Timeline for Geometrical meaning of the Ricci Tensor and its Symmetry
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Aug 26, 2012 at 13:56 | answer | added | S.B | timeline score: 1 | |
| Aug 4, 2011 at 12:10 | answer | added | Robert Bryant | timeline score: 20 | |
| Jul 5, 2011 at 2:30 | comment | added | Fly by Night | I disagree about having to rewrite anything. The original statement of my question was a very good illustration of my ack of understanding. Several patient and kind volunteers have identified problems with my original post, which identifies problems with my (lack of) understanding. This comment thread, and the one below, have identified and addressed problems in my understanding, and have suggested ways of combating that misunderstanding. I would have said that that was the aim of a question: to identify and to tackle one's misunderstanding. | |
| Jul 4, 2011 at 16:15 | comment | added | Deane Yang | I think you need to rewrite your question completely. If I understand your question, you are working with a Ricci tensor defined with respect to an affine connection, and you want a geometric interpretation of it. I still don't understand how the pseudo-Riemannian metric you mention is related to the connection. If it has no relationship and is arbitrary, then it is obvious that geodesic balls (by the way, what is a "geodesic ball" for metric with indefinite signature?) defined with this respect to this metric have nothing to do with the connection and its Ricci curvature at all. | |
| Jul 3, 2011 at 21:19 | comment | added | Fly by Night | That's what I thought: the trace sums the eigenvalues, and that should give an average. Although I think that's why the normalised Ricci tensor is defined (divide the ordinary one through by the dimension of M). I can understand the geometry behind $\mbox{Ric}(X,X)$. But then what does $\mbox{Ric}(X,Y)$ mean, and why should it differ from $\mbox{Ric}(Y,X)$ for $X≠Y$? I've been using all of these tensors to prove some nice little theorems. But one of my colleagues asked me: "But what does that mean, geometrically?" And I had to wave a white flag with a red face! | |
| Jul 3, 2011 at 18:38 | comment | added | Deane Yang | Indeed, you can view $Ricci(v,v)$, up to a constant factor, as the average sectional curvature of planes containing a unit vector $v$. And, yes, a trace is an average. If you have a symmetric $n$-by-$n$ matrix $A$, then the trace of $A$ is equal to $n$ times the average of $v\cdot Av$ over all unit vectors $v$. | |
| Jul 3, 2011 at 18:35 | comment | added | Deane Yang | Ben, there is another posting, mathoverflow.net/questions/68960/symmetric-ricci-tensor, where it's stated that the Ricci tensor is symmetric if and only if there is a parallel volume form. But this still leaves me rather from any explanation of what the Ricci tensor means geometrically. | |
| Jul 3, 2011 at 18:34 | comment | added | Fly by Night | The geodesic ball notion can from Wikipedia. I should have known better. I had to add a note the the Ricci Tensor article myself saying that the connexion was assumed to by Levi-Civita. The article says a lot of things, while secretly assuming Levi-Civita. I should know better. My apologies. But anyway, back to the non-Levi-Civita case. I'd still like to get some geometrical understanding of the Ricci Tensor. Some sources say that it "averages out" sectional curvatures. But why should that be non-symmetric? Does the trace "average things out"? Geometrically, what does symmetry mean? | |
| Jul 3, 2011 at 14:49 | comment | added | Ben McKay | I agree with Deane Yang that there has to be some relation between the connection and the metric in order to have any relation between the exponential balls of a connection and the metric. However, perhaps you mean to measure the volume of balls using a volume form which is parallel for the connection. Most connections have no such volume form, and even when a volume form is parallel for a connection, so is any constant multiple, so the question wouldn't make sense then either. | |
| Jul 3, 2011 at 11:11 | comment | added | Deane Yang | What relationship are you assuming between the pseudo-Riemannian metric and the connection? You appear to saying there is none. Then why is the metric needed at all? | |
| Jul 3, 2011 at 7:50 | answer | added | agt | timeline score: 2 | |
| Jul 3, 2011 at 0:59 | history | edited | Fly by Night | CC BY-SA 3.0 |
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| Jul 3, 2011 at 0:21 | history | asked | Fly by Night | CC BY-SA 3.0 |