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.

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