MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Regarding the sphere as complex projective line (take $(0,0,1)$ as the infinite point), the Gauss map of a smooth surface in the 3 dimensional space pulls a complex line bundle back on the surface.

My question is, what the bundle is? (In the trivial case, if the surface is sphere itself, the bundle is just the tautological line bundle.)

Does the chern class (Of course the first one) of this bundle depend on the embedding of the surface? (The Jacobian determinant of Gauss map is just the Gauss curvature, hence is intrinsic. Also its degree is the Euler $\chi$, so I ask for more...)

If yes, how much does the chern class/bundle reflect the geometry of embedding?

There may be something to make the question meaningless, such as there is no cannonical way to identify a sephere with the projective line... But as a beginner in learning geometry, I am still curious to it...

edit Thank Sergei for your answer, thank you

share|cite|improve this question
You forgot to divide $\chi$ by 2. – Sergei Ivanov Mar 28 '10 at 10:04
up vote 8 down vote accepted

I assume that your surface is closed. Suppose you have a fixed vector bundle $\xi$ over $S^2$ (no matter which one). You have an oriented surface $M$ embedded in $\mathbb R^3$, which defines the Gauss map $\nu:M\to S^2$, which defines the vector bundle $\nu^*\xi$ on $M$. You want to know whether $\nu^*\xi$ depends on the embedding.

No it does not. Indeed, $\deg\nu=\chi(M)/2$ regardless of the embedding. Two maps $f_1,f_2:M\to S^2$ having the same degree are homotopic. And homotopic maps induce the same bundle.

Concerning the Chern class, we have $c_1(\nu^*\xi)=\nu^*(c_1(\xi))$ by definition. So, if you identify the top cohomology with integers, then $c_1(\nu^*\xi)$ is (the same number as) $\deg(\nu) c_1(\xi)=\frac12\chi(M)c_1(\xi)$.

share|cite|improve this answer

Let me add something: As Sergei said, the pull-back bundle (without any geometric structure) will be the same. Nethertheless, there is more structure: Regrading your smooth embedded surface as a Riemann surface $M$ (induced from the metric or first fundamental form), you have the canonical bundle $K\to M$ on it (the bundle of complex linear 1-forms). This bundle is in a natural way a holomorphic bundle, holomorphic sections are exactly the closed complex linear forms. Now, on every Riemann surface, there are exactly $2^{2g}$ complex holomorphic line bundles $S\to M$ which satisfy $S^2=K$ holomorphically. These bundles are called spin-bundles.

It turns out that your pull-back bundle is a spin bundle of the Riemann surface $M.$ Moreover, the type of the spin-bundle tells you something about the way your surface is embedded. For example, the spin-bundle of an embedding has no global holomorphic section. Moreover two immersions are homotopic, iff the spin bundles are the same. There is a nice paper of Pinkall ("Regular homotopy classes of immersed surfaces") were all these questions are answered.

share|cite|improve this answer

Your Answer


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.