Talk:Tensor (intrinsic definition)

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Maintainability

I am now trying to make this page more easily maintainable, by replacing Wiki markup + Unicode with TeX markup for the math stuff. This makes this page visible in many more browsers, and should make the page more editable for experts; of which I am not one. Could any mathematicians here proof-read this article, please?

Can I protest (a) about calling things in mathematics 'formalisms' (which is too much like Serge Lang for me) - one might as well say 'unmotivated stuff'; and (b) calling things 'modern' - unlike calling things classical, which is fair enough?

Charles Matthews 09:11, 12 Nov 2003 (UTC)

Charles, what name would be suitable?

Tensor (differential geometry treatment) ?
Tensor (component-free treatment) ?
something else?

-- The Anome 22:44, 12 Nov 2003 (UTC)

Tensor (abstract algebra) is good. The old argument (six decades ago now) was: don't say modern algebra (perishable), say abstract algebra. In fact there is an argument now to go further to tensor (category theory) (as well, not instead of) for monoidal categories.

Charles Matthews 17:55, 13 Nov 2003 (UTC)

Tensor (differential geometry) is better, except that even the component-treatment is differential geometry... I don't quite think abstract algebra covers tensor bundles, differential manifolds, connections, sections, etc... Phys 17:25, 14 Nov 2003 (UTC)

Connections? Keep on topic, please. This is the old argument (for tensors here) that you have to introduce tensor fields at the same time as tensors. Why? There is a page for tensor fields.

Charles Matthews 18:50, 14 Nov 2003 (UTC)

I've looked around a bit - not hard to find ten 'tensor' pages. I think this tells me something a little more positive, namely that this isn't really a name-space problem any more. It's more a question of licking the hypertext into shape. I've not moved any pages up to now, and don't intend to start.

Charles Matthews 20:17, 14 Nov 2003 (UTC)

Wow, this page has mobilized! What is intrinsic definition supposed to mean? aren't tensors in general intrinsic definitions of a space? isn't that what riemannian geometry (classical tensors) is known for: intrinsic definition of curvature? how else would one use the phrase "intrinsic definition" other than refering to the topology of the space or curve? does one mean by this "component-free"? if so, why not just say "component-free" instead of the ambiguous, mysterious, and confusing (because of its allusion to the intrinisic definition of curvature in riemannian geometry (classical tensors)) "intrinsic definition".

regarding "abstract algebra": (and someone mentioned differential geometry) there was discusion earlier in the Talk:Tensor page regarding the topical arrangement of the mathematics page, and where in that tensors should go (why they are on top), and it was concluded that they belong in both the abstract algebra section (for the modern treatment) and the differential geometry section (for the classical treatment), which makes a lot of sense. maybe "abstract algebra approach" would be more fitting, informative, and helpfull, seeing that the approach is based on abstract algebra, comes from the perspective of abstract algebra, and one needs to know abstract algebra in order to learn it.

--Kevin Baas Still suggesting:

Tensor (absract algebra treatment)

-- Kevin Baas 14:41, 24 Feb 2004 (UTC)

This article needs re-hauling. The title is inadequate and suggests confusion with the notion of tensor fields in differential geometry. The header, as others have above suggested, ought to be: "Tensors (algebraic treatment)". While listing the correct properties of a tensor space, the standard explicit construction (in terms of a quotient of some free module) is not provided. Tensor products from the viewpoint of category theory (as covariant functors) should at least be briefly hinted at.

--Anon 14 July 2006.

(First talk post for me; experienced posters please correct me re: posting conventions.) Suggestion: Display a commutative diagram of VxW, Vx_FW, Q, Q', and X in the definition of the tensor product. I find commutative diagrams to be helpful, for both algebra-minded and non-algebra-minded readers. 24.155.243.76 21:30, 20 December 2006 (UTC), Nooj.[reply]

discussion at Wikipedia talk:WikiProject Mathematics/related articles

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)

This article seems to completely miss the point of the algebraic approach to tensors. A tensor is simply a multi-linear map. I think this article presents this idea in a manner that is more complicated than it needs to be.

-- Anon 2 Jan 2007 —Preceding unsigned comment added by 68.197.9.185 (talk) 03:20, 3 January 2008 (UTC)[reply]

Alternate notation

In the section "Alternate notation", what do the wavy approximate equal signs mean? Shouldn't they just be equal signs? —Preceding unsigned comment added by 128.103.54.116 (talk) 17:18, 20 May 2008 (UTC)[reply]

They aren't approximation signs but mean "is naturally isomorphic to", as the article states, hence the link to natural isomorphism. Dependent Variable (talk) 20:19, 24 September 2009 (UTC)[reply]

Tensor products

Is this correct, and if so, what does it mean?

"An element of this tensor product is referred to as a tensor (but this is not the notion of tensor discussed in this article)"?

The link to the "Tensor product of vector spaces" section of Tensor product" leads me to think that perhaps whoever made the link misread it as "This tensor product (i.e. the tensor product of vector spaces) is referred to as a tensor" or that, at the time when the link was made, the sentence actually said something equivalent to "This tensor product is referred to as a tensor", as opposed to "An element of this tensor product ..." As it stands - as far as I can see - the sentence refers to exactly the notion of tensor that the rest of the article goes on to define: "A tensor on the vector space V is then defined to be an element of (i.e., a vector in) a vector space of the form ..." Dependent Variable (talk) 20:43, 17 September 2009 (UTC)[reply]

Rank: Disambiguation

I suggest that Tensor Rank direct to a disambiguation page, and not directly to the subsection, so that inline disambiguation is not necessary and conflicting terminology can be accounted for by an explicitly defining link. Either that, or the definition of the term used on Wikipedia should be standardized. LokiClock (talk) 13:15, 23 January 2010 (UTC)[reply]

Contravariance and covariance

Since this is the second time I've needed to revert the wording of "contravariant" and "covariant" in the definition, I thought I'd include an explanation of my edit here. I think the confusion comes from the fact that one way of defining a type (m,n) tensor T is as the multilinear map

{\begin{matrix}T:&\underbrace {V^{*}\times \dots \times V^{*}} &\times &\underbrace {V\times \dots \times V} &\rightarrow \mathbf {R} .\\&{\text{m copies}}&&{\text{n copies}}&&\end{matrix}}

However, the definition used in this article is, instead, that a type (m,n) tensor is an element of the tensor product of vector spaces

{\begin{matrix}T\in &\underbrace {V\otimes \dots \otimes V} &\otimes &\underbrace {V^{*}\otimes \dots \otimes V^{*}} .\\&{\text{m copies}}&&{\text{n copies}}\end{matrix}}

In either case, though, a type (m,n) tensor is said to be contravariant of order m (denoted with upper indices) and covariant of order n (denoted with lower indices). I hope that clarifies the naming convention.

- Rundquist (talk) —Preceding undated comment added 03:53, 14 July 2011 (UTC).[reply]

This confusion should not occur if wording is corrected: it is the components of the tensor that are correctly called contravariant and covariant in the given number of indices, not the tensor as a whole. I see the wording is still poor in this respect. — Quondum^☏ 21:47, 23 August 2012 (UTC)[reply]

Components may exist without coordinate systems

A local section of the Frame bundle allows for components of a tensor without reference to a coordinate systems, and there are descriptions of General relativity that are written on that basis. Shmuel (Seymour J.) Metz Username:Chatul (talk) 23:09, 12 November 2013 (UTC)[reply]

The section Tensor (intrinsic definition)#Basis does create the incorrect impression that tensor components, and indeed a basis, might only be defined in the case of tangent spaces and then only in relation to coordinates of a smooth manifolds. Such a basis is a holonomic basis, but there are non-holonomic bases even for this case. In the case of a fibre bundle over a manifold, a relating a basis to the manifold coordinates may be impossible. This section deals with bases, components and transforms of components, which belongs in the article Tensor, and not here, anyway. I'm removing the entire section as not belonging here. —Quondum 04:45, 13 November 2013 (UTC)[reply]

Definition via multilinear functions?

Shouldn't there be a section on definitions via multilinear functions, with a wikilink to Tensor#Infinite-dimensional case from both sections? Note that the two definitions are equivalent in the finite dimensional case, or when continuity is not relevant. There is a discussion of such multilinear functions in Tensor (intrinsic definition)#Universal property, although some details are different for the infinite dimensional case. Shmuel (Seymour J.) Metz Username:Chatul (talk) 21:34, 13 November 2013 (UTC)[reply]

Intrinsic definition of tangent and cotangent bundles?

Are the coordinate-free definitions for the Tangent space and Cotangent space within scope of this article? If not, should there at least be wikilinks? The relevant definitions that I'm familiar with are

Differential operators at x
Equivalence classes of curves through x
germs of functions at x
Duals Shmuel (Seymour J.) Metz Username:Chatul (talk) 17:27, 14 November 2013 (UTC)[reply]

IMO (and I might be alone on this), tensors fields on manifolds are not directly within the scope of this article, but since some fields essentially exclusively use tensors on the tangent vector space of a manifold, generally with an associated differential structure, it is sensible to make reference to this concept. A coordinate-free formulation of differential geometry does have merit, though ideally not in this article. The coordinate-free approach to tensor algebra alone should be rich enough for this article. —Quondum 03:52, 15 November 2013 (UTC)[reply]

Lede - needs edit

The lede seems to me to have multiple problems, mostly about the use of language rather than the (non-existent) mathematics. Here is what I think are problems: "In mathematics, the modern component-free approach to the theory of a tensor views a tensor as an abstract object, expressing some definite type of multi-linear concept." Wow. Let me diagram the main clause: [the] approach→views→tensor. Where is ANY explanation of what is intrinsic about this view? Is "the theory of a tensor" correct? or should it be "the (intrinsic) theory of tensors"? Does "definite type..of concept" have ANY meaning? "Multi-linear concept"??! I think this entire sentence is vague to the point of meaninglessness. I challenge anyone to point out a tensor which is NOT an abstract object. This statement fails utterly in distinguishing the component-free approach from more traditional ones. Is it necessary to explicitly state that this is "in mathematics"? What does that mean? Are other disciplines prohibited from using it? While it is a mathematical theory, stating that it is "in" mathematics is problematical, I think. If someone knows what a "multilinear concept" is, they should let us know.

"Their well-known properties can be derived from their definitions, as linear maps or more generally; and the rules for manipulations of tensors arise as an extension of linear algebra to multilinear algebra." Wow, again. Their "definitions"? What definitions is this referring to? Shouldn't there be a clear explanation of why and how there is more than one definition for these objects BEFORE their properties are discussed? Which "mathematical objects" have properties that are NOT derived from their definitions? Why was this (tautology) added? It sheds NO light. Does this mean that their not well-know properties can't be derived from their definitions?? Who wrote this?! The rules "arise from" or is it "can be derived from" ?? So, given a mathematical object (yet to be defined), rules arise for them as linear algebra is extended to multi-linear algebra? What?! They are linear maps? (I thought they are multilinear maps!) What does "as linear maps or more generally" MEAN?!? To me, it means "A or not-A", and is (again) without meaning.

"In differential geometry an intrinsic geometric statement may be described by a tensor field on a manifold, and then doesn't need to make reference to coordinates at all." How does a description of a statement change the statement? If I understand this correctly, it states that a intrinsic geometric statement (which is coordinate free by definition) is somehow able to be described using tensors and there-by making it coordinate-free. What??!!

"The same is true in general relativity, of tensor fields describing a physical property." I submit that since whatever the author is talking about is obfuscated, saying that "the same is true" simply adds obfuscation on top of obfuscation.

"The component-free approach is also used heavily in abstract algebra and homological algebra, where tensors arise naturally." But previously the claim was made that they arose via extending linear algebra to multilinear algebra. Of course, they arise in MANY contexts, but scattering all this around leaves me with no more understanding (after I finished reading the lede) than before I started.

Why not say that this approach to the theory of tensors starts by considering them as mathematical, specifically geometric and algebraic, objects with properties based on a coordinate-free, intrinsic, definition. ? (And then move on from there.) The lede should do much better in answering "what, why and how?", imho.Abitslow (talk) 22:30, 11 May 2014 (UTC)[reply]

Added statement on rank

The statement "The rank of a non-zero order 2 or higher tensor is less than or equal to the product of all but the longest of the lengths of the vectors in (a sum of products of) which the tensor can be expressed." was added in this edit. This seems confusingly stated. Is this saying that if one takes any of the products in the sum, removes the vector of the space of highest dimension, and then multiplies the dimensions of the remaining vector spaces, one gets an upper limit on the rank? Why then restrict it to rank 2 or higher, since the trivial case of the rank of a vector is covered too (in the sense of an empty product)? Why also mention "non-zero", since the upper limit applies in that case too? —Quondum 15:55, 5 June 2015 (UTC)[reply]

I have no idea about the truth value of the statement, but if true, then it is too complicated to be expressed in words. Equation needed. YohanN7 (talk) 14:17, 6 June 2015 (UTC)[reply]

In the context of this article, which deals only with a finite-dimensional vector space and its dual, I think that it is saying that an upper limit on the rank is n^k–1, where n = dim V and k is the order of the tensor. I'd be inclined to trim it to that. Having a reference would be good. It seems to be valid except for k = 0, where I expect the upper limit to be 1 rather than 1/n. —Quondum 16:56, 6 June 2015 (UTC)[reply]

Last isomorphism in Universal property section

You can't just throw in that last isomorphism in the Universal property section with no context, flipping the V to the other side of the semicolon, sending us all to StackExchange for an explanation. — Preceding unsigned comment added by 173.127.192.166 (talk) 00:33, 30 March 2019 (UTC)[reply]

Shortdesc needed

This article could use a shortdesc. Thatsme314 (talk) 13:11, 17 October 2023 (UTC)[reply]

I have created the short description, "Coordinate-free definition of a tensor", after reading the comments, which suggest that "intrinsic" means "coordinate free", by @Kevin Baas and @Abitslow. Please let me know if this description is flawed in any way, as I am new to tensors. Thatsme314 (talk) 20:01, 17 October 2023 (UTC)[reply]

The tensors of order zero are just the scalars (elements of the field F)

The article says, "The tensors of order zero are just the scalars (elements of the field F)". Why is that so? Thatsme314 (talk) 16:13, 17 October 2023 (UTC)[reply]

The definition of tensor product requires separate statements for

n=0

and

n\neq 0

; we define the tensor product of no vector spaces to be the base field, F. https://math.stackexchange.com/questions/1491787/0th-tensor-power-v-otimes-0-bbb-f-definition-or-mathematical-con Thatsme314 (talk) 19:45, 17 October 2023 (UTC)[reply]

Equivalence of the two intrinsic definitions

The article claims equivalence of the two intrinsic definitions: $\underbrace {V\otimes \dots \otimes V} _{m}\otimes \underbrace {V^{*}\otimes \dots \otimes V^{*}} _{n}\cong L^{m+n}(\underbrace {V^{*},\ldots ,V^{*}} _{m},\underbrace {V,\ldots ,V} _{n};F).$

However, this result may not hold when $\dim V$ is infinite, for example, when $m=0$ and $n=1$ . Thatsme314 (talk) 20:16, 18 October 2023 (UTC)[reply]