Geometry

Differential geometry

Uses calculus to study curvature, geodesics and other local properties of curves and surfaces.

IntuitionIntuition: measuring curvature with calculus

A flat sheet of paper can be rolled into a cylinder without stretching, but it can never be wrapped smoothly onto a ball without tearing or wrinkling it. Differential geometry explains this difference with a single number attached to every point of a curve or surface: its curvature. By applying calculus — derivatives and integrals — to a parametrized curve or surface, we can measure exactly how much it bends, twists, and departs from being flat, entirely from measurements you could make while standing on the surface itself.

The Gaussian curvature

K

of a torus changes sign as you move around the tube: positive on the outer rim, negative on the inner rim, and zero along the two circles where the surface is momentarily flat.

SchoolFrom plane curves to the Frenet–Serret frame

Definition: Curvature of a curve

For a smooth curve $\mathbf{r}(t)$ in space, the curvature $\kappa$ measures how fast the unit tangent vector turns per unit of arc length. A straight line has $\kappa = 0$ everywhere; a circle of radius $R$ has constant curvature $\kappa = 1/R$ .

\kappa(t) = \frac{\lVert \mathbf{r}'(t) \times \mathbf{r}''(t) \rVert}{\lVert \mathbf{r}'(t) \rVert^{3}}

Formula $\kappa(t) = \frac{\lVert \mathbf{r}'(t) \times \mathbf{r}''(t) \rVert}{\lVert \mathbf{r}'(t) \rVert^{3}}$ lets you compute curvature directly from any parametrization, without first reparametrizing by arc length. At every point of the curve we can also build a moving frame of three orthonormal vectors: the unit tangent $\mathbf{T}$ , the principal normal $\mathbf{N}$ (pointing toward the center of the curve's bend), and the binormal $\mathbf{B} = \mathbf{T}\times\mathbf{N}$ . How this frame rotates as it travels along the curve is recorded by the curvature $\kappa$ and the torsion $\tau$ , which measures how far the curve twists out of its osculating plane.

\mathbf{T}' = \kappa\mathbf{N}, \quad \mathbf{N}' = -\kappa\mathbf{T} + \tau\mathbf{B}, \quad \mathbf{B}' = -\tau\mathbf{N}

UndergraduateThe two fundamental forms and Gaussian curvature

Definition: First fundamental form

For a parametrized surface $\mathbf{r}(u,v)$ , the first fundamental form $I = E\,du^2 + 2F\,du\,dv + G\,dv^2$ , with $E = \mathbf{r}_u\cdot\mathbf{r}_u$ , $F = \mathbf{r}_u\cdot\mathbf{r}_v$ , $G = \mathbf{r}_v\cdot\mathbf{r}_v$ , tells you how to measure lengths, angles and areas using only the coordinates $(u,v)$ — it is entirely intrinsic, meaning an ant living on the surface could measure it without ever leaving the surface.

I = E\,du^2 + 2F\,du\,dv + G\,dv^2

Definition: Second fundamental form

The second fundamental form $II = e\,du^2 + 2f\,du\,dv + g\,dv^2$ , built from $e = \mathbf{r}_{uu}\cdot\mathbf{n}$ , $f = \mathbf{r}_{uv}\cdot\mathbf{n}$ , $g = \mathbf{r}_{vv}\cdot\mathbf{n}$ where $\mathbf{n}$ is the unit normal, measures how fast the surface pulls away from its own tangent plane — it is extrinsic, since it depends on how the surface sits in space.

II = e\,du^2 + 2f\,du\,dv + g\,dv^2

Combining the two forms gives the single most important local invariant of a surface, the Gaussian curvature $K = \frac{eg-f^2}{EG-F^2}$ . When $K>0$ the surface curves the same way in every direction (like a sphere); when $K<0$ it curves oppositely in different directions (like a saddle); when $K=0$ the surface is developable and can be unrolled flat (like a cylinder or cone).

K = \frac{eg-f^2}{EG-F^2}

Gaussian curvature by surface type
Surface	Sign of $K$	Local shape
Sphere of radius $R$	$K = 1/R^2 > 0$	Bowl-shaped in every direction; the surface curves away from the tangent plane on the same side everywhere.
Plane or cylinder	$K = 0$	Developable: can be flattened without stretching.
Saddle or hyperboloid	$K < 0$	Curves upward in one direction and downward in the perpendicular direction; the surface crosses its tangent plane.

UndergraduateKey theorems: Theorema Egregium and Gauss–Bonnet

Theorem: Theorema Egregium (Gauss's Remarkable Theorem)

The Gaussian curvature $K$ of a surface can be computed entirely from the first fundamental form $E, F, G$ and its derivatives — it does not depend on the second fundamental form or on how the surface is embedded in space. Consequently, $K$ is preserved by any isometry (a map that preserves the first fundamental form, hence all lengths and angles measured on the surface).

Why is it true?

This is surprising because $K$ was originally defined using the second fundamental form, i.e. using how the surface bends within the ambient space. Gauss's theorem says curvature is secretly intrinsic: a two-dimensional being confined to the surface, unable to see the surrounding space, could still compute $K$ by measuring lengths and angles alone.

Proof

Work in coordinates where $F=0$ at the point of interest (an orthogonal parametrization, which always exists locally). Differentiating the identities $\mathbf{r}_u\cdot\mathbf{n}=0$ and $\mathbf{r}_v\cdot\mathbf{n}=0$ and using $e = \mathbf{r}_{uu}\cdot\mathbf{n}$ , $g = \mathbf{r}_{vv}\cdot\mathbf{n}$ expresses $eg-f^2$ in terms of the second derivatives $\mathbf{r}_{uu}, \mathbf{r}_{uv}, \mathbf{r}_{vv}$ projected onto $\mathbf{n}$ .

Because $\{\mathbf{r}_u, \mathbf{r}_v, \mathbf{n}\}$ form a basis of space at each point, every second derivative such as $\mathbf{r}_{uu}$ can be written as a combination of $\mathbf{r}_u$ , $\mathbf{r}_v$ and $\mathbf{n}$ , with the tangential coefficients being the Christoffel symbols — functions built only from $E, F, G$ and their derivatives with respect to $u$ and $v$ , obtained by solving the linear system that comes from differentiating $E, F, G$ .

Substituting these expressions into $\mathbf{r}_{uu}\cdot\mathbf{r}_{vv} - \mathbf{r}_{uv}\cdot\mathbf{r}_{uv}$ and comparing with $eg-f^2$ from the normal components, then simplifying with the compatibility (Gauss) equations, produces Brioschi's formula: $K$ expressed purely as a rational function of $E, F, G$ and their first and second partial derivatives with respect to $u$ and $v$ , with no reference to $e, f, g$ left in the final expression.

Since an isometry between two surfaces is, by definition, a map that carries the first fundamental form of one surface exactly onto the first fundamental form of the other (the same $E, F, G$ as functions of the parameters), and Brioschi's formula computes $K$ from $E, F, G$ alone, the two surfaces must have equal Gaussian curvature at corresponding points. This proves the theorem.

Theorem: Gauss–Bonnet theorem

For a compact, orientable surface $M$ without boundary, $\int_M K\,dA = 2\pi\chi(M)$ , where $\chi(M)$ is the Euler characteristic, $\chi(M) = V - E + F$ for any triangulation of $M$ (also equal to $2 - 2g$ for a surface of genus $g$ ).

Why is it true?

The theorem links a purely local, geometric quantity (curvature, which can change from point to point) to a purely global, topological quantity (the Euler characteristic, which only depends on how the surface is connected, not on its shape). No matter how you bend, stretch or dent a surface without tearing it, the total curvature $\int_M K\,dA$ stays the same.

Proof

Triangulate $M$ using geodesic triangles (triangles whose sides are shortest paths on the surface), obtaining $V$ vertices, $E$ edges and $F$ faces, related by $\chi(M) = V - E + F$ .

The local Gauss–Bonnet formula for a single geodesic triangle $T$ with interior angles $\alpha,\beta,\gamma$ states $\int_T K\,dA = \alpha+\beta+\gamma-\pi$ : the amount the angle sum exceeds the Euclidean value $\pi$ equals exactly the integral of curvature over that triangle, a fact obtained by applying Stokes' theorem to the rotation of a tangent vector parallel-transported around the triangle's boundary.

Summing $\int_T K\,dA = \alpha+\beta+\gamma-\pi$ over all $F$ triangles gives $\int_M K\,dA = 2\pi V - \pi F$ : the triangles meeting at each of the $V$ vertices sweep out exactly one full turn $2\pi$ there, so all their angles together sum to $2\pi V$ , while each of the $F$ triangles contributes a $-\pi$ .

Each triangle has $3$ edges and each edge is shared by exactly $2$ triangles, so $3F = 2E$ . Substituting $E = \frac{3F}{2}$ into $\chi(M) = V - E + F$ gives $2\pi\chi(M) = 2\pi V - \pi F$ , which matches the previous step exactly. Hence $\int_M K\,dA = 2\pi\chi(M)$ , proving the theorem.

UndergraduateReal-World Applications and Worked Examples

Differential geometry is the mathematical backbone of any technology that must represent, measure or move along curved shapes. Cartography lives with the Theorema Egregium every day: because the sphere has $K = 1/R^2 \neq 0$ and the plane has $K=0$ , no flat map of the Earth can show all distances and angles correctly at once, forcing mapmakers to choose which distortion — area, angle, or distance — to accept. Computer graphics and CAD systems compute Gaussian and mean curvature at every vertex of a 3D mesh to decide where to add detail, smooth a surface, or detect a manufacturing defect. GPS navigation and orbital mechanics compute geodesics — the straightest possible paths — on the curved reference ellipsoid of the Earth rather than on a flat map. Architects use surfaces of negative Gaussian curvature, such as hyperbolic paraboloid roofs, because they are doubly ruled and can be built from straight beams while still carrying loads efficiently.

Example: Curvature and torsion of a helix

A wire is bent into the helix $\mathbf{r}(t) = (3\cos t, 3\sin t, 4t)$ . Find its curvature $\kappa$ and torsion $\tau$ .

Solution

Differentiate the position vector: $\mathbf{r}'(t) = (-3\sin t, 3\cos t, 4)$ , $\mathbf{r}''(t) = (-3\cos t, -3\sin t, 0)$ , and $\mathbf{r}'''(t) = (3\sin t, -3\cos t, 0)$ . Note that $\lVert \mathbf{r}'(t)\rVert = \sqrt{9+16} = 5$ , a constant, independent of $t$ .

Compute the cross product $\mathbf{r}'(t)\times\mathbf{r}''(t) = (12\sin t, -12\cos t, 9)$ , which has length $\sqrt{144+81}=15$ . By the curvature formula, $\kappa = \dfrac{15}{5^3} = \dfrac{15}{125} = \dfrac{3}{25}$ .

For torsion, use $\tau = \dfrac{(\mathbf{r}'\times\mathbf{r}'')\cdot \mathbf{r}'''}{\lVert \mathbf{r}'\times\mathbf{r}'' \rVert^2}$ . The dot product $(12\sin t, -12\cos t, 9)\cdot(3\sin t,-3\cos t,0) = 36\sin^2 t + 36\cos^2 t = 36$ is constant, and $\lVert\mathbf{r}'\times\mathbf{r}''\rVert^2 = 225$ , so $\tau = \dfrac{36}{225} = \dfrac{4}{25}$ .

Both $\kappa=\dfrac{3}{25}$ and $\tau=\dfrac{4}{25}$ are constant along the whole curve, which is exactly why a helix looks the same at every point: it is (up to a rigid motion) the unique curve with constant nonzero curvature and torsion.

Example: Gaussian curvature of a sphere and a Gauss–Bonnet check

Using the parametrization $\mathbf{r}(u,v) = (R\cos u\sin v, R\sin u\sin v, R\cos v)$ of a sphere of radius $R$ , verify that $K=1/R^2$ everywhere, then check the Gauss–Bonnet theorem for the whole sphere.

Solution

Compute the tangent vectors $\mathbf{r}_u = (-R\sin u\sin v, R\cos u\sin v, 0)$ and $\mathbf{r}_v = (R\cos u\cos v, R\sin u\cos v, -R\sin v)$ , giving first fundamental form coefficients $E = R^2\sin^2 v$ , $F=0$ , $G=R^2$ .

The unit normal is $\mathbf{n} = -\mathbf{r}/R$ (pointing inward), and computing the second derivatives against $\mathbf{n}$ gives second fundamental form coefficients $e = R\sin^2 v$ , $f = 0$ , $g = R$ .

By the Gaussian curvature formula, $K = \dfrac{eg-f^2}{EG-F^2} = \dfrac{R\sin^2 v \cdot R}{R^2\sin^2 v\cdot R^2} = \dfrac{1}{R^2}$ , confirming that every point of a sphere of radius $R$ has the same curvature.

For Gauss–Bonnet, the area element is $dA = \sqrt{EG-F^2}\,du\,dv = R^2\sin v\,du\,dv$ , so $\displaystyle\int_M K\,dA = \frac{1}{R^2}\int_0^{2\pi}\!\!\int_0^\pi R^2\sin v\,dv\,du = 4\pi$ . Since a sphere has Euler characteristic $\chi(S^2)=2$ , the theorem predicts $2\pi\chi(M) = 4\pi$ , which matches exactly.

At a point on a surface, the first and second fundamental form coefficients are $E=1, F=0, G=1$ and $e=1, f=0, g=-1$ . What is the Gaussian curvature $K$ at that point?

Why can no single flat map of the Earth preserve every distance and angle at once?

A sphere $S^2$ has Euler characteristic $\chi(S^2)=2$ . According to the Gauss–Bonnet theorem, what is $\int_{S^2} K\,dA$ ?

In the Frenet–Serret frame of a space curve, which vector points toward the center of the curve's instantaneous bend, perpendicular to the direction of motion?

References

Manfredo P. do Carmo (2016). Differential Geometry of Curves and Surfaces
Kristopher Tapp (2016). Differential Geometry of Curves and Surfaces