Chapter X. R n \mathbb{R}^n R n
Section X. 1. R n \mathbb{R}^n R n
定义
Def. 设 S ⊆ R n S\subseteq\mathbb{R}^n S ⊆ R n S S S k k k ∀ p ∈ S \forall p\in S ∀ p ∈ S p p p U U U U ∩ S U\cap S U ∩ S R k \mathbb{R}^k R k D D D φ : D → U ∩ S , t = ( t 1 , … , t k ) ↦ φ ( t ) . \varphi:D\to U\cap S,\quad\mathbf{t}=(t^1,\dots,t^k)\mapsto\varphi(\mathbf{t}). φ : D → U ∩ S , t = ( t 1 , … , t k ) ↦ φ ( t ) .
同胚指的是连续且反函数也连续的双射.
此时称 ( U ∩ S , φ ) (U\cap S,\varphi) ( U ∩ S , φ ) p p p ( t 1 , … , t k ) (t^1,\dots,t^k) ( t 1 , … , t k ) φ ( t ) \varphi(\mathbf{t}) φ ( t ) S S S U = { ( U α , φ α ) ∣ α ∈ A } \mathcal U=\left\{(U_\alpha,\varphi_\alpha)\mid\alpha\in\mathcal A\right\} U = { ( U α , φ α ) ∣ α ∈ A } S = ⋃ α ∈ A U α S=\bigcup_{\alpha\in\mathcal A}U_\alpha S = ⋃ α ∈ A U α U \mathcal U U S S S
Exm. S 1 = { ( x , y ) ∣ x 2 + y 2 = 1 } S^1=\left\{(x,y)\mid x^2+y^2=1\right\} S 1 = { ( x , y ) ∣ x 2 + y 2 = 1 } U 1 : = { ( x , y ) ∈ S 1 ∣ y > 0 } , φ 1 : ( − 1 , 1 ) → U 1 , t ↦ ( t , 1 − t 2 ) U 2 : = { ( x , y ) ∈ S 1 ∣ y < 0 } , φ 2 : ( − 1 , 1 ) → U 2 , t ↦ ( t , − 1 − t 2 ) U 3 : = { ( x , y ) ∈ S 1 ∣ x > 0 } , φ 3 : ( − 1 , 1 ) → U 3 , t ↦ ( 1 − t 2 , t ) U 4 : = { ( x , y ) ∈ S 1 ∣ x < 0 } , φ 4 : ( − 1 , 1 ) → U 4 , t ↦ ( − 1 − t 2 , t ) U_1:=\left\{(x,y)\in S^1\mid y>0\right\},\quad\varphi_1:(-1,1)\to U_1,\quad t\mapsto\left(t,\sqrt{1-t^2}\right)\\
U_2:=\left\{(x,y)\in S^1\mid y<0\right\},\quad\varphi_2:(-1,1)\to U_2,\quad t\mapsto\left(t,-\sqrt{1-t^2}\right)\\
U_3:=\left\{(x,y)\in S^1\mid x>0\right\},\quad\varphi_3:(-1,1)\to U_3,\quad t\mapsto\left(\sqrt{1-t^2},t\right)\\
U_4:=\left\{(x,y)\in S^1\mid x<0\right\},\quad\varphi_4:(-1,1)\to U_4,\quad t\mapsto\left(-\sqrt{1-t^2},t\right)\\ U 1 := { ( x , y ) ∈ S 1 ∣ y > 0 } , φ 1 : ( − 1 , 1 ) → U 1 , t ↦ ( t , 1 − t 2 ) U 2 := { ( x , y ) ∈ S 1 ∣ y < 0 } , φ 2 : ( − 1 , 1 ) → U 2 , t ↦ ( t , − 1 − t 2 ) U 3 := { ( x , y ) ∈ S 1 ∣ x > 0 } , φ 3 : ( − 1 , 1 ) → U 3 , t ↦ ( 1 − t 2 , t ) U 4 := { ( x , y ) ∈ S 1 ∣ x < 0 } , φ 4 : ( − 1 , 1 ) → U 4 , t ↦ ( − 1 − t 2 , t )
则上述几个 ( U i , φ i ) (U_i,\varphi_i) ( U i , φ i ) S 1 S^1 S 1 S 1 S^1 S 1 1 1 1
Exm. S 2 = { ( x , y , z ) ∣ x 2 + y 2 + z 2 = 1 } S^2=\left\{(x,y,z)\mid x^2+y^2+z^2=1\right\} S 2 = { ( x , y , z ) ∣ x 2 + y 2 + z 2 = 1 }
类似 S 1 S^1 S 1
球极投影, 即将每个点投影到它和北极点的连线与 x O y xOy x O y
Exm. S : = { ( x , y ) ∣ x 3 = y 2 } S:=\left\{(x,y)\mid x^3=y^2\right\} S := { ( x , y ) ∣ x 3 = y 2 } φ : R → S , t ↦ ( t 2 , t 3 ) , \varphi:\mathbb{R}\to S,\quad t\mapsto(t^2,t^3), φ : R → S , t ↦ ( t 2 , t 3 ) ,
是一个 1 1 1
Exm. 曲线φ : ( − 1 , 1 ) → R 2 , t ↦ ( t ( t 2 − 4 ) , t 2 − 4 ) \varphi:(-1,1)\to\mathbb{R}^2,\quad t\mapsto(t(t^2-4),t^2-4) φ : ( − 1 , 1 ) → R 2 , t ↦ ( t ( t 2 − 4 ) , t 2 − 4 )
不是 1 1 1 ( 0 , 0 ) (0,0) ( 0 , 0 ) ( − 1 , 1 ) (-1,1) ( − 1 , 1 ) φ ( 2 ) = φ ( − 2 ) = ( 0 , 0 ) \varphi(2)=\varphi(-2)=(0,0) φ ( 2 ) = φ ( − 2 ) = ( 0 , 0 )
Exm. 曲线γ : ( − π / 2 , 3 π / 2 ) → R 2 , t ↦ ( sin 2 t , cos t ) \gamma:(-\pi/2,3\pi/2)\to\mathbb{R}^2,\quad t\mapsto(\sin 2t,\cos t) γ : ( − π /2 , 3 π /2 ) → R 2 , t ↦ ( sin 2 t , cos t )
不是 1 1 1
光滑曲面
Def. 设 S ⊆ R n S\subseteq\mathbb{R}^n S ⊆ R n k k k ∀ p ∈ S \forall p\in S ∀ p ∈ S ( U , φ ) (U,\varphi) ( U , φ ) φ : D → U \varphi:D\to U φ : D → U C m C^m C m D D D rank φ ≡ k \operatorname{rank}\varphi\equiv k rank φ ≡ k S S S C m C^m C m
Exm. S : = { ( x , y ) ∣ x 3 = y 2 } S:=\left\{(x,y)\mid x^3=y^2\right\} S := { ( x , y ) ∣ x 3 = y 2 } φ ′ ( 0 ) = ( 0 , 0 ) \varphi'(0)=(0,0) φ ′ ( 0 ) = ( 0 , 0 ) 0 0 0
设 S S S R n \mathbb{R}^n R n k k k C m C^m C m p ∈ S p\in S p ∈ S p p p ( U 1 , φ 1 ) , ( U 2 , φ 2 ) (U_1,\varphi_1),(U_2,\varphi_2) ( U 1 , φ 1 ) , ( U 2 , φ 2 ) φ 2 − 1 ∘ φ 1 : φ 1 − 1 ( U 1 ∩ U 2 ) → φ 2 − 1 ( U 1 ∩ U 2 ) , ( ∗ ) \varphi_2^{-1}\circ\varphi_1:\varphi_1^{-1}(U_1\cap U_2)\to\varphi_2^{-1}(U_1\cap U_2),\tag{$\ast$} φ 2 − 1 ∘ φ 1 : φ 1 − 1 ( U 1 ∩ U 2 ) → φ 2 − 1 ( U 1 ∩ U 2 ) , ( ∗ )
这给出了 U 1 ∩ U 2 U_1\cap U_2 U 1 ∩ U 2
Q: 这一映射的性质如何?
Lm. 设 φ : R k → R n \varphi:\mathbb{R}^k\to\mathbb{R}^n φ : R k → R n C m C^m C m t 0 \mathbf{t}_0 t 0 k k k R n \mathbb{R}^n R n ( t 0 , 0 ) (\mathbf{t}_0,\mathbf{0}) ( t 0 , 0 ) V = V t × V 0 ⊆ R k × R n − k V=V_t\times V_0\subseteq\mathbb{R}^k\times\mathbb{R}^{n-k} V = V t × V 0 ⊆ R k × R n − k C m C^m C m ψ : V → ψ ( V ) ⊆ R n , \psi:V\to\psi(V)\subseteq\mathbb{R}^n, ψ : V → ψ ( V ) ⊆ R n ,
使得 ψ ∣ V t × { 0 } = φ ∣ V t \psi|_{V_t\times\left\{\mathbf{0}\right\}}=\varphi|_{V_t} ψ ∣ V t × { 0 } = φ ∣ V t
Pf. 不妨设 det ( ∂ φ i ∂ x j ( t ) ) 1 ≤ i , j ≤ k ≠ 0 \det\left({\partial\varphi^i\over\partial x^j}(\mathbf{t})\right)_{1\leq i,j\leq k}\neq 0 det ( ∂ x j ∂ φ i ( t ) ) 1 ≤ i , j ≤ k = 0 t 0 \mathbf{t}_0 t 0 ψ : R n → R n , ( t 1 , … , t n ) ↦ ( φ 1 ( t ) , … , φ k ( t ) , φ k + 1 ( t ) + t k + 1 , … , φ n ( t ) + t n ) , \psi:\mathbb{R}^n\to\mathbb{R}^n,\quad(t^1,\dots,t^n)\mapsto(\varphi^1(\mathbf{t}),\dots,\varphi^k(\mathbf{t}),\varphi^{k+1}(\mathbf{t})+t^{k+1},\dots,\varphi^n(\mathbf{t})+t^n), ψ : R n → R n , ( t 1 , … , t n ) ↦ ( φ 1 ( t ) , … , φ k ( t ) , φ k + 1 ( t ) + t k + 1 , … , φ n ( t ) + t n ) ,
则此时 det J ψ ≠ 0 \det\mathbf{J}\psi\neq 0 det J ψ = 0 ( t 0 , 0 ) (\mathbf{t}_0,\mathbf{0}) ( t 0 , 0 ) ( t 0 , 0 ) (\mathbf{t}_0,\mathbf{0}) ( t 0 , 0 ) V = U × V 0 ⊆ R k × R n − k V=U\times V_0\subseteq\mathbb{R}^k\times\mathbb{R}^{n-k} V = U × V 0 ⊆ R k × R n − k ψ ∣ V \psi|_V ψ ∣ V C m C^m C m ψ ∣ U × { 0 } = φ ∣ U \psi|_{U\times\left\{\mathbf{0}\right\}}=\varphi|_U ψ ∣ U × { 0 } = φ ∣ U □ \Box □
Prop. 式 ( ∗ ) (\ast) ( ∗ ) C m C^m C m
Pf. 设 p p p ( U 2 , φ 2 ) (U_2,\varphi_2) ( U 2 , φ 2 ) t 0 \mathbf{t}_0 t 0 U : = U 1 ∩ U 2 U:=U_1\cap U_2 U := U 1 ∩ U 2 p p p W ⊆ U W\subseteq U W ⊆ U ( t 0 , 0 ) (\mathbf{t}_0,\mathbf{0}) ( t 0 , 0 ) V = V t × V 0 ⊆ R k × R n − k V=V_t\times V_0\subseteq\mathbb{R}^k\times\mathbb{R}^{n-k} V = V t × V 0 ⊆ R k × R n − k C m C^m C m ψ : V → ψ ( V ) = W . \psi:V\to\psi(V)=W. ψ : V → ψ ( V ) = W .
使得 ψ ∣ V t × { 0 } = φ 2 ∣ V t \psi|_{V_t\times\left\{\mathbf{0}\right\}}=\varphi_2|_{V_t} ψ ∣ V t × { 0 } = φ 2 ∣ V t π 1 : R k × R n − k → R k \pi_1:\mathbb{R}^k\times\mathbb{R}^{n-k}\to\mathbb{R}^k π 1 : R k × R n − k → R k φ 2 − 1 ∘ φ 1 ∣ φ 1 − 1 ( W ) = π 1 ∘ ψ − 1 ∘ φ 1 ∣ φ 1 − 1 ( W ) , \varphi_2^{-1}\circ\varphi_1|_{\varphi_1^{-1}(W)}=\pi_1\circ\psi^{-1}\circ\varphi_1|_{\varphi_1^{-1}(W)}, φ 2 − 1 ∘ φ 1 ∣ φ 1 − 1 ( W ) = π 1 ∘ ψ − 1 ∘ φ 1 ∣ φ 1 − 1 ( W ) ,
也即 φ 2 − 1 ∘ φ 1 \varphi_2^{-1}\circ\varphi_1 φ 2 − 1 ∘ φ 1 W W W C m C^m C m p p p U U U φ 2 − 1 ∘ φ 1 ∈ C m ( φ 1 − 1 ( U ) , φ 2 − 1 ( U ) ) . \varphi_2^{-1}\circ\varphi_1\in C^m(\varphi_1^{-1}(U),\varphi_2^{-1}(U)). φ 2 − 1 ∘ φ 1 ∈ C m ( φ 1 − 1 ( U ) , φ 2 − 1 ( U )) .
同理可证 φ 1 − 1 ∘ φ 2 \varphi_1^{-1}\circ\varphi_2 φ 1 − 1 ∘ φ 2 C m C^m C m C m C^m C m □ \Box □
Def. 称定义在 k k k C m C^m C m S S S f f f C m C^m C m ∀ p ∈ S \forall p\in S ∀ p ∈ S p p p ( U , φ ) (U,\varphi) ( U , φ ) f ∘ φ ∈ C m ( φ − 1 ( U ) ) f\circ\varphi\in C^m(\varphi^{-1}(U)) f ∘ φ ∈ C m ( φ − 1 ( U )) f ∈ C m ( S ) f\in C^m(S) f ∈ C m ( S )
由上一个命题可知这一定义的良定性.
Exm. 设 D ⊆ R n − 1 D\subseteq\mathbb{R}^{n-1} D ⊆ R n − 1 f : U → R f:U\to\mathbb{R} f : U → R C m C^m C m Γ f : = { ( x , f ( x ) ) ∣ x ∈ D } \Gamma_f:=\left\{(\mathbf{x},f(\mathbf{x}))\mid\mathbf{x}\in D\right\} Γ f := { ( x , f ( x )) ∣ x ∈ D }
为 n − 1 n-1 n − 1 C m C^m C m
Pf. 定义映射φ : D → Γ f , x ↦ ( x , f ( x ) ) , \varphi:D\to\Gamma_f,\quad\mathbf{x}\mapsto(\mathbf{x},f(\mathbf{x})), φ : D → Γ f , x ↦ ( x , f ( x )) ,
则 φ \varphi φ C m C^m C m rank φ = r ( I n − 1 ∇ f ) = n − 1 , \operatorname{rank}\varphi=\operatorname{r}\begin{pmatrix}\mathbf{I}_{n-1}\\ \nabla f\end{pmatrix}=n-1, rank φ = r ( I n − 1 ∇ f ) = n − 1 ,
故 Γ f \Gamma_f Γ f n − 1 n-1 n − 1 C m C^m C m □ \Box □
Prop. 设 Ω ⊆ R n \Omega\subseteq\mathbb{R}^n Ω ⊆ R n f ∈ C m ( Ω ) , a ∈ f ( Ω ) f\in C^m(\Omega),a\in f(\Omega) f ∈ C m ( Ω ) , a ∈ f ( Ω ) ∀ x ∈ f − 1 ( a ) \forall\mathbf{x}\in f^{-1}(a) ∀ x ∈ f − 1 ( a ) ∇ f ( x ) ≠ 0 \nabla f(\mathbf{x})\neq\mathbf{0} ∇ f ( x ) = 0 f − 1 ( a ) f^{-1}(a) f − 1 ( a ) R n \mathbb{R}^n R n n − 1 n-1 n − 1
此时称这样的 a a a f f f
Pf. 对 ∀ x 0 ∈ f − 1 ( a ) \forall x_0\in f^{-1}(a) ∀ x 0 ∈ f − 1 ( a ) ∇ f ( x 0 ) ≠ 0 \nabla f(\mathbf{x}_0)\neq\mathbf{0} ∇ f ( x 0 ) = 0 ∂ f ∂ x n ( x 0 ) ≠ 0 {\partial f\over\partial x^n}(\mathbf{x}_0)\neq 0 ∂ x n ∂ f ( x 0 ) = 0 t 0 = ( x 0 1 , … , x 0 n − 1 ) \mathbf{t}_0=(x_0^1,\dots,x_0^{n-1}) t 0 = ( x 0 1 , … , x 0 n − 1 ) D t D_t D t x 0 n x_0^n x 0 n D x D_x D x C m C^m C m ψ : D t → D x \psi:D_t\to D_x ψ : D t → D x
使得 ( t , x n ) ∈ D t × D x (\mathbf{t},x^n)\in D_t\times D_x ( t , x n ) ∈ D t × D x f ( t , x n ) = a f(\mathbf{t},x^n)=a f ( t , x n ) = a x n = ψ ( t ) x^n=\psi(\mathbf{t}) x n = ψ ( t ) φ : D t → ( D t × D x ) ∩ f − 1 ( a ) , t ↦ ( t , φ ( t ) ) , \varphi:D_t\to (D_t\times D_x)\cap f^{-1}(a),\quad\mathbf{t}\mapsto(\mathbf{t},\varphi(\mathbf{t})), φ : D t → ( D t × D x ) ∩ f − 1 ( a ) , t ↦ ( t , φ ( t )) ,
则 ( φ ( D t ) , φ ) (\varphi(D_t),\varphi) ( φ ( D t ) , φ ) f − 1 ( a ) f^{-1}(a) f − 1 ( a ) x 0 \mathbf{x}_0 x 0 □ \Box □
Exm. 设 f ( x , y ) = x 2 + y 2 f(x,y)=x^2+y^2 f ( x , y ) = x 2 + y 2 f − 1 ( 1 ) = S 1 f^{-1}(1)=S^1 f − 1 ( 1 ) = S 1 1 1 1
Exm. 设 f ( x , y , z ) = x 2 + y 2 + z 2 f(x,y,z)=x^2+y^2+z^2 f ( x , y , z ) = x 2 + y 2 + z 2 f − 1 ( 1 ) = S 2 f^{-1}(1)=S^2 f − 1 ( 1 ) = S 2 2 2 2
Exm. 设 f ( x , y , z ) = x 2 + y 2 + 1 − z 2 f(x,y,z)=x^2+y^2+1-z^2 f ( x , y , z ) = x 2 + y 2 + 1 − z 2 f − 1 ( 0 ) f^{-1}(0) f − 1 ( 0 ) 2 2 2
Exm. 圆周 ( y − a ) 2 + z 2 = r 2 (y-a)^2+z^2=r^2 ( y − a ) 2 + z 2 = r 2 z z z z 2 + ( x 2 + y 2 − a ) 2 = r 2 . z^2+\left(\sqrt{x^2+y^2}-a\right)^2=r^2. z 2 + ( x 2 + y 2 − a ) 2 = r 2 .
令 f ( x , y , z ) = z 2 + ( x 2 + y 2 − a ) 2 f(x,y,z)=z^2+(\sqrt{x^2+y^2}-a)^2 f ( x , y , z ) = z 2 + ( x 2 + y 2 − a ) 2 f − 1 ( r 2 ) f^{-1}(r^2) f − 1 ( r 2 ) 2 2 2
Prop. 设 F : R n → R n − k F:\mathbb{R}^n\to\mathbb{R}^{n-k} F : R n → R n − k C m C^m C m F − 1 ( 0 ) F^{-1}(\mathbf{0}) F − 1 ( 0 ) F F F n − k n-k n − k F − 1 ( 0 ) F^{-1}(0) F − 1 ( 0 ) k k k
Pf. 仿照 k = 1 k=1 k = 1 □ \Box □
Exm. 设 F ( x , y , z ) = ( x 2 + y 2 + z 2 − 1 , x + y + z ) F(x,y,z)=(x^2+y^2+z^2-1,x+y+z) F ( x , y , z ) = ( x 2 + y 2 + z 2 − 1 , x + y + z ) S : = F − 1 ( 0 , 0 ) S:=F^{-1}(0,0) S := F − 1 ( 0 , 0 ) 1 1 1 S S S
Section X. 2. 曲面的定向
R 3 \mathbb{R}^3 R 3
设 S S S R 3 \mathbb{R}^3 R 3 p ∈ S p\in S p ∈ S ( U , φ ) (U,\varphi) ( U , φ ) φ : ( − 1 , 1 ) → R 3 , φ ( 0 ) = p . \varphi:(-1,1)\to\mathbb{R}^3,\quad\varphi(0)=p. φ : ( − 1 , 1 ) → R 3 , φ ( 0 ) = p .
则此时 φ \varphi φ S S S U U U ∀ q = φ ( t ) ∈ U \forall q=\varphi(t)\in U ∀ q = φ ( t ) ∈ U S S S φ ′ ( t ) \varphi'(t) φ ′ ( t ) S S S p p p ( V , ψ ) (V,\psi) ( V , ψ ) ∀ q = ψ ( s ) ∈ V \forall q=\psi(s)\in V ∀ q = ψ ( s ) ∈ V S S S ψ ′ ( s ) \psi'(s) ψ ′ ( s )
由之前的引理可知, 两个局部坐标系之间有变换 s = ψ − 1 ∘ φ ( t ) s=\psi^{-1}\circ\varphi(t) s = ψ − 1 ∘ φ ( t ) S S S q q q φ ′ ( t ) = ψ ′ ( s ) d s d t . \varphi'(t)=\psi'(s){\operatorname{}\!\mathrm{d}s\over\operatorname{}\!\mathrm{d}t}. φ ′ ( t ) = ψ ′ ( s ) d t d s .
因此, 只要 d s d t > 0 {\operatorname{}\!\mathrm{d}s\over\operatorname{}\!\mathrm{d}t}>0 d t d s > 0 φ ′ ( t ) ∥ J φ ( t ) ∥ = ψ ′ ( s ) ∥ J ψ ( s ) ∥ , {\varphi'(t)\over\|\mathbf{J}\varphi(t)\|}={\psi'(s)\over\|\mathbf{J}\psi(s)\|}, ∥ J φ ( t ) ∥ φ ′ ( t ) = ∥ J ψ ( s ) ∥ ψ ′ ( s ) ,
此时称两个坐标系为相容的.
Def. 设 S S S R 3 \mathbb{R}^3 R 3 { ( U α , φ α ) ∣ α ∈ A } \left\{(U_\alpha,\varphi_\alpha)\mid\alpha\in\mathcal A\right\} { ( U α , φ α ) ∣ α ∈ A } U α ∩ U β ≠ ∅ U_\alpha\cap U_\beta\neq\emptyset U α ∩ U β = ∅ α , β \alpha,\beta α , β φ β − 1 ∘ φ α : φ α − 1 ( U α ∩ U β ) → φ β − 1 ( U α ∩ U β ) \varphi_\beta^{-1}\circ\varphi_\alpha:\varphi_\alpha^{-1}(U_\alpha\cap U_\beta)\to\varphi_\beta^{-1}(U_\alpha\cap U_\beta) φ β − 1 ∘ φ α : φ α − 1 ( U α ∩ U β ) → φ β − 1 ( U α ∩ U β )
满足 ( φ β − 1 ∘ φ α ) ′ > 0 (\varphi_\beta^{-1}\circ\varphi_\alpha)'>0 ( φ β − 1 ∘ φ α ) ′ > 0 S S S
Rmk. 可以证明, 一维曲线总是可定向的.
设 S S S R 3 \mathbb{R}^3 R 3 p ∈ S p\in S p ∈ S ( U , φ ) (U,\varphi) ( U , φ ) φ : ( − 1 , 1 ) 2 → S , φ ( 0 , 0 ) = p . \varphi:(-1,1)^2\to S,\quad\varphi(0,0)=p. φ : ( − 1 , 1 ) 2 → S , φ ( 0 , 0 ) = p .
则此时在 ∀ q = φ ( t ) ∈ U \forall q=\varphi(\mathbf{t})\in U ∀ q = φ ( t ) ∈ U e 1 = ∂ φ ∂ t 1 = ( ∂ φ 1 ∂ t 1 ( t ) , ∂ φ 2 ∂ t 1 ( t ) , ∂ φ 3 ∂ t 1 ( t ) ) ⊺ , e 2 = ∂ φ ∂ t 2 = ( ∂ φ 1 ∂ t 2 ( t ) , ∂ φ 2 ∂ t 2 ( t ) , ∂ φ 3 ∂ t 2 ( t ) ) ⊺ . \mathbf{e}_1={\partial\varphi\over\partial t^1}=\left({\partial\varphi^1\over\partial t^1}(\mathbf{t}),{\partial\varphi^2\over\partial t^1}(\mathbf{t}),{\partial\varphi^3\over\partial t^1}(\mathbf{t})\right)^\intercal,\\
\mathbf{e}_2={\partial\varphi\over\partial t^2}=\left({\partial\varphi^1\over\partial t^2}(\mathbf{t}),{\partial\varphi^2\over\partial t^2}(\mathbf{t}),{\partial\varphi^3\over\partial t^2}(\mathbf{t})\right)^\intercal. e 1 = ∂ t 1 ∂ φ = ( ∂ t 1 ∂ φ 1 ( t ) , ∂ t 1 ∂ φ 2 ( t ) , ∂ t 1 ∂ φ 3 ( t ) ) ⊺ , e 2 = ∂ t 2 ∂ φ = ( ∂ t 2 ∂ φ 1 ( t ) , ∂ t 2 ∂ φ 2 ( t ) , ∂ t 2 ∂ φ 3 ( t ) ) ⊺ .
由于 rank φ = 2 \operatorname{rank}\varphi=2 rank φ = 2 e 1 , e 2 \mathbf{e}_1,\mathbf{e}_2 e 1 , e 2 q q q n = e 1 × e 2 ∥ e 1 × e 2 ∥ . \mathbf{n}={\mathbf{e}_1\times\mathbf{e}_2\over\|\mathbf{e}_1\times\mathbf{e}_2\|}. n = ∥ e 1 × e 2 ∥ e 1 × e 2 .
设 S S S p p p ( V , ψ ) (V,\psi) ( V , ψ ) q q q e ~ 1 , e ~ 2 , n ~ \mathbf{\widetilde e}_1,\mathbf{\widetilde e}_2,\mathbf{\widetilde n} e 1 , e 2 , n ψ − 1 ∘ φ : φ − 1 ( U ∩ V ) → ψ − 1 ( U ∩ V ) , \psi^{-1}\circ\varphi:\varphi^{-1}(U\cap V)\to\psi^{-1}(U\cap V), ψ − 1 ∘ φ : φ − 1 ( U ∩ V ) → ψ − 1 ( U ∩ V ) ,
记它的 Jocabian 为 J \mathbf{J} J ( e 1 , e 2 ) = ( e ~ 1 , e ~ 2 ) J . (\mathbf{e}_1,\mathbf{e}_2)=(\mathbf{\widetilde e}_1,\mathbf{\widetilde e}_2)\mathbf{J}. ( e 1 , e 2 ) = ( e 1 , e 2 ) J .
由于 N , N ~ \mathbf{N},\mathbf{\widetilde N} N , N q q q N = ± N ~ \mathbf{N}=\pm\mathbf{\widetilde N} N = ± N N = det J ∣ det J ∣ N ~ , \mathbf{N}={\det\mathbf{J}\over|\det\mathbf{J}|}\mathbf{\widetilde N}, N = ∣ det J ∣ det J N ,
于是只要 det J > 0 \det\mathbf{J}>0 det J > 0
Def. 设 S S S R 3 \mathbb{R}^3 R 3 { ( U α , φ α ) ∣ α ∈ A } \left\{(U_\alpha,\varphi_\alpha)\mid\alpha\in\mathcal A\right\} { ( U α , φ α ) ∣ α ∈ A } U α ∩ U β ≠ ∅ U_\alpha\cap U_\beta\neq\emptyset U α ∩ U β = ∅ α , β \alpha,\beta α , β φ β − 1 ∘ φ α : φ α − 1 ( U α ∩ U β ) → φ β − 1 ( U α ∩ U β ) \varphi_\beta^{-1}\circ\varphi_\alpha:\varphi_\alpha^{-1}(U_\alpha\cap U_\beta)\to\varphi_\beta^{-1}(U_\alpha\cap U_\beta) φ β − 1 ∘ φ α : φ α − 1 ( U α ∩ U β ) → φ β − 1 ( U α ∩ U β )
满足 det J ( φ β − 1 ∘ φ α ) > 0 \det\mathbf{J}(\varphi_\beta^{-1}\circ\varphi_\alpha)>0 det J ( φ β − 1 ∘ φ α ) > 0 S S S
Prop. 对 R 3 \mathbb{R}^3 R 3 S S S S S S S S S N \mathbf{N} N
单位法向量场即每个点到该点的一个单位法向量的映射.
Pf. 必要性显然, 下证充分性. 如果存在这样的单位法向量场 N \mathbf{N} N S S S ( U , φ ) (U,\varphi) ( U , φ ) U U U ⟨ N , φ t 1 × φ t 2 ∥ φ t 1 × φ t 2 ′ ∥ ⟩ \left\langle\mathbf{N},{\varphi_{t^1}\times\varphi_{t^2}\over\|\varphi_{t^1}\times\varphi_{t^2}'\|}\right\rangle ⟨ N , ∥ φ t 1 × φ t 2 ′ ∥ φ t 1 × φ t 2 ⟩
在 U U U ± 1 \pm 1 ± 1 t 1 , t 2 t^1,t^2 t 1 , t 2 U U U N = φ t 1 × φ t 2 ∥ φ t 1 × φ t 2 ′ ∥ , \mathbf{N}={\varphi_{t^1}\times\varphi_{t^2}\over\|\varphi_{t^1}\times\varphi_{t^2}'\|}, N = ∥ φ t 1 × φ t 2 ′ ∥ φ t 1 × φ t 2 ,
于是得到了 S S S □ \Box □
Exm (Möbius 带). 设 A B = { ( x , y , z ) ∈ R 3 ∣ x = 0 , y = 2 , ∣ z ∣ < 1 } AB=\left\{(x,y,z)\in\mathbb{R}^3\mid x=0,y=2,|z|<1\right\} A B = { ( x , y , z ) ∈ R 3 ∣ x = 0 , y = 2 , ∣ z ∣ < 1 } y O z yOz y O z A B AB A B x y xy x y x 2 + y 2 = 4 x^2+y^2=4 x 2 + y 2 = 4 z z z A B AB A B θ \theta θ 1 2 θ \frac 12\theta 2 1 θ
它有参数表示:φ ( u , v ) = ( ( 2 − u sin v 2 ) sin v , ( 2 − u sin v 2 ) cos v , u cos v 2 ) , \varphi(u,v)=\left(\left(2-u\sin\frac v2\right)\sin v,\left(2-u\sin\frac v2\right)\cos v,u\cos\frac v2\right), φ ( u , v ) = ( ( 2 − u sin 2 v ) sin v , ( 2 − u sin 2 v ) cos v , u cos 2 v ) ,
其中 u ∈ ( − 1 , 1 ) , v ∈ [ 0 , 2 π ) u\in(-1,1),v\in[0,2\pi) u ∈ ( − 1 , 1 ) , v ∈ [ 0 , 2 π )
Exm. 设 U ⊆ R 3 U\subseteq\mathbb{R}^3 U ⊆ R 3 F ∈ C m ( U ) , a F\in C^m(U),a F ∈ C m ( U ) , a F F F S : = F − 1 ( a ) S:=F^{-1}(a) S := F − 1 ( a )
Pf. 只需取 N : = ∇ F ∥ ∇ F ∥ N:={\nabla F\over\|\nabla F\|} N := ∥∇ F ∥ ∇ F S S S □ \Box □
R n \mathbb{R}^n R n Def. 称 R n \mathbb{R}^n R n S S S { ( U α , φ α ) ∣ α ∈ A } \left\{(U_\alpha,\varphi_\alpha)\mid\alpha\in\mathcal A\right\} { ( U α , φ α ) ∣ α ∈ A } U α ∩ U β U_\alpha\cap U_\beta U α ∩ U β α , β \alpha,\beta α , β φ β − 1 ∘ φ α : φ α − 1 ( U α ∩ U β ) → φ β − 1 ( U α ∩ U β ) \varphi_\beta^{-1}\circ\varphi_\alpha:\varphi_\alpha^{-1}(U_\alpha\cap U_\beta)\to\varphi_\beta^{-1}(U_\alpha\cap U_\beta) φ β − 1 ∘ φ α : φ α − 1 ( U α ∩ U β ) → φ β − 1 ( U α ∩ U β )
的 Jocabian 的行列式 > 0 >0 > 0
这里的一个 "定向" 的含义是 k k k k k k S S S k = n − 1 k=n-1 k = n − 1
Exm. 设 Ω ⊆ R n \Omega\subseteq\mathbb{R}^n Ω ⊆ R n F : Ω → R n − k F:\Omega\to\mathbb{R}^{n-k} F : Ω → R n − k C m C^m C m a a a F F F S : = F − 1 ( a ) S:=F^{-1}(a) S := F − 1 ( a )
Pf. 对 ∀ p ∈ S \forall p\in S ∀ p ∈ S ∇ F 1 ( p ) , … , ∇ F n − k ( p ) \nabla F^1(p),\dots,\nabla F^{n-k}(p) ∇ F 1 ( p ) , … , ∇ F n − k ( p ) ( T p S ) ⊥ (\mathrm{T}_pS)^\perp ( T p S ) ⊥ ( U , φ ) (U,\varphi) ( U , φ ) p p p φ : ( − 1 , 1 ) k → U ⊆ S \varphi:(-1,1)^k\to U\subseteq S φ : ( − 1 , 1 ) k → U ⊆ S
为 C m C^m C m e 1 : = ( ∂ φ ∂ t 1 ) ⊺ , … , e k : = ( ∂ φ ∂ t k ) ⊺ \mathbf{e}_1:=\left(\partial\varphi\over\partial t^1\right)^\intercal,\quad\dots,\quad\mathbf{e}_k:=\left(\partial\varphi\over\partial t^k\right)^\intercal e 1 := ( ∂ t 1 ∂ φ ) ⊺ , … , e k := ( ∂ t k ∂ φ ) ⊺
张成了 U U U det ( ∇ F 1 ( p ) , … , ∇ F n − k ( p ) , e 1 ( p ) , … , e k ( p ) ) , \det\left(\nabla F^1(p),\dots,\nabla F^{n-k}(p),\mathbf{e}_1(p),\dots,\mathbf{e}_k(p)\right), det ( ∇ F 1 ( p ) , … , ∇ F n − k ( p ) , e 1 ( p ) , … , e k ( p ) ) ,
则我们知道这一行列式在 U U U ( U , φ ) (U,\varphi) ( U , φ ) t 1 , t 2 t^1,t^2 t 1 , t 2 > 0 >0 > 0 ( U , φ ) , ( V , ψ ) (U,\varphi),(V,\psi) ( U , φ ) , ( V , ψ ) ( e 1 φ , … , e k φ ) = ( e 1 ψ , … , e k ψ ) ( ∂ s i ∂ t j ) 1 ≤ i , j ≤ k . (\mathbf{e}_1^\varphi,\dots,\mathbf{e}_k^\varphi)=(\mathbf{e}_1^\psi,\dots,\mathbf{e}_k^\psi)\left(\partial s^i\over\partial t^j\right)_{1\leq i,j\leq k}. ( e 1 φ , … , e k φ ) = ( e 1 ψ , … , e k ψ ) ( ∂ t j ∂ s i ) 1 ≤ i , j ≤ k .
注意到( J F , e φ ) = ( J F , e ψ ) ( I n − k O O ( ∂ s i ∂ t j ) ) , (\mathbf{J} F,\mathbf{e}^\varphi)=(\mathbf{J} F,\mathbf{e}^\psi)\begin{pmatrix}
\mathbf{I}_{n-k} & \mathbf{O}\\
\mathbf{O} & \left(\partial s^i\over\partial t^j\right)
\end{pmatrix}, ( J F , e φ ) = ( J F , e ψ ) ( I n − k O O ( ∂ t j ∂ s i ) ) ,
两边取行列式可知 det ( ∂ s i ∂ t j ) > 0 \det\left(\partial s^i\over\partial t^j\right)>0 det ( ∂ t j ∂ s i ) > 0 □ \Box □
Section X. 3. 曲面的边界及其定向
曲面的边界
考虑闭圆盘 B ‾ ( 0 , 1 ) : = { ( x , y ) ∈ R 2 ∣ x 2 + y 2 ≤ 1 } \overline B(0,1):=\left\{(x,y)\in\mathbb{R}^2\mid x^2+y^2\leq 1\right\} B ( 0 , 1 ) := { ( x , y ) ∈ R 2 ∣ x 2 + y 2 ≤ 1 } R 2 \mathbb{R}^2 R 2 B ‾ ( 0 , 1 ) \overline B(0,1) B ( 0 , 1 ) 2 2 2 2 2 2
为了刻画这一边界, 我们定义H k : = { t = ( t 1 , … , t k ) ∣ t 1 ≤ 0 } \mathbb H^k:=\left\{\mathbf{t}=(t^1,\dots,t^k)\mid t_1\leq 0\right\} H k := { t = ( t 1 , … , t k ) ∣ t 1 ≤ 0 }
为下半平面, 则∂ H k = { t = ( t 1 , … , t k ) ∣ t 1 = 0 } . \partial\mathbb H^k=\left\{\mathbf{t}=(t^1,\dots,t^k)\mid t_1=0\right\}. ∂ H k = { t = ( t 1 , … , t k ) ∣ t 1 = 0 } .
Def. 设 S ⊆ R n S\subseteq\mathbb{R}^n S ⊆ R n ∀ p ∈ S \forall p\in S ∀ p ∈ S p p p U U U φ : R k → U ∩ S \varphi:\mathbb{R}^k\to U\cap S φ : R k → U ∩ S φ : H k → U ∩ S \varphi:\mathbb H^k\to U\cap S φ : H k → U ∩ S S S S k k k
特别地, 如果 φ \varphi φ C m C^m C m rank φ ≡ k \operatorname{rank}\varphi\equiv k rank φ ≡ k S S S k k k C m C^m C m
Rmk. 定义在 H k \mathbb H^k H k φ \varphi φ C m C^m C m φ \varphi φ H k \mathbb H^k H k C m C^m C m H k \mathbb H^k H k
Def. 设 S S S R n \mathbb{R}^n R n k k k C m C^m C m p ∈ S p\in S p ∈ S S S S S S S p p p ( U , φ ) (U,\varphi) ( U , φ ) φ − 1 ( U ) = H k \varphi^{-1}(U)=\mathbb H^k φ − 1 ( U ) = H k φ − 1 ( p ) ∈ ∂ H k \varphi^{-1}(p)\in\partial\mathbb H^k φ − 1 ( p ) ∈ ∂ H k S S S ∂ S \partial S ∂ S
Rmk. 可以证明, 对 S S S p p p p p p ( U , φ ) (U,\varphi) ( U , φ ) U U U R k \mathbb{R}^k R k
Prop. 设 S S S k k k C m C^m C m ∂ S ≠ ∅ \partial S\neq\emptyset ∂ S = ∅ ∂ S \partial S ∂ S k − 1 k-1 k − 1 C m C^m C m
Pf. 设 S S S { ( U α , φ α ) ∣ α ∈ A } ∪ { ( V β , ψ β ) ∣ β ∈ B } , \left\{(U_\alpha,\varphi_\alpha)\mid\alpha\in\mathcal A\right\}\cup\left\{(V_\beta,\psi_\beta)\mid\beta\in\mathcal B\right\}, { ( U α , φ α ) ∣ α ∈ A } ∪ { ( V β , ψ β ) ∣ β ∈ B } ,
其中 φ α : R k → U α , ψ β : H k → V β \varphi_\alpha:\mathbb{R}^k\to U_\alpha,\psi_\beta:\mathbb H^k\to V_\beta φ α : R k → U α , ψ β : H k → V β C m C^m C m ∂ S \partial S ∂ S { ( V β ∩ ∂ S , ψ b ∣ ∂ H k ) ∣ β ∈ B } . □ \left\{(V_\beta\cap\partial S,\psi_b|_{\partial\mathbb H^k})\mid\beta\in\mathcal B\right\}. \tag*{$\Box$} { ( V β ∩ ∂ S , ψ b ∣ ∂ H k ) ∣ β ∈ B } . □
Exm. 设 U ⊆ R n U\subseteq\mathbb{R}^n U ⊆ R n f ∈ C m ( U ) f\in C^m(U) f ∈ C m ( U ) 0 0 0 f f f M : = { x ∈ S ∣ f ( x ) ≤ 0 } , S : = { x ∈ U ∣ f ( x ) = 0 } , M:=\left\{x\in S\mid f(x)\leq 0\right\},\quad S:=\left\{x\in U\mid f(x)=0\right\}, M := { x ∈ S ∣ f ( x ) ≤ 0 } , S := { x ∈ U ∣ f ( x ) = 0 } ,
则 M M M R n \mathbb{R}^n R n n n n ∂ M = S \partial M=S ∂ M = S
Pf.
若 x ∈ M ∖ S x\in M\setminus S x ∈ M ∖ S x x x n n n V ⊆ U V\subseteq U V ⊆ U f ∣ V < 0 f|_V<0 f ∣ V < 0 ( V , id V ) (V,\operatorname{id}_V) ( V , id V ) x x x
若 x ∈ S x\in S x ∈ S ∇ f ( x 0 ) ≠ 0 \nabla f(x_0)\neq\mathbf{0} ∇ f ( x 0 ) = 0 x x x W ⊆ U W\subseteq U W ⊆ U C m C^m C m φ : R n → W \varphi:\mathbb{R}^n\to W φ : R n → W f ∘ φ ( t 1 , … , t n ) = t 1 . f\circ\varphi(t^1,\dots,t^n)=t^1. f ∘ φ ( t 1 , … , t n ) = t 1 .
记 ψ : = φ ∣ H n \psi:=\varphi|_{\mathbb H^n} ψ := φ ∣ H n ( W ∩ M , ψ ) (W\cap M,\psi) ( W ∩ M , ψ ) x x x □ \Box □
曲面边界的诱导定向
作为一张可定向曲面, 我们可以直接取 H n \mathbb H^n H n e 1 , … , e n \mathbf{e}_1,\dots,\mathbf{e}_n e 1 , … , e n H n \mathbb H^n H n ∂ H n ⊆ H n \partial\mathbb H^n\subseteq\mathbb H^n ∂ H n ⊆ H n e 2 , … , e n \mathbf{e}_2,\dots,\mathbf{e}_n e 2 , … , e n ∂ H n \partial\mathbb H^n ∂ H n H k \mathbb H^k H k
Prop. 设 S ⊆ R n S\subseteq\mathbb{R}^n S ⊆ R n k k k C m C^m C m k ≥ 2 k\geq 2 k ≥ 2 ∂ S \partial S ∂ S k − 1 k-1 k − 1 C m C^m C m
Pf. 设 { ( U α , φ α ) ∣ α ∈ A } \left\{(U_\alpha,\varphi_\alpha)\mid\alpha\in\mathcal A\right\} { ( U α , φ α ) ∣ α ∈ A } S S S U α ∩ ∂ S ≠ ∅ U_\alpha\cap\partial S\neq\emptyset U α ∩ ∂ S = ∅ φ α − 1 ( U α ) = H k \varphi_\alpha^{-1}(U_\alpha)=\mathbb H^k φ α − 1 ( U α ) = H k { ( U α ∩ ∂ S , φ α ∣ ∂ H k ) ∣ α ∈ A } \left\{(U_\alpha\cap\partial S,\varphi_\alpha|_{\partial\mathbb H^k})\mid\alpha\in\mathcal A\right\} { ( U α ∩ ∂ S , φ α ∣ ∂ H k ) ∣ α ∈ A }
为 ∂ S \partial S ∂ S ψ α : = φ α ∣ ∂ H k \psi_\alpha:=\varphi_\alpha|_{\partial\mathbb H^k} ψ α := φ α ∣ ∂ H k ψ β − 1 ∘ ψ α : φ α − 1 ( U α ∩ U β ∩ ∂ S ) → φ β − 1 ( U α ∩ U β ∩ ∂ S ) \psi_\beta^{-1}\circ\psi_\alpha:\varphi_\alpha^{-1}(U_\alpha\cap U_\beta\cap\partial S)\to\varphi_\beta^{-1}(U_\alpha\cap U_\beta\cap\partial S) ψ β − 1 ∘ ψ α : φ α − 1 ( U α ∩ U β ∩ ∂ S ) → φ β − 1 ( U α ∩ U β ∩ ∂ S )
的 Jocabian 的行列式 > 0 >0 > 0 t = ( t 1 , … , t k ) \mathbf{t}=(t^1,\dots,t^k) t = ( t 1 , … , t k ) s = ( s 1 , … , s k ) \mathbf{s}=(s^1,\dots,s^k) s = ( s 1 , … , s k ) t ∈ φ α − 1 ( U α ∩ U β ) ∩ ∂ H k \mathbf{t}\in\varphi_\alpha^{-1}(U_\alpha\cap U_\beta)\cap\partial\mathbb H^k t ∈ φ α − 1 ( U α ∩ U β ) ∩ ∂ H k s 1 ( t 1 − δ , t 2 , … , t n ) < 0 = s 1 ( t ) , ∀ δ > 0 s 1 ( t 1 , … , t j + δ , … , t n ) = 0 = s 1 ( t ) , ∀ δ ∈ R , 1 < j ≤ k s^1(t^1-\delta,t^2,\dots,t^n)<0=s^1(\mathbf{t}),\quad\forall\delta>0\\
s^1(t^1,\dots,t^j+\delta,\dots,t^n)=0=s^1(\mathbf{t}),\quad\forall\delta\in\mathbb{R},1<j\leq k\\ s 1 ( t 1 − δ , t 2 , … , t n ) < 0 = s 1 ( t ) , ∀ δ > 0 s 1 ( t 1 , … , t j + δ , … , t n ) = 0 = s 1 ( t ) , ∀ δ ∈ R , 1 < j ≤ k
对上下两式分别取 δ → 0 \delta\to 0 δ → 0 ∂ s 1 ∂ t 1 ( t ) > 0 , ∂ s 1 ∂ t j ( t ) = 0 , j = 2 , … , n . {\partial s^1\over\partial t^1}(\mathbf{t})>0,\quad {\partial s^1\over\partial t^j}(\mathbf{t})=0,\quad j=2,\dots,n. ∂ t 1 ∂ s 1 ( t ) > 0 , ∂ t j ∂ s 1 ( t ) = 0 , j = 2 , … , n .
于是将 det ( ∂ s i ∂ t j ) \det\left(\partial s^i\over\partial t^j\right) det ( ∂ t j ∂ s i ) det ( ∂ s i ∂ t j ) = ( ∂ s 1 ∂ t 1 ) det ( ∂ s i ∂ t j ) i , j > 1 > 0 ⟹ det ( ∂ s i ∂ t j ) i , j > 1 > 0 , \det\left(\partial s^i\over\partial t^j\right)=\left(\partial s^1\over\partial t^1\right)\det\left(\partial s^i\over\partial t^j\right)_{i,j>1}>0\\
\implies\det\left(\partial s^i\over\partial t^j\right)_{i,j>1}>0, det ( ∂ t j ∂ s i ) = ( ∂ t 1 ∂ s 1 ) det ( ∂ t j ∂ s i ) i , j > 1 > 0 ⟹ det ( ∂ t j ∂ s i ) i , j > 1 > 0 ,
于是就有结论成立. □ \Box □
Rmk. k = 1 k=1 k = 1 H 1 = R − \mathbb H^1=\mathbb{R}_- H 1 = R − R + \mathbb R_+ R + [ 0 , 1 ] [0,1] [ 0 , 1 ]
Rmk. 实际上我们可以证明, 1 1 1 R 1 , H 1 , [ 0 , 1 ] , S 1 . \mathbb{R}^1,\quad\mathbb H^1,\quad [0,1],\quad S^1. R 1 , H 1 , [ 0 , 1 ] , S 1 .
Rmk. R n \mathbb{R}^n R n n n n > 0 >0 > 0
Exm. 考虑 B ‾ ( 0 , 1 ) : = { ( x , y ) ∣ x 2 + y 2 ≤ 1 } \overline B(0,1):=\left\{(x,y)\mid x^2+y^2\leq 1\right\} B ( 0 , 1 ) := { ( x , y ) ∣ x 2 + y 2 ≤ 1 } φ : ( r , θ ) ↦ ( r cos θ , r sin θ ) , \varphi:(r,\theta)\mapsto(r\cos\theta,r\sin\theta), φ : ( r , θ ) ↦ ( r cos θ , r sin θ ) ,
此时 ∂ B ‾ ( 0 , 1 ) \partial\overline B(0,1) ∂ B ( 0 , 1 ) ψ : θ ↦ ( cos θ , sin θ ) \psi:\theta\mapsto(\cos\theta,\sin\theta) ψ : θ ↦ ( cos θ , sin θ )
Exm. 考虑 R 3 \mathbb{R}^3 R 3 S S S φ : ( 0 , π / 2 ] × ( 0 , 2 π ) → R 3 , ( θ 1 , θ 2 ) ↦ ( sin θ 1 cos θ 2 , sin θ 1 sin θ 2 , cos θ 1 ) . \varphi:(0,\pi/2]\times(0,2\pi)\to\mathbb{R}^3,\quad(\theta_1,\theta_2)\mapsto(\sin\theta_1\cos\theta_2,\sin\theta_1\sin\theta_2,\cos\theta_1). φ : ( 0 , π /2 ] × ( 0 , 2 π ) → R 3 , ( θ 1 , θ 2 ) ↦ ( sin θ 1 cos θ 2 , sin θ 1 sin θ 2 , cos θ 1 ) .
此时在 ∂ S \partial S ∂ S ψ : ( 0 , 2 π ) → R 3 , θ 2 ↦ ( cos θ 2 , sin θ 2 , 0 ) . \psi:(0,2\pi)\to\mathbb{R}^3,\quad\theta_2\mapsto(\cos\theta_2,\sin\theta_2,0). ψ : ( 0 , 2 π ) → R 3 , θ 2 ↦ ( cos θ 2 , sin θ 2 , 0 ) .
一般地, 对 R 3 \mathbb{R}^3 R 3 2 2 2 S S S ∂ S \partial S ∂ S ∂ S \partial S ∂ S S S S ∂ S \partial S ∂ S
Section X. 4. R n \mathbb{R}^n R n
Cauchy-Binet 公式
设 ξ 1 , ξ 2 \xi_1,\xi_2 ξ 1 , ξ 2 R 3 \mathbb{R}^3 R 3 V ( ξ 1 , ξ 2 ) V(\xi_1,\xi_2) V ( ξ 1 , ξ 2 ) V 2 ( ξ 1 , ξ 2 ) = ∥ ξ 1 ∥ 2 ∥ ξ 2 ∥ 2 sin 2 θ = ∥ ξ 1 ∥ 2 ∥ ξ 2 ∥ 2 ( 1 − ⟨ ξ 1 , ξ 2 ⟩ 2 ∥ ξ 1 2 ∥ ∥ ξ 2 ∥ 2 ) = ∥ ξ 1 ∥ 2 ∥ ξ 2 ∥ 2 − ⟨ ξ 1 , ξ 2 ⟩ 2 = det ( ⟨ ξ 1 , ξ 1 ⟩ ⟨ ξ 1 , ξ 2 ⟩ ⟨ ξ 2 , ξ 1 ⟩ ⟨ ξ 2 , ξ 2 ⟩ ) = det ( ξ ⊺ ξ ) , ξ = ( ξ 1 , ξ 2 ) . \begin{align*}
V^2(\xi_1,\xi_2)&=\|\xi_1\|^2\|\xi_2\|^2\sin^2\theta\\
&=\|\xi_1\|^2\|\xi_2\|^2\left(1-{\left\langle\xi_1,\xi_2\right\rangle^2\over\|\xi_1^2\|\|\xi_2\|^2}\right)\\
&=\|\xi_1\|^2\|\xi_2\|^2-\left\langle\xi_1,\xi_2\right\rangle^2\\
&=\det\begin{pmatrix}
\left\langle\xi_1,\xi_1\right\rangle & \left\langle\xi_1,\xi_2\right\rangle\\
\left\langle\xi_2,\xi_1\right\rangle & \left\langle\xi_2,\xi_2\right\rangle
\end{pmatrix}\\
&=\det(\xi^\intercal\xi),\quad\xi=(\xi_1,\xi_2).
\end{align*} V 2 ( ξ 1 , ξ 2 ) = ∥ ξ 1 ∥ 2 ∥ ξ 2 ∥ 2 sin 2 θ = ∥ ξ 1 ∥ 2 ∥ ξ 2 ∥ 2 ( 1 − ∥ ξ 1 2 ∥∥ ξ 2 ∥ 2 ⟨ ξ 1 , ξ 2 ⟩ 2 ) = ∥ ξ 1 ∥ 2 ∥ ξ 2 ∥ 2 − ⟨ ξ 1 , ξ 2 ⟩ 2 = det ( ⟨ ξ 1 , ξ 1 ⟩ ⟨ ξ 2 , ξ 1 ⟩ ⟨ ξ 1 , ξ 2 ⟩ ⟨ ξ 2 , ξ 2 ⟩ ) = det ( ξ ⊺ ξ ) , ξ = ( ξ 1 , ξ 2 ) .
同理, 对 R n \mathbb{R}^n R n k k k ξ 1 , … , ξ k \xi_1,\dots,\xi_k ξ 1 , … , ξ k V 2 ( ξ 1 , … , ξ k ) = det ( ⟨ ξ i , ξ j ⟩ ) ⟹ V ( ξ 1 , … , ξ k ) = det ( ⟨ ξ i , ξ j ⟩ ) . V^2(\xi_1,\dots,\xi_k)=\det\left(\left\langle\xi_i,\xi_j\right\rangle\right)
\implies V(\xi_1,\dots,\xi_k)=\sqrt{\det\left(\left\langle\xi_i,\xi_j\right\rangle\right)}. V 2 ( ξ 1 , … , ξ k ) = det ( ⟨ ξ i , ξ j ⟩ ) ⟹ V ( ξ 1 , … , ξ k ) = det ( ⟨ ξ i , ξ j ⟩ ) .
Prop. 根据 Cauchy-Binet 公式, 我们有V ( ξ 1 , … , ξ k ) = ∑ I ∈ A det ξ I ⊺ ξ I , V(\xi_1,\dots,\xi_k)=\sum_{I\in\mathcal A}\det\xi_I^\intercal\xi_I, V ( ξ 1 , … , ξ k ) = I ∈ A ∑ det ξ I ⊺ ξ I ,
其中ξ I = ( ξ j i i ) , A = { ( i 1 , … , i k ) ∣ 1 ≤ i 1 < ⋯ < i k ≤ n } . \xi_I=\left(\xi_j^{i_i}\right),\quad\mathcal A=\left\{(i_1,\dots,i_k)\mid 1\leq i_1<\cdots<i_k\leq n\right\}. ξ I = ( ξ j i i ) , A = { ( i 1 , … , i k ) ∣ 1 ≤ i 1 < ⋯ < i k ≤ n } .
Lm. 正交变换不改变 V ( ξ 1 , … , ξ k ) V(\xi_1,\dots,\xi_k) V ( ξ 1 , … , ξ k )
Prop. 选取一组合适的标准正交基, 使得在这组基下ξ i = ( ζ i , 0 , … , 0 ) , ζ i ∈ R k − 1 , i = 1 , … , k − 1 ξ k = ( ζ k , 1 , … , ζ k , k , 0 , … , 0 ) , \xi_i=(\zeta_i,0,\dots,0),\quad\zeta_i\in\mathbb{R}^{k-1},i=1,\dots,k-1\\
\xi_k=(\zeta_{k,1},\dots,\zeta_{k,k},0,\dots,0), ξ i = ( ζ i , 0 , … , 0 ) , ζ i ∈ R k − 1 , i = 1 , … , k − 1 ξ k = ( ζ k , 1 , … , ζ k , k , 0 , … , 0 ) ,
则我们知道V ( ξ 1 , … , ξ k ) = ∣ det ( ζ 1 , … , ζ k − 1 ) ∣ ∣ ζ k k ∣ , V(\xi_1,\dots,\xi_k)=|\det(\zeta_1,\dots,\zeta_{k-1})||\zeta_{kk}|, V ( ξ 1 , … , ξ k ) = ∣ det ( ζ 1 , … , ζ k − 1 ) ∣∣ ζ kk ∣ ,
也即 V ( ξ 1 , … , ξ k ) V(\xi_1,\dots,\xi_k) V ( ξ 1 , … , ξ k ) ξ 1 , … , ξ k − 1 \xi_1,\dots,\xi_{k-1} ξ 1 , … , ξ k − 1 k − 1 k-1 k − 1 ξ k \xi_k ξ k
k k k Prop. 设曲面 S S S φ : D → S \varphi:D\to S φ : D → S D ⊆ R k D\subseteq\mathbb{R}^k D ⊆ R k φ \varphi φ t 0 ∈ D \mathbf{t}_0\in D t 0 ∈ D φ ( t 0 ) \varphi(\mathbf{t}_0) φ ( t 0 ) ξ i = ∂ φ ∂ t i ( t 0 ) , i = 1 , … , k , \xi_i={\partial\varphi\over\partial t^i}(\mathbf{t}_0),\quad i=1,\dots,k, ξ i = ∂ t i ∂ φ ( t 0 ) , i = 1 , … , k ,
此时考虑一个充分小的 Δ t ∈ R k \Delta\mathbf{t}\in\mathbb{R}^k Δ t ∈ R k I I I t 0 \mathbf{t}_0 t 0 t 0 + Δ t \mathbf{t}_0+\Delta\mathbf{t} t 0 + Δ t φ ( I ) \varphi(I) φ ( I ) det ( ⟨ ξ i , ξ j ⟩ ) ∏ i = 1 k ∣ Δ t i ∣ \sqrt{\det(\left\langle\xi_i,\xi_j\right\rangle)}\prod_{i=1}^k|\Delta t^i| det ( ⟨ ξ i , ξ j ⟩ ) i = 1 ∏ k ∣Δ t i ∣
给出. 此时 g : = ⟨ ξ i , ξ j ⟩ \mathbf{g}:=\left\langle\xi_i,\xi_j\right\rangle g := ⟨ ξ i , ξ j ⟩ S S S R n \mathbb{R}^n R n g i j = ⟨ ξ i , ξ j ⟩ g_{ij}=\left\langle\xi_i,\xi_j\right\rangle g ij = ⟨ ξ i , ξ j ⟩
Def. 设 R n \mathbb{R}^n R n k k k φ : D → S \varphi:D\to S φ : D → S D ⊆ R k D\subseteq\mathbb{R}^k D ⊆ R k φ \varphi φ S S S V ( S ) : = ∫ D det ( ⟨ ∂ φ ∂ t i , ∂ φ ∂ t j ⟩ ) d t . V(S):=\int_D\sqrt{\det\left(\left\langle{\partial\varphi\over\partial t^i},{\partial\varphi\over\partial t^j}\right\rangle\right)}\operatorname{}\!\mathrm{d}\mathbf{t}. V ( S ) := ∫ D det ( ⟨ ∂ t i ∂ φ , ∂ t j ∂ φ ⟩ ) d t .
上述定义与坐标系的选取无关. 这是因为如果 S S S ψ : Ω → S \psi:\Omega\to S ψ : Ω → S s = g ( t ) = ψ − 1 ∘ φ ( t ) ⟹ ∂ φ i ∂ t j = ∂ ( ψ i ∘ g ) ∂ t j = ∑ l = 1 k ∂ g l ∂ t j ∂ ψ i ∂ s l . \mathbf{s}=g(\mathbf{t})=\psi^{-1}\circ\varphi(\mathbf{t})\\
\implies{\partial\varphi^i\over\partial t^j}={\partial(\psi^i\circ g)\over\partial t^j}=\sum_{l=1}^k{\partial g^l\over\partial t^j}{\partial\psi^i\over\partial s^l}. s = g ( t ) = ψ − 1 ∘ φ ( t ) ⟹ ∂ t j ∂ φ i = ∂ t j ∂ ( ψ i ∘ g ) = l = 1 ∑ k ∂ t j ∂ g l ∂ s l ∂ ψ i .
此时由积分的变量代换公式有∫ Ω det ( ⟨ ∂ ψ ∂ s i , ∂ ψ ∂ s j ⟩ ) d s = ∫ D det ( ⟨ ∂ ψ ∂ s i , ∂ ψ ∂ s j ⟩ ) ⋅ ∣ det J g ( t ) ∣ d t = ∫ D det ( ∂ g i ∂ s j ) ⊺ det ( ⟨ ∂ ψ ∂ s i , ∂ ψ ∂ s j ⟩ ) det ( ∂ g i ∂ s j ) d t = ∫ D det ( ⟨ ∂ φ ∂ t i , ∂ φ ∂ t j ⟩ ) d t \begin{align*}
\int_\Omega\sqrt{\det\left(\left\langle{\partial\psi\over\partial s^i},{\partial\psi\over\partial s^j}\right\rangle\right)}\operatorname{}\!\mathrm{d}\mathbf{s}&=\int_D\sqrt{\det\left(\left\langle{\partial\psi\over\partial s^i},{\partial\psi\over\partial s^j}\right\rangle\right)}\cdot|\det\mathbf{J} g(\mathbf{t})|\operatorname{}\!\mathrm{d}\mathbf{t}\\
&=\int_D\sqrt{\det\left(\partial g^i\over\partial s^j\right)^\intercal\det\left(\left\langle{\partial\psi\over\partial s^i},{\partial\psi\over\partial s^j}\right\rangle\right)\det\left(\partial g^i\over\partial s^j\right)}\operatorname{}\!\mathrm{d}\mathbf{t}\\
&=\int_D\sqrt{\det\left(\left\langle{\partial\varphi\over\partial t^i},{\partial\varphi\over\partial t^j}\right\rangle\right)}\operatorname{}\!\mathrm{d}\mathbf{t}\\
\end{align*} ∫ Ω det ( ⟨ ∂ s i ∂ ψ , ∂ s j ∂ ψ ⟩ ) d s = ∫ D det ( ⟨ ∂ s i ∂ ψ , ∂ s j ∂ ψ ⟩ ) ⋅ ∣ det J g ( t ) ∣ d t = ∫ D det ( ∂ s j ∂ g i ) ⊺ det ( ⟨ ∂ s i ∂ ψ , ∂ s j ∂ ψ ⟩ ) det ( ∂ s j ∂ g i ) d t = ∫ D det ( ⟨ ∂ t i ∂ φ , ∂ t j ∂ φ ⟩ ) d t
Rmk. 当 k = n k=n k = n det ( ⟨ ∂ φ ∂ t i , ∂ φ ∂ t j ⟩ ) = det ( ( ∂ i φ ∂ t j ) ⊺ ( ∂ i φ ∂ t j ) ) = d e t 2 ( ∂ i φ ∂ t j ) = d e t 2 J φ . \det\left(\left\langle{\partial\varphi\over\partial t^i},{\partial\varphi\over\partial t^j}\right\rangle\right)=\det\left(\left({\partial^i\varphi\over\partial t^j}\right)^\intercal\left({\partial^i\varphi\over\partial t^j}\right)\right)={\rm det}^2\left({\partial^i\varphi\over\partial t^j}\right)={\rm det}^2\mathbf{J}\varphi. det ( ⟨ ∂ t i ∂ φ , ∂ t j ∂ φ ⟩ ) = det ( ( ∂ t j ∂ i φ ) ⊺ ( ∂ t j ∂ i φ ) ) = det 2 ( ∂ t j ∂ i φ ) = det 2 J φ .
于是V ( S ) = ∫ D det ( ⟨ ∂ φ ∂ t i , ∂ φ ∂ t j ⟩ ) d t = ∫ D ∣ det J φ ∣ d t = ∫ S d x . V(S)=\int_D\sqrt{\det\left(\left\langle{\partial\varphi\over\partial t^i},{\partial\varphi\over\partial t^j}\right\rangle\right)}\operatorname{}\!\mathrm{d}\mathbf{t}=\int_D\left|\det\mathbf{J}\varphi\right|\operatorname{}\!\mathrm{d}\mathbf{t}=\int_S\operatorname{}\!\mathrm{d}\mathbf{x}. V ( S ) = ∫ D det ( ⟨ ∂ t i ∂ φ , ∂ t j ∂ φ ⟩ ) d t = ∫ D ∣ det J φ ∣ d t = ∫ S d x .
Exm. 设 S ⊆ R n S\subseteq\mathbb{R}^n S ⊆ R n V ( S ) = ∫ α β ( ∂ φ 1 ∂ t ) 2 + ⋯ + ( ∂ φ n ∂ t ) 2 d t ⏟ 弧长微元 . V(S)=\int_\alpha^\beta\underbrace{\sqrt{\left(\partial\varphi^1\over\partial t\right)^2+\cdots+\left(\partial\varphi^n\over\partial t\right)^2}\operatorname{}\!\mathrm{d}t}_{\text{弧长微元}}. V ( S ) = ∫ α β 弧长微元 ( ∂ t ∂ φ 1 ) 2 + ⋯ + ( ∂ t ∂ φ n ) 2 d t .
Exm. 设 S ⊆ R n S\subseteq\mathbb{R}^n S ⊆ R n E : = ⟨ ∂ φ ∂ t 1 , ∂ φ ∂ t 1 ⟩ ( = g 11 ) , F : = ⟨ ∂ φ ∂ t 1 , ∂ φ ∂ t 2 ⟩ ( = g 12 ) , G : = ⟨ ∂ φ ∂ t 2 , ∂ φ ∂ t 2 ⟩ ( = g 22 ) , E:=\left\langle{\partial\varphi\over\partial t^1},{\partial\varphi\over\partial t^1}\right\rangle(=g_{11}),\quad F:=\left\langle{\partial\varphi\over\partial t^1},{\partial\varphi\over\partial t^2}\right\rangle(=g_{12}),\quad G:=\left\langle{\partial\varphi\over\partial t^2},{\partial\varphi\over\partial t^2}\right\rangle(=g_{22}), E := ⟨ ∂ t 1 ∂ φ , ∂ t 1 ∂ φ ⟩ ( = g 11 ) , F := ⟨ ∂ t 1 ∂ φ , ∂ t 2 ∂ φ ⟩ ( = g 12 ) , G := ⟨ ∂ t 2 ∂ φ , ∂ t 2 ∂ φ ⟩ ( = g 22 ) ,
则其面积V ( S ) = ∫ D E G − F 2 d t 1 d t 2 . V(S)=\int_D\sqrt{EG-F^2}\operatorname{}\!\mathrm{d}t^1\operatorname{}\!\mathrm{d}t^2. V ( S ) = ∫ D EG − F 2 d t 1 d t 2 .
Exm. 计算 R 3 \mathbb{R}^3 R 3 R R R x = R sin θ cos φ , y = R sin θ sin φ , z = R cos θ , x=R\sin\theta\cos\varphi,\quad y=R\sin\theta\sin\varphi,\quad z=R\cos\theta, x = R sin θ cos φ , y = R sin θ sin φ , z = R cos θ ,
其中 0 < θ < π , 0 < φ < 2 π 0<\theta<\pi,0<\varphi<2\pi 0 < θ < π , 0 < φ < 2 π ξ 1 = ( R cos θ cos φ , R cos θ sin φ , − R sin θ ) , ξ 2 = ( − R sin θ sin φ , R sin θ cos φ , 0 ) , \xi_1=(R\cos\theta\cos\varphi,R\cos\theta\sin\varphi,-R\sin\theta),\\
\xi_2=(-R\sin\theta\sin\varphi,R\sin\theta\cos\varphi,0), ξ 1 = ( R cos θ cos φ , R cos θ sin φ , − R sin θ ) , ξ 2 = ( − R sin θ sin φ , R sin θ cos φ , 0 ) ,
则E = ⟨ ξ 1 , ξ 1 ⟩ = R 2 , F = ⟨ ξ 1 , ξ 2 ⟩ = 0 , G = ⟨ ξ 2 , ξ 2 ⟩ = R 2 sin 2 θ , E=\left\langle\xi_1,\xi_1\right\rangle=R^2,\quad F=\left\langle\xi_1,\xi_2\right\rangle=0,\quad G=\left\langle\xi_2,\xi_2\right\rangle=R^2\sin^2\theta, E = ⟨ ξ 1 , ξ 1 ⟩ = R 2 , F = ⟨ ξ 1 , ξ 2 ⟩ = 0 , G = ⟨ ξ 2 , ξ 2 ⟩ = R 2 sin 2 θ ,
于是V ( S ) = ∫ 0 2 π d φ ∫ 0 π R 2 sin θ d θ = 4 π R 2 . V(S)=\int_0^{2\pi}\operatorname{}\!\mathrm{d}\varphi\int_0^\pi R^2\sin\theta\operatorname{}\!\mathrm{d}\theta=4\pi R^2. V ( S ) = ∫ 0 2 π d φ ∫ 0 π R 2 sin θ d θ = 4 π R 2 .
Exm. R 3 \mathbb{R}^3 R 3 S S S y = f ( x ) y=f(x) y = f ( x ) x x x a < x < b , f ( x ) > 0 a<x<b,f(x)>0 a < x < b , f ( x ) > 0 S S S φ : ( a , b ) × ( 0 , 2 π ) → R 3 , ( u , v ) ↦ ( u , f ( u ) cos v , f ( u ) sin v ) . \varphi:(a,b)\times(0,2\pi)\to\mathbb{R}^3,\quad (u,v)\mapsto(u,f(u)\cos v,f(u)\sin v). φ : ( a , b ) × ( 0 , 2 π ) → R 3 , ( u , v ) ↦ ( u , f ( u ) cos v , f ( u ) sin v ) .
此时∂ φ ∂ u = ( 1 , f ′ ( u ) cos v , f ′ ( u ) sin v ) , ∂ φ ∂ v = ( 0 , − f ( u ) sin v , f ( u ) cos v ) , {\partial\varphi\over\partial u}=(1,f'(u)\cos v,f'(u)\sin v),\quad{\partial\varphi\over\partial v}=(0,-f(u)\sin v,f(u)\cos v), ∂ u ∂ φ = ( 1 , f ′ ( u ) cos v , f ′ ( u ) sin v ) , ∂ v ∂ φ = ( 0 , − f ( u ) sin v , f ( u ) cos v ) ,
于是E = 1 + f ′ ( u ) 2 , F = 0 , G = f 2 ( u ) , ⟹ V ( S ) = ∫ a b d u ∫ 0 2 π f ( u ) 1 + f ′ ( u ) 2 d v = 2 π ∫ a b f ( u ) 1 + f ′ ( u ) 2 d u . E=1+f'(u)^2,\quad F=0,\quad G=f^2(u),\\
\implies V(S)=\int_a^b\operatorname{}\!\mathrm{d}u\int_0^{2\pi}f(u)\sqrt{1+f'(u)^2}\operatorname{}\!\mathrm{d}v=2\pi\int_a^bf(u)\sqrt{1+f'(u)^2}\operatorname{}\!\mathrm{d}u. E = 1 + f ′ ( u ) 2 , F = 0 , G = f 2 ( u ) , ⟹ V ( S ) = ∫ a b d u ∫ 0 2 π f ( u ) 1 + f ′ ( u ) 2 d v = 2 π ∫ a b f ( u ) 1 + f ′ ( u ) 2 d u .
Exm. 计算环面φ ( u , v ) = ( ( 2 + cos u ) cos v , ( 2 + cos u ) sin v , sin u ) , 0 < u , v < 2 π \varphi(u,v)=((2+\cos u)\cos v,(2+\cos u)\sin v,\sin u),\quad 0<u,v<2\pi φ ( u , v ) = (( 2 + cos u ) cos v , ( 2 + cos u ) sin v , sin u ) , 0 < u , v < 2 π
的面积.
Sol I. 直接计算得到E = 1 , F = 0 , G = ( 2 + cos u ) 2 , ⟹ V ( S ) = ∫ 0 2 π d u ∫ 0 2 π ( 2 + cos u ) d v = 8 π 2 . E=1,\quad F=0,\quad G=(2+\cos u)^2,
\implies V(S)=\int_0^{2\pi}\operatorname{}\!\mathrm{d}u\int_0^{2\pi}(2+\cos u)\operatorname{}\!\mathrm{d}v=8\pi^2. E = 1 , F = 0 , G = ( 2 + cos u ) 2 , ⟹ V ( S ) = ∫ 0 2 π d u ∫ 0 2 π ( 2 + cos u ) d v = 8 π 2 .
Sol II. 将环面看做f ( z ) = 2 + 1 − z 2 , g ( z ) = 2 − 1 − z 2 f(z)=2+\sqrt{1-z^2},\quad g(z)=2-\sqrt{1-z^2} f ( z ) = 2 + 1 − z 2 , g ( z ) = 2 − 1 − z 2
绕着 z z z 2 π 2\pi 2 π V ( S ) = 2 π ∫ − 1 1 f ( z ) 1 + f ′ ( z ) 2 + g ( z ) 1 + g ′ ( z ) 2 d z = 2 π ∫ − 1 1 ( f ( z ) + g ( z ) ) 1 1 − z 2 d z = 8 π ∫ − 1 1 1 1 − z 2 d z = 8 π 2 . \begin{align*}
V(S)&=2\pi\int_{-1}^1f(z)\sqrt{1+f'(z)^2}+g(z)\sqrt{1+g'(z)^2}\operatorname{}\!\mathrm{d}z\\
&=2\pi\int_{-1}^1(f(z)+g(z)){1\over\sqrt{1-z^2}}\operatorname{}\!\mathrm{d}z\\
&=8\pi\int_{-1}^1{1\over\sqrt{1-z^2}}\operatorname{}\!\mathrm{d}z=8\pi^2.
\end{align*} V ( S ) = 2 π ∫ − 1 1 f ( z ) 1 + f ′ ( z ) 2 + g ( z ) 1 + g ′ ( z ) 2 d z = 2 π ∫ − 1 1 ( f ( z ) + g ( z )) 1 − z 2 1 d z = 8 π ∫ − 1 1 1 − z 2 1 d z = 8 π 2 .
Section X. 5. 第一型曲线, 曲面积分
Def. 设 S S S R n \mathbb{R}^n R n k k k f ∈ C ( S ) f\in C(S) f ∈ C ( S ) supp ( f ) \operatorname{supp}(f) supp ( f ) S S S ( U , φ ) (U,\varphi) ( U , φ ) f f f S S S ∫ S f d v = ∫ φ − 1 ( U ) f ∘ φ ( t ) ⋅ det ( ⟨ ∂ φ ∂ t i , ∂ φ ∂ t j ⟩ ) d t . \int_S f\operatorname{}\!\mathrm{d}v=\int_{\varphi^{-1}(U)}f\circ\varphi(\mathbf{t})\cdot\sqrt{\det\left(\left\langle{\partial\varphi\over\partial t^i},{\partial\varphi\over\partial t^j}\right\rangle\right)}\operatorname{}\!\mathrm{d}\mathbf{t}. ∫ S f d v = ∫ φ − 1 ( U ) f ∘ φ ( t ) ⋅ det ( ⟨ ∂ t i ∂ φ , ∂ t j ∂ φ ⟩ ) d t .
Rmk.
当 k = 1 k=1 k = 1 S S S γ \gamma γ d v \operatorname{}\!\mathrm{d}v d v d s \operatorname{}\!\mathrm{d}s d s
当 k = 2 k=2 k = 2 d v \operatorname{}\!\mathrm{d}v d v d σ \operatorname{}\!\mathrm{d}\sigma d σ
Rmk. 当 n = k n=k n = k
Exm. 设曲线 γ \gamma γ x 2 + y 2 + z 2 = R 2 x^2+y^2+z^2=R^2 x 2 + y 2 + z 2 = R 2 x + y + z = 0 x+y+z=0 x + y + z = 0 ∫ γ x 2 d s \int_\gamma x^2\operatorname{}\!\mathrm{d}s ∫ γ x 2 d s
Sol I. 由于 x , y , z x,y,z x , y , z ∫ γ x 2 d s = 1 3 ∫ γ ( x 2 + y 2 + z 2 ) d s = 1 3 R 2 ∫ γ d s = 2 3 π R 3 . \int_\gamma x^2\operatorname{}\!\mathrm{d}s=\frac 13\int_{\gamma}(x^2+y^2+z^2)\operatorname{}\!\mathrm{d}s=\frac 13R^2\int_{\gamma}\operatorname{}\!\mathrm{d}s=\frac 23\pi R^3. ∫ γ x 2 d s = 3 1 ∫ γ ( x 2 + y 2 + z 2 ) d s = 3 1 R 2 ∫ γ d s = 3 2 π R 3 .
Sol II. 我们直接给出 γ \gamma γ x + y + z = 0 x+y+z=0 x + y + z = 0 E w = e x + e y + e z \mathbf{E}_w=\mathbf{e}_x+\mathbf{e}_y+\mathbf{e}_z E w = e x + e y + e z E u = e x − e y , E v = e x + e y − 2 e z . \mathbf{E}_u=\mathbf{e}_x-\mathbf{e}_y,\quad\mathbf{E}_v=\mathbf{e}_x+\mathbf{e}_y-2\mathbf{e}_z. E u = e x − e y , E v = e x + e y − 2 e z .
对其单位化有e u = 1 2 ( e x − e y ) , e v = 1 6 ( e x + e y − 2 e z ) , e w = 1 3 ( e x + e y + e z ) . \mathbf{e}_u={1\over\sqrt 2}(\mathbf{e}_x-\mathbf{e}_y),\quad\mathbf{e}_v={1\over\sqrt 6}(\mathbf{e}_x+\mathbf{e}_y-2\mathbf{e}_z),\quad\mathbf{e}_w={1\over\sqrt 3}(\mathbf{e}_x+\mathbf{e}_y+\mathbf{e}_z). e u = 2 1 ( e x − e y ) , e v = 6 1 ( e x + e y − 2 e z ) , e w = 3 1 ( e x + e y + e z ) .
于是有坐标的正交替换( x y z ) = ( 1 2 1 6 1 3 − 1 2 1 6 1 3 0 − 2 6 1 3 ) ( u v w ) , \begin{pmatrix}
x\\[5pt] y\\[5pt] z
\end{pmatrix}=
\begin{pmatrix}
1\over\sqrt 2 & 1\over\sqrt 6 & 1\over\sqrt 3\\[5pt]
-{1\over\sqrt 2} & 1\over\sqrt 6 & 1\over\sqrt 3\\[5pt]
0 & -{2\over\sqrt 6} & 1\over\sqrt 3\\
\end{pmatrix}
\begin{pmatrix}
u\\[5pt] v\\[5pt] w
\end{pmatrix}, x y z = 2 1 − 2 1 0 6 1 6 1 − 6 2 3 1 3 1 3 1 u v w ,
在 γ \gamma γ w = 0 w=0 w = 0 u , v , w u,v,w u , v , w u = R cos t , v = R sin t , w = 0 , u=R\cos t,\quad v=R\sin t,\quad w=0, u = R cos t , v = R sin t , w = 0 ,
于是可以直接得到 γ \gamma γ x = 1 2 R cos t + 1 6 R sin t , y = − 1 2 R cos t + 1 6 R sin t , z = − 2 6 R sin t , 0 ≤ t ≤ 2 π . x={1\over\sqrt 2}R\cos t+{1\over\sqrt 6}R\sin t,\quad y=-{1\over\sqrt 2}R\cos t+{1\over\sqrt 6}R\sin t,\quad z=-{2\over\sqrt 6}R\sin t,\quad 0\leq t\leq 2\pi. x = 2 1 R cos t + 6 1 R sin t , y = − 2 1 R cos t + 6 1 R sin t , z = − 6 2 R sin t , 0 ≤ t ≤ 2 π .
随后用第一曲线积分的定义计算可得相同的结果.
Exm. 曲面 S = { ( x , y , z ) ∣ x 2 + y 2 + z 2 = R 2 , z ≥ k } S=\left\{(x,y,z)\mid x^2+y^2+z^2=R^2,z\geq k\right\} S = { ( x , y , z ) ∣ x 2 + y 2 + z 2 = R 2 , z ≥ k } 0 < k < R 0<k<R 0 < k < R ∫ S 1 z d v . \int_S{1\over z}\operatorname{}\!\mathrm{d}v. ∫ S z 1 d v .
Sol. 考虑曲面的参数化( x , y , z ) = φ ( θ 1 , θ 2 ) = ( R sin θ 1 sin θ 2 , R sin θ 1 cos θ 2 , R cos θ 1 ) , (x,y,z)=\varphi(\theta_1,\theta_2)=(R\sin\theta_1\sin\theta_2,R\sin\theta_1\cos\theta_2,R\cos\theta_1), ( x , y , z ) = φ ( θ 1 , θ 2 ) = ( R sin θ 1 sin θ 2 , R sin θ 1 cos θ 2 , R cos θ 1 ) ,
于是∫ S 1 z d v = ∫ 0 tan − 1 ( k / R ) ∫ 0 2 π R 2 sin θ 1 R cos θ 1 d θ 1 d θ 2 = 2 π R log R k . \int_S{1\over z}\operatorname{}\!\mathrm{d}v=\int_0^{\tan^{-1}(k/R)}\int_0^{2\pi}{R^2\sin\theta_1\over R\cos\theta_1}\operatorname{}\!\mathrm{d}\theta_1\operatorname{}\!\mathrm{d}\theta_2=2\pi R\log\frac Rk. ∫ S z 1 d v = ∫ 0 t a n − 1 ( k / R ) ∫ 0 2 π R cos θ 1 R 2 sin θ 1 d θ 1 d θ 2 = 2 π R log k R .
Claim (单位分解). 设 S ⊆ R n S\subseteq\mathbb{R}^n S ⊆ R n k k k S S S { ( U α , φ α ) ∣ α ∈ A } \left\{(U_\alpha,\varphi_\alpha)\mid\alpha\in\mathcal A\right\} { ( U α , φ α ) ∣ α ∈ A } { h i ∣ i = 1 , … , m } \left\{h_i\mid i=1,\dots,m\right\} { h i ∣ i = 1 , … , m }
h i : R n → R h_i:\mathbb{R}^n\to\mathbb{R} h i : R n → R 存在 α ∈ A \alpha\in\mathcal A α ∈ A supp h i ∩ S ⊆ U α \operatorname{supp}h_i\cap S\subseteq U_\alpha supp h i ∩ S ⊆ U α
∑ i = 1 m h i ( x ) = 1 , ∀ x ∈ S \sum_{i=1}^m h_i(x)=1,\forall x\in S ∑ i = 1 m h i ( x ) = 1 , ∀ x ∈ S
Rmk. 如果 supp ( f ) \operatorname{supp}(f) supp ( f ) f f f S S S ∫ S f d v = ∑ i = 1 m ∫ S f ⋅ h i d v , \int_S f\operatorname{}\!\mathrm{d}v=\sum_{i=1}^m\int_S f\cdot h_i\operatorname{}\!\mathrm{d}v, ∫ S f d v = i = 1 ∑ m ∫ S f ⋅ h i d v ,
这一定义与之前的定义相容.
Rmk. 在实际计算中, 我们也常将曲面分解为若干曲面片进行计算.
Section X. 6. 微分形式
张量
Def. 设 V V V R \mathbb{R} R n n n f : V × V × ⋯ × V ⏟ k → R f:\underbrace{V\times V\times\cdots\times V}_k\to\mathbb{R} f : k V × V × ⋯ × V → R
为 k k k
对 i = 1 , … , k i=1,\dots,k i = 1 , … , k f ( v 1 , … , v i + w i , … , v k ) = f ( v 1 , … , v k ) + f ( v 1 , … , w i , … , v k ) f(v_1,\dots,v_i+w_i,\dots,v_k)=f(v_1,\dots,v_k)+f(v_1,\dots,w_i,\dots, v_k) f ( v 1 , … , v i + w i , … , v k ) = f ( v 1 , … , v k ) + f ( v 1 , … , w i , … , v k )
对 i = 1 , … , k i=1,\dots,k i = 1 , … , k f ( v 1 , … , λ v i , … , v k ) = λ f ( v 1 , … , v k ) f(v_1,\dots,\lambda v_i,\dots,v_k)=\lambda f(v_1,\dots,v_k) f ( v 1 , … , λ v i , … , v k ) = λ f ( v 1 , … , v k )
记 V V V k k k T k ( V ) \mathrm{T}^k(V) T k ( V ) V V V k k k
Def. 对 f ∈ T k ( V ) , g ∈ T l ( V ) f\in\mathrm{T}^k(V),g\in\mathrm{T}^l(V) f ∈ T k ( V ) , g ∈ T l ( V ) ( f ⊗ g ) ( v 1 , … , v k + l ) = f ( v 1 , … , v k ) g ( v k + 1 , … , v k + l ) (f\otimes g)(v_1,\dots,v_{k+l})=f(v_1,\dots,v_k)g(v_{k+1},\dots,v_{k+l}) ( f ⊗ g ) ( v 1 , … , v k + l ) = f ( v 1 , … , v k ) g ( v k + 1 , … , v k + l )
为 f f f g g g v 1 , … , v k + l ∈ V v_1,\dots,v_{k+l}\in V v 1 , … , v k + l ∈ V
Prop. 张量积有如下性质:
( f 1 + f 2 ) ⊗ g = f 1 ⊗ g + f 2 ⊗ g (f_1+f_2)\otimes g=f_1\otimes g+f_2\otimes g ( f 1 + f 2 ) ⊗ g = f 1 ⊗ g + f 2 ⊗ g ( λ f ) ⊗ g = f ⊗ ( λ g ) = λ f ⊗ g (\lambda f)\otimes g=f\otimes(\lambda g)=\lambda f\otimes g ( λ f ) ⊗ g = f ⊗ ( λ g ) = λ f ⊗ g f ⊗ ( g 1 + g 2 ) = f ⊗ g 1 + f ⊗ g 2 f\otimes(g_1+g_2)=f\otimes g_1+f\otimes g_2 f ⊗ ( g 1 + g 2 ) = f ⊗ g 1 + f ⊗ g 2 f ⊗ g ⊗ h = f ⊗ ( g ⊗ h ) f\otimes g\otimes h=f\otimes(g\otimes h) f ⊗ g ⊗ h = f ⊗ ( g ⊗ h )
Prop. 设 { e i } \left\{\mathbf{e}_i\right\} { e i } V V V { ω i } \left\{\omega_i\right\} { ω i } V ∗ V^* V ∗ { ω i 1 ⊗ ⋯ ⊗ ω i k ∣ 1 ≤ i 1 , … , i k ≤ n } \left\{\omega_{i_1}\otimes\cdots\otimes\omega_{i_k}\mid 1\leq i_1,\dots,i_k\leq n\right\} { ω i 1 ⊗ ⋯ ⊗ ω i k ∣ 1 ≤ i 1 , … , i k ≤ n }
为 T k ( V ) \mathrm{T}^k(V) T k ( V ) dim T k ( V ) = n k \dim \mathrm{T}^k(V)=n^k dim T k ( V ) = n k
Def. 对 n n n π ∈ S n \pi\in\mathfrak S_n π ∈ S n π \pi π sgn π = 1 \operatorname{sgn}\pi=1 sgn π = 1 sgn π = − 1 \operatorname{sgn}\pi=-1 sgn π = − 1
Def.
对 k k k f f f ∀ σ ∈ S k \forall\sigma\in\mathfrak S_k ∀ σ ∈ S k f ( v σ 1 , … , v σ k ) = f ( v 1 , … , v k ) , f(v_{\sigma_1},\dots,v_{\sigma_k})=f(v_1,\dots,v_k), f ( v σ 1 , … , v σ k ) = f ( v 1 , … , v k ) ,
则称 f f f
对 k k k f f f ∀ σ ∈ S k \forall\sigma\in\mathfrak S_k ∀ σ ∈ S k f ( v σ 1 , … , v σ k ) = sgn σ ⋅ f ( v 1 , … , v k ) , f(v_{\sigma_1},\dots,v_{\sigma_k})=\operatorname{sgn}\sigma\cdot f(v_1,\dots,v_k), f ( v σ 1 , … , v σ k ) = sgn σ ⋅ f ( v 1 , … , v k ) ,
则称 f f f
记 V V V k k k A k ( V ) \mathrm{A}^k(V) A k ( V )
外积
Def. 定义反称化算子 A : T k ( V ) → A k ( V ) , f ↦ A f A:\mathrm{T}^k(V)\to\mathrm{A}^k(V),f\mapsto Af A : T k ( V ) → A k ( V ) , f ↦ A f A f ( v 1 , … , v k ) = 1 k ! ∑ σ ∈ S k sgn σ ⋅ f ( v σ 1 , … , v σ k ) Af(v_1,\dots,v_k)={1\over k!}\sum_{\sigma\in\mathfrak S_k}\operatorname{sgn}\sigma\cdot f(v_{\sigma_1},\dots,v_{\sigma_k}) A f ( v 1 , … , v k ) = k ! 1 σ ∈ S k ∑ sgn σ ⋅ f ( v σ 1 , … , v σ k )
为对 f f f
Def. 设 f ∈ A k ( V ) , g ∈ A l ( V ) f\in\mathrm{A}^k(V),g\in\mathrm{A}^l(V) f ∈ A k ( V ) , g ∈ A l ( V ) f f f g g g f ∧ g = ( k + l ) ! k ! l ! A ( f ⊗ g ) . f\wedge g={(k+l)!\over k!l!}A(f\otimes g). f ∧ g = k ! l ! ( k + l )! A ( f ⊗ g ) .
Note. 陈大广: 外积可以想象为 wedge 的音译.
Prop. 外积有如下性质:
由于 A A A
特别地, 外积满足 f ∧ g = ( − 1 ) k l g ∧ f f\wedge g=(-1)^{kl}g\wedge f f ∧ g = ( − 1 ) k l g ∧ f
Def. 设 V , W V,W V , W m , n m,n m , n φ ∈ L ( V , W ) \varphi\in\mathcal L(V,W) φ ∈ L ( V , W ) φ ∗ : T k ( W ) → T k ( V ) , φ ∗ : A k ( W ) → A k ( V ) , φ ∗ ( f ) ( v 1 , … , v k ) = f ( φ ( v 1 ) , … , φ ( v k ) ) . \varphi^*:\mathrm{T}^k(W)\to\mathrm{T}^k(V),\quad\varphi^*:\mathrm{A}^k(W)\to\mathrm{A}^k(V),\\
\varphi^*(f)(v_1,\dots,v_k)=f(\varphi(v_1),\dots,\varphi(v_k)). φ ∗ : T k ( W ) → T k ( V ) , φ ∗ : A k ( W ) → A k ( V ) , φ ∗ ( f ) ( v 1 , … , v k ) = f ( φ ( v 1 ) , … , φ ( v k )) .
Prop. 拉回映射有如下性质:
φ ∗ ( f ⊗ g ) = φ ∗ ( f ) ⊗ φ ∗ ( g ) , ∀ f ∈ T k ( V ) , g ∈ T l ( V ) \varphi^*(f\otimes g)=\varphi^*(f)\otimes\varphi^*(g),\quad\forall f\in\mathrm{T}^k(V),g\in\mathrm{T}^l(V) φ ∗ ( f ⊗ g ) = φ ∗ ( f ) ⊗ φ ∗ ( g ) , ∀ f ∈ T k ( V ) , g ∈ T l ( V ) φ ∗ ( f ∧ g ) = φ ∗ ( f ) ∧ φ ∗ ( g ) , ∀ f ∈ A k ( V ) , g ∈ A l ( V ) \varphi^*(f\wedge g)=\varphi^*(f)\wedge\varphi^*(g),\quad\forall f\in\mathrm{A}^k(V),g\in\mathrm{A}^l(V) φ ∗ ( f ∧ g ) = φ ∗ ( f ) ∧ φ ∗ ( g ) , ∀ f ∈ A k ( V ) , g ∈ A l ( V )
Exm. 设 V = R n V=\mathbb{R}^n V = R n v ∈ V v\in V v ∈ V f ∈ T 1 ( V ) = V ∗ f\in\mathrm{T}^1(V)=V^* f ∈ T 1 ( V ) = V ∗ f ( ξ ) = ⟨ ξ , v ⟩ , ξ ∈ V , f(\xi)=\left\langle\xi,v\right\rangle,\quad\xi\in V, f ( ξ ) = ⟨ ξ , v ⟩ , ξ ∈ V ,
从而可知 V ≅ V ∗ V\cong V^* V ≅ V ∗ f = v ♭ f=v^\flat f = v ♭
Exm. V = R 2 V=\mathbb{R}^2 V = R 2 f ∈ A 2 ( V ) f\in\mathrm{A}^2(V) f ∈ A 2 ( V ) f ( ξ , η ) = det ( ξ , η ) , ξ , η ∈ V . f(\xi,\eta)=\det(\xi,\eta),\quad\xi,\eta\in V. f ( ξ , η ) = det ( ξ , η ) , ξ , η ∈ V .
Exm. V = R 3 V=\mathbb{R}^3 V = R 3 v ∈ V v\in V v ∈ V f ∈ A 2 ( V ) f\in\mathrm{A}^2(V) f ∈ A 2 ( V ) f ( ξ , η ) = det ( v , ξ , η ) , ξ , η ∈ V . f(\xi,\eta)=\det(v,\xi,\eta),\quad\xi,\eta\in V. f ( ξ , η ) = det ( v , ξ , η ) , ξ , η ∈ V .
也可以定义 g ∈ A 3 ( V ) g\in\mathrm{A}^3(V) g ∈ A 3 ( V ) g ( ξ , η , ζ ) = det ( ξ , η , ζ ) , ∀ ξ , η , ζ ∈ V . g(\xi,\eta,\zeta)=\det(\xi,\eta,\zeta),\quad\forall\xi,\eta,\zeta\in V. g ( ξ , η , ζ ) = det ( ξ , η , ζ ) , ∀ ξ , η , ζ ∈ V .
Prop. 设 { e 1 , … , e n } \left\{\mathbf{e}_1,\dots,\mathbf{e}_n\right\} { e 1 , … , e n } V V V { ω i } \left\{\omega_i\right\} { ω i } V ∗ V^* V ∗ { ω i 1 ⊗ ⋯ ⊗ ω i k ∣ 1 ≤ i 1 < ⋯ < i k ≤ n } \left\{\omega_{i_1}\otimes\cdots\otimes\omega_{i_k}\mid 1\leq i_1<\dots<i_k\leq n\right\} { ω i 1 ⊗ ⋯ ⊗ ω i k ∣ 1 ≤ i 1 < ⋯ < i k ≤ n }
为 A k ( V ) \mathrm{A}^k(V) A k ( V ) dim A k ( V ) = ( n k ) \dim\mathrm{A}^k(V)={n\choose k} dim A k ( V ) = ( k n )
Hint. 为证明这组基线性无关, 对∑ 1 ≤ i 1 < ⋯ < i k ≤ n α i 1 , … , i k ω i 1 ⊗ ⋯ ⊗ ω i k = 0 , \sum_{1\leq i_1<\cdots<i_k\leq n}\alpha_{i_1,\dots,i_k}~\omega_{i_1}\otimes\cdots\otimes\omega_{i_k}=0, 1 ≤ i 1 < ⋯ < i k ≤ n ∑ α i 1 , … , i k ω i 1 ⊗ ⋯ ⊗ ω i k = 0 ,
我们需要证明每个 α i 1 , … , i k = 0 \alpha_{i_1,\dots,i_k}=0 α i 1 , … , i k = 0 ω i 1 ∧ ⋯ ∧ ω i k = k ! A ( ω i 1 ⊗ ⋯ ⊗ ω i k ) , \omega_{i_1}\wedge\cdots\wedge\omega_{i_k}=k!A(\omega_{i_1}\otimes\cdots\otimes\omega_{i_k}), ω i 1 ∧ ⋯ ∧ ω i k = k ! A ( ω i 1 ⊗ ⋯ ⊗ ω i k ) ,
从而可得求值公式( ω i 1 ∧ ⋯ ∧ ω i k ) ( v 1 , … , v k ) = det ( ω i i ( v j ) ) ⟹ ( ω i 1 ∧ ⋯ ∧ ω i k ) ( e j 1 , … , e j k ) = δ i 1 , … , i k j 1 , … , j k , \boxed{(\omega_{i_1}\wedge\cdots\wedge\omega_{i_k})(v_1,\dots,v_k)=\det(\omega_{i_i}(v_j))}\\
\implies(\omega_{i_1}\wedge\cdots\wedge\omega_{i_k})(\mathbf{e}_{j_1},\dots,\mathbf{e}_{j_k})=\delta_{i_1,\dots,i_k}^{j_1,\dots,j_k}, ( ω i 1 ∧ ⋯ ∧ ω i k ) ( v 1 , … , v k ) = det ( ω i i ( v j )) ⟹ ( ω i 1 ∧ ⋯ ∧ ω i k ) ( e j 1 , … , e j k ) = δ i 1 , … , i k j 1 , … , j k ,
其中δ i 1 , … , i k j 1 , … , j k : = { 1 i 1 , … , i k 互不相同, j 1 , … , j k 为 i 1 , … , i k 的偶置换 − 1 i 1 , … , i k 互不相同, j 1 , … , j k 为 i 1 , … , i k 的奇置换 0 o t h e r w i s e . \delta_{i_1,\dots,i_k}^{j_1,\dots,j_k}:=\begin{cases}
1 & {i_1,\dots,i_k}\text{ 互不相同, }j_1,\dots,j_k\text{ 为 }i_1,\dots,i_k\text{ 的偶置换}\\
-1 & {i_1,\dots,i_k}\text{ 互不相同, }j_1,\dots,j_k\text{ 为 }i_1,\dots,i_k\text{ 的奇置换}\\
0 & \rm otherwise.
\end{cases} δ i 1 , … , i k j 1 , … , j k := ⎩ ⎨ ⎧ 1 − 1 0 i 1 , … , i k 互不相同 , j 1 , … , j k 为 i 1 , … , i k 的偶置换 i 1 , … , i k 互不相同 , j 1 , … , j k 为 i 1 , … , i k 的奇置换 otherwise.
为广义 Kronecker 符号.
微分形式
Def. 设 D ⊆ R n D\subseteq\mathbb{R}^n D ⊆ R n ω \omega ω D D D k k k k k k k k k ∀ x ∈ D \forall x\in D ∀ x ∈ D T x D ≅ R n \mathrm{T}_xD\cong\mathbb{R}^n T x D ≅ R n k k k ω ( x ) ∈ A k ( T x D ) \omega(x)\in\mathrm{A}^k(\mathrm{T}_xD) ω ( x ) ∈ A k ( T x D ) ω \omega ω D D D A k ( R n ) \mathrm{A}^k(\mathbb{R}^n) A k ( R n )
Note. 在上下文明确时, 也可以用 ω ( ξ 1 , … , ξ k ) \omega(\xi_1,\dots,\xi_k) ω ( ξ 1 , … , ξ k ) ω ( x ) ( ξ 1 , … , ξ k ) \omega(\mathbf{x})(\xi_1,\dots,\xi_k) ω ( x ) ( ξ 1 , … , ξ k )
Note. 对形如 ω x y \omega_x^y ω x y x x x y y y
Exm. 设 D ⊆ R n D\subseteq\mathbb{R}^n D ⊆ R n f ∈ C ∞ ( D ) f\in C^\infty(D) f ∈ C ∞ ( D ) d f \operatorname{}\!\mathrm{d}f d f D D D 1 1 1 ξ = ξ 1 e 1 + ⋯ + ξ n e n ∈ T x D \xi=\xi^1\mathbf{e}_1+\cdots+\xi^n\mathbf{e}_n\in\mathrm{T}_xD ξ = ξ 1 e 1 + ⋯ + ξ n e n ∈ T x D d f ( x ) ( ξ ) = ∑ i = 1 n ∂ f ∂ x i ( x ) ⋅ ξ i . \operatorname{}\!\mathrm{d}f(\mathbf{x})(\xi)=\sum_{i=1}^n{\partial f\over\partial x^i}(\mathbf{x})\cdot\xi^i. d f ( x ) ( ξ ) = i = 1 ∑ n ∂ x i ∂ f ( x ) ⋅ ξ i .
当 f f f x i x^i x i d x i ( x ) ( ξ ) = ξ i \operatorname{}\!\mathrm{d}x^i(\mathbf{x})(\xi)=\xi^i d x i ( x ) ( ξ ) = ξ i d x i ( ξ ) \operatorname{}\!\mathrm{d}x^i(\xi) d x i ( ξ ) D D D k k k ω \omega ω ω ( x ) = ∑ 1 ≤ i 1 < ⋯ < i k ≤ n a i 1 , … , i k ( x ) d x i 1 ∧ ⋯ ∧ d x i k = 1 k ! ∑ 1 ≤ i 1 , … , i k ≤ n a i 1 , … , i k ( x ) d x i 1 ∧ ⋯ ∧ d x i k . \begin{align*}
\omega(\mathbf{x})&=\sum_{1\leq i_1<\cdots<i_k\leq n}a_{i_1,\dots,i_k}(\mathbf{x})\operatorname{}\!\mathrm{d}x^{i_1}\wedge\cdots\wedge\operatorname{}\!\mathrm{d}x^{i_k}\\\tag{$\ast$}
&={1\over k!}\sum_{1\leq i_1,\dots,i_k\leq n}a_{i_1,\dots,i_k}(\mathbf{x})\operatorname{}\!\mathrm{d}x^{i_1}\wedge\cdots\wedge\operatorname{}\!\mathrm{d}x^{i_k}.
\end{align*} ω ( x ) = 1 ≤ i 1 < ⋯ < i k ≤ n ∑ a i 1 , … , i k ( x ) d x i 1 ∧ ⋯ ∧ d x i k = k ! 1 1 ≤ i 1 , … , i k ≤ n ∑ a i 1 , … , i k ( x ) d x i 1 ∧ ⋯ ∧ d x i k . ( ∗ )
Def.
若每个 a i 1 , … , i k ∈ C r a_{i_1,\dots,i_k}\in C^r a i 1 , … , i k ∈ C r ω \omega ω C r C^r C r
若每个 a i 1 , … , i k ∈ C ∞ a_{i_1,\dots,i_k}\in C^\infty a i 1 , … , i k ∈ C ∞ ω \omega ω C ∞ C^\infty C ∞
称 D D D C r C^r C r k k k Ω r k ( D ) \Omega_r^k(D) Ω r k ( D ) Ω k ( D ) = Ω ∞ k ( D ) \Omega^k(D)=\Omega_\infty^k(D) Ω k ( D ) = Ω ∞ k ( D )
Rmk. Ω 0 ( D ) = C ∞ ( D ) \Omega^0(D)=C^\infty(D) Ω 0 ( D ) = C ∞ ( D )
Def. 定义外微分算子 d : Ω k ( D ) → Ω k + 1 ( D ) , k ≥ 0 \operatorname{}\!\mathrm{d}:\Omega^k(D)\to\Omega^{k+1}(D),k\geq 0 d : Ω k ( D ) → Ω k + 1 ( D ) , k ≥ 0 d ω = ∑ i = 1 n ∑ 1 ≤ i 1 < ⋯ < i k ≤ n ∂ a i 1 , … , i k ∂ x i d x i ∧ d x i 1 ∧ ⋯ ∧ d x i k = ∑ 1 ≤ i 1 < ⋯ < i k ≤ n d a i 1 , … , i k ∧ d x i 1 ∧ ⋯ ∧ d x i k \begin{align*}
\operatorname{}\!\mathrm{d}\omega&=\sum_{i=1}^n\sum_{1\leq i_1<\cdots<i_k\leq n}{\partial a_{i_1,\dots,i_k}\over\partial x^i}\operatorname{}\!\mathrm{d}x^i\wedge\operatorname{}\!\mathrm{d}x^{i_1}\wedge\cdots\wedge\operatorname{}\!\mathrm{d}x^{i_k}\\
&=\sum_{1\leq i_1<\cdots<i_k\leq n}\operatorname{}\!\mathrm{d}a_{i_1,\dots,i_k}\wedge\operatorname{}\!\mathrm{d}x^{i_1}\wedge\cdots\wedge\operatorname{}\!\mathrm{d}x^{i_k}
\end{align*} d ω = i = 1 ∑ n 1 ≤ i 1 < ⋯ < i k ≤ n ∑ ∂ x i ∂ a i 1 , … , i k d x i ∧ d x i 1 ∧ ⋯ ∧ d x i k = 1 ≤ i 1 < ⋯ < i k ≤ n ∑ d a i 1 , … , i k ∧ d x i 1 ∧ ⋯ ∧ d x i k
其中 ω ∈ Ω k ( D ) \omega\in\Omega^k(D) ω ∈ Ω k ( D ) a i 1 , … , i k a_{i_1,\dots,i_k} a i 1 , … , i k ( ∗ ) (\ast) ( ∗ )
Prop. 外微分算子有如下的性质:
Leibniz 公式: 对 ω k ∈ Ω k ( D ) , ω l ∈ Ω l ( D ) \omega^k\in\Omega^k(D),\omega^l\in\Omega^l(D) ω k ∈ Ω k ( D ) , ω l ∈ Ω l ( D ) d ( ω k ∧ ω l ) = d ω k ∧ ω l + ( − 1 ) k ω k ∧ d ω l \operatorname{}\!\mathrm{d}(\omega^k\wedge\omega^l)=\operatorname{}\!\mathrm{d}\omega^k\wedge\omega^l+(-1)^k\omega^k\wedge\operatorname{}\!\mathrm{d}\omega^l\\ d ( ω k ∧ ω l ) = d ω k ∧ ω l + ( − 1 ) k ω k ∧ d ω l
对 f ∈ Ω 0 ( D ) f\in\Omega^0(D) f ∈ Ω 0 ( D ) d f \operatorname{}\!\mathrm{d}f d f f f f
d 2 ω = d ( d ω ) = 0 , ∀ ω ∈ Ω k ( D ) , k ≥ 0 \operatorname{}\!\mathrm{d}^2\omega=\operatorname{}\!\mathrm{d}(\operatorname{}\!\mathrm{d}\omega)=0,\forall\omega\in\Omega^k(D),k\geq 0 d 2 ω = d ( d ω ) = 0 , ∀ ω ∈ Ω k ( D ) , k ≥ 0
第三条是因为 d 2 ω = ∑ i = 1 n ∑ j = 1 n ∑ 1 ≤ i 1 < ⋯ < i k ≤ n ∂ 2 a i 1 , … , i k ∂ x i ∂ x j d x i ∧ d x j ∧ d x i 1 ∧ ⋯ ∧ d x i k , \operatorname{}\!\mathrm{d}^2\omega=\sum_{i=1}^n\sum_{j=1}^n\sum_{1\leq i_1<\cdots<i_k\leq n}{\partial^2 a_{i_1,\dots,i_k}\over\partial x^i\partial x^j}\operatorname{}\!\mathrm{d}x^i\wedge\operatorname{}\!\mathrm{d}x^j\wedge\operatorname{}\!\mathrm{d}x^{i_1}\wedge\cdots\wedge\operatorname{}\!\mathrm{d}x^{i_k}, d 2 ω = i = 1 ∑ n j = 1 ∑ n 1 ≤ i 1 < ⋯ < i k ≤ n ∑ ∂ x i ∂ x j ∂ 2 a i 1 , … , i k d x i ∧ d x j ∧ d x i 1 ∧ ⋯ ∧ d x i k ,
从而由外积的反称性 ( d x i ∧ d x j = − d x j ∧ d x i ) \operatorname{}\!\mathrm{d}x^i\wedge\operatorname{}\!\mathrm{d}x^j=-\operatorname{}\!\mathrm{d}x^j\wedge\operatorname{}\!\mathrm{d}x^i) d x i ∧ d x j = − d x j ∧ d x i ) ∂ 2 a ∂ x i ∂ x j = ∂ 2 a ∂ x j ∂ x i {\partial^2 a\over\partial x^i\partial x^j}={\partial^2 a\over\partial x^j\partial x^i} ∂ x i ∂ x j ∂ 2 a = ∂ x j ∂ x i ∂ 2 a
光滑曲面上的微分形式
Def. 设 D ⊆ R n , G ⊆ R m D\subseteq\mathbb{R}^n,G\subseteq\mathbb{R}^m D ⊆ R n , G ⊆ R m φ : D → G \varphi:D\to G φ : D → G C ∞ C^\infty C ∞ ∀ x ∈ D \forall\mathbf{x}\in D ∀ x ∈ D φ ∗ : T x D → T φ ( x ) G , ξ ↦ d φ ( x ) ( ξ ) . \varphi_*:\mathrm{T}_\mathbf{x} D\to\mathrm{T}_{\varphi(\mathbf{x})}G,\quad\xi\mapsto\operatorname{}\!\mathrm{d}\varphi(\mathbf{x})(\xi). φ ∗ : T x D → T φ ( x ) G , ξ ↦ d φ ( x ) ( ξ ) .
Rmk. 此时对一条过 x ∈ D \mathbf{x}\in D x ∈ D γ : ( − 1 , 1 ) → D , γ ( 0 ) = x \gamma:(-1,1)\to D,\gamma(0)=\mathbf{x} γ : ( − 1 , 1 ) → D , γ ( 0 ) = x G G G Γ = φ ∘ γ \Gamma=\varphi\circ\gamma Γ = φ ∘ γ x \mathbf{x} x γ ′ ( 0 ) \gamma'(0) γ ′ ( 0 ) y = φ ( x ) \mathbf{y}=\varphi(\mathbf{x}) y = φ ( x ) Γ ′ ( 0 ) = ( ∂ ( φ i ∘ γ ) ( t ) ∂ t ) ⊺ = ( ∑ j = 1 m ∂ y i ∂ x j ∂ x j ∂ t ) ⊺ = d φ ( x ) ( γ ′ ( 0 ) ) = φ ∗ ( γ ′ ( 0 ) ) , \Gamma'(0)=\left(\partial(\varphi^i\circ\gamma)(t)\over\partial t\right)^\intercal=\left(\sum_{j=1}^m{\partial y^i\over\partial x^j}{\partial x^j\over\partial t}\right)^\intercal=\operatorname{}\!\mathrm{d}\varphi(\mathbf{x})(\gamma'(0))=\varphi_*(\gamma'(0)), Γ ′ ( 0 ) = ( ∂ t ∂ ( φ i ∘ γ ) ( t ) ) ⊺ = ( j = 1 ∑ m ∂ x j ∂ y i ∂ t ∂ x j ) ⊺ = d φ ( x ) ( γ ′ ( 0 )) = φ ∗ ( γ ′ ( 0 )) ,
也即 φ ∗ \varphi_* φ ∗
Rmk. 我们之前定义过对张量的拉回. 利用切映射我们也可以定义微分形式之间的拉回:φ ∗ : Ω k ( G ) → Ω k ( D ) , φ ∗ ( ω ) ( x ) ( v 1 , … , v k ) = ω ( φ ( x ) ) ( φ ∗ ( v 1 ) , … , φ ∗ ( v k ) ) . \varphi^*:\Omega^k(G)\to\Omega^k(D),\quad\varphi^*(\omega)(\mathbf{x})(v_1,\dots,v_k)=\omega(\varphi(\mathbf{x}))(\varphi_*(v_1),\dots,\varphi_*(v_k)). φ ∗ : Ω k ( G ) → Ω k ( D ) , φ ∗ ( ω ) ( x ) ( v 1 , … , v k ) = ω ( φ ( x )) ( φ ∗ ( v 1 ) , … , φ ∗ ( v k )) .
对一个 d y i ∈ Ω 1 ( G ) \operatorname{}\!\mathrm{d}y^i\in\Omega^1(G) d y i ∈ Ω 1 ( G ) φ ∗ ( d y i ) ( x ) ( ξ ) = d y i ( y ) ( φ ∗ ( ξ ) ) = d y i ( y ) ( d φ ( x ) ( ξ ) ) = ∑ j = 1 n ∂ y i ∂ x j d x j ( x ) ( ξ ) , \varphi^*(\operatorname{}\!\mathrm{d}y^i)(\mathbf{x})(\xi)=\operatorname{}\!\mathrm{d}y^i(\mathbf{y})(\varphi_*(\xi))=\operatorname{}\!\mathrm{d}y^i(\mathbf{y})(\operatorname{}\!\mathrm{d}\varphi(\mathbf{x})(\xi))=\sum_{j=1}^n{\partial y^i\over\partial x^j}\operatorname{}\!\mathrm{d}x^j(\mathbf{x})(\xi), φ ∗ ( d y i ) ( x ) ( ξ ) = d y i ( y ) ( φ ∗ ( ξ )) = d y i ( y ) ( d φ ( x ) ( ξ )) = j = 1 ∑ n ∂ x j ∂ y i d x j ( x ) ( ξ ) ,
从而实际上φ ∗ ( d y i ) = ∑ j = 1 n ∂ y i ∂ x j d x j ( = d φ ∗ ( y i ) ) . \varphi^*(\operatorname{}\!\mathrm{d}y^i)=\sum_{j=1}^n{\partial y^i\over\partial x^j}\operatorname{}\!\mathrm{d}x^j\left(=\operatorname{}\!\mathrm{d}\varphi^*(y^i)\right). φ ∗ ( d y i ) = j = 1 ∑ n ∂ x j ∂ y i d x j ( = d φ ∗ ( y i ) ) .
进一步对任意一个 ω ∈ Ω k ( G ) \omega\in\Omega^k(G) ω ∈ Ω k ( G ) φ ∗ ( d ω ) = d φ ∗ ( ω ) \varphi^*(\operatorname{}\!\mathrm{d}\omega)=\operatorname{}\!\mathrm{d}\varphi^*(\omega) φ ∗ ( d ω ) = d φ ∗ ( ω )
根据定义容易证明 φ ∗ ( ω 1 ∧ ω 2 ) = φ ∗ ω 1 ∧ φ ∗ ω 2 \varphi^*(\omega_1\wedge\omega_2)=\varphi^*\omega_1\wedge\varphi^*\omega_2 φ ∗ ( ω 1 ∧ ω 2 ) = φ ∗ ω 1 ∧ φ ∗ ω 2 k k k ω = ∑ 1 ≤ i 1 < ⋯ < i k ≤ m a i 1 , … , i k d y i 1 ∧ ⋯ ∧ d y i k , \omega=\sum_{1\leq i_1<\cdots<i_k\leq m}a_{i_1,\dots,i_k}\operatorname{}\!\mathrm{d}y^{i_1}\wedge\cdots\wedge\operatorname{}\!\mathrm{d}y^{i_k}, ω = 1 ≤ i 1 < ⋯ < i k ≤ m ∑ a i 1 , … , i k d y i 1 ∧ ⋯ ∧ d y i k ,
设 y = φ ( x ) \mathbf{y}=\varphi(\mathbf{x}) y = φ ( x ) φ ∗ ω = ∑ 1 ≤ i 1 < ⋯ < i k ≤ m a i 1 , … , i k ( ∑ j = 1 n φ x j i 1 d x j ) ∧ ⋯ ∧ ( ∑ j = 1 n φ x j i k d x j ) . \varphi^*\omega=\sum_{1\leq i_1<\cdots<i_k\leq m}a_{i_1,\dots,i_k}\left(\sum_{j=1}^n\varphi^{i_1}_{x^j}\operatorname{}\!\mathrm{d}x^j\right)\wedge\cdots\wedge\left(\sum_{j=1}^n\varphi^{i_k}_{x^j}\operatorname{}\!\mathrm{d}x^j\right). φ ∗ ω = 1 ≤ i 1 < ⋯ < i k ≤ m ∑ a i 1 , … , i k ( j = 1 ∑ n φ x j i 1 d x j ) ∧ ⋯ ∧ ( j = 1 ∑ n φ x j i k d x j ) .
Def. 设 S S S R n \mathbb{R}^n R n k k k ω \omega ω S S S p p p ∀ x ∈ S , ω ( x ) ∈ A p ( T x S ) \forall x\in S,\omega(x)\in\mathrm{A}^p(\mathrm{T}_xS) ∀ x ∈ S , ω ( x ) ∈ A p ( T x S )
Rmk. 对 ∀ x ∈ S \forall x\in S ∀ x ∈ S x x x ( U , φ ) (U,\varphi) ( U , φ ) ω \omega ω R k \mathbb{R}^k R k φ ∗ ω \varphi^*\omega φ ∗ ω φ ∗ ω \varphi^*\omega φ ∗ ω φ ∗ ω ( t ) = ∑ 1 ≤ i 1 < ⋯ < i p ≤ k a i 1 , … , i p ( t ) d t i 1 ∧ ⋯ ∧ d t i p . ( ∗ ) \varphi^*\omega(\mathbf{t})=\sum_{1\leq i_1<\dots<i_p\leq k}a_{i_1,\dots,i_p}(\mathbf{t})\operatorname{}\!\mathrm{d}t^{i_1}\wedge\cdots\wedge\operatorname{}\!\mathrm{d}t^{i_p}.\tag{$\ast$} φ ∗ ω ( t ) = 1 ≤ i 1 < ⋯ < i p ≤ k ∑ a i 1 , … , i p ( t ) d t i 1 ∧ ⋯ ∧ d t i p . ( ∗ )
Def. 称 ω \omega ω S S S ( ∗ ) (\ast) ( ∗ ) a i 1 , … , i p a_{i_1,\dots,i_p} a i 1 , … , i p S S S p p p Ω p ( S ) \Omega^p(S) Ω p ( S )
Exm. 在 R 2 \mathbb{R}^2 R 2 ω = ( x d y − y d x ) \omega=(x\operatorname{}\!\mathrm{d}y-y\operatorname{}\!\mathrm{d}x) ω = ( x d y − y d x ) S 1 S^1 S 1 ( x , y ) = φ ( θ ) = ( cos θ , sin θ ) (x,y)=\varphi(\theta)=(\cos\theta,\sin\theta) ( x , y ) = φ ( θ ) = ( cos θ , sin θ )
下, 可以得到φ ∗ ω ( θ ) = cos θ d sin θ − sin θ d cos θ = d θ . \varphi^*\omega(\theta)=\cos\theta\operatorname{}\!\mathrm{d}\sin\theta-\sin\theta\operatorname{}\!\mathrm{d}\cos\theta=\operatorname{}\!\mathrm{d}\theta. φ ∗ ω ( θ ) = cos θ d sin θ − sin θ d cos θ = d θ .