매개변수화된 곡면
📂기하학 매개변수화된 곡면 정의 열린집합 U ⊂ R 2 U \subset \mathbb{R}^{2} U ⊂ R 2 연결집합 A ⊂ R 2 A\subset \mathbb{R}^{2} A ⊂ R 2
U ⊂ A ⊂ U ‾ and ∂ A is piecewise differentiable
U \subset A \subset \overline{U} \quad \text{and} \quad \partial A \text{ is piecewise differentiable}
U ⊂ A ⊂ U and ∂ A is piecewise differentiable
s : A → M s : A \to M s : A → M 미분다양체 M M M 매개변수화된 곡면 parameterized surface in M M M 이라 한다.
s s s 벡터필드 V V V vector field V V V s s s 란, 각각의 q ∈ A q \in A q ∈ A V ( q ) ∈ T s ( q ) M V(q) \in T_{s(q)}M V ( q ) ∈ T s ( q ) M
만약 f f f M M M q ∈ A ↦ V ( q ) f ∈ R q \in A \mapsto V(q)f \in \mathbb{R} q ∈ A ↦ V ( q ) f ∈ R
설명 ( u , v ) (u, v) ( u , v ) R 2 \mathbb{R}^{2} R 2 v 0 v_{0} v 0 u ↦ s ( u , v 0 ) u \mapsto s(u,v_{0}) u ↦ s ( u , v 0 ) M M M R 2 \mathbb{R}^{2} R 2 탄젠트 공간 의 표준 기저를 생각하면 { ∂ ∂ u , ∂ ∂ v } \left\{ \frac{\partial }{\partial u}, \frac{\partial }{\partial v} \right\} { ∂ u ∂ , ∂ v ∂ } s s s 미분 인 d s ds d s d s ( ∂ ∂ u ) ds(\frac{\partial }{\partial u}) d s ( ∂ u ∂ )
∂ s ∂ u
\dfrac{\partial s}{\partial u}
∂ u ∂ s
라고 표기하자. 그러면 ∂ s ∂ u \dfrac{\partial s}{\partial u} ∂ u ∂ s 곡선 u ↦ s ( u , v 0 ) u \mapsto s(u,v_{0}) u ↦ s ( u , v 0 ) 이다. ∂ s ∂ v \dfrac{\partial s}{\partial v} ∂ v ∂ s s s s 이와 비슷한 것 을 생각해준다.
s s s V V V V V V 공변도함수 D V ∂ u \dfrac{D V}{\partial u} ∂ u D V D V ∂ v \dfrac{D V}{\partial v} ∂ v D V u ↦ s ( u , v 0 ) u \mapsto s(u, v_{0}) u ↦ s ( u , v 0 ) V V V 축소사상 을 생각하자. 그러면 D V d u ( u , v 0 ) \dfrac{DV}{du}(u,v_{0}) d u D V ( u , v 0 ) v 0 v_{0} v 0 D V ∂ u ( u , v ) \dfrac{DV}{\partial u}(u, v) ∂ u D V ( u , v ) ( u , v ) ∈ A (u,v) \in A ( u , v ) ∈ A D V ∂ v \dfrac{D V}{\partial v} ∂ v D V
대칭성 M M M 대칭 접속 을 가지는 미분다양체, s : A → M s : A \to M s : A → M
D ∂ v ∂ s ∂ u = D ∂ u ∂ s ∂ v
\dfrac{D}{\partial v}\dfrac{\partial s}{\partial u} = \dfrac{D}{\partial u}\dfrac{\partial s}{\partial v}
∂ v D ∂ u ∂ s = ∂ u D ∂ v ∂ s
증명 x : U ⊂ R n → M \mathbf{x} : U \subset \mathbb{R}^{n} \to M x : U ⊂ R n → M s ( A ) s(A) s ( A ) 좌표계 라고 하자. 다시말해 s ( A ) ⊂ x ( U ) s(A) \subset \mathbf{x}(U) s ( A ) ⊂ x ( U ) x − 1 ∘ s ( u , v ) \mathbf{x}^{-1} \circ s(u,v) x − 1 ∘ s ( u , v ) R n \mathbb{R}^{n} R n
x − 1 ∘ s ( u , v ) = ( s 1 ( u , v ) , … , s n ( u , v ) )
\mathbf{x}^{-1} \circ s (u,v) = \left( s^{1}(u,v), \dots, s^{n}(u,v) \right)
x − 1 ∘ s ( u , v ) = ( s 1 ( u , v ) , … , s n ( u , v ) )
또한 s : A → M s : A \to M s : A → M s s s 미분 은 d s : T ( u , v ) A → T s ( u , v ) M ds : T_{(u,v)}A \to T_{s(u,v)}M d s : T ( u , v ) A → T s ( u , v ) M A A A ( u , v ) (u, v) ( u , v ) M M M x − 1 = ( s 1 , … , s n ) \mathbf{x}^{-1} = \left( s^{1}, \dots, s^{n} \right) x − 1 = ( s 1 , … , s n )
d s = [ ∂ s 1 ∂ u ∂ s 1 ∂ v ⋮ ∂ s n ∂ u ∂ s n ∂ v ]
ds = \begin{bmatrix} \dfrac{\partial s^{1}}{\partial u} & \dfrac{\partial s^{1}}{\partial v} \\
\vdots \\
\dfrac{\partial s^{n}}{\partial u} & \dfrac{\partial s^{n}}{\partial v}
\end{bmatrix}
d s = ∂ u ∂ s 1 ⋮ ∂ u ∂ s n ∂ v ∂ s 1 ∂ v ∂ s n
T ( u , v ) A T_{(u,v)}A T ( u , v ) A { ∂ ∂ u , ∂ ∂ v } \left\{ \dfrac{\partial }{\partial u}, \dfrac{\partial }{\partial v} \right\} { ∂ u ∂ , ∂ v ∂ } 좌표벡터 는 [ 1 0 ] \begin{bmatrix} 1 \\ 0\end{bmatrix} [ 1 0 ]
∂ s ∂ u = d s ( ∂ ∂ u ) = [ ∂ s 1 ∂ u ∂ s 1 ∂ v ⋮ ⋮ ∂ s n ∂ u ∂ s n ∂ v ] [ 1 0 ] = [ ∂ s 1 ∂ u ⋮ ∂ s n ∂ u ] = ∑ i ∂ s i ∂ u ∂ ∂ x i
\dfrac{\partial s}{\partial u} = ds(\dfrac{\partial }{\partial u}) = \begin{bmatrix} \dfrac{\partial s^{1}}{\partial u} & \dfrac{\partial s^{1}}{\partial v} \\
\vdots & \vdots \\
\dfrac{\partial s^{n}}{\partial u} & \dfrac{\partial s^{n}}{\partial v}
\end{bmatrix} \begin{bmatrix} 1 \\ 0\end{bmatrix} = \begin{bmatrix}
\dfrac{\partial s^{1}}{\partial u} \\
\vdots \\
\dfrac{\partial s^{n}}{\partial u}
\end{bmatrix} = \sum_{i} \dfrac{\partial s^{i}}{\partial u}\dfrac{\partial }{\partial x_{i}}
∂ u ∂ s = d s ( ∂ u ∂ ) = ∂ u ∂ s 1 ⋮ ∂ u ∂ s n ∂ v ∂ s 1 ⋮ ∂ v ∂ s n [ 1 0 ] = ∂ u ∂ s 1 ⋮ ∂ u ∂ s n = i ∑ ∂ u ∂ s i ∂ x i ∂
이제 D ∂ v ( ∂ s ∂ u ) \dfrac{D }{\partial v}\left( \dfrac{\partial s}{\partial u} \right) ∂ v D ( ∂ u ∂ s ) 공변도함수의 성질 에 의해,
D ∂ v ( ∂ s ∂ u ) = D ∂ v ( ∑ j ∂ s j ∂ u ∂ ∂ x j ) = ∑ j ∂ 2 s j ∂ v ∂ u ∂ ∂ x j + ∑ j ∂ s j ∂ u D ∂ v ∂ ∂ x j = ∑ j ∂ 2 s j ∂ v ∂ u ∂ ∂ x j + ∑ i , j ∂ s j ∂ u ∇ ∂ s / ∂ u ∂ ∂ x j = ∑ j ∂ 2 s j ∂ v ∂ u ∂ ∂ x j + ∑ i , j ∂ s j ∂ u ∇ ∂ s i ∂ u ∂ ∂ x i ∂ ∂ x j = ∑ j ∂ 2 s j ∂ v ∂ u ∂ ∂ x j + ∑ i , j ∂ s j ∂ u ∂ s i ∂ u ∇ ∂ ∂ x i ∂ ∂ x j = ∑ k ∂ 2 s k ∂ v ∂ u ∂ ∂ x k + ∑ i , j , k ∂ s j ∂ u ∂ s i ∂ u Γ i j k ∂ ∂ x k
\begin{align*}
\dfrac{D }{\partial v} \left( \dfrac{\partial s}{\partial u} \right) &= \dfrac{D }{\partial v} \left( \sum_{j} \dfrac{\partial s^{j}}{\partial u}\dfrac{\partial }{\partial x_{j}} \right) \\
&= \sum_{j} \dfrac{\partial^{2} s^{j}}{\partial v \partial u}\dfrac{\partial }{\partial x_{j}} + \sum_{j} \dfrac{\partial s^{j}}{\partial u}\dfrac{D }{\partial v}\dfrac{\partial }{\partial x_{j}} \\
&= \sum_{j} \dfrac{\partial^{2} s^{j}}{\partial v \partial u}\dfrac{\partial }{\partial x_{j}} + \sum_{i,j} \dfrac{\partial s^{j}}{\partial u} \nabla_{{\partial s}/{\partial u}}\dfrac{\partial }{\partial x_{j}} \\
&= \sum_{j} \dfrac{\partial^{2} s^{j}}{\partial v \partial u}\dfrac{\partial }{\partial x_{j}} + \sum_{i,j} \dfrac{\partial s^{j}}{\partial u} \nabla_{\frac{\partial s^{i}}{\partial u}\frac{\partial }{\partial x_{i}}}\dfrac{\partial }{\partial x_{j}} \\
&= \sum_{j} \dfrac{\partial^{2} s^{j}}{\partial v \partial u}\dfrac{\partial }{\partial x_{j}} + \sum_{i,j} \dfrac{\partial s^{j}}{\partial u} \frac{\partial s^{i}}{\partial u}\nabla_{\frac{\partial }{\partial x_{i}}}\dfrac{\partial }{\partial x_{j}} \\
&= \sum_{k} \dfrac{\partial^{2} s^{k}}{\partial v \partial u}\dfrac{\partial }{\partial x_{k}} + \sum_{i,j,k} \dfrac{\partial s^{j}}{\partial u} \frac{\partial s^{i}}{\partial u}\Gamma_{ij}^{k} \dfrac{\partial }{\partial x_{k}} \\
\end{align*}
∂ v D ( ∂ u ∂ s ) = ∂ v D ( j ∑ ∂ u ∂ s j ∂ x j ∂ ) = j ∑ ∂ v ∂ u ∂ 2 s j ∂ x j ∂ + j ∑ ∂ u ∂ s j ∂ v D ∂ x j ∂ = j ∑ ∂ v ∂ u ∂ 2 s j ∂ x j ∂ + i , j ∑ ∂ u ∂ s j ∇ ∂ s / ∂ u ∂ x j ∂ = j ∑ ∂ v ∂ u ∂ 2 s j ∂ x j ∂ + i , j ∑ ∂ u ∂ s j ∇ ∂ u ∂ s i ∂ x i ∂ ∂ x j ∂ = j ∑ ∂ v ∂ u ∂ 2 s j ∂ x j ∂ + i , j ∑ ∂ u ∂ s j ∂ u ∂ s i ∇ ∂ x i ∂ ∂ x j ∂ = k ∑ ∂ v ∂ u ∂ 2 s k ∂ x k ∂ + i , j , k ∑ ∂ u ∂ s j ∂ u ∂ s i Γ ij k ∂ x k ∂
그러면 비슷하게 D ∂ u ( ∂ s ∂ v ) \dfrac{D }{\partial u}\left( \dfrac{\partial s}{\partial v} \right) ∂ u D ( ∂ v ∂ s )
D ∂ v ( ∂ s ∂ u ) = ∑ k ∂ 2 s k ∂ u ∂ v ∂ ∂ x k + ∑ i , j , k ∂ s j ∂ v ∂ s i ∂ v Γ i j k ∂ ∂ x k
\dfrac{D }{\partial v} \left( \dfrac{\partial s}{\partial u} \right) = \sum_{k} \dfrac{\partial^{2} s^{k}}{\partial u \partial v}\dfrac{\partial }{\partial x_{k}} + \sum_{i,j,k} \dfrac{\partial s^{j}}{\partial v} \frac{\partial s^{i}}{\partial v}\Gamma_{ij}^{k} \dfrac{\partial }{\partial x_{k}}
∂ v D ( ∂ u ∂ s ) = k ∑ ∂ u ∂ v ∂ 2 s k ∂ x k ∂ + i , j , k ∑ ∂ v ∂ s j ∂ v ∂ s i Γ ij k ∂ x k ∂
이때 s k : R 2 → R s^{k} : \mathbb{R}^{2} \to \mathbb{R} s k : R 2 → R ∂ 2 ∂ u ∂ v = ∂ 2 ∂ v ∂ u \dfrac{\partial ^{2}}{\partial u \partial v} = \dfrac{\partial ^{2}}{\partial v \partial u} ∂ u ∂ v ∂ 2 = ∂ v ∂ u ∂ 2
D ∂ v ( ∂ s ∂ u ) = D ∂ v ( ∂ s ∂ u )
\dfrac{D }{\partial v} \left( \dfrac{\partial s}{\partial u} \right) = \dfrac{D }{\partial v} \left( \dfrac{\partial s}{\partial u} \right)
∂ v D ( ∂ u ∂ s ) = ∂ v D ( ∂ u ∂ s )
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