Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $F=N\circ \phi$?

share|improve this question
4  
What about $N=\operatorname{id}$ and $\phi=F$? –  Alex Degtyarev Feb 12 at 18:33
add comment

2 Answers

up vote 5 down vote accepted

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

share|improve this answer
    
thank you for your interesting answer and useful information. –  Ali Taghavi Feb 14 at 14:03
    
But I have some questions:what is the restriction on singularities? Do these restriction invariant after diffeo $\phi$? Is there a degree-1 map on $\mathbb{S}^{2}$ which singularities does not satisfy the restriction- condition? –  Ali Taghavi Feb 14 at 14:07
1  
The answer to your questions is rather involved. The book I mentioned has many more details. The nature of the singularities is independent of the choice of local coordinates and in particular independent of the diffeomorphism $\phi$. –  Liviu Nicolaescu Feb 15 at 13:26
add comment

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.

share|improve this answer
    
thank you very much for your interesting answer. why there is an interior point which tangent space is horizontal? –  Ali Taghavi Feb 14 at 14:17
1  
Take a point with the highest $z$-coordinate. If it is on the boundary, take one with the lowest $z$-coordinate. Since the entire boundary is on the same level, either a highest or lowest point is in the interior. –  Sergei Ivanov Feb 14 at 16:27
    
thanks for the comment. what is a degree-1 map with the property which you mentioned? –  Ali Taghavi Feb 15 at 9:56
1  
For example, contract a hemisphere to a segment by contracting each radial circle to a point. The resulting space is a bouquet of a sphere and a circle. Place the segment to the sphere by an injective map. This can be made smooth. –  Sergei Ivanov Feb 17 at 16:24
    
Thank you very much for your very beautiful and intuitive answer. It was difficult for me to choose an answer to accept. I hope that this restriction in MO would be removed:meta.mathoverflow.net/questions/1491/… –  Ali Taghavi Feb 23 at 13:44
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.