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.


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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.

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