# Ngô Quốc Anh

## January 31, 2010

### R-G: Gauss and Codazzi equations in Riemannian geometry

Filed under: Riemannian geometry — Ngô Quốc Anh @ 10:29

So far, we have discussed Codazzi equations in classical differential geometry. Today, we discuss the Gauss and Codazzi equations in Riemannian geometry. Later on, we will discuss in a new entry the Gauss and Codazzi equations in general relativity.

A classical problem in geometry is to determine whether a Riemannian manifold $M$ can be isometrically immersed in another Riemaniann manifold $\overline M$. We will restrict ourselves to the case of codimension $1$ immersions, i.e., $M$ has dimension $n$ and $\overline M$ has dimension $n + 1$.

It is well known that the Gauss and Codazzi equations are necessary conditions relating the Riemann curvature tensor $R$ of $M$, the Riemann curvature tensor $\overline R$ of $\overline M$ and the shape operator $S$ of $M$ (or the second fundamental form). Denoting by $\nabla$ the Riemannian connection of $M$, these equations are the following

$\displaystyle\left\langle {R\left( {X,Y} \right)Z,W} \right\rangle- \left\langle {\overline R \left( {X,Y} \right)Z,W} \right\rangle= \left\langle {SX,Z} \right\rangle \left\langle {SY,W} \right\rangle- \left\langle {SY,Z} \right\rangle \left\langle {SX,W} \right\rangle$

and

$\displaystyle {\nabla _X}SY - {\nabla _Y}SX - S\left[ {X,Y} \right] = \overline R \left( {X,Y} \right)N$

for all vector fields $X$, $Y$, $Z$ and $W$ on $M$.

## January 27, 2010

### Maple: tensor Package

Filed under: Nghiên Cứu Khoa Học, Riemannian geometry — Ngô Quốc Anh @ 17:52

The tensor package (tensor) contains commands that deal with tensors, their operations, and their use in General Relativity both in the natural basis and in a moving frame.

The following is a list of available commands.

 act apply an operation on the elements of a tensor, spin or curvature table antisymmetrize fully antisymmetrize tensor change_basis change basis Christoffel1 Christoffel symbols of the first kind Christoffel2 Christoffel symbols of the second kind commutator commutator of two vectors compare compare two tensors, spin or curvature tables conj complex conjugation connexF connection coefficients for a rigid frame contract contract indices convertNP convert connection or Riemann tensor to the NP formalism cov_diff covariant differentiation create create a tensor object d1metric first partial derivatives of the metric d2metric second partial derivatives of the metric directional_diff directional derivative display_allGR display the objects used in General Relativity displayGR display one object used in General Relativity dual perform the dual operation on the indices of a tensor Einstein Einstein tensor entermetric facility for the input of metric tensor components exterior_diff exterior differentiation exterior_prod exterior product frame compute the frame that brings the metric to the diagonal signature metric geodesic_eqns Euler-Lagrange equations for geodesic curves get_char get the character (covariant/contravariant) of an object get_compts get the components of an object get_rank get the rank of an object invars invariants of the Riemann curvature tensor (General Relativity) invert inverse of a second rank tensor Jacobian Jacobian of a coordinate transformation Killing_eqns Killing’s equation (related to symmetries of the space) Levi_Civita Levi-Civita pseudo-tensors Lie_diff Lie derivative with respect to a vector lin_com linear combination of tensor objects lower lower indices npcurve Newmann-Penrose curvature component in Debever formalism (G.R.) npspin Newmann-Penrose spin component in Debever formalism (G.R.) partial_diff partial derivative of a tensor permute_indices permutation of indices petrov classification of polynomials of degree 4 prod inner and outer tensor product raise raise indices Ricci Ricci tensor Ricciscalar Ricci scalar Riemann Riemann tensor RiemannF Riemann curvature tensor in a rigid frame symmetrize fully symmetrize a tensor tensorsGR compute the objects used in General Relativity transform Change coordinates systems Weyl Weyl tensor

Example.

1. Define the coordinate variables and the covariant Schwarzchild metric tensor components

> with(tensor):
coord := [t, r, th, ph]:
g_compts := array(symmetric,sparse, 1..4, 1..4):
g_compts[1,1] := 1-2*m/r:
g_compts[2,2]:= -1/g_compts[1,1]:
g_compts[3,3] := -r^2:
g_compts[4,4] := -r^2*sin(th)^2:
g := create( [-1,-1], eval(g_compts));

2. Now compute all of the quantities necessary to compute the Einstein tensor

> ginv := invert( g, ‘detg’ ):
D1g:=d1metric( g, coord ): D2g:=d2metric( D1g, coord ):
Cf1 := Christoffel1 ( D1g ):
RMN := Riemann( ginv, D2g, Cf1 ):
RICCI := Ricci( ginv, RMN ):
RS := Ricciscalar( ginv, RICCI ):

3. Compute the Einstein tensor

> Estn := Einstein( g, RICCI, RS );

Here is the picture

## January 26, 2010

### RG: Pointwise norm of an (r,s)-tensor relative to a metric g

Filed under: Nghiên Cứu Khoa Học, Riemannian geometry — Tags: — Ngô Quốc Anh @ 1:03

Today, let’s talk about the pointwise norm of an $(r,s)$-tensor $\alpha$ relative to a metric $g$. We assume

$\displaystyle\alpha= \alpha _{{j_1} \cdots {j_r}}^{{k_1} \cdots {k_s}}\frac{\partial }{{\partial {x^{{k_1}}}}} \otimes\cdots\otimes \frac{\partial }{{\partial {x^{{k_s}}}}} \otimes d{x^{{j_1}}} \otimes\cdots\otimes d{x^{{j_r}}}$.

Then

$\displaystyle \left| \alpha\right| = \sqrt {{g^{{j_1}{q_1}}}...{g^{{j_r}{q_r}}}{g_{{k_1}{p_1}}}...{g_{{k_s}{p_s}}}\alpha _{{j_1} \cdots {j_r}}^{{k_1} \cdots {k_s}}\alpha _{{q_1} \cdots {q_r}}^{{p_1} \cdots {p_s}}}$.

For example, if $f$ is a scalar function, then $\nabla f$ is an $(1,0)$-tensor and $\nabla^2 f$ is an $(2,0)$-tensor. To be precise,

$\displaystyle \nabla f = \frac{{\partial f}}{{\partial {x^i}}}d{x^i}, \quad {\nabla ^2}f = \left( {\frac{{{\partial ^2}f}}{{\partial {x^i}\partial {x^j}}} - \Gamma _{ij}^k\frac{{\partial f}}{{\partial {x^k}}}} \right)d{x^i} \otimes d{x^j}$.

Then

$\displaystyle\left| {\nabla f} \right| = \sqrt {{g^{ij}}\frac{{\partial f}}{{\partial {x^i}}}\frac{{\partial f}}{{\partial {x^j}}}}$

and

$\displaystyle\left| {{\nabla ^2}f} \right| = \sqrt {{g^{ip}}{g^{jq}}\left( {\frac{{{\partial ^2}f}}{{\partial {x^i}\partial {x^j}}} - \Gamma _{ij}^k\frac{{\partial f}}{{\partial {x^k}}}} \right)\left( {\frac{{{\partial ^2}f}}{{\partial {x^p}\partial {x^q}}} - \Gamma _{pq}^l\frac{{\partial f}}{{\partial {x^l}}}} \right)}$.

## January 23, 2010

### RG: High order covariant derivatives

Filed under: Nghiên Cứu Khoa Học, Riemannian geometry — Ngô Quốc Anh @ 21:47

Today we talk about high order covariant derivative. Recall from this topic that covariant derivative $\nabla$ acts on a $(r,s)$-tensor, the result is an $(r+1,s)$-tensor. For convenience, we mean

$\displaystyle\alpha= \alpha _{{j_1} \cdots {j_r}}^{{k_1} \cdots {k_s}}\frac{\partial }{{\partial {x^{{k_1}}}}} \otimes\cdots\otimes \frac{\partial }{{\partial {x^{{k_s}}}}} \otimes d{x^{{j_1}}} \otimes\cdots\otimes d{x^{{j_r}}}$.

Thus

$\displaystyle\nabla :{C^\infty }({ \bigotimes\nolimits^{r,s}}M) \to {C^\infty }({ \bigotimes\nolimits^{r + 1,s}}M)$.

Precisely, for an $(r,s)$-tensor $\alpha$, one defines $\nabla \alpha$ as the following

$\displaystyle\left( {\nabla \alpha } \right)\left( {X,{Z_1},{Z_2},...,{Z_r}} \right) = \left( {{\nabla _X}\alpha } \right)\left( {{Z_1},{Z_2},...,{Z_r}} \right)$

where the right hand side is nothing but

$\displaystyle \left( {{\nabla _X}\alpha } \right)\left( {{Z_1},{Z_2},...,{Z_r}} \right) = {\nabla _X}\alpha \left( {{Z_1},{Z_2},...,{Z_r}} \right) - \sum\limits_{k = 1}^n {\alpha \left( {{Z_1},{Z_2},...,{\nabla _X}{Z_k},...,{Z_r}} \right)}$.

Note that, the first term of the right hand side of the above identity is covarian derivative acting on an $(0,s)$-tensor. This action can be defined by the following rule

$\displaystyle {\nabla _X}\left( {{Z_1} \otimes {Z_2} \otimes\cdots\otimes {Z_s}} \right) = \sum\limits_{k = 1}^s {{Z_1} \otimes {Z_2} \otimes\cdots\otimes {\nabla _X}{Z_k} \otimes\cdots\otimes {Z_s}}$.

We are now in a position to define the second order covariant derivative, denoted by $\nabla^2$. To be exact, we define

$\displaystyle {\nabla ^2}:{C^\infty }({\bigotimes\nolimits^{r,s}}M) \to {C^\infty }({\bigotimes\nolimits^{r + 2,s}}M)$

where

$\displaystyle\begin{gathered}\left( {{\nabla ^2}\alpha } \right)\left( {X,Y,{Z_1},{Z_2},...,{Z_r}} \right) = {\nabla _X}\left( {\nabla \alpha } \right)\left( {Y,{Z_1},{Z_2},...,{Z_r}} \right) \hfill \\ \qquad= \left[ {{\nabla _X}\left( {\nabla \alpha (Y)} \right) - \nabla \alpha ({\nabla _X}Y)} \right]\left( {Y,{Z_1},{Z_2},...,{Z_r}} \right) \hfill \\ \qquad= {\nabla _X}\left( {{\nabla _Y}\alpha } \right)\left( {{Z_1},{Z_2},...,{Z_r}} \right) - {\nabla _{{\nabla _X}Y}}\alpha \left( {{Z_1},{Z_2},...,{Z_r}} \right). \hfill \\ \end{gathered}$

The higher order can be defined similarly. We shall use the following notation

$\displaystyle \nabla _{X,Y}^2\alpha \left( {{Z_1},..,{Z_r}} \right) = {\nabla ^2}\alpha \left( {X,Y,{Z_1},...,{Z_r}} \right)$.

So now the way to understand our definition for second order covariant derivative is the following

$\displaystyle\nabla _{X,Y}^2\alpha= {\nabla _X}{\nabla _Y}\alpha- {\nabla _{{\nabla _X}Y}}\alpha$.

We also have another notation, we refer the reader to this topic for further discussion. We define

$\displaystyle ({\nabla _i}{\nabla _j}\alpha )\left( {{Z_1},{Z_2},...,{Z_r}} \right) = ({\nabla ^2}\alpha )\left( {\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^j}}},{Z_1},{Z_2},...,{Z_r}} \right)$.

This notation can be defined for higher order covariant derivatives. We are now interested in computing coefficients. Having

$\displaystyle\alpha= \alpha _{{j_1} \cdots {j_r}}^{{k_1} \cdots {k_s}}\frac{\partial }{{\partial {x^{{k_1}}}}} \otimes\cdots\otimes \frac{\partial }{{\partial {x^{{k_s}}}}} \otimes d{x^{{j_1}}} \otimes\cdots\otimes d{x^{{j_r}}}$

one can see that

$\displaystyle {\nabla _i}\alpha _{{j_1} \cdots {j_r}}^{{k_1} \cdots {k_s}}\frac{\partial }{{\partial {x^{{k_1}}}}} \otimes\cdots\otimes \frac{\partial }{{\partial {x^{{k_s}}}}} = ({\nabla _i}\alpha )\left( {\frac{\partial }{{\partial {x^{{j_1}}}}},...,\frac{\partial }{{\partial {x^{{j_r}}}}}} \right)$.

Examples. We now compute ${\nabla _i}{\nabla _j}f$. Clearly,

$\displaystyle\begin{gathered}{\nabla _i}{\nabla _j}f = ({\nabla ^2}f)\left( {\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^j}}}} \right) \hfill \\ \qquad= {\nabla _i}{\nabla _j}f - {\nabla _{{\nabla _i}\frac{\partial }{{\partial {x^j}}}}}f \hfill \\ \qquad= \frac{{{\partial ^2}f}}{{\partial {x^i}\partial {x^j}}} - {\nabla _{\Gamma _{ij}^k\frac{\partial }{{\partial {x^k}}}}}f \hfill \\ \qquad= \frac{{{\partial ^2}f}}{{\partial {x^i}\partial {x^j}}} - \Gamma _{ij}^k{\nabla _{\frac{\partial }{{\partial {x^k}}}}}f \hfill \\ \qquad= \frac{{{\partial ^2}f}}{{\partial {x^i}\partial {x^j}}} - \Gamma _{ij}^k\frac{{\partial f}}{{\partial {x^k}}}. \hfill \\ \end{gathered}$

This is similar to the Hessian of a function discussed here.

## January 22, 2010

### Finite element method: A brief introduction via an 2-point boundary value problem of laplacian equation in 1D

Filed under: Các Bài Tập Nhỏ, Giải tích 9 (MA5265), Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 17:05

Followed by this topic where I discussed how to use Finite Difference Method in numerical analysis. Today, we will study a so-called Finite Element Method, an improved Finite Difference Method. We first consider the following problem

$-u''(x)=f(x)$ where $0 \leq x \leq 1$ with boundary conditions $u(0)=0=u(1)$.

The above problem is usually called 2-point boundary value problem of laplacian equation in 1D.

We first consider such problem by using Finite Difference Method. The idea of using the Finite Difference Method is to split equally the interval $[0,1]$ into $n$ pieces by $x_i=\frac{i}{n}$ where $i=\overline{0,n}$. We then approximate $u''$ by using the following

$\displaystyle u''({x_i}) = \frac{{{u_{i - 1}} - 2{u_i} + {u_{i + 1}}}}{{{h^2}}}, \quad i=\overline{1,n-1}$

where $h:=\frac{1}{n}$. Therefore, we have the following system of linear equations

$\displaystyle\frac{{ - 1}}{{{h^2}}}\left( {\begin{array}{*{20}{c}} { - 2} & 1 & {} & {} & {}\\ 1 & { - 2} & 1 & {} & {} \\ {} & {} & {} & {} & {}\\ {} & {} & {} &\ddots& 1\\ {} & {} & {} & 1 & 2 \\ \end{array} } \right) = \left( {\begin{array}{*{20}{c}} {f({x_1})}\\ {f({x_1})}\\ \vdots \\ {f({x_{n - 1}})}\\ \end{array} } \right).$

The idea of the Finite Element Method is to split the interval $[0,1]$ into $n$ pieces by $x_i$ where $i=\overline{0,n}$ but not necessarily equality. In order to deal with the problem, we shall look for weak solutions. In the sense of the distribution, $u$ is called a weak solution if

$\displaystyle\int_0^1 {\nabla u\nabla vdx} = \int_0^1 {f(x)vdx} ,\quad\forall v \in H_0^1([0,1])$.

Under some conditions, for example, if $f \in L^2([0,1])$, the existence of weak solution is well-understood via the Lax-Milgram lemma.

Let $V_h$ be a collection of piecewise linear map $g$ in $[0,1]$ so that $g$ is linear on every intervals $[x_{i-1},x_i]$ where $i=\overline{1,n}$. We also suppose that $g$ is such that $g(0)=g(1)=0$. Clearly the space $V_h$ is finite dimension with a basis $\{\phi_i\}_{i=1}^{n-1}$ constructed as follows

$\displaystyle {\phi _i}(x) = \left\{ \begin{gathered}1,\quad x = {x_i}, \hfill \\ 0 , \quad x \ne {x_i}. \hfill \\ \end{gathered}\right.$

It is easy to see that $\phi_i \in H_0^1([0,1])$.

We will find $u$ as a linear combination of these functions $\phi_i$ denoted by $u_h$, to be exact, we will approximate $u$ by $u_h$ such that at $u(x_i)=u_h(x_i)$ for every $i=\overline{1,n-1}$. Precisely, let

$\displaystyle {u_h}(x) = \sum\limits_{i = 1}^{n - 1} {{\xi _i}{\phi _i}(x)}$

we then have

${\xi _i}=\displaystyle u({x_i}), \quad i = \overline {1,n - 1}$.

From the equation defining the notion of weak solution we replace $v$ by $\phi _i$ and $u$ by $u_h$, we now get

$\displaystyle\int_0^1 {\left( {\sum\limits_{j= 1}^{n - 1} {u({x_j}){{\phi '}_j}(x)} } \right){{\phi '}_i}(x)dx} = \int_0^1 {f(x){\phi _i}(x)dx}$

which yields

$\displaystyle\sum\limits_{j = 1}^{n - 1} {u({x_j})\int_0^1 {{{\phi '}_j}(x){{\phi '}_i}(x)dx} } = \int_0^1 {f(x){\phi _i}(x)dx}$.

The above identity can be rewritten as a linear system $Ax=b$ as the following

$\displaystyle\left( {\begin{array}{*{20}{c}} {({\phi' _1},{\phi' _1})} & {({\phi '_2},{\phi '_1})} &\cdots& {} & {} &\cdots& {({\phi' _{n - 1}},{\phi' _1})}\\{({\phi '_1},{\phi' _2})} & {({\phi' _2},{\phi' _2})} & \cdots & {} & {} & {} & \vdots\\ \vdots&\vdots&\ddots& {} & {} & {} & {}\\{} & {} & {} & {} & {} & {} & {}\\{} & {} & {} & {} & {} & {} & {}\\ \vdots&\vdots& {} & {} & {} &\ddots&\vdots \\{({\phi '_1},{\phi' _{n - 1}})} & {({\phi' _2},{\phi' _{n - 1}})} &\cdots& {} & {} &\cdots& {({\phi' _{n - 1}},{\phi' _{n - 1}})}\\ \end{array} } \right)\left( {\begin{array}{*{20}{c}} {u({x_1})}\\{u({x_2})}\\ \vdots \\ \vdots \\ \vdots \\{u({x_{n - 2}})}\\{u({x_{n - 1}})}\\ \end{array} } \right) = \left( {\begin{array}{*{20}{c}} {(f,{\phi _1})}\\{(f,{\phi _2})}\\ \vdots \\ \vdots \\ \vdots \\{(f,{\phi _{n - 2}})}\\{(f,{\phi _{n - 1}})}\\ \end{array} } \right)$

where we use the following notation

$\displaystyle (f,g) = \int_0^1 {f(x)g(x)dx}$.

## January 20, 2010

### RG: Divergence of an (p,s)-tensor

Filed under: Nghiên Cứu Khoa Học, Riemannian geometry — Ngô Quốc Anh @ 14:24

Followed by this topic where we define the divergence of a vector field $X$ as follows

$\displaystyle {\rm div} X = dx^i \left( \nabla_{\frac{\partial}{\partial x^i}} X\right)$

that is, in local coordinates,

$\displaystyle {\rm div} X = \left\langle {\frac{{\partial {X^j}}}{{\partial {x^i}}}\frac{\partial }{{\partial {x^j}}} + {X^j}\Gamma _{ij}^k\frac{\partial }{{\partial {x^k}}},d{x^i}} \right\rangle = \frac{{\partial {X^i}}}{{\partial {x^i}}} + {X^j}\Gamma _{ij}^i$

it’s time to take opportunity to define the divergence of an $(p,s)$-tensor where $p\geq 1$. We assume $\alpha$ is an $(p,0)$-tensor, that is,

$\displaystyle \alpha= {\alpha _{{i_1}{i_2} \cdots {i_p}}}d{x^{{i_1}}} \otimes d{x^{{i_2}}} \otimes\cdots\otimes d{x^{{i_p}}}$.

We then defined the divergence of $\alpha$ to be an $(p-1,0)$-tensor given as the following

$\displaystyle {\rm div}{(\alpha)_{{i_1}{i_2} \cdots {i_{p - 1}}}} = {g^{\alpha \beta }}{\nabla _\alpha }{\alpha _{\beta {i_1}{i_2} \cdots {i_{p - 1}}}} = {\nabla _\alpha }{\alpha _{\alpha {i_1}{i_2} \cdots {i_{p - 1}}}}$.

In particular, for a $1$-form $X$, that is an $(1,0)$-tensor with the form $X=X_idx^i$ in local coordinates, one has

$\displaystyle {\rm div}(X) = {g^{\alpha \beta }}{\nabla _\alpha }{X_\beta }$.

In general, if $\alpha$ is an $(p,s)$-tensor with $p\geq 1$, then the divergence of $\alpha$ to be an $(p-1,s)$-tensor given as the following

$\displaystyle {\rm div}{(\alpha)_{{i_1}{i_2} \cdots {i_{p - 1}}}}^{j_1j_2\cdots j_s} = {g^{\alpha \beta }}{\nabla _\alpha }{\alpha _{\beta {i_1}{i_2} \cdots {i_{p - 1}}}}^{j_1j_2\cdots j_s} = {\nabla _\alpha }{\alpha _{\alpha {i_1}{i_2} \cdots {i_{p - 1}}}}^{j_1j_2\cdots j_s}$.

Note that, for a vector field $X=X^i\frac{\partial}{\partial x^i}$ which is an $(0,1)$-tensor, we need to change to an $(1,0)$-tensor, an $1$-form. To this purpose, we need to change $\frac{\partial}{\partial x^i}$ to $g_{ij}dx^j$. Therefore,

$\displaystyle X = {X^i}\frac{\partial }{{\partial {x^i}}} = {X^i}{g_{ij}}d{x^j}$.

Since

$\displaystyle {\nabla _\alpha }({X^i}{g_{i\beta }}d{x^\beta }) = \frac{{\partial {X^i}}}{{\partial {x^\alpha }}}{g_{i\beta }}d{x^\beta } + {X^i}{g_{i\beta }}{\nabla _\alpha }d{x^\beta } = \frac{{\partial {X^i}}}{{\partial {x^\alpha }}}{g_{i\beta }}d{x^\beta } - {X^i}{g_{i\beta }}\Gamma _{\alpha i}^\beta d{x^i}$

then

$\displaystyle {\nabla _\alpha }{X_\beta } = \frac{{\partial {X^i}}}{{\partial {x^\alpha }}}{g_{i\beta }} - {X^i}{g_{ij}}\Gamma _{\alpha \beta }^j$.

Thus

$\displaystyle {\rm div}(X) = {g^{\alpha \beta }}{\nabla _\alpha }{X_\beta } = {g^{\alpha \beta }}\left( {\frac{{\partial {X^i}}}{{\partial {x^\alpha }}}{g_{i\beta }} - {X^i}{g_{ij}}\Gamma _{\alpha \beta }^j} \right) = \frac{{\partial {X^\alpha }}}{{\partial {x^\alpha }}} - {X^\alpha }\Gamma _{\alpha \beta }^\beta$.

## January 19, 2010

### Why the Einstein field equations are hyperbolic equations?

Filed under: Nghiên Cứu Khoa Học, PDEs, Riemannian geometry — Tags: — Ngô Quốc Anh @ 1:24

Today we discuss a little about the Einstein field equations, that is, ${\rm Eins}_{\alpha\beta} := {\rm Ric}_{\alpha\beta}-\frac{1}{2}g_{\alpha\beta}R(g)=T_{\alpha\beta}$. We shall prove that in the Minkowski spaces, the Einstein field equations are nothing but hyperbolic equations.

We start with the Riemann curvature tensor $R$, an $(3,1)$-tensor, defined to be

$\displaystyle R: (X,Y,Z) \mapsto R(X,Y)Z$

where

$\displaystyle R(X,Y)Z = {\nabla _X}{\nabla _Y}Z - {\nabla _Y}{\nabla _X}Z - {\nabla _{\left[ {X,Y} \right]}}Z$.

Note that, the metric that we are using is Riemannian metric, therefore $\left[ {\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^j}}}} \right] = 0$. Therefore, in local coordinates,

$\displaystyle\begin{gathered}R_{ijk}^\alpha \frac{\partial }{{\partial {x^\alpha }}} = {\nabla _i}{\nabla _j}\frac{\partial }{{\partial {x^k}}} - {\nabla _j}{\nabla _i}\frac{\partial }{{\partial {x^k}}} \hfill \\ \qquad= {\nabla _i}\left( {\Gamma _{jk}^l\frac{\partial }{{\partial {x^l}}}} \right) - {\nabla _j}\left( {\Gamma _{ik}^l\frac{\partial }{{\partial {x^l}}}} \right) \hfill \\ \qquad= \Gamma _{jk}^l{\nabla _i}\left( {\frac{\partial }{{\partial {x^l}}}} \right) + \frac{{\partial \Gamma _{jk}^l}}{{\partial {x^i}}}\frac{\partial }{{\partial {x^l}}} - \Gamma _{ik}^l{\nabla _j}\left( {\frac{\partial }{{\partial {x^l}}}} \right) - \frac{{\partial \Gamma _{ik}^l}}{{\partial {x^j}}}\frac{\partial }{{\partial {x^l}}} \hfill \\ \qquad= \Gamma _{jk}^l\Gamma _{il}^\beta \frac{\partial }{{\partial {x^\beta }}} + \frac{{\partial \Gamma _{jk}^l}}{{\partial {x^i}}}\frac{\partial }{{\partial {x^l}}} - \Gamma _{ik}^l\Gamma _{jl}^\beta \frac{\partial }{{\partial {x^\beta }}} - \frac{{\partial \Gamma _{ik}^l}}{{\partial {x^j}}}\frac{\partial }{{\partial {x^l}}} \hfill \\ \qquad= \left( {\Gamma _{jk}^l\Gamma _{il}^\alpha+ \frac{{\partial \Gamma _{jk}^\alpha }}{{\partial {x^i}}} - \Gamma _{ik}^l\Gamma _{jl}^\alpha- \frac{{\partial \Gamma _{ik}^\alpha }}{{\partial {x^j}}}} \right)\frac{\partial }{{\partial {x^\alpha }}}. \hfill \\ \end{gathered}$

the $\alpha$-component of the Riemann curvature tensor is

$\displaystyle R_{ijk}^\alpha= \Gamma _{jk,i}^\alpha- \Gamma _{ik,j}^\alpha+ \left\{ {\rm no \; derivative} \right\}$.

We now compute Ricci tensor, to this purpose, we get

$\displaystyle {R_{\alpha \beta }} = {g^{ij}}{R_{\alpha i\beta j}} = {g^{ij}}{g_{\alpha \gamma }}R_{i\beta j}^\gamma= R_{\alpha \beta \gamma }^\gamma = \Gamma _{\beta \gamma ,\alpha }^\gamma- \Gamma _{\alpha \gamma ,\beta }^\gamma+ \left\{ {\rm no \; derivative} \right\}$.

In terms of metric tensor, one has

$\displaystyle\Gamma _{ij}^k = \frac{1}{2}{g^{kl}}\left( {\frac{{\partial {g_{jl}}}}{{\partial {x^i}}} + \frac{{\partial {g_{il}}}}{{\partial {x^j}}} - \frac{{\partial {g_{ij}}}}{{\partial {x^l}}}} \right)$

which implies

$\displaystyle\Gamma _{\beta \gamma ,\alpha }^\gamma- \Gamma _{\alpha \gamma ,\beta }^\gamma= \frac{1}{2}{g^{\gamma l}}\left( {{g_{\gamma l,\beta \alpha }} + {g_{\beta l,\gamma \alpha }} - {g_{\beta \gamma ,l\alpha }} - {g_{\gamma l,\alpha \beta }} - {g_{\alpha l,\gamma \beta }} + {g_{\alpha \gamma ,l\beta }}} \right)$.

Thus

$\displaystyle\Gamma _{\beta \gamma ,\alpha }^\gamma- \Gamma _{\alpha \gamma ,\beta }^\gamma= \frac{1}{2}\left( {g_{\beta ,\gamma \alpha }^\gamma+ g_{\alpha ,\gamma \beta }^\gamma- {g_{,\beta \alpha }} - \square {g_{\alpha \beta }}} \right)$.

where $\square= {g^{ij}}{\nabla _i}{\nabla _j}$ is the d’Alembert wave operator.To see this, clearly

$\displaystyle\begin{gathered}\square ({g_{\alpha \beta }}d{x^\alpha } \otimes d{x^\beta }) = {g^{ij}}{\nabla _i}{\nabla _j}({g_{\alpha \beta }}d{x^\alpha } \otimes d{x^\beta }) \hfill \\ \qquad= {g^{ij}}\left[ {{\nabla _i}{\nabla _j}({g_{\alpha \beta }}d{x^\alpha }) \otimes d{x^\beta } + {\nabla _i}(d{x^\alpha } \otimes {\nabla _j}({g_{\alpha \beta }}d{x^\beta }))} +\cdots\right] \hfill \\ \qquad= {g^{ij}}\left[ {{g_{\alpha \beta }}{\nabla _i}{\nabla _j}(d{x^\alpha }) \otimes d{x^\beta } + {g_{\alpha \beta }}{\nabla _i}(d{x^\alpha } \otimes {\nabla _j}(d{x^\beta }))} +\cdots\right] \hfill \\ \qquad= {g^{ij}}\left[ { - {g_{\alpha \beta }}{\nabla _i}(\Gamma _{jk}^\alpha d{x^k} \otimes d{x^\beta }) - {g_{\alpha \beta }}{\nabla _i}(d{x^\alpha } \otimes \Gamma _{jk}^\beta d{x^k})} +\cdots\right] \hfill \\ \qquad= {g^{ij}}\left[ { - {g_{\alpha \beta }}\Gamma _{jk,i}^\alpha d{x^k} \otimes d{x^\beta } - {g_{\alpha \beta }}\Gamma _{jk,i}^\beta d{x^\alpha } \otimes d{x^k} +\cdots } \right] \hfill \\ \qquad= {g^{ij}}\left[ { - {g_{\alpha \beta }}\Gamma _{jk,i}^\alpha d{x^k} \otimes d{x^\beta } - {g_{\alpha \beta }}\Gamma _{jk,i}^\beta d{x^\alpha } \otimes d{x^k} +\cdots } \right] \hfill \\ \qquad=- \frac{1}{2}{g^{ij}}\left[ {({g_{j\beta ,ki}} + {g_{k\beta ,ji}} - {g_{jk,\beta i}})d{x^k} \otimes d{x^\beta } + ({g_{j\alpha ,ki}} + {g_{k\alpha ,ji}} - {g_{jk,\alpha i}})d{x^\alpha } \otimes d{x^k} +\cdots } \right] \hfill \\ \end{gathered}$

which gives

$\displaystyle\square {g_{\alpha \beta }} =- \frac{1}{2}{g^{ij}}\left[ {({g_{j\beta ,\alpha i}} + {g_{\alpha \beta ,ji}} - {g_{j\alpha ,\beta i}}) + ({g_{j\alpha ,\beta i}} + {g_{\beta \alpha ,ji}} - {g_{j\beta ,\alpha i}})} \right] +\cdots$.

Thus

$\displaystyle\square {g_{\alpha \beta }} =-{g^{ij}}{g_{\beta \alpha ,ji}} +\cdots$.

Contracting once more with $g^{\alpha\beta}$ the scalar curvature is obtained as

$\displaystyle R = {g^{\alpha \beta }}{R_{\alpha \beta }} = \frac{1}{2}{g^{\alpha \beta }}\left( {g_{\beta ,\gamma \alpha }^\gamma+ g_{\alpha ,\gamma \beta }^\gamma- {g_{,\beta \alpha }} - \square {g_{\alpha \beta }}} \right) = g_{,\alpha \beta }^{\alpha \beta } - {g^{\alpha \beta }}\square {g_{\alpha \beta }}$.

Thus, the Einstein tensor is

$\displaystyle\begin{gathered}{\rm Ein}{s_{\alpha \beta }} = {\rm Ric}_{\alpha \beta } - \frac{1}{2}{g_{\alpha \beta }}R \hfill \\ \qquad= \frac{1}{2}\left[ {\left( {g_{\beta ,\gamma \alpha }^\gamma+ g_{\alpha ,\gamma \beta }^\gamma- {g_{,\beta \alpha }} - \square {g_{\alpha \beta }}} \right) - {g_{\alpha \beta }}\left( {g_{,ij}^{ij} - {g^{ij}}\square {g_{ij}}} \right)} \right]. \hfill \\ \end{gathered}$

If we introduce the following notation

$\displaystyle {\overline g _{\alpha \beta }} = {g_{\alpha \beta }} - \frac{1}{2}{g_{\alpha \beta }}{g^{ij}}\square {g_{ij}}$

then

$\displaystyle\overline g _{\beta ,\gamma \alpha }^\gamma+ \overline g _{\alpha ,\gamma \beta }^\gamma- \square {\overline g _{\alpha \beta }} - {g_{\alpha \beta }}\overline g _{,ij}^{ij} = {T_{\alpha \beta }}$.

In Minkowski space, if we require the Lorenz condition, or the Lorenz gauge, we then have

$\displaystyle\square {g _{\alpha \beta }} = - {T_{\alpha \beta }}$.

This is what we need. Coordinates that obey the Lorenz condition are called harmonic.

## January 18, 2010

### R-G: Gauss and Codazzi equations in general relativity

Filed under: Riemannian geometry — Tags: — Ngô Quốc Anh @ 0:01

For physical reasons, a spacetime continuum is mathematically defined as a four-dimensional, smooth, connected Lorentzian manifold $(V,g)$. This means the smooth Lorentz metric $g$ has signature $(3,1)$. We are interested in the Cauchy problem for the Einstein equations. For that reason, we will consider the initial data on a hypersurface $\Sigma$, a $3$-dimensional manifold which is a space-like.

The Einstein equations are just equations for a metric $g$ defined throughout $V$ such that the induced matric $\overline g$ onto $\Sigma$ and the second fundamental form $k$ of $\Sigma$ are identical to the initial data. Since $\Sigma$ has $1$-dimention less than that of $V$, the initial data can not chosen arbitrary, they must satisfy the Gauss and Codazzi equations.

The purpose of this entry is to derive the constraint equations for the Einstein equations. For simplicity, we consider the case when $n=3$, then the Gauss and Codazzi equations can be rewritten as the following

$\displaystyle ^{(4)}{R_{abcd}} = {R_{abcd}} + {k_{ac}}{k_{bd}} - {k_{ad}}{k_{bc}}$

and

$\displaystyle ^{(4)}{R_{\sigma bcd}}{n^\sigma } =- {\nabla _c}{k_{ab}}+{\nabla _b}{k_{ac}}$

where $n^\alpha$ the normal vector to the space-like $\Sigma$.

Further equations can be obtained from these by contraction. Indeed, from the Gauss equation, one has

$\displaystyle ^{(4)}{R_{ab}} = {R_{ab}} + {\rm trace } k{k_{ab}} - {k_{ac}}k_b^c{ - ^{(4)}}{R_{\sigma a\tau b}}{n^\sigma }{n^\tau }$

and contracting again gives

$\displaystyle ^{(4)}R + {2^{(4)}}{R_{\alpha \beta }}{n^\alpha }{n^\beta } = R - {\left( {\rm trace }k \right)^2} - {k_{ab}}{k^{ab}}$.

Similarly, contracting the Codazzi equation gives

$\displaystyle ^{(4)}{R_{\sigma a}}{n^\sigma } =- {\nabla ^a}{k_{ab}} + {\nabla _b}{\rm trace } k$.

Combining all equations with the Einstein equations gives the constrain equations

$\displaystyle\begin{gathered}R - {\left( {\rm trace }k \right)^2} - {k_{ab}}{k^{ab}} =\cdots\hfill \\{\nabla ^a}{k_{ab}} - {\nabla _b}{\rm trace } k =\cdots\hfill \\\end{gathered}$

## January 17, 2010

### New Inequalities of Ostrowski-like Type Involving n knots and L^p-norm of m-th derivative

Filed under: Nghiên Cứu Khoa Học — Tags: — Ngô Quốc Anh @ 22:58

Let me introduce my recent result with V.N. Huy published in Applied Mathematics Letters last year.

I want to start by recalling the following results due to Nenad Ujevic:

Let $I \subset \mathbb R$ be an open interval such that $[a,b] \subset I$ and let $f : I \to \mathbb R$ be a twice differentiable function such that $f''$ is bounded and integrable. Then we have

$\displaystyle\begin{gathered}\Bigg|\int\limits_a^b {f\left( x \right)dx}- \frac{{b - a}}{2}\Bigg(f\left( {\frac{{a + b}}{2} - \left( {2 - \sqrt 3 } \right)\left( {b - a} \right)} \right) \hfill \\ \qquad\qquad+ f\left( {\frac{{a + b}}{2} + \left( {2 - \sqrt 3 } \right)\left( {b - a} \right)} \right)\Bigg)\Bigg| \leqq \frac{{7 - 4\sqrt 3 }}{8}{\left\| {f''} \right\|_\infty }{\left( {b - a} \right)^3}. \hfill \\ \end{gathered}$

And

Let $I \subset \mathbb R$ be an open interval such that $[a,b] \subset I$ and let $f : I \to \mathbb R$ be a twice differentiable function such that $f'' \in L^2(a,b)$. Then we have

$\displaystyle\begin{gathered}\Bigg|\int\limits_a^b {f\left( x \right)dx}- \frac{{b - a}}{2}\Bigg(f\left( {\frac{{a + b}}{2} - \frac{{3 - \sqrt 6 }}{2}\left( {b - a} \right)} \right) \hfill \\ \qquad\qquad+ f\left( {\frac{{a + b}}{2} + \frac{{3 - \sqrt 6 }}{2}\left( {b - a} \right)} \right)\Bigg)\Bigg| \leqq \sqrt {\frac{{49}}{{80}} - \frac{1}{4}\sqrt 6 } {\left\| {f''} \right\|_2}{\left( {b - a} \right)^{\frac{5}{2}}}. \hfill \\ \end{gathered}$

In the above mentioned results, constants $\frac{{7 - 4\sqrt 3 }}{8}$ in the first and $\sqrt {\frac{{49}}{{80}} - \frac{1}{4}\sqrt 6 }$ in the second result are sharp in sense that these cannot be replaced by smaller ones. This leads us to strengthen them by enlarging the number of knots (2 knots in both results) and replacing the norms $\| \cdot \|_\infty$ in the first and $\| \cdot \|_2$ in the second.

Before stating our main result, let us introduce the following notation

$\displaystyle I\left( f \right) =\int\limits _a^b {f\left( x \right)dx }$.

Let $1 \leqq m, n<\infty$ and $1 \leqq p \leqq \infty$. For each $i = \overline{1,n}$, we assume $0 < x_i < 1$ such that

$\displaystyle\left\{ \begin{gathered}{x_1} + {x_2} +\cdots+ {x_n} = \frac{n}{2}, \hfill \\ \cdots\hfill \\x_1^j + x_2^j +\cdots+ x_n^j = \frac{n}{{j + 1}}, \hfill \\ \cdots\hfill \\x_1^{m - 1} + x_2^{m - 1} +\cdots+ x_n^{m - 1} = \frac{n}{m}. \hfill \\ \end{gathered}\right.$

Put

$\displaystyle Q\left( {f,n,m,x_1 ,...,x_n } \right) = \frac{{b - a}}{n}\sum\limits_{i = 1}^n {f\left( {a + x_i \left( {b - a} \right)} \right)}$.

We are in a position to state our main result.

Let $I \subset \mathbb R$ be an open interval such that $[a,b] \subset I$ and let $f : I \to \mathbb R$ be a $m$-th differentiable function such that $f^{(m)} \in L^p(a,b)$. Then we have

$\displaystyle\left| {I\left( f \right) - Q\left( {f,n,m,{x_1},...,{x_n}} \right)} \right| \leqq \frac{1}{{m!}}\left( {{{\left( {\frac{1}{{mq + 1}}} \right)}^{\frac{1}{q}}} + {{\left( {\frac{1}{{\left( {m - 1} \right)q + 1}}} \right)}^{\frac{1}{q}}}} \right){\left\| {{f^{\left( m \right)}}} \right\|_p}{\left( {b - a} \right)^{m + \frac{1}{q}}}$.

As can be seen the above result is not sharp. It will be very interesting if we can derive a sharp estimate. Note that, the results due to Nenad Ujevic can be rewritten as the following

$\displaystyle\left| {I\left( f \right) - Q\left( {f,2,2,\frac{1}{2} - \left( {2 - \sqrt 3 } \right),\frac{1}{2} + \left( {2 - \sqrt 3 } \right)} \right)} \right| \leqq \frac{{7 - 4\sqrt 3 }}{8}{\left\| {f''} \right\|_\infty }{\left( {b - a} \right)^3}$

and

$\displaystyle\left| {I\left( f \right) - Q\left( {f,2,2,\frac{1}{2} - \frac{{3 - \sqrt 6 }}{2},\frac{1}{2} + \frac{{3 - \sqrt 6 }}{2}} \right)} \right| \leqq \sqrt {\frac{{49}}{{80}} - \frac{1}{4}\sqrt 6 } {\left\| {f''} \right\|_2}{\left( {b - a} \right)^{\frac{5}{2}}}$.

## January 13, 2010

### R-G: A few words about writing covariant derivative of tensors in a short form

Filed under: Linh Tinh, Nghiên Cứu Khoa Học, Riemannian geometry — Ngô Quốc Anh @ 14:21

Before going further, I would like to mention a convention in writing covariant derivative of tensors in a short form. What I mean is how to understand the following notation $\nabla_i g_{\alpha\beta}$ where $g$ is a Riemannian metric.

We all know that we can apply covariant derivative to a scalar function, for example, $\nabla_i f$ is nothing but the partial derivative with respect to $x^i$. However, the notation $\nabla_i g_{\alpha\beta}$ is a little bit different. For $g$, a Riemannian metric, which is also a $(0,2)$ tensor, in the full form, we can write

$\displaystyle g=g_{\alpha\beta}dx^\alpha \otimes dx^\beta$.

The sub indexes $\alpha\beta$ in the notation $\nabla_i g_{\alpha\beta}$ tell us that we are talking about the $\alpha\beta$-component of the covariant derivative of $(0,2)$ tensor $g$ with respect to the vector field $\frac{\partial}{\partial x^i}$.

If you have $\nabla_i h^{\alpha\beta}$, then you are working on some $(2,0)$-tensor $h$ of the form

$\displaystyle h=h^{\alpha\beta} \frac{\partial}{\partial x^\alpha} \otimes \frac{\partial}{\partial x^\beta}$.

We now go back to the case $\nabla_i g_{\alpha\beta}$. Precisely, one should write

$\displaystyle {\nabla _i}{g^{\alpha \beta }} = ({\nabla _{\frac{\partial }{{\partial {x^i}}}}}g)\left( {\frac{\partial }{{\partial {x^\alpha }}},\frac{\partial }{{\partial {x^\beta }}}} \right)$.

Since $\{dx^\alpha\}$ is a basis for the dual space of a space spanned by the basis $\{\frac{\partial}{\partial x^\beta}\}$, it is clear to see that the right hand side of the above convention is just the $\alpha\beta$-component, that is, the coefficient of the term

$\displaystyle {\nabla _i}{g^{\alpha \beta }}d{x^\alpha } \otimes d{x^\beta }$.

Now we show that $\nabla_i g^{\alpha\beta}=0$ whenever $g$ is a Riemannian metric. This is equivalent to show that every coefficients of $\nabla_i g$ equal to zero, in other word, $\nabla_i g=0$. Since $g$ is a $(0,2)$-tensor, then we can compute covariant derivative of $g$ as follows

$\displaystyle\begin{gathered}{\nabla _i}g = {\nabla _i}\left( {{g_{\alpha \beta }}d{x^\alpha } \otimes d{x^\beta }} \right) \hfill \\\qquad = \left( {\frac{\partial }{{\partial {x^i}}}{g_{\alpha \beta }}} \right)d{x^\alpha } \otimes d{x^\beta } + {g_{\alpha \beta }}{\nabla _i}\left( {d{x^\alpha } \otimes d{x^\beta }} \right) \hfill \\ \qquad= \left( {\frac{\partial }{{\partial {x^i}}}{g_{\alpha \beta }}} \right)d{x^\alpha } \otimes d{x^\beta } + {g_{\alpha \beta }}{\nabla _i}\left( {d{x^\alpha }} \right) \otimes d{x^\beta } + {g_{\alpha \beta }}d{x^\alpha } \otimes {\nabla _i}\left( {d{x^\beta }} \right) \hfill \\ \qquad= \left( {\frac{\partial }{{\partial {x^i}}}{g_{\alpha \beta }}} \right)d{x^\alpha } \otimes d{x^\beta } - \left( {{g_{\alpha \beta }}\Gamma _{ik}^\alpha+ {g_{\alpha \beta }}\Gamma _{ik}^\beta } \right)d{x^k} \otimes d{x^\beta } \hfill \\ \qquad= \left( {\frac{\partial }{{\partial {x^i}}}{g_{\alpha \beta }} - \Gamma _{i\alpha }^k{g_{k\beta }} - \Gamma _{i\beta }^k{g_{k\alpha }}} \right)d{x^\alpha } \otimes d{x^\beta }. \hfill \\ \end{gathered}$

Since $g$ is a metric connection, i.e.

$\displaystyle Xg\left( {Y,Z} \right) = g\left( {{\nabla _X}Y,Z} \right) + g\left( {Y,{\nabla _X}Z} \right)$

one has

$\displaystyle \frac{\partial }{{\partial {x^i}}}g\left( {\frac{\partial }{{\partial {x^\alpha }}},\frac{\partial }{{\partial {x^\beta }}}} \right) = g\left( {{\nabla _i}\frac{\partial }{{\partial {x^\alpha }}},\frac{\partial }{{\partial {x^\beta }}}} \right) + g\left( {\frac{\partial }{{\partial {x^\alpha }}},{\nabla _i}\frac{\partial }{{\partial {x^\beta }}}} \right)$.

Thus

$\displaystyle\frac{\partial }{{\partial {x^i}}}{g_{\alpha \beta }} = g\left( {\Gamma _{i\alpha }^k\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^\beta }}}} \right) + g\left( {\frac{\partial }{{\partial {x^\alpha }}},\Gamma _{i\beta }^k\frac{\partial }{{\partial {x^k}}}} \right) = \Gamma _{i\alpha }^k{g_{k\beta }} + \Gamma _{i\beta }^k{g_{k\alpha }}$.

In other word the $\alpha\beta$-component of $\nabla_i g$ equals to zero. At the last word, in order to calculate covariant derivative of some $(p,q)$-tensors, you need to use the following three properties

1. ${\nabla _X}\left( {fT} \right) = X(f)T + f{\nabla _X}T$.
2. ${\nabla _X}\left( {T \otimes Q} \right) = {\nabla _X}T \otimes Q + T \otimes {\nabla _X}Q$.
3. For $(1,0)$- or $(0,1)$-tensors:

$\displaystyle {\nabla _{\frac{\partial }{{\partial {x^i}}}}}\frac{\partial }{{\partial {x^j}}} = \Gamma _{ij}^k\frac{\partial }{{\partial {x^k}}}$

and

$\displaystyle {\nabla _{\frac{\partial }{{\partial {x^i}}}}}d{x^j}=-\Gamma _{ik}^jd{x^k}$.

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