Prove that $\int |K|\geq 8\pi$ for a compact surface with genus $>0$

Question

Let $M$ be a compact orientable smooth surface in $\mathbb R^3$ with $g>0$, and $K$ be its gaussian curvature. Prove that $$\int_M |K|\geq 8\pi.$$

One finds that, if $M$ is a $g$-torus, the two "extremal" regions with positive curvature are such that the Gauss map $N$ is a diffeomorphism from each of them onto $S^2\setminus\{N,S\}$. Then, since $K=\det\ dN$, the change of variables theorem gives us that $\int_M|K|\geq 4\pi$. Furthermore, applying the same argument to the most extremal negative-curvature regions (where you "put your fingers" when you grab the $g$-torus) we obtain another $4\pi$-term.

Therefore, in a $g$-torus $$\int_M |K|\geq 8\pi.$$

Now, we know that every compact, orientable surface of genus $g$ is diffeomorphic to a $g$ torus. Can this assure us the same property on the integral?

Thank you in advance.

Answer 1

This is not true for abstract surfaces (i.e. $2$-dimensional manifolds). Indeed the flat torus satisfies all the assumptions and has zero curvature everywhere.

For surfaces in $\mathbb{R}^3$, it is still not true if you allow less than perfectly smooth surfaces: the flat torus can be isometrically $C^1$-embedded in $\mathbb{R}^3$. Because of this, I would expect the statement to be false also for $C^\infty$-manifolds, but I don't have a proof.

Answer 2

Here is some heuristic: By the Gauss-Bonnet Theorem the total curvature of such a surface $S$ is $$\int_SK\>{\rm d}\omega=4\pi(1-g)\ .\tag{1}$$ Consider now the convex hull $C$ of this surface. Its boundary $\partial C$ is homeomorphic to a sphere and is lying on $S$ where it is not flat. Let $S':=S\cap \partial C$. Then $K\geq0$ on $S'$, and $$\int_{S'} |K|\>{\rm d}\omega=\int_{S'} K\>{\rm d}\omega=\int_{\partial C}K\>{\rm d}\omega=4\pi\ .\tag{2}$$ From $(1)$ and $(2)$ we therefore get $$\int_{S\setminus S'}|K|\>{\rm d}\omega\geq - \int_{S\setminus S'}K\>{\rm d}\omega=-\left(\int_S K\>{\rm d}\omega-\int_{S'}K\>{\rm d}\omega\right)=-\bigl(4\pi(1-g)-4\pi\bigl)=4\pi g\ .$$ It follows that $$\int_S|K|\>{\rm d}\omega\geq4\pi (g+1)\geq8\pi\ .$$

Answer 3

You can deduce this for smooth surfaces, using basic Morse theory, from Proposition 2.14 in Langevin-Shifrin, Polar Varieties and Integral Geometry, Am. J. Math. 104, No. 3 (1982), pp. 553-605. This is an integro-geometric version of the Gauss-Bonnet Theorem, both with and without absolute values, which says (in your case) that $$\int_M |K|\,dA = \int_{\Bbb RP^2} |\mu|(M,L)\,dL,$$ where $|\mu|(M,L)$ is the number of critical points of the orthogonal projection of $M$ onto the line $L$ through the origin in $\Bbb R^3$. (This number is finite for almost all $L$.) Since Morse theory tells us that for $g\ge 1$ we must have at least $4$ critical points, we get $\int_M |K|\,dA \ge 4\cdot 2\pi = 8\pi$.

Even better, it follows from Theorem 1 of Chern-Lashof's On the Total Curvature of Immersed Manifolds, Am. J. Math., 79 (1957), pp. 306-318, which as a simple corollary implies that for a smooth compact surface $M$ immersed in $\Bbb R^3$ the total absolute curvature is at least $2\cdot 4\pi = 8\pi$.

Both theorems apply in much greater generality (both to higher codimensions, and the former to compact submanifolds with boundary).