# Ngô Quốc Anh

## 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

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

## March 26, 2014

### Two pointwise conformal metrics having the same Ricci tensor must be homothetic

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 13:06

The aim of this note is to recall the following interesting result by X. Xu published  in Proc. AMS in 1992, here.

Theorem. Suppose $(M,g)$ is a compact, oriented Riemannian manifold without boundary of dimension $n \geq 2$. If $\widehat g=e^{2\varphi} g$ and $\text{Ric}(\widehat g)=\text{Ric}(g)$, then $\varphi$ is constant. In other words, two pointwise conformal metrics that have the same Ricci tensor must be homothetic.

A proof for this result is quite simple. First, we recall the following conformal change

$\displaystyle\widehat{\text{Ric}}_{ij}=\text{Ric}_{ij} -(n-2)\big( \text{Hess}(\varphi)_{ij}-\nabla_i\varphi \nabla_j\varphi \big) - \big( \Delta_g \varphi + (n-2) |\nabla \varphi|^2 \big) g_{ij}$

where $\Delta_g u = \text{div}(\nabla u)$. Therefore, if $\text{Ric}(\widehat g)=\text{Ric}(g)$, then we obtain the following fact

$\displaystyle (n-2)\big( \text{Hess}(\varphi)_{ij}-\nabla_i\varphi \nabla_j\varphi \big)+ \big( \Delta_g \varphi + (n-2) |\nabla \varphi|^2 \big) g_{ij}=0.$

However, the term $\nabla_i\varphi \nabla_j\varphi$ appearing in the preceding identity seems to be bad. To avoid it, the author used the following conformal change

$\displaystyle \widehat g = \frac{1}{u^2} g$

for some positive function $u$, i.e. $\varphi = -\log u$ or $u=e^{-\varphi}$. Then we calculate to obtain

$\displaystyle \nabla_i u =-e^\varphi \nabla_i \varphi$

## March 12, 2014

### H^1 boundedness of a sequence of solutions to the prescribing Gaussian curvature problem in the negative case

Filed under: Uncategorized — Tags: — Ngô Quốc Anh @ 22:31

In a previous note, we showed how to prove an $H^1$ boundedness of a sequence of solutions $\{u_k\}_k$ to the following PDE

$\displaystyle -\Delta u_k +\alpha_k = R(x)e^{u_k}$

in the negative case, i.e. $\alpha _k\searrow \alpha$ for some $\alpha <0$ as $k \to \infty$. In this note, we do not change $\alpha_0$, instead, we are going to change $R$. More precisely, we are interested in some $H^1$ boundedness of a sequence of solutions $\{u_k\}_k$ to the following PDE

$\displaystyle -\Delta u_k +\alpha = R_k(x)e^{u_k}$

for some $R_k = R+\lambda_k$ with $\lambda_k \searrow 0$ as $k \to \infty$. This note is adapted from a recent preprint by Michael Struwe et al., see this.

As always, we assume that $\alpha <0$ is constant and $(M,g)$ is a closed, connected Riemannian surface with smooth background metric $g$. We further assume for the sake of simplicity that $\text{vol}(M,g)=1$.

Step 1. We claim for sufficiently large $k$ that

$\displaystyle \int_M R_k^4 e^{u_k} \leqslant C(R)$

for some constant $C$ depending only on $\|R\|_{C^1}$ and on $(M,g)$.

Proof of Step 1. To prove this, we observe that

$\displaystyle \int_M R_k =\int_M e^{-u_k} \big( -\Delta u_k +\alpha\big) = -\int_M |\nabla u_k|^2 e^{-u_k} + \alpha\int_M e^{-u_k},$

which implies, by $\alpha <0$, for large $k$ that

$\displaystyle\int_M |\nabla u_k|^2 e^{-u_k}\leqslant \int_M R_k < 1+\int_M R.$

## March 5, 2014

### Uniformly upper Bound for Positive Smooth Solutions To The Lichnerowicz Equation In R^N

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 16:16

In this note, we are interested in the following Lichnerowicz type equation in $\mathbb R^n$

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

As we have already seen from the previous note that solutions for the above equation are always bounded from below for certain $q$.

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

Remarkably, we are able to prove that solutions for the Lichnerowicz type equation are also bounded from above if we require $q>1$ instead of $q>0$. This result is basically due to L. Ma and X. Xu, see this paper.

Theorem (Ma-Xu). Any solution of the Lichnerowicz type equation with $q>1$ is uniformly bounded from above in $\mathbb R^n$.

Combining the two theorem above, we conclude that any solution of the Lichnerowicz equation, i.e. $q=(n+2)/(n-2)$, is uniformly bounded in $\mathbb R^n$.

The idea of the proof for Ma-Xu’s theorem is as follows: Denote

$\displaystyle f(u)=u^{-q-2} - u^q.$

Fix $x_0 \in \mathbb R^n$ but arbitrary, we then look for a positive radial super-solution $v(x)=v(|x|)>0$ of the Lichnerowicz type equation on the ball $B_R(x_0)$ with positive infinity boundary condition for some $R$ to be specified. This is equivalent to finding $v$ in such a way that

$\displaystyle\begin{array}{rcl}\displaystyle-\Delta v &\geqslant& f(v), \qquad\text{ in } B_R(x_0),\\ v&\equiv& +\infty, \qquad\text{ on }\partial B_R(x_0).\end{array}$