Timeline for Geometrical meaning of the Ricci Tensor and its Symmetry
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7 events
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| Jul 6, 2011 at 17:16 | comment | added | Robert Bryant | One more comment: If you assume that the Ricci curvature is symmetric, so that there is a (local) $\nabla$-parallel volume form, say $\Upsilon$, then the Ricci curvature has the following interpretation: Let $\exp_p:T_pM\to M$ be the exponential map of $\nabla$ based at $p$. Then $\exp^\ast_p(\Upsilon)$ can be written as $(1 - \frac13 R_{ij} x^ix^j + \cdots)$ times a constant coefficient volume form on the vector space $T_pM$, where $\exp^\ast_p\bigl($Ric$(\nabla)\bigr)_p = R_{ij}\, dx^idx^j$. Thus, Ric gives the deviation of the parallel volume form from the exponentially flat one. | |
| Jul 5, 2011 at 12:13 | comment | added | Robert Bryant | In indices: First Bianchi is $R^i_{jkl}+R^i_{klj}+R^i_{ljk}=0$. Set $i=j$ and sum to get $R^i_{ikl}+R^i_{kli}+R^i_{lik}=0$, which becomes $R^i_{ikl}=R^i_{kil}-R^i_{lik}$. Now $\Omega = \frac12 R^i_{ikl}\ dx^k\wedge dx^l$ is the curvature of the induced connection on the top exterior power, and $\frac12(R^i_{kil}-R^i_{lik})dx^k\wedge dx^l$ is the skew-symmetric part of the Ricci tensor. I guess the universal constant is $1$. | |
| Jul 4, 2011 at 15:49 | comment | added | Fly by Night | This universal constant, is it a number, or is it an exact differential one-form? If I recall, the curvature tensor is a type (3,1)-tensor. How is this related to a connexion on $\bigwedge^n T^*M$, when n can be anything I choose? Have you got some references on this, and can you recommend a nice book? I appreciate you taking the time to answer. But your answer has replaced one confusion with an even bigger one. I don't properly understand the curvature tensor, so I've no hope of understanding curvatures of induced connexions on $\bigwedge^n T^*M$. I need to start from the beginning | |
| Jul 4, 2011 at 2:08 | comment | added | Robert Bryant | You already have the answer about the non-symmetric part. The first Bianchi identity shows that the skew-symmetric part of the Ricci tensor is a 2-form $\Omega$ that is equal (up to a universal constant) to the trace of the full curvature tensor. This $\Omega$ is the curvature of the induced connection on the top exterior power of the (co)tangent bundle, and hence this vanishes if and only if there is a $\nabla$-parallel volume form. Alternatively, the integral of this 2-form over a compact oriented surface $S$ in $M$ is the holonomy of the connection around $\partial S$. | |
| Jul 3, 2011 at 18:36 | comment | added | Fly by Night | Thanks for your comment Giuseppe. Please see my comment above. My source does (secretly) assume Levi-Civita. So I guess we should forget the geodesic ball example. However, I'd till like to have some geometrical interpretation of the trace: does it "average things out"? What, geometrically, does the Ricci Tensor mean? What, geometrically, does its (non-)symmetry mean? | |
| Jul 3, 2011 at 13:27 | comment | added | Willie Wong | For the Levi-Civita connection of a pseudo-Riemannian metric, you can get the above formula by just expanding the the metric in normal coordinates and compute. | |
| Jul 3, 2011 at 7:50 | history | answered | agt | CC BY-SA 3.0 |