Surface Tangents and Derivatives

Tangent Plane

The tangent vector to a surface \(\Sigma\) at point \(p\) are tangent vector to all smooth curves on \(\Sigma\) through \(p\) at \(p\).

The tangent plane to a surface \(\Sigma\) at point \(p\) is a 2-dim linear subspace of \(T_p\Sigma\subset\mathbb R^3\), a.k.a. a plane in \(\mathbb R^3\) pass through the origin. its points are precisely all tangent vectors to \(\Sigma\) at \(p\).

Spanning Tangent Plane

Claim Let \(\sigma:U\rightarrow V\cap \Sigma\) be a surface path containing \(p\). If \(U\) is spanned by vectors \(u, v\), then \(T_p\Sigma\) is spanned by \(d_u\sigma, d_v\sigma\).

proof. Consider a surface patch \(\sigma: U\rightarrow V\cap \Sigma\) containing \(p\). Let \(\gamma\) be a smooth curve in \(S\) with some \(\gamma(t_0) = p\). Then, consider \(\mathbf a:[t_0-\epsilon, t_0+\epsilon]\rightarrow U\) s.t. \(\gamma(t) =\sigma(\mathbf a(t)))\), a.k.a. \(\mathbf a\) is a path on \(U\) that is mapped to \(\gamma\). Then, we must have that \(\mathbf a\) is a smooth curve on \(U\).

Let \(a(t) = (u(t), v(t))\), then by chain rule, we have

\[\frac{d\gamma}{dt} = \frac{d\sigma}{d\mathbf a}\frac{d\mathbf a}{dt} = \frac{d\sigma}{du}\frac{du}{dt} + \frac{d\sigma}{dv}\frac{dv}{dt}\]

Therefore, \(\frac{d\gamma}{dt}\) is spanned by \(d_u\sigma,d_v\sigma\).

Conversely, define \(\tilde \gamma(t) = \sigma(u(t_0)+\lambda t, v(t_0) + \mu t)\) for some \(\lambda, \mu\). Then, \(\tilde \gamma\) is a smooth curve in \(\Sigma\) and \(\gamma(0) = p\) we have that

\[\frac{d\tilde \gamma}{dt} = \frac{d\sigma}{du}\lambda + \frac{d\sigma}{dv}\mu\]

Therefore, every vector in \(\text{span}\{d_u\sigma, d_v\sigma\}\) is the tangent vector at \(p\) of \(\Sigma\).

By regularity assumption of \(\sigma\), we have that \(\text{span}\{d_u\sigma, d_v\sigma\}\) are 2-dim, hence spans a plane, in which we call it the tangent plane.

Independence of surface patch choices

Claim For some surface path \(\sigma(u,v)\) at \(p\), the spanned subset space of \(d_u\sigma, d_v\sigma\) independent of the choice of \(\sigma\).

proof. Suppose that \(\tilde \sigma: \tilde U\rightarrow V\cap \Sigma\) is a reparameterization of the surface patch. Then, note that both \(\sigma, \tilde\sigma\) are bijective and smooth. Therefore, exists a bijective smooth map \(\phi: U\rightarrow\tilde U\) s.t. \((\tilde u, \tilde v) = \phi(u,v)\). Then, we have

\[\begin{align*} \sigma(u,v) &= \tilde\sigma(\tilde u(u, v), \tilde v(u, v))\\ \frac{d\sigma}{du} &= \frac{d\tilde \sigma}{d\tilde u}\frac{d\tilde u}{du} + \frac{d\tilde \sigma}{d\tilde v}\frac{d\tilde v}{du}\\ \frac{d\sigma}{dv} &= \frac{d\tilde \sigma}{d\tilde u}\frac{d\tilde u}{dv} + \frac{d\tilde \sigma}{d\tilde v}\frac{d\tilde v}{dv} \end{align*}\]

Therefore, \(d_u\sigma\) and \(d_v\sigma\) are both linear combinations of \(d_{\tilde u}\tilde \sigma, d_{\tilde v}\tilde \sigma\).

Surface Derivatives

Implicit Function Theorem

By inverse function theorem, for non-linear function \(F\)
IF \(F(\mathbf x_0) = \mathbf y_0 ,DF(\mathbf x)\) is invertible, and \(y_0 + DF(x_0)(x-x_0)\) is invertible
THEN, the inverse exists, \(F^{-1}(\mathbf y_0)= \mathbf x_0, \mathbf y = \mathbf x_0 + DF^{-1}|_{\mathbf y_0}(\mathbf y-\mathbf y_0)\)

Regularity Conditions

Claim for any point \(p\) on the regular surface \(\Sigma\), there exists an open set \(V\subset \Sigma\) s.t. \(p\in V\) and \(V\) is the graph of a smooth function \(z=\phi(x,y), y=\varphi(x,z), x=\psi(y,z)\).

In other words, for any points on a surface. Locally, the surface patch can be mapped from a standard plane.

proof. Let \(\sigma^{-1}(p) = (u_0, v_0)\). Consider the surface patch mapping \(\sigma\), by regularity we have that

\[D\sigma = \begin{bmatrix}\partial_u x&\partial_vx\\\partial_uy&\partial_vy\\\partial_u z&\partial_vz\end{bmatrix}\]

such that each of the \(2\times 2\) submatrix are invertible.

WLOG, assume that

\[ \begin{bmatrix} \frac{\partial x}{\partial u} (u_0, v_0)&\frac{\partial v}{\partial u} (u_0, v_0)\\ \frac{\partial z}{\partial u} (u_0, v_0)&\frac{\partial z}{\partial u} (u_0, v_0) \end{bmatrix} \]

is invertible, then consider the orthogonal projection \(\Pi_{XZ}\) of \(V\) to XZ plane.
Consider the composition \(\Pi_{XZ} \circ \sigma: U\rightarrow V\rightarrow W:= (u, v)\rightarrow (x, 0, z)\) where \(W\) is an open subset of XZ plane.

Therefore, the claim is equivalent to that \(\Pi_{XZ}\circ\sigma\) is locally invertible, a.k.a.

\[\exists f: W \rightarrow U. \forall p\in V. f\circ(\Pi_{XZ}\circ\sigma)(u, v) = (u, v)\]

Consider the Taylor expansion of \(\Pi_{XZ}\circ \sigma\) near \((u_0, v_0)\)

\[\begin{align*} \Pi_{XZ}\circ \sigma(u,v) &= \Pi_{XZ}\circ \sigma (u_0, v_0) + D[\Pi_{XZ}\circ \sigma]\vert_{(u_0, v_0)}\begin{bmatrix}u-u_0\\v-v_0\end{bmatrix} + rem.\\ &= \Pi_{XZ}\circ \sigma (u_0, v_0) + \Pi_{XZ}(D\sigma\vert_{(u_0, v_0)}) \begin{bmatrix}u-u_0\\v-v_0\end{bmatrix}+ rem.\\ &= \Pi_{XZ}\circ \sigma (u_0, v_0) + \begin{bmatrix}\frac{\partial x}{\partial u} (u_0, v_0)&\frac{\partial v}{\partial u} (u_0, v_0)\\\frac{\partial z}{\partial u} (u_0, v_0)&\frac{\partial z}{\partial u} (u_0, v_0)\end{bmatrix}\begin{bmatrix}u-u_0\\v-v_0\end{bmatrix} + rem. \end{align*}\]

Therefore, we are approximating the composition by the invertible linear map. Then, by implicit funciton theorem, there exists a unique solution \(y = \psi(x, z)\) near \((x_0, z_0)\) where \(\psi\) is differentiable.

Corollary Regularity condition satisfies IFF the tangent plane at \(p\) exists, and indeed, is a plane.

Smooth Maps

Consider \(\Sigma_1, \Sigma_2\) to be smooth surfaces, and let \(f: \Sigma_1\rightarrow \Sigma_2\) be a map.

For some point \(p_1\in \Sigma_1\), take \(p_2 = f(p_1) \in \Sigma_2\). Let \(\sigma_1: U_1\rightarrow V_1\cap \Sigma_1, p_1\in V_1\) and \(\sigma_2: U_2\rightarrow V_2\cap \Sigma_2, p_2\in V_2\).
Consider the composition function \(g:=\sigma_2^{-1}\circ f\circ \sigma_1: U_1\subset \mathbb R^2\rightarrow U_2\subset \mathbb R^2\), i.e.

\[g(u,v) = \sigma_2^{-1}\circ f\circ \sigma_1 (u,v) = (a(u,v), \beta(u,v))\]

Then, we define \(f\) is differentiable (smooth) at \(p = \sigma(u,v)\) if \(g\) is differentiable as \(Dg = \begin{bmatrix}\partial_u a&\partial_va \\\partial_u \beta&\partial_v\beta\end{bmatrix}\), and \(f\) is a differentiable (smooth) map if \(\forall p\) is differentiable.

If we linearize \(f\) near a point \(p_1\), then we have that

\[Df : T_{p_1}\Sigma_1\rightarrow T_{p_2}\Sigma_2\]

maps from a a tangent plane \(T_{p_1}\Sigma_1 = \text{span}\{\partial_u\sigma_1, \partial_v\sigma_1\}\) to \(T_{p_2}\Sigma_2 = \text{span}\{\partial_u\sigma_2, \partial_v\sigma_2\}\)