I am studying about the Gauss–Bonnet Theorem in context of Riemannian Manifolds. I'm following the book "Introduction to Riemannian Manifolds" of John Lee. In the text, is defined the signed curvature of $\gamma$ (a curved polygon) as $$k_N (t) = g(D_t \gamma'(t), N(t)),$$ where $D_t \gamma'(t)$ is the covariant derivative of $\gamma'(t)$ over $\gamma$ and $N$ is a unique unit normal vector field along the smooth portions of $\gamma$ such that $(\gamma'(t), N(t))$ is an oriented orthonormal basis for $T_{\gamma(t)}M$ for each $t$ .
Theorem (The Gauss–Bonnet Formula): Let $(M,g)$ be an oriented Riemannian 2-manifold. Suppose $\gamma$ is a positively oriented curved polygon in $M$, and $\Omega$ is its interior. Then $$\int_{\Omega} K dA + \int_{\gamma}k_{N} ds + \sum_{i=1}^k \varepsilon_i = 2\pi,$$ where $K$ is the Gaussian curvature of $g$, $dA$ is its Riemannian volume form, $\varepsilon_1, \ldots, \varepsilon_r$ are the exterior angles of $\gamma$, and the second integral is taken with respect to arc length.
The author claims that from the Gauss–Bonnet formula follow as easy corollaries:
Corollary 1 (Angle-Sum Theorem). The sum of the interior angles of a Euclidean triangle is $\pi$.
Corollary 2 (Circumference Theorem). The circumference of a Euclidean circle of radius $R$ is $2\pi R$.
As objects are flat, then $K = 0$ in two cases. My difficulty lies in proving these corollaries, especially regarding the calculation of $k_N$.
For the Corollary 1, denoted by $\theta_i$ the interior angle, we have: $$\sum_{i=1}^3 \theta_i = \sum_{i=1}^3(\pi - \varepsilon_i) = 3 \pi - \sum_{i=1}^3 \varepsilon_i = 3 \pi - (2\pi + \int_{\gamma}k_{N} ds) = \pi + \int_{\gamma}k_{N} ds.$$ I'm having trouble showing that this integral is indeed $0$.
For the Corollary 2 we have: the circumference not have angles then: $$2\pi = \int_{\gamma}k_{N} ds,$$ how to get the result from that?
Thank you for your help.
It is due to Hopf's Umlaufsatz. It states that the rotation of the unit normal about a closed curved is zero for the Euclidean plane.