Talk:Tensor product: Difference between revisions

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Ack! Too hard!!! My brain hurts! Can someone rewrite this so that a mere physicist like myself can understand it? -- [[User:Tim Starling|Tim Starling]] 04:15 Mar 4, 2003 (UTC)

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== inconsistent font usage ==

hmm... there seems to be a section under [[matrix multiplication]] on Kronecker product/direct product, which is the same thing as tensor product as far as i know, with a slightly different definition.(the rank is neglected) I know that my definition is correct. (the rank is relevant). The question is, how do we remedy this duplicity? [[User:Kevin_baas|Kevin Baas]] 2003.03.14

I made the fonts consistent in this edit https://en.wikipedia.org/w/index.php?title=Tensor_product&type=revision&diff=1111854170&oldid=1111816282 now undone by an editor of the article.

How's that for a start? See people, like this. Why are mathematicians bad prose writers? Sigh... I don't mean to invade on this. It just needs to be written clearly. [[User:Kevin_baas|Kevin Baas]] 2003.03.14

My reason was that the inconsistent font uses that have been restored by that undoing editor of the page are confusing. The weak point is where the Latin lower case letter vee in the undoing editor's inconsistently preferred math markup <nowiki>{{math|·}}</nowiki> looks like the Greek lower case letter nu in fonts that are clearer, such as that of <nowiki><math>·</math></nowiki>. This is evident in this snip from the article: "An element of the form <math>v \otimes w</math> is called the '''tensor product''' of {{mvar|v}} and {{mvar|w}}." The problem is that the two different math markup formats, <nowiki><math>·</math></nowiki> and <nowiki>{{math|·}}</nowiki>, use different fonts. It is true that, often enough, the font usage that is inconsistently preferred by the undoing editor is slightly quicker to type. My experience is that the LaTeX format <nowiki><math>·</math></nowiki> is to be preferred as it often is in the article. [[User:Chjoaygame|Chjoaygame]] ([[User talk:Chjoaygame|talk]]) 11:22, 23 September 2022 (UTC)

:Thank you, yes it's certainly a good start. I like the split format ("pretentious part starts here") - it's ugly enough to encourage contributors to properly integrate the article.

== Definition ==

:As for duplicity, personally I like repetition in Wikipedia. After all, there's essentially no space limit and a huge pool of contributors, so you may as well give the reader the most specific, tailor made information possible. I think it would be enough to explain the connection in both articles, with links of course. Since [[matrix multiplication]] is a more general subject, the Kronecker product section should be kept brief, perhaps indicating that more information is to be found here. See [[meta:Consolidating v/s breaking up]]. -- [[User:Tim Starling|Tim Starling]] 06:57 Mar 15, 2003 (UTC)

In the definition from bases, it says:

----

The statement about the [[rank]] can't be possibly true:

:<math>\begin{bmatrix}a\end{bmatrix} \otimes \begin{bmatrix}b\end{bmatrix} = \begin{bmatrix}ab\end{bmatrix}</math>

or still worse:

:<math>\begin{bmatrix}0\end{bmatrix} \otimes \begin{bmatrix}b\end{bmatrix} = \begin{bmatrix}0\end{bmatrix}</math>

Also, I think the example should be given as a tensor product of matrices from which the vector case can be derived easily. -- [[User:Looxix|looxix]] 14:12 Mar 15, 2003 (UTC)

The ''tensor product'' <math>V \otimes W</math> of {{mvar|V}} and {{mvar|W}} is a vector space which has as a basis the set of all <math>v\otimes w</math> with <math>v\in B_V</math> and <math>w \in B_W.</math>

---

Out of the blue appears the tensorproduct <math>v\otimes w</math>, just the thing that should be defined. [[User:Madyno|Madyno]] ([[User talk:Madyno|talk]]) 12:33, 25 November 2022 (UTC)

Maybe a tensor with 1 dimension has undefined rank? this is no disproof. i have very authoritave sources that verify the fact that the ranks are summed by a tensor product. can you tell me whether [0] has rank 1 or 2, or even 7? it's a 1x1x1x1x1x1.... tensor. But isn't 1 also <math>e^{i*2\pi}</math>?

:One usage of the symbol <math>\otimes</math> is for vector spaces; the other is for vectors. For basis vectors <math> v\in B_V </math> and <math> w\in B_W </math> the expression <math>v\otimes w</math>, which is a basis vector for <math> V\otimes W </math>, should be thought of as a primitive symbol that doesn't reduce to anything simpler. Elements of <math> V\otimes W </math> are sums of such symbols with numerical coefficients. You can of course take the tensor product of non-basis vectors—the result will be written in terms of tensor products of basis vectors, as described in the paragraphs following the one containing the passage you quoted. I think it would be helpful to many readers if some elementary examples could be added to the article showing how the notation works and what it is useful for. [[User:Will Orrick|Will Orrick]] ([[User talk:Will Orrick|talk]]) 03:48, 26 November 2022 (UTC)

The 'rank' of a tensor is actually not defined by the columns, rows, ect., but by an equation such as:

== Tensor product of n vector spaces for n>2 ==

<math>\bar{T}^{ij} = T^{rs}\frac{\partial \bar{x}^i}{\partial x^r}\frac{\partial \bar{x}^j}{\partial x^s}</math>

The article defines tensor products of a finite nonnegative integer <math>n</math> vector spaces for only the case <math>n=2</math>. However, the "General tensors" section uses tensor products for general <math>n</math>, so we need to define somewhere tensor products in general (finite nonnegative integer). [[User:Thatsme314|Thatsme314]] ([[User talk:Thatsme314|talk]]) 19:31, 17 October 2023 (UTC)

where the rank of such ''T'' is 2 because it has order 2, regardless of how many <math>x^i</math>'s or <math>x^j</math>'s there are.

:The case ''n'' > 2 is defined in section {{alink|Associativity}}. [[User:D.Lazard|D.Lazard]] ([[User talk:D.Lazard|talk]]) 20:48, 17 October 2023 (UTC)

[[user:kevin_baas|kevin]] -2003.03.15

::Good catch! Now, we just need to define it for <math>n=0</math>. Also, it might be nice to re-organize the article to emphasize the general definition. [[User:Thatsme314|Thatsme314]] ([[User talk:Thatsme314|talk]]) 01:00, 18 October 2023 (UTC)

== Linearly disjoint subsection ==

:Nothing is wrong if the rank is the [[rank_(tensor)]], but it is linked to [[Rank of a matrix]], which is (loosely speaking) the degree of linear dependency of a matrix and was what I talked about in my remark. -- [[User:Looxix|looxix]] 17:25 Mar 15, 2003 (UTC)

The [[Tensor product#Linearly disjoint]] section is for no apparent reason specialising to the case of complex vector spaces, when all sibling sections are over a general field <math>F</math>, and this definition likewise should work for a general field as well.

::That's my fault -- Kevin linked to [[rank]] and I disambiguated. I haven't studied tensors before, so I wouldn't know one from the other. -- [[User:Tim Starling|Tim Starling]] 23:02 Mar 15, 2003 (UTC)

The meaning of <math>\mathbb{C}^{mn}</math> needs to be clarified. If the intent is <math> m \times n </math> matrices (the formula given for <math>T</math> would suggest so), then <math>\mathbb{C}^{m \times n}</math> is better. <math>\mathbb{C}^{mn}</math> is rather vectors with <math>mn</math> elements, which would also work, but then you need to explain how you map the indices — straightforward if you've already grasped the concept of tensor product, but far from straightforward for those who don't. [[Special:Contributions/130.243.94.123|130.243.94.123]] ([[User talk:130.243.94.123|talk]]) 16:47, 16 April 2025 (UTC)

:::i fixed that. looxix, i don't understand what you mean by "giving the example as a tensor prodcut of matrices from which the vector result can be derived easily" - how would i express a tensor with rank>2 without using embedded matrices, which would be potentially confusing? -- [[User:Kevin_baas|Kevin Baas]] -2003.03.15

:Since {{tmath|1=m\times n=mn}}, your suggestion does not changes anything. However, the phrasing was awful, and I have inproved it. Note that your post suggests that the basis elements of {{tmath|\C^{mn} }} are necessarily indexed by the first {{tmath|mn}} natural numbers. After my edit, it becomes clear that this is not the case here [[User:D.Lazard|D.Lazard]] ([[User talk:D.Lazard|talk]]) 17:54, 16 April 2025 (UTC)

:I agree that the section must be rewritten over a general field {{tmath|F}}. Indeed, the only usage of linear disjunction that I have ever encountered is the case of two finite [[field extension]]s of the rational numbers or of an [[algebraic number field]]. I have never heard of linear disjunction over {{tmath|\C}}. [[User:D.Lazard|D.Lazard]] ([[User talk:D.Lazard|talk]]) 20:44, 16 April 2025 (UTC)

== Tensor products of modules ==

The [[Tensor product#Tensor products of modules over a ring]] section is doing the non-commutative case twice: once starting with “More generally …”, and then (immediately after) again as the subsection “Tensor product of modules over a non-commutative ring”. [[Special:Contributions/130.243.94.123|130.243.94.123]] ([[User talk:130.243.94.123|talk]]) 16:53, 16 April 2025 (UTC)

:I agree to either merge the two sections, or remove the noncommutative case from the first subsection. [[User:D.Lazard|D.Lazard]] ([[User talk:D.Lazard|talk]]) 17:13, 16 April 2025 (UTC)