Curvature of an Embedded Torus

Problem

Let $T$ be a smooth, closed surface embedded in $\mathbb{R}^3$ that is topologically a torus. Prove that $T$ must have:

At least one point where the Gaussian curvature $K > 0$ , and
At least one point where the Gaussian curvature $K < 0$ .

You may use the Gauss-Bonnet theorem: for a closed surface $S$ , $\iint_S K \, dA = 2\pi \chi(S)$ where $\chi(S)$ is the Euler characteristic.

Field

Differential Geometry

Why It's Beautiful

You can feel this result geometrically: the outer equator of a doughnut curves like a sphere (positive curvature), while the inner ring curves like a saddle (negative curvature). But making this rigorous requires a global theorem — Gauss-Bonnet — which converts local curvature into a topological invariant.

The result shows that topology constrains geometry: the torus cannot be "non-negatively curved throughout," no matter how you embed it. A sphere can be made uniformly positively curved (as a round sphere in $\mathbb{R}^3$ ), but a torus cannot — the topology forbids it.

Key Idea / Trick

Two ingredients:

Compactness gives a point of positive $K$ : the point on $T$ farthest from the origin lies on a sphere that $T$ touches from inside, forcing $K > 0$ there.
Gauss-Bonnet forces a point of negative $K$ : since $\chi(T^2) = 0$ , we have $\iint K \, dA = 0$ . But $K > 0$ somewhere means the integral can only be zero if $K < 0$ somewhere else.

Difficulty

3 / 5

Curvature of an Embedded Torus — Answer

Setup

Let $T \subset \mathbb{R}^3$ be a smooth closed surface homeomorphic to the torus $T^2$ . Recall:

Euler characteristic: $\chi(T^2) = 0$ (since $V - E + F = 0$ for any triangulation).
Gauss-Bonnet: $\iint_T K \, dA = 2\pi \chi(T^2) = 0$ .

Part 1: There exists a point with $K > 0$

Since $T$ is compact, it is bounded in $\mathbb{R}^3$ . Let $p \in T$ be the point farthest from the origin, i.e., $|p| = \max_{q \in T} |q|$ .

At $p$ , the surface $T$ is tangent (from inside) to the sphere $S_r$ of radius $r = |p|$ centered at the origin.

Since $T$ lies entirely inside $S_r$ and touches it at $p$ , the surface $T$ curves at least as much as $S_r$ at $p$ in every direction. The sphere $S_r$ has Gaussian curvature $K = 1/r^2 > 0$ , so:

$K(p) \geq \frac{1}{r^2} > 0$

More precisely: the principal curvatures $\kappa_1, \kappa_2$ of $T$ at $p$ both satisfy $\kappa_i \geq 1/r > 0$ , giving $K(p) = \kappa_1 \kappa_2 > 0$ . $\checkmark$

Part 2: There exists a point with $K < 0$

By Gauss-Bonnet: $\iint_T K \, dA = 2\pi \cdot \chi(T^2) = 2\pi \cdot 0 = 0$

From Part 1, $K > 0$ on some open neighborhood of $p$ (by continuity), so: $\iint_T K \, dA \geq \iint_{U} K \, dA > 0 \quad \text{if } K \geq 0 \text{ everywhere}$

This contradicts $\iint_T K \, dA = 0$ . Therefore $K < 0$ at some point. $\blacksquare$

Geometric Picture

On a standard torus of revolution (donut), you can see both signs explicitly:

Outer equator (farthest from the axis): the surface curves away from you in both directions, like a sphere — $K > 0$ .
Inner equator (closest to the axis): the surface curves toward you in one direction, away in the other, like a saddle — $K < 0$ .
Top and bottom circles: $K = 0$ (one principal curvature is zero).

The Gauss-Bonnet theorem guarantees these positive and negative regions balance out exactly.

Contrast with the Sphere

For $S^2$ : $\chi(S^2) = 2$ , so $\iint K \, dA = 4\pi > 0$ . The sphere can be embedded with $K > 0$ everywhere (the round sphere has $K \equiv 1/R^2$ ).

For the torus: $\chi = 0$ forces the total curvature to vanish — positive and negative regions must perfectly cancel. No matter how cleverly you embed a torus in $\mathbb{R}^3$ , you cannot avoid saddle points.

Key Takeaway

Topology constrains geometry. The Euler characteristic — a purely topological quantity — dictates the sign of the total curvature. A torus, having $\chi = 0$ , must have mixed curvature in any smooth embedding in $\mathbb{R}^3$ .

Type: Differential GeometryEdit on GitHub ↗