Talk:Intermediate treatment of tensors

message: It can't be true to say that a rank n tensor is 'simply' a tensor product of rank 1 tensors. In some or other sense it may be a linear combination of that kind of product.
Charles Matthews 09:41, 5 Nov 2003 (UTC)
thread_title:  Rank of a tensor 

message: There is a small, recurrent contradiction in this article. If a (p,q) tensor has contravariant rank p and covariant rank q, then the subscript and superscript indices have been mixed up.  It should be [i1, ... , ip] in the superscript and [j1, ... , jq] in the subscript, right?
--Anon 19:00, 10 Apr 2004 (EST)
:I've always understood the prefixes contra- and co- to refer to a relationship between two things.  I've understood contravariant to be a relationship between two tensors (not an absolute property of one), such that one of the tensors is a superscripted and the other is subscripted, resulting in an inner product operation (which would involve a metric tensor in Riemmanian geometry).  Covariant, on the other hand, would be where both tensors are superscripted or both are subscripted, resulting in something like a diffeomorphism.  Or an outer product, depending on whether indices are shared between (or among) the tensors. Kevin Baas 18:05, 11 May 2004 (UTC)
thread_title:  Superscripts and subscripts 

message: Is "the total number of degrees of freedom required for the specification of a
particular tensor" 
:"the dimension of the tensor raised to the power of the tensor's rank."
or
:"the dimension of the tensor times the tensor's rank."?
Surely if the tensor is the tensor product of lots of rank-1 tensors, each with d.o.f. = dimension, it's the latter? On the other hand, if we think about the representing array, it's the former, but am I missing something here? -- The Anome 16:22, Jun 2, 2005 (UTC)
Rereading the article, it surely must be the latter whichever way we view it, that is, n(p+q), not n^{(p+q)}? Or have I completely gone round the twist? -- The Anome 16:35, Jun 2, 2005 (UTC)
:So, in relativity say, dimension = 4 (space-time dimensions). A curvature tensor has four indices. What does that mean? It means we have 4x4x4x4 components to account for. The exponential is right, therefore. Charles Matthews 16:52, 2 Jun 2005 (UTC)
thread_title:  Degrees of freedom of a tensor 

message: 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 14:11, 12 Jun 2005 (UTC)
thread_title:  discussion at [[Wikipedia talk:WikiProject Mathematics/related articles]] 

message: I suspect that I am the target audience for this (and similar) articles.  So, I thought I would share my perspective with you.
Before I get into all of that, let me say how much I appreciate that "you people" (the smart guys who are writing all of this stuff) have taken the time to share their knowledge in this forum.  This is truly a fantastic resource, and demonstrates how the Internet can be used for a great and constructive purpose.
So that you can understand where I am coming from, I will give you a little background on myself.  Many years ago, I got a BS in engineering, but I don't work professionaly as an engineer.  Since I was a kid, I have been fascinated by science (primarily physics) and have read many popular accounts of physical theories (relativity, QED, etc).  In fact, I am currently reading Roger Penrose's book (Road to Reality).  Trying to understand some of the concepts in that book is what brought me to this page.
In any event, I have a basic background in math and science (a couple of years of calculus, diff. eq., classical physics, etc.) but nothing very advanced.  For example, while I have heard the terms, I never studied manifolds, eigenvalues, tensors, groups, etc.  So, that is where I am.  Would I not be your target audience?
Now, on to my comments (which apply to all of the entries on tensors):
(1) I would urge you to present some very simple and "intuitive" example of tensors (or whatever concept you are trying to explain) early in the non-technical introduction.  For example, the page on eigenvectors presents a marvelous and very intuitive example (the "face") early in the presentation.  If a tensor is "like" a matrix, show me a tensor, and the corresponding matrix.  How do I get from the tensor to the matrix, etc.
(2) I would suggest that you gently introduce other highly technical supporting subject matter by limiting the number of cross links early in the discussion.  Of course, when the subject itself is complex (like tensors), there is obviously going to be certain prerequisite areas of knowledge.  Perhaps it would be best to explicitly state that the reader should be familiar with x, y, and z (with supporting links) in order to understand the following discussion.  I would keep this list as short as possible.
(3) I know that this is a somewhat nebulous concept, but I would try to cultivate a "beginners mind" when sharing your knowlege.  In other words, keep in mind that a lot of us have no idea of what you are talking about once you get beyond basic algebra (especially in reference to mathematical formalisms).  For example, the "dual vector space U*" is introduced with no explanation and no cross-link.  I still don't know what this is.  
(4) Flesh out the sections on notation. Half the battle is just understanding the notation.   For example, the main tensor page has a section on the various notations.  I assume that the "Abstract Index Notation" is the "main" notation used now days, because it is listed first.  But when you go to that stub, it doesn't actually show you the notation at all.  It just says that Penrose invented it and it is "better."  Similarly, if you follow the next link to the Einstein notation, you learn about Einstein's summation convention, but nothing about the basic tensor notation.  Finally, after a lot of jumping around, I found the "intermediate" page, which gives the best explanation of the basic notation (although I am still scratching my head after following the links to the covariant and contravariant links).
(5) I do think that the whole tensor section has gotten a little too difuse and disconnected (with the whole "classical," "intermediate" and "modern" approaches).  Perhaps this is necessary, but it makes things rather confusing when you are first approaching the subject like I am. 
I know that these must be difficult concepts to convey, and I appreciate your efforts. —The preceding unsigned comment was added by Kyletownsend (talk • contribs) 05:10, 2006 November 2.
:There was plenty of noisy discussion years ago. This and the other pages formed would now be called 'POV forks', a somewhat deprecated notion. We might revisit the whole issue. (Please sign with ~~~~). Charles Matthews 10:10, 2 November 2006 (UTC)
replies: text: This page (the Classical Treatment of Tensors) is impossible to understand from the perspective of someone (me) with a B.S. in Math but who does not already know what a tensor is.  Also, why does the discussion button for the "classical treatment of tensors" redirect to a page titled "talk:intermediate treatment of tensors"? Halberdo (talk) 03:37, 16 January 2009 (UTC)
thread_title:  "Beginners Mind" 

message: I notice the indices are enclosed in brackets. In the classical notation square bracked means skew symmetrize and round brackets symmetrize. Why are they here? Billlion 16:01, 13 May 2007 (UTC)
replies: text: I image it's to be meta-syntactical.  I suppose they can be just taken out because people probably get that they can put anything there.  A better question: why does the classical treatment of tensors's talk page redirect to this one? Kevin Baastalk 16:56, 27 January 2009 (UTC)
thread_title: Brackets in indices

message: It seems to me there is a mistake here, a general tensor is not always the product of rank 1 tensors. Ylebru 11:23, 27 July 2007 (UTC)
replies: text: Really??  I suppose a tensor of rank 0 would not be the product of rank 1 tensors, but can you name any other examples?   Saying it's not always the product is like saying that a natural number is not always the sum of adding 1 with itself multiple times.  And that directly contradicts the definition of the set of natural numbers. Kevin Baastalk 16:53, 27 January 2009 (UTC)
replies: text: It could be the sum of such tensors, couldn't it (as opposed to being simply an outer product of vectors and 1-forms)?Dependent Variable (talk) 17:13, 17 September 2009 (UTC)
thread_title:  General tensor 

message: :"In general, the value of a tensor field at an event in spacetime is an element of a vector space which is the tensor product of multiple copies of the tangent space (contravariant vectors) and multiple copies of the cotangent space (covariant vectors). As such, it is a smooth (C∞) mapping from the base space of a vector bundle to the total space which when projected back onto the base space has returned to its starting point."
Umm, yeah...  what he said.
I think what he means to say is that you look nice today.  (Forgive him, he doesn't know the dialect very well.)
Can we put things in english or not at all please?  Thank you.  Kevin Baastalk 17:00, 27 January 2009 (UTC)
thread_title:  gobbledy-gook 

message: Shouldn't the equation in the section Transformation Rules
:\hat{T}^{i_1\ldots i_p}_{\,j_1\ldots j_q} =
A^{i_1} {}_{k_1}\cdots A^{i_q} {}_{k_q}
B^{l_1} {}_{j_1}\cdots B^{l_p} {}_{j_p}
T^{k_1\ldots k_p}_{l_1\ldots l_q}
include a typo in the indices?  Shouldn't it rather be
:\hat{T}^{i_1\ldots i_p}_{\,j_1\ldots j_q} =
A^{i_1} {}_{k_1}\cdots A^{i_p} {}_{k_p}
B^{l_1} {}_{j_1}\cdots B^{l_q} {}_{j_q}
T^{k_1\ldots k_p}_{l_1\ldots l_q}?
Jamesmelody (talk) 09:30, 26 March 2009 (UTC)
replies: text: Also A and B are swapped. Has been corrected. -- Theowoll (talk) 18:35, 9 January 2010 (UTC)
thread_title:  Typo in indices of equation?