Javascript bị vô hiệu

Javascript bị vô hiệu, một vài chức năng sẽ không hoạt động. Vui lòng bật lại Javascript để sử dụng đầy đủ các chức năng.

Một số bài tập hình học vi phân

Bắt đầu bởi toilachinhtoi, 18-09-2006 - 08:38

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#1
toilachinhtoi

Đã gửi 18-09-2006 - 08:38

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Trong topic này tôi xin tóm tắt một số kĩ thuật tính toán và số bài tập HHVP.

Bài 3a/chapter 0 trong do Carmo: Một mặt S là định hướng được nếu và chỉ nếu nó có một trường (smooth) vécto pháp tuyến đơn vị. Giải

Step 1: Assume that http://dientuvietnam...n/mimetex.cgi?S is orientable. We choose charts http://dientuvietnam...n/mimetex.cgi?M such that http://dientuvietnam.net/cgi-bin/mimetex.cgi?(U_{\alpha},u_{\alpha}) of http://dientuvietnam...n/mimetex.cgi?M http://dientuvietnam.net/cgi-bin/mimetex.cgi?T_{x,y}u_{\alpha} is injective) it is easy to see that http://dientuvietnam.net/cgi-bin/mimetex.cgi?v_{\alpha},w_{\alpha} is a local basis for http://dientuvietnam...mimetex.cgi?TM. Now for http://dientuvietnam...mimetex.cgi?N(p) as follow http://dientuvietnam.net/cgi-bin/mimetex.cgi?(U_{\alpha},u_{\alpha}) for which http://dientuvietnam...mimetex.cgi?|v| is the modul of a vector. Note that here http://dientuvietnam.net/cgi-bin/mimetex.cgi?v_{\alpha},w_{\alpha} is a local basis for http://dientuvietnam.net/cgi-bin/mimetex.cgi?TM.

Claim 1: http://dientuvietnam.net/cgi-bin/mimetex.cgi?N(p) is well-defined (i.e. it does not depent on the chart we choose), and http://dientuvietnam.net/cgi-bin/mimetex.cgi?N(p) is orthogonal to http://dientuvietnam.net/cgi-bin/mimetex.cgi?T_p(M), and http://dientuvietnam.net/cgi-bin/mimetex.cgi?|N(p)|=1 and http://dientuvietnam.net/cgi-bin/mimetex.cgi?N(p) is diffferentiable.

First we prove a formula relating overlap charts: Let http://dientuvietnam.net/cgi-bin/mimetex.cgi?(U_{\alpha},u_{\alpha}) and http://dientuvietnam.net/cgi-bin/mimetex.cgi?(U_{\beta},u_{\beta}) be charts of http://dientuvietnam.net/cgi-bin/mimetex.cgi?M such that http://dientuvietnam.net/cgi-bin/mimetex.cgi?x and http://dientuvietnam.net/cgi-bin/mimetex.cgi?y we get the following formula

http://dientuvietnam.net/cgi-bin/mimetex.cgi?(U_{\alpha},u_{\alpha}) or http://dientuvietnam.net/cgi-bin/mimetex.cgi?(U_{\beta},u_{\beta}). Now it is clear from definition that http://dientuvietnam.net/cgi-bin/mimetex.cgi?|N(p)|=1 and http://dientuvietnam.net/cgi-bin/mimetex.cgi?T_p(M) (http://dientuvietnam.net/cgi-bin/mimetex.cgi?v_{\alpha},w_{\beta} of http://dientuvietnam.net/cgi-bin/mimetex.cgi?T_p(M)). That http://dientuvietnam.net/cgi-bin/mimetex.cgi?f_{\alpha},g_{\alpha},h_{\alpha} are differentiable (http://dientuvietnam.net/cgi-bin/mimetex.cgi?(f_{\alpha},g_{\alpha},h_{\alpha}) is a chart).

Step 2: Now assume conversely that http://dientuvietnam.net/cgi-bin/mimetex.cgi?M$ has a differentiable vector field http://dientuvietnam.net/cgi-bin/mimetex.cgi?N(p) with http://dientuvietnam.net/cgi-bin/mimetex.cgi?|N(p)|=1 and http://dientuvietnam.net/cgi-bin/mimetex.cgi?N(p) is orthogonal to http://dientuvietnam.net/cgi-bin/mimetex.cgi?T_p(M) for each http://dientuvietnam.net/cgi-bin/mimetex.cgi?M by the charts http://dientuvietnam.net/cgi-bin/mimetex.cgi?(U_{\alpha},u_{\alpha}). Now we take a differentiable structure of http://dientuvietnam.net/cgi-bin/mimetex.cgi?M as follow: For each http://dientuvietnam.net/cgi-bin/mimetex.cgi?(U_\alpha,u_{\alpha}) around http://dientuvietnam.net/cgi-bin/mimetex.cgi?p of http://dientuvietnam.net/cgi-bin/mimetex.cgi?M. Now we have either of the following: http://dientuvietnam.net/cgi-bin/mimetex.cgi?N_{\alpha}(p)=N(p) or http://dientuvietnam.net/cgi-bin/mimetex.cgi?N_{\alpha}(p)=-N(p) (where http://dientuvietnam.net/cgi-bin/mimetex.cgi?N_{\alpha} is as in Step 1). If the former occurs, then by continuty of http://dientuvietnam.net/cgi-bin/mimetex.cgi?N_{\alpha}(q) and http://dientuvietnam.net/cgi-bin/mimetex.cgi?N(q) we have that there exists an open subset http://dientuvietnam.net/cgi-bin/mimetex.cgi?\widetilde{U_{\alpha}} of http://dientuvietnam.net/cgi-bin/mimetex.cgi?U_{\alpha} such that http://dientuvietnam.net/cgi-bin/mimetex.cgi?N_{\alpha}(q)=N(q) for http://dientuvietnam.net/cgi-bin/mimetex.cgi?N_{\alpha}(p)=-N(p) then we replace http://dientuvietnam.net/cgi-bin/mimetex.cgi?U_{\alpha} by the set http://dientuvietnam.net/cgi-bin/mimetex.cgi?u_{\alpha} by http://dientuvietnam.net/cgi-bin/mimetex.cgi?(f_{\alpha}(-x,y),g_{\alpha}(-x,y),h_{\alpha}(-x,y)). Then for this new chart http://dientuvietnam.net/cgi-bin/mimetex.cgi?(U_{\alpha},u_{\alpha}) we have http://dientuvietnam.net/cgi-bin/mimetex.cgi?N_{\alpha}(p)=N(p). So we can proceed as before to get an open subset http://dientuvietnam.net/cgi-bin/mimetex.cgi?\widetilde{U_{\alpha}}.

We now show that the charts http://dientuvietnam.net/cgi-bin/mimetex.cgi?(\widetilde{U_{\alpha}},u_{\alpha}) is an orientation of http://dientuvietnam.net/cgi-bin/mimetex.cgi?M. Indeed, if http://dientuvietnam.net/cgi-bin/mimetex.cgi?N_{\alpha}(p)=N(p)=N_{\beta}(p). Now apply formula in Step 1 we have

$N(p)=sign ~det J(u_{\beta}^{-1}\circ u_{\alpha}).N(p).$

Because $|N(p)|=1\not= 0$ , we get that $sign ~det J(u_{\beta}^{-1}\circ u_{\alpha})=1$ or $det J(u_{\beta}^{-1}\circ u_{\alpha})>0$ .

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

#2
xuongrong

Đã gửi 18-09-2006 - 22:53

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anh toilachinhtoi co the noi y nghia~ ba`i 12 trong do Carmo duoc khong? Ong ta xay dung ca'i Manifold dinh huong duoc tu ca'i thang khong dinh huong duoc de lam gi vay? Noi mot cach bay ba, khong biet co phai ham y' la` moi Manifold neu co' ga'ng de^`u dinh huong duoc?

Chém dao xuống nước, nước càng chảy. Nâng chén tiêu sầu, càng sầu thêm.

#3
toilachinhtoi

Đã gửi 25-09-2006 - 07:01

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To xuongrong: Ý nghĩa của bài 12 tôi nghĩ là như sau: Xét http://dientuvietnam.net/cgi-bin/mimetex.cgi?\{(U_{\alpha},u_{\alpha})\} là atlas của http://dientuvietnam.../mimetex.cgi?M. Xét một chart http://dientuvietnam.net/cgi-bin/mimetex.cgi?(U_{\alpha},u_{\alpha}) bất kì. Khi đó tất cả các chart khác chia làm 2 nhóm: nhóm A gồm http://dientuvietnam.net/cgi-bin/mimetex.cgi?(U_{\beta},u_{\beta}) mà http://dientuvietnam...imetex.cgi?N=-M mà những chart của nhóm B sẽ là phủ mở của http://dientuvietnam...mimetex.cgi?-M. Bây giờ atlas của M sẽ tạo thành một sự định hướng của http://dientuvietnam...n/mimetex.cgi?X là một (local) vector field trên http://dientuvietnam.../mimetex.cgi?M. Khi đó http://dientuvietnam...n/mimetex.cgi?X nếu
http://dientuvietnam.net/cgi-bin/mimetex.cgi?X_1,X_2 là những vector field trên M và $d\varphi .X_1=Y_1,d\varphi .X_2=Y_2$ . Chứng minh $d\varphi .[X_1,X_2]=[Y_1,Y_2]$ . Giải:
Fixed $p\in M$ and put $q=\varphi (p)$ . To show that
$d\varphi \circ [X_1,X_2]=[Y_1,Y_2]$ , it is sufficient to show that
for any differentiable function $f:N\rightarrow \RR$ we have
$(d\varphi \circ [X_1,X_2]).f|_q=[Y_1,Y_2].f|_q$ .

By definition we have that
$\begin{array}{l}(d\varphi \circ [X_1,X_2]).f|_q=(d\varphi \circ [X_1,X_2]).f|_{\varphi (p)}=[X_1,X_2].(f\circ \varphi )|_p\\=X_1(X_2.(f\circ \varphi ))|_p-X_2(X_1.(f\circ \varphi ))|_p.\end{array}$
Now if we put $\psi (z)=Y_2.f|_z$ we have by definition and
assumptions
$\begin{array}{l}X_2.(f\circ \varphi )|_z=(d\varphi X_2).f|_{\varphi (z)}\\=Y_2.f|_{\varphi (z)}=\psi \circ \varphi (z).\end{array}{l}$ Hence
$\begin{array}{l}X_1(X_2.(f\circ \varphi ))|_p=X_1.(\psi \circ \varphi )|_p\\ =(d\varphi .X_1)\psi |_{\varphi (p)}=Y_2Y_1f|_q.\end{array}{l}$ Similarly $X_2(X_1.(f\circ \varphi ))|_p=Y_2Y_1f|_q$ . From all above
we are done.

Bài viết đã được chỉnh sửa nội dung bởi toilachinhtoi: 25-09-2006 - 07:10

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

#4
toilachinhtoi

Đã gửi 25-09-2006 - 07:32

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Mẹo 2: Sử dụng đạo hàm riêng nhiều biến để tính đạo hàm một biến.
Ví dụ: Cho http://dientuvietnam...mimetex.cgi?X,Y là hai vector field trên M với các flow là http://dientuvietnam...metex.cgi?u,v,w fixed, http://dientuvietnam...mimetex.cgi?g(p)=-Yf we have $[-Yf](\psi _{-s}\circ \phi _{-u}\circ \psi _{v}\circ \phi _{w}(p))=g\circ \psi _{-s}\circ \phi _{-u}\circ \psi _{v}\circ \phi _{w}(p)$ we get
$\begin{array}{l}\dfrac{\partial ^2}{\partial s^2}H(s,u,v,w)=\dfrac{\partial }{\partial s}[-Yf](\psi _{-s}\circ \phi _{-u}\circ \psi _{v}\circ \phi _{w}(p))=\dfrac{\partial }{\partial s}g(\psi _{-s}\circ \phi _{-u}\circ \psi _{v}\circ \phi _{w}(p))\\=[Y(Yf)]((\psi _{-s}\circ \phi _{-u}\circ \psi _{v}\circ \phi _{w}(p))),\\\dfrac{\partial ^2}{\partial s\partial u}H(s,u,v,w)=\dfrac{\partial }{\partial u}[-Yf](\psi _{-s}\circ \phi _{-u}\circ \psi _{v}\circ \phi _{w}(p))=\dfrac{\partial }{\partial u}g(\psi _{-s}\circ \phi _{-u}\circ \psi _{v}\circ \phi _{w}(p))\\=[X(Yf)]((\psi _{-s}\circ \phi _{-u}\circ \psi _{v}\circ \phi _{w}(p))),\\\dfrac{\partial ^2}{\partial s\partial v}H(s,u,v,w)=\dfrac{\partial }{\partial v}[-Yf](\psi _{-s}\circ \phi _{-u}\circ \psi _{v}\circ \phi _{w}(p))=\dfrac{\partial }{\partial v}g(\psi _{-s}\circ \phi _{-u}\circ \psi _{v}\circ \phi _{w}(p))\\=-[Y(Yf)]((\psi _{-s}\circ \phi _{-u}\circ \psi _{v}\circ \phi _{w}(p))),\\\dfrac{\partial ^2}{\partial s\partial w}H(s,u,v,w)=\dfrac{\partial }{\partial w}[-Yf](\psi _{-s}\circ \phi _{-u}\circ \psi _{v}\circ \phi _{w}(p))=\dfrac{\partial }{\partial w}g(\psi _{-s}\circ \phi _{-u}\circ \psi _{v}\circ \phi _{w}(p))\\=-[X(Yf)]((\psi _{-s}\circ \phi _{-u}\circ \psi _{v}\circ \phi _{w}(p))).\end{array}$ Similarly, putting $g=-X(f\circ \psi _{-u})$ we have
$\begin{array}{l}\dfrac{\partial ^2}{\partial u\partial s}H(s,u,v,w)=\dfrac{\partial ^2}{\partial s\partial u}H(s,u,v,w)=[X(Yf)]((\psi _{-s}\circ \phi _{-u}\circ \psi _{v}\circ \phi _{w}(p)))\\\dfrac{\partial ^2}{\partial u^2}H(s,u,v,w)=\dfrac{\partial }{\partial u}[-X(f\circ \psi _{-u})](\phi _{-u}\circ \psi _{v}\circ \phi _{w}(p))=\dfrac{\partial }{\partial u}g(\phi _{-u}\circ \psi _{v}\circ \phi _{w}(p))\\=[X(X(f\circ \psi _{-u}))](( \phi _{-u}\circ \psi _{v}\circ \phi _{w}(p))),\\\dfrac{\partial ^2}{\partial u\partial v}H(s,u,v,w)=\dfrac{\partial }{\partial v}[-X(f\circ \psi _{-u})](\phi _{-u}\circ \psi _{v}\circ \phi _{w}(p))=\dfrac{\partial }{\partial v}g(\phi _{-u}\circ \psi _{v}\circ \phi _{w}(p))\\=-[Y(X(f\circ \psi _{-u}))](( \phi _{-u}\circ \psi _{v}\circ \phi _{w}(p))),\\\dfrac{\partial ^2}{\partial u\partial w}H(s,u,v,w)=\dfrac{\partial }{\partial w}[-X(f\circ \psi _{-u})](\phi _{-u}\circ \psi _{v}\circ \phi _{w}(p))=\dfrac{\partial }{\partial w}g(\phi _{-u}\circ \psi _{v}\circ \phi _{w}(p))\\=-[X(X(f\circ \psi _{-u}))](( \phi _{-u}\circ \psi _{v}\circ \phi _{w}(p))).\end{array}$ Similarly
$\begin{array}{l}\dfrac{\partial ^2}{\partial v\partial s}H(s,u,v,w)=\dfrac{\partial ^2}{\partial s\partial v}H(s,u,v,w)=-[Y(Yf)]((\psi _{-s}\circ \phi _{-u}\circ \psi _{v}\circ \phi _{w}(p))),\\\dfrac{\partial ^2}{\partial v\partial u}H(s,u,v,w)=\dfrac{\partial ^2}{\partial u\partial v}H(s,u,v,w)=-[Y(X(f\circ \psi _{-u}))](( \phi _{-u}\circ \psi _{v}\circ \phi _{w}(p))),\\\dfrac{\partial ^2}{\partial v^2}H(s,u,v,w)=[Y(Y(f\circ \psi _{-s}\circ \phi _{-u}))](\psi _v\circ \phi _w(p)),\\\dfrac{\partial ^2}{\partial v\partial w}H(s,u,v,w)=[X(Y(f\circ \psi _{-s}\circ \phi _{-u}))](\psi _v\circ \phi _w(p)),\end{array}$ and
$\begin{array}{l}\dfrac{\partial ^2}{\partial w\partial s}H(s,u,v,w)=\dfrac{\partial ^2}{\partial s\partial w}H(s,u,v,w)=--[X(Yf)]((\psi _{-s}\circ \phi _{-u}\circ \psi _{v}\circ \phi _{w}(p))),\\\dfrac{\partial ^2}{\partial w\partial u}H(s,u,v,w)=\dfrac{\partial ^2}{\partial u\partial w}H(s,u,v,w)=--[X(X(f\circ \psi _{-u}))](( \phi _{-u}\circ \psi _{v}\circ \phi _{w}(p))),\\\dfrac{\partial ^2}{\partial w\partial v}H(s,u,v,w)=\dfrac{\partial ^2}{\partial v\partial w}H(s,u,v,w)=[X(Y(f\circ \psi _{-s}\circ \phi _{-u}))](\psi _v\circ \phi _w(p)),\\\dfrac{\partial ^2}{\partial w^2}H(s,u,v,w)=[X(X(f\circ \psi _{-s}\circ \phi _{-u}\circ \psi _v))](\phi _w(p)).\end{array}$ Now
$\begin{array}{l}\dfrac{d^2}{dt^2}|_{t=0}f\circ \alpha (t)=(\dfrac{\partial ^2}{\partial s^2}+2\dfrac{\partial ^2}{\partial s\partial u}+2\dfrac{\partial ^2}{\partial s\partial v}+2\dfrac{\partial ^2}{\partial s\partial w}+\dfrac{\partial ^2}{\partial u^2}+2\dfrac{\partial ^2}{\partial u\partial v}+2\dfrac{\partial ^2}{\partial u\partial w}\\+\dfrac{\partial ^2}{\partial v^2}+2\dfrac{\partial ^2}{\partial v\partial w}+\dfrac{\partial ^2}{\partial w^2})|_{(0,0,0,0)}H(s,u,v,w)\\=(YYf+2XYf-2YYf-2XYf+XXf\\-2YXf-2XXf+YYf+2XYf+XXf)(p)\\=2(XYf-YXf)(p)=2[X,Y]f(p), \end{array}$ and we are done.

Bài viết đã được chỉnh sửa nội dung bởi toilachinhtoi: 25-09-2006 - 07:44

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

#5
xuongrong

Đã gửi 27-09-2006 - 06:40

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tính toán nhìn ù cả tai hoa cả mắt. nhưng công nhận mấy cái mẹo này hay!

Bài viết đã được chỉnh sửa nội dung bởi bookworm_vn: 27-09-2006 - 08:49

Chém dao xuống nước, nước càng chảy. Nâng chén tiêu sầu, càng sầu thêm.

#6
xuongrong

Đã gửi 27-09-2006 - 11:58

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toichinhlatoi computed

$\dfrac{\partial ^2H}{\partial s\partial u}(s,u,v,w) = \dfrac{\partial}{\partial u}\dfrac{\partial}{\partial s} H(s,u,v,w) = \dfrac{\partial}{\partial u} [-Yf]q = X(Yf)q$
But if I compute
$\dfrac{\partial ^2H}{\partial u\partial s}(s,u,v,w) = \dfrac{\partial}{\partial s}\dfrac{\partial}{\partial u} H(s,u,v,w) = \dfrac{\partial}{\partial s} [-Xf]q = Y(Xf)q$
What's wrong? They are supposed to be the same shit, aren't they?

Q: làm sao lúc đánh bài trong diễn đàn mà tắt chế cái tiếng việt được vậy? thanks.

Chém dao xuống nước, nước càng chảy. Nâng chén tiêu sầu, càng sầu thêm.

#7
xuongrong

Đã gửi 27-09-2006 - 12:02

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biết vì sao rồi. thằng sau phải là: $[-X(f \psi_{-s})]$ blah blah...

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#8
xuongrong

Đã gửi 09-10-2006 - 08:10

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(Problem 7 p.47, do Carmo) Let $G$ be a compact connected Lie
group. Prove that $G$ has a bi-invariant Riemannian metric.

Prove by three steps:
a. Let w be a differentiable n-form on G invariant on the left,
that is, http://dientuvietnam...ex.cgi?R^*_aw=w, for any a in G.

Recall the definition of http://dientuvietnam...mimetex.cgi?f^*, for any p in G,

http://dientuvietnam.net/cgi-bin/mimetex.cgi?R^*_aw is left invariant by definition.

\underline{Show that http://dientuvietnam...etex.cgi?R^*_aw and w are n-forms and both left invariant, we must have that http://dientuvietnam.net/cgi-bin/mimetex.cgi?(0,\infty) which contradicts with the compactness of
f(G) (since G is compact and f is continuous.) That is,
$f(a)=1$, for any $a\in G$. Thus, http://dientuvietnam...metex.cgi?<<,>> on G by

http://dientuvietnam.net/cgi-bin/mimetex.cgi?\{(U_k,\phi_k)\} is a partition of unity of $G$ with its subordinate $f_k$ and $x=\phi_k(x_1,...,x_n)$. It is
clear that $<<,>>$ is bilinear, symmetric ,positive definite.

Note that we have http://dientuvietnam...metex.cgi?<<,>> is right invariant.

Thus http://dientuvietnam...metex.cgi?<<,>> is bi-invariant.

có gì sai anh toilachinhtoi hoac moi nguoi sua dum. thanks.

Bài viết đã được chỉnh sửa nội dung bởi xuongrong: 09-10-2006 - 08:11

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#9
toilachinhtoi

Đã gửi 13-10-2006 - 08:48

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To xuongrong: I just read your post today (and not very carefully :leq

). Your solution is good, just one thing: You may use http://dientuvietnam...mimetex.cgi?S^n is the restriction of some linear isometries of http://dientuvietnam...mimetex.cgi?U,V be connected open sets of http://dientuvietnam...mimetex.cgi?f(x)=Ax+b where http://dientuvietnam.../mimetex.cgi?Ax is a linear map from http://dientuvietnam...mimetex.cgi?s>0 such that
http://dientuvietnam.net/cgi-bin/mimetex.cgi?B(a,r) is the open ball of center http://dientuvietnam...n/mimetex.cgi?a and radius http://dientuvietnam...n/mimetex.cgi?r in http://dientuvietnam...n/mimetex.cgi?f is isometry we have that http://dientuvietnam...Df.u,Df.v>_{f(p)}=<u,v>_p
for any http://dientuvietnam.net/cgi-bin/mimetex.cgi?|Df.u|_{f(p)}=|u|_{p}.
For http://dientuvietnam.net/cgi-bin/mimetex.cgi?a=x_0 we get that http://dientuvietnam.net/cgi-bin/mimetex.cgi?f(B(x_0,r))=B(f(x_0),r). Now
http://dientuvietnam.net/cgi-bin/mimetex.cgi?|a-b|=|f(b)-f(a)| for all http://dientuvietnam.net/cgi-bin/mimetex.cgi?f(a+t(b-a)) is of the segment, and moreover since
the lines are the unique shortest curves linking two points in
http://dientuvietnam.net/cgi-bin/mimetex.cgi?\RR^n, http://dientuvietnam.net/cgi-bin/mimetex.cgi?f(a+t(b-a))=f(a)+t(f(b)-f(a)). This means that http://dientuvietnam.net/cgi-bin/mimetex.cgi?f restricted to http://dientuvietnam.net/cgi-bin/mimetex.cgi?B(x_0,r) is an affine map.

Now for each http://dientuvietnam.net/cgi-bin/mimetex.cgi?B(x,r_x) such that
http://dientuvietnam.net/cgi-bin/mimetex.cgi?f|_{B(x,r_x)} is an affine http://dientuvietnam.net/cgi-bin/mimetex.cgi?A_x. If http://dientuvietnam.net/cgi-bin/mimetex.cgi?A_x|_{W}=A_y|_{W} hence
http://dientuvietnam.net/cgi-bin/mimetex.cgi?U is
connected and open, by element arguments of topology, we can show
that http://dientuvietnam.net/cgi-bin/mimetex.cgi?f is affine on http://dientuvietnam.net/cgi-bin/mimetex.cgi?U.

Now we solve Problem 5. Let http://dientuvietnam.net/cgi-bin/mimetex.cgi?f is locally the restriction of some linear
map. Then we can cover http://dientuvietnam.net/cgi-bin/mimetex.cgi?S^n by open sets http://dientuvietnam.net/cgi-bin/mimetex.cgi?W_i such that http://dientuvietnam.net/cgi-bin/mimetex.cgi?f|_{W_i} is linear for each $i$, and use that http://dientuvietnam.net/cgi-bin/mimetex.cgi?S^n is connected to have that http://dientuvietnam.net/cgi-bin/mimetex.cgi?f is the restriction of a linear map from http://dientuvietnam.net/cgi-bin/mimetex.cgi?A,B of http://dientuvietnam.net/cgi-bin/mimetex.cgi?A(Spoint)=x_0 and http://dientuvietnam.net/cgi-bin/mimetex.cgi?B(f(x_0))=Npoint where http://dientuvietnam.net/cgi-bin/mimetex.cgi?Spoint=(1,0,0,...,0) and http://dientuvietnam.net/cgi-bin/mimetex.cgi?Npoint=(0,1,...,0). Note that if http://dientuvietnam.net/cgi-bin/mimetex.cgi?f is isometric of http://dientuvietnam.net/cgi-bin/mimetex.cgi?S^n then also http://dientuvietnam.net/cgi-bin/mimetex.cgi?Spoint and http://dientuvietnam.net/cgi-bin/mimetex.cgi?Npoint (as in do Carmo's book page 21).

We have shown in problem 1 page 46 that http://dientuvietnam.net/cgi-bin/mimetex.cgi?k,l. Hence because
http://dientuvietnam.net/cgi-bin/mimetex.cgi?F is an isometry, we have http://dientuvietnam.net/cgi-bin/mimetex.cgi?dF is represented by the Jacobian http://dientuvietnam.net/cgi-bin/mimetex.cgi?W of http://dientuvietnam.net/cgi-bin/mimetex.cgi?Spoint such that http://dientuvietnam.net/cgi-bin/mimetex.cgi?W is contained in the chart http://dientuvietnam.net/cgi-bin/mimetex.cgi?F(W) is contained in the chart http://dientuvietnam.net/cgi-bin/mimetex.cgi?W is connected and http://dientuvietnam.net/cgi-bin/mimetex.cgi?\varphi is diffeomorphic) to a connected open set http://dientuvietnam.net/cgi-bin/mimetex.cgi?A_0 is an affine map in http://dientuvietnam.net/cgi-bin/mimetex.cgi?A_0(0,...,0)=(0,...,0), http://dientuvietnam.net/cgi-bin/mimetex.cgi?A_0 is a linear map. Now we have
http://dientuvietnam.net/cgi-bin/mimetex.cgi?F|_W is the restriction of some linear map from
http://dientuvietnam.net/cgi-bin/mimetex.cgi?k=1,l=2. Then
http://dientuvietnam.net/cgi-bin/mimetex.cgi?x_k=\sqrt{1-x_1^2-...-x_{k-1}^2-x_{k+1}^2-...-x_{n+1}^2} and
http://dientuvietnam.net/cgi-bin/mimetex.cgi?x_1^2+...+x_{k-1}^2+x_{k+1}^2+...+x_{n+1}^2=y_1^2+...+y_n^2 since http://dientuvietnam.net/cgi-bin/mimetex.cgi?A_0(x_1,...,x_{k-1},x_{k+1},...,x_n,x_{n+1})=(y_1,...,y_n) and
http://dientuvietnam.net/cgi-bin/mimetex.cgi?A_0 is a linear isometric map. Hence http://dientuvietnam.net/cgi-bin/mimetex.cgi?F is the restriction of a
linear isometric map and also for $f=B^{-1}\circ F\circ A^{-1}$

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

#10
toilachinhtoi

Đã gửi 17-10-2006 - 07:59

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343 Bài viết

Problem 6/Chapter 1 (do Carmo): Prove that locally isometry is not a symmetric relation.

Proof:

Claim: There is no open sets http://dientuvietnam...n/mimetex.cgi?U has hyperbolic metric
(see problem 4 above) and http://dientuvietnam...n/mimetex.cgi?V has Euclidean metric.

Proof of Claim: Assume that the claim is not true. Then there exists
open sets http://dientuvietnam...n/mimetex.cgi?U has
hyperbolic metric (see problem 4 above) and http://dientuvietnam...n/mimetex.cgi?V has Euclidean
metric.

Then we will have the following system of equations:
http://dientuvietnam.net/cgi-bin/mimetex.cgi?U and http://dientuvietnam...n/mimetex.cgi?V are defined by http://dientuvietnam...tex.cgi?e_1,e_2 are the standard basis of http://dientuvietnam.net/cgi-bin/mimetex.cgi?\varphi is an isometry we get that http://dientuvietnam...?dy^2 y^2dx^2=0, which has
solutions http://dientuvietnam...e^{ix}=constant and http://dientuvietnam.net/cgi-bin/mimetex.cgi?ye^{-ix}=constant. Using the new variables http://dientuvietnam.net/cgi-bin/mimetex.cgi?v(x,y)=G_1(ye^{ix})+H_1(ye^{-ix}) for some functions
http://dientuvietnam.net/cgi-bin/mimetex.cgi?G_1,H_1 of one variable. Similarly
http://dientuvietnam.net/cgi-bin/mimetex.cgi?u(x,y)=G_2(ye^{ix})+H_2(ye^{-ix}). Now use above formular for http://dientuvietnam.net/cgi-bin/mimetex.cgi?u and http://dientuvietnam.net/cgi-bin/mimetex.cgi?v and that
http://dientuvietnam.net/cgi-bin/mimetex.cgi?G'H'=\dfrac{1}{y}. Substituting
this formula into
http://dientuvietnam.net/cgi-bin/mimetex.cgi?1-2y-\dfrac{2}{y}=-(\dfrac{1}{y^2}-\dfrac{2}{y}-2y), which is a
contradiction with the assumption http://dientuvietnam.net/cgi-bin/mimetex.cgi?U is open in http://dientuvietnam.net/cgi-bin/mimetex.cgi?H^2 the hyperbolic space (problem 4). Then the inclusion http://dientuvietnam.net/cgi-bin/mimetex.cgi?M into http://dientuvietnam.net/cgi-bin/mimetex.cgi?N. But we don't have a locally isometry at http://dientuvietnam.net/cgi-bin/mimetex.cgi?M by the Claim above.

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

#11
Kakalotta

Đã gửi 17-10-2006 - 15:25

Thèm lấy vợ

Thành viên
805 Bài viết

Phải công nhận topic này xứng đáng được giải vô địch trong việc luyện đánh Latex. Trao đổi toán học thì quan trọng là ý tưởng chính bề sâu chứ cứ Latex thế này thì xin vái cả nón. Ví dụ cái bài nhóm Lie compact, chỉ cần nói một câu đơn giản là trên đại số Lie compact thì có dạng song tuyến tính sinh bởi Killing form, và nó xác định dương. Tịnh tiến nó đi khắp nhóm Lie thì ta thu được Metric, thế thôi mà nhìn phát sợ lên được. Cái bài về tích Lie của hai vector field thì chỉ cần nói đơn giản là địa phương hóa rồi khai triển taylor của mấy cái hàm exp của cái véctor field được xem như là các toán tử tác động lên đại số hàm, thế là xong chứ làm gì mà toàn đạo hàm với chả vi phân, nhức cả mắt.

PhDvn.org

#12
Kakalotta

Đã gửi 17-10-2006 - 15:40

Thèm lấy vợ

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805 Bài viết

Một mặt S là định hướng được nếu và chỉ nếu nó có một trường (smooth) vécto pháp tuyến đơn vị.

Bài này thì hiển nhiên véc tơ pháp tuyến cùng với tích ngoài trên không gian tiếp xúc (cảm sinh từ dạng Kahler trên R4) sẽ sinh ra dạng volume form , nên định hướng được, thế là xong.

PhDvn.org

#13
phtung

Đã gửi 18-10-2006 - 02:21

Trung sĩ

Thành viên
166 Bài viết

Cái môn này mỗi lần đọc sách nhìn toét cả mắt, học xong cũng cóc hiểu mình vừa học được cái gì, mỗi lần cố gắng tưởng tượng là lại chả ra được cái quái gì cả.
Thế mà có người ngồi type cả đống thế kia, kinh thật.

#14
toilachinhtoi

Đã gửi 18-10-2006 - 06:34

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In fact, these problems are solved using the material upto their points in do Carmo books, so it gets that long. Also, these are embeded from Tex files on my computer, whose soure are the homeworks that I solved, hence it takes just a short time for me to transform from the signs $$, or \begin{eqnarray*}\end{eqnarray*} in to the signs $$ here.

Anyway, this topic is to help people who are the beginners to Differential Geometry, and I think it is useful for those ones.

Also, according to your level, you will have the corresponding problems to solve. Anyone who feels his knowledge is much, should try to do the problems that the whole world are trying to solve, and should skip this topic.

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

#15
xuongrong

Đã gửi 18-10-2006 - 07:06

Hạ sĩ

Thành viên
89 Bài viết

According to my level which is so beginer, i found in this topic useful to me (but obviously not to my eyes, by the way). I vote this topic to be continued!

Chém dao xuống nước, nước càng chảy. Nâng chén tiêu sầu, càng sầu thêm.

#16
xuongrong

Đã gửi 18-10-2006 - 07:21

Hạ sĩ

Thành viên
89 Bài viết

In fact, these problems are solved using the material upto their points in do Carmo books, so it gets that long. Also, these are embeded from Tex files on my computer, whose soure are the homeworks that I solved, hence it takes just a short time for me to transform from the signs $$, or \begin{eqnarray*}\end{eqnarray*} in to the signs $$ here.

Anyway, this topic is to help people who are the beginners to Differential Geometry, and I think it is useful for those ones.

Also, according to your level, you will have the corresponding problems to solve. Anyone who feels his knowledge is much, should try to do the problems that the whole world are trying to solve, and should skip this topic.

có một cách không phải copy từ Tex trong máy bỏ vào Tex của diễn đàn đó là upload file ảnh dvi lên luôn. kĩ thuật này thấy bookworm_vn hôm nọ làm. hy vọng bookworm_vn ghé qua chỗ này chỉ vài chiêu. tui chịu.

Chém dao xuống nước, nước càng chảy. Nâng chén tiêu sầu, càng sầu thêm.

#17
Kakalotta

Đã gửi 18-10-2006 - 23:56

Thèm lấy vợ

Thành viên
805 Bài viết

According to my level which is so beginer, i found in this topic useful to me (but obviously not to my eyes, by the way). I vote this topic to be continued!

.
Không ai phản đối việc tiếp tục cả.

Có ai phản đối việc tiếp tục đâu cơ chứ, có điều thay đổi chiều hướng phát triển, gõ ít latex thôi, học cái môn này quan trọng là ý tưởng chứ không phải là sử dụng công cụ giải tích để hiếp dâm hình học.
Phản đối việc phá hoại vẻ đẹp toán học bằng việc sử dụng Latex tràn lan.

NangLuong: Anh Hạnh dùng từ sợ quá. Em giấu mấy từ ấy đi nhé không thì các em gái đang học hình học sẽ không dám chơi với các anh học giải tích nữa đâu.

PhDvn.org

#18
toilachinhtoi

Đã gửi 22-10-2006 - 10:45

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343 Bài viết

Problem 2/Chapter 2: Let http://dientuvietnam...metex.cgi?c(t_0)=p,~\dfrac{d}{dt}c(t)=X(c(t)). Prove that
http://dientuvietnam.net/cgi-bin/mimetex.cgi?P_{c,t_0,t} is the parallel transport along http://dientuvietnam...mimetex.cgi?c(t) (see problem 1/Chapter 2).

Proof:

We define http://dientuvietnam...metex.cgi?Y(c(t))=P_{c,t_0,t}Z(t). Since http://dientuvietnam...mimetex.cgi?c(t) is the integra curve of http://dientuvietnam...n/mimetex.cgi?X, and the connection is Riemann we have
http://dientuvietnam.net/cgi-bin/mimetex.cgi?e_1,...e_n an orthonormal basis of http://dientuvietnam...imetex.cgi?T_pM we have
http://dientuvietnam.net/cgi-bin/mimetex.cgi?P_{c,t_0,t}e_1,...,P_{c,t_0,t}e_n is also an orthornormal basis of
http://dientuvietnam.net/cgi-bin/mimetex.cgi?T_{c(t)}M by problem 1 above. Hence
http://dientuvietnam.net/cgi-bin/mimetex.cgi?\dfrac{D}{dt}(P_{c,t_0,t}e_k)=0 for all http://dientuvietnam.net/cgi-bin/mimetex.cgi?k=1,2,...,n.
Hence
$\nabla _XY(p)=\dfrac{D}{dt}P_{c,t_0,t}Z(t)|_{t=t_0}=\sum _{k=1}^n\dfrac{d}{dt}(<Z(t),e_k>)|_{t=t_0}P_{c,t_0,t_0}e_k=\sum _{k=1}^n\dfrac{d}{dt}(<Z(t),e_k>)|_{t=t_0}e_k=\dfrac{d}{dt}(\sum _{k=1}^n<Z(t),e_k>e_k)|_{t=t_0}=\dfrac{d}{dt}Z(t)|_{t=t_0}, $ and we are done.

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

#19
Doraemon

Đã gửi 26-10-2006 - 00:51

Mèo Ú

Hiệp sỹ
239 Bài viết

Hỏi chen vào một chút: quyển do Carmo này đầu đề cụ thể là gì đấy các bác? Cứ bàn mà em chả biết quyển gì?

Thân lừa ưa cử tạ !

#20
xuongrong

Đã gửi 26-10-2006 - 02:30

Hạ sĩ

Thành viên
89 Bài viết

Hỏi chen vào một chút: quyển do Carmo này đầu đề cụ thể là gì đấy các bác? Cứ bàn mà em chả biết quyển gì?

do Carmo, Riemannian Geometry. Translated by Flaherty. It's free on www.lookforbook.com

Chém dao xuống nước, nước càng chảy. Nâng chén tiêu sầu, càng sầu thêm.

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#1 toilachinhtoi Đã gửi 18-09-2006 - 08:38

#2 xuongrong Đã gửi 18-09-2006 - 22:53

#3 toilachinhtoi Đã gửi 25-09-2006 - 07:01

#4 toilachinhtoi Đã gửi 25-09-2006 - 07:32

#5 xuongrong Đã gửi 27-09-2006 - 06:40

#6 xuongrong Đã gửi 27-09-2006 - 11:58

#7 xuongrong Đã gửi 27-09-2006 - 12:02

#8 xuongrong Đã gửi 09-10-2006 - 08:10

#9 toilachinhtoi Đã gửi 13-10-2006 - 08:48

#10 toilachinhtoi Đã gửi 17-10-2006 - 07:59

#11 Kakalotta Đã gửi 17-10-2006 - 15:25

#12 Kakalotta Đã gửi 17-10-2006 - 15:40

#13 phtung Đã gửi 18-10-2006 - 02:21

#14 toilachinhtoi Đã gửi 18-10-2006 - 06:34

#15 xuongrong Đã gửi 18-10-2006 - 07:06

#16 xuongrong Đã gửi 18-10-2006 - 07:21

#17 Kakalotta Đã gửi 18-10-2006 - 23:56

#18 toilachinhtoi Đã gửi 22-10-2006 - 10:45

#19 Doraemon Đã gửi 26-10-2006 - 00:51

#20 xuongrong Đã gửi 26-10-2006 - 02:30

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#1
toilachinhtoi

Đã gửi 18-09-2006 - 08:38

#2
xuongrong

Đã gửi 18-09-2006 - 22:53

#3
toilachinhtoi

Đã gửi 25-09-2006 - 07:01

#4
toilachinhtoi

Đã gửi 25-09-2006 - 07:32

#5
xuongrong

Đã gửi 27-09-2006 - 06:40

#6
xuongrong

Đã gửi 27-09-2006 - 11:58

#7
xuongrong

Đã gửi 27-09-2006 - 12:02

#8
xuongrong

Đã gửi 09-10-2006 - 08:10

#9
toilachinhtoi

Đã gửi 13-10-2006 - 08:48

#10
toilachinhtoi

Đã gửi 17-10-2006 - 07:59

#11
Kakalotta

Đã gửi 17-10-2006 - 15:25

#12
Kakalotta

Đã gửi 17-10-2006 - 15:40

#13
phtung

Đã gửi 18-10-2006 - 02:21

#14
toilachinhtoi

Đã gửi 18-10-2006 - 06:34

#15
xuongrong

Đã gửi 18-10-2006 - 07:06

#16
xuongrong

Đã gửi 18-10-2006 - 07:21

#17
Kakalotta

Đã gửi 18-10-2006 - 23:56

#18
toilachinhtoi

Đã gửi 22-10-2006 - 10:45

#19
Doraemon

Đã gửi 26-10-2006 - 00:51

#20
xuongrong

Đã gửi 26-10-2006 - 02:30