Timeline for Formalising the principle of general covariance in differential geometry
Current License: CC BY-SA 2.5
32 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 30, 2010 at 23:47 | comment | added | Deane Yang | "smooth tensor bundle" has a standard rigorous definition that every differential geometer I know uses. It would be nice, if you could confirm that that's the definition you're using. Apparently not, though. | |
| Dec 30, 2010 at 20:06 | answer | added | Deane Yang | timeline score: 3 | |
| Dec 30, 2010 at 16:20 | comment | added | Zhen Lin | @Deane: Actually, you may as well assume I'm using my own definitions; as I said, my understanding of this formalism is incomplete. I can say things like "the tensor bundle is the vector bundle over the manifold of the tensor algebra of the tangent and cotangent spaces at each point" but I'm not sure that helps. I've realised the phrase "smooth tensor bundle" is horribly confusing — perhaps "the tensor algebra of smooth sections of the tangent and cotangent bundles, algebraic operations defined pointwise" would have been better. | |
| Dec 30, 2010 at 16:04 | comment | added | Deane Yang | I don't want to hear your ideas. I want to know the definitions of the words you are using. I presume you are not using your own definitions. | |
| Dec 30, 2010 at 16:03 | comment | added | Deane Yang | How can we answer your question if you do not know the meaning of the words you are using? | |
| Dec 30, 2010 at 16:01 | comment | added | Zhen Lin | [continuation] The intuitive idea in my head is to compare a tensor field with its pushforward (using the term loosely) under an diffeomorphism $\phi: M \to M$. I've scribbled some equations on paper which try to capture this, but I'm not sure if I've written down tautologies (or blatant falsehoods) rather than anything interesting. | |
| Dec 30, 2010 at 16:00 | comment | added | Deane Yang | I am unable to follow the conversation here. But I don't understand the statement that the difference between a tensor and a non-tensor can be subtle. Could you give an example of this? | |
| Dec 30, 2010 at 16:00 | comment | added | Zhen Lin | @Deane and others: I would like to make my ideas more explicit, but unfortunately the notation and formalism is not familiar to me. Any help in that area would be much appreciated. I notice that the question has been upvoted twice already — I hope that's because someone thinks there's an interesting idea here, and if so, I'd like to appeal to those people for help. | |
| Dec 30, 2010 at 15:54 | comment | added | Zhen Lin | @Andres: I apologise. I was frustrated by the various comments which suggested that the $\Gamma$ I had defined was not a tensor — my comment in reply to José was nothing more than a restatement of what I had written in my original post. I realise the notation was confusing, but that was also the point I intended to make — that the difference between a tensor and a non-tensor can be quite subtle at times. In any case, I have redacted the comment. | |
| Dec 30, 2010 at 15:43 | history | edited | Zhen Lin | CC BY-SA 2.5 |
Redacted comment
|
| Dec 30, 2010 at 3:56 | history | edited | Zhen Lin | CC BY-SA 2.5 |
Fixed a thinko.
|
| Dec 30, 2010 at 3:50 | comment | added | Deane Yang | Could you rewrite your question and explicitly define what you mean by a "tensor bundle", a "tensor field", what it means for a "tensor field to be contained in a subspace", and, most importantly, what it means for tensor field to be "diffeomorphism-invariant"? It appears that you are using these terms in ways unfamiliar to many of us (perhaps because you are using definitions from a rather old treatise on general relativity?), so many of us are completely confused. | |
| Dec 30, 2010 at 3:46 | comment | added | Zhen Lin | @Spiro: Oops. I had a thinko and assumed the metric was mapped to the metric of the target space. I'll fix that. The question still stands though — the isometry subgroup of the diffeomorphism group is, well, a subgroup, so falls under the scope of question 3. Another subgroup which we might ask about, for example, is the subgroup of conformal diffeomorphisms. | |
| Dec 30, 2010 at 3:20 | comment | added | Spiro Karigiannis | -1. The question still does not make sense. If, by a diffeomorphism invariant tensor $T$, you mean that $f^* T = T$, then the metric and curvature tensors are not in general diffeomorphism invariant. They are only isometry invariant. If you want tensor fields (sections of the tensor bundles) to "transform" as tensors, as physicists would say, then they ALL do. That's the definition of a tensor field. You don't need a (pseudo-) Riemannian metric to discuss tensor fields or diffeomorphisms anyway. | |
| Dec 30, 2010 at 3:09 | history | edited | Zhen Lin | CC BY-SA 2.5 |
Rewrote the question
|
| Dec 29, 2010 at 19:14 | comment | added | José Figueroa-O'Farrill | @Zhen: So I suppose you are defnining $\Gamma$ as in your comment and extending it $C^\infty(M)$-linearly, right? If so, my original comment is not relevant. But then notice that your $\Gamma$ is not defined everywhere, since unless $M$ is parallelisable, it will not have a global frame. | |
| Dec 29, 2010 at 16:59 | comment | added | Zhen Lin | @Jeff: Perhaps I should have been more clear about that. I did say "classical characterisation". It's also certainly the case that it's still taught that way to undergraduates in some universities. Personally I don't like it very much, but it appears (to me) to capture a principle which is lost when moving the modern coordinate-free picture — hence my question today. | |
| Dec 29, 2010 at 16:43 | comment | added | Jeff Harvey | First, this isn't how physicists define tensors (maybe some did 40 years ago) and second, I suspect your confusion could be cleared up by reading the relevant sections of Wald, "General Relativity." | |
| Dec 29, 2010 at 16:35 | comment | added | Zhen Lin | [continuation] I admit the preliminary woffle was, well, woffle. But I don't think I could make my question any clearer without narrowing the scope of it more than I would like. For instance, I could rephrase it as "Is there an algebraic characterisation of tensors which are diffeomorphism-invariant, i.e. an algebraic construction starting from a reasonable generating set which yields only invariant tensors, including the tensors of interest e.g. the Riemann tensor?" However, this assumes diffeomorphism-invariance is the correct formalisation of the principle of general covariance. | |
| Dec 29, 2010 at 16:31 | comment | added | Zhen Lin | @unknown: Let me define two functions f and g by f(x) = x + x and g(x) = 2x. They are extensionally the same function, because they are equal at every point. But they are intensionally different, because I defined them in two different ways. This is a concept I have borrowed from philosophy / logic. If I say a tensor is "manifestly covariant", this is a claim based on the intensional definition of the tensor. The question is, is there a way to detect this extensionally, i.e. in a way that is independent of how I explicitly define the tensor? [continued] | |
| Dec 29, 2010 at 16:26 | comment | added | Zhen Lin | @Sam: That is precisely the point I'm making. Once we have a global frame it's trivial to make tensors of any type we like. However, intuitively, almost all of them are meaningless because the construction is unnatural. | |
| Dec 29, 2010 at 16:17 | comment | added | Qfwfq | btw, what does "extensionally" mean in that context? | |
| Dec 29, 2010 at 16:15 | comment | added | Qfwfq | -1 for lack of clarity. I have the impression that understanding the content of a basic differential geometry course would clarify many doubts that physicists have (or doubts coming from physics) about "coordinate covariance" etc. Anyway, if the question will be improved i'll remove my downvote. | |
| Dec 29, 2010 at 16:14 | comment | added | Sam Gunningham | I am not sure if this is what you were getting at, but here is my interpretation of your question: If you have a manifold $M$ with a global frame $e_1 , \ldots , e_n$ of pointwise linearly independant vector fields, then it is very easy to make tensors - you just pick functions $a^i _j$ (where there could be many $i$'s and $j$'s), and you have a tensor $ \sum a^i _j e_i \otimes \epsilon ^j $. Because the frame was global, we don't have to check anything when changing coordinates - we can just define the tensor as above. This works with any collection of functions (e.g. Christoffel symbols). | |
| Dec 29, 2010 at 13:22 | answer | added | Patrick I-Z | timeline score: 4 | |
| Dec 29, 2010 at 12:58 | comment | added | Zhen Lin | I don't see how though — yes, the map $D: TM \times TM \times T^*M \to C^\infty(M)$ given by $(X, Y, Z) \mapsto Z(\nabla_X Y)$ is not linear in the second argument, but that's not how I defined $\Gamma$. Rather, I defined $\Gamma$ as the unique multilinear map such that in a fixed basis, $\Gamma(e_\nu, e_\mu, \epsilon^\sigma) = \epsilon^\sigma(\nabla_{e_\mu} e_\nu)$. This certainly is $C^\infty(M)$-linear, no? | |
| Dec 29, 2010 at 11:04 | comment | added | José Figueroa-O'Farrill | The confusion is that the $\Gamma$ you have defined is not a tensor. As Yakov mentions in his comment, the modern differential-geometric definition of a tensor is a section of a tensor bundle $\bigotimes^r TM \otimes \bigotimes^s T^*M$. In particular it has to be $C^\infty(M)$-linear in all entries. Your $\Gamma$, from its very definition, is not. | |
| Dec 29, 2010 at 10:40 | comment | added | Kelly Davis | Like Tim I am not sure I understand your exact question. But, I've found that J.D. Norton's "General covariance and the foundations of general relativity: eight decades of dispute"[1] helpful when I've ventured to think about general covariance. [1] tinyurl.com/34ahxtg | |
| Dec 29, 2010 at 9:53 | answer | added | Tim van Beek | timeline score: 3 | |
| Dec 29, 2010 at 9:21 | comment | added | Zhen Lin | As I noted, this space contains things which are "obviously" not "tensors" — e.g. the $\Gamma$ I defined above. Of course, the fact is that it is formally a tensor by the way I constructed it, but there's something intuitively non-tensorial about the construction. The problem is formalising this intuition. | |
| Dec 29, 2010 at 8:40 | comment | added | Yakov Shlapentokh-Rothman | Is there something wrong with the standard "invariant" characterization of a tensor, i.e. as a map from $T^1(M) \times \cdots T^1(M) \times T(M) \times \cdots \times T(M) \to C^{\infty}(M)$ which is multilinear over $C^{\infty}(M)$? $T(M)$ denotes the space of vector fields and $T^1(M)$ the space of $1$-forms. | |
| Dec 29, 2010 at 8:31 | history | asked | Zhen Lin | CC BY-SA 2.5 |