What is the unit normal to this radial graph?

Suppose is a smooth local frame on . Let be the covariant derivative on . Tangent space of consists of which are

In order to get the unit normal, we need some simplification. Let us assume are orthonormal basis of the tangent space of and . Then

Then we obtain an orthonormal basis of the tangent of

We are able to get the normal by projecting to this subspace

After normalization, the (outer)unit normal can be written

**Remark: **Guan, Bo and Spruck, Joel. *Boundary-value Problems on \mathbb{S}^n for Surfaces of Constant Gauss Curvature.*

Choose a local orthonormal frame such that , and in the neighborhood of some point . We want to change (1) to some expression on or . To do that, we need to apply both sides of (1) to , getting

where we have used . Similarly

Now let us calculate the Laplacian of second fundamental form

Then

Combining all the above estimates to (1), we get

where we write . Using this equation and the other one on , one can derive that if is mean convex then it is actually convex.

]]>Denote for short. If we pull the metric of back to , denote as , then

where and . Then one can use the local coordinate to calculate

also one can see from another definition of Laplacian

By using the expression of stated above, we can calculate

It follows from the definition of tangential derivative on , see, that

then

where is the mean curvature of the . Combining all the above calculations,

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Denote and is the blow down map. On (near A), takes the form

where is the boundary defining function for ff and is boundary defining function for rb. Similarly on (near B), takes the form

where is a bdf for lb and is a bdf for ff. One can verify that is a diffeomorphism . Let . Then is a polyhomogeneous conormal function on . Its index can be denoted correspond to lb, ff and rb.

Suppose is the projection to coordinate. Consider , is actually a fibration. Then the push forward map maps to a polyhomogeneous function on .

In order to make is integrable, let us assume support and . What is the index for on ?

For the first integral, letting

For the second integral, letting

Since the Taylor series

Consequently

Combining all the above analysis, the index for is

From another point of view, the vanishing order of on each boundary hypersurface of are , and . maps lb and ff to the boundary of . Therefore the index of is contained in

**Remark:** Daniel Grieser, Basics of Calculus.

where . Then one can see , . Under this stereographic projection, a strip will be equivalent to some lens domain on the sphere.

Let us pull the standard metric of to the . For the following statement, we will always omit and . Calculation shows,

therefore

here we used the fact that . Suppose is the connection on equipped with the standard metric. We want to calculate . To that end, it is better to use the coordinates in

Since we know

where means the tangential part to and is the unit normal to . Using the connection in , we get

One can verify from the above equalities that

Similarly

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What is the critical point of this area functional? To see that,

From some basic calculation in minimal surface(see colding minicozzi’s book)

Stokes’ theorem implies

Combining these above fact, we get

where is the unit normal of . Therefore the critical point of will satisfy

This is so called translating soliton.

Now consider the second variation at a translation soliton,

While

Suppose , where , then

]]>First approach, the cylinder is diffeomorphic to the puncture plane,

It is easy to verify that . One can calculate from the conformal change formula

Second approach, remember the Ricci curvature splits on product manifold with product metric. Therefore

From the above, we know that if , then while if . I always get confused with this two cases. There is a nice way to see this result for . can be covered locally isometrically by with Euclidean metric. Therefore in this case. However, should cover instead of .

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This means

then is a conformal diffeomorphism from to . Suppose is a conic metric such that on . Thus is isometric to through . Therefore . It is easy to see that because

]]>**1. Preliminary **

Suppose is a Riemannian manifold with . is the Schouten tensor

and denote . Define

where . It is well known that is conformally invariant.

Suppose where is a symmetric 2-tensor. We want to calculate the first derivative of at . To that end, let us list some basic facts (see the book of Toppings). Firstly denote the divergence operator and

where is the Lichnerowicz Laplacian. Then the first variation of Ricci curvature and scalar curvature are

where we were using upper dot to denote the derivative with respect to .

**2. First variation of the sigma2 functional **

*Proof:*

where . Since we have

Plugging this into the derivative of to get

In order to simplify the above equation, we recall the definition of Lichnerowicz Laplacian

Apply (1) to get

Therefore we can simplify it to be

Let us denote . Using the fact and the definition of ,

where

Remark 1It is easy to verify , this is equivalent to say is invariant under conformal change. More precisely, letting , then

Remark 2If is an Einstein metric with , then , and

It is easy to verify that . In other words, Einstein metrics are critical points of .

**Are there any non Einstein metric which are critical points of ?**

Here is one example. Suppose , where is the sphere with standard round metric and is a two dimensional compact manifolds with sectional curvature . is endowed with the product metric. We can prove , , , and consequently .

Note that the above example is a locally conformally flat manifold. For this type of manifold, we have the following lemma which can say

Lemma 2Suppose is locally conformally flat and , then

*Proof:* When is locally conformally flat,

is equivalent to

Actually this is equivalent to the Bach tensor is zero.

**3. Another point of view **

We have the Euler Characteristic formula for four dimensional manifolds

therefore the critical points for will be the same as the critical points of . However, the functional

is well studied by Bach. The critical points of this functional satisfy Bach tensor equal to 0.

Obviously, for Einstein metric, but not all Bach flat metrics are Einstein. For example for any locally conformally flat manifolds.

]]>- For every and , define

It is well know that . Moreover is isometric to the standard sphere minus one point.

- For any and define

where is the geodesic distance of and on . Actually

- For each , define by

Both the second and third one satisfy

If we make , the third one will be changed to the second one.

To get the second one from the first one, let us deonte be the stereographic projection from point . Then

It should be able to see the third one from hyperbolic translation directly.

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