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#21 toilachinhtoi

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Đã gửi 04-11-2006 - 09:11

Problem 7/Chapter 2: Describe geometrically the parallel transport along a parallel
of latitude on http://dientuvietnam...imetex.cgi?S^2.

Proof:

No loss of generality, we can consider [the latitude tex]c(t)$ to be the
curve http://dientuvietnam...mimetex.cgi?c(t) is the intersection of http://dientuvietnam...mimetex.cgi?S^2
with the hyperplane z=r).

Let http://dientuvietnam...mimetex.cgi?V(t)=(a(t),b(t),c(t)) to
be the parallel transport of http://dientuvietnam...mimetex.cgi?V_0 on http://dientuvietnam...mimetex.cgi?S^2 with http://dientuvietnam...mimetex.cgi?V(0)=V_0. We know
that the tangent vector of http://dientuvietnam...mimetex.cgi?c(t) is http://dientuvietnam...mimetex.cgi?S^2 at the
point http://dientuvietnam...mimetex.cgi?c(t) is http://dientuvietnam.net/cgi-bin/mimetex.cgi?n(t) and http://dientuvietnam.net/cgi-bin/mimetex.cgi?c'(t), we can choose a basis for http://dientuvietnam.net/cgi-bin/mimetex.cgi?T_{c(t)}S^2 as http://dientuvietnam.net/cgi-bin/mimetex.cgi?V(t)=a(t)u(t)+b(t)w(t). By Problem 4a) we have http://dientuvietnam.net/cgi-bin/mimetex.cgi?\dfrac{dV}{dt}
must have the same direction as n(t) so we have the equation
V'(t)=K(t)n(t) or equivalently
http://dientuvietnam.net/cgi-bin/mimetex.cgi?-b'(t)\sqrt{1-r^2}=K(t)r. Comparing the
x and y coordinates we get
http://dientuvietnam.net/cgi-bin/mimetex.cgi?rb'(t)-a(t)=K(t)\sqrt{1-r^2},a'(t)+rb(t)=0.
Substituting http://dientuvietnam.net/cgi-bin/mimetex.cgi?K(t)=-\dfrac{\sqrt{1-r^2}}{r}b'(t) we get a linear
system of ODEs:
http://dientuvietnam.net/cgi-bin/mimetex.cgi?a'(t)=-rb(t),b'(t)=ra(t).
Solving this system we get http://dientuvietnam.net/cgi-bin/mimetex.cgi?V(0)=V_0 we get
http://dientuvietnam.net/cgi-bin/mimetex.cgi?B(0,1,0)-A(r,0,-\sqrt{1-r^2})=V_0.

The geometric meaning of the "từ cấm" is clear from the above
decompose of http://dientuvietnam.net/cgi-bin/mimetex.cgi?V(t) and http://dientuvietnam.net/cgi-bin/mimetex.cgi?V_0 in to the directions of u(t) and
w(t).
There is no way leading to happiness. Happiness is just the way.
The Buddha

#22 toilachinhtoi

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Đã gửi 10-11-2006 - 08:46

(vi de bai dai qua nen toi khong chep lai. Neu ban nao quan tam thi moi xem trong
do Carmo.)

Problem 1/Chapter 3: We have
http://dientuvietnam.net/cgi-bin/mimetex.cgi?X the first column of http://dientuvietnam...ex.cgi?f'(v)\not=0 and http://dientuvietnam...-f^2=|f'|^2 which
is a contradiction to http://dientuvietnam.../mimetex.cgi?U. If cos u=0 then
http://dientuvietnam.net/cgi-bin/mimetex.cgi?u"=-2\dfrac{ff'}{f^2}u'v' and
http://dientuvietnam.net/cgi-bin/mimetex.cgi?v"(|f'|^2+|g'|^2)=ff'|u'|^2-(f'f"+g'g")|v'|^2 we can see that
http://dientuvietnam.net/cgi-bin/mimetex.cgi?r=|f(v)| hence $r\cos (\beta (t))=\dfrac{u'(t)f^2}{|\gamma
'(t)|}$. Hence since we showed that http://dientuvietnam...9;f^2=constant. But we have
http://dientuvietnam.net/cgi-bin/mimetex.cgi?\dfrac{d}{dt}(u'f^2(v(t)))=u"f^2+2u'v'ff'=0.
d) From the proof of c) we have in this case
http://dientuvietnam.net/cgi-bin/mimetex.cgi?u'(t)>0.

Remark: Here do Carmo require for a geosedic with time t in http://dientuvietnam.net/cgi-bin/mimetex.cgi?\dfrac{C_1^2}{v^2}+v'^2(1+4v^2)=C_2,u'=\dfrac{C_1}{v^2}.

From the first equation we see that except the case where v(t) is
a constant there exists K>0 such that for |t|>K we have v'(t)
does not change its sign. In particular we may assume that v(t) is
strictly increasing for |t|>K (in fact it can be shown from the equation that
). We can show also along with |t|. Henc the two branches for t>0 and t<0 cut infinitely many of times.
There is no way leading to happiness. Happiness is just the way.
The Buddha

#23 xuongrong

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Đã gửi 10-11-2006 - 13:48

toilachinhtoi hay quá. tiếp đi. trong câu d. hy vọng để nhìn rõ hơn chút, có thể xem những geodesic (not horizontal) chia làm 2 nhánh khi t >0 lớn và t<0 lớn. mỗi nhánh sẽ đi vòng ôm lấy hình nón ngược vì r cos( beta ) luôn là hằng số. r=u tiến ra vô cùng. ở đây beta là góc giữa đường geodesic và những horizontal. 2 nhánh này buộc phải cắt nhau vô hạn lần.
Chém dao xuống nước, nước càng chảy. Nâng chén tiêu sầu, càng sầu thêm.

#24 toilachinhtoi

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Đã gửi 13-11-2006 - 07:44

Problem 5/Chapter 3: About Killing field

d) Since http://dientuvietnam.net/cgi-bin/mimetex.cgi?\rightarrow) We just choose in the condition above t=0. Then
since http://dientuvietnam.net/cgi-bin/mimetex.cgi?\leftarrow) If http://dientuvietnam...a_{ij} a_{ji}=0 for every i,j iff A is
anti-symmetric.

c) Since f is an isometry, there is a 1-1 correspondence between
vector fields in M and vector fields in N. Moreover, we have
(see proof of Problem 3/Chapter 2)
http://dientuvietnam.net/cgi-bin/mimetex.cgi?&#091;X_n,X_i]=&#091;X_n,X_j]=0, above equation is equivalent to
, and we are
done.

b) By definition (see page 70 of do Carmo), showing that X is
tangent to the geodesic spheres centered at q is equivalent to
showing that :P\gamma (t))>=0" [/tex] for any geodesics
with . We have
:P\gamma (t))>_{\gamma
(t)}=<\dfrac{D}{dt}\gamma '(t),:lol:\gamma (t))>_{\gamma (t)}+<\gamma
'(t),\dfrac{D}{dt}:lol:\gamma (t))>_{\gamma (t)}=0.
" [/tex]
This is because since is a geodesics, and d).

Hence . At t=0 we have
There is no way leading to happiness. Happiness is just the way.
The Buddha

#25 xuongrong

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Đã gửi 13-11-2006 - 23:30

có thể mô tả Killing field trên vài manifold cụ thể được không? trên S^2, Torus, Mobious band? thanks.
Chém dao xuống nước, nước càng chảy. Nâng chén tiêu sầu, càng sầu thêm.

#26 toilachinhtoi

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Đã gửi 19-11-2006 - 02:49

To XR: The easiest to construct a Killing field is finding a family of ismetries and taking the derivatives, just as in definition of a Killing field.

Problem 6/Chapter 3: If M is connected, http://dientuvietnam...n/mimetex.cgi?X is a Killing field, and X(q)= 0, http://dientuvietnam.net/cgi-bin/mimetex.cgi?\dfrac{d}{ds}&#091;A.B(s)]=A.\dfrac{d}{ds}B(s). This
thing shows that http://dientuvietnam...mimetex.cgi?p_n to show that http://dientuvietnam...tex.cgi?X|_V=0. In particular,


Now A is not empty and both open and closed in a connected space
M, hence A=M.

Bài viết đã được chỉnh sửa nội dung bởi toilachinhtoi: 19-11-2006 - 02:53

There is no way leading to happiness. Happiness is just the way.
The Buddha

#27 toilachinhtoi

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Đã gửi 21-11-2006 - 11:48

The answer to this problem will be waited some days.

Problem: Let M be a compact manifold of dimension n. Let N_1 and N_2 be two closed totally geodesic submanifolds of M, with dimensions n_1 and n_2 satisfying n_1+n_2>n. Prove that N_1 and N_2 has a non-empty intersection.
There is no way leading to happiness. Happiness is just the way.
The Buddha

#28 toilachinhtoi

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Đã gửi 02-01-2007 - 03:19

Problem 11/Chap 3: For each fixed $p\in M$ we will show that
$d(ins(X)\nu )(p)=(div~X)\nu (p)$. For this end we choose a geodesic
frame $E_1,...,E_n$ at p. We also arrange that $E_1,...,E_n$ is
the (local) orientation for M . Then
$\nu (E_1,...,E_n)\equiv 1.$
Now we can write $X(q)=\sum _{i=1}^nf_i(q)E_i(q)$. Then for
$Z_i=(E_1,\ldots ,E_{i-1},\widehat{E_i},E_{i+1},\ldots E_n)$ we have
$ins(X)\nu (Z_i)(q)=\sum _{j=1}^nf_j\nu (E_j,Z)(q)=\sum
_{j=1}^{n}(-1)^{j-1}f_j\nu (E_1,\ldots ,E_{i-1},E_j,E_{i+1},\ldots
,E_n)(q)=(-1)^{i-1}f_i(q).$
The formula above shows that $ins(X)\nu =\sum
_{i=1}^n(-1)^{i+1}f_i\theta _i$ as hinted by do Carmo (we use
$(-1)^{i-1}=(-1)^{i+1}$). Then
$d(ins (X)\nu )&=&\sum _{i=1}^n(-1)^{i+1}df_i\wedge \theta _i+\sum
_{i=1}^n(-1)^{i+1}f_i\wedge d\theta _i=\sum _{i=1}^n(E_if_i)\nu +\sum _{i=1}^n(-1)^{i+1}f_i d\theta _i.
$
Now we show that $d\theta _i(p)=0$ for any i. We need to show this
only for $\theta _n$. We have
$d\theta _n=\sum _{j=1}^{n}(-1)^{j-1}\omega _1\wedge \ldots \wedge
d\omega _i\wedge \ldots \wedge \omega _{n-1}.$
Hence to show that $d\theta _n(p)=0$ it suffices to show that
$d\omega _i(p)=0$ for any i.

Claim: If $\omega$ is a 1-form then $[d\omega ](X,Y)(q)=X[\omega
(Y)]-Y[\omega (X)]-\omega ([X,Y])$. Proof of Claim: We write locally
$\omega =\sum _{i=1}^nf_idu _i$ where $du _1,...,du _n$ is the dual
basis for $\dfrac{\partial}{\partial u_1},...,\dfrac{\partial
}{\partial u_n}$. Then $d\omega =\sum _{i=1}^n\sum
_{j=1}^n\dfrac{\partial f_i}{\partial x_j}du _j\wedge du _i$. Now we
need only to check the Claim for the case
$X=g\dfrac{\partial}{\partial u_i},Y=h\dfrac{\partial}{\partial u_j}$.
For this case we have
$[d\omega ](g\dfrac{\partial}{\partial u_i},h\dfrac{\partial}{\partial
u_j})&=&gh[\dfrac{\partial f_j}{\partial x_i}-\dfrac{\partial
f_i}{\partial x_j}],g\dfrac{\partial}{\partial u_i}[\omega (h\dfrac{\partial}{\partial
u_j})]&=&g\dfrac{\partial}{\partial u_i}[hf_j]=gh\dfrac{\partial
f_j}{\partial u_i}-gf_j\dfrac{\partial h}{\partial u_i},
h\dfrac{\partial}{\partial u_j}[\omega (g\dfrac{\partial}{\partial
u_i})]&=&h\dfrac{\partial}{\partial u_j}[gf_i]=gh\dfrac{\partial
f_i}{\partial u_j}-hf_i\dfrac{\partial g}{\partial u_j},
\omega ([g\dfrac{\partial}{\partial u_i},h\dfrac{\partial}{\partial
u_j}])&=&\omega (g\dfrac{\partial h}{\partial
u_i}\dfrac{\partial}{\partial u_j}-h\dfrac{\partial g}{\partial
u_j}\dfrac{\partial}{\partial u_i})=gf_j\dfrac{\partial h}{\partial
u_i}-hf_i\dfrac{\partial g}{\partial u_j}.
$
These equations complete the proof of the Claim.

Now we continue the proof of Problem 11. By the Claim, the
properties of Riemannian connection and geodesic frame at p we
have
$[d\omega _i](E_j,E_k)(p)&=&E_j[\omega _i(E_k)]-E_k[\omega
_i(E_j)]-\omega [E_j,E_k](p)=E_j[\delta _{ik}]-E_k[\delta
_{ij}]-\omega (\nabla _{E_j}E_k(p)
&&-\nabla _{E_k}E_j(p))=0-0-\omega (0)=0.
$
Hence by Problem 8a), at p
$d(ins(X)\nu )(p)=(\sum _{i=1}E_i(f_i)(p))\nu =div~X(p)\nu .
$

Problem 12/Chapter 3 (Maximum principle): If $f:M\rightarrow R $ is subharmonic, where M is connected, compact without boundary then f is a constant.

By Stokes theorem, since M has no boundary, we have
$\int _{M}d\omega =\int _{\partial M}\omega =\int _{\emptyset}\omega
=0$
for any $(n-1)$-form.

For each function g if we define $Y=grad~g$ then by definition and
Problem 11 we have $(\Delta g)\nu =(div ~Y)\nu =d(ins(Y)\nu )$.
Hence by above observation we have
$\int _M(\Delta g)\nu =\int _Md(ins(Y)\nu )=0$
for any function g.

Now if f is a smooth function such that $\Delta f\geq 0$, then
since
$\int _M(\Delta f)\nu =0$
by above observation, we have that $\Delta f\equiv 0$. Substituting
this into formula of Problem 9b) we have $\Delta (f^2)=2f\Delta
f+2|grad f|^2= 2|grad f|^2$. Hence
$2\int _M|grad f|^2\nu =\int _M\Delta f^2\nu =0,$
which implies that $|grad f|\equiv 0$. Since M is connected, we
have that f must be a constant.
There is no way leading to happiness. Happiness is just the way.
The Buddha

#29 toilachinhtoi

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Đã gửi 06-01-2007 - 08:57

Problem 14/Chapter 3: Prove that if $G$ is the geodesic field on $TM$ then $divG=0$.

Proof: Fixed $p\in M$. We will prove that $divG(p)=0.$

First we construct the normal coordinates as hinted by do Carmo. We
choose a normal neighborhood $U=exp_p(B_{\epsilon }(0))$ of p.
Then since we have the exponential map $exp_{p}:~B_{\epsilon
}(0)\subset T_pM\rightarrow U$ is a diffeomorphism, we can use
$(u_1,...,u_n)$ to be the coordinates for a point
$q=exp_p(u_1e_1+...+u_ne_n)$ where $e_1,...e_n$ is a normal basis of
$T_pM.$ We compute the vector fields $\dfrac{\partial }{\partial
u_i}.$ To this end, we choose a function $f:~U\rightarrow \RR.$ Then
we have, where $q=exp _p(v),$
$\dfrac{\partial }{\partial u_i}=\dfrac{\partial }{\partial u_i}(f\circ
exp _{p}(v+te_i))|_{t=0}=Tf\circ (Texp _p)_{v}\circ e_i.$
This shows that $\dfrac{\partial }{\partial u_i}|_q=(Texp
_p)_{v}\circ e_i.$

Now we can see that $\Gamma _{ij}^k(p)=0$ for this coordinate
because any geodesic starting from p will be written by a linear
system of ODEs (the geodesic will be a straight line in this
coordinate)
$\dfrac{d^2x_k}{dt^2}=0.$
Comparing this with the equation (1) page 62 at the start point p
we see that
$\sum _{i,j}\Gamma _{ij}^k(p)v_iv_j=0$
for any initial velocity $(v_1,...,v_n)=\dfrac{dx}{dt}(0).$ Then we
must have $\Gamma _{ij}^k(p)=0$ for any i,j,k.

$\Gamma _{ij}^k(p)=0$ (If $\Gamma _{ij}^k(p)$ were not 0 then the
solution to this system $(1)$ will not be a straight line, for we
can choose an initial $dx(0)=v$ such that ).

We denote by $X_i=\dfrac{\partial }{\partial u_i},Y_j=\dfrac{\partial
}{\partial v_j}$ as tangent vectors of TM (We abuse the same
$X_1,...,X_n$ for either vectors in $T_pM$ and tangent vectors of
$TM$). Consider a point $(q,w)\in T_qM$ with $w=w_1X_1+...+w_nX_n.$

If we take the curve $\alpha (t)=(q(t),w(t))$ with $q(0)=q,
q'(0)=X_i,w(0)=v,w'(0)=0$ then $\alpha '(0)=X_i.$ Moreover we have
$\dfrac{Dw}{dt}(0)=\nabla _{X_i}v.$

If we take the curve $\alpha (t)=(q,w(t))$ with $w(0)=v,w'(0)=X_j$
then $\alpha '(0)=Y_j.$ Moreover we have $\dfrac{Dw}{dt}(0)=X_j.$

Hence for the natural metric we have
$<X_i,X_j>^{S}_{q,v}=<X_i,X_j>_q+<\nabla _{X_i}v,\nabla _{X_j}v>_q,<X_i,Y_j>^{S}_{q,v}=<\nabla _{X_i}v,X_j>_q,
<Y_i,Y_j>^{S}_{q,v}=<X_i,X_j>_q.$
For the product metric we have
$<X_i,X_j>^{P}_{q,v}=<Y_i,Y_j>_{q,v}=<X_i,X_j>_q,<X_i,Y_j>^{P}_{q,v}=0.$
If we define $\tilde{X_i}=(X_i,\nabla _{X_i}v),\tilde{Y_j}=(0,X_j)$
then we have
$<X_i,X_j>^{S}_{q,v}=<\tilde{X_i},\tilde{X_j}>^{P}_{q,v},<X_i,Y_j>^{S}_{q,v}=<\tilde{X_i},\tilde{Y_j}>^{P}_{q,v},
<Y_i,Y_j>^{S}_{q,v}&=&<\tilde{Y_i},\tilde{Y_j}>^{P}_{q,v}.$
(Here we use super-script S and P for natural and product
metric). We can find some $n\times n$ matrix J (depending on
q,v) such that $\tilde{X_i}=A.X_i,\tilde{Y_i}=A.Y_i$ where A is
the linear map represented by the matrix $(2n)\times (2n)$ with the
form
$A=\left (\begin{array}{ll}I_n&J\&I_n\end{array}\right ),$
with $I_n$ the $n\times n$ identity matrix. We see that $det (A)=1.$
This show that in the product metric
$vol(X_1,...,X_n,Y_1,...,Y_n)=vol(\tilde{X_i},...,\tilde{X_n},\tilde{Y_1},...,\tilde{Y_n}).$
Thus we have the volume form in natural metric and in product metric
is the same.

Now we can proceed as in do Carmo. We compute divG in the product
metric. Using formula (1') page 62 in the text book, we see that
$G(u_i)=v_i,~G(v_j)=-\sum _{ik}\Gamma _{ik}^jv_iv_k.$

Since in product metric, $\nabla _{X_i}Y_j\equiv 0,$ and at the
point p we has shown that $\nabla _{X_i}X_j(p)=\nabla
_{Y_iY_j}=0(p),$ we have $\Gamma _{ij}^k(p)=0$ in the product
metric. Hence, since $\Gamma _{ij}^k$ independent of $v_1,...,v_n$
we have as computed in do Carmo that $divG=0.$

Now fix p as above and let $f (t,u,v)$ be the geodesic flow. If we
write $f (t,u,v)=(f_1(t,u,v),...,f_{2n}(t,u,v),)$ then we have
$0=div G(t,u,v)=\sum _{i=1}^{2n}\dfrac{\partial\dfrac{df_i}{dt}}{\partial x_i}=\dfrac{d}{dt}[div ~f].$
Hence $div ~f$ is a constant with respect to t. At $t=0$ we have
f is the identity map, thus $div ~f(0,u,v)=2n$ for all u,v.
Hence $div~f(t,u,v)=2n=\mbox{dimension of }TM$ for all $t,u,v.$
There is no way leading to happiness. Happiness is just the way.
The Buddha

#30 toilachinhtoi

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Đã gửi 13-01-2007 - 10:20

Problem 8/Chapter 5:

a) Let $\varphi (s,p)$ be the flow of X, that is $\dfrac{d\varphi }{ds}=X\circ \varphi$. From definition, since X is a Killing field
we have that $\varphi (s,.)$ is an isomorphism for every s. Now we consider $f(t,s)=\varphi (s,\gamma (t))$. Since $\gamma$ is a
geodesic, for any s we have $f(s,.)$ is a geodesic. Hence
$X\circ \gamma (t)=X\circ \varphi (0,\gamma (t))=\dfrac{\partial f}{\partial s}(0,t)$
is a Jacobian field.

b) Since M is connected, to show that X=0 we need only to show that for every $p\in M$, X=0 in a neighborhood of p. Fixed $p\in M$.
Choose U to be a totally normal neighborhood of p. To show
that X=0 in U, we need to show only that X=0 on any geodesic
emanating from p. Define $J(t)=X\circ \gamma (t)$. Then J is a
Jacobian field with $J(0)=X\circ \gamma (0)=X(p)=0$, and
$J'(0)=\nabla _{\gamma '(0)}X(p)=0$. Hence by the uniqueness of
Jacobian fields we have $J\equiv 0$.
There is no way leading to happiness. Happiness is just the way.
The Buddha




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