Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Can anyone explain me as to why Tensor Products are important, and what makes Mathematician's to define them in such a manner. We already have Direct Product, Semi-direct products, so after all why do we need Tensor Product?

The Definition of Tensor Product at Planet Math is confusing.

Definition: Let $R$ be a commutative ring, and let $A, B$ be $R$-modules. There exists an $R$-module $A\otimes B$, called the tensor product of $A$ and $B$ over $R$, together with a canonical bilinear homomorphism $$\otimes: A\times B\rightarrow A\otimes B,$$ distinguished, up to isomorphism, by the following universal property. Every bilinear $R$-module homomorphism $$\phi: A\times B\rightarrow C,$$ lifts to a unique $R$-module homomorphism $$\tilde{\phi}: A\otimes B\rightarrow C,$$ such that $$\phi(a,b) = \tilde{\phi}(a\otimes b)$$ for all $a\in A,\; b\in B.$

The tensor product $A\otimes B$ can be constructed by taking the free $R$-module generated by all formal symbols $$a\otimes b,\quad a\in A,\;b\in B,$$ and quotienting by the obvious bilinear relations: \begin{align*} (a_1+a_2)\otimes b &= a_1\otimes b + a_2\otimes b,\quad &&a_1,a_2\in A,\; b\in B \\ a\otimes(b_1+b_2) &= a\otimes b_1 + a\otimes b_2,\quad &&a\in A,\;b_1,b_2\in B \\ r(a\otimes b) &= (ra)\otimes b= a\otimes (rb)\quad &&a\in A,\;b\in B,\; r\in R \end{align*}

Also what do is the meaning of this statement:(Why do we need this?)

  • Every bilinear $R$-module homomorphism $$\phi: A\times B\rightarrow C,$$ lifts to a unique $R$-module homomorphism $$\tilde{\phi}: A\otimes B\rightarrow C,$$ such that $$\phi(a,b) = \tilde{\phi}(a\otimes b)$$ for all $a\in A,\; b\in B.$

Me and my friends have planned to study this topic today. So i hope i am not wrong in asking such questions.

share|improve this question
The answers to math.stackexchange.com/questions/10282/… might be of some help. –  Hans Lundmark Jan 25 '11 at 11:10
We already have hamburgers and hot dogs, so after all why do we need pizza? –  Pete L. Clark Jan 25 '11 at 18:15
@Pete: Thats a nice joke pete. –  anonymous Mar 7 '11 at 17:59
dpmms.cam.ac.uk/~wtg10/tensors3.html is a good exposition of tensor products. –  Donkey_2009 Nov 6 '12 at 19:33
add comment

9 Answers

up vote 17 down vote accepted

Tensor products are useful because of two reasons:

  • they allow you to study certain non linear maps (bilinear maps) by transforming them first into linear ones, to which you can apply linear algebra;
  • they allow you to change the ring over which a module is defined.

There are many, many ways in which these two abilities show up in nature.

share|improve this answer
add comment

There are literally dozens of independent reasons to invent the tensor product, and just about every area of mathematics needs the tensor product for its own reasons (often several reasons). Here are a couple examples.

  • Suppose $X$ and $Y$ are topological spaces (metric spaces are fine if you like them better) and consider the rings $C(X)$ and $C(Y)$ of continuous real-valued functions. If you are convinced that products are worthy of consideration, then perhaps you are convinced that it is useful to look at $C(X \times Y)$. It is natural to ask if this can be expressed in terms of $C(X)$ and $C(Y)$; the answer (modulo largely irrelevant technical details) is that $C(X \times Y) = C(X) \otimes C(Y)$.

  • Let $V$ be a vector space over $\mathbb{R}$. It is often desirable to construct a complex vector space naturally associated to $V$ (the "complexificiation" of $V$). Here by "naturally" I mean in a way which is coordinate free and transparently compatible with linear maps. The solution is to set $V_{\mathbb{C}} = V \otimes \mathbb{C}$ (tensor product over $\mathbb{R}$). This is a special case of the more general phenomenon of "extension of scalars". As a fancy example demonstrating that this really is as useful as I claim, you might check out the wikipedia page on "pontryagin classes" (though it might be over your head if you haven't learned much algebraic topology).

  • One of the reasons why direct sums are important is that they let you turn strange objects into groups. For example, if $G$ is a group and $V$ and $W$ are two representations of $G$ (vector spaces on which $G$ acts nicely), then $V \oplus W$ is also a representation of $V$. So the set of all representations of $G$ has an additive structure, and with a little algebraic magic one can upgrade this structure to a group (don't spend too much time worrying about how you subtract representations). Groups are nice and have lots of their own invariants, but rings are even nicer and have even more invariants. So it would be great if we could define a natural product of representations. You guessed it: the product of $V$ and $W$ is just $V \otimes W$. The set of all representations of $G$ with this structure is the infamous "representation ring" of $G$. This product structure is apparently of paramount importance in quantum mechanics (I don't know why). As another example where the tensor product turns a group into a ring, you might check out the Wikipedia page on "topological K-theory".

There are many more examples. If you know about functional analysis, the Schwartz kernel theorem is a tool used to investigate existence questions and regularity properties of partial differential equations, and it can be formulated purely in terms of Grothendeick's theory of topological tensor products. I can't give you any deep reason why the same algebraic gadget has such a diverse array of applications, but I guess that's the way it is. You'll undoubtedly learn more as you keep studying math.

ADDED: I just noticed the other part of your question, in which you ask about the "lifting" property of the tensor product. If I were forced to give a one sentence explanation of what the tensor product really is, it would be the following sentence. Given two $R$-modules $A$ and $B$, we want to convert $R$-bilinear maps on $A \times B$ into linear maps on some other object. We want to do this because for many purposes it reduces the structure theory of bilinear maps to the (extensive!) structure theory for linear maps. The lifting property that you describe tells us that the tensor product does the job.

But it more than just "does the job" - it does the job in the absolute best way possible. When you learn about most mathematical objects, such as the direct sum of two vector spaces, it is typical to define the object as some set equipped with some structure and then prove that it has certain nice properties. With the tensor product, you should go about it backwards: you should think of the tensor product as an object with certain nice properties and then prove that there actually is an object with all of those properties. This is because the actual construction of the tensor product of two modules is completely unenlightening and completely irrelevant to how you actually use the idea in practice.

I'll be a little less vague and outline how the tensor product should be developed from scratch. Given two $R$-modules $A$ and $B$, define a tensor product of $A$ and $B$ to be a pair $T, t$ where $T$ is a $R$-module and $t: A \times B \to T$ is a bilinear map with the property that given any bilinear map $Q: A \times B \to C$ there exists a unique linear map $L: T \to C$ such that $B = L \circ t$.

Lemma 1: If the tensor product exists, it is unique up to unique isomorphism.

Lemma 2: The tensor product exists.

share|improve this answer
It is not true that $C(X \times Y) = C(X)\otimes C(Y)$. –  Georges Elencwajg Jan 25 '11 at 20:08
If we're being technical about it, they ARE isomorphic for the tensor product in the category of, say, C* algebras (or most other categories that remember the topology). But perhaps it's worth mentioning that the algebraic tensor product of $C(X)$ and $C(Y)$ must be completed to get something sensible... –  Paul Siegel Jan 26 '11 at 5:28
C(X) is not a $C^\ast$-algebra in general. –  Georges Elencwajg Jan 26 '11 at 8:33
Is this really the only imprecise claim you spotted in my answer? I'll stand by my loophole "or most other categories that remember the topology". –  Paul Siegel Jan 26 '11 at 13:15
add comment

We define tensor products for the same reason we define any other abstract mathematical structure: it's a structure that shows up a lot in mathematics, so it's worth having a name for. I don't see why this reason applies any less to tensor products than to direct sums. As the Wikipedia article says,

Tensor products are important in areas of abstract algebra, homological algebra, algebraic topology and algebraic geometry

and tensor products of vector spaces are also important in differential geometry and physics. I think it is better to learn about these applications thoroughly than to have someone attempt to summarize them.

Tim Gowers has written a short introduction to tensor products here, but it does not, in my opinion, give a good sense of the wide range of applicability of this notion.

share|improve this answer
"...but it does not, in my opinion, give a good sense of the wide range of applicability of this notion." Well, sure. But nor does it intend to: rather it wants to give as gentle as possible a route to understanding one (basic and important) thing tensor products are used for. I actually think this article could be very helpful to the OP. –  Pete L. Clark Jan 25 '11 at 18:13
add comment

The tensor product is the categorization of the multiplication in rings, so you can now multiply objects (vector spaces, rings, algebras, etc), and it is "linear" in each of the term.

Suppose for example you have an algebra over a ring, say $M_n(\mathbb{Z})$ the matrices with integer entries (the ring is the integers), but you want to be able to multiply these matrices with polynomials from $\mathbb{Q}[x]$. So you can define formally a multiplication of a polynomial f with matrix A by (f,A).

Now you want several nice properties you are used to when multiplying, meaning, it is linear in each of the terms (distributivity) $$ (f+g,A)=(f,A)+(g,A) $$ $$ (f,A+B)=(f,A)+(f,B) $$
Taking the tensor product gives you these properties among others. (actually, a multiplication in a ring $R$ is a function $\mu: R\otimes R \rightarrow R$ that satisfies associativity - $\mu (1\otimes \mu)= \mu (\mu \otimes 1)$)

If for example you start with a vector space $V$ over a field $F$ and you tensor it with a field $K$ which is an extension of $F$, then you get a new vector field $K \otimes V$, but over the bigger field $K$ and with the same dimension(and the multiplication is $\alpha(\beta \otimes v)= (\alpha \beta \otimes v)$. This operation is called extension of scalars.

Of course there are other reasons why tensor product is important, but I think that this is a good place to start.

share|improve this answer
In the third last sentence, do you mean "a new vector space"? –  Pandora Jan 25 '11 at 16:00
add comment

Let $R$ be a ring. We suppose $R$ is a commutative ring just for simplicity. Let Mod($R$) be the category of $R$-modules. Let $F$ be an $R$-module. By assigning Hom($F, X$) for each $R$-module $X$ we get a functor $T_F$:Mod($R$) $\to$ Mod($R$). Suppose there exists a left adjoint functor $S_F$ of $T_F$. Then there exists a functorial isomorphism: Hom($S_F(E), X) \cong$ Hom($E, T_F(X)$) for $E, X \in$ Mod($R$). In fact, $S_F(E) = E\otimes F$ satisfies this condition. In other words, $-\otimes F$ is a left adjoint functor of Hom($F, -$).

EDIT[Aug 26, 2012]

In short, tensor products are important because they are (sort of) duals of Hom functors.

share|improve this answer
Not that I'm complaining, but I'm surprised that this had no upvotes so far. Recognizing tensor products as ajoints of Hom functors is very fundamental and important IMO. –  Makoto Kato Aug 27 '12 at 5:07
What happens if $R$ is not commutative? –  Uday Reddy Apr 4 '13 at 20:43
@UdayReddy Similarly $-\otimes_R F$ is a left adjoint functor of Hom$_\mathbb{Z}(F, -$). –  Makoto Kato Apr 5 '13 at 10:11
I don't think so. The collection $Hom_R(X,Y)$ of $R$-modules for a non-commutative ring $R$ does not have the structure of an $R$-module. See the paragraph 7.5 of my Notes on Semigroups for example. –  Uday Reddy Apr 5 '13 at 14:40
@UdayReddy $Hom_\mathbb{Z}(E\otimes_R F, G) \cong Hom_R(E, Hom_\mathbb{Z}(F, G)$) where $E$ is a right $R$-module and $G$ is a $\mathbb{Z}$-module. –  Makoto Kato Apr 5 '13 at 20:44
show 2 more comments

I won't try to convince you of anything with the following, I just think this reason should be recorded. It is the right "product" in the category of Rings. It gives you the right symmetric monoidal structure. Essentially, it is important and useful because it actually is what we should be thinking about. The reasons essentially involve the universal property it has that is mentioned above: a linear map out of a tensor product is a multilinear map out of the cartesian product, and when doing algebra this is what we should be concerning ourselves with.

Please note the comment of elgeorges below, the tensor product is the categorical coproduct in the category of Commutative Rings, hence my use of quotation marks.

share|improve this answer
In the category of commutative rings! :) –  Mariano Suárez-Alvarez Jan 25 '11 at 15:08
The tensor product of two commutative rings is their $\textit co$product. And their product is their product! –  Georges Elencwajg Jan 25 '11 at 22:52
I am sorry, quite right elgeorges! –  Sean Tilson Jan 25 '11 at 22:55
One way one could naturally be led to wonder about this coproduct comes from classical algebraic geometry, where the product of two affine algebraic varieties over an algebraically closed field $k$ can be computed if you know the coproduct in the dual category of finitely-generated reduced $k$-algebras. (That's a mouthful, but the equivalence of these two categories is the essential geometric statement of the Nullstellensatz.) This allows one to immediately deduce that the product of two varieties is the maximal spectrum of the tensor product of their coordinate rings. –  Zach Conn Jan 25 '11 at 23:05
Since the coproduct in other related categories (i.e. the category of rings, or $R$-modules) is not the tensor product, doesn't the fact that the coproduct of commutative rings happens to match the tensor product seem more like an accident? Is there way to make that accident enlightening? –  Joe Hannon Feb 22 '12 at 22:19
add comment

What helped me see the charm of Tensor Products is the following: calculate Z/100Z ⊗ Z/101Z. Direct Product is a solution for one universal problem. Tensor Products is a solution for another one.

share|improve this answer
I don't see how this helps if you aren't convinced that tensor products are worth calculating. –  Qiaochu Yuan Jan 25 '11 at 12:07
It is surprising, consequently worth studding. –  user6223 Jan 25 '11 at 12:19
add comment

All it's saying is that we have a special operator that takes the two spaces A and B and gives us a new space C' ($A \otimes B$). And that every homomorphism from A x B to C gives us another homomorphism from C' to C. (and it is unique) and it's very nice because calculation in C' correspond to calculations in C.

This is the common generalization of mappings. What it does is let us construct a sort of special space, C' that directly relates in a unique way to C.


See the first diagram. Note how similar it is to many other diagrams found in group theory such as from quotient spaces. What is different is that we are working on Cartesian products. In some sense you can think of the tensor product as reducing the Cartesian product of a space to a new space that has a direct relation to the old. This is useful in that the new space may be isomorphic to a simpler space and/or features of the new space can be used to explore features of the old space.

See also: http://www.dpmms.cam.ac.uk/~wtg10/tensors3.html

share|improve this answer
Friends do not let friends use \bigotimes (a large operator, in $\TeX$speak) where they should have used \otimes (a binary operation)! –  Mariano Suárez-Alvarez Jan 25 '11 at 17:59
add comment

The reason for defining tensor products of $R$-modules (or of vector spaces) is the same as the same as that of defining products of sets. We do the latter in high school, probably without realizing what we are doing. If we have an expression $E[x,y]$ with two variables, we say that we can make it into a "function of two variables" $f(x,y) = E[x,y]$. What that means is that for any particular value of $x$, say $x_0$, $E[x_0,y]$ is a function of $y$ and, for any particular value $y = y_0$, $E[x,y_0]$ is a function of $x$. If $x$ and $y$ range over sets $A$ and $B$ respectively, and $E[x,y]$ is in $C$, we might right the type of this "function of two variables" as $f : A, B \to C$. But such "functions of two variables" are quite inconvenient to work with. So, we invent a set called $A \times B$, whose elements are ordered pairs like $(x,y)$ and has this property:

  • functions of two variables $A,B \to C$ are one-to-one with functions of type $A \times B \to C$

Then we never have to write strange types like "$A,B \to C$". "Functions of two variables" are now the same thing as ordinary functions $A \times B \to C$.

The same idea applied to $R$-modules leads to tensor products. If you have an expression $E[x,y]$ that is linear in $x$ and $y$ separately, i.e., for any fixed value $x = x_0$, $E[x_0, y]$ is linear in $y$ and, for any fixed value $y = y_0$, $E[x, y_0]$ is linear in $x$, then we have a "bilinear function" of two variables $A, B \to C$. In the same way as we did for sets, we invent an $R$-module called $A \otimes_R B$, which has this property:

  • bilinear functions $A, B \to C$ are one-to-one with linear functions $A \otimes_R B \to C$.

The Planet Math definition is saying exactly this except that it is using the notation "$A \times B \to C$" instead of my strange notation "$A, B \to C$", and it is telling you what kind of one-to-one correspondence we are looking for: "for every $\phi: A, B \to C$ there is a unique $\tilde{\phi} : A \otimes_R B \to C$ such that ...".

I avoided using the notation "$A \times B \to C$" because it is highly misleading. What is meant by $A \times B$ here? $A$ and $B$ are $R$-modules, which have a notion of product. Is that what $A \times B$ means? Not really. If you take the $R$-module $A \times B$ and look at linear maps from there to $C$, you don't get bilinear maps (maps that are linear in $A$ and $B$ independently). I will let you think about why such misleading notation is used (almost universally in mathematics). But if you try to type that kind of thing into an automated theorem prover, which requires you to be precise about what you write, you won't get away with it.

(The strange notation "$A,B \to C$" that I have used actually comes from a well-studied system of multicategories in Category Theory. Even though it is inconvenient to use, it is sometimes necessary because the tensor products in some branches of mathematics may be even more inconvenient to use.)

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.