Ngô Quốc Anh

August 14, 2012

Almost-Schur lemma on 4-dimensional closed Riemannian manifolds

Filed under: Riemannian geometry — Tags: — Ngô Quốc Anh @ 11:49

Let us continue our previous post regarding to the Schur lemma, i.e., the following estimate

$\displaystyle \int_M {|\text{Ric} - \frac{{\overline R}}{n}g{|^2}} \leqslant \frac{{{n^2}}}{{{{(n - 2)}^2}}}\int_M {|\text{Ric} - \frac{R}{n}g{|^2}}$

holds provided $\text{Ric} \geqslant 0$ and $n\geqslant 5$ where $R$ is the scalar curvature and $\overline R=\text{vol(M)}^{-1}\int_M R$ is the average of $R$.

It was proved by De Lellis and Topping that the condition $\text{Ric} \geqslant 0$ cannot be relaxed. Also, the condition $n\geqslant 5$ plays an important role in their argument.

Very recently, in their paper, Ge and Wang proved the following

Theorem. If $n = 4$ and if $(M, g)$ is a closed Riemannian manifold with nonnegative scalar curvature, then

$\displaystyle \int_M {|\text{Ric} - \frac{{\overline R}}{n}g{|^2}} \leqslant \frac{{{n^2}}}{{{{(n - 2)}^2}}}\int_M {|\text{Ric} - \frac{R}{n}g{|^2}}$

olds. Moreover, equality holds if and only if $(M, g)$ is an Einstein manifold.

Also, if we denote by $\sigma_k(g)$ the $k$-scalar curvature of metric $g$, they found that the above inequality is equivalent to the following

$\displaystyle {\left( {\int_M {{\sigma _1}(g)} } \right)^2} \geqslant \frac{{2n}}{{n - 2}}\text{vol}(M,g)\int_M {{\sigma _2}(g)}.$

As such, instead of proving the former inequality, they aimed to prove the latter one. In order to mention their proof, let us recall the definition of the $k$-scalar curvature, which was first introduced by Viaclovsky in his PhD thesis and has been intensively studied by many mathematicians.

Let

$\displaystyle S_g=\frac{1}{n-2}\left( \text{Ric}-\frac{R}{2(n-1)}g\right)$

be the Schouten tensor of $g$. For an integer $k$ with $1 \leqslant k \leqslant n$ let $\sigma_k$ be the $k$-th elementary symmetric function in $\mathbb R^n$. The $k$-scalar curvature is

$\sigma_k(g)=\sigma_k(\Lambda_g)$

where $\Lambda_g$ is the set of eigenvalues of the matrix $g^{-1}S$. In particular, $\sigma_1(g)=\text{trace}(S_g)$ and $\sigma_2(g)=\frac{1}{2}((\text{trace}S_g)^2-|S_g|^2)$.

Clearly,

$\displaystyle \sigma_1(g)=\frac{R}{2(n-1)}$

and

$\displaystyle \sigma_2(g)=\frac{1}{2(n-2)^2}\left(-|\text{Ric}|^2+\frac{n}{4(n-1)}R^2\right).$

In order to prove the inequality, they proved the following

Lemma. For any $n \geqslant 3$ and any closed Riemannian manifold $(M, g)$, thereexists a conformal metric $g_1 \in [g]$ satisfying

$\displaystyle\frac{{2n}}{{n - 2}}\frac{{\int_M {{\sigma _2}(g_1)} }}{{{{\left( {\text{vol}(M,{g_1})} \right)}^{\frac{{n - 4}}{n}}}}} \leqslant {Y}{([g])^2}$

where the term on the right hand side is usually called the Yamabe invariant, i.e.,

$\displaystyle Y([g])= \mathop {\inf }\limits_{{g_2} \in [g]} \frac{{\int_M {{\sigma _1}({g_2})} }}{{{{(\text{vol}(M,{g_2}))}^{\frac{{n - 2}}{n}}}}}.$

The proof of the lemma depends on the following: for any $n \times n$ symmetric matrix $A$, there holds

$\displaystyle (\sigma_1(A))^2 \geqslant \frac{2n}{n-1}\sigma_2(A).$

The metric $g_1$ is chosen in such a way that the scalar curvature of $g_1$ is constant. This is the so-called Yamabe problem.

In order to prove the main result, they observed that in the case of  dimension $n = 4$, it is well known that $\int_M \sigma_2(g)$ is constant in any given conformal class. Therefore,

$\displaystyle\frac{{2n}}{{n - 2}}\int_M {{\sigma _2}(g)} = \frac{{2n}}{{n - 2}}\int_M {{\sigma _2}({g_1})} \leqslant Y{([g])^2} \leqslant {\left( {\frac{{\int_M {{\sigma _1}(g)} }}{{{{(\text{vol}(M,g))}^{\frac{1}{2}}}}}} \right)^2}.$

Recently, Ye and Wang also extended the above result to the case $n=3$. We shall discuss this result later.