# The integral of the absolute value of the Gaussian curvature of a compact surface

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.

• 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. Commented Mar 20 at 21:18
• 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. Commented Mar 20 at 21:19
• Thanks for your comments, it should be a local constant Commented Mar 20 at 21:41
• @quarague Not true. You can easily draw examples where the number is $1$ on an open set and $2$ on an open set. Commented Mar 21 at 0:41

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 :

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