Talk:Curvature invariant

Order versus degree of an invariant
What is a zeroth order invariant. From the def it is a scalar constructed from covariant derivatives, but zero Riemannn tensors.--MarSch 15:52, 19 October 2005 (UTC)


 * Hi, Marsch, somewhere in one of these pages I said that n-th order invariant uses at most n covariant derivatives. So the Kretschmann scalar is a zeroth order quadratic curvature invariant, for example.---CH  (talk) 22:11, 19 October 2005 (UTC)


 * You also call the number of Riemann tensors the order, you call the other differential order, so I've changed it accordingly. Perhaps the first should be degree (see polynomial)? --MarSch 09:01, 20 October 2005 (UTC)


 * Huh? The number of Riemann tensors?!!!  I propose to call
 * $$R_{abcd} \, R^{abcd}$$ a zeroth order second degree invariant,
 * $$R_{abcd;e} \, R^{abcd;e}$$ a first order second degree invariant,
 * and so forth. OK?


 * I reverted your edit because you turned sense into nonsense. You wrote An important unsolved problem in general relativity is to give a basis (and any syzygies) for the zero-th order differential invariant).  That doesn't make sense, because differential invariants have order one or higher.  Zeroth order invariants involve no covariant differentiations.---CH  (talk) 18:48, 21 October 2005 (UTC)


 * It does make sense. The problem is that you use order for two different things currently. I've already thought of and mentioned two solutions and implemented one of those. I'm going to revert back to my version pending your decision on a solution. --MarSch 09:42, 23 October 2005 (UTC)


 * What you wrote is vacuous because there are no zeroth order differential invariants (see order versus degree in examples above). So I changed it back to what I had written previously.  Are we approaching a resolution of this yet?  Is there some disagreement over what the open problem is, or is the disagreement purely one of consistent notation/terminology in the article? ---CH  (talk) 00:53, 24 October 2005 (UTC)

One issue is resolved now that you've changed order to degree. Unfortunately now you define "nth order differential invariant" and further on you talk about "zeroth order invariants" which is apparently a special case of this, but you don't want to call it differential because it is a degenerate special case.

The invariants most often considered are polynomial invariants. These are polynomials constructed from contractions such as traces. Second degree examples are called quadratic invariants, and so forth. Invariants constructed using covariant derivatives up to order n are called n-th order differential invariants.

It is now clear to me what you mean, but the text is not clear yet. Also I would like the definition of degree to be a bit more precise. Maybe: "A polynomial invariant of degree n or equivalently an nth degree polynomial invariant is a scalar formed by summing contractions of n Riemann tensors and less.". Same for differential invariants. Also don't forget the combined case: an nth order differential mth degree polynomial invariant. You probably want to call this an nth order mth degree invariant, but you have to say that.


 * I've been too distracted by other concerns to pay as much attention to all this as I should have done, but yes, I agree with "degenerate" and with your definition of "nth order mth degree invariant". In fact, this is what I had in mind all along :-/ but I was trying to avoid fighting to the whole darn world over this degree of pedantry.  (Er... no pun intended, not that anyone will believe me.) Obviously, I need not have assumed that I would have no allies in attempting to make and impose on others such terminology nicities!  And I have to agree that it is not helpful to try to weasel out of insistence upon a sensible terminology (and I think the one we have agreed upon here is the best choice).


 * I am still too distracted to read over these articles to make sure they all agree with this terminology, but since I think we agree on the essentials, I'd be grateful if you can find time to read through them and make sure they use degree and order consistently. I think you are arguing that "differential invariant" is more trouble than its worth, and I tend to agree, as long as one clearly says somewhere that this term is widely used rather loosely for any invariant of positive order, i.e. involving at least one covariant differentiation.  And one needs to warn that unfortunately, many authors carelessly use "order" where we would use "degree", which of course leads to all kinds of unneccessary confusion. ---CH  (talk) 20:03, 24 October 2005 (UTC)


 * Glad we sorted this out. I guess by these articles you mean other GR stubs. I'll try and fix this one, maybe I'll also look at the other stubs. Right now I'm working my way through geometry stubs though.--MarSch 13:58, 8 November 2005 (UTC)

Students beware
I created the original version of this article and had been monitoring it, 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 in your seach for information, regardless!---CH 23:13, 30 June 2006 (UTC)

Suggested merge from Curvature invariant (general relativity)
I don't think there is enough difference in the usage of this term between general (pseudo)-Riemannian geometry and general relativity to warrant two separate articles.TimothyRias (talk) 12:50, 10 November 2010 (UTC)