Total absolute curvature of a compact surface

[Goklayeh]

Total "absolute" curvature of a compact surface

Hi! Someone could help me resolving the following problem? Let $$\Sigma \subset \mathbb{R}^3$$ be a compact surface: show that
$$\int_{\Sigma}{|K|\mathrm{d}\nu} \ge 4\pi$$

where $$K$$ is the gaussian curvature of $$\Sigma$$. The real point is that I want to prove this using weaker results as possible (in particular, the Gauss-Bonnet theorem). I tried using that
$$\int_{\Sigma}{|K|\mathrm{d}\nu} = \text{Area}\left(N(\Sigma^+)\right) + \text{Area}(N(\Sigma^-))$$
where
$$\Sigma^{\pm}:=\left\{p \in \Sigma \biggr| \text{sign}\left(K(p)\right)=\pm 1\right\}$$
$$N$$ is the Gauss application
and the formula is to be understood as "counted with multiplicity", i.e. we don't consider overlaps
(as follows from, e.g., the proof of Proposition 2 on page 167 of do Carmo's Differential Geometry). I'm pretty sure that this way will bring to some result, but i don't have any idea how to proceed from here! Thanks in advance for your help!

 

[lavinia]

Hi! Someone could help me resolving the following problem? Let $$\Sigma \subset \mathbb{R}^3$$ be a compact surface: show that
$$\int_{\Sigma}{|K|\mathrm{d}\nu} \ge 4\pi$$

where $$K$$ is the gaussian curvature of $$\Sigma$$. The real point is that I want to prove this using weaker results as possible (in particular, the Gauss-Bonnet theorem). I tried using that
$$\int_{\Sigma}{|K|\mathrm{d}\nu} = \text{Area}\left(N(\Sigma^+)\right) + \text{Area}(N(\Sigma^-))$$
where
$$\Sigma^{\pm}:=\left\{p \in \Sigma \biggr| \text{sign}\left(K(p)\right)=\pm 1\right\}$$
$$N$$ is the Gauss application
and the formula is to be understood as "counted with multiplicity", i.e. we don't consider overlaps
(as follows from, e.g., the proof of Proposition 2 on page 167 of do Carmo's Differential Geometry). I'm pretty sure that this way will bring to some result, but i don't have any idea how to proceed from here! Thanks in advance for your help!

I think you are on the right track.

Here is another approach.
The integral of the Gauss curvature is 2pi times the Euler characteristic of the surface. The sphere has Euler characteristic 2 and all other surfaces except the torus have Euler characteristic a negative even integer. So for these the theorem is immediate. The only hard case is the torus whose Euler characteristic is zero. I am not sure how to do this.

 

[lavinia]

Here are some more more thoughts. But not an answer.

If the Gauss mapping is surjective then the integral of the absolute Gauss curvature must equal at least the volume of the unit sphere which is 4pi. This I think follows from the change of variables formula for integration since the image of the critical values of the Gauss map has measure zero by Sard's theorem.

For any surface other than the torus, the Gauss map must be surjective since if not the tangent bundle of the surface would be trivial - I think - but check this.

For the torus one would need to show that the Gauss map at least covers a hemisphere.

 

[Goklayeh]

I think I've solved: if $$\Sigma$$ is a compact surface, then the Gauss map restricted to parabolic and elliptic points
$$N: \Sigma^+_0:=\left\{p \in \Sigma\biggr| K(p) \ge 0\right\} \longrightarrow S^2$$
is surjective (this follows from a simple geometric argument), so
$$\int_{\Sigma}{|K|} = \int_{\Sigma^+_0}{K} - \int_{\Sigma^-}{K} = \text{Area}\left(N(\Sigma^+_0)\right) + \text{Area}(N(\Sigma^-)) = \text{Area}\left(S^2\right) + \text{Area}(N(\Sigma^-)) = 4\pi + \text{Area}(N(\Sigma^-)) \ge 4\pi$$
Is it right?

 

[lavinia]

I think I've solved: if $$\Sigma$$ is a compact surface, then the Gauss map restricted to parabolic and elliptic points
$$N: \Sigma^+_0:=\left\{p \in \Sigma\biggr| K(p) \ge 0\right\} \longrightarrow S^2$$
is surjective (this follows from a simple geometric argument), so
$$\int_{\Sigma}{|K|} = \int_{\Sigma^+_0}{K} - \int_{\Sigma^-}{K} = \text{Area}\left(N(\Sigma^+_0)\right) + \text{Area}(N(\Sigma^-)) = \text{Area}\left(S^2\right) + \text{Area}(N(\Sigma^-)) = 4\pi + \text{Area}(N(\Sigma^-)) \ge 4\pi$$
Is it right?

I don't think this is right because you don't know a priori that the area covered by elliptic points is the whole sphere unless I am really missing the point. I agree with you that this approach feels right and that in fact the Gauss map must be surjective for any surface other than the torus which is still stumping me.

 

[Goklayeh]

I'm saying that
$$\Sigma \subset \mathbb{R}^3$$ compact $$\Rightarrow$$ Gauss map restricted to $$\Sigma^+_0$$ is surjective

and that's for sure! If this statement is the problem, I can post the proof. Otherwise, I didn't understand the objection! (Anyway, thanks for your time lavinia!)

 

[lavinia]

I'm saying that
$$\Sigma \subset \mathbb{R}^3$$ compact $$\Rightarrow$$ Gauss map restricted to $$\Sigma^+_0$$ is surjective

and that's for sure! If this statement is the problem, I can post the proof. Otherwise, I didn't understand the objection! (Anyway, thanks for your time lavinia!)

Give the proof. I don't see why the Gauss map is surjective for the case of the torus. And you are saying more - that the restriciton of the Gauss map to the elliptic points is surjective.

The arguments I gave, the Euler characteristic argument and the non-triviality of the tangent bundle argument both fail for the torus but work for every other surface.

 

[lavinia]

I guess if you take any plane and parallel translate it until it first touches the surface then it must be tangent to the surface at that point and the curvature must also be positive at that point. This is like Hilbert's proof that every surface must have a point of positive curvature. Very cool.

 

[Goklayeh]

Ok, here is the proof. Recall that if $$\Sigma \subset \mathbb{R}^3$$ is a compact surface and $$\Pi$$ is a plane, we can define the function "height from a plane"
$$\begin{matrix} h:& \Sigma & \longrightarrow & \mathbb{R}\\ & p & \mapsto & \text{d}\left(p, \Pi\right) \:\:(\text{ with sign}) \end{matrix}$$
and $$dh = 0 \Leftrightarrow T_p \Sigma \parallel \Pi$$. Furthermore, the hessian is defined only on crital points, and $$\text{He}_h(p) = II_{ff}(h)$$. So, if $$p$$ is a min/max for $$h$$, then the matrix $$\text{He}_h(p) = II_{ff}(h)$$ is semidefinite, and then $$K(p)\ge 0$$.
Now, if $$\Sigma$$ is compact, in particular is closed and then orientable, so $$N:\Sigma\rightarrow S^2$$ is well defined.. Furthermore, by Jordan's Theorem, $$\Sigma$$ splits $$\mathbb{R}^3$$ in two connected components, one "external" and unbounded, and the other "internal" and bounded. Because of the boundedness of $$\Sigma$$, there exists an $$R>0$$ s.t. $$\Sigma \subset B_R$$. Let $$S_R:=\partial B_R$$. Now, let $$q \in S^2$$ and $$R_q$$ be the point of $$S_R$$ obtained extending the radious vector to $$q$$ up to $$S_R$$. Let now $$\Pi:=T_{R_q}S_R$$, and consider $$h:\Sigma \rightarrow \mathbb{R}$$, the distance from $$\Pi$$. $$\Sigma$$ compact $$\Rightarrow\:\: \exists M \in \Sigma$$ s.t. $$h(M)=\max{h}$$. So, $$M$$ is a critic point for $$h$$, and thus $$T_M\Sigma \parallel \Pi$$. Choosing as $$N$$ the normal external vector, we'll have $$N(M)=q$$. Because of the arbitrariness of $$q$$, we get the thesis (or at least, we should, if I'm right!)

Is the argument valid in your opinion?

 

[Goklayeh]

Oops! I haven't seen your reply!

 

[lavinia]

Ok, here is the proof. Recall that if $$\Sigma \subset \mathbb{R}^3$$ is a compact surface and $$\Pi$$ is a plane, we can define the function "height from a plane"
$$\begin{matrix} h:& \Sigma & \longrightarrow & \mathbb{R}\\ & p & \mapsto & \text{d}\left(p, \Pi\right) \:\:(\text{ with sign}) \end{matrix}$$
and $$dh = 0 \Leftrightarrow T_p \Sigma \parallel \Pi$$. Furthermore, the hessian is defined only on crital points, and $$\text{He}_h(p) = II_{ff}(h)$$. So, if $$p$$ is a min/max for $$h$$, then the matrix $$\text{He}_h(p) = II_{ff}(h)$$ is semidefinite, and then $$K(p)\ge 0$$.
Now, if $$\Sigma$$ is compact, in particular is closed and then orientable, so $$N:\Sigma\rightarrow S^2$$ is well defined.. Furthermore, by Jordan's Theorem, $$\Sigma$$ splits $$\mathbb{R}^3$$ in two connected components, one "external" and unbounded, and the other "internal" and bounded. Because of the boundedness of $$\Sigma$$, there exists an $$R>0$$ s.t. $$\Sigma \subset B_R$$. Let $$S_R:=\partial B_R$$. Now, let $$q \in S^2$$ and $$R_q$$ be the point of $$S_R$$ obtained extending the radious vector to $$q$$ up to $$S_R$$. Let now $$\Pi:=T_{R_q}S_R$$, and consider $$h:\Sigma \rightarrow \mathbb{R}$$, the distance from $$\Pi$$. $$\Sigma$$ compact $$\Rightarrow\:\: \exists M \in \Sigma$$ s.t. $$h(M)=\max{h}$$. So, $$M$$ is a critic point for $$h$$, and thus $$T_M\Sigma \parallel \Pi$$. Choosing as $$N$$ the normal external vector, we'll have $$N(M)=q$$. Because of the arbitrariness of $$q$$, we get the thesis (or at least, we should, if I'm right!)

Is the argument valid in your opinion?

I believe that this is right but can't convince myself that the tangent space at the maximum must be parallel to the given plane. Pardon my stupidity.

 

[Goklayeh]

I was looking for a rigorous proof of this fact on my notes, but I think that is intuitively trivial (I was trying arguing by absurdum...). Maybe a look at this on a sphere could be useful to convince yourself (or, at least, I guess!)

 

[lavinia]

I was looking for a rigorous proof of this fact on my notes, but I think that is intuitively trivial (I was trying arguing by absurdum...). Maybe a look at this on a sphere could be useful to convince yourself (or, at least, I guess!)

Right. But this is the same as the touching argument. Rest the surface on a plane then the minimum of the height function occurs at the touching point where the tangent plane is equal to the given plane.