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考试题
习题 1. 叙述$C^\infty$-微分流形的定义. 设若$M$和$N$分别是$m$维和$n$维的$C^\infty$-微分流形, 证明$M\times N$是$m+n$维$C^\infty$-微分流形.
证: 称$M$为拓扑流形, 若$M$为一个拓扑空间, 满足$C_2$与$T_2$条件,
即有可数拓扑基, 且$\,\forall\,p,q\in M,$
$\,\exists\,U,V$分别为$p,q$的开邻域, 满足$U\cap V=\varnothing.$ 另外,
还需是局部欧氏的, 即$\,\forall\,p\in M,$ $\,\exists\,U$为$p$的开邻域,
$\varphi:U\approx \varphi(U)\subset \mathbb{R}^m$为同胚.
这样的拓扑空间称为$m$维拓扑流形.
给定$M$的开覆盖$\{U_\alpha\},$ 以及同胚到欧氏空间的$\{\varphi_\alpha\}.$
称$\{(U_\alpha,\varphi_\alpha)\}$为$M$的坐标图册.
若其满足$\,\forall\,U_\alpha\cap U_\beta\neq \varnothing,$
$\varphi_\alpha\circ\varphi_{\beta}^{-1}:\varphi_{\beta}(U_\alpha\cap U_\beta)\rightarrow \varphi_{\alpha}(U_\alpha\cap U_\beta)$为$C^\infty$微分同胚,
即作为同胚的同时, 映射与逆映射都是$C^\infty$的,
则称$\{(U_\alpha,\varphi_\alpha)\}$为$C^\infty$坐标图册,
赋予了$M$一个微分结构.
具$C^\infty$微分结构的拓扑流形称为$C^\infty$微分流形.
设$\{U_\alpha\},\{V_\beta\}$分别是$M,N$的拓扑基,
则$\{U_\alpha\times V_\beta\}$为$M\times N$的拓扑基.
$M\times N$取积拓扑. 同时, 对$(p_1,q_1)\neq (p_2,q_2),$
不妨设$p_1\neq p_2,$ 取$p_1,p_2,q_1,q_2$的开邻域$U_1,U_2,V_1,V_2,$
能保证$U_1\cap U_2=\varnothing.$
那么此时$U_1\times V_1\cap U_2\times V_2=\varnothing,$
为$(p_1,q_1),(p_2,q_2)$的无交开邻域.
若$\{(U_\alpha,\varphi_\alpha)\},\{(V_\beta,\psi_\beta)\}$为$M,N$的$C^\infty$-坐标图册,
下证$\{(U_\alpha\times V_\beta,\varphi_\alpha\times \psi_\beta)\}$为$M\times N$的$C^\infty$-坐标图册.
显然$\varphi_\alpha\times \psi_\beta:U_\alpha\times V_\beta\approx \varphi_\alpha(U_\alpha)\times \psi_\beta(V_\beta)\subset \mathbb{R}^{m+n},$
同时由$(f\times g)'=f'\times g',$ 转移函数也都是$C^\infty$微分同胚,
确实是$C^\infty$-坐标图册. 因此$M\times N$是$m+n$维$C^\infty$-微分流形.
习题 2. 设若$M$和$N$分别是$m$维和$n$维的$C^\infty$-微分流形, $F:M\rightarrow N$是$C^\infty$-微分流形之间的光滑映照. 对于$p\in M$以及$q = F(p) \in N$, $F$自然地确定了一个余切空间的拉回映照$F^\ast _q:T^\ast _qN\rightarrow T^\ast _pM.$ 如果$p$处有局部坐标$(U, u^i),$ $q$处有局部坐标$(V,v^\alpha),$ 它们分别给出了$T^\ast _pM$和$T^\ast _qN$的自然基$\{(du^i)_p\}$和$\{(dv^\alpha)_q\}.$ 请写出$F^\ast _q$在该基下的矩阵表示, 并证明, 当$F$是微分同胚时, 上述$F^\ast _q$还 是线性同构.
证: 设$F^\ast _q(d v^\alpha)=a^\alpha_i du^i,$
作用到$X_i=\frac{\partial {} }{\partial {}u^i}$上,
得到$a_i^\alpha=X_i(v^\alpha\circ F).$
记$\widehat{F}=\psi\circ F\circ \varphi^{-1}$为局部表示,
则矩阵为$J(\widehat{F})$的转置,
$F_q^\ast (b_\alpha dv^\alpha)=b_\alpha a_i^\alpha du^i.$ 当$F$为微分同胚时,
记$\widehat{F}$为微分同胚, 此时$\det J(\widehat{F})\neq 0,$
因此$F_q^\ast $的矩阵表示是满秩的, 从而是线性同构.
习题 3. 设映照$f : \mathbb{R}^2\rightarrow \mathbb{R}^2$的具体形式是$y_1 = x_1e^{x_2} + x_2,$ $y_2 = x_1e^{x_2}-x_2.$ 试写出$f$在欧式空间自然基底下, 诱导出切空间之间映照$f_\ast $的矩阵表示, 并证明$f$是个光滑同胚.
证:
回忆$f_\ast (\frac{\partial {} }{\partial {}x_i})=a_i^j \frac{\partial {} }{\partial {}y_j},$
作用$dy^j,$ 得到$a_i^j=\frac{\partial {}y^j\circ f}{\partial {}x_i}.$
因此$f^\ast (b^i\frac{\partial {} }{\partial {}x_i})=b^ia_i^j \frac{\partial {} }{\partial {}y_j},$
矩阵表示$a_i^j$为局部表示$\widehat{f}=\mathrm{id}\circ f\circ \mathrm{id}^{-1}=f$的Jacobi矩阵.
计算得到, $J(f)=\begin{bmatrix}
e^{x_2} & x_1e^{x_2}+1\\
e^{x_2} & x_1e^{x_2}-1
\end{bmatrix}.$ 此即为$f_*$的矩阵表示. 易见$f\in C^\infty,$
且$\det J(f)=-2e^{x_2}\neq 0,$ 因此$f$处处为局部微分同胚, 且为单射.
注意到$\frac{y_1-y_2}{2}=x_2,$
$y_1=x_1e^{\frac{y_1-y_2}{2} }+\frac{y_1-y_2}{2},$
$x_1=\frac{y_1+y_2}{2}e^{\frac{y_2-y_1}{2} }.$
这就给出了$f^{-1}$的具体形式, 易见它没有定义域的限制,
且也是$C^\infty$的, 因此$f$为满射, 且是光滑同胚,
即作为同胚的同时映射与逆映射都是光滑的.
习题 4. 设$M$是$C^\infty$-微分流形, $f, g \in C^\infty(M),$ $X, Y \in X (M),$ $[-, -]$是Lie括号. 请展开$[fX, gY].$
证:
$$
\begin{aligned}
[fX,gY]&=fX(gY)-gY(fX)\\
&=fX(g)Y+fgXY-gY(f)X-fgYX\\
&=fX(g)Y-gY(f)X+fg[X,Y]
\end{aligned}
$$
这里$X(g)(p)=X_p(g),$ $[X,Y]_p(h)=X_p(Y_q(h))-Y_p(X_q(h)),$
其中$X_q(h)$为关于$q$的$C^\infty$函数.
习题 5. 设$M$是$C^\infty$-微分流形, $X, Y \in \mathfrak{X}(M),$ $\omega \in \Lambda^1(M),$ $[-, -]$是Lie括号. 证明公式:
$$
d\omega (X, Y ) = X\omega (Y ) - Y \omega(X) - \omega([X, Y ])
$$
证: 由于公式具线性性, 只需验证$\omega=fdg$的情形.
$d\omega=df\wedge dg.$ 此时
$$
d\omega(X,Y)=df\wedge dg(X,Y)=df(X)dg(Y)-df(Y)dg(X)=X(f)Y(g)-Y(f)X(g).
$$
$$
X\omega(Y)=X(fY(g))=X(f)Y(g)+fXY(g)
$$
$$
Y\omega(X)=Y(fX(g))=Y(f)X(g)+fYX(g)
$$
$$
\omega([X,Y])=f[X,Y] (g)=fXY(g)-fYX(g)
$$
因此,
$X\omega(Y)-Y\omega(X)-\omega([X,Y])=X(f)Y(g)-Y(f)X(g)=d\omega(X,Y).$
习题 6. 设$M$是$m$维$C^\infty$-紧致无边微分流形, $f : M \rightarrow \mathbb{R}^m$是光滑映照. 试证明存在$p\in M,$ 使得$f$在$p$处的秩小于$m.$
证: 不然, $f$处处秩等于$m$(显然不可能大于$m$). 由反函数定理,
$f$处处为局部微分同胚. 即$\,\forall\,p\in M,$
$\,\exists\,U$为$p$的开邻域, $f:U\approx f(U),$
且可以要求$f(U)$也是开集. 由秩定理,
事实上可以要求$U,f(U)$在适当坐标系下都是开方盒. 由$M$紧,
$\,\exists\,\{U_i\}_{i=1}^m,$ $f:U_i\approx f(U_i),$
$\bigcup_{i=1}^m U_i=M.$
设$M$为连通流形, 不然在每个连通分支上讨论即可. 连通分支既开又闭,
同时紧空间中闭集仍是紧集. 注意到我们有
$$
\bigcup^m_{i=1} f(U_i)= f(\bigcup^m_{i=1}U_i)=f(M),
$$
这产生了矛盾.
因为紧集连续像是紧的, $f(M)$为紧集, 从而是有界闭集.
但$f(M)=\bigcup_{i=1}^m f(U_i)$又是开集,
因此它只能是全空间$\mathbb{R}^m,$ 而这又会与紧性矛盾.
从而$f$在某点秩小于$m.$
题设中无边性用在了反函数定理中. 当$p$为带边流形边上点时,
不再有类似的反函数定理. 即使找到了邻域$U$上的局部微分同胚,
也无法说明像集在全空间中是开集, 从而无法进行类似的讨论.
习题 7. 设$M_1, M_2$是$C^\infty$-微分流形, $f : M_1 \rightarrow M_2$是局部微分同胚. 若$M_2$是可定向的, 证明$M_1$也是 可定向的.
证: 由于$f$是局部微分同胚, 它的局部表示是微分同胚,
从而$M_1,M_2$维数相同. 由$M_2$可定向,
存在$\Omega\in \mathcal{A}^{m}(M_2)$为处处非零的外微分形式.
断言$f^\ast \Omega\in \mathcal{A}^m(M_1)$也是处处非零的外微分形式,
从而是可定向的.
在局部坐标系中考虑. 设$f:U\approx V,$
$\Omega=hdy^1\wedge\cdots \wedge dy^m\in \mathcal{A}^m(V),$ $h\neq 0,$
则
$$
f^\ast \Omega=f^\ast h\cdot f^\ast (dy^1\wedge\cdots\wedge dy^m)=h\circ f \cdot \det J(f) {}dx^{1}\wedge\cdots\wedge{}dx^{ {}m}\in \mathcal{A}^m(M_1).
$$
由于$f$为局部微分同胚, $\det J(f)\neq 0,$ 同时$h\circ f\neq 0.$
因此$f^\ast \Omega$确实是处处非零的外微分形式.
另一种方法是, 取合适的加细, 使得$f:U_\alpha\approx V_\alpha,$
且$\{(V_\alpha,\psi_\alpha)\}$构成$M_2$上的保定向光滑坐标图册.
断言$\{(U_\alpha,\varphi_\alpha=\psi_\alpha\circ f)\}$构成$M_1$上的保定向光滑坐标图册.
只需注意到
$$
\varphi_\beta\circ\varphi_\alpha^{-1}=(\psi_\beta\circ f)\circ (f^{-1}\circ \psi_\alpha^{-1})=\psi_\beta\circ\psi_\alpha^{-1}.
$$
习题 8. 证明, 具有紧支集的形式的积分和单位分解的选取无关. 再叙述Stokes定理.
证: 首先说明紧支集形式的积分和坐标系选取无关.
若$\omega\in \mathcal{A}^m_c(U\cap V),$ 则
$$
\int_U \omega=\int_{\varphi(U\cap V)}(\varphi^{-1})^\ast \omega=\int_{\psi(U\cap V)}(\psi^{-1})^\ast \omega=\int_V\omega,
$$
中间的等号由$\mathbb{R}^m$间的积分变换公式得到,
注意$U,V$的坐标选取需要与$M$定向相符,
这样保证$\det J(\varphi\circ\psi^{-1})> 0.$
接下来, 对两组单位分解$\{f_i\},\{g_j\},$ $\{f_ig_j\}$也是一组单位分解.
因此,
$$
\sum_i \int_{U_i} f_i \omega=\sum_{i,j}\int_{U_i\cap V_j} f_ig_j \omega=\sum_{j}\int_{V_j} g_j\omega.
$$
从而积分与单位分解选取无关.
Stokes定理: 对定向$m$维光滑流形$M,$ 若$D$为其上(闭)带边区域,
即$D$由内点$p\in U\approx \varphi(U)\subset \mathbb{R}^m$与边界点$p\in U,$
$U\cap D\approx \varphi(U)\cap \mathbb{H}$组成.
$D$从$M$上诱导定向,
那么对于$M$上的$m-1$次的紧支撑$C^1$外微分形式$\omega$, 有
$$
\int_D d\omega=\int_{\partial D}\omega.
$$
当$D$无边界点时, 右端消失.
文章最后更新于 2021-12-28 19:33:06
Exam questions
Exercise 1. Describe the definition of $C^\infty$-differential manifold. Suppose $M$ and $N$ are $C^\infty$-differential manifolds of $m$ and $n$ dimensions respectively, prove that $M\times N$ is a $m+n$-dimensional $C^\infty$-differential manifold.
Certificate: Call $M$ a topological manifold. If $M$ is a topological space and satisfies the conditions $C_2$ and $T_2$,
That is, there are countable topological bases, and $\,\forall\,p,q\in M,$
$\,\exists\,U,V$ are the open neighborhoods of $p,q$ respectively, satisfying $U\cap V=\varnothing.$. In addition,
It also needs to be locally Euclidean, that is, $\,\forall\,p\in M,$ $\,\exists\,U$ is the open neighborhood of $p$,
$\varphi:U\approx \varphi(U)\subset \mathbb{R}^m$ is homeomorphism.
Such a topological space is called a $m$-dimensional topological manifold.
Given the open covering $\{U_\alpha\},$ of $M$ and the homeomorphism to the Euclidean space $\{\varphi_\alpha\}.$
Call $\{(U_\alpha,\varphi_\alpha)\}$ the coordinate atlas of $M$.
If it satisfies $\,\forall\,U_\alpha\cap U_\beta\neq \varnothing,$
$\varphi_\alpha\circ\varphi_{\beta}^{-1}:\varphi_{\beta}(U_\alpha\cap U_\beta)\rightarrow \varphi_{\alpha}(U_\alpha\cap U_\beta)$ is $C^\infty$ diffeomorphism,
That is, as homeomorphisms, both mapping and inverse mapping are $C^\infty$,
Then $\{(U_\alpha,\varphi_\alpha)\}$ is called $C^\infty$ coordinate atlas,
$M$ is given a differential structure.
The topological manifold with $C^\infty$ differential structure is called $C^\infty$ differential manifold.
Let $\{U_\alpha\},\{V_\beta\}$ be the topological basis of $M,N$ respectively,
Then $\{U_\alpha\times V_\beta\}$ is the topological basis of $M\times N$.
$M\times N$ takes the product topology. At the same time, for $(p_1,q_1)\neq (p_2,q_2),$
Let’s assume $p_1\neq p_2,$ and take the open neighborhood $U_1,U_2,V_1,V_2,$ of $p_1,p_2,q_1,q_2$.
Guaranteed $U_1\cap U_2=\varnothing.$
Then at this time $U_1\times V_1\cap U_2\times V_2=\varnothing,$
is the intersectionless open neighborhood of $(p_1,q_1),(p_2,q_2)$.
If $\{(U_\alpha,\varphi_\alpha)\},\{(V_\beta,\psi_\beta)\}$ is $C^\infty$ of $M,N$ - coordinate atlas,
It is proved that $\{(U_\alpha\times V_\beta,\varphi_\alpha\times \psi_\beta)\}$ is $C^\infty$ of $M\times N$ - coordinate atlas.
Obviously $\varphi_\alpha\times \psi_\beta:U_\alpha\times V_\beta\approx \varphi_\alpha(U_\alpha)\times \psi_\beta(V_\beta)\subset \mathbb{R}^{m+n},$
At the same time, the transfer functions of $(f\times g)'=f'\times g',$ are also $C^\infty$ diffeomorphisms,
It is indeed a $C^\infty$-coordinate atlas. Therefore $M\times N$ is a $m+n$-dimensional $C^\infty$-differential manifold.
Exercise 2. Suppose $M$ and $N$ are $m$-dimensional and $n$-dimensional $C^\infty$-differential manifolds respectively, $F:M\rightarrow N$ is a smooth mapping between $C^\infty$-differential manifolds. For $p\in M$ and $q = F(p) \in N$, $F$ naturally determines a pullback mapping of the cotangent space $F^\ast _q:T^\ast _qN\rightarrow T^\ast _pM.$ If there are local coordinates $(U, u^i),$ at $p$ There are local coordinates $(V,v^\alpha),$ at $q$, which give the natural basis $\{(du^i)_p\}$ and $\{(dv^\alpha)_q\}.$ of $T^\ast _pM$ and $T^\ast _qN$ respectively. Please write down the matrix representation of $F^\ast _q$ under this basis, and prove that when $F$ is a diffeomorphism, the above $F^\ast _q$ is also a linear isomorphism.
Certificate: Let $F^\ast _q(d v^\alpha)=a^\alpha_i du^i,$
Act on $X_i=\frac{\partial {} }{\partial {}u^i}$,
Get$a_i^\alpha=X_i(v^\alpha\circ F).$
Let $\widehat{F}=\psi\circ F\circ \varphi^{-1}$ be the local representation,
Then the matrix is the transpose of $J(\widehat{F})$,
$F_q^\ast (b_\alpha dv^\alpha)=b_\alpha a_i^\alpha du^i.$ When $F$ is diffeomorphism,
Note $\widehat{F}$ as diffeomorphism, then $\det J(\widehat{F})\neq 0,$
Therefore, the matrix representation of $F_q^\ast $ is full rank and thus linearly isomorphic.
Exercise 3. Assume that the specific form of the mapping $f : \mathbb{R}^2\rightarrow \mathbb{R}^2$ is $y_1 = x_1e^{x_2} + x_2,$ $y_2 = x_1e^{x_2}-x_2.$. Try to write $f$ under the natural basis of Euclidean space, induce the matrix representation of the mapping $f_\ast $ between tangent spaces, and prove that $f$ is a smooth homeomorphism.
Certificate:
Memories$f_\ast (\frac{\partial {} }{\partial {}x_i})=a_i^j \frac{\partial {} }{\partial {}y_j},$
Act on $dy^j,$ and get $a_i^j=\frac{\partial {}y^j\circ f}{\partial {}x_i}.$
Therefore $f^\ast (b^i\frac{\partial {} }{\partial {}x_i})=b^ia_i^j \frac{\partial {} }{\partial {}y_j},$
The matrix representation $a_i^j$ is the Jacobi matrix of the local representation $\widehat{f}=\mathrm{id}\circ f\circ \mathrm{id}^{-1}=f$.
Calculated, $J(f)=\begin{bmatrix}
e^{x_2} & x_1e^{x_2}+1\\
e^{x_2} & x_1e^{x_2}-1
\end{bmatrix}.$ is the matrix representation of $f_*$. It is easy to see $f\in C^\infty,$
And $\det J(f)=-2e^{x_2}\neq 0,$ therefore $f$ is locally diffeomorphism everywhere and is injective.
Notice$\frac{y_1-y_2}{2}=x_2,$
$y_1=x_1e^{\frac{y_1-y_2}{2} }+\frac{y_1-y_2}{2},$
$x_1=\frac{y_1+y_2}{2}e^{\frac{y_2-y_1}{2} }.$
This gives the specific form of $f^{-1}$. It is easy to see that it has no restriction on the definition domain.
And it is also $C^\infty$, so $f$ is surjective and smooth homeomorphism,
That is, both simultaneous mapping and inverse mapping as homeomorphisms are smooth.
Exercise 4. Let $M$ be the $C^\infty$-differential manifold, $f, g \in C^\infty(M),$ $X, Y \in X (M),$ $[-, -]$ be the Lie bracket. Please expand $[fX, gY].$
Certificate:
$$
\begin{aligned}
[fX,gY]&=fX(gY)-gY(fX)\\
&=fX(g)Y+fgXY-gY(f)X-fgYX\\
&=fX(g)Y-gY(f)X+fg[X,Y]
\end{aligned}
$$
Here $X(g)(p)=X_p(g),$ $[X,Y]_p(h)=X_p(Y_q(h))-Y_p(X_q(h)),$
Where $X_q(h)$ is the $C^\infty$ function about $q$.
Exercise 5. Let $M$ be the $C^\infty$-differential manifold, $X, Y \in \mathfrak{X}(M),$ $\omega \in \Lambda^1(M),$ $[-, -]$ be the Lie bracket. Proof formula:
$$
d\omega (X, Y ) = X\omega (Y ) - Y \omega(X) - \omega([X, Y ])
$$
Certificate: Since the formula is linear, only the case of $\omega=fdg$ needs to be verified.
$d\omega=df\wedge dg.$ At this time
$$
d\omega(X,Y)=df\wedge dg(X,Y)=df(X)dg(Y)-df(Y)dg(X)=X(f)Y(g)-Y(f)X(g).
$$
$$
X\omega(Y)=X(fY(g))=X(f)Y(g)+fXY(g)
$$
$$
Y\omega(X)=Y(fX(g))=Y(f)X(g)+fYX(g)
$$
$$
\omega([X,Y])=f[X,Y] (g)=fXY(g)-fYX(g)
$$
Therefore,
$X\omega(Y)-Y\omega(X)-\omega([X,Y])=X(f)Y(g)-Y(f)X(g)=d\omega(X,Y).$
Exercise 6. Suppose $M$ is a $m$-dimensional $C^\infty$-compact boundless differential manifold, $f : M \rightarrow \mathbb{R}^m$ is a smooth mapping. Prove that there exists $p\in M,$ such that the rank of $f$ at $p$ is smaller than $m.$
Certificate: Otherwise, the rank of $f$ is equal to $m$ everywhere (obviously it cannot be greater than $m$). According to the inverse function theorem,
$f$ is a local diffeomorphism everywhere. That is, $\,\forall\,p\in M,$
$\,\exists\,U$ is the open neighborhood of $p$, $f:U\approx f(U),$
And it can be required that $f(U)$ is also an open set. According to the rank theorem,
In fact, it can be required that $U,f(U)$ is a square box in the appropriate coordinate system. From $M$,
$\,\exists\,\{U_i\}_{i=1}^m,$ $f:U_i\approx f(U_i),$
$\bigcup_{i=1}^m U_i=M.$
Let $M$ be a connected manifold, otherwise we can discuss it on each connected component. A connected component is both open and closed,
At the same time, a closed set in a compact space is still a compact set. Note that we have
$$
\bigcup^m_{i=1} f(U_i)= f(\bigcup^m_{i=1}U_i)=f(M),
$$
This creates a contradiction.
Because the continuous image of a compact set is compact, $f(M)$ is a compact set and therefore a bounded closed set.
But $f(M)=\bigcup_{i=1}^m f(U_i)$ is an open set again,
Therefore it can only be the total space $\mathbb{R}^m,$ which would contradict compactness.
Therefore $f$ has a rank smaller than $m.$ at a certain point
The edgelessness in the problem is used in the inverse function theorem. When $p$ is a point on the edge of the manifold with edges,
There is no longer a similar inverse function theorem. Even if a local diffeomorphism on the neighborhood $U$ is found,
It is also impossible to explain that the image set is an open set in the entire space, so similar discussions cannot be carried out.
Exercise 7. Suppose $M_1, M_2$ is a $C^\infty$-differential manifold, and $f : M_1 \rightarrow M_2$ is a local diffeomorphism. If $M_2$ is orientable, prove that $M_1$ is also orientable.
Certificate: Since $f$ is a local diffeomorphism, its local representation is a diffeomorphism,
Therefore, the dimensions of $M_1,M_2$ are the same. It can be oriented by $M_2$,
There exists $\Omega\in \mathcal{A}^{m}(M_2)$ as an exterior derivative form that is non-zero everywhere.
Assert that $f^\ast \Omega\in \mathcal{A}^m(M_1)$ is also an exterior derivative form that is non-zero everywhere,
It is therefore orientable.
Consider in the local coordinate system. Let $f:U\approx V,$
$\Omega=hdy^1\wedge\cdots \wedge dy^m\in \mathcal{A}^m(V),$ $h\neq 0,$
rule
$$
f^\ast \Omega=f^\ast h\cdot f^\ast (dy^1\wedge\cdots\wedge dy^m)=h\circ f \cdot \det J(f) {}dx^{1}\wedge\cdots\wedge{}dx^{ {}m}\in \mathcal{A}^m(M_1).
$$
Since $f$ is a local diffeomorphism, $\det J(f)\neq 0,$ and $h\circ f\neq 0.$
Therefore $f^\ast \Omega$ is indeed an exterior derivative form that is non-zero everywhere.
Another method is to take appropriate thinning so that $f:U_\alpha\approx V_\alpha,$
And $\{(V_\alpha,\psi_\alpha)\}$ constitutes a direction-preserving smooth coordinate atlas on $M_2$.
Assert that $\{(U_\alpha,\varphi_\alpha=\psi_\alpha\circ f)\}$ constitutes a direction-preserving smooth coordinate atlas on $M_1$.
Just notice
$$
\varphi_\beta\circ\varphi_\alpha^{-1}=(\psi_\beta\circ f)\circ (f^{-1}\circ \psi_\alpha^{-1})=\psi_\beta\circ\psi_\alpha^{-1}.
$$
Exercise 8. Prove that integrals in the form of compact supports have nothing to do with the choice of partition of unity. Let's describe Stokes' theorem again.
Certificate: First, it is explained that the integral in the form of a compactly supported set has nothing to do with the selection of the coordinate system.
If $\omega\in \mathcal{A}^m_c(U\cap V),$ then
$$
\int_U \omega=\int_{\varphi(U\cap V)}(\varphi^{-1})^\ast \omega=\int_{\psi(U\cap V)}(\psi^{-1})^\ast \omega=\int_V\omega,
$$
The equal sign in the middle is obtained by the integral transformation formula between $\mathbb{R}^m$,
Note that the coordinate selection of $U,V$ needs to match the orientation of $M$.
This guarantees $\det J(\varphi\circ\psi^{-1})> 0.$
Next, the decomposition of two sets of units $\{f_i\},\{g_j\},$ $\{f_ig_j\}$ is also a set of partitions of unity.
Therefore,
$$
\sum_i \int_{U_i} f_i \omega=\sum_{i,j}\int_{U_i\cap V_j} f_ig_j \omega=\sum_{j}\int_{V_j} g_j\omega.
$$
Therefore, the integral has nothing to do with the choice of partition of unity.
Stokes' theorem: For the oriented $m$-dimensional smooth manifold $M,$, if $D$ is its upper (closed) edge region,
That is, $D$ consists of the inner point $p\in U\approx \varphi(U)\subset \mathbb{R}^m$ and the boundary point $p\in U,$
$U\cap D\approx \varphi(U)\cap \mathbb{H}$Composition.
$D$ from $M$ upward induced orientation,
Then for $m-1$ times on $M$ tight support $C^1$ Exterior derivative form $\omega$, there is
$$
\int_D d\omega=\int_{\partial D}\omega.
$$
When $D$ has no boundary point, the right end disappears.
The article was last updated on 2021-12-28 19:33:06
