I want to prove the following theorem:
Let $S$ be a compact surface, and $N:S\rightarrow \Bbb{S}^2$ the Gauss map, then we have
$$ \int_{S} |K| \,dA = \int_{\Bbb{S}^2} \#N^{-1} \,dA $$
where $K$ is the Gaussian curvature of the surface $S$, and $\#N^{-1}$ is the number of the preimages of the Gauss map.
I know that by the stack of record theorem and the fact that $\Bbb{S}^2$ is connected, the value $\#N^{-1}$ should be a constant locally. Also, I know that locally (within a local chart of $S$), we should have
$ \int_{U} |K| \,dA = \int_{N(U)} \,dA $
But I don't know how to prove the theorem globally by involving the number of preimages of the Gauss map. Any hints would be helpful.
Best Answer
You should look at the Area Formula applicable to Lipschitz functions.
Suppose that $S$ is the image of a compact set $K$ of $\Bbb{R}^2$ through the (just one chart) smooth coordinates $\varphi$. Consider $f = N \circ \varphi$, a function from $K$ to $\Bbb{S}^2$. In this case the Area Formula gives : $$\int_K J_f(x) \; d \mathcal{L}^2(x) =\int\limits_{f(K)} \# f^{-1}(y) \; d \mathcal{H}^2(y)$$ where $J_f(x)$ is the absolute value of the Jacobian determinant of $f$, $\; d \mathcal{L}^2(x)$ is the Lebesgue measure on $\Bbb{R}^2$ and $d \mathcal{H}^2(y)$ the Hausdorff measure of $\Bbb{S}^2$.
Since $J_f(x) = J_N(\varphi(x)) \cdot J_{\varphi}(x)$ and $J_{\varphi}(x) d \mathcal{L}^2(x)$ is the volume form on $S$ we get $$\int_{S} J_N(s) \; d A(s) = \int_{K} J_N(\varphi(x)) \cdot J_{\varphi}(x) \; d \mathcal{L}^2(x) =$$ $$=\int\limits_{f(K)} \#(N \circ \varphi)^{-1}(y) \; d \mathcal{H}^2(y)=\int\limits_{N(S)} \#N^{-1}(y) \; d \mathcal{H}^2(y)$$
We obtain the result since $J_N(x)$ (without absolute values) is the Gaussian curvature.
For the general case, consider a partition of the unity of $S$ subordinate to the charts. By compactness you can find $h_1, \ldots, h_n$ (n finite) positive, smooth with compact support included in a chart and such that $h_1+\ldots + h_n=1$ on $S$. Use the following corollary of the Area Formula for the integration of $h_i$ :
$$\int_{S} h_i(s) J_N(s) \; d A(s) = \int\limits_{N(S)} \sum_{s \in N^{-1}(y)} h_i(s) \; d \mathcal{H}^2(y)$$
Finally we get :
$$\int_{S} J_N(s) \; d A(s) =$$ $$=\int_{S} \sum_{i=1}^n h_i(s) J_N(s) \; d A(s) = \sum_{i=1}^n \int_{S} h_i(s) J_N(s) \; d A(s) = \sum_{i=1}^n \int\limits_{N(S)} \sum_{s \in N^{-1}(y)} h_i(s) \; d \mathcal{H}^2(y) = $$ $$=\int\limits_{N(S)} \sum_{s \in N^{-1}(y)} \sum_{i=1}^n h_i(s) \; d \mathcal{H}^2(y) = \int\limits_{N(S)} \sum_{s \in N^{-1}(y)} 1 \; d \mathcal{H}^2(y) = \int\limits_{N(S)} \#N^{-1}(y) \; d \mathcal{H}^2(y)$$