# Ngô Quốc Anh

## August 20, 2012

### A Note On The Almost-Schur Lemma On 4-dimensional Riemannian Closed Manifolds

Filed under: Riemannian geometry — Tags: — Ngô Quốc Anh @ 14:10

Let us continue our posts 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 $R$.

Recently, Ge and Wang improved the above inequality for the case $n=4$. They showed that the above estimate remains valid provided the scalar curvature is non-negative.

Today, we talk about a work by Ezequiel R. Barbosa recently published in Proc. Amer. Math. Soc. 2012 [here]. Following is his main result

Theorem. Let $(M,g)$ be a $4$-dimensinal closed Riemannian manifold. Then

$\displaystyle\int_M {|\text{Ric} - \frac{{\overline R}}{4}g{|^2}} \leqslant 4\int_M {|\text{Ric} - \frac{R}{4}g{|^2}} + 9\lambda _g^2 - \frac{{\overline R}}{4}\int_M R$

where $\overline R=\text{vol(M)}^{-1}\int_M R$ is the average of the scalar curvature $R$ of $g$ and $\lambda_g$ is the Yamabe invariant. Moreover, the equality holds if and only if there exists a metric $g_1 \in [g]$ such that $(M,g_1)$ is an Einstein manifold.

As can be seen, the only contribution of the above theorem is to assume no conditions on the Ricci tensor or the scalar curvature.

It is worth noticing that if the Yamabe invariant $\lambda_g$ is nonnegative, then

$\displaystyle 9\lambda _g^2 - \frac{{\overline R}}{4}\int_M R = 9{\left( {\mathop {\inf }\limits_{{g_2} \in [g]} \frac{{\int_M {\frac{{{R_{{g_2}}}}}{6}} }}{{{{(\text{vol}(M,{g_2}))}^{\frac{1}{2}}}}}} \right)^2} - \frac{{\overline R}}{4}\int_M R \leqslant 0.$

Since

$\displaystyle |\text{Ric} - \frac{{\overline R}}{4}g{|^2} = |\text{Ric}{|^2} - \frac{{\overline R}}{2}R + \frac{{{{\overline R}^2}}}{4}$

the above inequality is equivalent to

$\displaystyle\frac{8}{3}\int_M {{\sigma _2}(g)} \leqslant \lambda _g^2.$

We now choose the metric $g_1$ in such a way that $\sigma_1(g_1)$ is constant. Then

$\displaystyle\frac{8}{3}\text{vol}(M,{g_1})\int_M {{\sigma _2}({g_1})} \leqslant \text{vol}(M,{g_1})\int_M {{{({\sigma _1}({g_1}))}^2}} = {\left( {\int_M {{\sigma _1}({g_1})} } \right)^2}.$

Therefore,

$\displaystyle\frac{8}{3}\int_M {{\sigma _2}({g_1})} \leqslant {\left( {\frac{1}{{\text{vol}(M,{g_1})}}\int_M {{\sigma _1}({g_1})} } \right)^2} = \lambda _g^2$

since $g_1$ is a Yamabe solution. Making use of $n=4$, we get

$\displaystyle\frac{8}{3}\int_M {{\sigma _2}(g)} = \frac{8}{3}\int_M {{\sigma _2}({g_1})} \leqslant \lambda _g^2$

which proves the result.