Ngô Quốc Anh

Tháng Mười Hai 7, 2009

The QE – UCLA Department of Mathematics

Chuyên mục: Đề Thi — Ngô Quốc Anh @ 11:36

There are two types of qualifying exam: the Basic exam and the Area exams. The Basic exam is designed to be passed by well-trained students before they commence study at UCLA. It examines fundamental topics of the undergraduate mathematics curriculum. The Area exams are graduate level exams. For each Area exam there is a preparatory course sequence. There are Area exams in Algebra, Analysis, Applied Differential Equations, Numerical Analysis, Geometry /Topology, and Logic. Students may attempt any number of examinations in each examination period.

MA students must pass the Basic Exam only. PhD students must pass the Basic exam and two Area exams. MA students must pass the Basic by the beginning of the sixth quarter of study. PhD students must pass the Basic by the fourth quarter of graduate study. A PhD student must pass the one Area examination by the sixth quarter. A PhD student must pass the second Area examination by the seventh quarter of graduate study.

The exams are offered in the Fall and in the Spring, usually just before the beginning of those quarters. Precise dates and times are posted well in advance of the exams. Students must sign-up for the exams in the Graduate Office. Each exam lasts 4 hours. Copies of past exams may be downloaded from our website by clicking here: Download Exams. However, it is strongly recommended that students prepare for exams by studying their syllabi theme by theme, and by doing numerous exercises other than those on old exams. Experience shows that study organized around working old exams is not as efficacious as thematically organized study.

Each exam is written and graded by a committee created for that purpose. The Graduate Studies Committee approves exam results (passing or failing), taking into account recommendations of the examination committee. Shortly after the Graduate Studies Committee’s decision, students are notified of their exam results. Students are reminded that the grading of exams is a complex matter, and that final result (Pass or Fail) is not usually determined by the total score of all work on all problems. Students should read and follow carefully the instructions of an exam.

Graded exams are kept in the Graduate Office for six months and then destroyed. They may be examined in the Graduate Office during this time. After the results of the exams are announced, there is a one week appeal period during which students may petition, in writing, to a Qual Committee for regrading of problems. Appeals must be submitted via the Graduate Office. The Qual Committee will respond, usually in writing, to any appeal within one week.

Currently, most UCLA PhD students pass all their exams on schedule. However, the few students who fail to pass exams by the required deadlines are deemed not to be making Satisfactory Progress. Each such student is discussed individually by the Graduate Studies Committee at a meeting shortly after the above period of appeals is over. Students who have missed a deadline, or otherwise failed to make Satisfactory Progress, will receive a letter from the Graduate Vice Chair indicating any action that was taken, and detailing any schedule for performance that must be satisfied in order to continue in the program. Only in unusual circumstances will a PhD student who is more than six months behind the schedule of Satisfactory Progress be permitted to remain in the PhD program. Students who are facing negative actions are encouraged to write to GSC, and to speak to the Graduate Vice Chair, before GSC meets, to explain any extenuating circumstances that could positively influence it.

Here follow the syllabi for the Examinations. Each examination may test on any of the topics of its syllabus.

Link: http://www2.math.ucla.edu/grad/handbook/hbqex.shtml

Problems: http://www2.math.ucla.edu/grad/handbook/hbquals.shtml

Basic Exam Algebra Analysis Applied Differential Equations
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Geometry/Topology Logic Numerical Analysis French Language Exam
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Tháng Mười Hai 5, 2009

R-G: Laplace-Beltrami operator

Chuyên mục: Riemannian geometry — Ngô Quốc Anh @ 13:35

So far for a smooth function f : M \to \mathbb R, gradient of f is a vector field defined as follows

\displaystyle \nabla f = g^{kj} \dfrac{\partial f}{\partial x^j} \frac{\partial}{\partial x^k}.

Since the gradient of f is nothing but a vector field then it is reasonable to talk about the divergence of a vector field X. To be exact, we define

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

where X is a vector field. All above was discussed in this topic.

In order to go further, I need to spend some time talking about {\rm div} X. Precisely, if f is a smooth function, then can we write down explicitly {\rm div}(\nabla f) in local coordinates?

Having the definition of divergence operator yields

\displaystyle {\rm div} X= \left\langle {{\nabla _{\frac{\partial }{{\partial {x^i}}}}}X,d{x^i}} \right\rangle

where \left\langle \cdot , \cdot \right\rangle presents a pairing between TM and its dual space T^\star M. If, in local coordinates, X=X^i\frac{\partial}{\partial x^i}, we then have

\displaystyle {\nabla _{\frac{\partial }{{\partial {x^i}}}}}X = {\nabla _{\frac{\partial }{{\partial {x^i}}}}}\left( {{X^j}\frac{\partial }{{\partial {x^j}}}} \right) = \frac{{\partial {X^j}}}{{\partial {x^i}}}\frac{\partial }{{\partial {x^j}}} + {X^j}{\nabla _{\frac{\partial }{{\partial {x^i}}}}}\frac{\partial }{{\partial {x^j}}} = \frac{{\partial {X^j}}}{{\partial {x^i}}}\frac{\partial }{{\partial {x^j}}} + {X^j}\Gamma _{ij}^k\frac{\partial }{{\partial {x^k}}}.

Thus,

\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.

As a consequence,

\displaystyle {\rm div}(\nabla f)=\frac{\partial }{{\partial {x^i}}}\left( {{g^{ij}}\frac{{\partial f}}{{\partial {x^j}}}} \right) + {g^{jk}}\frac{{\partial f}}{{\partial {x^k}}}\Gamma _{ij}^i.

We now look at the Christoffel symbols. Clearly, by definition

\displaystyle\begin{gathered} \Gamma _{ij}^i = \frac{1}{2}{g^{ik}}\left( {\frac{{\partial {g_{ik}}}}{{\partial {x^j}}} + \frac{{\partial {g_{jk}}}}{{\partial {x^i}}} - \frac{{\partial {g_{ij}}}}{{\partial {x^k}}}} \right) = \frac{1}{2}{g^{ik}}\frac{{\partial {g_{ik}}}}{{\partial {x^j}}} \hfill \\ \qquad= \frac{1}{2}{\rm Trace}\left( {\left( {{g^{mn}}} \right)\frac{\partial }{{\partial {x^j}}}\left( {{g_{mn}}} \right)} \right) \hfill \\ \qquad= \frac{1}{2}{\rm Trace}\left( {{{\left( {{g_{mn}}} \right)}^{ - 1}}\frac{\partial }{{\partial {x^j}}}\left( {{g_{mn}}} \right)} \right) \hfill \\ \qquad = \frac{1}{2}\frac{{\dfrac{\partial }{{\partial {x^j}}}\det \left( {{g_{mn}}} \right)}}{{\det \left( {{g_{mn}}} \right)}} \hfill \\ \end{gathered}

where the main theorem posted here has been used. Thus,

\displaystyle \begin{gathered} {\rm div}X = \frac{{\partial {X^i}}}{{\partial {x^i}}} + {X^j}\Gamma _{ij}^i \hfill \\ \qquad= \frac{1}{{\sqrt {\det \left( {{g_{mn}}} \right)} }}\sqrt {\det \left( {{g_{mn}}} \right)} \frac{{\partial {X^i}}}{{\partial {x^i}}} + \frac{1}{{\sqrt {\det \left( {{g_{mn}}} \right)} }}\frac{1}{{2\sqrt {\det \left( {{g_{mn}}} \right)} }}\frac{{\partial \det \left( {{g_{mn}}} \right)}}{{\partial {x^j}}}{X^j} \hfill \\ \qquad= \frac{1}{{\sqrt {\det \left( {{g_{mn}}} \right)} }}\frac{\partial }{{\partial {x^j}}}\left( {\sqrt {\det \left( {{g_{mn}}} \right)} {X^i}} \right). \hfill \\ \end{gathered}

As a consequence,

\displaystyle\Delta f = \frac{1}{{\sqrt {\det \left( {{g_{mn}}} \right)} }}\frac{\partial }{{\partial {x^j}}}\left( {\sqrt {\det \left( {{g_{mn}}} \right)} {g^{ij}}\frac{{\partial f}}{{\partial {x^i}}}} \right).

The above \Delta acting on a smooth function f is called the Laplace-Beltrami operator.

Tháng Mười Hai 2, 2009

MuPAD

Chuyên mục: Linh Tinh — Ngô Quốc Anh @ 22:21

MuPAD was a Computer algebra system (CAS). Originally developed by the MuPAD research group at the University of Paderborn, Germany, it was developed by the company SciFace Software GmbH & Co. KG in cooperation with the MuPAD research group and partners from some other universities since 1997.

I found MuPAD a very cool tool for plotting graph of functions. I will show you in details some examples.

1. We start with the following function

f(x,y)=-(x^2+y^2)(\cos (3y) + \sin (7y)).

To plot its graph, I use the following

f := plot::Function3d(exp(-x^2-y^2)*(cos(3*y)+sin(7*x)), x=-2..2, y=-2..2, Submesh=[1,1], FillColorFunction=((x, y, z) -> [(z+2)/4, 0.5, (2-z)/4])): plot(f)

and this is what we get

exp(-x^2-y^2)*(cos(3*y)+sin(7*x))

To obtain the following picture

you need

f := plot::Function3d(exp(-x^2-y^2)*(cos(3*y)+sin(7*x)),
x=-2..2, y=-2..2,
Submesh=[1,1]):
graphLines:=plot::modify(f, Filled = FALSE, XLinesVisible = FALSE, YLinesVisible = FALSE,
ZContours = [Automatic, 15], LineColorFunction=((x, y, z) -> [(z+2)/4, 0.5, (2-z)/4])):
linesProjection := plot::Transform3d(
[0, 0, -1.5], // shift vector
[1, 0, 0, // transformation matrix
0, 1, 0,
0, 0, 0],
graphLines):
plot(f,linesProjection)

2. With the following function

{\rm Re}(\cos (x+iy).

To plot its graph, I use the following

f := plot::Function3d(Re(cos(x+I*y)), x=-2..2, y=-2..2, Submesh=[1,1], FillColorFunction=((x, y, z) -> [(z+2)/4, 0.5, (2-z)/4])): plot(f)

and this is what we getRe(cos(x+I*y)3. If our function has singularities, for example, Gamma function f(x,y)=|\Gamma (x+iy)|, we have

Re(cos(x+I*y)f := plot::Function3d(abs(gamma(x+I*y)), x=-4.5..4.5, y=-3..3, Submesh=[3,3], XSubmesh = 1, YSubmesh = 2): plot(f)

4. If you intend to plot sereval graphs, the following is a good example.

sin(x^2 + y^2), cos(x^2 - y^2)

Here we used functions \sin(x^2 + y^2), \cos(x^2 - y^2). The code I used is

plotfunc3d(sin(x^2 + y^2), cos(x^2 - y^2), x = 0..1, y = 0..2)

5. The Mobius band can be drawn by using the following

x := cos(alpha) * (1 + r * cos(alpha/2)):
y := sin(alpha) * (1 + r * cos(alpha/2)):
z := r * sin(alpha/2):
plot(plot::Surface([x,y,z], r = -0.5 .. 0.5, alpha = -PI .. PI,
Mesh = [35, 31], LineColor = RGB::Black.[0.2]),
Axes = None, Scaling = Constrained)

6. If you need a transparent picture, probably you need to add a factor like abs(sin(z)) in the FillColorFunction. Following is an example

K := (x,y,z)->[abs(sin(x)), abs(sin(y)), abs(sin(z)), abs(sin(z))]:
F1 := plot::Function3d(
1.7*x*exp(-1/2*(x^2+y^2)),
x=-4..4, y=-4..4,
FillColorFunction=K):
F2 := plot::Function3d(-1, x=-4..4, y=-4..4):
plot(F1, F2)

The original one is

7. I will end this topic by showing some more pictures obtained by zooming a little bit closed.

sin(x

sin(x

sin(x

sin(x

plotfunc3d(sin(x - -PI)*sin(y - -PI), Submesh = [4, 4], x = -PI .. PI, y = -PI .. PI)

sin(x

sin(x

In conclusion, if you want to plot a general surface in 3D, you need its parametrization in order to draw. For details, I prefer the reader to this paper.

R-G: A useful formula

Chuyên mục: Riemannian geometry — Ngô Quốc Anh @ 3:40

I intend to give an explicit formula for the Laplacian of a function f. So far what we have done is to provide the following

\displaystyle \Delta f = {\rm div}(\nabla f)

without any further information. Note that, we may think the Laplacian is an operator. In this case, our Laplacian is often called the Laplace-Beltrami operator. In general, our Laplacian is noting but the Connection Laplacian. We also note that there are lots of Laplacian in the literature, for examples, Hodge Laplacian, Bochner Laplacian, Lichnerowicz Laplacian, Conformal Laplacian. But sooner or later we need the following linear algebraic identity

Theorem. Assume M is a symmetric non-singular matrix whose elements depend on a parameter x. Then

\displaystyle\frac{d}{{dx}}\det M = \det M \cdot {\rm Trace}\left( {{M^{ - 1}}\frac{d}{{dx}}M} \right).

Proof. This formula follows by simply diagonalising the matrix by a similarity transformation

\displaystyle D=SMS^{-1}

where D is the diagonal matrix. The formula is obviously true for a diagonal non-singular matrix. If D is a matrix with non-zero diagonal elements

\displaystyle D = \left( {\begin{array}{*{20}{c}} {{\lambda _1}} & {} & 0 \\ {} & \ddots & {} \\ 0 & {} & {{\lambda _n}} \\ \end{array} } \right)

we then have

\displaystyle \begin{gathered} \frac{d}{{dx}}\det D = \sum\limits_i {{\lambda _1} \cdots \frac{{d{\lambda _i}}}{{dx}}} \cdots {\lambda _n} \hfill \\\qquad = \det D \cdot \sum\limits_i {\frac{1}{{{\lambda _i}}}\frac{{d{\lambda _i}}}{{dx}}} \hfill \\ \qquad = \det D \cdot {\rm Trace}\left( {{D^{ - 1}}\frac{d}{{dx}}D} \right). \hfill \\ \end{gathered}

Now we know that the determinant is unchanged, \det D = \det M. And in the trace

\displaystyle\begin{gathered} {\rm Trace}\left( {{D^{ - 1}}\frac{d}{{dx}}D} \right) = {\rm Trace}\left( {S{M^{ - 1}}{S^{ - 1}}\frac{d}{{dx}}\left( {SM{S^{ - 1}}} \right)} \right) \hfill \\ \qquad ={\rm Trace}\left( {S{M^{ - 1}}{S^{ - 1}}\frac{{dS}}{{dx}}\left( {M{S^{ - 1}}} \right)} \right) + {\rm Trace}\left( {S{M^{ - 1}}\frac{{dM}}{{dx}}{S^{ - 1}}} \right) + {\rm Trace}\left( {S\frac{{d{S^{ - 1}}}}{{dx}}} \right). \hfill \\\end{gathered}

In the first term on the right-hand side we can bring the matrix MS^{-1} as the first factor inside the trace using the property of the trace {\rm Trace}(AB) = {\rm Trace}(BA). Then the first term becomes

\displaystyle {\rm Trace}\left( {{S^{ - 1}}\frac{{dS}}{{dx}}} \right) ={\rm Trace}\left( {\frac{{dS}}{{dx}}{S^{ - 1}}} \right)

which can be combined with the third term to give the trace of

\displaystyle \frac{{dS}}{{dx}}{S^{ - 1}} + S\frac{{d{S^{ - 1}}}}{{dx}} = \frac{{d\left( {S{S^{ - 1}}} \right)}}{{dx}} = 0.

The middle term becomes the required expression in the formula when the last factor is brought to the left as first.

Reference: Pankaj Sharan, Spacetime, Geometry and Gravitation (Progress in Mathematical Physics, 56), Birkhäuser 2009.

Tháng Mười Một 30, 2009

A property of the essentially bounded function 2

Chuyên mục: Các Bài Tập Nhỏ, Giải Tích 6 (MA5205) — Ngô Quốc Anh @ 22:52

This topic is a companion to the following topic. In this topic, we consider the case when E is the whole space, i.e. E = \mathbb R^n. We also add an extra function g to a_n. To be precise, we have

Question. Suppose g>0 on \mathbb R^n is in L^1(\mathbb R^n) in Lebesgue sense. Let f \in L^\infty(\mathbb R^n) such that \| f\|_\infty > 0. Define

\displaystyle {a_n} = \int_{\mathbb R^n} {{{\left| f \right|}^ng}}

for n=1,2,3,... Show that

\displaystyle\mathop {\lim }\limits_{n \to \infty } \frac{{{a_{n + 1}}}} {{{a_n}}} = {\left\| f \right\|_\infty }.

Solution. For any \alpha with 0<\alpha < \|f\|_\infty, let

\displaystyle{E_\alpha } = \left\{ {x \in E: f\left( x \right) \geqslant \alpha } \right\}

and

\displaystyle {F_\alpha } = E\backslash {E_\alpha }

then \infty> |E_\alpha|>0. Clearly when \alpha is sufficiently closed to \|f\|_\infty, \int_{E_\alpha}g>0. For any k \in \mathbb N (k can be zero), note that

\displaystyle\int_{{E_\alpha }} {{{\left| f \right|}^n}g} \geqslant {\alpha ^n}\int_{{E_\alpha }} g

and

\displaystyle\int_{{F_\alpha }} {{{\left| f \right|}^{n + k}}g} \leqslant \left\| f \right\|_\infty ^k\int_{{F_\alpha }} {{{\left| f \right|}^n}g}.

Then

\displaystyle\frac{{\int_{{F_\alpha }} {{{\left| f \right|}^{n + k}}g} }}{{\int_{{E_\alpha }} {{{\left| f \right|}^n}g} }} \leqslant \frac{{\left\| f \right\|_\infty ^k\int_{{F_\alpha }} {{{\left| f \right|}^n}g} }}{{{\alpha ^n}\int_{{E_\alpha }} g }} = \frac{{\left\| f \right\|_\infty ^k}}{{\int_{{E_\alpha }} g }}\int_{{F_\alpha }} {{{\left| {\frac{f}{\alpha }} \right|}^n}g}.

By the Dominated Convergence Theorem, one gets

\displaystyle 0 \leqslant \mathop {\lim }\limits_{n \to \infty } \left( {\frac{{\int_{{F_\alpha }} {{{\left| f \right|}^{n + k}}g} }}{{\int_{{E_\alpha }} {{{\left| f \right|}^n}g} }}} \right) \leqslant \mathop {\lim }\limits_{n \to \infty } \left( {\frac{{\left\| f \right\|_\infty ^k}}{{\int_{{E_\alpha }} g }}\int_{{F_\alpha }} {{{\left| {\frac{f}{\alpha }} \right|}^n}g} } \right) = 0.

Hence

\displaystyle\begin{gathered}\mathop {\lim \inf }\limits_{n \to \infty } \left( {\frac{{\int\limits_E {{{\left| f \right|}^{n + 1}}g} }}{{\int\limits_E {{{\left| f \right|}^n}g} }}} \right) \geqslant \mathop {\lim \inf }\limits_{n \to \infty } \left( {\frac{{\int\limits_{{F_\alpha }} {{{\left| f \right|}^{n + 1}}g} + \int\limits_{{E_\alpha }} {{{\left| f \right|}^{n + 1}}g} }}{{\int\limits_{{E_\alpha }} {{{\left| f \right|}^n}g} + \int\limits_{{F_\alpha }} {{{\left| f \right|}^n}g} }}} \right) \hfill \\\qquad \geqslant \mathop {\lim \inf }\limits_{n \to \infty } \left( {\frac{{\int\limits_{{F_\alpha }} {{{\left| f \right|}^{n + 1}}g} + \alpha \int\limits_{{E_\alpha }} {{{\left| f \right|}^n}g} }}{{\int\limits_{{E_\alpha }} {{{\left| f \right|}^n}g} + \int\limits_{{F_\alpha }} {{{\left| f \right|}^n}g} }}} \right) \hfill \\\qquad = \mathop {\lim \inf }\limits_{n \to \infty } \left( {\frac{{\int\limits_{{F_\alpha }} {{{\left| f \right|}^{n + 1}}g} }}{{\int\limits_{{E_\alpha }} {{{\left| f \right|}^n}g} }} + \alpha } \right)/\left( {1 + \frac{{\int\limits_{{F_\alpha }} {{{\left| f \right|}^n}g} }}{{\int\limits_{{E_\alpha }} {{{\left| f \right|}^n}g} }}} \right) \hfill \\ \qquad = \alpha . \hfill \\ \end{gathered}

Letting \alpha \nearrow {\left\| f \right\|_\infty }, we get that

\displaystyle\mathop{\lim }\limits_{n\to\infty }\left({\frac{{\int_{E}{{{\left| f\right|}^{n+1}g}}}}{{\int_{E}{{{\left| f\right|}^{n}g}}}}}\right) ={\left\| f\right\|_\infty }.

As an application, if we put a_0 = 1, then from

\displaystyle {a_{n + 1}} = \frac{{{a_1}}} {{{a_0}}}.\frac{{{a_2}}} {{{a_1}}} \cdots\frac{{{a_{n + 1}}}} {{{a_n}}}

we deduce that

\displaystyle\mathop{\lim }\limits_{n\to\infty }\sqrt[n]{{{a_{n}}}}=\mathop{\lim }\limits_{n\to\infty }\frac{{{a_{n+1}}}}{{{a_{n}}}}={\left\| f\right\|_\infty }.

In other words,

\displaystyle\mathop{\lim }\limits_{n\to\infty }{\left({\int_{E}{{{\left| f\right|}^{n}g}}}\right)^{\frac{1}{n}}}={\left\| f\right\|_\infty }.

Tháng Mười Một 26, 2009

R-G: Scalar curvature

Chuyên mục: Riemannian geometry — Ngô Quốc Anh @ 20:51

In Riemannian geometry, the scalar curvature (or Ricci scalar) is the simplest curvature invariant of a Riemannian manifold. To each point on a Riemannian manifold, it assigns a single real number determined by the intrinsic geometry of the manifold near that point. Specifically, the scalar curvature represents the amount by which the volume of a geodesic ball in a curved Riemannian manifold deviates from that of the standard ball in Euclidean space. In two dimensions, the scalar curvature is twice the Gaussian curvature, and completely characterizes the curvature of a surface. In more than two dimensions, however, the curvature of Riemannian manifolds involves more than one functionally independent quantity.

In general relativity, the scalar curvature is the Lagrangian density for the Einstein–Hilbert action. The Euler–Lagrange equations for this Lagrangian under variations in the metric constitute the vacuum Einstein field equations, and the stationary metrics are known as Einstein metrics. The scalar curvature is defined as the trace of the Ricci tensor, and it can be characterized as a multiple of the average of the sectional curvatures at a point. Unlike the Ricci tensor and sectional curvature, however, global results involving only the scalar curvature are extremely subtle and difficult. One of the few is the positive mass theorem of Richard Schoen, Shing-Tung Yau and Edward Witten. Another is the Yamabe problem, which seeks extremal metrics in a given conformal class for which the scalar curvature is constant.

Definition. The scalar curvature is the function S defined as the trace of the Ricci tensor.

Since the Ricci tensor is an (2,0)-tensor field then in the local coordinates

S = {\rm Trace}( {\rm Ric}) = g^{jk}R_{jk}.

Theorem (Contracted Bianchi Identity). The covariant derivatives of the Ricci and scalar curvatures satisfy the following identity

\displaystyle {\rm div} {\rm Ric} = \frac{1}{2} \nabla S.

Examples 1. We still work on the two-dimensional spherical surface of radius R whose metric is

\displaystyle \left( {{g_{ij}}} \right) = {R^2}\left( {\begin{array}{*{20}{c}} 1 & 0 \\ 0 & {{{\sin }^2}\theta } \\ \end{array} } \right)

as the previous topic. Then

\displaystyle S = {g^{jk}}{R_{jk}} = {g^{11}}{R_{11}} + {g^{22}}{R_{22}} = \frac{2}{{{R^2}}}.

Examples 2. We now work for the two-dimensional space-like “upper hyperboloid” of the Minkowski space whose metric is

\displaystyle {\left( {ds} \right)^2} = \frac{{{R^2}}}{{{r^2} + {R^2}}}{\left( {dr} \right)^2} + {r^2}{\left( {d\phi } \right)^2}

that is

\displaystyle \left( {{g_{ij}}} \right) = \left( {\begin{array}{*{20}{c}} {\frac{{{R^2}}}{{{r^2} + {R^2}}}} & 0 \\ 0 & {{r^2}} \\ \end{array} } \right),\left( {{g^{ij}}} \right) = \left( {\begin{array}{*{20}{c}} {\frac{{{r^2} + {R^2}}}{{{r^2}}}} & 0 \\ 0 & {\frac{1}{{{r^2}}}} \\ \end{array} } \right).

Then

\displaystyle {R_{11}} = - \frac{1}{{{r^2} + {R^2}}},{R_{12}} = {R_{21}} = 0,{R_{22}} = - \frac{{{r^2}}}{{{R^2}}}

and

\displaystyle S = - \frac{2}{{{R^2}}}.

R-G: Ricci curvature

Chuyên mục: Riemannian geometry — Ngô Quốc Anh @ 19:56

In differential geometry, the Ricci curvature tensor, named after Gregorio Ricci-Curbastro, represents the amount by which the volume element of a geodesic ball in a curved Riemannian manifold deviates from that of the standard ball in Euclidean space. As such, it provides one way of measuring the degree to which the geometry determined by a given Riemannian metric might differ from that of ordinary Euclidean n-space. More generally, the Ricci tensor is defined on any pseudo-Riemannian manifold. Like the metric itself, the Ricci tensor is a symmetric bilinear form on the tangent space of the manifold.

The Ricci curvature is broadly applicable to modern Riemannian geometry and general relativity theory. In connection with the latter, it is up to an overall trace term, the portion of the Einstein field equation representing the geometry of spacetime, the other significant portion of which comes from the presence of matter and energy. In connection with the former, lower bounds on the Ricci tensor on a Riemannian manifold allow one to extract global geometric and topological information by comparison (cf. comparison theorem) with the geometry of a constant curvature space form. If the Ricci tensor satisfies the vacuum Einstein equation, then the manifold is an Einstein manifold, which have been extensively studied (cf. Besse 1987). In this connection, the Ricci flow equation governs the evolution of a given metric to an Einstein metric, the precise manner in which this occurs ultimately leads to the solution of the Poincaré conjecture.

Definition. Ricci curvature (or Ricci tensor) is an (2,0)-tensor field denoted by \rm Ric, that is {\rm Ric} : TM \times TM \to \mathbb R, defined by

{\rm Ric}(X,Y) = {\rm Trace}( x \to R(x, X)Y).

In local coordinates, \rm Ric is of the form

{\rm Ric} = R_{ij} dx^i \otimes dx^j.

We assume \frac{\partial}{\partial x^i} where i=1,2,...,n  is an orthonormal basis for T_pM, then

\displaystyle R\left( {\frac{\partial }{{\partial {x^i}}},X} \right)Y = {X^j}{Y^k}R\left( {\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^j}}}} \right)\frac{\partial }{{\partial {x^k}}} = {X^j}{Y^k}R_{kij}^l\frac{\partial }{{\partial {x^l}}}.

Thus,

\displaystyle {\rm Trace}\left( {x \mapsto R\left( {x,X} \right)Y} \right) = {X^j}{Y^k}R_{kij}^i.

In other words,

\displaystyle {\rm Ric}\left( {\frac{\partial }{{\partial {x^j}}},\frac{\partial }{{\partial {x^k}}}} \right) = {R_{ij}}d{x^i} \otimes d{x^j}\left( {\frac{\partial }{{\partial {x^j}}},\frac{\partial }{{\partial {x^k}}}} \right) = {R_{jk}} = R_{kij}^i.

To be exact, one should read

\displaystyle {R_{jk}} = \sum\limits_i {R_{jik}^i}.

A simple calculation shows us that

\displaystyle {R_{jk}} = R_{jik}^i = {g^{li}}{g_{li}}R_{jik}^i = {g^{li}}{R_{ljik}}.

Thus, Ricci tensor can be thought as the trace of curvature tensor R_{ljil}.

Example. For the two-dimensional spherical surface of radius R whose metric is

\displaystyle{\left( {ds} \right)^2} = {R^2}\left[ {{{\left( {d\theta } \right)}^2} + {{\sin }^2}\theta {{\left( {d\phi } \right)}^2}} \right]

we have

\displaystyle \left( {{g_{ij}}} \right) = {R^2}\left( {\begin{array}{*{20}{c}} 1 & 0 \\ 0 & {{{\sin }^2}\theta } \\ \end{array} } \right), \qquad \left( {{g^{ij}}} \right) = \frac{1}{{{R^2}}}\left( {\begin{array}{*{20}{c}} 1 & 0 \\ 0 & {\frac{1}{{{{\sin }^2}\theta }}} \\ \end{array} } \right).

Thus

\displaystyle\begin{gathered} {R_{11}} = R_{111}^1 + R_{121}^2 = {g^{22}}{R_{2121}} = {g^{22}}{R_{1212}} = 1, \hfill \\ {R_{12}} = R_{112}^1 + R_{122}^2 = 0, \hfill \\ {R_{22}} = R_{212}^1 + R_{222}^2 = {g^{11}}{R_{1212}} = {\sin ^2}\theta , \hfill \\ {R_{21}} = {R_{12}} = 0. \hfill \\ \end{gathered}

R-G: Sectional curvature

Chuyên mục: Riemannian geometry — Ngô Quốc Anh @ 14:40

In Riemannian geometry, the sectional curvature is one of the ways to describe the curvature of Riemannian manifolds.

Definition. The sectional curvature of the plane spanned by the (linearly independent) tangent vectors X, Y \in T_xM of the Riemannian manifold M is

\displaystyle K\left( {X,Y} \right) = \frac{{\left\langle {R\left( {X,Y} \right)Y,X} \right\rangle }}{{\left\langle {X,X} \right\rangle \left\langle {Y,Y} \right\rangle - {{\left\langle {X,Y} \right\rangle }^2}}}.

In local coordinates, if

\displaystyle X = {X^i}\frac{\partial }{{\partial {x^i}}}, \quad Y = {Y^j}\frac{\partial }{{\partial {x^j}}}

we then have

\displaystyle R\left( {X,Y} \right)Y = {X^i}{Y^j}{Y^k}R\left( {\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^j}}}} \right)\frac{\partial }{{\partial {x^k}}} = {X^i}{Y^j}{Y^k}R_{kij}^l\frac{\partial }{{\partial {x^l}}}

which implies

\displaystyle\begin{gathered} \left\langle {R\left( {X,Y} \right)Y,X} \right\rangle = {X^i}{Y^j}{Y^k}R_{kij}^l\left\langle {\frac{\partial }{{\partial {x^l}}},{X^m}\frac{\partial }{{\partial {x^m}}}} \right\rangle \hfill \\ \qquad= {X^i}{Y^j}{X^m}{Y^k}R_{kij}^l{g_{lm}} \hfill \\ \qquad= {R_{mkij}}{X^i}{Y^j}{X^m}{Y^k} \hfill \\ \qquad = {R_{ijmk}}{X^i}{Y^j}{X^m}{Y^k}. \hfill \\ \end{gathered}

Besides

\displaystyle\begin{gathered} \left\langle {X,X} \right\rangle \left\langle {Y,Y} \right\rangle - {\left\langle {X,Y} \right\rangle ^2} = {X^i}{X^m}{g_{im}}{Y^j}{Y^k}{g_{jk}} - {\left( {{X^\alpha }{Y^\beta }{g_{\alpha \beta }}} \right)^2} \hfill \\ \qquad= {X^i}{X^m}{g_{im}}{Y^j}{Y^k}{g_{jk}} - {X^\alpha }{Y^\beta }{g_{\alpha \beta }}{X^\gamma }{Y^\delta }{g_{\gamma \delta }} \hfill \\ \qquad= \left( {{g_{im}}{g_{jk}} - {g_{ij}}{g_{mk}}} \right){X^i}{X^m}{Y^j}{Y^k}. \hfill \\\end{gathered}

Thus

\displaystyle K\left( {X,Y} \right) = \frac{{{R_{ijmk}}{X^i}{Y^j}{X^m}{Y^k}}}{{\left( {{g_{im}}{g_{jk}} - {g_{ij}}{g_{mk}}} \right){X^i}{X^m}{Y^j}{Y^k}}}.

To be exact, without using Einstein summation convention, one reads the above identity as following

\displaystyle K\left( {X,Y} \right) = \frac{{\sum\limits_{ijmk} {{R_{ijmk}}{X^i}{Y^j}{X^m}{Y^k}} }}{{\sum\limits_{ijmk} {\left( {{g_{im}}{g_{jk}} - {g_{ij}}{g_{mk}}} \right){X^i}{X^m}{Y^j}{Y^k}} }}.

We refer the reader to this topic for examples. In addition, if we choose

\displaystyle {g_{ij}} = {\left( {\displaystyle\frac{2}{{1 - {{\left| y \right|}^2}}}} \right)^2}{\delta _{ij}}

then the sectional curvature of g is -1.

R-G: Hessian and Laplacian

Chuyên mục: Riemannian geometry — Ngô Quốc Anh @ 1:17

For a given smooth function f on manifold M, the gradient of f is given by

\displaystyle \nabla f = g^{kj} \dfrac{\partial f}{\partial x^j} \frac{\partial}{\partial x^k}.

Note that gradient of f is also a vector field on M. Thus, for each X \in TM, it is reasonable to talk about \nabla_X \nabla f.

Definition 1. Hessian of f, denoted by {\rm Hess}, is defined as the symmetric (0,2)-tensor

{\rm Hess} f (X,Y)=g(\nabla_X \nabla f, Y).

We also denote by f_{ij} the following

\displaystyle {\rm Hess} f \left(\frac{\partial}{\partial x^i}, \frac{\partial}{\partial x^j}\right).

Thus

\displaystyle\begin{gathered} {f_{ij}} = g\left( {{\nabla _{\frac{\partial }{{\partial {x^i}}}}}\nabla f,\frac{\partial }{{\partial {x^j}}}} \right) = g\left( {{\nabla _{\frac{\partial }{{\partial {x^i}}}}}\left( {{g^{kl}}\frac{{\partial f}}{{\partial {x^l}}}\frac{\partial }{{\partial {x^k}}}} \right),\frac{\partial }{{\partial {x^j}}}} \right) \hfill \\ \quad\; = g\left( {\frac{\partial }{{\partial {x^i}}}\left( {{g^{kl}}\frac{{\partial f}}{{\partial {x^l}}}} \right)\frac{\partial }{{\partial {x^k}}} + {g^{kl}}\frac{{\partial f}}{{\partial {x^l}}}{\nabla _{\frac{\partial }{{\partial {x^i}}}}}\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^j}}}} \right) \hfill \\ \quad\; = \frac{\partial }{{\partial {x^i}}}\left( {{g^{kl}}\frac{{\partial f}}{{\partial {x^l}}}} \right)g\left( {\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^j}}}} \right) + {g^{kl}}\frac{{\partial f}}{{\partial {x^l}}}g\left( {{\nabla _{\frac{\partial }{{\partial {x^i}}}}}\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^j}}}} \right) \hfill \\\quad\; = \frac{\partial }{{\partial {x^i}}}\left( {{g^{kl}}\frac{{\partial f}}{{\partial {x^l}}}} \right){g_{kj}} + {g^{kl}}\frac{{\partial f}}{{\partial {x^l}}} \left[ \frac{1}{2}\left( {-\frac{{\partial {g_{ki}}}}{{\partial {x^j}}} + \frac{{\partial {g_{ij}}}}{{\partial {x^k}}} + \frac{{\partial {g_{kj}}}}{{\partial {x^i}}}} \right)\right]. \hfill\end{gathered}

Note that

\displaystyle \frac{\partial }{{\partial {x^i}}}\left( {{g^{kl}}\frac{{\partial f}}{{\partial {x^l}}}} \right){g_{kj}} = \frac{{\partial {g^{kl}}}}{{\partial {x^i}}}\frac{{\partial f}}{{\partial {x^l}}}{g_{kj}} + {g^{kl}}\frac{\partial }{{\partial {x^i}}}\left( {\frac{{\partial f}}{{\partial {x^l}}}} \right){g_{kj}} = \frac{{\partial {g^{kl}}}}{{\partial {x^i}}}\frac{{\partial f}}{{\partial {x^l}}}{g_{kj}} + \frac{{{\partial ^2}f}}{{\partial {x^i}\partial {x^j}}}.

Since 0=\frac{\partial}{\partial x^i}(g^{kl}g_{kj}) then

\displaystyle\frac{{\partial {g^{kl}}}}{{\partial {x^i}}}\frac{{\partial f}}{{\partial {x^l}}}{g_{kj}} = - \frac{{\partial {g_{kj}}}}{{\partial {x^i}}}\frac{{\partial f}}{{\partial {x^l}}}{g^{kl}}

which implies

\displaystyle f_{ij} =\frac{{{\partial ^2}f}}{{\partial {x^i}\partial {x^j}}} - \Gamma _{ij}^m\frac{{\partial f}}{{\partial {x^m}}}.

Definition 2. Laplacian of f, denoted by \Delta f, is defined as the trace of {\rm Hess} f.

Note that {\rm Hess} f is a (0,2)-tensor, then in local coordinates, one has

\displaystyle \Delta f = {g^{ij}}{f_{ij}}.

It is clear that \nabla X is a (1,1)-tensor field. To see this fact, one can assume X=X^i \frac{\partial}{\partial x^i} then from

\displaystyle {\nabla _{\frac{\partial }{{\partial {x^j}}}}}\left( {{X^i}\frac{\partial }{{\partial {x^i}}}} \right) = \frac{{\partial {X^i}}}{{\partial {x^j}}}\frac{\partial }{{\partial {x^i}}} + {X^i}\Gamma _{ji}^l\frac{\partial }{{\partial {x^l}}}

one has

\displaystyle\nabla X = \left[ {\frac{{\partial {X^i}}}{{\partial {x^j}}}\frac{\partial }{{\partial {x^i}}} + {X^i}\Gamma _{ji}^l\frac{\partial }{{\partial {x^l}}}} \right] \otimes d{x^j}

since {\nabla _Y}X = \left\langle {Y,\nabla X} \right\rangle which is exactly an (1,1)-tensor. Then we can define divergence of a vector field X as following

Definition 3. Divergence of vector field X is given by

\displaystyle {\rm div} X = {\rm Trace}(\nabla X).

In coordinates, this is

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

and with respect to an orthornormal basis

\displaystyle {\rm div} X =g\left( {{\nabla _{\frac{\partial }{{\partial {x^i}}}}}X,\frac{\partial }{{\partial {x^i}}}} \right).

Thus \Delta f = {\rm Trace}(\nabla(\nabla f)) = {\rm div}(\nabla f).

Tháng Mười Một 17, 2009

R-G: Bianchi identities

Chuyên mục: Riemannian geometry — Ngô Quốc Anh @ 0:22

Recall that R_{ikl}^j is defined to be

\displaystyle R_{ikl}^j = \left\langle {R\left( {\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^l}}}} \right)\frac{\partial }{{\partial {x^i}}},d{x^j}} \right\rangle .

Let

\displaystyle R_{ijkl}=g_{hj} R_{ikl}^h.

The way to understand R_{ijkl} is to look at the following 4-covariant tensor

R(X,Y,Z,T) = g(R(X,Y)Z, T).

As can be seen, the components of R(X,Y,Z,T) are R_{ijkl}.

We first obtain the following result.

Theorem 1. The curvature tensor R_{ijkl}  satisfies the following property

{R_{ijkl}} = - {R_{ijlk}} = - {R_{jikl}} .

Proof.

The proof relies on the definition of the 4-covariant tensor above. To be precise, one has

\displaystyle g\left( {R\left( {\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^l}}}} \right)\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^j}}}} \right) = g\left( {R_{ikl}^h\frac{\partial }{{\partial {x^h}}},\frac{\partial }{{\partial {x^j}}}} \right) = {g_{hj}}R_{ikl}^h = {R_{ijkl}}

and

\displaystyle g\left( {R\left( {\frac{\partial }{{\partial {x^l}}},\frac{\partial }{{\partial {x^k}}}} \right)\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^j}}}} \right) = {R_{ijlk}}.

Since

\displaystyle R\left( {\frac{\partial }{{\partial {x^l}}},\frac{\partial }{{\partial {x^k}}}} \right)\frac{\partial }{{\partial {x^i}}} = - R\left( {\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^l}}}} \right)\frac{\partial }{{\partial {x^i}}}

then {R_{ijkl}} = - {R_{ijlk}}. This comes from the definition of curvature tensor and the fact that

\displaystyle\left[ {\frac{\partial }{{\partial {x^m}}},\frac{\partial }{{\partial {x^n}}}} \right] = 0.

Similarly, for the latter case, one can argue as follows

\displaystyle \begin{gathered} g\left( {R\left( {\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^l}}}} \right)\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^i}}}} \right) \hfill \\ \qquad\qquad= g\left( {{\nabla _{\frac{\partial }{{\partial {x^k}}}}}{\nabla _{\frac{\partial }{{\partial {x^l}}}}}\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^i}}}} \right) - g\left( {{\nabla _{\frac{\partial }{{\partial {x^l}}}}}{\nabla _{\frac{\partial }{{\partial {x^k}}}}}\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^i}}}} \right) - g\left( {{\nabla _{\left[ {\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^l}}}} \right]}}\frac{\partial }{\partial x^i},\frac{\partial }{{\partial {x^i}}}} \right) \hfill \\ \end{gathered}.

We now use the fact that \nabla is a metric connection. Indeed,

\displaystyle \begin{gathered} \;\;\; g\left( {{\nabla _{\frac{\partial }{{\partial {x^k}}}}}{\nabla _{\frac{\partial }{{\partial {x^l}}}}}\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^i}}}} \right) = \frac{\partial }{{\partial {x^k}}}g\left( {{\nabla _{\frac{\partial }{{\partial {x^l}}}}}\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^i}}}} \right) - g\left( {{\nabla _{\frac{\partial }{{\partial {x^l}}}}}\frac{\partial }{{\partial {x^i}}},{\nabla _{\frac{\partial }{{\partial {x^k}}}}}\frac{\partial }{{\partial {x^i}}}} \right) \hfill \\ - g\left( {{\nabla _{\frac{\partial }{{\partial {x^l}}}}}{\nabla _{\frac{\partial }{{\partial {x^k}}}}}\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^i}}}} \right) = - \frac{\partial }{{\partial {x^l}}}g\left( {{\nabla _{\frac{\partial }{{\partial {x^k}}}}}\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^i}}}} \right) + g\left( {{\nabla _{\frac{\partial }{{\partial {x^k}}}}}\frac{\partial }{{\partial {x^i}}},{\nabla _{\frac{\partial }{{\partial {x^l}}}}}\frac{\partial }{{\partial {x^i}}}} \right) \hfill \\ - g\left( {{\nabla _{\left[ {\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^l}}}} \right]}}\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^i}}}} \right) = - \left[ {\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^l}}}} \right]g\left( {\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^i}}}} \right) + g\left( {\frac{\partial }{{\partial {x^i}}},{\nabla _{\left[ {\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^l}}}} \right]}}\frac{\partial }{{\partial {x^i}}}} \right). \hfill \\\end{gathered}

Thus

\displaystyle \begin{gathered} g\left( {R\left( {\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^l}}}} \right)\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^i}}}} \right) \hfill \\ \qquad\qquad= \frac{\partial }{{\partial {x^k}}}g\left( {{\nabla _{\frac{\partial }{{\partial {x^l}}}}}\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^i}}}} \right) - \frac{\partial }{{\partial {x^l}}}g\left( {{\nabla _{\frac{\partial }{{\partial {x^k}}}}}\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^i}}}} \right) - \frac{1}{2}\left[ {\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^l}}}} \right]g\left( {\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^i}}}} \right) = 0. \hfill \\\end{gathered}

Hence

\displaystyle g\left( {R\left( {\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^l}}}} \right)\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^i}}}} \right) = 0.

The above identity also holds if we replace \frac{\partial}{\partial x^i} by a vector field X. Thus

\displaystyle g\left( {R\left( {\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^l}}}} \right)\left( {\frac{\partial }{{\partial {x^i}}} + \frac{\partial }{{\partial {x^j}}}} \right),\frac{\partial }{{\partial {x^i}}} + \frac{\partial }{{\partial {x^j}}}} \right) = 0

which implies

\displaystyle g\left( {R\left( {\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^l}}}} \right)\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^j}}}} \right) = - g\left( {R\left( {\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^l}}}} \right)\frac{\partial }{{\partial {x^j}}},\frac{\partial }{{\partial {x^i}}}} \right).

Therefore, {R_{ijkl}} = - {R_{jikl}} .

Corollary 1. R(X,Y)Z=-R(Y,Z)Z and \left\langle {R\left( {X,Y} \right)Z,W} \right\rangle = - \left\langle {R\left( {X,Y} \right)W,Z} \right\rangle.

Theorem 2 (the first Bianchi identity). The curvature tensor R_{ijkl}  satisfies the following property

{R_{ijlk}} + {R_{iklj}} + {R_{iljk}} = 0.

Proof. Since

\displaystyle R\left( {\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^l}}}} \right)\frac{\partial }{{\partial {x^i}}} = {\nabla _{\frac{\partial }{{\partial {x^k}}}}}{\nabla _{\frac{\partial }{{\partial {x^l}}}}}\frac{\partial }{{\partial {x^i}}} - {\nabla _{\frac{\partial }{{\partial {x^l}}}}}{\nabla _{\frac{\partial }{{\partial {x^k}}}}}\frac{\partial }{{\partial {x^i}}} - \underbrace {{\nabla _{\left[ {\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^l}}}} \right]}}\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^i}}}}_0

then

\displaystyle R\left( {\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^l}}}} \right)\frac{\partial }{{\partial {x^i}}} = {\nabla _{\frac{\partial }{{\partial {x^k}}}}}{\nabla _{\frac{\partial }{{\partial {x^l}}}}}\frac{\partial }{{\partial {x^i}}} - {\nabla _{\frac{\partial }{{\partial {x^l}}}}}{\nabla _{\frac{\partial }{{\partial {x^k}}}}}\frac{\partial }{{\partial {x^i}}}.

Similarly,

\displaystyle \begin{gathered} R\left( {\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^k}}}} \right)\frac{\partial }{{\partial {x^l}}} = {\nabla _{\frac{\partial }{{\partial {x^i}}}}}{\nabla _{\frac{\partial }{{\partial {x^k}}}}}\frac{\partial }{{\partial {x^l}}} - {\nabla _{\frac{\partial }{{\partial {x^k}}}}}{\nabla _{\frac{\partial }{{\partial {x^i}}}}}\frac{\partial }{{\partial {x^l}}}, \hfill \\ R\left( {\frac{\partial }{{\partial {x^l}}},\frac{\partial }{{\partial {x^i}}}} \right)\frac{\partial }{{\partial {x^k}}} = {\nabla _{\frac{\partial }{{\partial {x^l}}}}}{\nabla _{\frac{\partial }{{\partial {x^i}}}}}\frac{\partial }{{\partial {x^k}}} - {\nabla _{\frac{\partial }{{\partial {x^i}}}}}{\nabla _{\frac{\partial }{{\partial {x^l}}}}}\frac{\partial }{{\partial {x^k}}}. \hfill \\ \end{gathered}.

Since \nabla is torsion free, one gets

\displaystyle {\nabla _{\frac{\partial }{{\partial {x^l}}}}}\frac{\partial }{{\partial {x^i}}} = {\nabla _{\frac{\partial }{{\partial {x^i}}}}}\frac{\partial }{{\partial {x^l}}}, \quad {\nabla _{\frac{\partial }{{\partial {x^k}}}}}\frac{\partial }{{\partial {x^i}}} = {\nabla _{\frac{\partial }{{\partial {x^i}}}}}\frac{\partial }{{\partial {x^k}}}, \quad {\nabla _{\frac{\partial }{{\partial {x^k}}}}}\frac{\partial }{{\partial {x^l}}} = {\nabla _{\frac{\partial }{{\partial {x^l}}}}}\frac{\partial }{{\partial {x^k}}}.

As a consequence,

\displaystyle R\left( {\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^l}}}} \right)\frac{\partial }{{\partial {x^i}}} + R\left( {\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^k}}}} \right)\frac{\partial }{{\partial {x^l}}} + R\left( {\frac{\partial }{{\partial {x^l}}},\frac{\partial }{{\partial {x^i}}}} \right)\frac{\partial }{{\partial {x^k}}} = 0.

Now

\displaystyle g\left( {R\left( {\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^l}}}} \right)\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^j}}}} \right) + g\left( {R\left( {\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^k}}}} \right)\frac{\partial }{{\partial {x^l}}},\frac{\partial }{{\partial {x^j}}}} \right) + g\left( {R\left( {\frac{\partial }{{\partial {x^l}}},\frac{\partial }{{\partial {x^i}}}} \right)\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^j}}}} \right) = 0

which implies

\displaystyle {R_{ijkl}} + {R_{ljik}} + {R_{kjli}} = 0.

If we change i by j, j by i we then obtain

\displaystyle \underbrace {{R_{ijkl}}}_{{R_{jikl}}} + \underbrace {{R_{ljik}}}_{{R_{lijk}}} + \underbrace {{R_{kjli}}}_{{R_{kilj}}} = 0

which implies, by using Theorem 1,

\displaystyle - {R_{ijkl}} - {R_{iljk}} - {R_{iklj}} = 0.

Corollary 2. R\left( {X,Y} \right)Z + R\left( {Z,X} \right)Y + R\left( {Y,Z} \right)X = 0.

Corollary 3. Followed from the proof of Theorem 2, by pairing with dx^m to the both sides one has

\displaystyle R_{ikl}^m + R_{lki}^m + R_{kil}^m = 0.

Theorem 3 . The curvature tensor R_{ijkl}  satisfies the following property

R_{ijkl}=R_{klij}.

Proof. By the first Bianchi indentity,

\displaystyle \begin{gathered} {R_{ijkl}} + {R_{iljk}} + {R_{iklj}} = 0, \hfill \\ {R_{jikl}} + {R_{jlik}} + {R_{jkli}} = 0, \hfill \\ \end{gathered}

which implies

\displaystyle 2{R_{ijkl}} + {R_{iljk}} - {R_{jlik}} + {R_{iklj}} - {R_{jkli}} = 0.

Thus

\displaystyle 2{R_{ijkl}} + {R_{iljk}} + {R_{ikjl}} + {R_{iklj}} + {R_{lijk}} = 0.

Similarly, by changing i \to k, j \to l, k \to i and l \to j one gets

\displaystyle 2{R_{klij}} + {R_{kjli}} + {R_{kilj}} + {R_{kijl}} + {R_{jkli}} = 0.

Hence R_{ijkl}=R_{klij} by using Theorem 1.

Theorem 4 (the second Bianchi identity). The curvature tensor R_{ijkl}  satisfies the following property

{R_{ijkl,h}} + {R_{ijlh,k}} + {R_{ijhk,l}} = 0.

Proof. One can use the normal coordinates in order to simplify the calculation. Indeed, normal coordinates tell us at a given point that

g_{ij}=\delta_{ij} and g_{ij,k}=\Gamma_{ij}^k=0

for all i, j, k. Thus,

\displaystyle\begin{gathered} R_{ikl}^h\frac{\partial }{{\partial {x^h}}} \;= R\left( {\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^l}}}} \right)\frac{\partial }{{\partial {x^i}}} = {\nabla _{\frac{\partial }{{\partial {x^k}}}}}{\nabla _{\frac{\partial }{{\partial {x^l}}}}}\frac{\partial }{{\partial {x^i}}} - {\nabla _{\frac{\partial }{{\partial {x^l}}}}}{\nabla _{\frac{\partial }{{\partial {x^k}}}}}\frac{\partial }{{\partial {x^i}}} \hfill \\ \qquad\qquad = {\nabla _{\frac{\partial }{{\partial {x^k}}}}}\left( {\Gamma _{li}^m\frac{\partial }{{\partial {x^m}}}} \right) - {\nabla _{\frac{\partial }{{\partial {x^l}}}}}\left( {\Gamma _{ki}^n\frac{\partial }{{\partial {x^n}}}} \right) = \frac{{\partial \Gamma _{li}^m}}{{\partial {x^k}}}\frac{\partial }{{\partial {x^m}}} - \frac{{\partial \Gamma _{ki}^n}}{{\partial {x^l}}}\frac{\partial }{{\partial {x^n}}} \hfill \\\qquad\qquad = \left( {\frac{{\partial \Gamma _{li}^m}}{{\partial {x^k}}} - \frac{{\partial \Gamma _{ki}^m}}{{\partial {x^l}}}} \right)\frac{\partial }{{\partial {x^m}}} \hfill \\ \end{gathered}

which implies

\displaystyle \begin{gathered}R_{ikl}^h \;= \frac{{\partial \Gamma _{li}^h}}{{\partial {x^k}}} - \frac{{\partial \Gamma _{ki}^h}}{{\partial {x^l}}} \hfill \\ \qquad= \frac{\partial }{{\partial {x^k}}}\left( {\frac{1}{2}{g^{hm}}\left( {{g_{ml,i}} + {g_{mi,l}} - {g_{li,m}}} \right)} \right) - \frac{\partial }{{\partial {x^l}}}\left( {\frac{1}{2}{g^{hn}}\left( {{g_{nk,i}} + {g_{ni,k}} - {g_{ki,n}}} \right)} \right) \hfill \\ \qquad= \frac{1}{2}{g^{hm}}\left( {{g_{ml,ik}} + {g_{mi,lk}} - {g_{mk,il}} - {g_{mi,kl}} - {g_{li,mk}} + {g_{ki,ml}}} \right) \hfill \\ \qquad = \frac{1}{2}{g^{hm}}\left( {{g_{ml,ik}} - {g_{mk,il}} - {g_{li,mk}} + {g_{ki,ml}}} \right). \hfill \\ \end{gathered}

Hence

\displaystyle {R_{ijkl}} = {g_{jh}}R_{ikl}^h = \frac{1}{2}\left( {{g_{jl,ik}} - {g_{jk,il}} - {g_{li,jk}} + {g_{ki,jl}}} \right)

which implies

\displaystyle{R_{ijkl,h}} = \frac{1}{2}\left( {{g_{jl,ikh}} - {g_{jk,ilh}} - {g_{li,jkh}} + {g_{ki,jlh}}} \right).

Similarly, we can write down R_{ijlh,k} and R_{ijhk,l}. Summing up we get the desired result.

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