where is the normal vector of , that is . What is the mean curvature of ?

We can parametrize the cylinder by where , , , . One can write as

Then the induced metric on is

Since

The normal vector of is

The second fundamental form is

Then the mean curvature is

If and , then

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where . is called the angle of the conical singularity. We will always assume .

For example equipped with the metric is isometric to an Euclidean cone of total angle .

Suppose now is a closed Riemann surface with conical metric . **Assume that the Gauss curvature extends on as a H\”{o}lder continuous function.**

Suppose is a conformal change of the conical metric. Then necessarily we have . In order to prescribe the Gauss curvature , we need to solve the equation

.

Any reasonable solution of (0) will satisfy

where follows from the Gauss-Bonnet formula for conical surface. We will use variational method to attack it. To that end, define to be Sobolev space of functions with

and functionals

It is easy to verify that the minimizer of the following variation problem will be a weak solution of (0)

As one can see, there are two immediate questions we need to answer,

(a)

(b) minimizer exists

The first question follows from the Trudinger inequality. Namely, define . Then

**Lemma:** For any fixed , there exist constant such that,

for any and and .

**Lemma:** Suppose and . Let . Then for any fixed , there exists constants such that

for any and . Here .

From this key lemma, one can derive that is lower bounded on provided and . To do that, choose and such that , then it follows from the above lemma that

For the second question, we need to prove the embedding is compact(actually it is true for any for any ). Note that this is true if is some smooth metric, however is conical one. Therefore, we want to compare with of some smooth metric.

Suppose is equipped with smooth metric such that . Here is smooth and positive outside the support of and near . If we use the denote the gradient of with respect to , then and

This is only true in two dimension. Now and have inner product

From the above analysis, we need to examine the difference of and . It follows from the singular behavior of that

for any and some . Then for any ,

Then . On the contrary, we need the following inequality

**Lemma: **For conical metric

Then we have . Therefore is compact for . Furthermore, after some effort, one can prove

is also compact for .

Remark:

[1] M. Troyanov, Prescribing curvature on compact surfaces with conical singularities, Trans. Am. Math. Soc. 324 (1991) 793–821. doi:10.1090/S0002-9947-1991-1005085-9.

]]>But if is minimal surface, then must be simply connected. The reason is has convex haul property.

Learn this example from Jacob Bernstein.

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therefore

If we use the polar coordinates , and and the following fact

then one can calculate the Hessian of under this coordinates

Then

If the metric is , then we will have

]]>**Propsotion 2.4** Let be a domain in . is a smooth function on . Then

**Proof:** Take an orthonormal frame field such that is tangent to . Notice

where . It follows from Remark 2.3 that

where is the outward unit normal to . Changing the coordinates to and , we can get

It is easy to see

and

Therefore the proposition is established.

Remark: Robert Reilly, On the hessian of a function and the curvature of its graph

]]>The principle curvatures are taken from the eigenvalues of the Jacobian matrix . One can define its mean curvature using the notation of above

Now let us consider general matrix ,

where is the generalized Kronecker delta.

Then we define the Newton tensor

For any vector field on , is a matrix, where and , denote , we have

Since for any

because is skew-symmetric in . Then

Applying the formula (Check [1] Propsition 1.2) and , we get

It follows from the result of Reilly, Remark 2.3(a), that

Suppose is a vector field normal to , then

where is the outer ward unit normal to .

**Remark:** [1] R.C. Reilly, On the Hessian of a function and the curvatures of its graph., Michigan Math. J. 20 (1974) 373â€“383. doi:10.1307/mmj/1029001155.

[2] N. Trudinger, Apriori bounds for graphs with prescribed curvature.

We probably have to assume is a symmetric matrix in order to use the formula of Reilly. Not sure about this.

]]>This formula has more general forms.

]]>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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