Machine-translated from Chinese.
外代数
推论 1. 若$r>n,$ 则$\Lambda^r(V^\ast )=\{0\}.$ 若$0\le r\le n,$ 则$\dim \Lambda^r(V^\ast )=\binom{n}{r},$ 以$\{\omega^{i_1}\wedge\cdots\wedge \omega^{i_r}|1\le i_1<\cdots<i_r\le n\}$为基.
证:
$\,\forall\,\Phi\in \Lambda^r(V^\ast )\subset V^\ast \otimes \cdots\otimes V^\ast =\mathcal{L}(V,\cdots,V,\mathbb{R}),$
$\Phi=\Phi_{i_1\cdots i_r}\omega^{i_1}\otimes\cdots\otimes\omega^{i_r}.$
那么
$$\Phi=A(\Phi)=\frac{1}{r!}\Phi_{i_1\cdots i_r} \omega^{i_1}\wedge\cdots\wedge \omega^{i_n}=\sum_{1\le i_1<\cdots<i_r\le n}a_{i_1\cdots i_r}\omega^{i_1}\wedge\cdots\wedge \omega^{i_r}.$$
因此$r>n$时, 求和集为空集, $\Phi\equiv 0.$ $0\le r\le n$时,
只需证$\{\omega^{i_1}\wedge\cdots\wedge \omega^{i_r}|1\le i_1<\cdots<i_r\le n\}$线性无关.
利用$\omega^{i_1}\wedge\cdots\wedge \omega^{i_r}(e_{i_1},\cdots,e_{i_r})=\det(\omega^{i_\alpha}e_{i_\beta})=1,$
其它作用为$0$即可.
推论 2. 设$\{\theta^i\},\{\omega^i\}$是$V^\ast $两组基, $\theta^i=a^i_j\omega^j.$ 则$\theta^1\wedge\cdots\wedge \theta^n=\det(a^i_j)\omega^1\wedge\cdots\wedge\omega^n.$
证: 取$\{\omega^i\}$的对偶基$\{e_i\},$
往等式两侧作用$(e_1,\cdots,e_n)$即可.
$$\mathrm{LHS}=\det(\theta^ie_j)=\det(a^i_k\omega^ke_j)=\det(a^i_j)=\mathrm{RHS}.$$
引理 3 (Cartan引理). 设$\omega^i,\theta^i\in V^\ast ,$ $1\le i\le r\le \dim V.$ 设$\{\omega^i\}$线性无关, 则:
$$\sum_{i=1}^r\omega^i\wedge \theta^i=0\,\Leftrightarrow\, \theta^i=\sum_{j=1}^ra_{ij}\omega^j,\quad a_{ij}=a_{ji}.$$
证: 将$\{\omega^i\}$扩充为一组基. 那么$\theta^i=a_{ij}\omega^j,$
$\omega^i\wedge\theta^i=a_{ij}\omega^i\wedge\omega^j.$ 这样,
$$\sum_{i=1}^r \omega^i\wedge\theta^i=\sum_{1\le i<j\le r}(a_{ij}-a_{ji})\omega^i\wedge\omega^j+\sum_{1\le i\le r<j}a_{ij}\omega^i\wedge\omega^j.$$
注意右侧为用$\Lambda^2(V^\ast )$的基来表示. 由此证得等价性.
定理 4. 设$\omega^1,\cdots,\omega^r\in V^\ast $线性无关, $\Phi\in \Lambda^s(V^\ast ),$ 则$\,\exists\,\psi^1,\cdots,\psi^r\in \Lambda^{s-1}(V^\ast ),$ 使$\Phi=\omega^1\wedge\psi^1+\cdots+\omega^r\wedge\psi^r$ $\Leftrightarrow$ $\omega^1\wedge\cdots\wedge \omega^r\wedge \Phi=0.$
证: 只需证明充分性.
设有$\omega^1\wedge\cdots\wedge\omega^r\wedge \Phi=0,$ 将$\Phi$用基表示,
$$\Phi=c_{i_1\cdots i_s}\omega^{i_1}\wedge\cdots\wedge\omega^{i_s}=\omega^1\wedge\psi^1+\cdots+\omega^r\wedge\psi^r+\sum_{r+1\le \alpha_1<\cdots<\alpha_s\le n}a_{\alpha_1\cdots\alpha_s}\omega^{\alpha_1}\wedge\cdots\wedge\omega^{\alpha_s}.$$
当$r+s>n$时, 后侧求和集为空集; 当$r+s\le n$时, 利用条件,
与$\omega^1\wedge\cdots\wedge\omega^r$做外积.
由线性无关性即有后面的系数全为零.
命题 5. 设$f\in \mathcal{L}(V;W),$ $f^\ast :\Lambda^r(W^\ast )\rightarrow\Lambda^r(V^\ast )$为拉回映照, 则$f^\ast (\Phi\wedge\Psi)=f^\ast \Phi\wedge f^\ast \Psi.$
证: 利用$f^\ast $与$A,\otimes$的交换性即可:
$$\begin{aligned}
f^*(\Phi\wedge\Psi)(x_1,\cdots,x_{r+s})&=(\Phi\wedge\Psi)(f(x_1),\cdots,f(x_{r+s}))\\
&=\frac{(r+s)!}{r!s!}A(\Phi\otimes \Psi) (f(x_1),\cdots,f(x_{r+s}))\\
&=\frac{(r+s)!}{r!s!}A(f^*(\Phi\otimes \Psi))(x_1,\cdots,x_{r+s})\\
&=f^*\Phi \wedge f^*\Psi (x_1,\cdots,x_{r+s}).
\end{aligned}$$
做形式直和$\Lambda(V^\ast )=\bigoplus_{r=0}^n \Lambda^r(V^\ast )=\{\omega_0+\cdots+\omega_n|\omega_r\in \Lambda^r(V^\ast )\}.$
那么它成为$2^n$维向量空间, 并且以$\wedge$为满足分配律的乘法, 成为代数.
称之为$V^\ast $上的外代数. 类似地,
可以定义$\Lambda(V)=\bigoplus_{r=0}^n \Lambda^r(V),$
称为$V$上外代数.
张量丛
令$M^m$为$m$维$C^\infty$流形, $\,\forall\,p\in M,$ 有切空间$T_pM.$
我们定义$(r,s)$-型张量空间$T_s^r(p)=T_pM\otimes\cdots\otimes T_pM\otimes T_p^\ast M\otimes\cdots\otimes T_p^\ast M.$
$\dim T_s^r(p)=m^{r+s}.$ 定义丛空间$T_s^r(M)=\bigcup_{p\in M}T_s^r(p),$
可赋予自然拓扑与微分结构, 使之成为$m+m^{r+s}$维$C^\infty$流形.
$T_s^r(q)$有自然基$\{\frac{\partial {} }{\partial {}x^{i_1} }\otimes\cdots\otimes\frac{\partial {} }{\partial {}x^{i_r} }\otimes dx^{j_1}\otimes \cdots\otimes dx^{j_s}|1\le i_1,\cdots,i_r,j_1,\cdots,j_s\le n\}.$
$\Phi\in T_s^r$有坐标$(\Phi^{i_1\cdots i_r}_{j_1\cdots j_s}).$
取$\pi:T_s^r(M)\rightarrow M$为自然投影,
$\Psi:\pi^{-1}(U)\rightarrow U\times \mathbb{R}^N$将$\Phi$映至$(q,(\Phi^{i_1\cdots i_r}_{j_1\cdots j_s})).$
那么定义
$$\widetilde{\Psi}=(\varphi\times \mathrm{id}_{\mathbb{R}^N})\circ \Psi:\pi^{-1}(U)\rightarrow U\times \mathbb{R}^N\rightarrow \varphi(U)\times \mathbb{R}^N\subset \mathbb{R}^{m+N}.$$
赋予$T_s^r(M)$的拓扑使$\pi^{-1}(U)$是$T_s^r(M)$的开子集且$\widetilde{\Psi}$为同胚.
若$\{(U_\alpha,\varphi_\alpha)\}$为$M$的光滑坐标图册,
则$\{(\pi^{-1}(U_\alpha),\widetilde{\Psi}_\alpha)\}$为$T_s^r(M)$的光滑坐标图册.
它满足如下性质, $\pi:T_s^r(M)\rightarrow M$为$C^\infty$满射,
$\pi^{-1}(q)=T_s^r(q)$为$m^{r+s}$维向量空间,
$\Psi:\pi^{-1}(U)\rightarrow U\times \mathbb{R}^N$为$C^\infty$映照,
$\pi_1\circ \Psi=\pi,$ 并且$\Psi|_q$为线性同构.
这些要素使其成为向量丛, 称其为张量丛.
张量场
若有$\Phi:M\rightarrow T_s^r(M),$ 满足$\pi\circ\Phi=\mathrm{id}_M,$
则称$\Phi$为$M$上$(r,s)$-型张量场. 当$\Phi$为光滑映照时,
称其为光滑$(r,s)$-型张量场. 这当且仅当$\Phi$的每个分量为光滑函数.
$\Phi\in T_s^r(p)=\mathcal{L}(T_p^\ast M,\cdots,T_p^\ast M,T_pM,\cdots,T_pM;\mathbb{R})$可视为$(r+s)$-重线性函数.
考虑$(0,s)$-型张量场$\Phi.$ 任取$X_1,\cdots,X_s\in \mathfrak{X}(M),$
定义张量场在向量场上的作用:
$$[\Phi(X_1,\cdots,X_s)] (p):=\Phi(p)(X_1(p),\cdots,X_s(p)),$$
那么$\Phi(X_1,\cdots,X_s)\in C^\infty(M).$ 因为各个分量为光滑函数,
做线性组合或相乘后仍是光滑函数.
由于张量是多重线性函数, 任取$f_1,\cdots,f_s\in C^\infty(M),$
逐点考虑得到
$$\Phi(f_1X_1,\cdots,f_sX_s)=f_1\cdots f_s\Phi(X_1,\cdots,X_s)\in C^\infty(M).$$
反过来,
若有映照$\Phi:\mathfrak{X}(M)\times \cdots\times \mathfrak{X}(M)\rightarrow C^\infty(M)$是$C^\infty(M)$线性的,
则$\Phi$必是$(0,s)$-型$C^\infty$-张量场.
一般地, 记$\Gamma(T_s^r(M))$表示$C^\infty$的$(r,s)$-型张量场全体,
则它是无穷维向量空间, $f\Phi+g\Psi(p):=f(p)\Phi(p)+g(p)\Psi(p).$
类似前面, 它作用在余切向量场与向量场的组合上,
并且也是$C^\infty(M)$线性的. 反过来也对.
外微分形式
$M^m$为$C^\infty$-流形.
令$\Lambda^r(T^\ast M)=\bigcup_{p\in M}\Lambda^r(T_p^\ast M),$ 类似地,
赋予其自然拓扑结构与微分结构使其成为$m+\binom{m}{r}$维$C^\infty$流形,
称为$M$上$r$-次外形式丛.
若$\Phi:M\rightarrow \Lambda^rT^\ast M$是光滑映照,
$\pi\circ\Phi=\mathrm{id}_M,$ 则称$\Phi$是$r$-次外微分形式.
这当且仅当它的各个分量是光滑函数.
令$\mathcal{A}^r(M)=\Gamma(\Lambda^rT^\ast M)$为$r$-次外微分形式全体,
为无穷维向量空间. 类似地,
也可以定义$f\omega_1+g\omega_2\in \mathcal{A}^r(M).$
由于我们在$\Lambda^r T^\ast M$上有乘法$\wedge,$
我们可以逐点定义$\wedge:\mathcal{A}^r(M)\times \mathcal{A}^s(M)\rightarrow \mathcal{A}^{r+s}(M).$
即
$$(\omega\wedge\sigma)(p)=\omega(p)\wedge \sigma(p)\in \Lambda^{r+s}(T_p^\ast M).$$
它当然同样满足分配律, 反交换律与结合律.
设$F:M\rightarrow N$为光滑映照,
则${F_{\ast } }_p:T_pM\rightarrow T_{F(p)}N$为切映照,
$F^\ast :T^\ast _{F(p)}N\rightarrow T_p^\ast (M)$为拉回映照,
诱导到$\Lambda^r T_q^\ast N\rightarrow \Lambda^r T_p^\ast M$上,
再诱导了$\mathcal{A}^r(N)\rightarrow \mathcal{A}^r(M)$上的拉回,
满足$F^\ast (\omega\wedge\sigma)=F^\ast \omega\wedge F^\ast \sigma.$
文章最后更新于 2021-12-08 15:50:09
Exterior Algebra
Corollary 1. If $r>n,$ then $\Lambda^r(V^\ast )=\{0\}.$ if $0\le r\le n,$ then $\dim \Lambda^r(V^\ast )=\binom{n}{r},$ based on $\{\omega^{i_1}\wedge\cdots\wedge \omega^{i_r}|1\le i_1<\cdots<i_r\le n\}$.
Certificate:
$\,\forall\,\Phi\in \Lambda^r(V^\ast )\subset V^\ast \otimes \cdots\otimes V^\ast =\mathcal{L}(V,\cdots,V,\mathbb{R}),$
$\Phi=\Phi_{i_1\cdots i_r}\omega^{i_1}\otimes\cdots\otimes\omega^{i_r}.$
Then
$$\Phi=A(\Phi)=\frac{1}{r!}\Phi_{i_1\cdots i_r} \omega^{i_1}\wedge\cdots\wedge \omega^{i_n}=\sum_{1\le i_1<\cdots<i_r\le n}a_{i_1\cdots i_r}\omega^{i_1}\wedge\cdots\wedge \omega^{i_r}.$$
Therefore, when $r>n$, the summation set is the empty set, and when $\Phi\equiv 0.$ $0\le r\le n$,
Just prove that $\{\omega^{i_1}\wedge\cdots\wedge \omega^{i_r}|1\le i_1<\cdots<i_r\le n\}$ is linearly independent.
Utilize$\omega^{i_1}\wedge\cdots\wedge \omega^{i_r}(e_{i_1},\cdots,e_{i_r})=\det(\omega^{i_\alpha}e_{i_\beta})=1,$
Other functions can be $0$.
Corollary 2. Assume $\{\theta^i\},\{\omega^i\}$ is the two bases of $V^\ast $, $\theta^i=a^i_j\omega^j.$ then $\theta^1\wedge\cdots\wedge \theta^n=\det(a^i_j)\omega^1\wedge\cdots\wedge\omega^n.$
Certificate: Take the dual basis $\{e_i\},$ of $\{\omega^i\}$
Just apply $(e_1,\cdots,e_n)$ to both sides of the equation.
$$\mathrm{LHS}=\det(\theta^ie_j)=\det(a^i_k\omega^ke_j)=\det(a^i_j)=\mathrm{RHS}.$$
Lemma 3 (Cartan’s lemma). Assume $\omega^i,\theta^i\in V^\ast ,$ $1\le i\le r\le \dim V.$ and $\{\omega^i\}$ are linearly independent, then:
$$\sum_{i=1}^r\omega^i\wedge \theta^i=0\,\Leftrightarrow\, \theta^i=\sum_{j=1}^ra_{ij}\omega^j,\quad a_{ij}=a_{ji}.$$
Certificate: Expand $\{\omega^i\}$ to a set of basis. Then $\theta^i=a_{ij}\omega^j,$
$\omega^i\wedge\theta^i=a_{ij}\omega^i\wedge\omega^j.$ In this way,
$$\sum_{i=1}^r \omega^i\wedge\theta^i=\sum_{1\le i<j\le r}(a_{ij}-a_{ji})\omega^i\wedge\omega^j+\sum_{1\le i\le r<j}a_{ij}\omega^i\wedge\omega^j.$$
Note that the right side is represented by the basis of $\Lambda^2(V^\ast )$. This proves the equivalence.
Theorem 4. Assume $\omega^1,\cdots,\omega^r\in V^\ast $ is linearly independent, $\Phi\in \Lambda^s(V^\ast ),$ then $\,\exists\,\psi^1,\cdots,\psi^r\in \Lambda^{s-1}(V^\ast ),$ makes $\Phi=\omega^1\wedge\psi^1+\cdots+\omega^r\wedge\psi^r$ $\Leftrightarrow$ $\omega^1\wedge\cdots\wedge \omega^r\wedge \Phi=0.$
Certificate: Just prove sufficiency.
Let $\omega^1\wedge\cdots\wedge\omega^r\wedge \Phi=0,$ represent $\Phi$ in base,
$$\Phi=c_{i_1\cdots i_s}\omega^{i_1}\wedge\cdots\wedge\omega^{i_s}=\omega^1\wedge\psi^1+\cdots+\omega^r\wedge\psi^r+\sum_{r+1\le \alpha_1<\cdots<\alpha_s\le n}a_{\alpha_1\cdots\alpha_s}\omega^{\alpha_1}\wedge\cdots\wedge\omega^{\alpha_s}.$$
When $r+s>n$, the backside summation set is the empty set; when $r+s\le n$, using the condition,
Take the outer product with $\omega^1\wedge\cdots\wedge\omega^r$.
Due to linear independence, the following coefficients are all zero.
Proposition 5. Assume $f\in \mathcal{L}(V;W),$ $f^\ast :\Lambda^r(W^\ast )\rightarrow\Lambda^r(V^\ast )$ is the pullback mapping, then $f^\ast (\Phi\wedge\Psi)=f^\ast \Phi\wedge f^\ast \Psi.$
Certificate: Just use the commutativity of $f^\ast $ and $A,\otimes$:
$$\begin{aligned}
f^*(\Phi\wedge\Psi)(x_1,\cdots,x_{r+s})&=(\Phi\wedge\Psi)(f(x_1),\cdots,f(x_{r+s}))\\
&=\frac{(r+s)!}{r!s!}A(\Phi\otimes \Psi) (f(x_1),\cdots,f(x_{r+s}))\\
&=\frac{(r+s)!}{r!s!}A(f^*(\Phi\otimes \Psi))(x_1,\cdots,x_{r+s})\\
&=f^*\Phi \wedge f^*\Psi (x_1,\cdots,x_{r+s}).
\end{aligned}$$
Make formal sum $\Lambda(V^\ast )=\bigoplus_{r=0}^n \Lambda^r(V^\ast )=\{\omega_0+\cdots+\omega_n|\omega_r\in \Lambda^r(V^\ast )\}.$
Then it becomes a $2^n$-dimensional vector space, and with $\wedge$ as the multiplication that satisfies the distributive law, it becomes algebra.
Call it $V^\ast $ Exterior Algebra.Similarly,
Can define $\Lambda(V)=\bigoplus_{r=0}^n \Lambda^r(V),$
called $V$ on Exterior Algebra.
tensor bundle
Let $M^m$ be a $m$-dimensional $C^\infty$ manifold, $\,\forall\,p\in M,$ a tangent space $T_pM.$
We define $(r,s)$-type tensor space $T_s^r(p)=T_pM\otimes\cdots\otimes T_pM\otimes T_p^\ast M\otimes\cdots\otimes T_p^\ast M.$
$\dim T_s^r(p)=m^{r+s}.$ Define bundle space$T_s^r(M)=\bigcup_{p\in M}T_s^r(p),$
Natural topology and differential structure can be given to make it a $m+m^{r+s}$-dimensional $C^\infty$ manifold.
$T_s^r(q)$ Has natural base $\{\frac{\partial {} }{\partial {}x^{i_1} }\otimes\cdots\otimes\frac{\partial {} }{\partial {}x^{i_r} }\otimes dx^{j_1}\otimes \cdots\otimes dx^{j_s}|1\le i_1,\cdots,i_r,j_1,\cdots,j_s\le n\}.$
$\Phi\in T_s^r$ has coordinates $(\Phi^{i_1\cdots i_r}_{j_1\cdots j_s}).$
Take $\pi:T_s^r(M)\rightarrow M$ as the natural projection,
$\Psi:\pi^{-1}(U)\rightarrow U\times \mathbb{R}^N$ Map $\Phi$ to $(q,(\Phi^{i_1\cdots i_r}_{j_1\cdots j_s})).$
Then define
$$\widetilde{\Psi}=(\varphi\times \mathrm{id}_{\mathbb{R}^N})\circ \Psi:\pi^{-1}(U)\rightarrow U\times \mathbb{R}^N\rightarrow \varphi(U)\times \mathbb{R}^N\subset \mathbb{R}^{m+N}.$$
The topology assigned to $T_s^r(M)$ is such that $\pi^{-1}(U)$ is an open subset of $T_s^r(M)$ and $\widetilde{\Psi}$ is a homeomorphism.
If $\{(U_\alpha,\varphi_\alpha)\}$ is the smooth coordinate atlas of $M$,
Then $\{(\pi^{-1}(U_\alpha),\widetilde{\Psi}_\alpha)\}$ is the smooth coordinate atlas of $T_s^r(M)$.
It satisfies the following properties: $\pi:T_s^r(M)\rightarrow M$ is surjective to $C^\infty$,
$\pi^{-1}(q)=T_s^r(q)$ is $m^{r+s}$-dimensional vector space,
$\Psi:\pi^{-1}(U)\rightarrow U\times \mathbb{R}^N$ reflects $C^\infty$,
$\pi_1\circ \Psi=\pi,$ and $\Psi|_q$ are linear isomorphisms.
These elements make it vector bundle, call it tensor bundle.
tensor field
If there is $\Phi:M\rightarrow T_s^r(M),$, it satisfies $\pi\circ\Phi=\mathrm{id}_M,$
Then $\Phi$ is called $M$ on $(r,s)$-type tensor field. When $\Phi$ is smooth mapping,
Call it smooth $(r,s)$-type tensor field. This is if and only if each component of $\Phi$ is a smooth function.
$\Phi\in T_s^r(p)=\mathcal{L}(T_p^\ast M,\cdots,T_p^\ast M,T_pM,\cdots,T_pM;\mathbb{R})$ can be regarded as $(r+s)$-heavy linear function.
Consider $(0,s)$-type tensor field $\Phi.$ and choose $X_1,\cdots,X_s\in \mathfrak{X}(M),$ arbitrarily
Define the role of the tensor field on the vector field:
$$[\Phi(X_1,\cdots,X_s)] (p):=\Phi(p)(X_1(p),\cdots,X_s(p)),$$
Then $\Phi(X_1,\cdots,X_s)\in C^\infty(M).$ because each component is a smooth function,
After linear combination or multiplication, it is still a smooth function.
Since the tensor is a multilinear function, take $f_1,\cdots,f_s\in C^\infty(M),$ arbitrarily
Considered point by point
$$\Phi(f_1X_1,\cdots,f_sX_s)=f_1\cdots f_s\Phi(X_1,\cdots,X_s)\in C^\infty(M).$$
conversely,
If there is a mapping $\Phi:\mathfrak{X}(M)\times \cdots\times \mathfrak{X}(M)\rightarrow C^\infty(M)$ that is linear to $C^\infty(M)$,
Then $\Phi$ must be a $(0,s)$-type $C^\infty$-tensor field.
Generally, let $\Gamma(T_s^r(M))$ denote the entire $(r,s)$-type tensor field of $C^\infty$,
Then it is an infinite-dimensional vector space, $f\Phi+g\Psi(p):=f(p)\Phi(p)+g(p)\Psi(p).$
Similar to the previous one, it acts on the combination of cotangent vector field and vector field,
And it is also $C^\infty(M)$ linear. The reverse is also true.
exterior derivative form
$M^m$ is the $C^\infty$-manifold.
Let $\Lambda^r(T^\ast M)=\bigcup_{p\in M}\Lambda^r(T_p^\ast M),$ similarly,
Give it natural topological structure and differential structure to make it a $m+\binom{m}{r}$-dimensional $C^\infty$ manifold,
Called $M$ on $r$-times external form cluster.
If $\Phi:M\rightarrow \Lambda^rT^\ast M$ is a smooth mapping,
$\pi\circ\Phi=\mathrm{id}_M,$ then $\Phi$ is said to be $r$-times exterior derivative form.
This is if and only if its components are smooth functions.
Let $\mathcal{A}^r(M)=\Gamma(\Lambda^rT^\ast M)$ be the whole body of $r$-subexternal differential forms,
is an infinite-dimensional vector space. Similarly,
You can also define $f\omega_1+g\omega_2\in \mathcal{A}^r(M).$
Since we have multiplication $\wedge,$ on $\Lambda^r T^\ast M$
We can define $\wedge:\mathcal{A}^r(M)\times \mathcal{A}^s(M)\rightarrow \mathcal{A}^{r+s}(M).$ point by point
That is
$$(\omega\wedge\sigma)(p)=\omega(p)\wedge \sigma(p)\in \Lambda^{r+s}(T_p^\ast M).$$
Of course it also satisfies the distributive law, the anti-commutative law and the associative law.
Let $F:M\rightarrow N$ be smooth mapping,
Then ${F_{\ast } }_p:T_pM\rightarrow T_{F(p)}N$ is the cut mapping,
$F^\ast :T^\ast _{F(p)}N\rightarrow T_p^\ast (M)$ To pull back the reflection,
induced to $\Lambda^r T_q^\ast N\rightarrow \Lambda^r T_p^\ast M$,
Re-induced pullback on $\mathcal{A}^r(N)\rightarrow \mathcal{A}^r(M)$,
Satisfied$F^\ast (\omega\wedge\sigma)=F^\ast \omega\wedge F^\ast \sigma.$
The article was last updated on 2021-12-08 15:50:09
