Gauss's lemma (Riemannian geometry)

Summary

In Riemannian geometry, Gauss's lemma asserts that any sufficiently small sphere centered at a point in a Riemannian manifold is perpendicular to every geodesic through the point. More formally, let M be a Riemannian manifold, equipped with its Levi-Civita connection, and p a point of M. The exponential map is a mapping from the tangent space at p to M:

which is a diffeomorphism in a neighborhood of zero. Gauss' lemma asserts that the image of a sphere of sufficiently small radius in TpM under the exponential map is perpendicular to all geodesics originating at p. The lemma allows the exponential map to be understood as a radial isometry, and is of fundamental importance in the study of geodesic convexity and normal coordinates.

IntroductionEdit

We define the exponential map at   by

 

where   is the unique geodesic with   and tangent   and   is chosen small enough so that for every   the geodesic   is defined in 1. So, if   is complete, then, by the Hopf–Rinow theorem,   is defined on the whole tangent space.

Let   be a curve differentiable in   such that   and  . Since  , it is clear that we can choose  . In this case, by the definition of the differential of the exponential in   applied over  , we obtain:

 

So (with the right identification  ) the differential of   is the identity. By the implicit function theorem,   is a diffeomorphism on a neighborhood of  . The Gauss Lemma now tells that   is also a radial isometry.

The exponential map is a radial isometryEdit

Let  . In what follows, we make the identification  .

Gauss's Lemma states: Let   and  . Then,  

For  , this lemma means that   is a radial isometry in the following sense: let  , i.e. such that   is well defined. And let  . Then the exponential   remains an isometry in  , and, more generally, all along the geodesic   (in so far as   is well defined)! Then, radially, in all the directions permitted by the domain of definition of  , it remains an isometry.

 
The exponential map as a radial isometry

ProofEdit

Recall that

 


We proceed in three steps:

  •   : let us construct a curve

  such that   and  . Since  , we can put  . Therefore,

 

where   is the parallel transport operator and  . The last equality is true because   is a geodesic, therefore   is parallel.

Now let us calculate the scalar product  .

We separate   into a component   parallel to   and a component   normal to  . In particular, we put  ,  .

The preceding step implies directly:

 
 

We must therefore show that the second term is null, because, according to Gauss's Lemma, we must have:

 

  •   :
 
The curve chosen to prove lemma

Let us define the curve

 

Note that

 

Let us put:

 

and we calculate:

 

and

 

Hence

 

We can now verify that this scalar product is actually independent of the variable  , and therefore that, for example:

 

because, according to what has been given above:

 

being given that the differential is a linear map. This will therefore prove the lemma.

  • We verify that  : this is a direct calculation. Since the maps   are geodesics,
 

Since the maps   are geodesics, the function   is constant. Thus,

 

See alsoEdit

ReferencesEdit

  • do Carmo, Manfredo (1992), Riemannian geometry, Basel, Boston, Berlin: Birkhäuser, ISBN 978-0-8176-3490-2