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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.

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    $\begingroup$ No, $\#N^{-1}(q)$ is not a constant function of $q\in\Bbb S^2$. The number of preimages, counted with orientation sign, is independent of the regular value $q$, but this is a very different statement. Try examples of a deformed sphere (so that some $q$ have 3 or more preimages) or just a regular torus of revolution. You might want to think directly of the Gaussian curvature as the Jacobian of the Gauss map and then think about the change of variables theorem. $\endgroup$ Commented Mar 20 at 21:18
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    $\begingroup$ The number of preimages is not globally constant, otherwise you would get a homeomorphism from say the torus to some number of disjoint copies of the sphere. I think it is constant everyone except some set of measure zero where it has some singularities. $\endgroup$
    – quarague
    Commented Mar 20 at 21:19
  • $\begingroup$ Thanks for your comments, it should be a local constant $\endgroup$
    – vegetandy
    Commented Mar 20 at 21:41
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    $\begingroup$ @quarague Not true. You can easily draw examples where the number is $1$ on an open set and $2$ on an open set. $\endgroup$ Commented Mar 21 at 0:41

1 Answer 1

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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 :

$$\int_{S} J_N(s) \; d A(s) =$$ $$=\int_{S} \sum_{i=1}^n h_i(s) J_N(s) \; d A(s) = \sum_{i=1}^n \int_{S} h_i(s) J_N(s) \; d A(s) = \sum_{i=1}^n \int\limits_{N(S)} \sum_{s \in N^{-1}(y)} h_i(s) \; d \mathcal{H}^2(y) = $$ $$=\int\limits_{N(S)} \sum_{s \in N^{-1}(y)} \sum_{i=1}^n h_i(s) \; d \mathcal{H}^2(y) = \int\limits_{N(S)} \sum_{s \in N^{-1}(y)} 1 \; d \mathcal{H}^2(y) = \int\limits_{N(S)} \#N^{-1}(y) \; d \mathcal{H}^2(y)$$

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