# Area - Gaussian curvature

I am looking at the following exercise:

Let $S$ be the torus $$\sigma (\theta, \phi)=((a+b\cos\theta )\cos \phi , (a+b\cos \theta )\sin \phi , b\sin\theta)$$ Describe the parts $S^+$ and $S^−$ of $S$ where the Gaussian curvature $K$ of $S$ is positive and negative, respectively. Show, without calculation, that

$$\iint_{S^+}K d\mathcal{A} = −\iint_{S^−}K d\mathcal{A} = 4\pi$$



I have done the following:

$$K=\frac{LN-M^2}{EG-F^2}=\frac{\cos \theta}{b(a+b\cos\theta)}$$

Since $0<b<a$ we have that $a+b\cos\theta>0$.

Therefore, $$K>0 \Leftrightarrow \cos\theta>0 \Leftrightarrow 0<\theta<\frac{\pi}{2} \land \frac{3\pi}{2}<\theta<2\pi$$ and $$K<0 \Leftrightarrow \cos\theta<0 \Leftrightarrow \frac{\pi}{2}<\theta<\frac{3\pi}{2}$$

So, $$S^+=\{(\theta, \phi) \mid 0<\theta<\frac{\pi}{2} \land \frac{3\pi}{2}<\theta<2\pi, 0\leq \phi \leq 2\pi\}$$ and $$S^-=\{(\theta, \phi) \mid \frac{\pi}{2}<\theta<\frac{3\pi}{2}, 0\leq \phi \leq 2\pi\}$$

Is this correct?



In a previous exericse I have shown the following:

Let $\sigma : U \rightarrow \mathbb{R}^3$ be a patch of a surface $S$. The image under the Gauss map of the part $\sigma (R)$ of $S$ corresponding to a region $R \subseteq U$ has area $$\iint_R |K|d\mathcal{A}σ,$$ where $K$ is the Gaussian curvature of $S$.



Is it maybe as follows?

$$\iint_S |K|d\mathcal{A}=\iint_{S^+} Kd\mathcal{A}-\iint_{S^-} Kd\mathcal{A}$$

How could we continue to get the desired result?



According to

$$\iint_{S^+}|K|d\mathcal{A}$$ is the area of the image under the Gauss map of the part $\sigma (S^+)$ corresponding to a region $S^+$.

The image under the Gauss map is the values of $\textbf{N}$ at the points of $S$.

So is maybe the image under the Gauss map in this case a surface of which the area is known?

• You better parametrize $S_{+}$ as $\{ ( \theta, \phi ) : |\theta| < \pi/2, \phi \in [0,2\pi) \}$, the normal map $N$ is a bijection between the part corresponding to $S_{+}$ on the torus and the unit sphere excluding the north and south poles. Commented Jan 4, 2016 at 23:27
• How do we get that $|\theta |<\frac{\pi}{2}$ ? Isn't the interval of $\theta \in [0,2\pi]$ ? How do we get then $-\frac{\pi}{2}<\theta<\frac{\pi}{2}$ ? Could you explain to me further why $N$ is that bijection? @achillehui Commented Jan 5, 2016 at 0:17
• For any $(\theta,\phi) \in \mathbb{R}^2$, the map $\sigma$ defines a parametrization of the torus on some neighborhood of $\sigma(\theta,\phi)$. The choice of domain $[0,2\pi] \times [0,2\pi]$ is only one of them. The problem is in this parametrization, the description of $S_{+}$ isn't that clean. About $N$, the normal at $\sigma(\theta,\phi)$ is $(\cos\theta\cos\phi,\cos\theta\sin\phi,\sin\theta)$. This provides a parametrization of the unit sphere ( same as (latitude, longitude) for earth or ( declination, right ascension) for sky). Commented Jan 5, 2016 at 1:23
• I got it!! Is $$\iint_S |K|d\mathcal{A}=\iint_{S^+} Kd\mathcal{A}-\iint_{S^-} Kd\mathcal{A}$$ correct? Or should it be "plus" instead of "minus" ? @achillehui Commented Jan 5, 2016 at 1:55
• Isn't it $$N=(-\cos\theta \cos\phi, -\cos\theta\sin\phi , -\sin\theta)$$ Or have I calculated it wrong? @achillehui Commented Jan 5, 2016 at 1:59

If you know the Gauss-Bonnet theorem then, you know that

$$0= 4\pi*(2-2g) =\iint_S K \;\mathrm{d}\mathcal{A} = \iint_{S^+} K \;\mathrm{d}\mathcal{A}+\iint_{S^-} K \;\mathrm{d}\mathcal{A},$$

thus,

$$\iint_{S^+} K \;\mathrm{d}\mathcal{A}= -\iint_{S^-} K \;\mathrm{d}\mathcal{A}.$$

We also know from Gauss-Bonnet that,

$$\iint_{S^+}K \; \mathrm{d}\mathcal{A} = \int_{\delta S^+} \kappa_g \mathrm{d}\mathcal{s},$$

where $\kappa_g$ is the geodesic curvature, but the boundary of $S$ is two circles along which $K\equiv0$. (We know this since $K$ switches sign here). Thus, these circles are asymptotic lines, and the geodesic curvature is just the curvature of the circles. Since the total curvature of a circle is $2\pi$ and there are two of them, we are done.

EDIT: $\sigma(S^{+})$ is a sphere a radius 1. The normal traces out a circle along the "equator" of the torus that corresponds to the equator of the sphere, and for every point on this circle it traces out a perpendicular circle (half on each side of the torus). This is easy to understand if you just think of the directions that the normals point in. (And the radius is 1 since $|N|=1$).

EDIT 2: You can think of $S^+$ as a sphere with a pair of antipodal points poked out, and then stretched along its equatorial plane.

EDIT 3: If you're allowed to compute $N$, then the field of normals is clearly a unit sphere when restricted to $S^+$ or $S^-$.

If you're not allowed to compute $N$, then consider the (non-regular) torus with $a=0$, this is clearly just a sphere of radius $b$ when restricted to $S^+$ or $S^-$. You know that the Gauss map of this sphere is a sphere of radius 1.

Now consider $\sigma$ restricted to $S^+$. If you scale $a$ and $b$ simultaneously, then the Gauss map will remain unchanged since it is scale invariant.

Since $\sigma$ is continuous as a function of $a$ and $b$, the Gauss map approaches a unit sphere (which does not depend on $b$) as $a\to 0$. This means that $\forall \epsilon>0 \exists \delta$ s.t. $a\in B_\delta(0), b\in \mathbb{R^+} \Rightarrow |N -\sigma| < \epsilon$. But since the Gauss map is scale invariant, you can choose $a$ in the $\delta$-ball and any $b$ you want and then just scale to get any pair for $a$ and $b$ that yield Gauss maps that are arbitrarily close to a unit sphere so $\epsilon$ can be as large as you want. This implies that $\forall a,b\in \mathbb{R^+}$ the Gauss map is constant. In particular it is the same as it is for $a=0$ which is the unit sphere.

• We haven't done the Gauss-Bonnet theorem in class... Is this the only way to calculate it? Commented Jan 4, 2016 at 23:13
• I just edited the answer to explain that the image under the Gauss map of $S^+$ is a sphere of radius 1. Commented Jan 4, 2016 at 23:26
• I'm not exactly sure what it means to "show without calculation". I have a feeling that this means without directly computing the integral. If you compute the value of $N$, then it is immediately clear that $N$ as a function of $\phi$ and $\theta$ is just a unit sphere for $(\phi, \theta)\in S^+$. Commented Jan 7, 2016 at 18:44
• Yeah, I also added an explanation that requires zero calculation although it's a little complicated. Commented Jan 7, 2016 at 19:39
• Yup that's right. Commented Jan 7, 2016 at 19:50