# Does there exist a surface homemomorphic to a torus with positive Gaussian curvature?

This is a problem from the my last exam in Differential Geometry II and I didn't solve it. I'm studying again, but without success. So I need help.

Does there exist a surface $S \subset \mathbb{R}^3$ which is homeomorphic to the torus $\mathbb{T}^2$ and has Gaussian curvature $K \geq 0$?

What I have to work with: Differential forms, Gauss-Bonnet Theorem, Stokes Theorem, Euler characteristic, etc.

Can someone help me?

• What does Gauss Bonnet tell you? What is $\chi(\Bbb T^2)$? – Ted Shifrin Jan 27 '14 at 18:04
• Must be $S\subset{\Bbb R}^3$. – Martín-Blas Pérez Pinilla Jan 27 '14 at 18:13

Assume such a surface $S$ exists. A torus is a surface of genus $g=1$, therefore the Euler characteristic is $\chi(S)=\chi(\mathbb{T}^2)=2-2g=0$. By the Gauss-Bonnet theorem it follows that

$$\int_{S} K dA=0$$

where $K$ is the Gaussian curvature of $S$ and $dA$ the area element on $S$. Assume $K\ge 0$. Since $K$ cannot be identically zero (*), the left hand side is positive. Contradiction.

Therefore such a surface cannot exist.

Edit:

(*) Because the only surfaces in $\mathbb{R}^3$ with constant Gaussian curvature $0$ are planes and cylinders.

• The Gaussian curvature is invariant by homeomorphisms? – Felipe Jan 27 '14 at 18:17
• @Felipe: no (deform a little bit the tip of a sphere for example), but the Euler characteristic is. Therein lies the power of Gauss-Bonnet! It combines topological data (Euler characteristic) with geometrical data (global curvature). – Bruno Stonek Jan 27 '14 at 18:20
• $K$ not being identically 0 isn't something to take for granted. This is where you need to use the fact that you are in $\mathbb{R}^3$, and not in some other ambient space. – Nate Jan 27 '14 at 18:22
• I am satisfied with the part that if $K> 0$ the left side of the integral is positive and this results in a contradiction but I could not quite understand what happens when $K =$ 0 and where the absurdity of it. – Felipe Jan 27 '14 at 18:23
• @Felipe: any compact surface in $\mathbb{R}^3$ has a point with positive curvature. Indeed, it is bounded, thus it lies inside of a big enough sphere. Now reduce the sphere's radio until it first touches the surface in a point. That point of the surface must therefore have positive curvature. (You should fill in the details for this argument). – Bruno Stonek Jan 27 '14 at 18:27