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.

See also:

  1. Almost-Schur lemma by De Lellis and Topping.

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