For any compact surface (without boundary) in , it is well known that is orientable, i.e. it has a smooth (outward) normal vector field . (I think it’s not an easy theorem, anyone know where can I find an elementary proof, esp. without referring to classification theorem or additional (e.g. complex) structure on ? ) We then define the Gauss map by . We have the following result.
Theorem 1 The Gauss map is surjective if is compact.
Proof: (Sketch) The proof is very geometric in nature. Actually we prove more: the image of is the whole under the Gauss map. To see this, take any direction , push the plane perpendicular to from infinity towards until it touches at . Then the normal at that point is and as all “bends away” from the plane at that point, .
Exercise: Make the above geometric argument rigorous.
When I discussed this result with Bean, we discovered an alternative proof for the case is not a torus. Actually we have proved more:
As implies is surjective, we have
Corollary 3 If , then is onto.
Here is the degree of the Gauss map. For those who have attended Bean’s lectures on degree of maps, I think you know what a degree is. It is defined as , where is a regular value. Alternatively,
where is the standard area form (or area element) of .
Proof: Notice that the area form is given by , where , i.e. it is the oriented area spanned by . Thus under local coordinates of (assume is positively oriented ),
As is positively oriented, it is easy to see that
Thus by Gauss-Bonnet,