- Gauss's lemma (Riemannian geometry)
Riemannian geometry, Gauss's lemma asserts that any sufficiently small spherecentered at a point in a Riemannian manifoldis perpendicular to every geodesicthrough 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 spaceat "p" to "M"::which is a diffeomorphismin a neighborhood of zero. Gauss' lemma asserts that the image of a sphereof sufficiently small radius in "T"p"M" 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.
We define on the exponential map at by:where we have had to restrict the domain by definition of a ball of radius and centre to ensure that is well-defined, and where is the point reached by following the unique
geodesicpassing through the point with tangent for a distance . It is easy to see that is a local diffeomorphismaround . 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:
The fact that is a local diffeomorphism and that for all allows us to state that is a local isometry around , i.e.
This means in particular that it is possible to identify the ball with a small neighbourhood around . We can see that is a local isometry, but we would like it to be rather more than that. We assert that it is in fact possible to show that this map is a radial isometry !
The exponential map is a radial isometry
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. Moreover, 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.
We proceed in three steps:
* "" : let us construct a curve such that and . Since , we can put . We find that, thanks to the identification we have made, and since we are only taking equivalence classes of curves, it is possible to choose (these are exactly the same curves, but shifted (###décalées###), because of the domain of definition ; however, the identification allows us to gather them (###ramener###) around !!!). Hence,
Now let us calculate the scalar product .
We separate into a component tangent 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:
* "" : Let us define the curve
:with and . We remark in passing 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. We first take account of the fact that the maps are geodesics, i.e. . Therefore,
:Hence, in particular,:because, since the maps are geodesics, we have .
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