What have I missed?
I'm trying to show that $N(A)$ and $N(B)$ have the same area, under the following conditions:
- $N$ is the Gauss map $N: S \rightarrow S^2$.
- The regions $A$ and $B$ on $S\subset\mathbb{R}^3$ share $\alpha \subset S$, a simple closed geodesic on $S$, as a common boundary.
- $S$ is a regular compact surface with $K>0$ (Positive Gaussian curvature).
I've attempted the following:
given $K>0$, the map $N$ preserves orientation, therefore is a diffeomorphism from $S$ onto the sphere. Using the Gauss-Bonnet Theorem on region $A$, we have the following formula: $$\int_{\Gamma} \kappa_g(s) ds + \iint_A K d\sigma + \sum_{l=1}^p\theta_l = 2\pi\chi(A)$$ and since the geodesic curvature of $\Gamma$ vanishes on $S$, we have $k_g = 0 \implies \int_{\Gamma} k_g(s) ds =0$. Also, since region $A$ is homeomorphic to a disk, the Euler characteristic is 1, hence RHS = $2\pi$.
Angle sum term would be zero because $\Gamma$ is a simple closed curve that bounds a region homeomorphic to a disk. Therefore $\iint_A K d\sigma = 2\pi$.
Since Gaussian curvature can be written as the ratio $\frac{\mid N_u x N_v\mid}{\mid X_u x X_v \mid}$, it follows that $$\mid ~N_u \times N_v\mid ~= ~\mid K \mid \cdot\mid X_u \times X_v \mid.$$ By integration,
Area of $N(A) = \iint_{N(A)} 1 = \iint_A\frac{\mid N_u x N_v\mid}{\mid X_u x X_v \mid} = \iint_A \mid K\mid = 2\pi$.
By the same argument, I get that $N(B)$ has the exact same area, $2\pi$.
However, I have not used the fact that the regions $A$ and $B$ share a common boundary, and am not sure where this piece of information would apply. What does a common boundary indicate, and is this information relevant to my approach?
Thanks!