Hypersurfaces of Euclidean space

  • Thread starter lavinia
  • Start date
  • #1
lavinia
Science Advisor
Gold Member
3,236
623
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.

If this is true - which it seems to be - 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?
 

Answers and Replies

  • #2
lavinia
Science Advisor
Gold Member
3,236
623
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.

If this is true - which it seems to be - 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?
The tangent bundle of the hypersurface definitely is the unduced bundle under parallel translation. This construction generalizes to submanifolds of Euclidean space.

Further the induced bundle inherits the same metric as the embedded manifold and it seems that the induced connection is compatible with the metric - this because the connection 1 forms on the sphere are skew symmetric.

However it does not seem that the induced connection is necessarily symmetric. Or at least I can not see a reason why it would be.

In fact, for a surface in 3 space take 1 forms that are dual to an orthonormal basis that points along principal curvature directions. The symmetry equation is

dp1 = (dk1/k1)^p1 + (k2/k1)w*^p2

so the connection is not symmetric.
 
Last edited:

Related Threads on Hypersurfaces of Euclidean space

  • Last Post
Replies
1
Views
6K
Replies
8
Views
6K
  • Last Post
Replies
11
Views
4K
Replies
3
Views
2K
  • Last Post
Replies
11
Views
5K
Replies
14
Views
3K
Replies
13
Views
5K
Replies
13
Views
1K
Replies
9
Views
2K
Top