22

This is a beginner's question on what exactly is a tensor product, in laymen's term, for a beginner who has just learned basic group theory and basic ring theory.

I do understand from wikipedia that in some cases, the tensor product is an outer product, which takes two vectors, say $\textbf{u}$ and $\textbf{v}$, and outputs a matrix $\textbf{uv}^T$. ($\textbf{u}$ being a $m\times 1$ column vector and $\textbf{v}$ being a $n\times 1$ column vector)

How about more general cases of tensor products, e.g. in the context of quantum groups?

Sincere thanks.

Rudy the Reindeer
  • 48,301
yoyostein
  • 20,428
  • 1
    What kinds of objects are you taking the tensor product of? – Alex Becker May 13 '12 at 07:21
  • 8
    Look at this http://www.dpmms.cam.ac.uk/~wtg10/tensors3.html – Ehsan M. Kermani May 13 '12 at 07:24
  • 5
    +1 for ehsanmo's fantastic link. The explanations there are due to Timothy Gowers, a Fields medalist and one of the outstanding mathematicians of this century (the linked page modestly does not mention the author !) – Georges Elencwajg May 13 '12 at 07:48
  • 1
    I would add to the list of references two expository articles on the tensor product by Keith Conrad. The link to the web page containing all his expository stuff is this:http://www.math.uconn.edu/~kconrad/blurbs/. They are a great resource for algebra. The relevant articles are under the section titled Linear/ Multilinear Algebra. – Rankeya May 13 '12 at 15:06
  • As a believer in simplicity, I've posted what may be the simplest P.O.V. on this question below. Definitely my answer is not the last word though---i.e. there's lots more to know. – Michael Hardy May 13 '12 at 17:50
  • I don't think tensor products live in layman's terms, outside of seeing examples. For instance, ${\mathbf C} \otimes_{\mathbf R} {\mathbf R}[x] \cong {\mathbf C}[x]$ and ${\mathbf R}[x] \otimes_{\mathbf R} {\mathbf R}[y] \cong {\mathbf R}[x,y]$. Everybody finds tensor products hard the first (or second?) time through and I don't think there is a way around that. Someone once said tensor products are the hardest thing in mathematics. While that is definitely not true, I think it's fair to say they are the hardest thing in basic mathematics. I don't think you can "get" them (contd.) – KCd May 14 '12 at 09:13
  • just from having a layman's idea about what they are. I'm not trying to be discouraging, but rather to emphasize that you're going to have to put in a lot of work and time for tensor products to be something you are comfortable with. – KCd May 14 '12 at 09:14

3 Answers3

26

If you want to study a mathematical object, whether it is a set, manifold, group, vector space, whatever, it is often fruitful to look at natural collections of functions on that space.

Roughly, the purpose of the tensor product, $\otimes$, is to make the following statement true: $$\text{functions}(X \times Y) = \text{functions}(X)\otimes \text{functions}(Y)$$

The specific details about which spaces of functions to choose depend on the type of mathematical object you are interested in.

Here's a pdf that explains it better than I can, http://abel.math.harvard.edu/archive/25b_spring_05/tensor.pdf

Nick Alger
  • 19,977
  • 2
    The link does not work anymore. However, I feel that https://www.math3ma.com/blog/the-tensor-product-demystified by Tai-Danae Bradley will help a lot of people. – Jan Rothkegel Mar 28 '20 at 18:22
  • Is the cross inside the parenthesis on the left term a Cartesian product? – Gabriel Sandoval Apr 27 '20 at 06:05
  • 1
    A working link to the PDF given above can now be found at http://abel.math.harvard.edu/archive/25b_spring_05/tensor.pdf – jhu Aug 01 '21 at 04:59
  • @jhu Thanks, updated the post with the new link! – Nick Alger Aug 01 '21 at 05:10
  • @GabrielSandoval $\otimes$ is a generic symbol for an operator that is linear in both its arguments. Since it shares this property with the "standard" products one finds in other other context, it is denote as a product (other properties, like commutativity, make no sense when talking about functions of X and Y). – VictorZurkowski Jun 06 '25 at 08:19
4

The difference between an ordered pair of vectors and a tensor product of two vectors is this:

If you multiply one of the vectors by a scalar and the other by the reciprocal of that scalar, you get a different ordered pair of vectors, but the same tensor product of two vectors.

Similarly with an ordered triple of vectors and a tensor product of three vectors, etc.

Michael Hardy
  • 1
  • I've thought along these lines before (tensors are combos of vectors with special rules), but always ran into the same intuitive stumbling block. Why is it that some elements of a tensor space are pure and can be represented as $w=x \otimes y$, whereas others are composite and can only be built from sums of pure tensors? I sort of gave up on thinking of it that way, but maybe you have some insight..? – Nick Alger May 13 '12 at 21:36
  • I don't have a simple answer to that right now.... Maybe later? – Michael Hardy May 13 '12 at 22:55
  • 6
    There is a simple answer: just think about multivariable polynomials! Let's just look at polynomials in two variables. Some of them are monomials, or more generally products $f(x)g(y)$, but most multivariable polynomials are not of that (separable) form. The theory of polynomials in two variables is more than the theory of things that factor as $f(x)g(y)$. This in fact is a special instance of tensor products, since $F[x] \otimes_F F[y] \cong F[x,y]$ for any field $F$. – KCd May 14 '12 at 08:51
  • 6
    Another point on the subtlety of the phenomenon of tensors that are not pure (elementary, monomial, separable, whatever you want to call them) is that they are show up in the mathematical description of entanglement in quantum mechanics. That is, in QM the combined state space for two quantum systems is a (completed) tensor product of the original two spaces, and trying to wrap your head around entangled states of two particles is the conundrum of trying to wrap your head around the idea that some tensors are not pure. – KCd May 14 '12 at 08:54
  • 2
    There's a second way two unequal ordered pairs can correspond to the same pure tensor that I think is worth keeping in mind: for vector spaces $V$, $W$ and non-zero vectors $v\in V\setminus{0}$, $w\in W\setminus{0}$ the pairs $(0,0)$, $(v,0)$, and $(0,w)$ are all different, but the pure tensors $0\otimes0$, $v\otimes0$, and $0\otimes w$ are all the same. – Will Orrick Jan 25 '20 at 18:17
0

Personally I think the concept of tensor product is not well described, especially not friendly enough for a beginner to understand and often there is a little abuse of language. The official definition looks like this: Given two vector spaces A and B, the tensor product is a vector space C associated with a bilinear map $$A\times B \rightarrow C$$ such that C should also satisfy a unverisal property which you can find in text-book.

But this definition leads to a misunderstanding that tensor product of two vector space is a new vector space. But actually it is not, it a vector space associated with a blinear map, the blinear map is a component of the tensor product.

But I would define tensor product from basis vector, so given A, B with basis vector $B_A, B_B$ ,then we can construct a new vector space C with basis $B_C$, and there exists a bilinear map $f$ from $A\times B$ to $C$ such that, each basis c of $C$, satisfy : $c=f(a,b)$ where $a$ is a basis vector of $A$ and $b$ is a basis vector of $B$, so there is 1-1 corresponding between the set $\{(a,b)| a \in B_A, b \in B_B\}$ and the set $B_C$ and the 1-1 corresponding is just $f$ Then $(f,C)$ would be the tensor product of $A,B$ as you desired, then everything just follow exactly the same as the text-book said. You need to know that the tensor product is not something concretly, you can define a concrete space as you like and then you find the bilinear map, then you can say it is the tensor product, since all tensor product are isomorphic.

jones betty
  • 11