# 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

## 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.

## 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

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

## July 26, 2009

### On the positive definite property of the Schur complement

Filed under: Các Bài Tập Nhỏ, Linh Tinh, Nghiên Cứu Khoa Học — Tags: — Ngô Quốc Anh @ 0:48

The following question was proposed in the NUS Q.E. in 2009: Given matrices $A \in \mathbb R^{n \times n}$, $B \in \mathbb R^{n \times m}$ and $C \in \mathbb R^{m \times m}$. Suppose $A$ and $C$ are symmetric. Consider the following matrices

$\displaystyle H = \left( {\begin{array}{*{20}{c}} A&B \\ {{B^T}}&C \end{array}} \right), \qquad S = C - {B^T}{A^{ - 1}}B.$

Show that $H$ is positive definite if and only if $A$ and $S$ are positive definite.

In the literature, the matric $S$ is called the Schur complement, usually, it is denoted by $H|A$ with respect to $A$. In other word, $H|C$ is of the form $A-B^TC^{-1}B$. It is worth noting that the letter $H$ used in the above notation indicates the full matrix $H$, roughly speaking, by $H|A$ we mean the Schur complement of $H$ with respect to $A$.

Throughout this entry, by $A >0$ (resp. $A \geq 0$) we mean that $A$ is positive definite (resp. positive semi-definite). In order to solve the above problem, one needs the following matrix identity, the Aitken block-diagonalization formula,

$\displaystyle\left( {\begin{array}{*{20}{c}} I&0 \\ { - {B^T}{A^{ - 1}}}&I \end{array}} \right)\left( {\begin{array}{*{20}{c}} A&B \\ {{B^T}}&C \end{array}} \right)\left( {\begin{array}{*{20}{c}} I&{ - {A^{ - 1}}B} \\ 0&I \end{array}} \right) = \left( {\begin{array}{*{20}{c}} A&0 \\ 0&{H|A} \end{array}} \right).$

Now we assume $A >0$ and $H|A >0$. Then the following property

$\displaystyle \left( {\begin{array}{*{20}{c}} A & 0 \\ 0 & {C - {B^T}{A^{ - 1}}B} \\ \end{array} } \right) > 0$

holds true. Indeed, since

$\displaystyle\left( {\begin{array}{*{20}{c}} {{A^{n \times n}}}&0 \\ 0&{{C^{m \times m}} - {B^T}{A^{ - 1}}B} \end{array}} \right) \in \text{Mat}(n + m)$

then for every

$\displaystyle\left( {\begin{array}{*{20}{c}} x&y \end{array}} \right) \in \text{Mat}(n + m)$

one has

$\displaystyle\left( {\begin{array}{*{20}{c}} {{x^T}}&{{y^T}} \end{array}} \right)\left( {\begin{array}{*{20}{c}} A&0 \\ 0&{M|A} \end{array}} \right)\left( {\begin{array}{*{20}{c}} x&y \end{array}} \right) = {x^T}Ax + {y^T}(M|A)y.$

Note that at least $x$ or $y$ is not a zero vector so that

$\displaystyle {x^T}Ax + {y^T}\left( {C - {B^T}{A^{ - 1}}B} \right)y > 0$

which proves the positive definite property of

$\displaystyle \left( {\begin{array}{*{20}{c}} A & 0 \\ 0 & {C - {B^T}{A^{ - 1}}B} \\ \end{array} } \right).$

Now by means of the above matrix identity we claims that $H>0$.

Conversely, for every $x \in \mathbb R^n$, one has

$\displaystyle 0 < \left( {\begin{array}{*{20}{c}} {{x^T}}&0 \end{array}} \right)H\left( {\begin{array}{*{20}{c}} x&0 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {{x^T}}&0 \end{array}} \right)\left( {\begin{array}{*{20}{c}} A&B \\ {{B^T}}&C \end{array}} \right)\left( {\begin{array}{*{20}{c}} x&0 \end{array}} \right) = {x^T}Ax$

whenever $x \ne 0$. Thus, this and the fact that $A$ is symmetric implies that $A >0$. As a consequence, $A^{-1}$ exists which helps us to say that $S$ is well-defined.

Now with the help of the matrix identity, one gets

$\displaystyle\left( {\begin{array}{*{20}{c}} 0&{{y^T}} \end{array}} \right)\left( {\begin{array}{*{20}{c}} I&0 \\ { - {B^T}{A^{ - 1}}}&I \end{array}} \right)\left( {\begin{array}{*{20}{c}} A&B \\ {{B^T}}&C \end{array}} \right)\left( {\begin{array}{*{20}{c}} I&{ - {A^{ - 1}}B} \\ 0&I \end{array}} \right)\left( {\begin{array}{*{20}{c}} 0&y \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0&{{y^T}} \end{array}} \right)\left( {\begin{array}{*{20}{c}} A&0 \\ 0&{H|A} \end{array}} \right)\left( {\begin{array}{*{20}{c}} 0&y \end{array}} \right).$

Note that, the left left side of the above identity is nothing but

$\displaystyle\underbrace {\left( {\begin{array}{*{20}{c}} 0&{{y^T}} \end{array}} \right)\left( {\begin{array}{*{20}{c}} I&0 \\ { - {B^T}{A^{ - 1}}}&I \end{array}} \right)}_D\left( {\begin{array}{*{20}{c}} A&B \\ {{B^T}}&C \end{array}} \right){D^T}$

which is positive by the assumption provided $y \ne 0$. Hence, if $y \ne 0$ is arbitrary, the right hand side equals to

$\displaystyle {{y^T}\left( {C - {B^T}{A^{ - 1}}B} \right)y}$

which proves that $S>0$. The proof is complete.

If I have time, I will provide another proof using the Sylvester’s Law of Inertia. For your convenience regarding to the Schur complement, I prefer you to the book entitled THE SCHUR COMPLEMENT AND ITS APPLICATIONS due to Fuzhen Zhang (edt.) for details.