# Show that if G is a convex polygon in an open hemisphere or in a hyperbolic plane, then

Gauss-Bonnet Formula for Polygons:

Show that if $G$ is a convex polygon in an open hemisphere or in a hyperbolic plane, then

$H(G)$ = $2p - \sum a_{i}$ = $\sum b_i - (n - 2)p$ = $Area(G)\cdot K$,

where $\sum a_{i}$ is the sum of the exterior angles, $\sum b_i$ is the sum of the interior angles, $n$ is the number of sides, and $K$ is the Gaussian curvature. Also prove this on sphere.

I can divide the convex polygon into triangles and apply

$H(D) = 2p - (a_1 + a_2 + a_3) = (b_1 + b_2 + b_3) - p$

to triangle and add up the results ($H(G) = 2p - \sum a_i$).

I not sure what to do after, I was wondering if anyone can help me out?

• Is your convex polygon a geodesic polygon? – Upax Jan 18 '18 at 18:11
• I think he implied it. – Narasimham Jan 18 '18 at 20:53

A geodesic polygon is a polygon whose segments are geodesics. In this case if the geodesic polygon lies in a framed open set of an oriented Riemannian 2-manifold given $b_i$ he interior angle at vertex i, R is the closed region enclosed by the polygon, and K the Gaussian curvature on R,then $$\sum_{i=1}^{m} b_i = (m-2) \pi+\int_D K dS$$ As a matter of fact a geodesic polygon has geodesic curvature $k_g=0$, while the exterion angles $a_i$ are related to the interior angles $b_i$ by the relation $$a_i=\pi-b_i$$ Hence, the Gauss–Bonnet formula becomes $$2 \pi-\sum_{i=1}^{m} (\pi-b_i) = \int_D K dS$$ or $$\sum_{i=1}^{m} b_i = (m-2) \pi+\int_D K dS$$ An hyperbolic plane has constant Gaussian curvature −1, while for a sphere $K=\frac{1}{R^2}$