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



About the Gauss map of a surface in euclidean 3 space
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) $\deg(\nu) c\_1(\xi)=\frac12\chi(M)c\_1(\xi)$.


