# Ngô Quốc Anh

## November 16, 2011

### Conformal compactification

Filed under: Riemannian geometry — Ngô Quốc Anh @ 0:49

Start with a pseudo-Riemannian manifold $(M,g)$, let $\tilde{g}$ be another pseudo-Riemannian metric on $M$, we say that $g$ and $\tilde{g}$ are conformal if there exists a positive scalar function $\phi$ on $M$ such that $\tilde{g} = \phi g$ (sufficient smoothness of the relevant quantities are always assumed).

Observe that two conformal metrics measure angles the same way: recall that on a pseudo-Riemannian manifold $(M,g)$, given a point $p\in M$ and two non-null vectors $v,w\in T_pM$, the angle between the vectors can be defined by

$\displaystyle \frac{g(v,w)^2}{g(v,v) g(w,w) }.$

(Notice that on Euclidean space, if $v,w$ form an angle $\theta$, then $v\cdot w = |v||w| \cos\theta$.) Thus if $\tilde{g}$ is conformal to $g$, they define the same angles

$\displaystyle \frac{\tilde{g}(v,w)^2}{\tilde{g}(v,v)\tilde{g}(w,w)} = \frac{\phi^2 g(v,w)^2}{\phi g(v,v) \phi g(w,w)} = \frac{g(v,w)^2}{g(v,v)g(w,w)}$

In fact, this inference works the other way too. If $g$,$\tilde{g}$ are two pseudo-Riemannian metrics such that for any two vectors $v,w$ we have $g(v,w) = 0 \iff \tilde{g}(v,w) = 0$, then $g$,$\tilde{g}$ are conformal (up to a change of sign) by the above definition (see e.g. Exercise 14, Chapter 2 from B.O’Neill, Semi-Riemannian Geometry).

So, in plain English, two metrics are conformal if they measure angles the same way.

Now, let $(M,g)$ be a pseudo-Riemannian manifold that is non-compact. A conformal compactification of $(M,g)$ is a choice of a metric $\tilde{g}$ such that $(M,\tilde{g})$ can be isometrically embedded into a compact domain $\tilde{M}$ of a pseudo-Riemannian manifold $(M',g')$ (well, I am ignoring some regularity issues here). Let $\phi$ be the conformal factor as before. Then observe that any regular extension of $\phi$ to the conformal boundary $\partial\tilde{M} \subset M'$ must vanish on said boundary. This reflects the property of a conformal compactification that “brings infinity to a finite distance”.

The simplest example of conformal compactification is the one-point compactification of Euclidean space via the stereographic projection. In this case, the target manifold $(M',g')$ is compact itself, taken to be standard sphere. The source manifold $(M,g)$ is Euclidean space with the standard metric, and the image set $\tilde{M}$ is taken to be the sphere minus the north pole.

[Source]

## November 8, 2011

### A blowup proof of the Aubin theorem in the Yamabe problem

Filed under: PDEs, Riemannian geometry — Tags: — Ngô Quốc Anh @ 13:29

Yamabe’s approach was to consider first the perturbed functional

$\displaystyle Q_s(u)\doteqdot\frac{\displaystyle\int_M\Big(|\nabla u|^2+\frac{n-2}{4(n-1)}R_gu^2\Big)d\mu_g}{\left(\displaystyle\int_M|u|^sd\mu_g\right)^\frac{2}{s}}$

where

$\displaystyle s\in \left(0,\frac{2n}{n-2} \right] \quad \text{ and } \quad u\in H^1(M)\setminus\{0\}.$

Set

$\displaystyle \lambda_s\doteqdot\inf\big\{Q_s(u):u\in H^{1}(M)\setminus\{0\}\big\}\quad\text{and}\quad Y(M)=\lambda_{2^*}.$

By using a direct minimizing procedure, it can be shown that for $2 < s < 2^*$, there exists a smooth positive function $u_s$ such that its $L^s$-norm is equal to one, $Q_s(u_s) = \lambda_s$, and $u_s$ satisfies the equation

$\displaystyle \Delta_gu_s-\frac{n-2}{4(n-1)}R_gu_s+\lambda_su^{s-1}_s=0,\quad \text{in}\;M.$

The direct method does not work when $s=2^*$ because the Sobolev embedding $H^1(M) \hookrightarrow L^{2^*}(M)$  is continuous but not compact. However, if one can show that $u_s$ is uniformly bounded, i.e. there exists a positive constant $c$ such that $u_s \le c$ in $M$ for $2 < s < 2^*$, then there exists a sequence $\{s_i\} \subset (2, 2^*)$ such that and $u_{s_i}$ converges to a smooth positive function $u$ which satisfies the Yamabe equation .

## November 5, 2011

Filed under: Giải Tích 2, Giải Tích 5, Liên Kết — Tags: — Ngô Quốc Anh @ 0:26

This is not mathematics. I just found an equation so that we can draw a heart in 3D. Indeed, the following equation

$\displaystyle {\left( {{x^2} + \frac{9}{4}{y^2} + {z^2} - 1} \right)^3} - {x^2}{z^3} - \frac{9}{{80}}{y^2}{z^3} = 0$

will generate a heart. I have tried and the following pictures show that fact.

## November 1, 2011

### An ODE appearing in the Nirenberg problem

Filed under: PDEs, Riemannian geometry — Ngô Quốc Anh @ 20:21

It is well-known that the simplest form of the Nirenberg problem is equivalent to solving the following PDE

$-\Delta u + 2= e^u$

in $\mathbb S^2$. Using stereographic projection, one can see that the above PDE is equivalent to

$-\Delta u = e^u$

in $\mathbb R^2$. If we assume that the solution $u$ has finite energy in the sense that

$\displaystyle \int_{\mathbb R^2} u <+\infty,$

it is well-known that the preceding PDE has unique radial solution. In terms of ODE language, our PDE can be rewritten as

$\displaystyle -u''(r)-\frac{1}{r}u'(r)=e^{u(r)},\quad r\geqslant 0.$

The purpose of this note is to find solutions to the above ODE. Our approach consists of several steps as shown below.

Step 1. Let $r = e^z$. We then have $u(r)=u(e^z)=v(z)$ which implies that

$\displaystyle u'(r) = e^{-z}v'(z), \quad u''(r) = e^{-2z}(v''(z)-v'(z)).$