The Tangent Plane and Differential of a Map

The tangent plane and differential of a map are the entry point to essentially all of modern geometry. In the First Fundamental Form, the inner product on $T_pS$ inherited from $\mathbb R^3$ is used to measure lengths of curves and angles between tangent vectors directly in terms of the coefficients $E, F, G$ built from $\mathbf x_u, \mathbf x_v$ — this is impossible to even state without the tangent plane machinery developed here. The Gauss Map $N: S \to S^2$ sends each point to its unit normal, and its differential $dN_p : T_pS \to T_pS$ (using the natural identification $T_{N(p)}S^2 \cong T_pS$ for surfaces with unit normal, since both are the orthogonal complement of $N(p)$) is precisely the shape operator whose eigenvalues are the principal curvatures — the entire theory of curvature of surfaces is a direct descendant of the differential defined here. In physics, the tangent plane to a configuration space or constraint surface encodes the allowed instantaneous velocities of a constrained mechanical system (e.g., a bead on a wire, a rigid body's configuration space), and Lagrangian mechanics is built on exactly this linearization. In computer graphics and geometric modeling, tangent planes are used for normal-vector computation, shading (Phong/Gouraud shading interpolate normals derived from tangent-plane data), and mesh smoothing algorithms. In general relativity, the tangent space to spacetime at an event is where physical velocities and momenta of particles live, and the differential of coordinate changes (diffeomorphisms) is exactly the tensor transformation law taught in any GR course. In robotics, the tangent space to the configuration manifold of a robot arm gives the space of achievable joint velocities, and the differential of the forward-kinematics map is the classical robotic Jacobian used in velocity control and singularity analysis.

Ask the tutor: How does the differential of the Gauss map encode the principal curvatures of a surface?

Think of a smooth surface $S$ sitting in $\mathbb{R}^3$, like the surface of an egg. Near a point $p$, the surface looks almost flat if you zoom in enough — this is exactly the statement that $S$ has a well-defined tangent plane at $p$, the best flat approximation to $S$ near $p$. Concretely, if you parametrize a neighborhood of $p$ by $\mathbf{x}(u,v)$, then moving in the $u$-direction traces a curve on $S$ through $p$ whose velocity vector is $\mathbf{x}_u$, and likewise $\mathbf{x}_v$ for the $v$-direction. Because the parametrization is regular, $\mathbf{x}_u$ and $\mathbf{x}_v$ are linearly independent, so they span a genuine 2-dimensional plane of velocity vectors — this plane, translated to sit at the point $p$ itself (rather than at the origin), is the tangent plane. A cleaner and coordinate-free way to think about $T_pS$: it is exactly the set of all vectors $w \in \mathbb{R}^3$ such that $w = \alpha'(0)$ for some smooth curve $\alpha : (-\varepsilon,\varepsilon) \to S$ with $\alpha(0) = p$. Every such curve, when you have a parametrization $\mathbf{x}(u,v)$ around $p$, can be written as $\alpha(t) = \mathbf{x}(u(t), v(t))$ for smooth functions $u(t), v(t)$ with $\mathbf{x}(u(0),v(0)) = p$; the chain rule then gives

which is visibly a vector in $\operatorname{span}\{\mathbf{x}_u(q), \mathbf{x}_v(q)\}$. Conversely, any vector $a\,\mathbf{x}_u(q) + b\,\mathbf{x}_v(q)$ arises as $\alpha'(0)$ for the curve $\alpha(t) = \mathbf{x}(u_0 + at, v_0 + bt)$ (valid for $t$ near $0$, by openness of $U$). This equivalence — span of partials equals set of curve-velocities — is what makes the tangent plane a genuinely geometric, parametrization-independent object, proven rigorously below. Once we have tangent planes, a smooth map $\varphi: S_1 \to S_2$ between surfaces can be linearized at each point: instead of tracking what $\varphi$ does to all of $S_1$ near $p$, we track only what it does to velocity vectors of curves through $p$. Push a curve $\alpha$ through $p$ forward by $\varphi$ to get a curve $\varphi\circ\alpha$ through $\varphi(p)$ on $S_2$, and differentiate. This is exactly how the ordinary derivative (Jacobian) of a map $F: \mathbb{R}^n \to \mathbb{R}^m$ works, restricted to the curved setting.

Example 1: Tangent plane of a graph surface, at an asymmetric point. Let $S$ be the graph of $f(x,y) = x^2 y - 3xy$, i.e.

This is a regular surface, parametrized globally by

We compute the tangent plane at the point $p = \mathbf{x}(2,1) = (2,1, 4\cdot 1 - 3\cdot2\cdot 1) = (2,1,-2)$. Compute partials:

At $(u,v) = (2,1)$:

Check regularity: $\mathbf x_u, \mathbf x_v$ are linearly independent (neither is a multiple of the other, e.g. their cross product below is nonzero), so $p$ is a regular point and

A normal vector is $\mathbf x_u \times \mathbf x_v = (1,0,1)\times(0,1,-2) = (0\cdot(-2) - 1\cdot 1,\ 1\cdot 0 - 1\cdot(-2),\ 1\cdot1 - 0\cdot 0) = (-1, 2, 1)$. So $T_pS$ is the plane through the origin (as a vector space) with normal $(-1,2,1)$; equivalently, translated to $p$, the affine tangent plane to $S$ at $p$ is

This matches the general graph-surface formula $z = f(2,1) + f_x(2,1)(x-2) + f_y(2,1)(y-1)$, since $f_x = 2xy - 3y$ gives $f_x(2,1) = 4-3=1$ and $f_y = x^2-3x$ gives $f_y(2,1) = 4-6=-2$; indeed $z = -2 + 1(x-2) -2(y-1) = x - 2y -2$, i.e. $-x+2y+z+2=0$. The two computations agree, confirming the tangent-plane formula.

Example 2: The differential of the "square-and-normalize" map from a torus-like surface piece to the sphere. To exhibit the differential on a genuinely non-symmetric, non-graph example, consider the map

restricted to a regular surface $S_1$ not containing the origin, and record its differential at a specific point using a specific parametrization, to see the Jacobian-matrix computation concretely (this restriction of $\varphi$ is exactly how the Gauss map on an oriented surface with unit normal is often compared to a "radial projection," and is smooth by the quotient rule for smooth functions). Take $S_1$ to be (a piece of) the ellipsoid

parametrized near the relevant point by

Let $p = \mathbf x(u_0,v_0)$ with $u_0 = \pi/3$, $v_0 = \pi/6$. Numerically, $\cos(\pi/3) = 1/2$, $\sin(\pi/3) = \sqrt3/2$, $\cos(\pi/6) = \sqrt3/2$, $\sin(\pi/6)=1/2$, so

We compute $d\varphi_p$ on the basis $\{\mathbf x_u(u_0,v_0), \mathbf x_v(u_0,v_0)\}$ of $T_pS_1$ directly (this bypasses needing an explicit parametrization of the target $S^2$, since we can differentiate the curve $t\mapsto \varphi(\mathbf x(u_0+t,v_0))$ directly — this is legitimate because the differential is defined via any representative curve, per the theorem).

Tangent vectors. Differentiate $\mathbf x$:

At $(u_0,v_0) = (\pi/3,\pi/6)$:

Differential of $\varphi(v) = v/\|v\|$ at a point $p\ne 0$ in $\mathbb R^3$ (an ambient computation we use as a tool): for $\varphi:\mathbb R^3\setminus\{0\}\to \mathbb R^3$, $\varphi(x)=x/\|x\|$, the ordinary derivative is

(This is the standard quotient-rule/product-rule computation: $D\varphi_p(w) = \frac{d}{dt}\Big|_{t=0} \frac{p+tw}{\|p+tw\|}$, and one computes $\frac{d}{dt}\big|_0 \|p+tw\| = \frac{p\cdot w}{\|p\|}$, then applies the quotient rule.) Since $\varphi$ restricted to $S_1$, $\varphi|_{S_1}: S_1 \to S^2$, is the restriction of this ambient smooth map, and $S^2 \subset \mathbb R^3$ inherits its differential structure compatibly (a curve in $S_1$ maps to a curve in $S^2$, and differentiating the composite is the same whether we think of $\varphi$ as ambient or as a map of surfaces — this is a basic and standard compatibility that follows immediately from the curve-based definition), we have $d\varphi_p = D\varphi_p|_{T_pS_1}$. Here $\|p\|^2 = \tfrac34 + \tfrac9{16} + \tfrac14 = \tfrac{12}{16}+\tfrac9{16}+\tfrac4{16} = \tfrac{25}{16}$, so $\|p\| = \tfrac54$.

Compute $d\varphi_p(\mathbf x_u)$: with $w = \mathbf x_u = (-\tfrac32,\tfrac{\sqrt3}4,0)$,

Then

Numerically $\frac{4}{5}w = (-\tfrac65, \tfrac{\sqrt3}{5}, 0)$, and $\frac{36\sqrt3}{125}p = \frac{36\sqrt3}{125}\big(\tfrac{\sqrt3}2,\tfrac34,\tfrac12\big) = \big(\tfrac{54}{125}, \tfrac{27\sqrt3}{125}, \tfrac{18\sqrt3}{125}\big)$. Adding coordinate-wise (using $\tfrac65 = \tfrac{150}{125}$ and $\tfrac{\sqrt3}5 = \tfrac{25\sqrt3}{125}$):

This vector is by construction tangent to $S^2$ at $\varphi(p) = p/\|p\| = \tfrac45 p = \big(\tfrac{2\sqrt3}{5}, \tfrac35, \tfrac25\big)$ (one can check $d\varphi_p(\mathbf x_u)\cdot \varphi(p) = 0$, confirming it lies in $T_{\varphi(p)}S^2$, as it must since $d\varphi_p$ maps into $T_{\varphi(p)}S^2$ by the theorem). Indeed:

confirming orthogonality (using $-192+156+36=0$), exactly as the theorem guarantees: $d\varphi_p$ genuinely lands in the tangent plane to the sphere, not in some ambient 3-space at random. This computation, though numerically involved, exemplifies exactly the general mechanism: the differential of a map between surfaces is computed by differentiating the ambient formula (or the coordinate representative $F = \mathbf y^{-1}\circ\varphi\circ\mathbf x$) and restricting to the tangent plane, and the theorem guarantees the answer always lands back in the tangent plane of the target.