Editing Gauss–Bonnet theorem (section)

== For triangles ==
Sometimes the Gauss–Bonnet formula is stated as

: <math>\int_T K = 2\pi - \sum \alpha - \int_{\partial T} \kappa_g,</math>

where {{mvar|T}} is a [[geodesic triangle]]. Here we define a "triangle" on {{mvar|M}} to be a simply connected region whose boundary consists of three [[geodesic]]s. We can then apply GB to the surface {{mvar|T}} formed by the inside of that triangle and the piecewise boundary of the triangle.

The geodesic curvature the bordering geodesics is 0, and the Euler characteristic of {{mvar|T}} being 1.

Hence the sum of the turning angles of the geodesic triangle is equal to 2{{pi}} minus the total curvature within the triangle. Since the turning angle at a corner is equal to {{pi}} minus the interior angle, we can rephrase this as follows:<ref>{{Cite book |last=Weeks |first=Jeffrey R. |author-link=Jeffrey Weeks (mathematician) |date=2001-12-12 |title=The Shape of Space |url=https://www.taylorfrancis.com/books/9781135542665 |language=en |publisher=CRC Press |doi=10.1201/9780203912669 |isbn=9780203912669 |via=[[Taylor & Francis]] }}</ref>

: The sum of interior angles of a geodesic triangle is equal to {{pi}} plus the total curvature enclosed by the triangle: <math>\sum (\pi - \alpha) = \pi + \int_T K.</math>

In the case of the plane (where the Gaussian curvature is 0 and geodesics are straight lines), we recover the familiar formula for the sum of angles in an ordinary triangle. On the standard sphere, where the curvature is everywhere 1, we see that the angle sum of geodesic triangles is always bigger than {{pi}}.