# Possibility of a Gauss Bonnet theorem for the total mean curvature

For a compact, regular orientable surface $$S$$ in $$\mathbb{R}^3$$, Gauss-Bonnet states that its total Gaussian curvature (which a priori we know from theorema Egregium is a geometrical invariant) is determined by its Euler characteristic (i.e, it's also a topological invariant, which is even stronger). Explicitly, we have that: $$\int_{S} K \ \mathrm{d} A = 2 \pi \cdot \chi(S)$$

I wondered, could there be an expression like this replacing $$K$$ for $$H$$? In other words, could there exist a real function $$f$$ which works for all such surfaces $$S$$, such that: $$\int_{S} H \ \mathrm{d} A = f(\chi(S))$$

Obviously (since $$H$$ is not a geometrical invariant), it would seem the answer is no (and at first you'd think you'd need at least some condition on $$f$$ like it being a polynomial, but I think no such condition is necessary). I figured out a pretty simple proof but it's so simple I'm not sure it's correct, hence me asking this question here. It's enough to notice that the right hand side must necessarily be invariant by changes of scale (since the Euler characteristic is a topological invariant), while the left hand side is not. Indeed, if we take $$S$$ to be the sphere of radius $$1$$, we get that $$f(2) = 4\pi$$, while if $$S$$ is the sphere of radius $$\frac{1}{2}$$, we get that $$f(2) = 2 \pi$$, therefore no such $$f$$ can exist.

Is my reasoning correct or am I missing something here? If so, is there another argument for why there can be no such $$f$$?

• What happens to both sides of the equation if you change the orientation of $S$? Your argument is fine, too. Apr 28, 2022 at 0:22
• @TedShifrin since topology is independent of orientation, the right hand side would remain the same and (assuming connectedness, so that there are only two possible orientations) the left hand side would switch signs (because $H$ would switch signs), right? Then we'd get a contradiction that the total mean curvature of any surface is $0$. Is that what you were getting at? Apr 28, 2022 at 0:25
• That’s one contradiction you can choose to arrive at, yes. Apr 28, 2022 at 0:26
• @TedShifrin thanks a lot! Did you have some other contradiction in mind when you wrote the first comment? Apr 28, 2022 at 0:28
• Can’t you use a sphere again? Apr 28, 2022 at 0:30

I assume that this is not what you are looking for, but what you can do is consider the Willmore energy of a closed (compact without boundary) embedded surface $$\Sigma \subset S^3$$: $$\mathcal{W}(\Sigma)=\int_{\Sigma}(1+H^{2})d\Sigma.$$ Fernando Codá Marques and André Neves proved some years ago that for surfaces as before, with genus at least one, we have that $$\mathcal{W}(\Sigma)\geq 2 \pi^{2}$$ with equality if and only if $$\Sigma$$ is the Clifford torus ($$S^1(1/\sqrt{2}) \times S^{1}(1/\sqrt{2})$$) (up to conformal transformations). This works is famous because they proved the Willmore conjecture (open for almost 60 years) and because they used min-max methods (sweepouts) and Almgren-Pitts theory.