"리만 곡률 텐서"의 두 판 사이의 차이

2014년 2월 1일 (토) 02:33 판

개요

접속 (connection) $\nabla$이 정의되어 있다고 하자
세 개의 벡터장 X,Y,Z 가 주어지면, 새로운 벡터장 $R(X,Y)Z$ 를 얻는다

\[R(X,Y)Z=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z\]

covariant tensor

리만 곡률 텐서

성분

${R^\rho}_{\sigma\mu\nu} = dx^\rho(R(\partial_{\mu},\partial_{\nu})\partial_{\sigma})$
크리스토펠 기호 를 이용한 성분의 계산

\[{R^\rho}_{\sigma\mu\nu} = \partial_\mu\Gamma^\rho_{\nu\sigma}-\partial_\nu\Gamma^\rho_{\mu\sigma}+ \Gamma^\rho_{\mu\lambda}\Gamma^\lambda_{\nu\sigma}- \Gamma^\rho_{\nu\lambda}\Gamma^\lambda_{\mu\sigma}\] \[{R^l}_{kij} = \partial_i\Gamma^l_{jk} - \partial_j\Gamma^l_{ik} + \Gamma^l_{is}\Gamma^s_{jk} - \Gamma^l_{js}\Gamma^s_{ik}\] \[R_{\rho\sigma\mu\nu} = g_{\rho \zeta} {R^\zeta}_{\sigma\mu\nu} .\]

성질

리만 곡률 텐서는 다음의 대칭성을 갖는다

$R_{ijkl}=-R_{jikl}=-R_{ijlk}$
$R_{ijkl}=R_{klij}$
$R_{ijkl}+R_{iklj}+R_{iljk}=0$, 비앙키 항등식

곡률 2형식

$R(X,Y)\partial_{j}=\Omega_{j}^{s}(X,Y)\partial_s$
$\,\Omega=d\omega +\frac{1}{2}[\omega,\omega]=d\omega +\omega\wedge \omega$
$\Omega^i_{j}=d\omega^i_{j} +\sum_k \omega^i_{k}\wedge\omega^k_{j}$

곡면의 경우

제1기본형식이 $E=e(u,v),F=0,G=g(u,v)$ 로 주어진 경우, 리만 곡률 텐서는 다음과 같다 (이외의 $ R_{jkl}^i$는 0이다)\[ R_{212}^1 = \frac{e(u,v) \left(e^{(0,1)}(u,v) g^{(0,1)}(u,v)+g^{(1,0)}(u,v)^2\right)+g(u,v) \left(e^{(1,0)}(u,v) g^{(1,0)}(u,v)-2 e(u,v) \left(e^{(0,2)}(u,v)+g^{(2,0)}(u,v)\right)+e^{(0,1)}(u,v)^2\right)}{4 e(u,v)^2 g(u,v)}\]\[R_{112}^2 = -\frac{e(u,v) \left(e^{(0,1)}(u,v) g^{(0,1)}(u,v)+g^{(1,0)}(u,v)^2\right)+g(u,v) \left(e^{(1,0)}(u,v) g^{(1,0)}(u,v)-2 e(u,v) \left(e^{(0,2)}(u,v)+g^{(2,0)}(u,v)\right)+e^{(0,1)}(u,v)^2\right)}{4 e(u,v) g(u,v)^2}\]\[R_{221}^1 = -\frac{e(u,v) \left(e^{(0,1)}(u,v) g^{(0,1)}(u,v)+g^{(1,0)}(u,v)^2\right)+g(u,v) \left(e^{(1,0)}(u,v) g^{(1,0)}(u,v)-2 e(u,v) \left(e^{(0,2)}(u,v)+g^{(2,0)}(u,v)\right)+e^{(0,1)}(u,v)^2\right)}{4 e(u,v)^2 g(u,v)}\]\[R_{121}^2 = \frac{e(u,v) \left(e^{(0,1)}(u,v) g^{(0,1)}(u,v)+g^{(1,0)}(u,v)^2\right)+g(u,v) \left(e^{(1,0)}(u,v) g^{(1,0)}(u,v)-2 e(u,v) \left(e^{(0,2)}(u,v)+g^{(2,0)}(u,v)\right)+e^{(0,1)}(u,v)^2\right)}{4 e(u,v) g(u,v)^2}\]

역사

메모

매스매티카 파일 및 계산 리소스

https://docs.google.com/leaf?id=0B8XXo8Tve1cxN2ZmMGViMGQtMmI4Ny00MmI3LWE4ZTYtYmQyNjZiYWVhMTc5&sort=name&layout=list&num=50

사전 형태의 자료

리뷰, 에세이, 강의노트

Dunn, An Introduction to the Riemann Curvature Tensor and Differential Geometry
- 슬라이드

@@ 2번째 줄: / 2번째 줄: @@
 * [[접속 (connection)]] <math>\nabla</math>이 정의되어 있다고 하자
-*  세 개의 벡터장 X,Y,Z 가 주어지면, 새로운 벡터장 R(X,Y)Z 를 얻는다:<math>R(X,Y)Z=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z</math><br>
+*  세 개의 벡터장 X,Y,Z 가 주어지면, 새로운 벡터장 $R(X,Y)Z$ 를 얻는다
+:<math>R(X,Y)Z=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z</math>
 * covariant tensor
-==리만 곡률 텐서의 성분==
+==리만 곡률 텐서==
+===성분===
 * <math>{R^\rho}_{\sigma\mu\nu} = dx^\rho(R(\partial_{\mu},\partial_{\nu})\partial_{\sigma})</math>
-* [[크리스토펠 기호]] 를 이용한 성분의 계산:<math>{R^\rho}_{\sigma\mu\nu} = \partial_\mu\Gamma^\rho_{\nu\sigma}     - \partial_\nu\Gamma^\rho_{\mu\sigma}     + \Gamma^\rho_{\mu\lambda}\Gamma^\lambda_{\nu\sigma}     - \Gamma^\rho_{\nu\lambda}\Gamma^\lambda_{\mu\sigma}</math>:<math>{R^l}_{kij} = \partial_i\Gamma^l_{jk} - \partial_j\Gamma^l_{ik} + \Gamma^l_{is}\Gamma^s_{jk} - \Gamma^l_{js}\Gamma^s_{ik}</math>:<math>R_{\rho\sigma\mu\nu} = g_{\rho \zeta} {R^\zeta}_{\sigma\mu\nu} .</math><br>
+* [[크리스토펠 기호]] 를 이용한 성분의 계산
+:<math>{R^\rho}_{\sigma\mu\nu} = \partial_\mu\Gamma^\rho_{\nu\sigma}-\partial_\nu\Gamma^\rho_{\mu\sigma}+ \Gamma^\rho_{\mu\lambda}\Gamma^\lambda_{\nu\sigma}- \Gamma^\rho_{\nu\lambda}\Gamma^\lambda_{\mu\sigma}</math>
+:<math>{R^l}_{kij} = \partial_i\Gamma^l_{jk} - \partial_j\Gamma^l_{ik} + \Gamma^l_{is}\Gamma^s_{jk} - \Gamma^l_{js}\Gamma^s_{ik}</math>
+:<math>R_{\rho\sigma\mu\nu} = g_{\rho \zeta} {R^\zeta}_{\sigma\mu\nu} .</math>
+===성질===
+* 리만 곡률 텐서는 다음의 대칭성을 갖는다
+# $R_{ijkl}=-R_{jikl}=-R_{ijlk}$
+# $R_{ijkl}=R_{klij}$
+# $R_{ijkl}+R_{iklj}+R_{iljk}=0$, 비앙키 항등식
 ==곡률 2형식==
-* <math>R(X,Y)\partial_{j}=\Omega_{j}^{s}(X,Y)\partial_s</math><br>
+* <math>R(X,Y)\partial_{j}=\Omega_{j}^{s}(X,Y)\partial_s</math>
-* <math>\,\Omega=d\omega +\frac{1}{2}[\omega,\omega]=d\omega +\omega\wedge \omega</math><br>
+* <math>\,\Omega=d\omega +\frac{1}{2}[\omega,\omega]=d\omega +\omega\wedge \omega</math>
-* <math>\Omega^i_{j}=d\omega^i_{j} +\sum_k \omega^i_{k}\wedge\omega^k_{j}</math><br>
+* <math>\Omega^i_{j}=d\omega^i_{j} +\sum_k \omega^i_{k}\wedge\omega^k_{j}</math>
 ==곡면의 경우==
-*  제1기본형식이 <math>E=e(u,v),F=0,G=g(u,v)</math> 로 주어진 경우, 리만 곡률 텐서는 다음과 같다 (이외의 <math> R_{jkl}^i</math>는 0이다):<math> R_{212}^1 = \frac{e(u,v) \left(e^{(0,1)}(u,v) g^{(0,1)}(u,v)+g^{(1,0)}(u,v)^2\right)+g(u,v) \left(e^{(1,0)}(u,v) g^{(1,0)}(u,v)-2 e(u,v) \left(e^{(0,2)}(u,v)+g^{(2,0)}(u,v)\right)+e^{(0,1)}(u,v)^2\right)}{4 e(u,v)^2 g(u,v)}</math>:<math>R_{112}^2 = -\frac{e(u,v) \left(e^{(0,1)}(u,v) g^{(0,1)}(u,v)+g^{(1,0)}(u,v)^2\right)+g(u,v) \left(e^{(1,0)}(u,v) g^{(1,0)}(u,v)-2 e(u,v) \left(e^{(0,2)}(u,v)+g^{(2,0)}(u,v)\right)+e^{(0,1)}(u,v)^2\right)}{4 e(u,v) g(u,v)^2}</math>:<math>R_{221}^1 = -\frac{e(u,v) \left(e^{(0,1)}(u,v) g^{(0,1)}(u,v)+g^{(1,0)}(u,v)^2\right)+g(u,v) \left(e^{(1,0)}(u,v) g^{(1,0)}(u,v)-2 e(u,v) \left(e^{(0,2)}(u,v)+g^{(2,0)}(u,v)\right)+e^{(0,1)}(u,v)^2\right)}{4 e(u,v)^2 g(u,v)}</math>:<math>R_{121}^2 = \frac{e(u,v) \left(e^{(0,1)}(u,v) g^{(0,1)}(u,v)+g^{(1,0)}(u,v)^2\right)+g(u,v) \left(e^{(1,0)}(u,v) g^{(1,0)}(u,v)-2 e(u,v) \left(e^{(0,2)}(u,v)+g^{(2,0)}(u,v)\right)+e^{(0,1)}(u,v)^2\right)}{4 e(u,v) g(u,v)^2}</math><br>
+*  제1기본형식이 <math>E=e(u,v),F=0,G=g(u,v)</math> 로 주어진 경우, 리만 곡률 텐서는 다음과 같다 (이외의 <math> R_{jkl}^i</math>는 0이다):<math> R_{212}^1 = \frac{e(u,v) \left(e^{(0,1)}(u,v) g^{(0,1)}(u,v)+g^{(1,0)}(u,v)^2\right)+g(u,v) \left(e^{(1,0)}(u,v) g^{(1,0)}(u,v)-2 e(u,v) \left(e^{(0,2)}(u,v)+g^{(2,0)}(u,v)\right)+e^{(0,1)}(u,v)^2\right)}{4 e(u,v)^2 g(u,v)}</math>:<math>R_{112}^2 = -\frac{e(u,v) \left(e^{(0,1)}(u,v) g^{(0,1)}(u,v)+g^{(1,0)}(u,v)^2\right)+g(u,v) \left(e^{(1,0)}(u,v) g^{(1,0)}(u,v)-2 e(u,v) \left(e^{(0,2)}(u,v)+g^{(2,0)}(u,v)\right)+e^{(0,1)}(u,v)^2\right)}{4 e(u,v) g(u,v)^2}</math>:<math>R_{221}^1 = -\frac{e(u,v) \left(e^{(0,1)}(u,v) g^{(0,1)}(u,v)+g^{(1,0)}(u,v)^2\right)+g(u,v) \left(e^{(1,0)}(u,v) g^{(1,0)}(u,v)-2 e(u,v) \left(e^{(0,2)}(u,v)+g^{(2,0)}(u,v)\right)+e^{(0,1)}(u,v)^2\right)}{4 e(u,v)^2 g(u,v)}</math>:<math>R_{121}^2 = \frac{e(u,v) \left(e^{(0,1)}(u,v) g^{(0,1)}(u,v)+g^{(1,0)}(u,v)^2\right)+g(u,v) \left(e^{(1,0)}(u,v) g^{(1,0)}(u,v)-2 e(u,v) \left(e^{(0,2)}(u,v)+g^{(2,0)}(u,v)\right)+e^{(0,1)}(u,v)^2\right)}{4 e(u,v) g(u,v)^2}</math>
 ==역사==
 * http://www.google.com/search?hl=en&tbs=tl:1&q=
 * [[수학사 연표]]
 ==메모==
@@ 54번째 줄: / 63번째 줄: @@
 * http://users-phys.au.dk/fedorov/nucltheo/GTR/09/note6.pdf
 ==관련된 항목들==
@@ 64번째 줄: / 73번째 줄: @@
 * [[접속 (connection)]]
 ==매스매티카 파일 및 계산 리소스==
 * https://docs.google.com/leaf?id=0B8XXo8Tve1cxN2ZmMGViMGQtMmI4Ny00MmI3LWE4ZTYtYmQyNjZiYWVhMTc5&sort=name&layout=list&num=50
-* http://www.wolframalpha.com/input/?i=
-==사전 형태의 자료==
+==사전 형태의 자료==
 * http://ko.wikipedia.org/wiki/
 * http://en.wikipedia.org/wiki/Riemann_curvature_tensor
-* [http://eom.springer.de/default.htm The Online Encyclopaedia of Mathematics][http://dlmf.nist.gov/ ]
-==리뷰논문, 에세이, 강의노트==
-* http://www.math.csusb.edu/faculty/dunn/lecture1.pdf
+==리뷰, 에세이, 강의노트==
+* Dunn, [http://www.math.csusb.edu/faculty/dunn/lecture1.pdf An Introduction to the Riemann Curvature Tensor and Differential Geometry]
+** 슬라이드
 [[분류:미분기하학]]