Talk:Ricci decomposition

The term "Ricci decomposition"
I've recently seen this referred to by another name in papers on the arXiv. I am trying to find them again so that I can mention the alternative name. However, I believe the one I have given here is not only more accurate but is fairly standard in the mathematical literature--- I am also trying to locate the textbook(s) where I have seen it called by this name.---CH (talk) 19:12, 13 August 2005 (UTC)

Shouldn't
 * $$E_{abcd} = \frac{1}{n-2} \, \left( g_{ad} \, R_{bc} - g_{ac} \, R_{bd} + g_{bc} \, R_{ad} - g_{bd} \, R_{ac} \right)$$

read
 * $$E_{abcd} = \frac{1}{n-2} \, \left( g_{ad} \, S_{bc} - g_{ac} \, S_{bd} + g_{bc} \, S_{ad} - g_{bd} \, S_{ac} \right)$$

so that it really is constructed from S?

Terminology
During my time studying symmetries and working with the decomposition (I'm not giving 'the decomposition' a name yet), I was taught that:


 * $$\, S_{ab}$$ was called the Trace-free Ricci tensor and usually denoted by $$\, \tilde{R}_{ab}$$.


 * $$\, E_{abcd}$$ was referred to simply as the E-tensor (as did others who classified it - I forget the names, but I can find out).


 * The tensor $$\, H_{abcd}$$ denoted by $$\, G_{abcd}$$ and was sometimes called the G-tensor, but usually thought of as - and called - the bivector metric, as it behaves precisely like a metric on the space of bivectors.


 * The decomposition we often called the Géhéniau-Debever decomposition.

(At least) some of the above alternative terminologies are used in many research papers (certainly the trace-free Ricci tensor).

Well, I just seem to have complicated the whole issue of terminology in this article now! This may not be such a huge problem, and actually quite interesting, as the decomposition can obviously be defined mathematically (with it's own mathematical terminology) and it's applications to GR give alternative terminology. MP  (talk) 11:00, 29 January 2006 (UTC)

Students beware
I extensively edited the August 2005 version of this article and had been monitoring it for bad edits, but I am leaving the WP and am now abandoning this article to its fate.

Just wanted to provide notice that I am only responsible (in part) for the last version I edited; see User:Hillman/Archive. I emphatically do not vouch for anything you might see in more recent versions, although I hope for the best.

Good luck to User:Mpatel and to all students searching for information, regardless!---CH 03:27, 1 July 2006 (UTC)

See also http://mathworld.wolfram.com/RiemannTensor.html 24.56.66.131 (talk) 07:55, 27 May 2011 (UTC)Collin237

Weyl flat versus Ricci flat versus...
I recently "corrected" a statement in the "Physical interpretation" section. I had misread that to say something like "if a spacetime is Ricci-flat, then there is no light-bending", which is obviously not true: there is just no focusing. But the statement was actually about Weyl-flat (i.e., conformally flat) spacetimes. Now, it is not exactly true that there is no light bending in such spacetimes (there is matter, after all). Rather light doesn't distort astigmatically, so for instance one can observe a circular Einstein ring, but never an ellipsoidal one. I still think something more should be said (probably about both the Ricci-flat and conformally flat cases), but I lack the expertise to do it. In the meantime, I have removed the offending statement altogether. Sławomir Biały (talk) 22:10, 25 September 2010 (UTC)

"Depending on the signature of the metric tensor"?
Can someone please justify the statement "The decomposition works in slightly different ways depending on the signature of the metric tensor"? I'm confused, since whilst e.g. the sign of $$\det(g)$$ depends on this, the contractions $$g^{ab}g_{bc}=\delta^a_c$$ and $$g^{ab}g_{ba}=\delta^a_a = n$$ are, I hope, always true...?

(I've just fixed an error in the actual decomposition which has been around for a couple of years... Please do check that the Ricci scalar calculated from the decomposed Riemann tensor is now actually equal to the Ricci scalar.) Carl Turner (talk) 13:34, 7 August 2012 (UTC)


 * It doesn't depend on the signature of the metric, but it will depend on curvature sign conventions (so this wasn't an error, but a different sign convention). The formulas seem to have the wrong factors, though.  I will fix these.  Sławomir Biały  (talk) 15:05, 7 August 2012 (UTC)


 * OK, I've added a thing specifying the convention in use. Pretty sure the original factors were correct though, so I've reverted this. If you contract with $$g^{ac}$$ you get $$nS_{bd} - S_{bd} + 0 - S_{bd} = (n-2)S_{bd}$$ from the bracket in $$E_{abcd}$$. Carl Turner (talk) 15:57, 7 August 2012 (UTC)


 * Yes, sorry, the factors seem to have been correct before (up to sign). Sławomir Biały  (talk) 16:10, 7 August 2012 (UTC)