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It seems that the tangent bundle of a hypersurface of Euclidean space is the bundle induced from the tangent bundle of the unit sphere under Gauss mapping. Is this true?

The reason I think this is that tangent space at a point on the surface can be parallel translated to the tangent space on the sphere at the unit normal translated to the origin.

If this is true then there is an induced Levi-Cevita connection on the hypersurface under the Gauss mapping. What are the conditions under which this induced connection is the same as the connection that the hypersurface inherits from the embedding?

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In general the Gauss map is a map to the real projective space corresponding to the normal line -- take for example a Moebius band in $\mathbb R^3$. In this generality, yes the Gauss map is tautologically the classifying map for the tangent bundle, because the space of lines in $\mathbb R^n$ (by orthogonal complements) is the Grassmannian of $n-1$-dimensional subspaces. That answers your 1st question. For your 2nd question, how do you want to "pull back" a connection? – Ryan Budney Mar 2 '11 at 23:15

2 Answers 2

The pullback of the tangent bundle of the sphere by the Gauss map is indeed the tangent bundle of the hypersurface, and the pullback of the Levi-Civita connection is indeed the Levi-Civita connection. Let's make use of the old-fashioned notation of frames: write points of Euclidean space as $x$, and orthonormal frames as $e_1, \dots, e_{n+1}$. Let $e$ be the matrix whose columns are the $e_i$. Take the bundle of adapted frames $FM$ of a hypersurface $M$. The adapted frames are the orthonormal frames with $e_{n+1}$ perpendicular to $M$. Write the soldering 1-forms as $\omega_i=e_i \cdot dx$, and the connection 1-forms as $\omega_{ij}=e_i \cdot de_j$. Similarly write the points of the unit sphere as $X$ and the orthonormal frames as $E_1, E_2, \dots, E_{n+1}$. Let $E$ be the matrix whose columns are the $E_i$. Then the frame bundle of a hypersurface $M$ maps to the frame bundle of the sphere by taking $(e,x)$ to $(E,X)$ in the frame bundle of the sphere, where $E=e$ and $X=e_{n+1}$. We can then see that the connection 1-forms of $M$ are $\omega_{ij}=e_i \cdot de_j=E_i \cdot dE_j=\Omega_{ij}$, matching up the pullback connection 1-forms of the sphere. The tangent bundle is an associated vector bundle, so matching up the connections on the principal bundles ensures that the connections on the associated bundles agree. You have to get used to this old-fashioned notation, but it does unwind to a proof that this particular lift of the Gauss map is connection-preserving. This is surprising, because it says that the curvature 2-form pulls back to the curvature 2-form. But that doesn't force the sectional curvature, or even the scalar curvature, to agree.

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If by "the bundle induced from the tangent bundle of the unit sphere" you mean "the pull back of the tangent bundle..." then the answer to your first question is "no", because that would mean that the Gauss map is unramified. So if you start with, say, a compact torus in Euclidean $\mathbb R^3$, you get a counter-example.

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Can you explain what it means for the Gauss map to be unramified? Thinking about this more I now think that the tangent bundle of the hypersurface is the induced bundle under the Gauss map. The thing is that the bundle mapping is not the differential of the Gauss map but rather parallel translation of tangent vectors to the surface to tangent vectors to the sphere. This bundle mapping is also metric preserving. Further the induced connection 1 form matrix is skew symmetric since it is just the pull back of the connection 1 form matrix on the sphere. (cont.) – marc Mar 12 '11 at 12:26
In the case of a surface in three space the induced connection has the same Gauss curvature as the connection inherited from the embedding. – marc Mar 12 '11 at 12:26

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