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Motivating examples

I'm contemplating adding a section with "motivating examples" before the definition section. This section would discuss the objects that a tensor is supposed to generalize and maybe something like a linear map. It would be a good place the introduce the representation of vectors in components and how these transform. It would also allow the introduction of the concepts of dual vectors and the canonical dual basis.

It could also talk about linear maps and their representation as matrices. This could discuss the similarity transformation as the effect of a change of basis.

The current definitions section takes knowledge of these concepts somewhat for granted, or introduces them very quickly. Having a somewhat more extended section introducing these concepts could do a lot for accessibility of the article. Do others agree that this is a good idea? And if so, what should be covered in this section? TimothyRias (talk) 09:05, 2 June 2010 (UTC)

I support this. It would balance with the generalizations section. Perhaps as a subsection of history, rather than on its own? ᛭ LokiClock (talk) 10:21, 2 June 2010 (UTC)

Image

I think this is wrong:

whose columns are the forces acting on the ${\displaystyle \mathbf {e} _{1}}$, ${\displaystyle \mathbf {e} _{2}}$, and ${\displaystyle \mathbf {e} _{3}}$ faces of the cube.

it should be:

whose rows are the forces acting on the ${\displaystyle \mathbf {e} _{1}}$, ${\displaystyle \mathbf {e} _{2}}$, and ${\displaystyle \mathbf {e} _{3}}$ faces of the cube.

--Thomas gölles (talk) 15:35, 17 February 2011 (UTC)

maybe the vector on the left side should be a column vector? maybe the picture is wrong? a basis is usually a row vector, right? which is the basis, the e(space) or the o(force)? (i would think the e.) that would imply the left side should be a column vector and the wording should be changed as suggested. Kevin Baas 16:30, 17 February 2011 (UTC)

Tensors in machine learning and pattern recognition applications

The "tensors" referred to in the linked articles are just high-dimensional arrays erroneously referred to as tensors. Should this section be allowed to remain? Sławomir Biały (talk) 11:58, 1 March 2011 (UTC)

Are you sure? I've skimmed the cited Lu et al. article. I couldn't find any instance in which it used the "tensors" as multilinear maps or performed changes of basis on them. On the other hand, it does use some tensor products and contractions, and it does explicitly talk about tensor spaces. Is it possible that there really are tensorial things happening here, that are hidden? Mgnbar (talk) 14:08, 1 March 2011 (UTC)

A tensor product followed by a contraction is essentially a change of basis, whether it's explicitly called that or not is irrelevant. ("a rose by any other name...") Kevin Baas 14:35, 1 March 2011 (UTC)

Change of basis is a particular kind of tensor product followed by contraction. Not all examples of tensor product followed by contraction are changes of basis, right? Anyway, this is not a deal-breaker. A piece of mathematics can be tensorial without having a bunch of changes of basis in it. Change of basis was just a tell-tale sign that I was looking for as I skimmed the article. Mgnbar (talk) 16:22, 1 March 2011 (UTC)

Also, the applications of tensors to machine learning are fairly obvious. If they weren't being used in machine learning, that'd certainly be something to start doing right this instance! And really, we're generally not that slow. In today's day and age, if you've thought of something, chances are somebody's (if not 100's of people) already thought of it and done it. So it is quite unlikely that tensors are not being used in machine learning. Kevin Baas 14:40, 1 March 2011 (UTC)

I agree with your statement that "if you've thought of something, then chances are somebody's done it". I have no doubt that researchers in machine learning are smart people who are fully capable of using tensors. I don't know anything about machine learning, so it is not at all obvious to me that tensors are applicable, or that they are not applicable. I have to agree with Slawomir that it's hard to see how tensors are really being used in the linked material. Additionally, the Multilinear subspace learning article itself is in some dispute.

At this point I do not favor deletion of the material. My ideal outcome is for the editors who know and care about this material to shore it up and make it relevant to Tensor. Mgnbar (talk) 16:22, 1 March 2011 (UTC)

A lot of work has been done on Multilinear subspace learning over the past few days, but I am still ambivalent. That article says "The term tensor in MSL refers to multidimensional arrays. Examples of tensor data include images (2D/3D), video sequences (3D/4D)..." This supports Slawomir's point, that practitioners are not really using tensors. For example, I have trouble seeing how a 2D image (presumably a "2-tensor") is supposed to represent a bivector, linear transformation, or bilinear pairing (the only kinds of 2-tensors). On the other hand, if there really is an entire field using the word "tensor" in this way, then Misplaced Pages should at least mention that use. It's not our job to decide that some popular word usages are wrong. Mgnbar (talk) 16:50, 4 March 2011 (UTC)

New image.

I've done some editing for the picture in the lead. First of all, I've corrected the labeling. More importantly, the vectors ${\displaystyle \mathbf {T} ^{(\mathbf {e} _{i})}}$ now actually are the sum of their components ${\displaystyle \sigma _{ji}}$. Moreover, in the spirit of less is more, I've reduce the clutter by removing various elements which are not necessary on this page. Hopefully, this makes the image less confusing. (I know it had me confused for quite a while.) Any feedback before I insert it in the article?TR 10:23, 29 March 2011 (UTC)

Poorly written intro

The lead sentence says it all. "Higher orders"? Higher orders of what? Higher order functions (which at least have a wiki page)? The casual reader will never know. The intro carries on in a like vein without given the reader any clue at all as to what a tensor is. Then, having failed to give any clue as to what a tensor is, the intro provides an example. Of a stress tensor. A particular type of tensor. A particular type of thing that the reader knows nothing about because she/he has looked tensor up on Misplaced Pages.

Anyone who does not already know what a tensor is will find this intro utterly useless. Truly, this is one of the worst intros to a math-related article on Misplaced Pages. I'd start to fix it, but other than BOLD, I honestly don't know where to begin. Ross Fraser (talk) 02:58, 28 April 2011 (UTC)

(Why is it that people making remarks, always have to rant like teetering idiots? Is it so hard to take a deep breath, and write a mature comment?)

I agree that the current lede is not particularly good. It leaves the reader with the nagging question "But what is a tensor?". However, I'm still at a loss as how to best improve it.

The base of the problem is that there exist different views as to what a tensor is exactly. The main views can be summarized as

"A tensor is a multi-dimensional array of numbers with certain transformation rules."

and

"A tensor is a multilinear map."

With various variants existing.

This makes it hard start Tensor like you would most lemma's by saying something like "A tensor is XXX", where XXX is something rather concrete. Any choice of XXX however almost automatically picks one of the competing views over the other introducing a POV.TR 10:08, 28 April 2011 (UTC)

Calling another WP editor a ranting, "teetering idiot" is commensurate with your standard of what constitues a mature comment?

Ross Fraser (talk) 06:03, 29 April 2011 (UTC)

I've rearranged the lead a bit, to put the discussion of tensor order first. I think that for someone never before encountering the notion of tensor, the idea of a multidimensional array is probably the most likely to be attainable. I don't find the new lead altogether satisfactory either, but perhaps you can improve it. Sławomir Biały (talk) 13:16, 28 April 2011 (UTC)

(unindent) As the second line we now have,

"A tensor is a multi-dimensional array of numerical values."

That will make many people cringe. More importantly it will serve to confuse readers who have had limited exposure to tensors, who have been taught that tensors are not multi-dimensional arrays. I do, however, understand why you want to mention multi-dimensional arrays first. Would any of the following alternatives work:

1. "A tensor can be represent as a multi-dimensional array of numerical values." (this certainly is true, but leaves the reader in the dark as to what a tensor is. Feature rather than bug?)
2. "A tensor can be thought of as a multi-dimensional array of numerical values." (Also true, hints at the ambiguity in definition, but might also be interpreted as WP:WEASEL.

Not sure what's best so I'm suggesting it here first.TR 14:16, 28 April 2011 (UTC)

The second line makes me cringe. Originally I had something like your first suggestion, but I think the phrase "can be represented" is likely to be even more confusing to someone with limited background. As the original poster said, the lead failed to convey adequately a sense of what tensors "are". This at least says what they "are", even at the risk of telling a little white lie at first that we attempt to correct in the second paragraph. Just my 2c. Sławomir Biały (talk) 14:30, 28 April 2011 (UTC)

I know that this article is hard to write, but I vote against the "multidimensional array of numbers" definition. That would exacerbate the problem that novices have in distinguishing tensors from matrices.

How about this amalgam of the first sentences of the first and second paragraphs: "In mathematics and physics, a tensor is a scalar, a vector, or a certain kind of relationship among, or function of, vectors." And people learn by examples, so why not put basic ones early in the article? "Elementary examples include the dot product, the cross product, the determinant, and all linear transformations." And applications are good, like the stress tensor relating surface normal vector to the force vector on the surface. The second paragraph could be all about coordinates of tensors.

Lastly, there is no reason to restrict to geometric vectors; tensors don't require an inner product on the underlying vector space. Mgnbar (talk) 14:26, 28 April 2011 (UTC)

Your suggestion seems like a reasonable one as well. But ultimately some stylistic decisions need to be made, and I think the "multidimensional array" definition is both more concrete and perhaps the most commonly encountered first definition (despite its obvious limitations). Also, to my mind, geometric vectors don't assume an inner product, but are the naive coordinate-independent "directed line segment" definition of vectors. (It's true that our article discusses inner products, but it also discusses the generalized notion somewhat.) I think it's important to distinguish the vectors as geometrical objects somehow. Sławomir Biały (talk) 14:36, 28 April 2011 (UTC)

A compromise must be made, and sadly (for me) I may get outvoted, but I'd like to assert that the dot product (which anyone who is trying to learn tensors already knows) is more familiar than multidimensional arrays. Examples make abstract concepts concrete. Mgnbar (talk) 14:48, 28 April 2011 (UTC)

Regarding "vector" vs. "geometric vector": To me this seems related to the whole issue of Tensor vs. Tensor (intrinsic definition), which I've always found unsatisfactory. Maybe things would be clearer if we divided them up along these lines: algebra of tensors vs. geometry of tensors? Mgnbar (talk) 14:48, 28 April 2011 (UTC)

I don't think the two notions are easily separated, though. Sławomir Biały (talk) 14:52, 28 April 2011 (UTC)

Another suggestion

How about the following suggestion for the first paragraph,

Tensors are geometric objects that describe linear relations between vectors, scalars, and other tensors. Elementary examples include the dot product, the cross product, and linear maps. A tensor can be represented as a multi-dimensional array of numerical values. The order (also degree or rank) of a tensor is the dimensionality of the array needed to represent it. A number is a 0-dimensional array, so it is sufficient to represent a scalar, a 0th-order tensor. A coordinate vector, or 1-dimensional array, can represent a vector, a 1st-order tensor. A 2-dimensional array, or square matrix, is then needed to represent a 2nd-order tensor. In general, an order-k tensor can be represented as a k-dimensional array of components. The order of a tensor is the number of indices necessary to refer unambiguously to an individual component of a tensor.

This combines various aspects of Sławomir Biały's and Mgnbar's suggestions and some of my own. The first line gives as a concrete a statement of what tensors are as we can, they are "geometric objects". It also makes clear that they have to do with linear relations between other geometric objects such as scalars and vectors. Hopefully, it doesn't go too far in the direction of saying that tensors are multilinear maps. If not, maybe replace "... objects that describe ..." with "... objects used to describe ...".

I liked Mgnbar's suggestion of including some concrete examples in the second line. There is a good chance readers will have heard of things like the dot product. (Note that I didn't include determinant as an example. Although the determinant can be interpreted as a rank N tensor, most readers will only know it as a map on linear maps/matrices. In that context it is not linear, which will act as an unnecessary point of confusion (IMHO).)

We can then safely say that this objects can be represented as multi-dimensional arrays, and explain what is meant by order of a tensor in that context. Some additional tweaking may be needed since the first line no longer tells us that vectors and scalars are themselves tensors (at least not explicitly).TR 08:12, 29 April 2011 (UTC)

Seems fine to me, except there is some circularity in the first line saying that tensors describe linear relations between tensors. Would it be better to say something like "multilinear relations between vectors"? Sławomir Biały (talk) 12:06, 29 April 2011 (UTC)

I agree that this is a good attempt except for the first sentence. I prefer something along the lines of "scalars, vectors, and ... among vectors" (as I wrote above).

Now there is one other issue. This treatment emphasizes geometry, going so far as to define "order", which makes sense only in the presence of an inner product (or more generally a nondegenerate bilinear pairing). I know that this is common in physics and engineering, but there are whole swathes of mathematics where tensors are used without inner products. There is no canonical way to raise and lower indices, so there are real differences among (2, 0)-, (1, 1)- and (0, 2)-tensors, for example. This is why I was arguing for a clear distinction between algebra and geometry of tensors above; the latter would assume an inner product, and the former would not. Mgnbar (talk) 12:51, 29 April 2011 (UTC)

First off, I think you are wrong about the last part. In no way does the definition of covariant and contravariant tensors require the introduction of of an inner product. The inner product is only required if you want to be raise or lower indices. More to the point, the suggested paragraph does not even mention tensor type, but only the total order/rank/degree.

As to the perceived circularity in the first sentence. That was intentional. In general high order tensors, do describe linear relations between lower order tensors. The inverse metric for example, is a tensor that describes a linear relation between pairs of covectors (rank 1 tensors) and scalars. I don't think it is bad thing, to hint at this sort of thing up front, since it gives an indication of the generality of the concept. It nibs in the butt, any sort of lingering questions like "what do you can something that describes a linear relation between tensors?" (answer, a tensor)TR 14:32, 29 April 2011 (UTC)

Well, I guess I'm satisfied. By me, I'd say go ahead and be bold. Sławomir Biały (talk) 14:34, 29 April 2011 (UTC)

We have miscommunicated about this "geometry issue". I was certainly not saying that you need an inner product to define (p, q)-tensors. In fact, I was saying the opposite: Without an inner product (or more generally a nondegenerate bilinear pairing) you must keep upper and lower indices distinct, because you have no raising/lowering mechanism. For example, there are three kinds of "2-tensors" --- (2, 0), (1, 1), and (0, 2) --- and they all behave differently and mean different things.

So I have never been able to recognize any significance or value to the "total order" concept p + q, in the absence of an inner product. (Please correct me if I am wrong!) This is why I said that a discussion of total order implicitly slants the article toward geometry. Please note that you could completely satisfy me by replacing p + q with the more detailed notation (p, q). Mgnbar (talk) 15:18, 29 April 2011 (UTC)

I understood what you meant. I just don't think the technical distinction between covariant and contravariant indices is something that should be emphasized in the lead. The current lead is pitched at an absolute novice. Insisting on being strict about conceptually hard technical issues is going to make it far less accessible in my opinion. Sławomir Biały (talk) 15:39, 29 April 2011 (UTC)

Well, the significance of the total order is very simple. It is the dimension of the array needed to represent it/the number of indices, etc. This is the only thing that is currently mentioned in the first paragraph. Explaining, the (p,q) type of tensor is somewhat more involved, as it refers to concepts that most readers will have never heard of. Their definition is discussed in detail in the body of the article. For the above reasons, I don't think it is such a good idea to mention them in the lede.TR 15:45, 29 April 2011 (UTC)

Okay, I am still of the opinion that maintaining p and q separately clarifies the meaning of tensors rather than obscuring it, but we don't want to hit the reader with too much in the intro, so I concede that point.

TimothyRias, I know that p + q is the total dimension of the array; I just don't see what the point of computing this number is. If you were writing a computer program to do calculations with tensors, you would have to store the numbers p and q (rather than just the number p + q), and use them frequently, or else your computations would go wrong. (I have written such programs.) As further evidence, I note that Tensor (intrinsic definition) mentions the total order barely, as a tiny aside. But this is veering away from the editing of Tensor, so let's not argue about it here. Cheers. Mgnbar (talk) 17:09, 29 April 2011 (UTC)

This is indeed veering off-topic so this the last comment I'll make on the matter. If you want to store the components of a (2,2) tensor in a C program, what depth array would you declare? Exactly 2+2=4. If you are treating a tensor as multidimensional array the dimension of that array will always be very important. In fact the numbers p and q seperately, are somewhat useless since you will need to keep track of the valence of each index seperately. I.e. the knowledge that the 10-dimensional array A has 4 covariant and 6 controvariant indices is not that useful in an actual computation. What you need is the knowledge that indices 1,3,6,and 7 are covariant.TR 20:59, 29 April 2011 (UTC)

Salomon Bochner is reputed to have proclaimed at the beginning of a lecture at Princeton: "A tensor is something with indices!" This initial proclamation is perhaps too crude to be useful for most serious applications, but it is a very useful starting point. Sławomir Biały (talk) 17:25, 29 April 2011 (UTC)

Yeah, I'd like to hear the part of the anecdote where he talks about the Christoffel symbol. ;-) Mgnbar (talk) 19:25, 29 April 2011 (UTC)

Yeah, well he probably left that as a homework exercise...  ;-) Sławomir Biały (talk) 19:46, 29 April 2011 (UTC)

Spinor

Currently the article contains a section on spinors as generalisations of tensors. This seems a bit odd. I'm not sure in what sense spinors are supposed to generalize tensors. A spinor can of course be thought of as a generalisation of a vector. The generalization of a tensor would be a half-integer object, i.e. the tensor product of a tensor with a spinor. Would it be OK, if we just drop this section?TR 14:22, 28 April 2011 (UTC)

I don't know, and I would love to know. Here is some vague speculation, that perhaps you can clear up. One can build up tensors from vectors, in that the vectors form a vector space V and the (p, q)-tensors on V are elements of ${\displaystyle T_{q}^{p}:=V\otimes \cdots \otimes V\otimes V^{*}\otimes \cdots \otimes V^{*}}$ (p and q times, respectively). If only rotational coordinate changes are desired, then we consider the action of a special orthogonal group on V. If general linear coordinate changes are desired, then we consider the action of a general linear group. That is, we have some Lie group, and V is a representation of it, and this representation extends to ${\displaystyle T_{q}^{p}}$. So if the Lie group happens to be a spin group, instead of a special orthogonal group, then you consider representations V of the spin group, and these extend to ${\displaystyle T_{q}^{p}}$ and are called spinors? Honestly, does this make any sense? Mgnbar (talk) 13:32, 29 April 2011 (UTC)

Well, you can consider it like this (I think). An ordinary vector bundle, V, can be thought of as carrying a (spin) 1 representation of SO(n). A (p, q)-tensor field on V is an section of ${\displaystyle T_{q}^{p}:=V\otimes \cdots \otimes V\otimes V^{*}\otimes \cdots \otimes V^{*}}$ (p and q times, respectively), therefore carries ${\displaystyle 1\otimes \cdots \otimes 1\otimes {\bar {1}}\otimes \cdots \otimes {\bar {1}}}$ representation of SO(n), which may be viewed as a direct sum of integer spin representations. If the base space allows a spin structure, these representation may be lifted to integer spin representations of Spin(n). You can also introduce a vector bundle S, which carries a spin-1/2 representation of Spin(n). The sections of that bundle are commonly known as spinors. Obviously, you can consider arbitrary tensor product of S and V, which will decompose in various spin representations. For example, V⊗S has irreducible component with spin 3/2. But I'm not sure these are ever called spinors, that term seems to be reserved for the actual spin-1/2 fields.

(The above is written in the context of vector and tensor fields, since that is more familiar, but I think it more or less translates to the context of vector spaces as well.

I'm not sure that this really constitutes a generalization of the concept of a tensor. It seems more like the generalisation of a certain specialisation of the concept of tensor. It seems somewhat far fetched to discuss them in this article.TR 14:54, 29 April 2011 (UTC)

Thanks for your response. When I get the time and energy I will try to absorb it. Cheers. Mgnbar (talk) 17:10, 29 April 2011 (UTC)

In some areas, fields belonging to these tensor product bundles are indeed called spinors. For instance, in relativity theory the "Weyl spinor" is an element of the symmetric tensor product of four copies of an irreducible spin representation of the Lorentz group. As a module for the spin group, this is isomorphic to a certain irreducible tensor representation of the Lorentz group (the Weyl factor appearing in the Ricci decomposition), but the isomorphism is somewhat awkward, and the two objects are generally handled in completely different ways. Sławomir Biały (talk) 20:14, 29 April 2011 (UTC)

Basis and type lost

This webpage states, "When converting a tensor into a matrix one loses two pieces of information: the type of the tensor and the basis in which it was expanded." If this is true in general, and not just of TCC, it should be part of the section on multi-dimensional arrays because it clarifies the difference between the two. ᛭ LokiClock (talk) 05:13, 15 May 2011 (UTC)

I would never have said it that way, but it's a legitimate point that might help some readers, so I'm fine with your adding that. Mgnbar (talk) 14:01, 15 May 2011 (UTC)

Hmm, link doesn't work for me right now. However, the thing is; In the multi-dimensional array approach a tensor is a multi-dimension array satisfying a certain transformation law. The information about the type and the basis in which the tensor is expressed is contained in this last bit. That is, the data for a tensor (in this approach) is: a) a multi-dimensional array b)a transformation law. Without b) it is just a multidimensional array.TR 17:08, 15 May 2011 (UTC)

Right, whereas a matrix is just the multi-dimensional array, without the transformation law. So in converting a tensor to a matrix, you lose the transformation law or the type and basis, however you want to encode that extra information. I think that you two agree. Mgnbar (talk) 20:18, 15 May 2011 (UTC)

But don't we already adequately say that a tensor is a multidimensional array with a transformation law? Sławomir Biały (talk) 11:09, 16 May 2011 (UTC)

Not quite, because it isn't stated what information is contained in the transformation law. Though the transformation law is given importance, the language doesn't indicate that the basis and type are separate information from the matrix, and that the transformation law is what contains that information. As a definition, the information should be given the name "transformation law" right next to the fact that an array doesn't have the information by that name. ⁂ It almost says it here: Taking a coordinate basis or frame of reference and applying the tensor to it results in an organized multidimensional array representing the tensor in that basis, or as it looks from that frame of reference. The coordinate-independence of a tensor then takes the form of a "covariant" transformation law that relates the array computed in one coordinate system to that computed in another one. ⁂ My experience, for debugging purposes: I assumed that a transformation matrix would have its type and basis converted before being applied. I figured that the transformation law was a description of how to alter a transformation matrix to properly operate on another matrix, and saw that the basis and type would be the description required to make such a conversion. I thought basis and type converters were other matrices, so I was prevented from concluding that the transformation laws of the two tensors replaced the converter operator I was expecting. ᛭ LokiClock (talk) 13:44, 24 May 2011 (UTC)

The lead doesn't have a precise definition, and doesn't even discuss tensor type or bases: it's aimed at a much more basic level than that. If you're looking for a definition, you should look in the section prominently labelled "Definition". In that section, tensor type is defined and characterized in terms of the transformation law. Sławomir Biały (talk) 20:28, 24 May 2011 (UTC)

I wasn't talking about adding it to the lede, just showing that the lede almost says what needs to be said. Search for "transformation law" in the article, and you'll see that nowhere comes closer to explicitly stating what structure the transformation law introduces to the array. Instead, "As multidimensional arrays" concludes with, "The definition of a tensor as a multidimensional array satisfying a 'transformation law' traces back to the work of Ricci," after the specifications for the notation and without having previously indicated how what was being talked about is a description of "a multidimensional array satisfying a 'transformation law.'" I have read the article many times, and just before my last reply I payed particular attention to the section on multidimensional arrays, where this information is relevant. ᛭ LokiClock (talk) 17:03, 25 May 2011 (UTC)

At the very end of Tensor#As multidimensional arrays, it says explicitly what the transformation law is, and how this is related to the tensor valence. Sławomir Biały (talk) 17:28, 25 May 2011 (UTC)

Near the end, it says, "The "transformation law" for a rank m tensor with n contravariant indices and m-n covariant indices is thus given as..." The "thus" slips by the fact that you've just established a definition - it doesn't indicate that this location is that of the transformation law's definition, not a previously established object which is useful to derive but not necessary to understand the concept, like when demonstrating the discriminant of a polynomial form. Besides, that paragraph only describes properties of tensors, it doesn't contrast it with the array. There needs to be a constant alternation between the two if you hope to use the array analogy as a self-contained definition. ᛭ LokiClock (talk) 17:50, 25 May 2011 (UTC)

There's an entire paragraph explaining that there are two different kinds of indices and that these transform differently (one by R and the other by the inverse of R). Then it gives the explicit transformation law for a general tensor. I fail to see what is not clear about this. Sławomir Biały (talk) 18:00, 25 May 2011 (UTC)

Those statements are clear on their own, yes. They clearly describe properties of tensors, and the transformation law. The transformation law and those indices are not clearly contrasted with simple matrices. It gives the impression that the difference between a tensor and a matrix should have been clear by the time you start talking about the properties of tensors, and that from then on you should be considering properties of tensors that can be derived from the facts of this relationship. ᛭ LokiClock (talk) 18:05, 25 May 2011 (UTC)

I've added words to the effect that computing the array requires a basis, and attempted a formal definition emphasizing the basis-dependence. Sławomir Biały (talk) 18:32, 25 May 2011 (UTC)

Okay, I tried to add what I'm getting at. ᛭ LokiClock (talk) 20:53, 25 May 2011 (UTC)

Three comments about this "As multidimensional array..." subsection. First, in the definition, can we have the change-of-basis matrix R always on the left of vectors, rather than on the right, for consistency? Second, can we switch the notation (n, m - n) to the simpler notation (n, m)? Third, I feel that the Einstein summation convention (ESC) is a helpful time-saver for experts, but an unnecessary hindrance to beginners. Many people who encounter math have trouble getting past to the notation to the content. I've spoken with students who were under the mistaken impression that the ESC is somehow an essential aspect of tensors. So I vote to avoid ESC wherever possible. Mgnbar (talk) 12:55, 26 May 2011 (UTC)

On the last part. You will need to employ some sort of notation for the sums over indices. The Sigma notation for sums is also a notation convention, and I don't see it as helpful for most readers too use a more cumbersome notational convention. Put differently, how exactly were you planning on writting something like:

${\displaystyle {\hat {T}}_{i_{n+1},\ldots ,i_{m}}^{i_{1},\ldots ,i_{n}}=(R^{-1})_{j_{1}}^{i_{1}}\cdots (R^{-1})_{j_{n}}^{i_{n}}R_{i_{n+1}}^{j_{n+1}}\cdots R_{i_{m}}^{j_{m}}T_{j_{n+1},\ldots ,j_{m}}^{j_{1},\ldots ,j_{n}}.}$

in another notion? TR 13:07, 26 May 2011 (UTC)

The change of basis should be on the right. A basis is naturally identified with an invertible map from R to a vector space. Elements of GL(N) compose on the right with these. Put another way, if you arrange the elements of the basis into a row vector f=(X₁,...,X_N), then multiplication is naturally on the right f R. Sławomir Biały (talk) 13:28, 26 May 2011 (UTC)

Note that since the elements of the basis are indeed vectors, it is conventional to have linear transformations of them act on the left. This convention is used in the rest of the section. A bigger, issue at the moment though is that this "formal definition" cannot be sourced, and hence should not appear at all, IMHO.TR 13:58, 26 May 2011 (UTC)

A basis is not a vector. A change of basis is a numerical matrix (in GL(N)) that acts on the right. A physicist would call this a passive transformation. If you want to act on the vectors in the basis, then you can act on the left with an element of GL(V). A physicist would call this an active transformation; this is not what is meant by "change of basis". The definition should be easy to source. Off the top of my head I can recomment, for instance, Kobayashi and Nomizu or Richard Sharpe's "Differential geometry". I'll try to add references at some point. I don't see that the definition is in any way controversial (except possibly that I need to convince folks that the matrix multiplication really is in the correct order). Sławomir Biały (talk) 14:13, 26 May 2011 (UTC)

You are right. However, only very few of the readers of this article will be comfortable with things acting on the right. Note that for the part where R is written in components the actual order does not matter. Only a few lines above, the R components and basis vectors appear in the opposite order. (On a related note, it would be more consistent to use e_i for the basis vectors as in the text.) Notation wise this should all be made more consistent.TR 14:36, 26 May 2011 (UTC)

About the type (n,m-n), this is a lot more convenient in labelling the indices in the transformation law. If you say the type is (n,m) then the labels in the transformation law become even more complicated.TR 14:57, 26 May 2011 (UTC)

Most of these comments come down to how basic we're willing to get, to help the reader actually learn the thing. Misplaced Pages is not a textbook, but surely we can put in a little more detail, to make the notation less daunting.

Regarding ESC: I would put in one or two more examples of a change-of-basis formula, such as the one for (1, 1)-tensors, with the summation signs intact. For the fully general change-of-basis formula, I would put

${\displaystyle {\hat {T}}_{j_{1}\ldots j_{q}}^{i_{1}\ldots i_{p}}=\sum _{k_{1}=1}^{N}\cdots \sum _{k_{p}=1}^{N}\sum _{\ell _{1}=1}^{N}\cdots \sum _{\ell _{q}=1}^{N}Q_{k_{1}}^{i_{1}}\cdots Q_{k_{p}}^{i_{p}}T_{\ell _{1}\ldots \ell _{q}}^{k_{1}\ldots k_{p}}S_{j_{1}}^{\ell _{1}}\cdots S_{j_{q}}^{\ell _{q}}.}$

(for appropriate ${\displaystyle Q=S^{-1}}$). Yes, this is a monstrosity, but that's the point. It takes a lot of work to change the basis, and this formula shows that. Also, the summation signs are tedious; this motivates the ESC. I would show the ESC version of this formula immediately after it:

${\displaystyle {\hat {T}}_{j_{1}\ldots j_{q}}^{i_{1}\ldots i_{p}}=Q_{k_{1}}^{i_{1}}\cdots Q_{k_{p}}^{i_{p}}T_{\ell _{1}\ldots \ell _{q}}^{k_{1}\ldots k_{p}}S_{j_{1}}^{\ell _{1}}\cdots S_{j_{q}}^{\ell _{q}}.}$

Regarding the left/right action of the change of basis matrix: I understand Slawomir's math, but I disagree with it as pedagogy. At least on the first pass, it's easier for the uninitiated reader to see something like

${\displaystyle \{v_{1},\ldots v_{N}\}\mapsto \{Rv_{1},\ldots ,Rv_{N}\},}$

which shows each vector getting transformed as you would expect. Mgnbar (talk) 15:18, 26 May 2011 (UTC)

Yes, but the change-of-basis transformation is not of the form ${\displaystyle (Rv_{1},\dots ,Rv_{N})}$. That's clearly the sort of thinking that needs to be discouraged. Sławomir Biały (talk) 21:35, 26 May 2011 (UTC)

Regarding type (n, m - n): The transformation law I just posted doesn't seem any more complicated than the one using (n, m - n). Does it seem more complicated to you? Mgnbar (talk) 15:20, 26 May 2011 (UTC)

Sorry, Slawomir, I now realize that you were talking about taking linear sums of vectors, whereas I was working with their coordinate arrays (and sloppily writing R where the article would have R). In other words, you're using a formula like

${\displaystyle \mathbf {\hat {e}} _{i}=\sum _{j}R_{i}^{j}\mathbf {e} _{j}=R_{i}^{j}\mathbf {e} _{j},}$

whereas I'm talking about a formula like

${\displaystyle {\hat {v}}^{i}=(R^{-1})_{j}^{i}v^{j}}$

where v is the coordinate array for any ${\displaystyle \mathbf {e} _{i}}$. Or do I still misunderstand you? If we're going to do tensors as multidimensional arrays, it makes sense to me to treat the basis vectors in the same way as any other vector. Maybe this is what TR was getting at above. Mgnbar (talk) 23:07, 26 May 2011 (UTC)

New illustrations

Do any of the new illustrations have any value at all? The one on dyadic tensors seems a little bit offtopic for understanding the article, and it might be more suitable in the dyadic tensor article (if anywhere). I don't think it adds anything helpful or relevant to the Definitions section, so I have removed it.

The other recently-added graphics are more mystifying. The issues are numerous. First, the index conventions are wrong (what is ${\displaystyle A_{\rho }\mathbf {e} _{\rho }}$ supposed to mean, for instance?) Second, the vector itself should not be changed by a coordinate transformation: in fact, that's the whole point of having a covariant/contravariant transformation law, to ensure that the same vector is described with respect to the two different coordinate systems. Secondly, there appears to be some confusion here between the basis vectors and the ${\displaystyle x^{i}}$ coordinates: one of these should be, roughly, "dual" to the other, so it's not correct to have the basis vectors pointing along the "coordinate axes" in this way as far as I can tell. Sławomir Biały (talk) 01:01, 27 May 2011 (UTC)

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