Gaussian and Mean Curvature of Surfaces

Chapter 7: Differential Geometry — Curvature

Learning objectives

Define the principal curvatures $\kappa_1, \kappa_2$ at a point on a smooth surface
Compute the Gaussian curvature $K = \kappa_1\kappa_2$ and mean curvature $H = (\kappa_1+\kappa_2)/2$ for canonical examples
Classify a surface point as elliptic, hyperbolic, parabolic, or planar from the sign of $K$
State Gauss's Theorema Egregium and explain why $K$ is intrinsic while $H$ is extrinsic

A surface bends in two independent directions at every point. A pringle chip bends up along one axis and down along the perpendicular one; a sphere bends the same way in every direction; a cylinder bends one way around and not at all along the axis. Capturing this richer geometry needs two numbers per point, not one. The framework due to Gauss organises them as principal curvatures $\kappa_1, \kappa_2$ , and packages them into the Gaussian curvature $K$ and the mean curvature $H$ . These two scalars then drive almost every result in surface theory, from the Theorema Egregium to soap-film equations.

Normal sections and principal curvatures

Pick a point $p$ on a smooth oriented surface $S \subset \mathbb{R}^3$ . The unit normal $\mathbf{n}$ and any direction $\mathbf{v}$ in the tangent plane $T_pS$ together span a vertical plane. Slice $S$ with that plane — you get a plane curve, the normal section. It has a signed curvature $\kappa(\mathbf{v})$ at $p$ depending on the chosen direction.

As $\mathbf{v}$ rotates through the unit circle in $T_pS$ , the value $\kappa(\mathbf{v})$ traces out a function. Its maximum and minimum are the two principal curvatures $\kappa_1 \geq \kappa_2$ , and the directions where they are attained are the principal directions. They are always perpendicular (Euler's theorem on principal directions).

The two intrinsic combinations

From the two principal curvatures we form:

K = \kappa_1 \kappa_2 \qquad\text{(Gaussian curvature)}

H = \dfrac{\kappa_1 + \kappa_2}{2} \qquad\text{(mean curvature)}

The sign of $K$ classifies the local shape:

$K > 0$ — elliptic point. Both principal curvatures have the same sign; the surface curves the same way in every direction (sphere, ellipsoid, dome).
$K < 0$ — hyperbolic point. Principal curvatures have opposite signs; the surface curves up one way and down the perpendicular way (saddle, Pringle chip, mountain pass).
$K = 0$ , not all zero — parabolic point. One principal curvature is zero (cylinder, cone).
$K = 0$ , both zero — planar point. The surface is flat there to second order.

Canonical examples

Sphere of radius $R$ . By symmetry every direction is principal: $\kappa_1 = \kappa_2 = 1/R$ . So $K = 1/R^2 > 0$ and $H = 1/R$ .
Cylinder of radius $R$ . Curvature around the cylinder is $1/R$ ; curvature along the axis is $0$ . So $K = 0$ , $H = 1/(2R)$ . Cylinders are intrinsically flat.
Saddle $z = x^2 - y^2$ at the origin. Principal curvatures are $\kappa_1 = 2, \kappa_2 = -2$ . So $K = -4 < 0$ and $H = 0$ .
Pseudosphere of "radius" $a$ . Has constant $K = -1/a^2 < 0$ everywhere. The pseudosphere is the surface-theoretic shadow of the hyperbolic plane you will meet in chapter 8.

Theorema Egregium

Gauss's Theorema Egregium (Latin: "remarkable theorem") says that $K$ depends only on the intrinsic metric of the surface — on distances measured within the surface — and not on how the surface sits inside $\mathbb{R}^3$ . An ant living on the surface and knowing only the metric (the first fundamental form) can compute $K$ without ever leaving home. The mean curvature $H$ , in contrast, depends on how the surface curves in the ambient space; it is extrinsic. This is why a flat sheet of paper ( $K = 0$ ) cannot be wrapped onto a sphere ( $K > 0$ ) without stretching: any isometric deformation preserves $K$ , and the values disagree.

Where this shows up

- **Cartography:** The reason every flat world map distorts something is the Theorema Egregium:

K

1/R^2 > 0

on the Earth but

0

on paper. Projections (Mercator, Robinson, equal-area) trade off which property — angles, areas, distances — takes the hit. There is no isometric world map. - **Soap films and minimal surfaces:** A soap film stretched on a wire frame minimises area subject to the wire boundary; the Euler-Lagrange equation forces

H = 0

everywhere. Catenoids, helicoids, and the Costa surface are all minimal surfaces. Pixar and architectural firms simulate them by enforcing

H = 0

on a triangulated mesh. - **General relativity:** The Einstein field equations involve curvature tensors that are the higher-dimensional analogues of

K

and

H

. Black holes, gravitational lensing, and cosmological expansion are all consequences of the way mass-energy bends spacetime — quantified by Ricci curvature and Gaussian curvature's descendants. - **Architecture and shell structures:** Buildings like Frei Otto's tensile roofs and Gaudi's Sagrada Familia exploit negative Gaussian curvature for stiffness without weight. Pringle-chip geometry resists bending out of plane because the two principal curvatures cancel mean-curvature responses while doubling stiffness in shear. - **Brain mapping:** Neuroimaging packages flatten the cortical surface to a sphere or plane for region-of-interest comparison across subjects. Because the cortex has

K \neq 0

, exact flattening is impossible — algorithms minimise curvature distortion globally.

Pause and think: A cylinder has $K = 0$ everywhere, the same value as a flat plane. Can you, by bending alone (no stretching), unroll a cylinder into a flat strip? Yes — try it with a sheet of paper. Can you do the same with a half-sphere? Why does the Theorema Egregium guarantee you cannot?

Try it

Predict first: at a hyperbolic point ( $K < 0$ ), can the mean curvature $H$ be positive, negative, or zero? Show that all three are possible with explicit examples.
For the torus parametrized as $\mathbf{r}(u, v) = ((R + r\cos u)\cos v, (R + r\cos u)\sin v, r\sin u)$ with $R > r$ : identify which regions are elliptic, parabolic, and hyperbolic. (Hint: think about which directions the surface curves at the inner ring, top, and outer ring.)
True or false: a compact closed surface in $\mathbb{R}^3$ must have at least one elliptic point. Justify using the fact that such a surface, being compact, attains a maximum distance from the origin.
Suppose $K \equiv 0$ everywhere on a surface. List two qualitatively different surfaces matching this description (besides the plane).

A trap to watch for

The mean curvature $H$ depends on the choice of unit normal. Flip $\mathbf{n}\to -\mathbf{n}$ and every principal curvature switches sign, so $H\to -H$ . The Gaussian curvature $K = \kappa_1\kappa_2$ is unaffected because the two sign flips cancel. This is why " $H = 0$ " (minimal surface) is well-defined while " $H > 0$ " requires a chosen orientation. When you read a paper that quotes mean curvature, always check which normal convention is in force.

What you now know

You can compute the two principal curvatures of a surface, combine them into Gaussian curvature $K$ and mean curvature $H$ , classify points as elliptic / hyperbolic / parabolic / planar from the sign of $K$ , and state the Theorema Egregium that distinguishes intrinsic from extrinsic geometry. The next section combines these point-wise quantities into a single global statement — the Gauss-Bonnet theorem — where the integral of $K$ over a closed surface turns out to encode pure topology.

References

Garrity, T. (2002). All the Mathematics You Missed: But Need to Know for Graduate School. Cambridge University Press, ch. 7.
do Carmo, M. P. (2016). Differential Geometry of Curves and Surfaces (2nd ed.). Dover, ch. 3.
Pressley, A. (2010). Elementary Differential Geometry (2nd ed.). Springer, ch. 6-8.
Spivak, M. (1999). A Comprehensive Introduction to Differential Geometry (3rd ed., Vol. 3). Publish or Perish, ch. 3.
Misner, C. W., Thorne, K. S., Wheeler, J. A. (2017). Gravitation. Princeton University Press, ch. 11 (Riemannian curvature in higher dimensions).