Ngô Quốc Anh

June 28, 2011

The Yamabe problem: A Story

Filed under: PDEs, Riemannian geometry — Tags: — Ngô Quốc Anh @ 15:47

I want to write a short survey about the Yamabe problem. Long time ago, I introduced the problem in this blog [here] but it turns out that the note was not rich enough to perform the importance of the problem.

Hidehiko Yamabe, in his famous paper entitled On a deformation of Riemannian structures on compact manifolds, Osaka Math. J. 12 (1960), pp. 21-37,  wanted to solve the Poincaré conjecture

Conjecture. Every simply connected, closed 3-manifold is homeomorphic to the 3-sphere

For this he thought, as a first step, to exhibit a metric with constant scalar curvature. We refer the reader to this note for details. He considered conformal metrics (the simplest change of metric is a conformal one), and gave a proof of the following statement:

Theorem (Yamabe). On a compact Riemannian manifold (M, g) of dimension \geqslant 3, there exists a metric g' conformal to g, such that the corresponding scalar curvature \text{Scal}_{g'} is constant.

As can be seen, the Yamabe problem is a special case of the prescribing scalar curvature problem that can be completely solved. For the prescribing scalar curvature, we also solve it completely when the invariant is non-positive.

1. Conformal metrics.

Definition (conformal). Two pseudo-Riemannian metrics g and \widetilde g on a manifold M are said to be

  • (pointwise) conformal if there exists a C^\infty function f on M such that

    \displaystyle \widetilde g=e^{2f}g;

  • conformally equivalent if there exists a diffeomorphism \alpha of M such that \alpha^* \widetilde g and g are pointwise conformal.

Note that, if g and \widetilde g are conformally equivalent, then \alpha is an isometry from e^{2f}g onto \widetilde g. So we will only study below the case \widetilde g = e^{2f}g.

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June 18, 2011

A note on the equation involving the prescribing Gaussian curvature problem, 2

Filed under: PDEs, Riemannian geometry — Ngô Quốc Anh @ 17:28

Followed by this note, the trick use in that note also works for the following equation

\displaystyle -\Delta u +K_0(x)=K_1(x)e^{2u}+K_2(x)e^{-2u}, \quad x \in M.

In fact, by letting \overline u=2u, then our PDE becomes

\displaystyle -\Delta \overline u +2K_0(x)=2K_1(x)e^{\overline u}+2K_2(x)e^{-\overline u}.

Let v be a solution of the following PDE

\displaystyle -\Delta v=2K_0(x)-2\overline K_0

where \overline K_0 is the average of K_0 over M. We let w=\overline u+v. Then it is easy to verify that w solves the following

\displaystyle -\Delta w +2\overline K_0=2K_1(x)e^{-v}e^w+2K_2(x)e^ve^{-w}.

Finally, letting

\alpha=2\overline K_0, \quad R_1(x)=2K_1(x)e^{-v(x)}, \quad R_2(x)=2K_2(x)e^{v(x)}

we get that

\displaystyle -\Delta w +\alpha = R_1(x)e^w+R_2(x)e^{-w}.

Unfortunately, it is hard to see the relation between \alpha and \chi(M), the characteristic of M.

June 11, 2011

The Entropy-Logarithmic energy inequality

Filed under: PDEs, Riemannian geometry — Tags: — Ngô Quốc Anh @ 17:39

Recently, I have read a result by J. Demange published in J. Funct. Anal. in 2008 [here]. In that paper,  as a simple consequence, the author proves the following interesting inequality

Theorem. Let n\geqslant 2 be an integer and M be a compact connected n-dimensional Riemannian manifold with Ricci curvature bounded below by a positive constant \rho. The following inequality holds for d \in \mathbb R, d>n, d \geqslant 3, \alpha=1/2-1/d, and f , a smooth function mapping M onto \mathbb R_+

\displaystyle \left(\int_M f\right)^{2\alpha}-\int_M f^{2\alpha} \leqslant \frac{1}{K(n,d)}\int_M |\nabla f^\alpha|^2

and

\displaystyle \left(\int_M f\right)^{2\alpha}-\int_M f^{2\alpha} \leqslant K_1\log\left(1+K_2\int_M |\nabla f^\alpha|^2\right)

where

\displaystyle K(n,d)=\frac{\rho (d-2)}{4(1-\frac{1}{n})}, \quad L(n,d)=\frac{d-n}{n+2}\left( 4-9\frac{d-n}{d(n+2)}\right),

and

\displaystyle K_2=\frac{L(n,d)d}{4(d-2)K(n,d)(\int_M f)^{2\alpha}},\quad \frac{1}{K_1}=K_2K(n,d).

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June 8, 2011

Uniformly boundedness of positive smooth solutions to the Lichnerowicz equation in R^n

Filed under: PDEs — Tags: , — Ngô Quốc Anh @ 7:26

Let us consider the following so-called Lichnerowicz equation in \mathbb R^n

\displaystyle -\Delta u =-u^q+\frac{1}{u^{q+2}}, \quad q>0.

Recently, Brezis [here] proved the following

Theorem. Any solution of the Lichnerowicz equation with q>0 satisfies u\geqslant 1 in \mathbb R^n.

Let us study the trick used in his paper.

Proof. Let

f(t)=t^q-t^{q+2}, \quad t>0.

Then \Delta u =f(u). Fix any point x_0 \in \mathbb R^n and consider the function

u_\varepsilon(x)=u(x)+\varepsilon |x-x_0|^2, \quad \varepsilon>0, x \in \mathbb R^n.

Since u_\varepsilon(x) \to \infty as |x|\to \infty, \min_{\mathbb R^n}u_\varepsilon(x) is achieved at some x_1. We have

0\leqslant \Delta u_\varepsilon(x_1)=\Delta u(x_1)+2\varepsilon n=f(u(x_1))+2\varepsilon n.

By definition,

u(x_1)+\varepsilon |x_1-x_0|^2=u_\varepsilon(x_1) \leqslant u_\varepsilon(x_0)=u(x_0).

Thus,

u(x_1)\leqslant u(x_0).

Since f is increasing we deduce that

f(u(x_1))\leqslant f(u(x_0)).

Therefore,

0\leqslant \Delta u_\varepsilon(x_1) \leqslant f(u(x_1))+2\varepsilon n \leqslant f(u(x_0))+2\varepsilon n.

By sending \varepsilon \to 0 we deduce that f(u(x_0)) \geqslant 0. In other words, u(x_0) \geqslant 1.

As can be seen, he only uses the fact that f is monotone increasing in his argument, therefore, this approach can be used for a wider class of nonlinearity.

June 5, 2011

Variation of the determinant

Filed under: Riemannian geometry — Ngô Quốc Anh @ 6:05

Today, we shall prove the following identity

\displaystyle\boxed{\delta \det(g) = \delta \det (g_{\mu\nu}) = \det(g )g^{\mu\nu} \delta g_{\mu\nu} }.

First, by the Jacobi formula, we know that for any matrix A

\displaystyle d\mbox{det} (A) = \mbox{tr} (\mbox{adj}(A) \, dA)

where d is a differential of A. Since

\displaystyle A^{-1}=\frac{1}{\det (A)}\mbox{adj}(A),

we can rewrite the Jacobi formula as follows

\displaystyle d\mbox{det} (A) = \det (A)\mbox{tr} (A^{-1}dA).

We now make use this rule with A replaced by metric g and d replaced by \delta. Obviously,

\displaystyle \delta \det(g) =\det(g)\mbox{tr}(g^{-1}\delta g)= \det(g )g^{\mu\nu} \delta g_{\mu\nu}

where we have used the fact that g^{-1}=g^{\mu\nu} and that

\displaystyle \mbox{tr}(AB)=\sum_{i,j}A_{ij}B_{ij}.

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June 2, 2011

Variation of the Riemann tensor, the Ricci tensor, and the Ricci scalar

Filed under: Riemannian geometry — Ngô Quốc Anh @ 5:24

As an application of the principle of least action to the Einstein-Hilbert action, in this short note, we discuss a question:  the variation with respect to metric of the scalar curvature.

To calculate the variation of the scalar curvature we calculate first the variation of the Riemann curvature tensor, and then the variation of the Ricci tensor.

The variation of the Riemann curvature tensor. So, the Riemann curvature tensor is defined as,

\displaystyle {R^\rho}_{\sigma\mu\nu} = \partial_\mu\Gamma^\rho_{\nu\sigma} - \partial_\nu\Gamma^\rho_{\mu\sigma} + \Gamma^\rho_{\mu\lambda}\Gamma^\lambda_{\nu\sigma} - \Gamma^\rho_{\nu\lambda}\Gamma^\lambda_{\mu\sigma},

Since the Riemann curvature depends only on the Levi-Civita connection \Gamma^\lambda_{\mu\nu}, the variation of the Riemann tensor can be calculated as,

\displaystyle \delta{R^\rho}_{\sigma\mu\nu} = \partial_\mu\delta\Gamma^\rho_{\nu\sigma} - \partial_\nu\delta\Gamma^\rho_{\mu\sigma} + \delta\Gamma^\rho_{\mu\lambda} \Gamma^\lambda_{\nu\sigma} + \Gamma^\rho_{\mu\lambda} \delta\Gamma^\lambda_{\nu\sigma} - \delta\Gamma^\rho_{\nu\lambda} \Gamma^\lambda_{\mu\sigma} - \Gamma^\rho_{\nu\lambda} \delta\Gamma^\lambda_{\mu\sigma}.

Now, since \delta\Gamma^\rho_{\nu\mu} is the difference of two connections, it is a tensor and we can thus calculate its covariant derivative,

\displaystyle \nabla_\lambda (\delta \Gamma^\rho_{\nu\mu} ) = \partial_\lambda (\delta \Gamma^\rho_{\nu\mu} ) + \Gamma^\rho_{\sigma\lambda} \delta\Gamma^\sigma_{\nu\mu} - \Gamma^\sigma_{\nu\lambda} \delta \Gamma^\rho_{\sigma\mu} - \Gamma^\sigma_{\mu\lambda} \delta \Gamma^\rho_{\nu\sigma}.

We can now cleverly observe that the expression for the variation of Riemann curvature tensor above is equal to the difference of two such terms,

\displaystyle\boxed{\delta R^\rho{}_{\sigma\mu\nu} = \nabla_\mu (\delta \Gamma^\rho_{\nu\sigma}) - \nabla_\nu (\delta \Gamma^\rho_{\mu\sigma})}.

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