differential geometry w notes from teacher_Part_77

differential geometry w notes from teacher_Part_77 - 1 R i...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
6.2. TENSOR ANALYSIS 153 Proof : 1. 6.2.4 Properties of the Curvature Tensor Let ( M , g ) be an n -dimensional Riemannian manifold. We will restrict our- selves to the Levi-Civita connection below. We define some new curvature tensors. The Ricci tensor R i j = R k ik j . The scalar curvature R = g i j R i j = g i j R k ik j . The Einstein tensor G i j = R i j - 1 2 g i j R . The trace-free Ricci tensor E i j = R i j - 1 n g i j R . The Weyl tensor (for n > 2) C i j kl = R i j kl - 4 n - 2 R [ i [ k δ j ] l ] + 2 ( n - 1)( n - 2) R δ [ i [ k δ j ] l ] = R i j kl - 4 n - 2 E [ i [ k δ j ] l ] - 2 n ( n - 1) R δ [ i [ k δ j ] l ] di ff geom.tex; April 12, 2006; 17:59; p. 152
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
154 CHAPTER 6. CONNECTION AND CURVATURE Theorem 6.2.2 The Riemann curvature tensor of the Levi-Civita con- nection has the following symmetry properties
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1. R i jkl =-R i jlk 2. R i jkl =-R jikl 3. R i jkl = R kli j 4. R i [ jkl ] = R i jkl + R i kl j + R i l jk = 5. R i j = R ji Proof : 1. ± • Theorem 6.2.3 The Weyl tensor has the same symmetry properties as the Riemann tensor and all its contractions vanish, that is, C i jik = . Proof : 1. ± • Theorem 6.2.4 The number of algebraically independent components of the Riemann tensor of the Levi-Civita connection is equal to n 2 ( n 2-1) 12 . Proof : 1. ± di ff geom.tex; April 12, 2006; 17:59; p. 153...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern