I'm reading the John Lee's Introduction to Riemannian Manifold, p.273, proof of Theorem 9.3. and stuck at understanding some statement :
Theorem 9.3 ( 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
$$ \int_{\Omega}KdA + \int_{\gamma}\kappa_Nds +\Sigma_{i=1}^{k} \epsilon_i = 2\pi,$$
where $K$ is the Gaussian curvature of $g$, $dA$ is ita Riemannian volume form, $\epsilon_1 , \dots , \epsilon_k$ are the exterior angles of $\gamma$ ( c.f. his book p.271 ), and the second integral is taken with respect to arc length ( Problem 2-32 ).
In the proof of the theroem, he argues as follows :
Proof. Let $(a_0 , \dots ,a_k)$ be an admissible partition of $[a,b]$, and let $(x,y)$ be oriented smooth coordinates on an open set $U$ containing $\bar{\Omega}$ ( by the definition of $\Omega$, refer to definition of curved polygon ; his book p.271 ). Let $\theta : [a,b] \to \mathbb{R}$ be a tangent angle function for $\gamma$ ( c.f. his book p.272 ). Using the rotation index theorem ( his book Lemma 9.2. ) and the fundamental theorem of calculus, we can write
$$ 2\pi = \theta(b) – \theta(a) = \Sigma_{i=1}^{k}\epsilon_i + \Sigma_{i=1}^{k}\int_{a_{i-1}}^{a_i}\theta'(t)dt$$
Q. And why this is true? Why the term $\Sigma_{i=1}^{k}\epsilon_i$ appears ? An issue that makes me confusing is,
$$ \Sigma_{i=1}^{k}\int_{a_{i-1}}^{a_i}\theta'(t)dt = \Sigma_{i=1}^{k} ( \theta(a_i) – \theta(a_{i-1})) = \theta(b) – \theta (a) ?$$
What is a point that I made misunderstood? What is the definition of $\int_{a_{i-1}}^{a_i}\theta'(t)dt$ for each $i$ ?
In page p.272, the author defined a tangent angle function for $\gamma$ as a piecewise continuous function $\theta : [a,b] \to \mathbb{R}$ that satisfies
$$ T(t) = \cos\theta(t) E_1 |_{\gamma(t)} + \sin\theta(t) E_2|_{\gamma(t)}$$
( where $T(t)$ is the unit tangent vector field of $\gamma$ , c.f. p.271 ) at each $t$ where $\gamma'$ is continuous, and that is continuous from the right and satisfies (9.1) and (9.2) at vertices. Here the (9.1) and (9.2) in his book are as follows :
$$ \theta(a_i) = \lim_{t \nearrow a_i}\theta(t) + \epsilon_i,$$
$$ \theta(b) = \lim_{t \nearrow b}\theta(t) + \epsilon_k $$
Can we try to use these ? Can anyone helps?
Best Answer
I think you have all the ingredients to make this work: first, one writes $$\theta(b)-\theta(a)=\theta(a_k)-\theta(a_0)=\theta(a_k)-\theta(a_{k-1})+\theta(a_{k-1})-\theta(a_{k-2})+\theta(a_{k-2})-\ldots+\theta(a_{1})-\theta(a_0)$$ by adding and subtracting the angles. Now, if $\theta$ were a usual differentiable function, one could write, for each $0\leq j<k$, $$\theta(a_{j+1})-\theta(a_j)=\int_{a_j}^{a_{j+1}}\theta'(t)\operatorname{d}t.$$ The problem is that $\theta$ is only differentiable in the interior of each of the segments $[a_j,a_{j+1}]$. So, we can say that the previous identity holds for any segment $[x,y]$ such that $a_j<x<y<a_{j+1}$, namely $$\theta(y)-\theta(x)=\int_{x}^{y}\theta'(t)\operatorname{d}t.$$ Taking the limits for $x\to a_j$ and $y\to a_{j+1}$ then should give you the result you are looking for: the limit $\lim_{y\to {a_{j+1}}^{-}}\theta(y)$ equals $\theta(a_{j+1})-\epsilon_{j+1}$ by the equations $(9.1)$, $(9.2)$.