A Converse to the Gauss Bonnet Theorem

Question

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 $F=N\circ \phi$?

Answer 1

$\newcommand{\bR}{\mathbb{R}}$ 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.

Answer 2

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.