TITLE: A Converse to the Gauss Bonnet Theorem
QUESTION [5 upvotes]: Let $S$ be  a compact surface in $\mathbb{R}^{3}$ with the gauss normal map $N:S\to \mathbb{S}^{2}$. Assme that $\phi;\mathbb{S}^{2}\to S$ is a diffeomorphism. Put $F=N\circ \phi$ and represent $F:\mathbb{S}^{2}\to \mathbb{S}^{2}$ in the form $F=(f,g,h)$. then as a consequence of the Gauss Bonnet theorem we have \begin{equation}\iint_{\mathbb{S^{2}}}  fdgdh= 4/3 \; \pi\end{equation} (See page 14 of the(printed version) of the book, Non commutative geometry by Alain Connes
My question is that ;is the converse of the above statment true?
That is: 

Assume that $F:\mathbb{S}^{2}\to \mathbb{S}^{2}$ is a smooth map with $F=(f,g,h)$ such that the above integral equality hold. Is there a compact surface $S$ with Gauss normal map $N$ and a diffeomorphism $\phi: \mathbb{S}^{2} \to S$ such that $F=N\circ \phi$?

REPLY [11 votes]: First of all, the identity holds for any degree 1 map $F:\mathbb S^2\to\mathbb S^2$. Moreover, for any $F=(f,g,h):\mathbb S^2\to\mathbb S^2$,
$$
 \int_{\mathbb S^2} f\,dgdh = \frac43\pi \deg F.
$$
This follows from Stokes' formula. So basically the identity means that the Gauss map of a surface diffeomorphic to the sphere has degree 1. This is by the way not always true, you have to choose $N$ according to the orientation (that you use to define the degree), otherwise you get the value $-\frac43\pi$. In other words, $\phi$ must be orientation-preserving.
Back to the question, the answer is trivially yes if $F$ is a diffeomorphism, by Alex Degtyarev's comment. In general, the answer is no because certain types of singularities can not occur.
For example, consider $F$ that maps some circle $C\subset\mathbb S^2$ to the north pole but such that the $F$-image of an open disc $D\subset\mathbb S^2$ bounded by $C$ avoids the north and south poles. It is easy to construct such a map of degree 1. Suppose that the desired $S$ and $\phi$ exist and consider the disc $\phi(D)$ in $S$. On its boundary $\phi(C)$ the normal is vertical, hence $\phi(C)$ is contained in some horizontal plane $H$. Then there is a point in the interior of $\phi(D)$ where the tangent plane is horizontal (look at a point furthest from $H$). The normal at this point is vertical contrary to the fact that $F(D)$ avoids the north and south poles.