미분기하에서 풀백
📂기하학 미분기하에서 풀백 개요 미분다양체 위의 풀백 을 정의한다. 미분다양체가 어렵다면 M = R m M = \mathbb{R}^{m} M = R m N = R n N = \mathbb{R}^{n} N = R n
정의 두 미분다양체 M , N M, N M , N 미분가능한 함수 f : M → N f : M \to N f : M → N N N N k k k M M M k k k f ∗ f^{\ast} f ∗ ω \omega ω N N N k k k M M M k k k f ∗ ω f^{\ast}\omega f ∗ ω ω \omega ω 풀백 pull back, 당김 이라하고 다음과 같이 정의한다.
( f ∗ ω ) ( p ) ( v 1 , … , v k ) : = ω ( f ( p ) ) ( d f p v 1 , … , d f p v k ) , v i ∈ T p M
\begin{equation}
(f^{\ast}\omega)(p) (v_{1}, \dots, v_{k}) := \omega (f(p))\left( df_{p}v_{1}, \dots, df_{p}v_{k} \right),\quad v_{i} \in T_{p}M
\end{equation}
( f ∗ ω ) ( p ) ( v 1 , … , v k ) := ω ( f ( p )) ( d f p v 1 , … , d f p v k ) , v i ∈ T p M
설명 풀백 이라는 이름에는, (f f f M M M N N N f ∗ f^{\ast} f ∗ N N N M M M
f ∗ f^{\ast} f ∗ N N N k k k M M M k k k ω \omega ω N N N k k k f ∗ ω = f ∗ ( ω ) f^{\ast}\omega = f^{\ast}(\omega) f ∗ ω = f ∗ ( ω ) M M M k k k
f ∗ ω ( p ) f^{\ast}\omega (p) f ∗ ω ( p ) 다양체 M M M k k k p ∈ M p \in M p ∈ M Λ k ( T p ∗ M ) \Lambda^{k}(T_{p}^{\ast}M) Λ k ( T p ∗ M )
f ∗ ω : M → Λ k ( T p ∗ M )
f^{\ast}\omega : M \to \Lambda^{k}(T_{p}^{\ast}M)
f ∗ ω : M → Λ k ( T p ∗ M )
Λ k ( T p ∗ M ) : = { φ : T p M × ⋯ × T p M ⏟ k times → R ∣ φ is k-linear alternating map }
\Lambda^{k} (T_{p}^{\ast}M) := \left\{ \varphi : \underbrace{T_{p}M \times \cdots \times T_{p}M}_{k \text{ times}} \to \mathbb{R}\ | \ \varphi \text{ is k-linear alternating map} \right\}
Λ k ( T p ∗ M ) := ⎩ ⎨ ⎧ φ : k times T p M × ⋯ × T p M → R ∣ φ is k-linear alternating map ⎭ ⎬ ⎫
다시말해 f ∗ ω ( p ) ∈ Λ k ( T p ∗ M ) f^{\ast}\omega (p) \in \Lambda^{k} (T_{p}^{\ast}M) f ∗ ω ( p ) ∈ Λ k ( T p ∗ M ) Λ k ( T p ∗ M ) \Lambda^{k} (T_{p}^{\ast}M) Λ k ( T p ∗ M ) f ∗ ω ( p ) f^{\ast}\omega (p) f ∗ ω ( p ) p p p k k k ( 1 ) (1) ( 1 ) f ∗ ( p ) f^{\ast}(p) f ∗ ( p )
( f ∗ ω ) p = f ∗ ω ( p )
(f^{\ast}\omega)_{p} = f^{\ast}\omega (p)
( f ∗ ω ) p = f ∗ ω ( p )
ω ( f ( p ) ) \omega (f(p)) ω ( f ( p )) ω \omega ω N N N k k k N N N f ( p ) f(p) f ( p ) Λ k ( T f ( p ) ∗ N ) \Lambda^{k}(T_{f(p)}^{\ast}N) Λ k ( T f ( p ) ∗ N )
Λ k ( T f ( p ) ∗ N ) : = { φ : T f ( p ) N × ⋯ × T f ( p ) N ⏟ k times → R ∣ φ is k-linear alternating map }
\Lambda^{k} (T_{f(p)}^{\ast}N) := \left\{ \varphi : \underbrace{T_{f(p)}N \times \cdots \times T_{f(p)}N}_{k \text{ times}} \to \mathbb{R}\ | \ \varphi \text{ is k-linear alternating map} \right\}
Λ k ( T f ( p ) ∗ N ) := ⎩ ⎨ ⎧ φ : k times T f ( p ) N × ⋯ × T f ( p ) N → R ∣ φ is k-linear alternating map ⎭ ⎬ ⎫
Λ k ( T f ( p ) ∗ N ) \Lambda^{k} (T_{f(p)}^{\ast}N) Λ k ( T f ( p ) ∗ N ) ω ( f ( p ) ) \omega (f(p)) ω ( f ( p )) ω ( f ( p ) ) \omega (f(p)) ω ( f ( p )) f ( p ) f(p) f ( p ) k k k w ( f ( p ) ) w(f(p)) w ( f ( p ))
ω f ( p ) = ω ( f ( p ) )
\omega_{f(p)} = \omega (f(p))
ω f ( p ) = ω ( f ( p ))
d f p : T p M → T f ( p ) N
df_{p} : T_{p}M \to T_{f(p)}N
d f p : T p M → T f ( p ) N
f : M → N f : M \to N f : M → N f f f 미분 d f p df_{p} d f p v i ∈ T p M v_{i} \in T_{p}M v i ∈ T p M d f p v i = d f p ( v i ) df_{p}v_{i} = df_{p}(v_{i}) d f p v i = d f p ( v i ) T f ( p ) N T_{f(p)}N T f ( p ) N
이제 이를 종합하면 ( 1 ) (1) ( 1 )
( f ∗ ω ) p ( v 1 , … , v k ) : = ω f ( p ) ( d f p v 1 , … , d f p v k ) , v i ∈ T p M
(f^{\ast}\omega)_{p} (v_{1}, \dots, v_{k}) := \omega_{f(p)}\left( df_{p}v_{1}, \dots, df_{p}v_{k} \right),\quad v_{i} \in T_{p}M
( f ∗ ω ) p ( v 1 , … , v k ) := ω f ( p ) ( d f p v 1 , … , d f p v k ) , v i ∈ T p M
위 두 함수의 정의역을 보면 다음과 같은 차이가 있다.
( f ∗ ω ) p : T p M × ⋯ × T p M ⏟ k times → R ω f ( p ) : T f ( p ) N × ⋯ × T f ( p ) N ⏟ k times → R
\begin{align*}
(f^{\ast}\omega)_{p} : && \underbrace{T_{p}M \times \cdots \times T_{p}M}_{k \text{ times}} &\to \mathbb{R} \\
\omega_{f(p)} : && \underbrace{T_{f(p)}N \times \cdots \times T_{f(p)}N}_{k \text{ times}} &\to \mathbb{R}
\end{align*}
( f ∗ ω ) p : ω f ( p ) : k times T p M × ⋯ × T p M k times T f ( p ) N × ⋯ × T f ( p ) N → R → R
이를 미분 d f p : T p M → T f ( p ) N df_{p} : T_{p}M \to T_{f(p)}N d f p : T p M → T f ( p ) N d f p df_{p} d f p 푸쉬 포워드 push forward, 밂 라고도 한다. 1 1 1 φ \varphi φ
φ ( d f v ) = f ∗ φ ( v )
\begin{equation}
\varphi( dfv) = f^{\ast}\varphi(v)
\end{equation}
φ ( df v ) = f ∗ φ ( v )
0 0 0 f : M → N f : M \to N f : M → N g : N → R g : N \to \mathbb{R} g : N → R N N N 0 0 0 g g g f ∗ g : M → R f^{\ast}g : M \to \mathbb{R} f ∗ g : M → R M M M 0 0 0
f ∗ g : = g ∘ f
f^{\ast}g := g \circ f
f ∗ g := g ∘ f
좌표 변환 함수 f : R n → R m f : \mathbb{R}^{n} \to \mathbb{R}^{m} f : R n → R m x = ( x 1 , … , x n ) ∈ R n \mathbf{x} = (x_{1}, \dots ,x_{n}) \in \mathbb{R}^{n} x = ( x 1 , … , x n ) ∈ R n y = ( y 1 , … , y m ) ∈ R m \mathbf{y} = (y_{1}, \dots ,y_{m}) \in \mathbb{R}^{m} y = ( y 1 , … , y m ) ∈ R m
f ( x 1 , … , x n ) = ( f 1 ( x ) , … , f m ( x ) ) = ( y 1 , … , y m )
f(x_{1}, \dots, x_{n}) = (f_{1}(\mathbf{x}), \dots, f_{m}(\mathbf{x}) )= (y_{1}, \dots ,y_{m})
f ( x 1 , … , x n ) = ( f 1 ( x ) , … , f m ( x )) = ( y 1 , … , y m )
그리고 ω = ∑ I a I d y I \omega = \sum\limits_{I} a_{I} dy_{I} ω = I ∑ a I d y I R m \mathbb{R}^{m} R m k k k ω \omega ω f ∗ ω f^{\ast}\omega f ∗ ω
f ∗ ω = f ∗ ( ∑ a I d y I ) = ∑ f ∗ ( a I d y I ) = ∑ f ∗ a I f ∗ d y I = ∑ f ∗ a I f ∗ ( d y i 1 ∧ ⋯ ∧ d y i k ) = ∑ f ∗ a I ( f ∗ d y i 1 ∧ ⋯ ∧ f ∗ d y i k )
\begin{align*}
f^{\ast} \omega
&= f^{\ast} \left( \sum a_{I}dy_{I} \right) \\
&= \sum f^{\ast} \left( a_{I}dy_{I} \right) \\
&= \sum f^{\ast}a_{I} f^{\ast}dy_{I} \\
&= \sum f^{\ast}a_{I} f^{\ast}(dy_{i1} \wedge \cdots \wedge dy_{ik}) \\
&= \sum f^{\ast}a_{I} (f^{\ast}dy_{i1} \wedge \cdots \wedge f^{\ast}dy_{ik})
\end{align*}
f ∗ ω = f ∗ ( ∑ a I d y I ) = ∑ f ∗ ( a I d y I ) = ∑ f ∗ a I f ∗ d y I = ∑ f ∗ a I f ∗ ( d y i 1 ∧ ⋯ ∧ d y ik ) = ∑ f ∗ a I ( f ∗ d y i 1 ∧ ⋯ ∧ f ∗ d y ik )
이때 ( 2 ) (2) ( 2 ) f ∗ d y i 1 ( v ) = d y i 1 ( d f ( v ) ) = d ( y i 1 ∘ f ) ( v ) = d f i 1 ( v ) f^{\ast}dy_{i1}(v) = dy_{i1}(df(v)) = d(y_{i1}\circ f)(v) = df_{i1}(v) f ∗ d y i 1 ( v ) = d y i 1 ( df ( v )) = d ( y i 1 ∘ f ) ( v ) = d f i 1 ( v ) f ∗ a I = a I ∘ f f^{\ast}a_{I} = a_{I} \circ f f ∗ a I = a I ∘ f
f ∗ ω = ∑ a I ( f 1 , … f m ) d f i 1 ∧ ⋯ ∧ d f i k
\begin{equation}
f^{\ast} \omega = \sum a_{I}(f_{1}, \dots f_{m}) df_{i1} \wedge \cdots \wedge df_{ik}
\end{equation}
f ∗ ω = ∑ a I ( f 1 , … f m ) d f i 1 ∧ ⋯ ∧ d f ik
위 식은 좌표변환을 의미하는데, 구체적으로 어떻게 되는지 아래의 예제에서 보자.
예제 R 2 ∖ { 0 , 0 } \mathbb{R}^{2} \setminus \left\{ 0, 0 \right\} R 2 ∖ { 0 , 0 } 1 1 1 ω \omega ω
ω = − y x 2 + y 2 d x + x x 2 + y 2 d y = a 1 d x + a 2 d y
\omega = - \dfrac{y}{x^{2} + y^{2}}dx + \dfrac{x}{x^{2} + y^{2}}dy = a_{1}dx + a_{2}dy
ω = − x 2 + y 2 y d x + x 2 + y 2 x d y = a 1 d x + a 2 d y
이 직교좌표위의 1 1 1 U = { ( r , θ ) : 0 < r , 0 ≤ θ < 2 π } U = \left\{ (r,\theta) : 0 \lt r, 0 \le \theta \lt 2\pi \right\} U = { ( r , θ ) : 0 < r , 0 ≤ θ < 2 π } f : U → R 2 f : U \to \mathbb{R}^{2} f : U → R 2
f ( r , θ ) = ( r cos θ , r sin θ ) = ( f 1 , f 2 )
f(r,\theta) = (r\cos\theta, r\sin\theta) = (f_{1}, f_{2})
f ( r , θ ) = ( r cos θ , r sin θ ) = ( f 1 , f 2 )
이제 d f 1 , d f 2 df_{1}, df_{2} d f 1 , d f 2 f 1 = r cos θ , f 2 = r sin θ f_{1} = r\cos\theta, f_{2}=r\sin\theta f 1 = r cos θ , f 2 = r sin θ
d f 1 = ∂ f 1 ∂ r d r + ∂ f 1 ∂ θ d θ = cos θ d r − r sin θ d θ d f 2 = ∂ f 2 ∂ r d r + ∂ f 2 ∂ θ d θ = sin θ d r + r cos θ d θ
\begin{align*}
df_{1} &= \dfrac{\partial f_{1}}{\partial r}dr + \dfrac{\partial f_{1}}{\partial \theta}d\theta = \cos\theta dr - r \sin \theta d\theta \\
df_{2} &= \dfrac{\partial f_{2}}{\partial r}dr + \dfrac{\partial f_{2}}{\partial \theta}d\theta = \sin\theta dr + r \cos \theta d\theta \\
\end{align*}
d f 1 d f 2 = ∂ r ∂ f 1 d r + ∂ θ ∂ f 1 d θ = cos θ d r − r sin θ d θ = ∂ r ∂ f 2 d r + ∂ θ ∂ f 2 d θ = sin θ d r + r cos θ d θ
그러면 ( 3 ) (3) ( 3 )
f ∗ ω = a 1 ( f 1 , f 2 ) d f 1 + a 2 ( f 1 , f 2 ) d f 2 = − f 2 f 1 2 + f 2 2 ( cos θ d r − r sin θ d θ ) + f 1 f 1 2 + f 2 2 d f 2 ( sin θ d r + r cos θ d θ ) = − r sin θ r 2 cos 2 θ + r 2 sin 2 θ ( cos θ d r − r sin θ d θ ) + r cos θ r 2 cos 2 θ + r 2 sin 2 θ ( sin θ d r + r cos θ d θ ) = − sin θ cos θ r d r + sin 2 θ d θ + cos θ sin θ r d r + cos 2 θ d θ = d θ
\begin{align*}
f^{\ast} \omega
&= a_{1}(f_{1}, f_{2})df_{1} + a_{2}(f_{1}, f_{2})df_{2} \\
&= - \dfrac{f_{2}}{f_{1}^{2} + f_{2}^{2}}(\cos\theta dr - r \sin \theta d\theta) + \dfrac{f_{1}}{f_{1}^{2} + f_{2}^{2}}df_{2}(\sin\theta dr + r \cos \theta d\theta) \\
&= - \dfrac{r\sin\theta}{r^{2}\cos^{2}\theta + r^{2}\sin^{2}\theta}(\cos\theta dr - r \sin \theta d\theta) \\
&\quad + \dfrac{r\cos\theta}{r^{2}\cos^{2}\theta + r^{2}\sin^{2}\theta}(\sin\theta dr + r \cos \theta d\theta) \\
&= -\dfrac{\sin\theta \cos\theta}{r}dr + \sin^{2}\theta d\theta + \dfrac{\cos\theta \sin\theta}{r}dr + \cos^{2}\theta d\theta \\
&= d\theta
\end{align*}
f ∗ ω = a 1 ( f 1 , f 2 ) d f 1 + a 2 ( f 1 , f 2 ) d f 2 = − f 1 2 + f 2 2 f 2 ( cos θ d r − r sin θ d θ ) + f 1 2 + f 2 2 f 1 d f 2 ( sin θ d r + r cos θ d θ ) = − r 2 cos 2 θ + r 2 sin 2 θ r sin θ ( cos θ d r − r sin θ d θ ) + r 2 cos 2 θ + r 2 sin 2 θ r cos θ ( sin θ d r + r cos θ d θ ) = − r sin θ cos θ d r + sin 2 θ d θ + r cos θ sin θ d r + cos 2 θ d θ = d θ
따라서
∫ − y x 2 + y 2 d x + x x 2 + y 2 d y = ∫ d θ
\int - \dfrac{y}{x^{2} + y^{2}}dx + \dfrac{x}{x^{2} + y^{2}}dy = \int d\theta
∫ − x 2 + y 2 y d x + x 2 + y 2 x d y = ∫ d θ
■
성질 M , N M, N M , N m , n m, n m , n f : M → N f : M \to N f : M → N ω , φ \omega, \varphi ω , φ N N N k k k g g g N N N 0 0 0 φ i \varphi_{i} φ i N N N 1 1 1
f ∗ ( ω + φ ) = f ∗ ω + f ∗ φ f ∗ ( g ω ) = ( f ∗ g ) ( f ∗ ω ) f ∗ ( φ 1 ∧ ⋯ ∧ φ k ) = f ∗ ( φ 1 ) ∧ ⋯ ∧ f ∗ ( φ k )
\begin{align}
f^{\ast} (\omega + \varphi) =&\ f^{\ast}\omega + f^{\ast}\varphi \tag{a}
\\ f^{\ast} (g \omega) =&\ (f^{\ast}g) (f^{\ast}\omega) \tag{b}
\\ f^{\ast} (\varphi_{1} \wedge \cdots \wedge \varphi_{k}) =&\ f^{\ast}(\varphi_{1}) \wedge \cdots \wedge f^{\ast}(\varphi_{k}) \tag{c}
\end{align}
f ∗ ( ω + φ ) = f ∗ ( g ω ) = f ∗ ( φ 1 ∧ ⋯ ∧ φ k ) = f ∗ ω + f ∗ φ ( f ∗ g ) ( f ∗ ω ) f ∗ ( φ 1 ) ∧ ⋯ ∧ f ∗ ( φ k ) ( a ) ( b ) ( c )
이때 + + + ∧ \wedge ∧ k k k
ω , φ \omega, \varphi ω , φ N N N L L L l l l g : L → N g : L \to N g : L → N
f ∗ ( ω ∧ φ ) = ( f ∗ ω ) ∧ ( f ∗ φ ) ( f ∘ g ) ∗ ω = g ∗ ( f ∗ ω )
\begin{align*}
f^{\ast}(\omega \wedge \varphi) &= (f^{\ast}\omega) \wedge (f^{\ast}\varphi) \tag{d} \\
(f \circ g)^{\ast} \omega &= g^{\ast}(f^{\ast}\omega) \tag{e}
\end{align*}
f ∗ ( ω ∧ φ ) ( f ∘ g ) ∗ ω = ( f ∗ ω ) ∧ ( f ∗ φ ) = g ∗ ( f ∗ ω ) ( d ) ( e )
증명 증명 ( a ) (a) ( a ) ( f ∗ ( ω + φ ) ) p ( v 1 , … , v k ) = ( ω + φ ) f ( p ) ( d f p v 1 , … , d f p v k ) = ω f ( p ) ( d f p v 1 , … , d f p v k ) + φ f ( p ) ( d f p v 1 , … , d f p v k ) = ( f ∗ ω ) p ( v 1 , … , v k ) + ( f ∗ φ ) p ( v 1 , … , v k ) = ( f ∗ ω + f ∗ φ ) p ( v 1 , … , v k )
\begin{align*}
(f^{\ast}(\omega + \varphi))_{p} (v_{1}, \dots, v_{k})
=&\ (\omega + \varphi)_{f(p)} \left( df_{p}v_{1}, \dots, df_{p}v_{k} \right) \\
=&\ \omega_{f(p)} \left( df_{p}v_{1}, \dots, df_{p}v_{k} \right) + \varphi_{f(p)} \left( df_{p}v_{1}, \dots, df_{p}v_{k} \right) \\
=&\ (f^{\ast} \omega)_{p}(v_{1}, \dots, v_{k}) + (f^{\ast} \varphi)_{p}(v_{1}, \dots, v_{k}) \\
=&\ \left( f^{\ast}\omega + f^{\ast}\varphi \right)_{p}(v_{1}, \dots, v_{k})
\end{align*}
( f ∗ ( ω + φ ) ) p ( v 1 , … , v k ) = = = = ( ω + φ ) f ( p ) ( d f p v 1 , … , d f p v k ) ω f ( p ) ( d f p v 1 , … , d f p v k ) + φ f ( p ) ( d f p v 1 , … , d f p v k ) ( f ∗ ω ) p ( v 1 , … , v k ) + ( f ∗ φ ) p ( v 1 , … , v k ) ( f ∗ ω + f ∗ φ ) p ( v 1 , … , v k )
■
증명 ( b ) (b) ( b ) 0 0 0 g g g k k k ω \omega ω
( g ω ) ( p ) = g ( p ) ω ( p )
(g\omega)(p) = g(p) \omega (p)
( g ω ) ( p ) = g ( p ) ω ( p )
여기서 g ( p ) = g p g(p) = g_{p} g ( p ) = g p ω ( p ) = ω p \omega (p) = \omega_{p} ω ( p ) = ω p
( f ∗ ( g ω ) ) p ( v 1 , … , v k ) = g ω f ( p ) ( d f p v 1 , … , d f p v k ) = g f ( p ) ω f ( p ) ( d f p v 1 , … , d f p v k ) = g ∘ f ( p ) ω f ( p ) ( d f p v 1 , … , d f p v k ) = ( f ∗ g ) p ( f ∗ ω ) p ( v 1 , … , v k )
\begin{align*}
(f^{\ast} (g\omega))_{p} (v_{1}, \dots, v_{k})
=&\ g\omega_{f(p)} \left( df_{p}v_{1}, \dots, df_{p}v_{k} \right) \\
=&\ g_{f(p)} \omega_{f(p)} (df_{p}v_{1}, \dots, df_{p}v_{k}) \\
=&\ g\circ f(p) \omega_{f(p)} (df_{p}v_{1}, \dots, df_{p}v_{k}) \\
=&\ (f^{\ast}g)_{p} (f^{\ast}\omega)_{p} (v_{1}, \dots, v_{k})
\end{align*}
( f ∗ ( g ω ) ) p ( v 1 , … , v k ) = = = = g ω f ( p ) ( d f p v 1 , … , d f p v k ) g f ( p ) ω f ( p ) ( d f p v 1 , … , d f p v k ) g ∘ f ( p ) ω f ( p ) ( d f p v 1 , … , d f p v k ) ( f ∗ g ) p ( f ∗ ω ) p ( v 1 , … , v k )
■
증명 ( c ) (c) ( c ) ( f ∗ ( φ 1 ∧ ⋯ ∧ φ k ) ) p ( v 1 , … , v k ) = ( φ 1 ∧ ⋯ ∧ φ k ) f ( p ) ( d f 1 , … , d f k ) = det [ φ i d f ( v j ) ] = det [ f ∗ φ i ( v j ) ]
\begin{align*}
(f^{\ast}\left( \varphi_{1} \wedge \cdots \wedge \varphi_{k} \right))_{p} (v_{1}, \dots, v_{k})
=&\ (\varphi_{1} \wedge \dots \wedge \varphi_{k})_{f(p)} \left( df_{1}, \dots, df_{k} \right) \\
=&\ \det [\varphi_{i}df(v_{j})] \\
=&\ \det [ f^{\ast} \varphi_{i}(v_{j})] \\
\end{align*}
( f ∗ ( φ 1 ∧ ⋯ ∧ φ k ) ) p ( v 1 , … , v k ) = = = ( φ 1 ∧ ⋯ ∧ φ k ) f ( p ) ( d f 1 , … , d f k ) det [ φ i df ( v j )] det [ f ∗ φ i ( v j )]
■
