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几何 >> 微分几何 >> 外微分
Questions in category: 外微分 (Exterior differential).

1

对于一般的 $s$ 次微分形式, 证明 $d\omega$ 是函数线性的, 即为张量场.

Posted by haifeng on 2015-08-09 15:48:33 last update 2015-08-09 15:48:33 | Answers (0) | 收藏

对于一般的 $s$ 次微分形式, 证明 $d\omega$ 是函数线性的, 即为张量场.

2

$f^{*}\Omega=(\det A)\Omega$

Posted by haifeng on 2015-08-09 15:29:59 last update 2015-08-09 15:38:37 | Answers (0) | 收藏

如果 $f:\mathbb{R}^n\rightarrow\mathbb{R}^n$ 为线性映射, 则 $f^{*}\Omega=(\det A)\Omega$, 其中 $\Omega$ 为 $\mathbb{R}^n$ 的标准体积形式, 即 $\Omega=dx^1\wedge dx^2\wedge\cdots\wedge dx^n$, $A$ 为 $f$ 在标准基下的矩阵表示.

3

关于微分形式李导数的几个性质

Posted by haifeng on 2013-07-06 13:49:13 last update 2013-07-08 17:29:21 | Answers (8) | 收藏

设 $M$ 为 $n$ 维光滑黎曼流形, $A(M)$ 是 $M$ 上微分形式集合. $\omega\in A(M)$. $X,Y,Y_i$ 均是 $M$ 上的向量场.
 

(1) $L_X\circ i(Y)-i(Y)\circ L_X=i([X,Y])$

(2) $L_X\circ L_Y-L_Y\circ L_X=L_{[X,Y]}$

(3) $d\circ i(X)+i(X)\circ d=L_X$

(4) $d\circ L_X=L_X\circ d$

(5) $i(X)L_X=L_X i(X)$

(6) $L_{fX}\omega=fL_X \omega+df\wedge i(X)\omega$, 其中 $f\in A^0(M)$, $\omega\in A(M)$.

(7) $L_X i(Y)-L_Y i(X)-i([X,Y])=[d,i(X)\circ i(Y)]$


第一题要用到下面的结论

(0) 设 $\omega\in A^r(M)$, 则

\[
(L_X\omega)(Y_1,\ldots,Y_r)=X\bigl(\omega(Y_1,\ldots,Y_r)\bigr)-\sum_{i=1}^{r}\omega(Y_1,\ldots,L_X Y_i,\ldots,Y_r)
\]

这是基础.


以上一些结论可以推广到一般的 $(0,p)$-型张量.

4

光滑流形 $M$ 上的 $r$-次微分形式的微分运算

Posted by haifeng on 2013-07-06 13:22:20 last update 2013-07-06 13:35:20 | Answers (0) | 收藏

设 $\omega\in A^r(M)$, $X_1,\ldots,X_{r+1}$ 是 $M$ 上任意 $r+1$ 个光滑切向量场, 则

\[
\begin{split}
d\omega(X_1,\ldots,X_{r+1})&=\sum_{i=1}^{r+1}(-1)^{i+1}\bigl(\langle X_1\wedge\cdots\wedge\hat{X}_i\wedge\cdots\wedge X_{r+1},\omega\rangle\bigr)\\
&\quad +\sum_{1\leq i < j\leq r+1}\bigl\langle [X_i,X_j]\wedge\cdots\wedge\hat{X}_i\wedge\cdots\wedge\hat{X}_j\wedge\cdots\wedge X_{r+1},\omega\bigr\rangle
\end{split}
\]

5

外微分运算的宏包

Posted by haifeng on 2012-07-26 10:21:47 last update 2012-07-26 10:22:52 | Answers (0) | 收藏

http://www.inp.demokritos.gr/~sbonano/EDC/

Exterior Differential Calculus

and

Symbolic Matrix Algebra @ Mathematica

 

This package enables Mathematica to carry out calculations with differential forms. It defines the two basic operations - Exterior Product (Wedge) and Exterior Derivative (d)  - in such a way that:

  (1)  they can act on any valid Mathematica expression
  (2)  they allow the use of any symbols to denote differential forms
  (3)  input - output notation is as close as possible to standard usage

 

6

设 $\varphi_1,\ldots,\varphi_r\in V^*=\Lambda^1(V)$. 证明 $\varphi_1,\ldots,\varphi_r$ 线性相关当且仅当 $\varphi_1\wedge\cdots\wedge\varphi_r=0$.

Posted by haifeng on 2012-07-26 09:30:40 last update 2012-07-26 09:34:26 | Answers (1) | 收藏

设 $\varphi_1,\ldots,\varphi_r\in V^*=\Lambda^1(V)$. 证明 $\varphi_1,\ldots,\varphi_r$ 线性相关当且仅当

\[\varphi_1\wedge\cdots\wedge\varphi_r=0.\]


与之对偶的结论是:

矢量 $v_1,\ldots,v_r\in V$ 线性相关的充要条件是

\[v_1\wedge\cdots\wedge v_r=0.\]

(参见 [1,pp.59])


References:

[1] 陈省身、陈维桓 著 《微分几何讲义》第二版, 北京大学出版社, 2001.

7

设 $\varphi_i\in\Lambda^1(V)$, $i=1,2,\ldots,k$. 证明: $\varphi_1\wedge\cdots\wedge\varphi_k(v_1,\ldots,v_k)=\det(\varphi_i(v_j))$.

Posted by haifeng on 2012-07-26 09:15:30 last update 2013-07-05 16:41:13 | Answers (0) | 收藏

设 $\varphi_i\in\Lambda^1(V)$, $i=1,2,\ldots,k$. 证明:

\[\varphi_1\wedge\cdots\wedge\varphi_k(v_1,\ldots,v_k)=\det(\varphi_i(v_j)).\]


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