Talk:Tensor density

message: (one post copied from above thread, separating as above thread is getting unwieldy)
:::Do you still have access to Lee's book? You can find that kind of density in chapter 14 (old ed). It allows you to do invariant integration on non-orientable manifolds. Sorry for interrupting. YohanN7 (talk) 19:06, 5 July 2014 (UTC)
replies: text: I think Lee's density is a density in the sense of the present article with weight -1 (our convention). Haven't checked thoroughly though. YohanN7 (talk) 21:57, 5 July 2014 (UTC)

text: If a density in Lee's sense is something that can be integrated, it has weight +1 in our convention (reams of text explain this above).  I only have the old electronic version of his book on me.  I'm away from my dead tree copies at the moment. He doesn't discuss densities in that version unfortunately. On oriented manifolds, the bundle of densities in naturally isomorphic to the bundle of n-forms, where n is the dimension of the base manifold.  In general, a non-oriented manifold M has an oriented double cover, N→M, together with an action of the group with two elements Z_2.  The bundle of densities is  naturally isomorphic the bundle of Z_2 equivariant maps from N to the bundle of n-forms on M.  Such a thing has an invariant integral because it can be pulled back to an n-form on the oriented manifold N, integrated there, and then divided by two.  Notice that everything in this construction is tensorial: there are no absolute values appearing.  But the densities are tensors on the bundle N.  Sławomir Biały (talk) 23:09, 5 July 2014 (UTC)
replies: text: I find the double cover of a manifold idea a rather beautiful way of dealing with the integrability orientation issue – I was toying with it but was unsure; I'm glad you mentioned it. In any event, Lee's use of densities appears to be a way of dealing with the orientability problem.  Here is a link (or this newer edition) to a publicly viewable page in his book.  It should be clear that he is defining it in a coordinate-independent fashion. In a nutshell, Lee's density is defined to scale its result according to the determinant of a transformation applied to its input vectors, but this article's density is defined to scale its result according to the Jacobian determinant of the transformation of the basis (potentially in addition to the scaling as per Lee's density, if it is an n-form density), and cannot be made tensorial via your double cover construction. These are two very different scaling dependencies. Thus, the two meanings of "density" are entirely distinct. —Quondum 00:55, 6 July 2014 (UTC)
replies: text: Thanks for the link.  The usual bare word "density" refers to an even 1-density (in the terminology of this article).  That is, one for which the Jacobian appears in the absolute value.  This is probably worth mentioning.  Also, I wonder whether the distinction between even and odd densities is a useful one at all.  Sławomir Biały (talk) 02:41, 6 July 2014 (UTC)

text: @Q Lee's density measures volumes spanned by n vectors in the tangent space at each point of the manifold. That's, for instance, what every n-form does, some of them with a sign, this one is unsigned. These things can be integrated and that's the point. (The same is true for volume forms on orientable manifolds.) This article addresses what happens when the coordinates are changed. Edit: What I'm trying to say is that there aren't two meanings of density in Lee's density. YohanN7 (talk) 03:35, 6 July 2014 (UTC)

text: The article ought to address why 1-densities are invariantly integrable. As far as I can see it is because the "usual" volume element  has weight -1. When multiplied with something with weight +1, these weights add and the net result is a true scalar with weight 0 that is immune to coordinate changes. Also, Lee's density appears to be just an even 1-density. YohanN7 (talk) 03:09, 6 July 2014 (UTC)
thread_title:  Comparing to Lee's definition 

message: I don't think it is clear cut in our discussion what is required for an invariant integral to exists. It is hinted that objects with weight +1 are invariantly integrable, no other. I'd say that it depends on what pre-exists in the integrand. If there is an invariant volume element in place, then objects of weight 0 will be invariantly integrable with respect to it. If, on the other hand, only  is in place, then an object of weight +1 will be invariantly integrable. 
The default seems to be that  is in place in the integrand, no invariant element, i.e. not . 
By extension if the integrand has weight  (including everything!), then it can be multiplied by an object of weight  and then invariantly integrated. 
Are these observations correct? YohanN7 (talk) 04:31, 6 July 2014 (UTC)
replies: text: Yes, and you're hopefully starting to get my point. If we use the (invariant) weight-  as our volume element, the whole point of multiplying by the Jacobian determinant elsewhere disappears. The volume element becomes tensorial, and we need only tensors, never "weighted densities" in the sense of this article. In short, we never need to work with anything not of weight . The whole, complex treatment of weights appears to be an artefact of failing to identify an invariant volume element at the start.  (I've said this before above.)  Furthermore, the nontensorial even/odd/pseudo sign shebang can be abolished by using a double cover of the manifold, as per Sławomir. —Quondum 05:34, 6 July 2014 (UTC)
replies: text: But identifying a volume form at the start is not a canonical choice.  The bundle of densities is trivial, but not canonically trivial.  Sławomir Biały (talk) 13:30, 6 July 2014 (UTC)
thread_title:  Invariantly integrable quantities 

message: The introductory section and the Definition section profoundly fail to explain what a tensor density is to the reader.
Can someone who actually understands this concept please rewrite these sections?
Note: Merely using the concepts is not the same as understanding them. To understand them it is necessary to grasp their mathematical underpinnings. To define a tensor by referring only to "how it transforms" and without ever saying what it is that does the transforming shows a complete lack of understanding of what tensors are.2600:1700:E1C0:F340:21E2:140D:C505:D61A (talk) 00:45, 12 November 2017 (UTC)
thread_title:  Can someone who understands the subject please define it in the article?