot\sigma_{uu} = -L \end{align*}\]

Similarly, differentiate wrt. \(u, v\) on each of \(N\cdot \sigma_u, N\cdot\sigma_v\) will get the claim.

Theorem The inner product is a symmetric bilinear form

\[\langle\langle v, w\rangle\rangle_{p} = \begin{bmatrix}c\\d\end{bmatrix}\begin{bmatrix}L&M\\M&N\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix} = Lv_u w_u + M(v_u + w_v + v_v + w_v) + Nv_vw_v\]

For \(v = a \sigma_u + b\sigma_v, w = c \sigma_u + d\sigma_v\).

Note that both sides of the equation is bilinear form on \(T_p\Sigma\), we only need to prove on the set \(\{\sigma_u, \sigma_v\}\).
Also, note that \(W_p = -D_pG = D_p \mathbf N\), so that

\[\frac{\partial W_p}{\partial_u} = -\frac{\partial G}{\partial u} = -\frac{\partial\mathbf N}{\partial u}, \frac{\partial W_p}{\partial_v} = -\frac{\partial \mathbf N}{\partial_u}\]

Therefore, using the lemma above

\[\langle\langle \sigma_u, \sigma_u\rangle\rangle = \langle W_p(v), w\rangle = \langle -\mathbf N_u, \sigma_u\rangle = \mathbf N \cdot \sigma_{uu}\]

And we can obtain the similar results for

\[\langle\langle \sigma_u, \sigma_v\rangle\rangle = \langle\langle \sigma_v, \sigma_u\rangle\rangle = M, \langle\langle \sigma_v, \sigma_v\rangle\rangle = N\]

Corollary Weingarten map is self adjoint. \(\langle W_p(v), w\rangle = \langle v, W_p(w)\rangle\)

Normal and Geodesic Curvatures

The third attempt to define curvature is to define via the curvature of curves on the surface.

Let \(\gamma: \mathbb R\rightarrow\Sigma\) be a unit-speed curve on \(\Sigma\), parameterized by \(\sigma\). Let \(\mathbf t = \gamma', \mathbf n = \frac{\gamma''}{\|\gamma''\| }, \mathbf b = \mathbf t\times \mathbf n\) be the unit tangent, unit normal, and binormal, which forms the orthonormal basis.
Also, let \(\mathbf N\) be the unit normal, hence \(\{\mathbf N, \mathbf t, \mathbf N\times \mathbf t\}\) forms another orthonormal basis.
Then, note that \(\gamma\) is unit-speed, hence \(\gamma''\perp \mathbf t\), thus, \(\gamma''\) must resides on the plane spanned by \(\mathbf N, \mathbf N\times \mathbf t\), in other words

\[\gamma'' = \kappa_n \mathbf N + \kappa_g(\mathbf N\times\mathbf t)\]

Define \(\kappa_n\) be the normal curvature and \(\kappa_g\) be the geodesic curvature of \(\gamma\).

Since \(\{\mathbf N, \mathbf t, \mathbf N\times \mathbf t\}\) forms a orthonormal basis, by Pythagorean theorem or trig. identities, we can easily obtain

\[\kappa^2 = \kappa_n^2 + \kappa_g^2\]

\[\kappa_n = \gamma''\cdot \mathbf N = \kappa\cos\psi, \kappa_g = \gamma'' \cdot (\mathbf N\times \gamma') = \pm\kappa\sin\psi\]

for \(\psi = \arccos(\mathbf N\cdot\mathbf n)\)

Theorem \(\kappa_n\) is invariant of reparameterization of unit-speed \(\gamma\), while \(\kappa_g\) changes up to the sign.

proof. Let \(\gamma, \tilde\gamma\) be two unit-speed parameterizations of the same curve, then the map can only be \(\tilde t = \pm t + c\).

Theorem If \(\gamma:\mathbb R\rightarrow\Sigma\) is unit-speed on an oriented surface \(\Sigma\), then

\[\kappa_n = II(\gamma',\gamma') = Lu'^2 + 2Mu'v' + Nv'^2\]

proof. Let \(\gamma(t) = \sigma(u(t), v(t))\). Note that \(\mathbf t = \gamma'\) is a tangent vector to \(\Sigma\),

\[\begin{align*} \mathbf N\cdot \gamma' &= 0\\ \frac{d}{dt}(N\cdot \gamma') = \mathbf N\cdot \gamma'' + \mathbf N'\cdot \gamma' &= 0\\ \mathbf N\cdot \gamma'' &= -\mathbf N'\cdot \gamma' \end{align*}\]

Therefore, we have that

\[\kappa_n = \mathbf N\cdot \gamma'' = -\mathbf N'\cdot \gamma' = W(\gamma')\cdot\gamma' = II(\gamma',\gamma')\]

Theorem If \(\gamma\) is regular (not necessarily unit-speed), then

\[\kappa_n = \frac{II(\gamma',\gamma')}{I(\gamma',\gamma')}. \kappa_g = \frac{\gamma''\cdot (\mathbf N\times \gamma')}{I(\gamma',\gamma')^{3/2} }\]

proof. Let \(\hat\gamma\) be the unit-speed parameterization of \(\gamma, \gamma(t) = \hat\gamma(s(t))\) where \(s\) is the arc-length of \(\gamma\), and \(s'\) is the speed
Therefore, we have that

\[\frac{d\gamma}{dt} = \frac{d\hat\gamma}{ds}\frac{ds}{dt}\implies \hat\gamma' = \frac{\gamma'}{s'}\]

\[\frac{d^2\gamma}{dt^2} = \frac{d^2\hat\gamma}{ds^2}(\frac{ds}{dt})^2 + \frac{d\hat\gamma}{ds}\frac{d^2s}{dt^2}\implies \hat\gamma'' = s'^{-2}(\gamma'' - \hat\gamma's'')\]

Therefore,

\[\begin{align*} \kappa_n &= II(\hat\gamma',\hat\gamma') \\ &= II(\frac{d\gamma}{dt}(\frac{ds}{dt})^{-1}, \frac{d\gamma}{dt}(\frac{ds}{dt})^{-1}) \\ &= \frac{II(\gamma', \gamma')}{s'(t)^2} \\ &= \frac{II(\gamma',\gamma')}{I(\gamma',\gamma')}\\ \kappa_g &= \hat\gamma''\cdot (\mathbf N\times \hat\gamma') \\ &= s'^{-2}(\gamma'' - \hat\gamma's')\cdot (\mathbf N \times s'^{-1}\gamma')\\ &= s'^{-3}\gamma''\cdot (\mathbf N \times \hat\gamma')\\ &= I(\gamma',\gamma')^{-3/2}\gamma''\cdot (\mathbf N \times \hat\gamma') \end{align*}\]

Example: Sphere

Claim For any curve \(\gamma: \mathbb R\rightarrow S^2_R\) on the sphere of radius \(R\), \(\kappa_n = \pm R^{-1}\).

proof. WLOG assume that the sphere is centered at origin and \(\gamma\) is unit-speed. Then, the unit normal at point \(p = \gamma(t) \in S^2\) is \(\pm\frac{\gamma(t)}{R}\).
Also, we have that \(\gamma\cdot \gamma = R^2\), differentiate both side gives that

\[2\gamma'\cdot \gamma = 0\]

Differentiate twice gives that

\[\gamma''\cdot \gamma + \gamma'\cdot\gamma' = 0\implies \gamma''\cdot \gamma' = -1\]

Therefore, the normal curvature

\[\kappa_n = \mathbf N \cdot \gamma'' = \pm\frac{\gamma}{R}\cdot \gamma'' = \pm R^{-1}\]

Claim For any circle \(c\) on the sphere of radius \(R\), its geodesic curvature is

\[\kappa_g = \pm r^{-1}\sin(\frac{r}{R})\]

proof. Note that \(\kappa_g\) is invariant of rotations, WLOG assume that \(c(\theta) = (r\cos\theta, r\sin\theta, \sqrt{R^2-r^2})\) where \(0 < r\leq R\), then we have the unit normal for curve \(c\) being

\[\mathbf n(\theta) = \frac{c''}{\|c''\|} = \frac{(-r\cos\theta, -r\sin\theta, 0)}{r} = (-\cos\theta, -\sin\theta, 0)\]

The unit normal for sphere at \(c(\theta)\) is simply \(\pm (R^{-1} c(\theta)) = \pm(\frac{r}{R}\cos\theta, \frac{r}{R}\sin\theta, \sqrt{1-\frac{r^2}{R^2} })\).
Known that the curvature for a circle of radius \(r\) is \(r^{-1}\), we have that

\[\kappa_g = \pm \kappa \sin (\mathbf n\cdot\mathbf N) = \pm r^{-1}\sin(\frac{r}{R})\]

Asymptotic curves

A curve \(\gamma:\mathbb R\rightarrow\Sigma\) is called asymptotic if \(\kappa_n = 0\).

For example, any straight line on a surface is asymptotic, since \(\kappa_n = \mathbf N\cdot \gamma'' = \mathbf N \cdot 0 = 0\).

Theorem \(\gamma\) with positive curvature is aymptotic IFF its binormal \(\mathbf b\) is parallel to the unit normal of \(\Sigma\) for all points of \(\gamma\).

proof. Since \(\mathbf b = \mathbf t\times \mathbf n\), \(\mathbf b \parallel \mathbf N\) so that \(\mathbf N \perp \mathbf n\) implying that \(\kappa_n = \kappa (\mathbf N \cdot \mathbf n) = \kappa 0 = 0\)