## Surjectivity of Gauss map and its degree

For any compact surface ${M}$ (without boundary) in ${\mathbb{R}^3}$, it is well known that ${M}$ is orientable, i.e. it has a smooth (outward) normal vector field ${N}$. (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 ${M}$? ) We then define the Gauss map ${N: M\rightarrow \mathbb{S}^2}$ by ${p\mapsto N(p)}$. We have the following result.

Theorem 1 The Gauss map is surjective if ${M}$ is compact.

Proof: (Sketch) The proof is very geometric in nature. Actually we prove more: the image of ${\{p\in M: K(p)\geq 0\}}$ is the whole ${\mathbb{S}^2}$ under the Gauss map. To see this, take any direction ${v}$, push the plane perpendicular to ${v}$ from infinity towards ${M}$ until it touches ${M}$ at ${p}$. Then the normal at that point is ${v}$ and as ${M}$ all “bends away” from the plane at that point, ${K(p)\geq 0}$. $\Box$

Exercise: Make the above geometric argument rigorous.

When I discussed this result with Bean, we discovered an alternative proof for the case ${M}$ is not a torus. Actually we have proved more:

Theorem 2 (Bean’s theorem, 2010) If ${M}$ is of genus ${g}$, then ${deg(N)=1-g}$.

As ${deg(N)\neq 0}$ implies ${N}$ is surjective, we have

Corollary 3 If ${g\neq 1}$, then ${N: M\rightarrow \mathbb{S}^2}$ is onto.

Here ${deg (N)}$ 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 ${\displaystyle deg( N)=\sum_{N(p)=q}sign(dN_p)}$, where ${q}$ is a regular value. Alternatively,

$\displaystyle deg(N)=\frac{\int_M N^*\omega}{\int_{\mathbb{S}^2}\omega}=\frac{1}{4\pi}\int_M N^*\omega\in \mathbb{Z},$

where ${\omega }$ is the standard area form (or area element) of ${\mathbb{S}^2}$.

Proof: Notice that the area form ${\omega}$ is given by ${\omega_q(v_1, v_2)=\det(v_1, v_2, q)}$, where ${v_i\in T_q\mathbb{S}^2}$, i.e. it is the oriented area spanned by ${v_1, v_2}$. Thus under local coordinates ${u, v}$ of ${M}$ (assume ${X_u, X_v}$ is positively oriented ),

$\displaystyle (N^*\omega)_p(X_u, X_v)\stackrel{def}=\omega_{N(p)}(dN(X_u), dN(X_v))=\det (dN(X_u), dN(X_v), N(p))$

As ${X_u, X_v}$ is positively oriented, it is easy to see that

$\displaystyle \det (dN(X_u), dN(X_v), N(p))=K((X_u\times X_v)\cdot N)=K|X_u\times X_v|$

Thus by Gauss-Bonnet,

$\displaystyle deg(N)=\frac{1}{4\pi}\int_M N^*\omega=\frac{1}{4\pi}\int_M K |X_u\times X_v|dudv=\frac{2\pi (2-2g)}{4\pi}=1-g.$

$\Box$

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### 9 Responses to Surjectivity of Gauss map and its degree

1. Yeung Wai Kit says:

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.

2. KKK says:

Yes, of course you can prove it using Poincare duality. But I am interested to find an elementary proof which, I hope, at least could be understood by students having the first course in differential geometry.

3. marcomen says:

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.

4. KKK says:

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

5. antinatter says:

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

6. antinatter says:

Sorry, I forgot to add, connected.

7. KKK says:

No, consider $\mathbb{R}^2\subset \mathbb{R}^3$.

8. antinatter says:

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

9. KKK says:

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.