# Talk:Tensor

I was very surprised NOT to see anything about Cartesian Tensors in the Introduction. The word 'tensor' is commonly used to signify either the general or the Cartesian concepts, rarely both. Many engineering students are ONLY exposed to Cartesian tensors. Much of that literature - in my experience - uses the term without qualification, making many statements and conclusions that are NOT true in the general case. It would be worthwhile, IMHO, to add a paragraph detailing the differences between a general tensor and a Cartesian tensor. The subjects are approached very differently in the applied literature vs the mathematical (and theoretical Physics) literature. I also think a section dealing exclusively with the Cartesian tensor would be useful. Unfortunately, I lack the competence to do this myself.216.96.76.236 (talk) 16:13, 22 June 2013 (UTC)

A second talk thread in the same section. In an attempt to make the page more accessible to laypeople, I have done my best to untangle the explanations, moving each to its own paragraph and also adding simplifying interpretations where I felt understanding of the article depended way too much on the reader understanding other elements of mathematics - elements that are not, in themselves, crucial to understanding the article. I have, next to the previous text, added references to things a reader can be expected to know (most obvious is the addition of basis ≈ coordinate system analogy).

However, I have not been able to figure out how to break down the introductory text without removing information from it. The problem with the current text is that it assumes the reader knows a whole bunch of stuff the average person may very well have never heard of. Somehow, the rigorous mathematical terms should be replaced with their "common tongue" equivalents. Triklod (talk) 01:09, 13 August 2013 (UTC)

The problem is in the opening sentence, tensor are not necessarily geometric objects, and as a working mathematician the best way to grasp the concept is understanding them as a multilinear maps, and then make exemplification with determinants, metrics, change of basis in vector spaces, change of coordinates on manifolds, and so kmath (talk) 02:08, 13 August 2013 (UTC)
This point has been disputed for several years now. The article emphasizes the engineer's (?) perspective on tensors, and relegates the mathematician's perspective to other articles. As a mathematician, I'm okay with this. The engineer's perspective requires less background, and hence is probably a better introduction to the topic for an encyclopedia. I do agree with you that the "geometric" part is rather unnecessary and probably biased, although it may reflect the historical motivation for the concept.
In Wikipedia articles with heavily disputed intros, it is common to propose rewrites on the talk page, that other editors can critique. So give us a rewrite? Mgnbar (talk) 02:55, 13 August 2013 (UTC)
I disagree that the article emphasizes any particular perspective. The purpose is to discuss all significant perspectives, per WP:NPOV. I also don't see what's wrong with the word "geometrical". Even in pure mathematics, tensors are defined as equivariant functions on a G-torsor with values in a representation of G; that's an intrinsically geometrical perspective, a la Klein. Finally, I fail to see how the first sentence is in any way inconsistent with the view of tensors as multilinear maps. Indeed, it seems to accommodate this view specifically (without going into details). Sławomir Biały (talk) 11:39, 13 August 2013 (UTC)
The modern mathematical viewpoint is relegated to Tensor (intrinsic definition), and only briefly mentioned here. The "As multidimensional array" viewpoint is given first and in greatest detail, suggesting to the reader that it is primary. Tensor fields are introduced before the multilinear and intrinsic definitions of tensor non-fields. All of this is what I meant about "engineer's perspective". But something has to go first, and the emphasis is not as strong as it once was (if I recall correctly), and anyway I'm not upset about it.
A tensor can be defined on a vector space in a purely algebraic way, without reference to geometry or topology (although historically those subjects were major motivators). That's what I meant by "geometric". Again, i do not consider this a major problem.
I agree that the first sentence is consistent with "multilinear maps". Mgnbar (talk) 12:17, 13 August 2013 (UTC)
Could it be that in this context, what You call "algebraic" and others call "geometric" is actually the same thing, namely the coordinate-free resp. index-free treatment of tensors?--LutzL (talk) 13:09, 13 August 2013 (UTC)
That could be, although coordinates are also a purely algebraic concept (scalars arising in linear combinations, which are algebraic). I mean, coordinates and coordinate-freeness do not require an inner product, a topology, or any of the other core concepts of "geometry" in its various manifestations. But we're arguing over a term ("geometry") that has no precise definition anyway. So it's probably not worth our effort. Mgnbar (talk) 14:15, 13 August 2013 (UTC)
Re: rewrite, if I had an idea for a rewrite (of the introduction), I would have proposed the text. But I don't. However, since my edits to the meat of the article were reverted for introducing errors, for me the introduction is a moot point now. And so is the continuance of me editing Wikipedia since, if I make errors on simple articles like tensors, what will happen if I move to more complicated articles? Whelp, nevermind. There are other things to do. Triklod (talk) 00:46, 14 August 2013 (UTC)

Sorry for no suggestions for improvements, but I think the entire article (not just the lead) reads extremely well and coherently, and the article could at least be rated to B class (if not A class). M∧Ŝc2ħεИτlk 12:52, 13 August 2013 (UTC)

## Not All Vectors and Scalars are Tensors

I'm certainly no expert in tensors, but it seems to me that the article says that all vectors and scalars are tensors. I have been led to believe this is not true due to the tensor's rules concerning coordinate transformations, which, as far as I know, do not apply to the general vector and scalar.

No this is always true (for the appropriate notion of a tensor). A scalar is invariant under changes of the coordinate system (note that a scalar is not just a number, but a number times its unit of measurement: this product is invariant). Likewise, the components of a vector transform contravariantly under changes of coordinates in the vector space that it lives in. Sławomir Biały (talk) 19:58, 1 July 2013 (UTC)

Not position vectors. -- — Preceding unsigned comment added by 86.184.230.128 (talk) 16:58, 18 September 2013 (UTC)

Vectors are tensors of type (1, 0). Scalars are tensors of type (0, 0). The coordinate transformations operate exactly as described. Mgnbar (talk) 03:48, 2 July 2013 (UTC)
What might underlie the perception is that some sets of quantities are defined with a dependence on the choice of coordinate basis and that are incompatible with the coordinate transformation rules. In this context, however, the terms "scalar" and "vector" are reserved for concepts that in some sense are independent of the choice of basis and thus inherently fit the tensor transformation rules. There are therefore quantities that look like scalars and vectors, but the terms may not be used to refer to them in this context. As examples, you might want to look at pseudotensor, Levi-Civita symbol and Christoffel symbols. — Quondum 13:04, 2 July 2013 (UTC)
Thank you very much for the replies, guys. So does that mean texts such as this: https://www.grc.nasa.gov/WWW/k-12/Numbers/Math/documents/Tensors_TM2002211716.pdf on page 7 are wrong or am I simply reading it and interpreting it incorrectly? Perhaps it is more in line with what Quondum is saying and I need to get my contexts straight. --204.154.192.252 (talk) 14:04, 2 July 2013 (UTC)
The treatment there is very far from the mathematical idiom (and rigor), so I have trouble reading it. But the relevant example is actually on page 10: The frequency of a light source is a scalar, but two observers moving with nonzero relative velocity will observe different numerical values for this scalar, and hence this scalar does not obey the (0, 0)-tensor transformation law. I'll let someone with better physics explain this. Mgnbar (talk) 15:46, 2 July 2013 (UTC)
Frequency in Relativity is not a scalar; it is the time component of a 4-vector, and as such changes with coordinate changes. Shmuel (Seymour J.) Metz Username:Chatul (talk) 18:58, 2 July 2013 (UTC)
I guess you mean one divided by the time component, since time has the dimension T while frequency has the dimension T -1? —Kri (talk) 21:33, 10 November 2013 (UTC)
There are serveral definitions for tensors, e.g., Tensors#As multilinear maps and for vectors, and with the most common definition of tensors a vector is not a tensor. However, there is a natural map from vectors into tensors, and it is common to ignore the distinction in informal prose. Shmuel (Seymour J.) Metz Username:Chatul (talk) 18:58, 2 July 2013 (UTC)
The linked document appears to use the terms scalar, vector, dyad in the "wrong" sense in this context: to mean a collection of real numbers used to represent something. So, where they say "scalar", read "real number", where they say "vector", read "tuple of real numbers". There is nothing inherently wrong about different use of terminology (I should know, I've spent much energy on trying to decipher what the defining terms really mean in various WP articles), but it helps hugely to figure out the exact intended meaning of the author. This WP article takes the approach that each of the names (scalar, vector etc.) applies to a class of tensor, which is useful in a more abstract approach, whereas the linked text uses the terms in a more school-text approach. On a point of order: this is no longer discussing anything that might relate to editing the article, so if taken further, it should be at Wikipedia:Reference desk/Mathematics‎‎. — Quondum 02:45, 3 July 2013 (UTC)
Tensor#As multilinear maps defines tensors in terms of vectors, so it is not true that the article uses vector interchangeably with tensor of rank 1. Shmuel (Seymour J.) Metz Username:Chatul (talk) 19:50, 3 July 2013 (UTC)
I don't quite see what you are driving at. This section effectively defines two types of tensor of rank 1: as a vector, or as a covector. They are thus the same thing by definition, and are interchangeable except for the purpose of making this definition. Also, the concepts vector and covector being used here each refer to an element of an abstract vector space, independent of any concept of basis, so there is no room for confusing this with the "tuple" concept dealt with earlier. — Quondum 23:50, 3 July 2013 (UTC)
No, by the definition in Tensors#As multilinear maps, a tensor of type (1,0) is a map $T:V^{*}\rightarrow \mathbf {R}$ , not an element of V. Shmuel (Seymour J.) Metz Username:Chatul (talk) 17:38, 5 July 2013 (UTC)
... which is naturally isomorphic to a vector, and in the context of tensors, the two are identified (by defining the action of a vector on a covector, thus making V∗∗ = V). But, as I alluded, this aspect (i.e. a discussion about the abstract definitions) has no relevance in this thread. If this should be addressed, start a new thread so that it can be addressed as a topic on its own. — Quondum 19:01, 5 July 2013 (UTC)
Of course there is a natural map from V to V**; otherwise I wouldn't have written However, there is a natural map from vectors into tensors, and it is common to ignore the distinction in informal prose.
The subject of this thread is Not All Vectors and Scalars are Tensors ; the distinction between V and V** is most definitely relevant to that. Shmuel (Seymour J.) Metz Username:Chatul (talk) 15:20, 8 July 2013 (UTC)
If V and V** are not identified, then arguably it is wrong to define tensors as multilinear maps. From the point of view of linear algebra, these are the same space, even if they are strictly not the same set. Sławomir Biały (talk) 16:30, 8 July 2013 (UTC)
I think that this is understood by all. As far as the article is concerned, notability is primary. Sławomir, in your experience, are there any notable sources that do not make this identification in their treatment, and if so, are they relevant to the article? — Quondum 16:39, 8 July 2013 (UTC)
Yes, right. In the rare occurrence that these spaces are not identified, we are in infinite dimensions and the relevant notion of tensor product becomes considerably more nontrivial. The relevant notions were developed in the 1950s by Alexander Grothendieck, fwiw. Sławomir Biały (talk) 23:11, 8 July 2013 (UTC)
I guess that the identification is sometimes/often made might be worth a mention. Another point that could be improved is the assumption of a holonomic basis in Tensor#Tensor fields. I'm not about to tackle this at this point, but might at some stage. — Quondum 02:18, 9 July 2013 (UTC)

## Can a tensor have different dimensionality for different indices?

When reading this article, I can't' find whether a tensor has to have the same number of dimensions for every index. That is, can all indices go up to the same number or can some indices go up higher than others? For example, can a linear map between two-dimensional vectors and three-dimensional vectors be considered a (1,1)-tensor? If so, then the contravariant index of that tensor could go to three while the covariant index only could go to two. —Kri (talk) 21:22, 10 November 2013 (UTC)

In the abstract mathematical treatment of tensor products, it is certainly possible to take a tensor product of two vector spaces of differing dimension. The elements of this tensor product are "tensors", in which the indices do not all go up to the same number. However, there has been some argument on this talk page in the past, about whether the term "tensor" is really used in this generality.
This article tends to focus on tensors from the engineering/physics perspective. In that context (for example fluid dynamics textbooks), I have never seen the indices differ in dimension, because the tensors in question are constructed from a single vector space (and its dual). But someone more expert than me may comment differently. Cheers. Mgnbar (talk) 22:00, 10 November 2013 (UTC)
To make a more strict statement, of course you can have tensors of the type $(V\otimes V)\otimes (W\otimes W\otimes W)^{*}$ . But firstly, in the index-notation, this would be counted as a 5-fold tensor product, preferably with different types of letters for V and W. Which actually happens in physics for space and spinor indices. And "But" secondly, the (r,s) classification is, as Mgnbar says, only used for tensor product of a single vector space and its dual. So ${\mathcal {T}}^{(2,0)}V\otimes {\mathcal {T}}^{(0,3)}W$ could be used, if necessary.--LutzL (talk) 01:48, 11 November 2013 (UTC)
If one wants to be really pedantic, braiding can also be taken as significant, so if being thoroughly pedantic, one might put for example the type of a tensor as $V^{*}\otimes V\otimes W\otimes V\otimes W$ . Tensor (intrinsic definition) is more the place for this, though. — Quondum 02:29, 11 November 2013 (UTC)
Why can't the (r,s) classification be used for a tensor product of two or more vector spaces? What is wrong with the example I provided for example, where a linear map from two to three dimensions is written as a tensor — can't that be said to be a (1,1)-tensor? It takes one contravector and converts to another contravector, so there has to be equaly many covariant as contravariant indices in that tensor, doesn't it? By the way, do r and s in (r,s) classifications stand for anything, or are they just arbitrary variable names? In the article they use m and n. —Kri (talk) 21:21, 11 November 2013 (UTC)
The r and s are just variable names. They are positional placeholders in the "type-( , )" terminology. One reason why you cannot apply the "type-(r,s)" terminology for multiple independent vector spaces is that choosing which to call covariant and which to call contravariant is completely arbitrary, independently for each pairing of a vector space and its dual. Choosing V as "covariant" (and thus V* as "contravariant") doesn't stop you choosing W as "contravariant" (and thus and W* as "covariant"). You'd need a separate (r,s) pair for each new vector space in the product. — Quondum 23:11, 11 November 2013 (UTC)

## Raising and lowering indices

There is a problem with the description of Raising and lowering indices in Tensor and related articles. In the traditional braided index notation, you get the original tensor after you raise and lower a particular index, but with the definition in the article you get a braided equivalent instead. Using a definition of the tensor algebra that uses mixed products, e.g., $V\otimes V^{*}\otimes V^{*}\otimes V$ , with either the quotient vector space definition or the multilinear function definition of the tensor product, allows for clearer and unambiguous raising and lowering of indices of mixed tensors like $T_{\alpha \beta \gamma },\ T_{\alpha \beta }{}^{\gamma },\ T_{\alpha }{}^{\beta }{}_{\gamma },\ T_{\alpha }{}^{\beta \gamma },\ T^{\alpha }{}_{\beta \gamma },\ T^{\alpha }{}_{\beta }{}^{\gamma },\ T^{\alpha \beta }{}_{\gamma },\ T^{\alpha \beta \gamma },$ . at the expense of more complicated notation for the type of a tensor and a more complicated definition of the tensor algebra. I believe that this isue should be discussed at least briefly. Shmuel (Seymour J.) Metz Username:Chatul (talk) 19:27, 13 December 2013 (UTC)

I don't really follow your specific objection in relation to this article, but I do feel that the notational conventions for braiding in this context get glossed over unduly. This is in part because braiding is so naturally handled by the index notation, but also in part because so many people think of the set of tensor components as the tensor. It would be nice if these subtleties are made clearer. For example, I've seen a suggestion that lexicographical order of the indices can be significant in some conventions with respect to braiding, but this is rarely mentioned. This should also be addressed in more explicit detail in Ricci calculus. —Quondum 03:43, 15 December 2013 (UTC)
My concern is that the definitions given do not allow for braiding of indices, which is still common in the literature. Shmuel (Seymour J.) Metz Username:Chatul (talk)