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 .

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November 5, 2011

MuPad: Heart in 3D

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.

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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)).

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