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Hypersurface in Hyperbolic space and its tangent horospheres
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Hypersurface in hyperbolic space and conformal metric on Sphere
Suppose is a domain in , , where is the standard metric on . One can construct a hypersurface in hyperboloid by
where and . What are the induced metric on ? One can calculate as the following,
Suppose is a unit eigenvector associated to the eigenvalue of . We need to calculate . Firstly it is easy to see
and
Then
So this gives the following formula
where . Suppose is the eigenvector corresponding to and are orthonormal basis, then
where . Recall that Schouten tensor has the following formula under conformal change,
for and . Therefore the eigenvalues of are .
I am using the thesis of Dimas Percy Abanto Silva and some communication with him.
Hyperbolic translation in Poincare disc model
Hyperboloid model is hyperquadric in , with
Halfspace model . Denote ,
Poincaré ball model There are some transformation formula between these hyperbolic models, say
On , there are vertical scaling transformations which are isometries of
On , they are called translation in direction through the map
In component,
On ,
In component,
Conformal killing operator and divergence transformation under conformal change
Suppose is a Riemannian manifold. For each vector field , we can define the conformal killing operator to be the trace free part of Lie derivative , more precisely
Obviously maps vector field to trace free symmetric two tensors. Now suppose we have a conformal transformation , then what happen to the conformal killing operator ? Notice that
By using the identity for any vector field , here is the volume element, one can get the transfromation of divergence under confromal change
therefore
induces a formal adjoint on trace free 2tensors. Suppose we have a symmetric 2tensor , where are coordinates on . If one has
for some vector field and trace free 2tensor . What does this coorespond to under metric ? To see that, we first need a formula about symmtric 2tensors,
Now choose any vector field , then
This is equivalent to
Next consider the divergence operator and its adjoints .
It is well know that on 1forms
where operator turns 1form to a vector field by using metric . What is the relation of and ? To find that, choose any 1form ,
using the formula about , one gets
then using is symmetric, continue from (1)
In other words,
In the other way, we can calculate more directly
We get
therefore
One can compare this with (2).
Product metric on product manifold
Suppose we have two Riemannian manifolds and , what happened to their product manifold with product metric?
.
We will use and . Then by definition . Therefore
Similarly for all mixed indices on . This implies
Therefore the Riemannian curvature operator is
and Ricci tensor
When you have the warped product metric like with metric , the curvature is given like the following
BV function and its property involves translation
Theorem 1 Suppose , then if and only if such that
Moreover, one can take . Here is the tanslation operator.
Proof: Firstly suppose . Let us prove
where
it is easy to verify therefore (1) is proved. Next one can choose such with (it is easy to show by mollification). By dominating theorem, one get
The other direction need more analysis.
Automorphism of annulus
Suppose is the annulus for . Then
is not biholomorphic to whenever .
The proof goes like following: Suppose there exists a biholomorphic map between them, then this map can be extended to the boundary by the so called Kellog theorem, then one can reflect the annulus the inner part and outside part. Near the origin, one can use the removable singularity theorem to prove the extended map is holomophic. Thus we can get a biholomorphic map between and which must have the form . First , because the map keeps the origin. Second must have the unit length to keep the inner boundary to be . Thus the map can only be rotation around origin.
This actually tells us the automorphism of an annulus can only be the rotation. Let us verify this fact by another perspective.
As we all know, is biholomorphic to , where is the upper half plane and . Here . acts on the freely discontinuously.
Any automorphism of can be lift to an automphism of which commutes with the action . In the matrix language, these automphisms are the matrix in commutes with . Some elementary calculation shows that these matrix have the form where . Any one of these is corresponding to a rotation of the annulus.
The perturbation of and metric and standard bubble
Suppose , then it is well know that it satisfies
on . Now we want to perturb the Euclidean metric a little bit and an approximate solution of above equation. Suppose where is a symmetric 2tensor and is the Euclidean metric. The reasonable equation of must satisfy should be
Suppose . Taking derivative on both sides of the above equation and restrict to , one can get the equation must satisfy
where we have use the fact that
Dual spaces
Suppose and are Banach space, and are their dual spaces. If with , then
is an embedding. Let us consider the relation of two dual spaces. For any
Then will be a bounded linear functional on
is a bounded linear operator.
In a very special case that is a closed subset of under the norm , one can prove is surjective. In fact can be extended to on by HahnBanach thm such that . Then
Let us take and for example. Define the inner product to be
When is a bounded subset of , is a proper closed subset of . From the above explanation,
Define the continuous linear functional for . One can see that .
Should attribute this to Lun Zhang. I am always confused that .
Geodesic Normal Coordinates, Gauss lemma and Identity
Suppose is a point on dim manifold . Consider the exponential map near , at suitable neighborhood any point within it can be expressed uniquely as
where is an orthonormal basis. Then can be coordinate around . Let us deduce useful identities using this coordinate. Obviously, the geodesic in this coordinate will be
In particular ,
The geodesic equation will be
Multiplying , we get
Since the tangent vector has constant length along , i.e.
Multiplying , it is equivalent to say
Furthermore, recalling the definition of Christoffel symbol, one can get
Or equivalently
While on the left hand side
Combining this two facts,
This means along ,
Since at , , one get
on any point in the neighborhood. This identity is called Gauss Lemma.
Suppose , is a symmetric 2tensor. Then the above identity means .