# Ngô Quốc Anh

## August 7, 2012

### Almost-Schur lemma

Filed under: Nghiên Cứu Khoa Học, Riemannian geometry — Tags: — Ngô Quốc Anh @ 16:23

Following this note, today we talk about an almost-Schur lemma recently obtained by De Lellis and Topping, see here. If we denote by $\mathop {{\text{Ric}}}\limits^ \circ$ the traceless Ricci tensor, the main theorem of the paper is the following

Theorem. For any integer $n \geqslant 3$, if $(M, g)$ is a closed Riemannian manifold of  dimension $n$ with nonnegative Ricci curvature, then

$\displaystyle \int_M {{{(R - \overline R )}^2}} \leqslant \frac{{4n(n - 1)}}{{{{(n - 2)}^2}}}\int_M {| \mathop {{\text{Ric}}}\limits^ \circ {|^2}}$

where $\overline R$ is the average value of the scalar curvature $R$ over $M$. Moreover equality holds if and only if $(M, g)$ is Einstein.

For a proof of the theorem, recall that the contracted second Bianchi identity tells us that

$\displaystyle\delta \text{Ric} + \frac{1}{2}dR = 0$

where

$\displaystyle {(\delta \text{Ric})_j} = - {\nabla _i}{R_{ij}}.$

and hence that

$\displaystyle\delta\mathop{\text{Ric}}\limits^\circ = - \frac{{n - 2}}{{2n}}dR.$

Let $f : M \to \mathbb R$ be the unique solution to $\Delta f = R-\overline R$ with $\int_M f=0$. We may then compute

$\displaystyle\begin{gathered} \int_M {{{(R - \overline R )}^2}} = \int_M {(R - \overline R )\Delta f} \hfill \\ \qquad= \int_M {\left\langle {dR,df} \right\rangle } \hfill \\ \qquad= \frac{{2n}}{{n - 2}}\int_M {\left\langle {\delta \mathop {{\text{Ric}}}\limits^ \circ,df} \right\rangle } \hfill \\ \qquad= \frac{{2n}}{{n - 2}}\int_M {\left\langle {\mathop {{\text{Ric}}}\limits^ \circ,\text{Hess}f} \right\rangle } \hfill \\ \qquad= \frac{{2n}}{{n - 2}}\int_M {\left\langle {\mathop {{\text{Ric}}}\limits^ \circ,\text{Hess}f - \frac{{\Delta f}}{n}g} \right\rangle } \hfill \\ \qquad\leqslant \frac{{2n}}{{n - 2}}{\left\| {\mathop {{\text{Ric}}}\limits^ \circ} \right\|_{{L^2}}}{\left\| {\text{Hess}f - \frac{{\Delta f}}{n}g} \right\|_{{L^2}}}. \hfill \\ \end{gathered}$

Now by integration by parts (i.e. the Bochner formula) we know that

$\displaystyle\int_M {|\text{Hess}f{|^2}} = \int_M {{{(\Delta f)}^2}} - \int_M {\text{Ric}(\nabla f,\nabla f)}$

and therefore

$\displaystyle\begin{gathered} \int_M {|\text{Hess}sf - \frac{{\Delta f}}{n}g{|^2}} = \int_M {[|\text{Hess}f{|^2} - \frac{1}{n}{{(\Delta f)}^2}]} \hfill \\ \qquad= \frac{{n - 1}}{n}\int_M {{{(\Delta f)}^2}} - \int_M {\text{Ric}(\nabla f,\nabla f)} \hfill \\ \qquad= \frac{{n - 1}}{n}\int_M {{{(R - \overline R )}^2}} - \int_M {\text{Ric}(\nabla f,\nabla f)} \hfill \\ \end{gathered}$

and since the Ricci curvature is nonnegative, we have

$\displaystyle {\left\| {\text{Hess}f - \frac{{\Delta f}}{n}g} \right\|_{{L^2}}} \leqslant {\left( {\frac{{n - 1}}{n}\int_M {{{(R - \overline R )}^2}} } \right)^{\frac{1}{2}}},$

which concludes the proof. For the equality case, we refer the reader to the original paper.

Using the identity mentioned in this note, one has the following estimate

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

and equality holds if and only if $(M, g)$ is Einstein.

Finally, they proved that the condition $\text{Ric} \geqslant 0$ cannot be dropped as they constructed a counterexample for $\mathbb S^n$ with $n \geqslant 5$. The case $n=3$ was also considered in the paper. Unfortunately, the case $n=4$ was left as an open question.