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미분 형식의 연산: 합과 쐐기곱

정의¹

합 $+$

$\omega = \sum\limits_{I} a_{I} dx_{I}, \varphi = \sum\limits_{I} b_{I} dx_{I}$를 $k$형식이라고 하자. 이 둘의 합을 다음과 같이 정의한다.

$$\omega + \varphi := \sum\limits_{I}\left( a_{I} + b_{I} \right)dx_{I}$$

쐐기곱 $\wedge$

$\omega = \sum\limits_{I} a_{I}dx_{I}$, $\varphi = \sum\limits_{J} b_{J}dx_{J}$를 각각 $k$형식, $s$형식이라고 하자. 그러면 이 둘의 쐐기곱을 다음과 같이 정의한다.

$$\omega \wedge \varphi = \sum \limits_{I,J} a_{I}b_{J} dx_{I} \wedge dx_{J}$$

설명

$M$을 $n$차원 미분다양체라고 하자. (미분다양체가 어렵다면 $M=\mathbb{R}^{n}$이라 생각해도 좋다.) 쐐기곱 $\wedge$는 $T_{p}^{\ast}M$의 두 원소 $\varphi_{1}, \varphi_{2}$를 교대함수 $\varphi_{1} \wedge \varphi_{2}$로 만들어준다. 그런데 쐐기곱 자체도 교대함수이다. 즉 다음이 성립한다. $\varphi_{i} \in T_{p}^{\ast}M$에 대해서,

$$\varphi_{i} \wedge \varphi_{j} = - \varphi_{j} \wedge \varphi_{i} \quad \text{and} \quad \varphi_{i} \wedge \varphi_{i} = 0$$

정의대로 계산하면 쉽게 보일 수 있다.

$$\begin{align*} \left(\varphi_{1} \wedge \varphi_{2}\right)\left(v_{1}, v_{2}\right) =&\ \det \left[\varphi_{i}\left(v_{j}\right)\right] \\ =&\ \begin{vmatrix} \varphi_{1}\left(v_{1}\right) & \varphi_{1}\left(v_{2}\right) \\ \varphi_{2}\left(v_{1}\right) & \varphi_{2}\left(v_{2}\right) \end{vmatrix} \\ =&\ -\begin{vmatrix} \varphi_{2}\left(v_{1}\right) & \varphi_{2}\left(v_{L}\right) \\ \varphi_{1}\left(v_{1}\right) & \varphi_{1}\left(v_{2}\right) \end{vmatrix} \\ =&\ -\left(\varphi_{2} \wedge \varphi_{1}\right)\left(v_{1}, v_{1}\right) \\ \end{align*}$$

$$\begin{align*} \left(\varphi_{1}, \varphi_{1}\right)\left(v_{1}, v_{2}\right) =&\ \det \left[\varphi_{1}\left(v_{i}\right)\right] \\ =&\ \begin{vmatrix} \varphi_{1}\left(v_{1}\right) & \varphi_{1}\left(v_{2}\right) \\ \varphi_{1}\left(v_{1}\right) & \varphi_{1}\left(v_{2}\right) \end{vmatrix} \\ =&\ 0 \end{align*}$$

정의에 의해 $k$형식과 $s$형식의 쐐기곱은 $k+s$형식이다.

또한 모든 $i$에 대해서 $dx_{i} \wedge dx_{i} = 0$이지만, $\omega \wedge \omega = 0$이 성립하는 것은 아니다. 예를 들어 $\omega = x_{1}dx_{1}\wedge dx_{2} + x2_{2} dx_{3} \wedge dx_{4}$라고 하면,

$$\omega \wedge \omega = 2x_{1}x_{2} dx_{1} \wedge dx_{2} \wedge dx_{3} \wedge dx_{4}$$

예시

$\omega = x_{1}dx_{1} + x_{2}dx_{2}$를 $\mathbb{R}^{3}$의 1차형식, $\varphi = x_{1}dx_{1}\wedge dx_{2} + x_{2} dx_{1}\wedge dx_{3} + dx_{2} \wedge dx_{3}$을 $\mathbb{R}^{3}$의 2차형식이라고 하자. 그러면

$$\begin{align*} \omega \wedge \varphi =&\ (x_{1}dx_{1} + x_{2}dx_{2}) \wedge (x_{1}dx_{1}\wedge dx_{2} + x_{2} dx_{1}\wedge dx_{3} + dx_{2} \wedge dx_{3}) \\ =&\ x_{1}^{2} \cancel{dx_{1} \wedge dx_{1}\wedge dx_{2}} + x_{1}x_{2}\cancel{dx_{1} \wedge dx_{1}\wedge dx_{3}} + x_{1}dx_{1} \wedge dx_{2} \wedge dx_{3} \\ & + \cancel{x_{1}x_{2}dx_{2} \wedge dx_{1}\wedge dx_{2}} + x_{2}^{2} dx_{2} \wedge dx_{1}\wedge dx_{3} + x_{2}\cancel{dx_{2} \wedge dx_{2} \wedge dx_{3}} \\ =&\ (x_{1}-x_{2}^{2}) dx_{1} \wedge dx_{2} \wedge dx_{3} \end{align*}$$

성질

$\omega$를 $k$형식, $\varphi$를 $s$형식, $\theta$를 $r$형식이라고 하자. 그러면

$$\begin{align} (\omega \wedge \varphi) \wedge \theta =&\ \omega \wedge (\varphi \wedge \theta) \\ (\omega \wedge \varphi) =&\ (-1)^{ks} (\varphi \wedge \omega) \\ \omega \wedge (\varphi + \theta) =&\ \omega \wedge \varphi + \omega \wedge \theta,\quad \text{if } r=s \end{align}$$

증명

증명(1)

$\omega = \sum\limits_{I} a_{I} dx_{I}$, $\varphi = \sum\limits_{J} b_{J} dx_{J}$, $\vartheta = \sum\limits_{L} c_{L} dx_{L}$라고 하자.

$$\begin{align*} (\omega \wedge \varphi) \wedge \vartheta =&\ \left( \sum\limits_{I}a_{I}dx_{I} \wedge \sum\limits_{J}b_{J}dx_{J} \right) \wedge \sum\limits_{L} c_{L}dx_{L} \\ =&\ \left( \sum\limits_{I,J}a_{I}b_{J} dx_{I} \wedge dx_{J} \right) \wedge \sum\limits_{L} c_{L}dx_{L} & \text{by definition of } \wedge\\ =&\ \sum\limits_{I,J,L} (a_{I}b_{J})c_{L} (dx_{I} \wedge dx_{J}) \wedge dx_{L} & \text{by definition of } \wedge\\ =&\ \sum\limits_{I,J,L} (a_{I}b_{J})c_{L} dx_{I} \wedge (dx_{J} \wedge dx_{L}) & \text{by property of } \wedge\\ \end{align*}$$

이때 $a, b, c$들은 각각 함숫값이 실수인 함수이므로 곱에 대한 결합법칙이 성립한다. 따라서

$$\begin{align*} (\omega \wedge \varphi) \wedge \vartheta =&\ \sum\limits_{I,J,L} a_{I}(b_{J}c_{L}) dx_{I} \wedge (dx_{J} \wedge dx_{L}) \\ =&\ \sum\limits_{I} a_{I} dx_{I} \wedge \sum\limits_{J,L} b_{J}c_{L} dx_{J} \wedge dx_{L} & \text{by property of } \wedge\\ =&\ \omega \wedge ( \varphi \wedge \vartheta) & \text{by property of } \wedge\\ \end{align*}$$

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증명(2)

$\omega = \sum\limits_{I} a_{I} dx_{I}$, $\varphi = \sum\limits_{J} b_{J} dx_{J}$라고 하자.

$$\begin{align*} \omega \wedge \varphi =&\ \sum\limits_{I,J} a_{I}b_{J} dx_{I} \wedge dx_{J} \\ =&\ \sum\limits_{I,J} a_{I}b_{J} dx_{i_{1}} \wedge \dots \wedge dx_{i_{k}} \wedge dx_{j_{1}} \wedge \dots \wedge dx_{j_{s}} \\ =&\ \sum\limits_{I,J} a_{I}b_{J} (-1)^{k} dx_{j_{1}} \wedge dx_{i_{1}} \wedge \dots \wedge dx_{i_{k}} \wedge dx_{j_{2}} \wedge \dots \wedge dx_{j_{s}} \\ =&\ \sum\limits_{I,J} a_{I}b_{J} (-1)^{k}(-1)^{k} dx_{j_{1}} \wedge dx_{j_{2}} \wedge dx_{i_{1}} \wedge \dots \wedge dx_{i_{k}} \wedge dx_{j_{3}} \wedge \dots \wedge dx_{j_{s}} \\ &\ \vdots \\ =&\ \sum\limits_{I,J} a_{I}b_{J} (-1)^{ks} dx_{j_{1}} \wedge dx_{j_{2}} \wedge \cdots \wedge dx_{j_{s}} \wedge dx_{i_{1}} \wedge \dots \wedge dx_{i_{k}}\\ =&\ (-1)^{ks} \sum\limits_{I,J} a_{I}b_{J} dx_{I} \wedge dx_{J} \\ =&\ (-1)^{ks} \varphi \wedge \omega \end{align*}$$

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증명(3)

$\omega = \sum\limits_{I} a_{I} dx_{I}$, $\varphi = \sum\limits_{J} b_{J} dx_{J}$, $\vartheta = \sum\limits_{J} c_{J} dx_{J}$라고 하자.

$$\begin{align*} \omega =&\ \omega \wedge (\varphi + \vartheta) \\ =&\ \omega \wedge \left( \sum\limits_{J} (b_{J} + c_{J}) dx_{J} \right) & \text{by definition of } + \\ =&\ \left( \sum\limits_{I} a_{I} dx_{I} \right)\wedge \left( \sum\limits_{J} (b_{J} + c_{J}) dx_{J} \right) \\ =&\ \sum\limits_{I,J} a_{I}(b_{J} + c_{J})dx_{I}\wedge dx_{J} & \text{by definition of } \wedge\\ \end{align*}$$

이때 $a, b, c$들은 각각 함숫값이 실수인 함수이므로 분배법칙이 성립한다. 따라서

$$\begin{align*} \omega =&\ \sum\limits_{I,J} a_{I}b_{J}dx_{I}\wedge dx_{J} + \sum\limits_{I,J} a_{I}c_{J}dx_{I}\wedge dx_{J} \\ =&\ \omega \wedge \varphi + \omega \wedge \vartheta & \text{by definition of } \wedge\\ \end{align*}$$

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