Talk:Tensor

RFC: is V = V**?

The following discussion is closed. Please do not modify it. Subsequent comments should be made on the appropriate discussion page. No further edits should be made to this discussion.

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Should Tensor state that tensors of rank 1 are vectors, given that $V$ and $V^{**}$ are distinct vector spaces? Standard usage where the author does not wish to distinguish objects with canonical isomorphisms is to state initially that although they are different, he will ignore the differences for the sake of simplicity.

• Bowen, Ray M.; Wang, C.-C.; Ratiu, T. (2010), "Section 32. The Second Dual Space, Canonical Isomorphisms, Chapter 7 TENSOR ALGEBRA", INTRODUCTION TO VECTORS AND TENSORS, Linear and Multilinear Algebra (PDF), vol. Volume 1, p. 214, ISBN 0-306-37508-7, Since the isomorphism $J$ is defined without using any structure in addition to the vector space structure on $V$, its notation can often be suppressed without any ambiguity. We shall adopt such a convention here and identify any $v\in V$ as a linear function on $V^{*}$. {{cite book}}: |volume= has extra text (help); Cite has empty unknown parameters: |laydate=, |laysummary=, |trans_title=, |month=, |trans_chapter=, |chapterurl=, and |lastauthoramp= (help); Math stripmarker in |quote= at position 23 (help)

Shmuel (Seymour J.) Metz Username:Chatul (talk) 20:32, 12 April 2015 (UTC)

• Comment. This RfC is not a neutral summary of the dispute in question. Here, as I see it, is a summary of the dispute. A few weeks ago, Chatul committed this edit, stating that vectors and scalars are not tensors. This edit was reverted by User:Andy Dingley, with the edit summary "No, they really are. (first rank and zero rank tensors)" a sentiment that I (and most sources) agree with. Chatul reverted Dingley, with the edit summary "No, they are not the same - see talk page". Then, on the talk page, we see: "Scalars and vectors are distinct from tensors. While it is usually convenient to ignore the difference, scalars and vectors are not tensors, especially if you use the definition as a tensor as a multi-linear map." Dingley then restored the content, with a reference that very clearly and directly supported the statement. Chatul placed a tag on the article. I removed the tag. (Editors placing tags like these are expected, at a minimum, to come up with sources that clearly and directly support their point of view (particularly in the light of countervailing sources. These aren't to be used for something like "there is a point of view reflected in the article that I personally disagree with".)

Chatul was asked for a reference that supports his perspective. Instead of providing such a source, Chatul argued above that in mathematics generally, it is important not to identify objects that are canonically isomorphic. He used the example of homology and cohomology theories in algebraic topology that, in good cases, can sometimes be identified. I rebutted that this was a red herring, since whether homology or cohomology theories can be identified has nothing to do with the subject of this article.

Following that, I gave a long list of references that supported that the identification of a vector space and its double dual is indeed standard throughout the subject. For ease of reference, here are the statements as they appear in various books on the subject:

• Bowen and Wang "Introduction to vectors and tensors" p. 218 "$T^{1}(V)=V,\quad T_{1}(V)=V^{*}$ Here we have made use of the identification of V with V** as explained in Section 32."
• Borisenko and Taparov "Vector and tensor analysis with applications", p. 61: "2.3. First-order tensors (vectors)"
• Kobayashi and Nomizu, "Foundations of differential geometry, volume 1", p. 20: "$T^{1}$ is nothing but V"
• Lee, "Introduction to smooth manifolds", p. 180: "Clearly there are natural identifications $T^{0}M=T_{0}M=M\times R,T^{1}M=T^{*}M,T_{1}M=TM$.
• J Schouten, "Tensor analysis for physicists", p. 17: "p=0, q=0 gives a scalar; p=1, q=0 a contravariant and p=0, q=1 a covariant vector."
• L. P. Eisenhart, "An introduction to differential geometry with use of tensor calculus", p. 89: "A contravariant vector is a contravariant tensor of the first order."
• Marsden and Ratiu, "Manifolds, Tensor Analysis and Applications", p. 340: "$T_{0}^{1}(E)=E$... and make the convention $T_{0}^{0}(E;F)=F$"
• Arfken and Weber, "Mathematical methods for physicists", p. 131: "A scalar is specified by one real number and is a tensor of rank zero. ... In three dimensional space, a vector ... is a tensor of rank one."
• Hawking and Ellis, "The Large-scale structure of space-time", p. 18: "In particular, $T_{0}^{1}(p)=T_{p}$"

Now, to the crux of the matter, even if, in some models, it is true that a vector space and its double dual are not identified, overwhelmingly they are in practice. Even the book that Chatul cites does not directly support his views here: "We shall adopt such a convention here and identify any $v\in V$ as a linear function on $V^{*}$." This is the convention throughout linear algebra, mathematical physics, differential geometry, etc. There is no need to say that "vectors are not tensors". That just conveys a view that is unnecessarily confusing to the average reader, and not in agreement with standard practices in the literature. Anyone for whom such a statement is likely to be meaningful is presumably also clever enough to read the discussion of such nuances in the text of the article, especially the footnote that reads: "The double duality isomorphism, for instance, is used to identify V with the double dual space V^**, which consists of multilinear forms of degree one on V^*. It is typical in linear algebra to identify spaces that are naturally isomorphic, treating them as the same space." But for the average reader, such minutiae are extremely unimportant marginalia (as evidence by the distinct lack of discussion in the literature). Sławomir Biały (talk) 21:41, 12 April 2015 (UTC)

• Opinion – In this article, it is sufficient to make the reader aware of any such distinction in a footnote. Text highlighting the distinction between V and V^∗∗ should not occur in the lead, since it is distracting and unnecessarily confusing for the average reader of this article. Any discussion of the matter belongs elsewhere, e.g in Dual space. —Quondum 00:23, 13 April 2015 (UTC)

To clarify, are you saying that this sentence in the lead should have a footnote but not be reworded?

A vector can be represented as a 1-dimensional array and is a 1st-order tensor. Scalars are single numbers and are thus 0th-order tensors.

My preference would be to change is to can be treated as, but a footnote certainly satisfies my objection. Shmuel (Seymour J.) Metz Username:Chatul (talk) 15:31, 13 April 2015 (UTC)

I would not be happy with the footnote as you've given here. Firstly, the distinction between V and V^∗∗ is not the same thing, because if one is making this sort of distinction, "elements of V" and "vectors" are not the same thing, and "scalars" in the context of tensors is not the same thing as "the underlying field of V". Secondly, all that this article should do, even in a footnote, is to say that the tensors of order 0 T⁰(V) are referred to as "scalars", and the tensors of order 1 T¹(V) and referred to as "vectors". It could also mention that there is a canonical embedding of the underlying field and of V respectively that identifies them with these tensors. I think that perhaps this hinges around the precise way in which the words "scalar" and "vector" are to be interpreted, and I'm suggesting a particular widely understood meaning in this context. —Quondum 16:29, 13 April 2015 (UTC)

That might be reasonable for Tensors#Using tensor products, but with Tensors#As multilinear maps the elements of $V^{**}$ are tensors of order 1 distinct from the elements of $V$. Changing the text to treated as or adding a footnote would resolve the issue.

Shmuel (Seymour J.) Metz Username:Chatul (talk) 21:28, 14 April 2015 (UTC)

There is already a footnote in the article that explains this. See above, where I quote the footnote at length. Sławomir Biały (talk) 21:42, 14 April 2015 (UTC)

I'm sorry, I seem to have confused myself above. The objection in previous comment seems to have been in reaction to something that I misread or read over-hastily; I'm having difficulty understanding what I wrote.

Starting again: I see no benefit to the rewording from "is" to "can be treated as", since implies a particular definition of the word "vector", which has not been pinned down that precisely, and in the context I do not see any benefit in doing so. Even using the definition as a linear map, we run into problems without assuming the "is", such as that vectors are identifiable with, but are different objects, when regarded as (a) maps from scalars to vectors, (b) maps from covectors to scalars or (c) objects mapped by a covector onto scalars. Add all the variations for tensors as linear maps (according to what they are acting on), and you end up with a real mess. So it is simplest to define tensors as elements of spaces that map to linear maps, rather than as being linear maps. For example, a tensor product is not in general a linear map in this sense at all, but there is a canonical isomorphism. So, if one wishes to approach this in a rigorously consistent way, it is far more complicated than some wording or a footnote change. —Quondum 22:24, 14 April 2015 (UTC)

I'm trying to adhere to Wikipedia:Neutral point of view by crafting text that acknowledges the multiple definitions. I can live with any text that differentiates between rigorous usage and pragmatic usages. The issue is not which issue to use, but rather whether to briely acknowledge the pragmatic status of the identifications at issue. Shmuel (Seymour J.) Metz Username:Chatul (talk) 16:28, 28 May 2015 (UTC)

Wikipedia needs an article about tensors over finite-dimensional vector spaces. In the future that might be something like "Tensor (finite dimensional case)", but currently this article is it.

For finite-dimensional vector spaces this identification V = V** is canonical. As I understand it, the preponderance of references make use of this identification without a lot of discussion. So this article should mention, but not dwell upon or emphasize, the identification. A footnote is adequate. That said, the current text that references the V = V** footnote should be improved. Mgnbar (talk) 00:10, 15 April 2015 (UTC)

The primary topic is this article. An approach more compliant with our guidelines would be to have an article tensors in infinite dimensions, rather than disambiguate a title that is not actually ambiguous. There are sections of this article that address tensors in infinite dimensions that point to relevant subtopics on Wikipedia. Sławomir Biały (talk) 11:58, 15 April 2015 (UTC)

I was talking about this article. My point was that this article is primarily about tensors over finite-dimensional vector spaces (despite its name, which might suggest something more general to some editors). Therefore this article should use the V = V** identity liberally. I think that you and I agree. Mgnbar (talk) 13:57, 15 April 2015 (UTC)

I too am in agreement. I like Mgnbar's characterization. Making clear in the article (in footnotes or otherwise) what its scope is and necessary identifications seem like a good idea. —Quondum 14:05, 15 April 2015 (UTC)

The question is: should the lead be written for users who don't know what tensors are and are hoping to find out, or for users who already know and are hoping to read something pedantically correct? As a user closer to the first type, I agree with Mgnbar. Maproom (talk) 06:48, 20 April 2015 (UTC)

• Opinion- The bot summoned me. I agree with User:Sławomir Biały regarding the article title, i.e. don't disambiguate an unambiguous title. Remember, this is about Wikipedia naming conventions. Your frame of reference is within-Wikipedia, and Wikipedia readers are your audience. Next, after reading about half of the entire article talk page, I came to the conclusion that other editors and IP commenting readers are clamoring for a basic, not a nuanced, description of tensors. That is why it I also agree with Slawomir Bialy regarding the identification of finite-dimensional vector spaces as V = V** AND with what Quondum said here:

In this article, it is sufficient to make the reader aware of any such distinction in a footnote. Text highlighting the distinction between V and V^∗∗ should not occur in the lead, since it is distracting and unnecessarily confusing for the average reader of this article. Any discussion of the matter belongs elsewhere, e.g in Dual space.

Finally, I would not agree with the prior proposed rewording from "is" to "can be treated as", as it will more likely confuse rather increase understanding of the intended audience for this Wikipedia article. Vectors are tensors in finite dimensional space, which is what this article is about. I'm sorry that my wording has such a brusque tone. That isn't my intent (I'm in a hurry).--FeralOink (talk) 01:37, 25 April 2015 (UTC)

• Opinion- The bot summoned me too and I substantially agree with FeralOink. I am however no mathematician and my view of this matter is partly different.

Firstly, the lede is far too extensive. I realise that it is popular to say that the lede should have such and such a ratio to the size of the article, but that is hooey; the point of the lede is to enable the reader to tell within a short time whether he might expect to find it rewarding to read on. If his reaction is "Huuuhhh???" he knows not to, and the same is true if he says "Oh that!", or even just "Oh." No time lost; the purpose of the lede has been achieved economically and neatly. The body of the article is intended for readers who reacted to the lede with any variation on "Oh really? Well then, let's see...!" or "Hmmm... yes. Then how did that relate to..." or even "Rubbish, how could that be?" To clutter the lede with details and arguments that belong in the body sections or that confuse the issue by duplicating or contradicting body text and demanding maintenance in step with the body, have no place in the lede. In practice that means that most ledes, decidedly including this one, are far too cumbersome and are obstacles to readers rather than aids. IMO the current first paragraph, perhaps plus the first sentence of the second paragraph, with a bit of editing, would be quite adequate for a lede. The rest of the material in the present lede could be dispersed through the body text or put into an introductory section.

Compressing the lede seems reasonable. Shmuel (Seymour J.) Metz Username:Chatul (talk) 16:28, 28 May 2015 (UTC)

As for the question of whether a tensor is a vector or not, that is partly a matter of semantics and partly a matter of empirical logic. It has precious little to do with the topic and nothing to do with the lede. Semantically, if a mathematician wishes to regard a given string of values as a (first-order) tensor, that might suit his personal semantic purposes because of the contexts in which he uses the string, but logically it does not put him into any coordinate set calculated to forbid another practitioner to call it a vector. If I encounter (3,1,4) or indeed (a, 1,c) and say "Ah, a vector! It might represent say, a (class of) 3D position," then who is to forbid me to use it to represent a position, and on what basis? The basis that someone else wants to use it to represent a direction? I don't THINK so! And the fact that the same string represents certain classes of order-1 tensor similarly does not disqualify it from being a vector, or FTM vice versa.

Now, all this about footnotes etc is hardly better. If there indeed are certain contexts in which it is important to remember in working with order-1 tensors, that certain classes of vectorial operations would not be valid, then that needs to be said explicitly and in the place in the text where it is relevant. A footnote is for saying things that don't belong in the text and as such is to be treated as a last resort, not as a lazy substitute for a parenthetical remark. Most footnotes simply need to be properly put into context in the text and clearly stated. Many footnotes are hangovers from conventions suited to hard-copy books and have no functional place in soft copy. Nothing in what the constraints on operations on tensors might be has anything to do with what else a given string of values might be used for, whether those happen to be numbers, symbols, names, or any other positional values meaningful in context, such as in a state vector, and hence the constraints on what may be (called) a tensor do not put any constraints on whether it may be a vector. To argue the contrary would make as much sense as saying that an elongated wooden object taken from a tree is not a stick because I have chosen to use it as a pointer. JonRichfield (talk) 07:44, 28 April 2015 (UTC)

"the point of the lede is to enable the reader to tell within a short time whether he might expect to find it rewarding to read on." Wrong. Please see WP:LEAD. "In practice that means that most ledes, decidedly including this one, are far too cumbersome and are obstacles to readers rather than aids..." Well, many leads follow our guidelines. You seem to be saying that the guidelines are wrong, and should be changed. You're entitled to such a view, of course, but until you have successfully changed the guidelines, this article also should adhere to them. Sławomir Biały (talk) 11:20, 28 April 2015 (UTC)

@User:Sławomir Biały wrong again. I said and implied nothing of the kind. Guidelines are for when one needs guidance; they are not prescriptions for how to serve readers' needs. That is reflected in this very guideline: The lead should be able to stand alone as a concise overview. It should define the topic, establish context, explain why the topic is notable, and summarize the most important points... The lead is the first part of the article most people read, and many only read the lead... it should ideally contain no more than four well-composed paragraphs... (my emphasis)
Too many editors seem to think that means that you need four paragraphs (after all, it is the guideline; you can't argue with that...!) even if you have to cram ten topics into the four paragraphs though that means four full pages, or split one topic into four if one topic is all you can think of to say. If an editor cannot tell what "well-composed" means, perhaps an appeal for assistance might help. Similarly if one cannot tell what would be most valuable to the readers who ...only read the lead... a little help from one's friends might be in order.
Any user who wants to read more will come to no harm in reading past the end of a lede sufficient for such purposes, if he finds himself in a full introductory section that goes far beyond what a casually uninterested reader could or should want to follow. Many a lede could adequately fit into a single clear sentence. Not many should need more than half a screenful.
I agree with you that "this article also should adhere to" the guidelines; anyone reading it may see that although it has four carefully-titrated paragraphs, the current lede does not so adhere; it certainly violates the functional intentions. JonRichfield (talk) 02:44, 2 May 2015 (UTC)

Nonsense. The lead of this article is four well composed paragraphs. They define the subject for a nonspecialist and summarize its most important aspects. That's exactly what a lead is meant to do per our guidelines. Sławomir Biały (talk) 13:21, 2 May 2015 (UTC)

• Yet another Opinion A footnote in a lede is a gesture of futility. The very notion of a footnote is questionable in a soft-copy medium; it is an inconvenience to the reader at best and it is hard to think of an example even outside a lede where it is helpful, let alone necessary. If it is a short footnote it belongs in a simple competently written parenthetical remark. If it is a relevant and long footnote that would be inappropriate to include in context, it should be the target of a link. For example consider: "Vectors and scalars themselves are also tensors.[Note 1]". In a lede that is counter-functional; it adds nothing to the reader's understanding and breaks the coherence of the text. To add insult, the added footnote is less than a line and wouldn't even have been out of place in the mutilated statement with the footnote. If (as seems reasonable in context) the author feels that the material is relevant, it could better read something like: Tensors may be regarded as a generalisation of vectors, much as vectors may be regarded as a generalisation of scalars; in suitable contexts one may regard a scalar as a zero rank tensor, and a vector as a first rank tensor. A reader who understands that is not inconvenienced and can see what the intention of the terminology is, and a reader who does not yet understand and wishes to know, now has the links accessible that can clarify technical points obscure to him. Or he can press the information overload button and go somewhere else. After all, not everyone who reads an encyclopaedia and does not know what a tensor (or indeed a vector) is, is equipped to know it. If you don't realise that, you have been living too sheltered an academic life.
  Furthermore, I have had a look at all the footnotes in the article, and on similar grounds, not a solitary one is justified, in or out of the lede. Please re-think! Rattle my cage if you would like participation. JonRichfield (talk) 03:28, 2 May 2015 (UTC)

Your suggested wording would satisfy my objections. Shmuel (Seymour J.) Metz Username:Chatul (talk) 16:28, 28 May 2015 (UTC)

Here is not the place to discuss the merits of footnotes, but briefly, they fill the same function as in hard media. I agree that the content of some of the footnotes should be inline. To be precise, notes 1-4 could well be inline while 5-8 could remain as footnotes. (Part of their purpose, by the way, is to relieve the reader from information overload.)— Preceding unsigned comment added by YohanN7 (talk • contribs) 06:08, 2 May 2015

The discussion above is closed. Please do not modify it. Subsequent comments should be made on the appropriate discussion page. No further edits should be made to this discussion.

Ambiguous terminology: Rank and Order

Rank and Order are the same thing. The article doesn't make clear this and it's confusing. — Preceding unsigned comment added by 190.24.189.75 (talk) 12:55, 29 May 2015 (UTC)

From the article:

The total number of indices required to uniquely select each component is equal to the dimension of the array, and is called the order, degree or rank of the tensor. Footnote: This article uses the term order, since the term rank has a different meaning in the context of matrices and tensors.

--Sławomir Biały (talk) 12:58, 29 May 2015 (UTC)

Also from the article:

There is a plethora of different terminology for this around. The terms "order", "type", "rank", "valence", and "degree" are in use for the same concept. This article uses the term "order" or "total order" for the total dimension of the array (or its generalisation in other definitions) m in the preceding example, and the term "type" for the pair giving the number contravariant and covariant indices. A tensor of type (n, m − n) will also be referred to as a "(n, m − n)" tensor for short.

--Sławomir Biały (talk) 13:01, 29 May 2015 (UTC)

Intoductory definition

"Tensors are geometric objects that describe linear relations between geometric vectors, scalars, and other tensors."

I am not a mathematician, so may be missing some nuance here, but this definition seems to be an infinite loop and says, in effect:

"Tensors are geometric objects that describe linear relations between geometric vectors, scalars, and other geometric objects that describe linear relations between geometric vectors, scalars, and other geometric objects that describe linear relations between geometric vectors, scalars, and other..." etc. ad infinitum.

If it's my lack of mathematics that is the problem, could this be re-written so that I and other non-mathematicians can understand it? Hundovir (talk) 17:19, 11 October 2015 (UTC)

The first sentence is not intended as a formal definition. Several different definitions of tensors appear in the definition section of the article. Sławomir
Biały 17:26, 11 October 2015 (UTC)

This is a frequent complaint. The opening sentence for this article is difficult to write. For what it's worth, the easily-overlooked word "other" is crucial to avoiding the infinite loop. And the core content of the sentence is the easily-overlooked word "linear". Mgnbar (talk) 18:06, 11 October 2015 (UTC)

I am not sure why you think that such and infinite loop would be problematic. Linear maps between space of tensors are again tensors. It seems to me that you (as a self-described non-mathematician) understood the sentence just fine. TR 15:43, 12 October 2015 (UTC)

it is problematic because you can't define a word with the word itself. It doesn't matter of the word "other" is added or not. Watch: a cube is a shape that is similar to other cube-like objects. ... Not really helpful. — Preceding unsigned comment added by 74.197.144.134 (talk) 17:06, 28 December 2015 (UTC)

Who is defining the word in terms of itself? The first sentence is not the definition of "tensor". Various actual definitions appear in the section of the article labelled "Definition". Sławomir
Biały 17:41, 28 December 2015 (UTC)

ok, since people are taking the '#%*× you' approach to any critisims of this let me ask who is this written for? The "definition" seems to be written for someone that knows what it is already. When you go to an article like this and people get brushed off for commenting that it is written assuming a degree in the subject.. It seems like a wiki-circle jerk instead of anything remotly helpful. — Preceding unsigned comment added by 74.197.144.134 (talk) 15:06, 29 December 2015 (UTC)

Your criticism was that the first sentence was a circular definition. I was just pointing out that it is not a definition. If you want to read any of the three (!) definitions in the article, they are there for you. If you don't want to read them, that's fine too. But then your criticism is that: "The definition is too hard for me too understand, and the first sentence is not a definition." Yes, we already know this.

As for the intended readership, unfortunately, this is an encyclopedia article, not a tutorial. If you want a tutorial on the subject, there are textbooks on linear algebra where you can learn the basics of vectors and transformations. Then you can grab a book on tensor analysis. It is quite typical that mathematical definitions require some background in order to be fully appreciated. Sławomir
Biały 15:16, 29 December 2015 (UTC)

74.197.144.134, the editors here do try to write clear, understandable articles. And we appreciate constructive feedback from readers. If we could write a brief introduction that satisfied rigor, motivation, and applications, then we would. But we can't. So instead we have separate sections for those. Many Wikipedia articles on advanced math topics are like this. To learn more you might read the policies Wikipedia:Wikipedia is not a textbook and Wikipedia:Manual of Style/Lead section. Regards -- Mgnbar (talk) 22:08, 29 December 2015 (UTC)

Suggestion: Transpose the table in Examples section

For those of us familiar with linear algebra, and trying to understand more general tensors, paired indices (n,m) always index first the rows and then the columns of a matrix. When viewing matrices as operators / linear functions, one reads from right to left as one does with functions and sees m inputs and n outputs. Putting these together, the columns are always associated with the input and the rows associated with the output.

If I'm understanding the type of a tensor correctly, a (0,2) tensor (like an inner product) takes in an input with 2 indices (like 2 or more vectors, or one matrix) and gives an output with 0 indices (a scalar). This is compatible with reading from right to left, going from input to output as is should be. But this table is laid out as the transpose to such (ingrained) perspectives. When reading the table, the right most index (m) iterates over the rows, and the left-most index (n) iterates over the rows. When looking for a (0,2) tensor, I would look for it in row 0 and column 2. Unless there is some strict convention where this table has always been laid out this way (in the literature, in texts), which I doubt there is, I suggest it be transposed. --Yoda of Borg (✉) 16:34, 15 November 2015 (UTC)

I was the editor who added the (first version of the) table long ago. And I can't recall any clear rationale for having it in that orientation. So I just transposed it for you. Please check it for errors. Mgnbar (talk) 17:55, 15 November 2015 (UTC)

Matrix actions

In the section "As multidimensional arrays", numerical matrices act on the right of lists of basis vectors, because we tend to think of abstract vectors as "columns". Thus, to change the basis $[\mathbf {e} _{1}\ \mathbf {e} _{2}\ \cdots \mathbf {e} _{n}]$, we must multiply this on the right by a numerical matrix. This is consistent with viewing a basis as an isomorphism from $\mathbb {R} ^{n}$ to the vector space, with the action of numerical matrices given by right-composing with an element of GL(n). It is also consistent with the description given an covariance and contravariance of vectors. tl;dr, the correct order for the change of basis action on a linear transformation is ${\hat {T}}=R^{-1}TR$. Sławomir
Biały 11:14, 3 March 2016 (UTC)

It was requested that I give a reference for writing the change of basis in this way, with the GL(n) action on the right instead of the left. I refer, for instance, to the books "Lectures on differential geometry" by Shlomo Sternberg, or Kobayashi and Nomizu, "Foundations of differential geometry". Sławomir
Biały 18:41, 3 March 2016 (UTC)

• According to my book, for another frame field $S'$, the change of basis is

$S'=A\cdot S$,

where

$A={\begin{bmatrix}a_{1}^{1}&\cdots &a_{1}^{q}\\\vdots &&\vdots \\a_{q}^{1}&\cdots &a_{q}^{q}\end{bmatrix}},\quad S={\begin{bmatrix}s_{1}\\\vdots \\s_{q}\end{bmatrix}},$

$\{s_{i}\}$ are vectors, ie. a frame field. Under this setting, for a (1,1)-Tensor $T$, after change of basis, we get a new (1,1)-tensor ${\hat {T}}$, if we write in a matrix form, we should have

${\hat {T}}=A\cdot T\cdot A^{-1}.$

The is in chapter 4 of the book "Lectures on Differential Geometry" by Shiing-Shen Chern. In his book, pp108, he also writes clearly, for a curvature matrix $\Omega =d\omega -\omega \wedge \omega$, there is

$\Omega '=A\cdot \Omega \cdot A^{-1}.$

And $\Omega (X\wedge Y)=D_{X}D_{Y}-D_{Y}D_{X}-D_{[X,Y]}$ is a well-known as a (1,1)-tensor.