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 $F=N\circ \phi$?
> 
> 
> 



A Converse to the Gauss Bonnet Theorem
$\newcommand{\bR}{\mathbb{R}}$ If $\Sigma\subset \bR^3$ is a cooriented surface then its Gauss map $\Gamma:\Sigma\to S^2$ has a symplectic nature. Its graph, viewed as a submanifold of $\bR^3\times S^2$ is a Legendrian submanifold with respect to the canonical contact structure on $\bR^3\times S^2$. 


The Legendrian condition imposes restrictions on the types of singularities of this map. In other words, there are also local obstructions to the converse. 


This is similar with the fact that a smooth map $F:\bR^n\to\bR^n$ need not be the differential of a function $U:\bR^n\to\bR$. If it where, the graph of $F$ would be a lagrangian submanifold of $\bR^{2n}$ with the canonical symplectic structure. For details see Volume 1 of 



> 
> Arnold, Gusein-Zade, Varchenko : Singularities of Differentiable Maps.
> 
> 
> 


