Talk:Cartan's equivalence method

message: I need help outsourcing this to other pages as much as possible.  Silly rabbit 04:48, 10 November 2005 (UTC)
replies: text: As I understand it, the 'method' is in his hands a bit less than algorithmic? It does however have the key idea that the graph of the 'diffeomorphism to construct' is to be constructed by setting up differential forms and thus a differential system for which the graph is a solution manifold (locally ...). Forgive me if this sounds a naive description. As for things on other pages, yes, it is often easier to work on some of the prerequisites first, and then write higher-level pages as hypertext quite dependent on others things. This tends to take a little longer to do than one might expect. On the other hand the alternative is to write pages dense with red links and hope to fill those out later. I don't think there is a single, right way to do it; in the end one has to nduce some sort of 'organic growth' in the coverage here so the top-level contributions don't seem too isolated. Charles Matthews 10:16, 10 November 2005 (UTC)

text: You are right, it is less than algorithmic because the process may not always terminate.  As for a description of the method, or "algorithm" (although the terminology is not my own), it is hideously complex.  Silly rabbit 15:45, 10 November 2005 (UTC)

text: I dipped into Cartan's book on his Travaux, which is always interesting if only to try to get the language down. Only half-a-dozen papers on 'equivalence' as such. One of those did sound a bit like early work on CR structures. At the outset it is (in his view) mainly derived from Lie's work, i.e. something like deriving a Lie group action by considering tuples of Pfaffian forms. Which ties up with saying 'coframe fields'. He doesn't really claim novelty for that, though he possibly constructed some of the exceptional Lie groups that way. I also dipped into Chern's Selected Papers, but (although Chern digested as much of it as anyone, AFAICS) the introduction doesn't really offer a general description. Charles Matthews 16:15, 10 November 2005 (UTC)
replies: text: Cartan solved the equivalence problem for 3-dimensional CR manifolds, although I don't have the paper on hand.  Chern solved the equivalence problem in general.  (Chern, Moser Real hypersurfaces in complex manifolds  Acta Math., 133 (1974) 219-271.)  The original Cartan paper is referenced there.  Silly rabbit 16:56, 10 November 2005 (UTC)

text: Right - the Phillip Griffiths introduction in the Chern papers book discusses that at some length (I was looking at Chern's own introduction). Charles Matthews 17:06, 10 November 2005 (UTC)

message: There are precise circumstances under which Cartan's equivalence method is an algorithm, given in the Cartan-Kuranishi prolongation theorem.  This theorem is seldom used in applications because it is usually just about as hard to solve the equivalence problem as it is to check the hypotheses of the theorem.  Nevertheless the theorem is an interesting result, since it is essentially a "finite-order" result much like the Peetre theorem (an article I am also working on at the moment).  Best, Silly rabbit 23:09, 11 November 2005 (UTC)
replies: text: Yes; it's an 'exercise' in Dieudonné's 'Treatise on Analysis'. I probably set too much store by those books; but they're an excellent source of things to include in WP, Dieudonné being something of a one-man encyclopedia himself. Charles Matthews 23:31, 11 November 2005 (UTC)
replies: text: No kidding?  I have dipped into Dieudonné's treatise in the past.  But I really hope you meant Peetre's theorem!  Cartan-Kuranishi is no joke.  Out of curiosity, do you have a volume and page number so I can check it out.  I'd like to see the Bourbakist perspective on finite-order theorems.  Undoubtedly his mere exercise is better than my actual article.  ;-)  (No sarcasm intended.)  Silly rabbit 00:30, 12 November 2005 (UTC)

text: Clearly I should pay more attention to Dieudonné.  However, although it is difficult to state the Peetre theorem in a correct and applicable way (as I'm currently trying to do), once the thing is settled, it ought to be a relatively easy theorem to prove.  Cartan-Kähler is significantly more complicated, but also can be proven readily once all the bits have been correctly assembled.  But you haven't said anything about Cartan-Kuranishi: Kuranishi, On E. Cartan's prolongation theorem of exterior differential systems, Amer. J. Math., 79 (1957), 1-47.  It's truly no joke.  To be honest, I don't even understand the hypotheses of the theorem, let alone the proof.  Silly rabbit 09:17, 12 November 2005 (UTC)

text: Clarification: Volume IV, Ch.18.13 is the section with Cartan-Kuranishi in the exercises, straight after Cartan-Kähler in the text. (Dieudonné clearly was a hard taskmaster on homework). This is all very interesting, by the way. I wouldn't call Dieudonné a geometer, and on most of these Cartan-legacy things one can believe that what is being stated is abstractly true, without having a good picture. I certainly appreciate what you are doing here. Charles Matthews 10:24, 12 November 2005 (UTC)

text: Sorry for my earlier incredulity.  This is worth reading up on.  ;-)  Silly rabbit 14:52, 12 November 2005 (UTC)
thread_title:  Algorithm 

message: I found this article very interesting, Cartan's program was something I wanted to learn about, so I am planning to read this in more detail later, but just a quick question for now: I've seen this passage: 
"Although the answer to this particular question was known in dimension 2 to Gauss and in higher dimensions to Christoffel and perhaps Riemann as well, Élie Cartan and his intellectual heirs developed a technique..." 
and wondered what the solution is, in the case of higher dimensional Riemannian manifolds. If the curvature tensor is zero, then the Riem. manifold is locally isometric to Euclidean space, but this doesn't help with deciding whether there exists an isometry between two given metrics, neither of which are necessarily flat. You can't just compare the components of the curvature tensor, and I am unaware of a set of invariants of the metric that characterize it completely in arbitrary dimensions. Can anybody give some pointers to the "answer" cited in the passage I quoted?  —Preceding unsigned comment added by 128.61.118.134 (talk) 16:59, 20 May 2008 (UTC)
thread_title:  The case of Riemannian geometry