流形上的微分形式与Stokes定理

Proof. \(M\) 的图册中的局部图 \(\varphi_i:H^n\to U_i\) 同样是 \(\partial M\) 的局部图, 因为 \(\varphi_i:\partial H^n \to U'\subset \partial M\), 而 \(\partial H^n = \mathbb{R}^{n-1}\). 所有这样的局部图构成了 \(\partial M\) 的图册, 所以 \(\dim \partial M= n-1\).

由于 \(\partial M\) 的所有局部图的参数域都是 \(\mathbb{R}^{n-1}\), 因此 \(\partial M\) 是无边流形. ◻

Proof. 只须证明,图集 \(A(\partial M)=\{ (R^{n-1}, \varphi_i \vert_{\partial H^n}, \partial U_i) \}\) 是两两相容的.

设坐标变换函数 \(\psi:H^n\to H^n\) 是微分同胚,且具有正的Jacobi行列式. 只要证明 \(\psi \vert_{\partial H^n}: R^{n-1}\to R^{n-1}\) 也具有正的Jacobi行列式.

根据已知的相容性条件,设 \(x\in H^n\),有 \[\begin{aligned} J_n(x)= \begin{vmatrix} \frac{\partial\psi^1}{\partial x^1} & \frac{\partial\psi^1}{\partial x^2} & \cdots & \frac{\partial\psi^1}{\partial x^n} \\ \frac{\partial\psi^2}{\partial x^1} & \frac{\partial\psi^2}{\partial x^2} & \cdots & \frac{\partial\psi^2}{\partial x^n} \\ \vdots & \vdots & \ddots & \vdots \\ \frac{\partial\psi^n}{\partial x^1} & \frac{\partial\psi^n}{\partial x^2} & \cdots & \frac{\partial\psi^n}{\partial x^n} \\ \end{vmatrix}(x) >0 \end{aligned}\]

当 \(x^1=0\), 由于 \(\psi(\partial H^n)\subset\partial H^n\), 则当 \(k \ge 2\) 时, \(\frac{\partial\psi^1}{\partial x^k} (x)=0\). 从而知 \[\begin{aligned} J_n(x) = \begin{vmatrix} \frac{\partial\psi^1}{\partial x^1} & 0 & \cdots & 0 \\ \frac{\partial\psi^2}{\partial x^1} & \frac{\partial\psi^2}{\partial x^2} & \cdots & \frac{\partial\psi^2}{\partial x^n} \\ \vdots & \vdots & \ddots & \vdots \\ \frac{\partial\psi^n}{\partial x^1} & \frac{\partial\psi^n}{\partial x^2} & \cdots & \frac{\partial\psi^n}{\partial x^n} \\ \end{vmatrix}(x) = \frac{\partial\psi^1}{\partial x^1}(x) \begin{vmatrix} \frac{\partial\psi^2}{\partial x^2} & \cdots & \frac{\partial\psi^2}{\partial x^n} \\ \vdots & \ddots & \vdots \\ \frac{\partial\psi^n}{\partial x^2} & \cdots & \frac{\partial\psi^n}{\partial x^n} \\ \end{vmatrix}(x) > 0 \end{aligned}\]

取 \(y\in\partial H^n=\mathbb{R}^{n-1}\), 不妨记 \(y\) 的坐标 \(y=(0, y^2,\cdots,y^n)\), 则 \[\begin{aligned} J_{n-1}(y) = \begin{vmatrix} \frac{\partial\psi^2}{\partial y^2} & \cdots & \frac{\partial\psi^2}{\partial y^n} \\ \vdots & \ddots & \vdots \\ \frac{\partial\psi^n}{\partial y^2} & \cdots & \frac{\partial\psi^n}{\partial y^n} \\ \end{vmatrix}(y) = \frac{J_n(y)}{\frac{\partial\psi^1}{\partial y^1}(y)} \end{aligned}\]

由于 \(J_n(x)>0\),当 \(x^1=0\), 我们知道 \(\frac{\partial\psi^1}{\partial x^1}(x)\) 不等于 \(0\), 只需说明 \(\frac{\partial\psi^1}{\partial x^1}(x)\) 不是负数.

假设 \(\frac{\partial\psi^1}{\partial x^1}(x)<0\), 对 \(\psi^1\) 应用泰勒展开,固定 \(x_2,x_3,\cdots,x_n\), 记 \(x'=(\Delta x^1, x^2,\cdots,x^n)\), \[\psi^1(x') = \frac{\partial\psi^1}{\partial x^1}(x) \Delta x^1 + o(\Delta x^1)\] 所以存在 \(\Delta x^1 < 0\), \(\psi(x') > 0\), 也就是说, \(\psi(x') \not\in H^n\), 与 \(\psi\) 的定义矛盾.

这样,我们就说明了将 \(\psi\) 限制在 \(\partial H^n\) 后的行列式 \(J_{n-1}>0\), 也就是说, \(M\) 的定向图册限制到 \(\partial M\) 上之后,仍然是两两相容的,从而 \(\partial M\) 也是可定向的.

由于使用的是同一套图册,限制参数域后光滑性不变. ◻

Proof. 构造法.首先 \(\forall x\in M\), 取 \(x\) 的邻域 \(V_x\) 使得 \(\overline{V_x}\) 是紧致的. 因为微分流形 \(M\) 有可数拓扑基, 因此存在至多可数个 \(x_i\) 使得 \(\{V_{x_i}\}\) 是 \(M\) 的开覆盖. 令 \(G_0=\varnothing\), \(G_1=V_{x_1}\).假设 \(G_1,\cdots,G_i\) 均定义好, 必存在 \(I\) 使得 \[\bigcup_{k\le i} \overline{G_k} \subset \bigcup_{j\le I} V_{x_j}\] 令 \(G_{i+1}=\bigcup_{j\le I} V_{x_j}\) 即可. ◻

Proof. 令 \(\{G_i\}\) 为 \(M\) 的穷竭列(由引理知穷竭存在), 令 \(A_i = \overline{G_i}-G_{i-1}\). 易知 \(A_i\) 是紧致闭集, 且 \(\{A_i\}\) 是 \(M\) 的一个覆盖.

任给 \(x\in A_i\), 选取 \(x\) 的局部坐标系 \(U_x,\varphi_x\) 使得

1. 存在 \(\alpha(x)\in \Gamma\), 使得 \(U_x \subset U_{\alpha(x)}\).

2. \(\varphi_x(U_x)=B_2(0)\)

3. \(\forall \vert j-i \vert > 1\), \(U_x \cap A_j = \varnothing\).

设 \(f\) 是 \(\mathbb{R}^n\) 上的鼓包函数, 定义函数 \(f_x:M\to \mathbb{R}\) \[f_x(p) = \begin{cases} f(\varphi_x(p)), & p\in U \\ 0, & p\in M-U \end{cases}\] 令 \(V_x=\varphi^{-1}(B_{\frac{1}{2}}(0))\), 则 \(\{V_x \vert x\in A_i\}\) 构成 \(A_i\) 的开覆盖, 可选出有限个点 \(x^i_1,\cdots,x^i_{k(i)}\) 使得 \(\{V_{x^i_j \vert 1\le j\le k(i)}\}\) 覆盖 \(A_i\).

这样的选择保证了每个点 \(x\in A\), \(\sum_{j\le k(i)} f_{x^i_j}(x)>0\), 且 \(\{\operatorname{supp}f_{x^i_j}\}\) 是局部有限的. 最后, 进行归一化.和函数 \[\psi(x)=\sum_{i}\sum_{j\le k(i)} f_{x^i_j}(x)\] 是 \(M\) 上的光滑函数, 且恒为正.从而 \(\{ \frac{f_{x^i_j}}{\psi}\}\) 为所求的从属于开覆盖的单位分解. ◻

Proof. 设 \(\operatorname{supp}\omega\) 含于另一局部坐标 \(U_\beta\), 则在 \(U_\beta\) 中, \(\omega\) 可以表示为 \[\omega=b_\beta dx_\beta^1\wedge\cdots\wedge dx_\beta^n\] 则有 \(b_\beta=\det J(\varphi_\alpha \circ \varphi_\beta^{-1}) a_\alpha\), 因此积分值 \[\begin{aligned} \int_{\varphi_\beta(U_\beta)} b_\beta \circ \varphi_\beta^{-1} dx_\beta^1\cdots dx_\beta^n & = \int_{\varphi_\beta(U_\alpha \cap U_\beta)} b_\beta \circ \varphi_\beta^{-1} dx_\beta^1\cdots dx_\beta^n \\ & = \int_{\varphi_\alpha(U_\alpha \cap U_\beta)} \vert\det J(\varphi_\alpha \circ \varphi_\beta^{-1})\vert b_\alpha \circ \varphi_\alpha^{-1} dx_\alpha^1\cdots dx_\alpha^n \\ & = \int_{\varphi_\alpha(U_\alpha)} a_\alpha \circ \varphi_\alpha^{-1} dx_\alpha^1\cdots dx_\alpha^n \end{aligned}\] ◻

Proof. 通过单位分解, 不妨设 \(\operatorname{supp}\omega\) 含于坐标邻域 \(U\) 中, \(\varphi\) 为 \(U\) 上的坐标映射. 由坐标映射的连续性, 有 \[\varphi(U\cap\partial M)=\varphi(U)\cap \partial H^n\] \(n-1\) 次微分形式 \(\omega\) 在 \(U\) 中可以表示为 \[\omega = \sum_{i=1}^{n} (-1)^{i-1} a_i dx^1\wedge\cdots\wedge\widehat{dx^i}\wedge dx^n\] 从而 \[d\omega = \sum_{i=1}^{n} \frac{\partial a_i}{\partial x^i} dx^1\wedge\cdots\wedge dx^n\]

下面分两种情况讨论

1. \(\partial M \cap U = \varnothing\). \[\begin{aligned} \int_M d\omega = \int_U d\omega & = \sum_{i=1}^n \int_{\varphi(U)} \frac{\partial a_i}{\partial x^i} dx^1\cdots dx^n = \sum_{i=1}^n \int_{\mathbb{R}^n} \frac{\partial a_i}{\partial x^i} dx^1\cdots dx^n \\ & = \sum_{i=1}^n \int_{\mathbb{R}^{n-1}} \left(\int_{-\infty}^{+\infty}\frac{\partial a_i}{\partial x^i} dx^i\right) dx^1\cdots\widehat{dx^i}\cdots dx^n \\ & = \sum_{i=1}^n \int_{\mathbb{R}^{n-1}} a_i \vert_{x_i=-\infty}^{x_i=+\infty} dx^1\cdots\widehat{dx^i}\cdots dx^n \\ & = \sum_{i=1}^n \int_{\mathbb{R}^{n-1}} 0 dx^1\cdots\widehat{dx^i}\cdots dx^n \\ & = 0 \\ \end{aligned}\] 其中, 因为 \(\operatorname{supp}\omega\) 紧支, 因此 \(a_i\) 在 \(\pm\infty\) 处取值为 \(0\).

   另一方面, 在边界上, 由于 \(\partial M \cap U = \varnothing\), 所以 \(\int_{\partial M} \omega =0\). 所以 \(\partial M \cap U = \varnothing\) 时有 \[\int_{\partial M} \omega =0 = \int_M d\omega\]

2. \(\partial M \cap U \not= \varnothing\).此时, \[\begin{aligned} \int_M d\omega & = \int_{\varphi(U)} \sum_{i=1}^n \frac{\partial a_i}{\partial x^i} dx^1\cdots dx^n \\ & = \sum_{i=1}^n \int_{H^n} \frac{\partial a_i}{\partial x^i} dx^1\cdots dx^n \\ & = \int_{H^n} \frac{\partial a_1}{\partial x^i} dx^1\cdots dx^n + \sum_{i=2}^n \int_{H^n} \frac{\partial a_i}{\partial x^i} dx^1\cdots dx^n \\ & = \int_{\mathbb{R}^{n-1}} \left(\int_{-\infty}^{0}\frac{\partial a_1}{\partial x^i} dx^1\right) dx^2\cdots dx^n + 0 \\ & = \int_{\mathbb{R}^{n-1}} a_1\vert_{x^1=-\infty}^{x^1=0} dx^2\cdots dx^n \\ & = \int_{\partial H^n} a_1(0,x^2,\cdots,x^n) dx^2\cdots dx^n \\ \end{aligned}\]

   另一方面, \(\forall p\in \partial M, X\in T_p \partial M\), \(dx^1(p)(X)=0\), 因此对于 \(i\not=1\), \[\begin{aligned} \int_{\partial M} dx^1 \wedge \cdots \wedge \widehat{dx^i} \wedge dx^n = 0 \end{aligned}\] 从而 \[\begin{aligned} \int_{\partial M}\omega & = \int_{\partial M} a_1 dx^2 \wedge \cdots \wedge dx^n \\ & = \int_{\partial H^n} a_1 dx^2 \cdots dx^n \\ \end{aligned}\] 因此当 \(\partial M \cap U \not= \varnothing\) 时, \[\int_M d\omega=\int_{\partial M}\omega\]

综上两种情况, 结合单位分解, 对任意 \[\int_M d\omega=\int_{\partial M}\omega\] ◻