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The section on "twisting by a line bundle" and "half-density" is quite opaque as it currently is written. linas 00:21, 5 May 2005 (UTC)
I was redirected to this page from half-form (by clicking the link in the article on geometric quantization) but on this page there is no explanation to what a half-form is, if it is the same as a tensor field or something else. I have not found information on this anyewhere else. Erik J 10:41, 5 October 2005 (UTC)
- I'm planning on fixing this, but at the rate I'm going, it'll take another 6 months. Mean time line bundle and Hopf bundle may help. linas 14:02, 5 October 2005 (UTC)
- OK, a half-form would be something like a square root of an n-form on an n-dimensional manifold. The question is, what is the geometric meaning of such a thing? It is not obvious that geometrically taking square roots is harmless (it isn't for complex numbers). But you are right that it is time to support this all by references (Guillemin-Sternberg, Geometric Asymptotics, p. 251). Charles Matthews 21:28, 5 October 2005 (UTC)
This article is part of a series of closely related articles for which I would like to clarify the interrelations. Please contribute your ideas at Wikipedia talk:WikiProject Mathematics/related articles. --MarSch 14:12, 12 Jun 2005 (UTC)
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This article was automatically assessed because at least one WikiProject had rated the article as start, and the rating on other projects was brought up to start class. BetacommandBot 10:04, 10 November 2007 (UTC)
Tensor field graphic
By analogy with the tensor field arrows, shouldn't the vector field illustration have 2 arrows for each cell. One arrow pointing out of the cell and one in the cell and their product is the resulting arrow? I think that is the sense of the final illustration. Each pair of arrows defines a product. YouRang? (talk) 02:23, 23 February 2010 (UTC)
- My understanding (and this is more from intuition) is that in a plot of tensors any number of arrows greater than one represent a 2nd-order tensor of that dimension. Note that there are three arrows with three degrees of freedom; the tensors in question can be represented by 3x3 matrices, with each vector representing a separate three indices. To represent a 3rd-order tensor requires groups of such arrows (multiple 2nd-order tensors). To differentiate you could use color coding or make them the vertices of a polytope (with multiple polytopes being assigned to each point). Further methods are available, but every visualization method must make compromises, and in general higher-order tensors are notoriously difficult to visualize. I can understand why this would be the case much easier if the geometric complexity of the units of representation grows as the order grows. But if a 3rd-order tensor can be parametrized by three vectors, what then is the point of using the higher-order form? ᛭ LokiClock (talk) 23:46, 14 September 2010 (UTC)
- I agree with User:YouRang?, and as you can see I have removed the graphic. For one thing, 3rd-order tensors are not parametrized by three vectors (you are confusing the tensor product with the direct sum). Anyway, the graphic was supposed to illustrate a 0th order, 1st order, and 2nd order tensor in two dimensions. But 2nd order tensors in two dimensions are represented in a basis by 2×2 matrices, not 3×3 matrices, and so it would be more natural to display a pair of vectors at each point. In any event, if there is to be such a graphic, then it would be better to also explain the meaning of the representation of a 2nd order tensor as a pair of vectors, lest readers leave with the false impression that an nth order tensor field is nothing more than an association of n vectors to each point of space. Sławomir Biały (talk) 13:04, 21 September 2010 (UTC)
I am trying to teach myself the basics of fluid mechanics in order to implement wind flow in a video game, and I have been intimidated all night by the constant references to tensors. The sentence in the introduction about how a tensor is a generalization of a scalar field and a vector field was exactly what I needed. I'm not scared anymore. That is a much simpler and more concise explanation than is found in the actual article on tensors. —Preceding unsigned comment added by 188.8.131.52 (talk) 05:31, 7 September 2010 (UTC)
The C∞(M) module explanation a bit conservative?
Is there any particular reason that it only assumes that is a module over a ring, instead of a vector space over a field? It seems to me that C∞(M) would definitely qualify as a field (just using point-wise addition and multiplication), and that could just as easily be a vector space. It could be that I'm overlooking some subtleties, but if so, I'd love to hear what they are. (Perhaps it would be good in general to add one or two references.) --Jaspervdg (talk) 10:29, 28 October 2011 (UTC)