# Ngô Quốc Anh

## October 3, 2017

### Stereographic projection not at the North pole and an example of conformal transformation on S^n

Filed under: Riemannian geometry — Tags: , — Ngô Quốc Anh @ 12:09

Denote $\pi_P : \mathbb S^n \to \mathbb R^n$ the stereographic projection performed with $P$ as the north pole to the equatorial plane of $\mathbb S^n$. Clearly when $P$ is the north pole $N$, i.e. $N = (0,...,0,1)$, then $\pi_N$ is the usual stereographic projection.

Clearly, for arbitrary $x \in \mathbb S^n$, the image of $x$ is

$\displaystyle \pi_P : x \mapsto y = P+\frac{x-P}{1-x \cdot P}.$

For the inverse map, it is not hard to see that

$\displaystyle \pi_P^{-1} : y \mapsto x =\frac{|y|^2-1}{|y|^2+1}P+\frac 2{|y|^2+1}y.$

Derivation of $\pi_P$ and $\pi_P^{-1}$ are easy, for interested reader, I refer to an answer in . Let us now define the usual conformal transformation $\varphi_{P,t} : \mathbb S^n \to \mathbb S^n$ given by

$\displaystyle \varphi_{P,t} : x \mapsto \pi_P^{-1} \big( t \pi_P ( x) \big)$

## September 9, 2016

### Benefits of “complete” and “compact” for analysis on Riemannian manifolds

Filed under: Riemannian geometry — Tags: — Ngô Quốc Anh @ 10:08

When working on Riemannian manifolds, it is commonly assumed that the manifold is complete and compact. (The case of non-compactness is also of interest too.) In this entry, let us review the role of completeness and compactness in this setting.

How important the completeness is? Let us recall that for given a Riemannian manifold $(M,g)$, what we have is a nice structure as well as an appropriate analysis on any tangent space $T_pM$. For a $C^1$-curve $\gamma : [a,b] \to M$ on $M$, the length of $\gamma$ is

$\displaystyle L(\gamma) = \int_a^b \sqrt{g(\gamma (t)) \langle \partial_t \gamma \big|_t, \partial_t \gamma \big|_t\rangle} dt$

where $\partial_t \gamma\big|_t \in T_{\gamma (t)}M$ a tangent vector. (Note that by using curves, the tangent vector $\partial_t \gamma\big|_t$ is being understood as follows

$\displaystyle \partial_t \gamma\big|_t (f) = (f \circ \gamma)'(t)$

for any differentiable function $f$ at $\gamma(t)$.) Length of piecewise $C^1$ curves can be defined as the sum of the lengths of its pieces. From this a distance on $M$ whose topology coincides with the old one on $M$ is given as follows

$\displaystyle d_g(x,y) = \inf_\gamma L(\gamma)$

where the infimum is taken on all over the set of all piecewise $C^1$ curves connecting $x$ and $y$.

## April 20, 2016

### Stereographic projection, 6

Filed under: PDEs, Riemannian geometry — Tags: — Ngô Quốc Anh @ 1:08

I want to propose an alternative way to calculate the Jacobian of the stereographic projection $\mathcal S$. In Cartesian coordinates  $\xi=(\xi_1, \xi_2,...,\xi_{n+1})$ on the sphere $\mathbb S^n$ and $x=(x_1,x_2,...,x_n)$ on the plane, the projection and its inverse are given by the formulas

$\displaystyle\xi _i = \begin{cases} \dfrac{{2{x_i}}}{{1 + {{\left| x \right|}^2}}},&1 \leqslant i \leqslant n, \hfill \\ \dfrac{{{{\left| x \right|}^2} - 1}}{{1 + {{\left| x \right|}^2}}},&i = n + 1. \hfill \\ \end{cases}$

and

$\displaystyle {x_i} = \frac{{{\xi _i}}}{{1 - {\xi _{n + 1}}}}, \quad 1 \leqslant i \leqslant n$.

It is well-known that the Jacobian of the stereographic projection $\mathcal S: \xi \mapsto x$ is

$\displaystyle \frac{\partial \xi}{\partial x} = {\left( {\frac{2}{{1 + {{\left| x \right|}^2}}}} \right)^n}.$

The way to calculate its Jacobian is to compare the ratio of volumes. First pick two arbitrary points $x, y \in \mathbb R^n$ and denote $\xi = \mathcal S(x)$ and $\eta = \mathcal S(y)$.

The Euclidean distance between $\xi$ and $\eta$ is

$\displaystyle |\xi -\eta|^2 = \sum_{i=1}^{n+1} |\xi_i - \eta_i|^2 =\sum_{i=1}^n |\xi_i - \eta_i|^2+|\xi_{n+1} - \eta_{n+1}|^2.$

## August 29, 2014

### Prescribed Q-curvature and scalar curvature problems in the null case

Filed under: PDEs, Riemannian geometry — Tags: — Ngô Quốc Anh @ 20:24

On a 2-dimensional compact Riemannian manifold $(M, g)$ without boundary, the prescribed scalar curvature problem in the flat case is equivalent to solving the following PDE

$\displaystyle -\Delta_g u = fe^{2u}$

with $f$ is a given non-constant smooth function on $M$ and $\Delta_g$ is the Laplace-Beltrami operator associated with the metric $g$.

Simply by integrating both sides of the PDE, it is immediate to see that if $u$ solves the PDE, it would satisfy $\int_M f e^{2u} dv =0$; hence the candidate function $f$ must change sign in $M$. In their elegant paper published in 1974, Kazdan and Warner showed that in addition to the sign-changing property of $f$, it must also satisfy the following inequality

$\displaystyle \int_M f dv <0.$

This is just a simple observation from integration by parts if we multiply both sides of the PDE by $e^{-2u}$. Interestingly, Kazdan and Warner were able to show that the above two properties are also sufficient in the sense that it is enough to prove that the PDE is solvable.

In higher dimensions, perhaps, the most natural generalization of the operator $\Delta_g$ is the well-known Paneitz operator and its corresponding Q-curvature, see this link.

Assume that $(M,g)$ is a compact Riemannian 4-manifold without boundary. We denote by $P_g^4$ the so-called Paneitz operator acting on any smooth function $u$ via the following rule

$\displaystyle P_g^4(u) = \Delta _g^2u + {\rm div}\left( {\frac{2}{3}{R_g} - 2{\rm Ric}_g} \right)du ,$

where by ${\rm Ric}$ and $R$ we mean the Ricci tensor and the scalar curvature of $g$, respectively.

## July 15, 2014

### Why should we call ” f=φg ” conformal change?

Filed under: Riemannian geometry — Tags: — Ngô Quốc Anh @ 1:07

I asked this question to Professor Alice Chang when I met her during a conference in the University of Notre Dame this June. Loosely speaking, given a Riemannian manifold $(M,g)$, why frequently we follow the rule $f = \varphi g$ to change our metric in conformal geometry?

Professor Chang told me that it is because under the new metric, angles are preserved. The aim of this note is to make her answer clearer.

The best way to see this is to make use of the vector formulation of the law of cosines. Indeed, let us take two vectors $X$ and $Y$ sitting in the same tangent space, say $T_pM$ for some $p \in M$. Then the angel between these two vectors can be estimated as follows

$\displaystyle \cos_g(X,Y)=\frac{g(X,Y)}{\|X\|_g\|Y\|_g}$

where $\|\cdot\|_g$ is the norm evaluated with respect to the metric $g$. Under the new metric $f$ given by $\varphi g$, we first obtain

$\displaystyle \cos_f(X,Y)=\frac{f(X,Y)}{\|X\|_f\|Y\|_f}.$

Clearly, $f(X,Y) = \varphi g(X,Y)$. Moreover, if we write $X$ in local coordinates as $X=X^i \frac{\partial}{\partial x^i}$, we then have

$\displaystyle {\left\| X \right\|_f} = \sqrt {{{\left\langle {{X^i}\frac{\partial }{{\partial {x^i}}},{X^j}\frac{\partial }{{\partial {x^j}}}} \right\rangle }_f}} = \sqrt {{X^i}{X^j}} \sqrt {{f_{ij}}} = \sqrt \varphi \sqrt {{X^i}{X^j}} \sqrt {{g_{ij}}} .$

Therefore,

$\displaystyle {\left\| X \right\|_f}{\left\| Y \right\|_f} = \varphi {\left\| X \right\|_g}{\left\| Y \right\|_g},$

which immediately shows that $\cos_f(X,Y)=\cos_g(X,Y)$. In other words, the angle between the two vectors $X$ and $Y$ is preserved under the change of metrics.

## April 15, 2014

### Locally H^1-bounded for the Palais-Smale sequences in the region when the Gaussian curvature candidate is negative

Filed under: PDEs, Riemannian geometry — Tags: — Ngô Quốc Anh @ 16:22

In 1993, Ding and Liu announced their result about the multiplicity of solutions to the prescribed Gaussian curvature on compact Riemannian $2$-manifolds of negative Euler characteristic. Their interesting result then published in the journal Trans. Amer. Math. Soc. in 1995, see this link.

Roughly speaking, by starting with the prescribed Gaussian curvature equation, i.e.

$\displaystyle -\Delta u + k = Ke^{2u}$

when the Euler characteristic $\chi(M)$ is negative, i.e.

$\displaystyle 2\pi \chi (M) = \int_M k e^{2u}dv_g <0,$

they perturbed $K$ using

$\displaystyle K_\lambda = K+\lambda$

where $\lambda$ is a real number and the candidate function $K$ is assumed to be

$\displaystyle \max_{x \in M} K(x)=0$.

Then they were interested in solving the following PDE

$\displaystyle -\Delta u + k = K_\lambda e^{2u}.$

Their main result can be stated as follows:

Theorem (Ding-Liu). There exists a $\lambda^\star > 0$ such that

• the PDE has a unique solution for $\lambda \leqslant 0$;
• the PDE has at least two solutions if $0<\lambda<\lambda^\star$; and
• the PDE has at least one solution when $\lambda = \lambda^\star$.

## March 29, 2014

### A new Rayleigh-type quotient for the conformal Killing operator on manifolds with boundary

Filed under: Riemannian geometry — Tags: — Ngô Quốc Anh @ 15:22

In the previous note, I showed a Rayleigh-type quotient for the conformal Killing operator $\mathbb L$ on manifolds $(M,g)$ with boundary $\partial M$, i.e. the following result holds:

Whenever $M$ admits no non-zero conformal Killing vector fields, the following holds

$\displaystyle C_g(M)=\inf \frac{{{{\left( {\int_M {|\mathbb LX|^2 d{v_g}} } \right)}^{1/2}}}}{{{{\left( {\int_M {|X|^{2n/(n - 2)}d{v_g}} } \right)}^{(n - 2)/(2n)}}}} > 0$

where the infimum is taken over all smooth vector fields $X$ on $M$ with $X \not\equiv 0$.

Today, I am going to prove a slightly stronger version of the above inequality, namely, when some terms on the boundary $\partial M$ take part in. Precisely, we shall prove

Whenever $M$ admits no non-zero conformal Killing vector fields, the following holds

$\displaystyle C_g(M,\partial M)=\inf \frac{{{{\left( {\int_M {|\mathbb LX|^2 d{v_g}} } \right)}^{1/2}}}}{{{{\left( {\int_M {|X|^\frac{2n}{n - 2}d{v_g}} } \right)}^\frac{n - 2}{2n}}} + \left( \int_{\partial M}|X|^\frac{2(n-1)}{n-2}ds_g\right)^\frac{n-2}{2(n-1)}} > 0$

where the infimum is taken over all smooth vector fields $X$ on $M$ with $X \not\equiv 0$.

However, a proof for this new inequality remains the same. To do so, we first make use of some Sobolev embeddings as follows:

## March 28, 2014

### Lower bound for conformal factor in terms of upper bound of L^4-norm of the scalar curvature in the negative case

Filed under: Riemannian geometry — Tags: — Ngô Quốc Anh @ 18:01

In this note, we talk about the following interesting result obtained by X. Xu in Nagoya Mathematical Journal in 1995, here.

To state his result, let us fix $(M,g_0)$ a compact manifold of dimension $4$. We shall denote by $\text{Scal}_g$ the scalar curvature computed with respect to the metric $g$. It is clear that under the conformal change $g=u^2g_0$, $\text{Scal}_{g_0}$ and $\text{Scal}_g$ are related by the following rule

$\displaystyle \text{Scal}_g = u^{-3}\big(u\text{Scal}_{g_0}-6\Delta_{g_0}u \big).$

Furthermore, we may also assume that $\text{Scal}_{g_0}$ is a negative constant. We are now able to state his result.

Theorem. If $\text{Scal}_{g_0} < 0$ and $\int_M |\text{Scal}_g|^4 dv_g< C_o$, then there exists a constant $C_1 > 0$ such that $u > C_1$ where $g = u^2 g_0$.

Proof. Using the conformal change rule shown above, we obtain

$\begin{array}{lcl}{C_0} &\ge& \displaystyle\int_M {|\text{Scal}_g{|^4}d{v_g}} \\ &=& \displaystyle\int_M {{{[{u^{ - 3}}(u \text{Scal}_{g_0} - 6{\Delta _{{g_0}}}u)]}^4}({u^4}d{v_{{g_0}}})} \\ &=& \displaystyle\int_M {{{[{u^{ - 1}}\text{Scal}_{g_0} - 6{u^{ - 2}}{\Delta _{{g_0}}}u]}^4}d{v_{{g_0}}}} \\ &=& \displaystyle\int_M {{u^{ - 4}}|\text{Scal}_{g_0}|^4 d{v_{{g_0}}}} + {\rm{1296}}\int_M {{u^{ - 8}}{{({\Delta _{{g_0}}}u)}^4}d{v_{{g_0}}}} + \\&& \displaystyle + {\rm{216}}\int_M {{u^{ - 6}}|\text{Scal}_{g_0}|^2{{({\Delta _{{g_0}}}u)}^2}d{v_{{g_0}}}} - 24\int_M {{u^{ - 5}}|\text{Scal}_{g_0}|^3{\Delta _{{g_0}}}ud{v_{{g_0}}}} \\ &&\displaystyle - 864\int_M {{u^{ - 7}} \text{Scal}_{g_0} {{({\Delta _{{g_0}}}u)}^3}d{v_{{g_0}}}} \\ &=& \displaystyle |\text{Scal}_{g_0}|^4\int_M {{u^{ - 4}}d{v_{{g_0}}}} - 24|\text{Scal}_{g_0}|^3 \int_M {{u^{ - 5}}{\Delta _{{g_0}}}ud{v_{{g_0}}}} \\ &&\displaystyle +432\int_M {{u^{ - 8}}{{({\Delta _{{g_0}}}u)}^4}d{v_{{g_0}}}} + 216\int_M {{u^{ - 8}}{{({\Delta _{{g_0}}}u)}^2}{{[\text{Scal}_{g_0} - 2{\Delta _{{g_0}}}u]}^2}d{v_{{g_0}}}} .\end{array}$

## February 22, 2014

### An application of the Brezis-Li-Shafrir estimate to the limiting case of the prescribing Gaussian curvature problem in the negative case

Filed under: PDEs, Riemannian geometry — Ngô Quốc Anh @ 14:56

On a Riemannian surface $M$, let consider the following PDE

$\displaystyle -\Delta u +\alpha = R(x)e^u$

naturally arising from the prescribed Gaussian curvature problem. A simple variable change, one can assume that $\alpha$ is a negative constant, see this. It follows from a very well-known result due to Kazdan and Warner that it is necessary to have

$\displaystyle \int_M R(x) dx<0.$

In addition, Kazdan and Warner also showed that if $\int_M R(x) dx<0$ and $R$ changes sign, then there exists a number $\alpha_0 \in (-\infty, 0)$ such that the above PDE is solvable for all $\alpha > \alpha_0$ but not if $\alpha<\alpha_0$. In fact, the number $\alpha_0$ can be characterized as follows

$\displaystyle \alpha_0 = \inf\{\alpha : \text{the PDE is solvable}\}.$

This can be easily seen from the the following comprising property: If the PDE is solvable for some $\alpha_1$, it is also solvable for any $\alpha_2 \geqslant \alpha_1$.

However, Kazdan and Warner did not tell us what happens when $\alpha=\alpha_0$. In an attempt to see what really happens when $\alpha=\alpha_0$, Chen and Li made use of the Brezis-Li-Shafrir estimate to answer in the following way: The PDE is also solvable even when $\alpha=\alpha_0$. The purpose of this note is to talk about the beautiful Chen-Li argument, see this.

The idea is to approximate the equation for $\alpha_0$ by a sequence $\{\alpha_k\}_k$ of negative real numbers in the following sense $\alpha _k\searrow a_0$ as $k \to \infty$. Their proof consists of three steps as follows:

## January 8, 2014

### Conformal metric having strictly negative scalar curvature in a given region

Filed under: PDEs, Riemannian geometry — Ngô Quốc Anh @ 0:39

Let $(M,g)$ be a closed (compact without boundary) Riemannian manifold of dimension $3$ with a $C^2$ metric $g$ of arbitrary Yamabe class. Suppose $\Omega \subset M$ be an open subset of $M$ with regular boundary $\partial\Omega$ and with $M\backslash (\Omega \cup \partial \Omega )$ non-empty and open in $M$.

In this note, we mention a very interesting result basically due to O’Murchadha-York from here and Isenberg from here. The result says that there exists a conformal metric $\widehat g \in [g]$ such that

$\displaystyle \text{Scal}_{\widehat g} < -\xi<0$

in $\Omega$ for some constant $\xi>0$ to be specify later. The novelty of this result is that although the metric $g$ may be of positive Yamabe class which tells us that it is impossible to construct a conformal metric which is everywhere negative, it is possible to make it negative in a proper subset of $M$. A proof for this result goes as follows:

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