A closed orientable surface S has Euler characteristic χ(S) = 2 - 2g, where g is the genus (number of handles). By Gauss-Bonnet, ∫_S K dA = 2π(2 - 2g). For the torus (g = 1), this gives...
A∫ K dA = 4π, so the torus must have everywhere positive curvature
B∫ K dA = 0, so any metric on the torus must have regions of both positive and negative curvature (unless K = 0 everywhere)
C∫ K dA = -4π, so the torus must have everywhere negative curvature
D∫ K dA = 2π, so the torus has exactly half the total curvature of a sphere
For g = 1: ∫ K dA = 2π(2-2) = 0. The total curvature vanishes. If K is not identically zero, it must be positive somewhere and negative somewhere (by the intermediate value theorem on a connected manifold, if K is continuous and integrates to zero but is not identically zero, it must change sign). The flat torus (K = 0 everywhere) is the special case where the curvature vanishes pointwise. The embedded torus in ℝ³ (doughnut shape) has positive curvature on the outside and negative curvature on the inside, integrating to zero.
Question 2 True / False
The Gauss-Bonnet theorem implies that the total curvature ∫_S K dA is unchanged if you smoothly deform the metric on S.
TTrue
FFalse
Answer: True
The Euler characteristic χ(S) is a topological invariant — it depends only on the homeomorphism type of S, not on the metric. Since ∫_S K dA = 2πχ(S), the total curvature is also a topological invariant. You can bend, stretch, and deform the surface in any smooth way, and the integral of curvature will not change. This is remarkable because K and dA separately depend on the metric — only their product integrates to a topological constant.
Question 3 Short Answer
Apply the Gauss-Bonnet theorem to the sphere S² with any Riemannian metric. What can you conclude about the Gaussian curvature?
Think about your answer, then reveal below.
Model answer: The sphere has χ(S²) = 2, so ∫_{S²} K dA = 4π for any metric on S². This means the total Gaussian curvature is always 4π, regardless of how you deform the sphere. In particular, K must be positive somewhere — a sphere cannot carry a metric with everywhere non-positive curvature. For the round sphere of radius r, K = 1/r² and Area = 4πr², giving ∫ K dA = (1/r²)(4πr²) = 4π, confirming the theorem.
More generally, Gauss-Bonnet constrains which curvature conditions are compatible with a given topology. A surface of genus g ≥ 2 has χ < 0, so it must have negative curvature somewhere (and cannot have a metric of everywhere non-negative curvature). These topological obstructions to curvature conditions are a central theme in Riemannian geometry.
Question 4 True / False
The Gauss-Bonnet theorem has a version for surfaces with boundary: ∫_S K dA + ∫_{∂S} κg ds = 2πχ(S), where κg is the geodesic curvature of the boundary.
TTrue
FFalse
Answer: True
When the surface has boundary, the boundary's geodesic curvature contributes a correction term. For a geodesic triangle (boundary consists of three geodesic segments, κg = 0 on each), the formula gives ∫_T K dA + (sum of exterior angles) = 2π, which is equivalent to the angle excess formula: (sum of interior angles) - π = ∫_T K dA. On a sphere, the angles of a geodesic triangle sum to more than π, with the excess equal to the enclosed area times the curvature.