The Global Gauss-Bonnet Theorem states:
Let $R\subset S$ be a regular region and $C_1,\ldots,C_r$ be closed, simple, piecewise regular curves forming the boundary of $R$. Suposse $C_i$ is positively oriented and $\theta_1,\ldots,\theta_n$ be the external angles of the curves. Then: \begin{equation*} \sum\limits_{i=1}^{r} \int_{C_i} k_g^{C_i}(s) ds + \iint_R K d\sigma + \sum\limits_{j=1}^{n} \theta_j = 2\pi\chi(R) \end{equation*}
My questions are:
- What is the easiest way to orientate positively the curves?
- Should we compute the external angles before or after the orientation of the curves?
- How do we compute easily the sign of the external angles?
- Suppose we have a region $R$ with is homeomorphic to a square like this:
Is it true that $\chi(R)=1-n$, where $n$ denote the number of holes?
Answer to 1: Use the right hand rule, with respect to an orientation on the surface. The local model for using the right hand rule is that, when applied in the upper half plane of $\mathbb{R}^2$, it gives you the positive orientation on the $x$-axis.
Answer to 2: The external angles are not dependent on orientation. In your example, each of the four external angles on the outer rectangular boundary is equal to $+\pi/2$. And on each of the three inner rectangular boundaries, each of the four external angles is equal to $-\pi/2$.
Answer to 3: At a vertex $V$, if the internal angle equals $\phi$ then the external angle equals $\pi-\phi$. So, if the internal angle is $<\pi$ then the external angle is positive, whereas if the internal angle is $>\pi$ then the external angle is negative (see the answer to 2).
Answer to 4: Yes (assuming that your intention is that $R$ is the region between the outer rectangle and the three inner rectangles).