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 1The 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 3If , 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,

As regards the problem that an embedded compact submanifold K in M=R^3 is orientable, this can be proved using Poincare duality. Please refer to Glen E. Bredon’s Topology and Geometry. (Section VI.8 Problem 1)

The idea is as follows:

Note that a compact (topological) n-manifold K satisfy:

H^n (K)=Z if K is orientable; and H^n (K)=Z_2 if it is not.

So it suffices to prove that H^2(K) is free in our case. But this follows easily from the very general version of Poincare duality presented in Bredon’s book.

The version it says is as follows:

Suppose K,L are compact subsets of an orientable topological n-manifold M.

Let H#^p(K,L)=direct lim ( H^p(U,V) | (U,V) contain (K,L)),

then H#^p(K,L) is isomorphic to H_(n-p) (M-L,M-K), for p=0,1,…,n.

(the isomorphism is provided by capping with the fundamental class [M] if we consider [M] also as some direct limit)

If we apply it to our case M=R^3 and K=embedded 2-manifold. Then we have:

1) Since K has a tubular neighborhood, we have H#^2(K)=H^2(K).

2) Hence H^2(K)=H_1 (M,M-K).

3) Consider the long exact sequence for (M,M-K), we have

H_1(M,M-K) = (reduced) H_0(M-K), which is a free abelian group.

Thus we have

1)H^2(K) is free, and hence must be Z, which means K is orientable

2) (reduced) H_0(M-K) must be Z, and hence M-K has two components.

Yes, of course you can prove it using Poincare duality. But I am interested to find an

elementaryproof which, I hope, at least could be understood by students having the first course in differential geometry.Take a look at the book “Curves and Surfaces” by Sebastián Montiel and Antonio Ros, there is a proof of that fact which I think is sort of what you want. It is an application of the Jordan-Brower Theorem (for which they also present an elementary proof). The book is for a first course on the subject.

Great! Thank you! That’s exactly what I’m looking for :-)

Can the Gauss map be surjective for a noncompact (orientable) surface?

Sorry, I forgot to add, connected.

No, consider .

You’ve misunderstood my question. What I meant was, is there a connected noncompact surface the Gauss map on which is onto?

Take a revolutionary torus whose axis of rotation is the z-axis, Delete from it one point on the “top”, i.e. whose z value is maximum. This surface is non-compact whose Gauss map is onto.