Tensor product

From bases

Let V and W be two vector spaces over a field F, with respective bases ${\displaystyle B_{V}}$ and ⁠${\displaystyle B_{W}}$⁠.

The tensor product ${\displaystyle V\otimes W}$ of V and W is a vector space that has as a basis the set of all ${\displaystyle v\otimes w}$ with ${\displaystyle v\in B_{V}}$ and ⁠${\displaystyle w\in B_{W}}$⁠, regarded purely as symbols with no further meaning. This definition can be formalized in the following way (this formalization is rarely used in practice, as the preceding informal definition is generally sufficient): ${\displaystyle V\otimes W}$ is the set of functions from the Cartesian product ${\displaystyle B_{V}\times B_{W}}$ to F that have a finite number of nonzero values. The pointwise operations make ${\displaystyle V\otimes W}$ a vector space. The function that maps ${\displaystyle (v,w)}$ to 1 and the other elements of ${\displaystyle B_{V}\times B_{W}}$ to 0 is denoted ⁠${\displaystyle v\otimes w}$⁠.

The set ${\displaystyle \{v\otimes w\mid v\in B_{V},w\in B_{W}\}}$, called the tensor product of the bases ${\displaystyle B_{V}}$ and ⁠${\displaystyle B_{W}}$⁠, is straightforwardly a basis of ⁠${\displaystyle V\otimes W}$⁠.

Equivalently, we can define ${\displaystyle V\otimes W}$ to be the set of bilinear forms on ${\displaystyle V\times W}$ that are nonzero at only a finite number of elements of ⁠${\displaystyle B_{V}\times B_{W}}$⁠. To see this, given ${\displaystyle (x,y)\in V\times W}$ and a bilinear form ⁠${\displaystyle B:V\times W\to F}$⁠, we can decompose ${\displaystyle x}$ and ${\displaystyle y}$ in the bases ${\displaystyle B_{V}}$ and ${\displaystyle B_{W}}$ as: $${\displaystyle x=\sum _{v\in B_{V}}x_{v}\,v\quad {\text{and}}\quad y=\sum _{w\in B_{W}}y_{w}\,w,}$$ where only a finite number of ${\displaystyle x_{v}}$'s and ${\displaystyle y_{w}}$'s are nonzero, and find by the bilinearity of ${\displaystyle B}$ that: $${\displaystyle B(x,y)=\sum _{v\in B_{V}}\sum _{w\in B_{W}}x_{v}y_{w}\,B(v,w)}$$

Hence, we see that the value of ${\displaystyle B}$ for any ${\displaystyle (x,y)\in V\times W}$ is uniquely and totally determined by the values that it takes on ⁠${\displaystyle B_{V}\times B_{W}}$⁠. This lets us extend the maps ${\displaystyle v\otimes w}$ defined on ${\displaystyle B_{V}\times B_{W}}$ as before into bilinear maps ${\displaystyle v\otimes w:V\times W\to F}$ , by letting: $${\displaystyle (v\otimes w)(x,y):=\sum _{v'\in B_{V}}\sum _{w'\in B_{W}}x_{v'}y_{w'}\,(v\otimes w)(v',w')=x_{v}\,y_{w}.}$$

Then we can express any bilinear form ${\displaystyle B}$ as a (potentially infinite) formal linear combination of the ${\displaystyle v\otimes w}$ maps according to: $${\displaystyle B=\sum _{v\in B_{V}}\sum _{w\in B_{W}}B(v,w)(v\otimes w)}$$ making these maps similar to a Schauder basis for the vector space ${\displaystyle {\text{Hom}}(V,W;F)}$ of all bilinear forms on ⁠${\displaystyle V\times W}$⁠. To instead have it be a proper Hamel basis, it only remains to add the requirement that ${\displaystyle B}$ is nonzero at an only a finite number of elements of ⁠${\displaystyle B_{V}\times B_{W}}$⁠, and consider the subspace of such maps instead.

In either construction, the tensor product of two vectors is defined from their decomposition on the bases. More precisely, taking the basis decompositions of ${\displaystyle x\in V}$ and ${\displaystyle y\in W}$ as before: $${\displaystyle {\begin{aligned}x\otimes y&={\biggl (}\sum _{v\in B_{V}}x_{v}\,v{\biggr )}\otimes {\biggl (}\sum _{w\in B_{W}}y_{w}\,w{\biggr )}\\[5mu]&=\sum _{v\in B_{V}}\sum _{w\in B_{W}}x_{v}y_{w}\,v\otimes w.\end{aligned}}}$$

This definition is quite clearly derived from the coefficients of ${\displaystyle B(v,w)}$ in the expansion by bilinearity of ${\displaystyle B(x,y)}$ using the bases ${\displaystyle B_{V}}$ and ⁠${\displaystyle B_{W}}$⁠, as done above. It is then straightforward to verify that with this definition, the map ${\displaystyle {\otimes }:(x,y)\mapsto x\otimes y}$ is a bilinear map from ${\displaystyle V\times W}$ to ${\displaystyle V\otimes W}$ satisfying the universal property that any construction of the tensor product satisfies (see below).

If arranged into a rectangular array, the coordinate vector of ${\displaystyle x\otimes y}$ is the outer product of the coordinate vectors of ${\displaystyle x}$ and ⁠${\displaystyle y}$⁠. Therefore, the tensor product is a generalization of the outer product, that is, an abstraction of it beyond coordinate vectors.

A limitation of this definition of the tensor product is that, if one changes bases, a different tensor product is defined. However, the decomposition on one basis of the elements of the other basis defines a canonical isomorphism between the two tensor products of vector spaces, which allows identifying them. Also, contrarily to the two following alternative definitions, this definition cannot be extended into a definition of the tensor product of modules over a ring.

As a quotient space

A construction of the tensor product that is basis independent can be obtained in the following way.

Let V and W be two vector spaces over a field F.

One considers first a vector space L that has the Cartesian product ${\displaystyle V\times W}$ as a basis. That is, the basis elements of L are the pairs ${\displaystyle (v,w)}$ with ${\displaystyle v\in V}$ and ⁠${\displaystyle w\in W}$⁠. To get such a vector space, one can define it as the vector space of the functions ${\displaystyle V\times W\to F}$ that have a finite number of nonzero values and identifying ${\displaystyle (v,w)}$ with the function that takes the value 1 on ${\displaystyle (v,w)}$ and 0 otherwise.

Let R be the linear subspace of L that is spanned by the relations that the tensor product must satisfy. More precisely, R is spanned by the elements of one of the forms:

${\displaystyle {\begin{aligned}(v_{1}+v_{2},w)&-(v_{1},w)-(v_{2},w),\\(v,w_{1}+w_{2})&-(v,w_{1})-(v,w_{2}),\\(sv,w)&-s(v,w),\\(v,sw)&-s(v,w),\end{aligned}}}$

where ⁠${\displaystyle v,v_{1},v_{2}\in V}$⁠, ${\displaystyle w,w_{1},w_{2}\in W}$ and ⁠${\displaystyle s\in F}$⁠.

Then, the tensor product is defined as the quotient space:

${\displaystyle V\otimes W=L/R,}$

and the image of ${\displaystyle (v,w)}$ in this quotient is denoted ⁠${\displaystyle v\otimes w}$⁠.

It is straightforward to prove that the result of this construction satisfies the universal property considered below. (A very similar construction can be used to define the tensor product of modules.)

Universal property

In this section, the universal property satisfied by the tensor product is described. As for every universal property, two objects that satisfy the property are related by a unique isomorphism. It follows that this is a (non-constructive) way to define the tensor product of two vector spaces. In this context, the preceding constructions of tensor products may be viewed as proofs of existence of the tensor product so defined.

A consequence of this approach is that every property of the tensor product can be deduced from the universal property, and that, in practice, one may forget the method that has been used to prove its existence.

The "universal-property definition" of the tensor product of two vector spaces is the following (recall that a bilinear map is a function that is separately linear in each of its arguments):

The tensor product of two vector spaces V and W is a vector space denoted as ⁠${\displaystyle V\otimes W}$⁠, together with a bilinear map ${\displaystyle {\varphi }:(v,w)\mapsto v\otimes w}$ from ${\displaystyle V\times W}$ to ⁠${\displaystyle V\otimes W}$⁠, such that, for every bilinear map ⁠${\displaystyle h:V\times W\to Z}$⁠, there is a unique linear map ⁠${\displaystyle {\tilde {h}}:V\otimes W\to Z}$⁠, such that ${\displaystyle h={\tilde {h}}\circ {\varphi }}$ (that is, ${\displaystyle h(v,w)={\tilde {h}}(v\otimes w)}$ for every ${\displaystyle v\in V}$ and ⁠${\displaystyle w\in W}$⁠).

Linearly disjoint

Like the universal property above, the following characterization may also be used to determine whether or not a given vector space and given bilinear map form a tensor product.^[2]

For example, it follows immediately that if ⁠${\displaystyle X=\mathbb {C} ^{m}}$⁠ and ⁠${\displaystyle Y=\mathbb {C} ^{n}}$⁠, where ${\displaystyle m}$ and ${\displaystyle n}$ are positive integers, then one may set ${\displaystyle Z=\mathbb {C} ^{mn}}$ and define the bilinear map as $${\displaystyle {\begin{aligned}T:\mathbb {C} ^{m}\times \mathbb {C} ^{n}&\to \mathbb {C} ^{mn}\\(x,y)=((x_{1},\ldots ,x_{m}),(y_{1},\ldots ,y_{n}))&\mapsto (x_{i}y_{j})_{\stackrel {i=1,\ldots ,m}{j=1,\ldots ,n}}\end{aligned}}}$$ to form the tensor product of ${\displaystyle X}$ and ⁠${\displaystyle Y}$⁠.^[3] Often, this map ${\displaystyle T}$ is denoted by ${\displaystyle \,\otimes \,}$ so that ${\displaystyle x\otimes y=T(x,y).}$

As another example, suppose that ${\displaystyle \mathbb {C} ^{S}}$ is the vector space of all complex-valued functions on a set ${\displaystyle S}$ with addition and scalar multiplication defined pointwise (meaning that ${\displaystyle f+g}$ is the map ${\displaystyle s\mapsto f(s)+g(s)}$ and ${\displaystyle cf}$ is the map ⁠${\displaystyle s\mapsto cf(s)}$⁠). Let ${\displaystyle S}$ and ${\displaystyle T}$ be any sets and for any ${\displaystyle f\in \mathbb {C} ^{S}}$ and ⁠${\displaystyle g\in \mathbb {C} ^{T}}$⁠, let ${\displaystyle f\otimes g\in \mathbb {C} ^{S\times T}}$ denote the function defined by ⁠${\displaystyle (s,t)\mapsto f(s)g(t)}$⁠. If ${\displaystyle X\subseteq \mathbb {C} ^{S}}$ and ${\displaystyle Y\subseteq \mathbb {C} ^{T}}$ are vector subspaces then the vector subspace ${\displaystyle Z:=\operatorname {span} \left\{f\otimes g:f\in X,g\in Y\right\}}$ of ${\displaystyle \mathbb {C} ^{S\times T}}$ together with the bilinear map: $${\displaystyle {\begin{alignedat}{4}\;&&X\times Y&&\;\to \;&Z\\[0.3ex]&&(f,g)&&\;\mapsto \;&f\otimes g\\\end{alignedat}}}$$ form a tensor product of ${\displaystyle X}$ and ⁠${\displaystyle Y}$⁠.^[3]

Dimension

If V and W are vector spaces of finite dimension, then ${\displaystyle V\otimes W}$ is finite-dimensional, and its dimension is the product of the dimensions of V and W.

This results from the fact that a basis of ${\displaystyle V\otimes W}$ is formed by taking all tensor products of a basis element of V and a basis element of W.

Associativity

The tensor product is associative in the sense that, given three vector spaces ⁠${\displaystyle U,V,W}$⁠, there is a canonical isomorphism:

${\displaystyle (U\otimes V)\otimes W\cong U\otimes (V\otimes W),}$

that maps ${\displaystyle (u\otimes v)\otimes w}$ to ⁠${\displaystyle u\otimes (v\otimes w)}$⁠.

This allows omitting parentheses in the tensor product of more than two vector spaces or vectors.

Commutativity as vector space operation

The tensor product of two vector spaces ${\displaystyle V}$ and ${\displaystyle W}$ is commutative in the sense that there is a canonical isomorphism:

${\displaystyle V\otimes W\cong W\otimes V,}$

that maps ${\displaystyle v\otimes w}$ to ⁠${\displaystyle w\otimes v}$⁠.

On the other hand, even when ⁠${\displaystyle V=W}$⁠, the tensor product of vectors is not commutative; that is ⁠${\displaystyle v\otimes w\neq w\otimes v}$⁠, in general.

The map ${\displaystyle x\otimes y\mapsto y\otimes x}$ from ${\displaystyle V\otimes V}$ to itself induces a linear automorphism that is called a braiding map. More generally and as usual (see tensor algebra), let ${\displaystyle V^{\otimes n}}$ denote the tensor product of n copies of the vector space V. For every permutation s of the first n positive integers, the map:

${\displaystyle x_{1}\otimes \cdots \otimes x_{n}\mapsto x_{s(1)}\otimes \cdots \otimes x_{s(n)}}$

induces a linear automorphism of ⁠${\displaystyle V^{\otimes n}\to V^{\otimes n}}$⁠, which is called a braiding map. This determines a right action of the symmetric group ${\displaystyle S_{n}}$ on ${\displaystyle V^{\otimes n}}$.

Evaluation map and tensor contraction

For tensors of type (1, 1) there is a canonical evaluation map: $${\displaystyle V\otimes V^{*}\to K}$$ defined by its action on pure tensors: $${\displaystyle v\otimes f\mapsto f(v).}$$

More generally, for tensors of type ⁠${\displaystyle (r,s)}$⁠, with r, s > 0, there is a map, called tensor contraction: $${\displaystyle T_{s}^{r}(V)\to T_{s-1}^{r-1}(V).}$$ (The copies of ${\displaystyle V}$ and ${\displaystyle V^{*}}$ on which this map is to be applied must be specified.)

On the other hand, if ${\displaystyle V}$ is finite-dimensional, there is a canonical map in the other direction (called the coevaluation map): $${\displaystyle {\begin{cases}K\to V\otimes V^{*}\\\lambda \mapsto \sum _{i}\lambda v_{i}\otimes v_{i}^{*}\end{cases}}}$$ where ${\displaystyle v_{1},\ldots ,v_{n}}$ is any basis of ⁠${\displaystyle V}$⁠, and ${\displaystyle v_{i}^{*}}$ is its dual basis. This map does not depend on the choice of basis.^[7]

The interplay of evaluation and coevaluation can be used to characterize finite-dimensional vector spaces without referring to bases.^[8]

Adjoint representation

The tensor product ${\displaystyle T_{s}^{r}(V)}$ may be naturally viewed as a module for the Lie algebra ${\displaystyle \mathrm {End} (V)}$ by means of the diagonal action: for simplicity let us assume ⁠${\displaystyle r=s=1}$⁠, then, for each ⁠${\displaystyle u\in \mathrm {End} (V)}$⁠, $${\displaystyle u(a\otimes b)=u(a)\otimes b-a\otimes u^{*}(b),}$$ where ${\displaystyle u^{*}\in \mathrm {End} \left(V^{*}\right)}$ is the transpose of u, that is, in terms of the obvious pairing on ⁠${\displaystyle V\otimes V^{*}}$⁠, $${\displaystyle \langle u(a),b\rangle =\langle a,u^{*}(b)\rangle .}$$

There is a canonical isomorphism ${\displaystyle T_{1}^{1}(V)\to \mathrm {End} (V)}$ given by: $${\displaystyle (a\otimes b)(x)=\langle x,b\rangle a.}$$

Under this isomorphism, every u in ${\displaystyle \mathrm {End} (V)}$ may be first viewed as an endomorphism of ${\displaystyle T_{1}^{1}(V)}$ and then viewed as an endomorphism of ⁠${\displaystyle \mathrm {End} (V)}$⁠. In fact it is the adjoint representation ad(u) of ⁠${\displaystyle \mathrm {End} (V)}$⁠.

Tensor product of modules over a non-commutative ring

Let A be a right R-module and B be a left R-module. Then the tensor product of A and B is an abelian group defined by: $${\displaystyle A\otimes _{R}B:=F(A\times B)/G}$$ where ${\displaystyle F(A\times B)}$ is a free abelian group over ${\displaystyle A\times B}$ and G is the subgroup of ${\displaystyle F(A\times B)}$ generated by relations: $${\displaystyle {\begin{aligned}&\forall a,a_{1},a_{2}\in A,\forall b,b_{1},b_{2}\in B,{\text{ for all }}r\in R:\\&(a_{1},b)+(a_{2},b)-(a_{1}+a_{2},b),\\&(a,b_{1})+(a,b_{2})-(a,b_{1}+b_{2}),\\&(ar,b)-(a,rb).\\\end{aligned}}}$$

The universal property can be stated as follows. Let G be an abelian group with a map ${\displaystyle q:A\times B\to G}$ that is bilinear, in the sense that: $${\displaystyle {\begin{aligned}q(a_{1}+a_{2},b)&=q(a_{1},b)+q(a_{2},b),\\q(a,b_{1}+b_{2})&=q(a,b_{1})+q(a,b_{2}),\\q(ar,b)&=q(a,rb).\end{aligned}}}$$

Then there is a unique map ${\displaystyle {\overline {q}}:A\otimes B\to G}$ such that ${\displaystyle {\overline {q}}(a\otimes b)=q(a,b)}$ for all ${\displaystyle a\in A}$ and ⁠${\displaystyle b\in B}$⁠.

Furthermore, we can give ${\displaystyle A\otimes _{R}B}$ a module structure under some extra conditions:

1. If A is a (S,R)-bimodule, then ${\displaystyle A\otimes _{R}B}$ is a left S-module, where ⁠${\displaystyle s(a\otimes b):=(sa)\otimes b}$⁠.
2. If B is a (R,S)-bimodule, then ${\displaystyle A\otimes _{R}B}$ is a right S-module, where ⁠${\displaystyle (a\otimes b)s:=a\otimes (bs)}$⁠.
3. If A is a (S,R)-bimodule and B is a (R,T)-bimodule, then ${\displaystyle A\otimes _{R}B}$ is a (S,T)-bimodule, where the left and right actions are defined in the same way as the previous two examples.
4. If R is a commutative ring, then A and B are (R,R)-bimodules where ${\displaystyle ra:=ar}$ and ⁠${\displaystyle br:=rb}$⁠. By 3), we can conclude ${\displaystyle A\otimes _{R}B}$ is a (R,R)-bimodule.

Computing the tensor product

For vector spaces, the tensor product ${\displaystyle V\otimes W}$ is quickly computed since bases of V of W immediately determine a basis of ⁠${\displaystyle V\otimes W}$⁠, as was mentioned above. For modules over a general (commutative) ring, not every module is free. For example, Z/nZ is not a free abelian group (Z-module). The tensor product with Z/nZ is given by: $${\displaystyle M\otimes _{\mathbf {Z} }\mathbf {Z} /n\mathbf {Z} =M/nM.}$$

More generally, given a presentation of some R-module M, that is, a number of generators ${\displaystyle m_{i}\in M,i\in I}$ together with relations: $${\displaystyle \sum _{j\in J}a_{ji}m_{i}=0,\qquad a_{ij}\in R,}$$ the tensor product can be computed as the following cokernel: $${\displaystyle M\otimes _{R}N=\operatorname {coker} \left(N^{J}\to N^{I}\right)}$$

Here ⁠${\displaystyle N^{J}=\oplus _{j\in J}N}$⁠, and the map ${\displaystyle N^{J}\to N^{I}}$ is determined by sending some ${\displaystyle n\in N}$ in the jth copy of ${\displaystyle N^{J}}$ to ${\displaystyle a_{ij}n}$ (in ⁠${\displaystyle N^{I}}$⁠). Colloquially, this may be rephrased by saying that a presentation of M gives rise to a presentation of ⁠${\displaystyle M\otimes _{R}N}$⁠. This is referred to by saying that the tensor product is a right exact functor. It is not in general left exact, that is, given an injective map of R-modules ⁠${\displaystyle M_{1}\to M_{2}}$⁠, the tensor product: $${\displaystyle M_{1}\otimes _{R}N\to M_{2}\otimes _{R}N}$$ is not usually injective. For example, tensoring the (injective) map given by multiplication with n, n : Z → Z with Z/nZ yields the zero map 0 : Z/nZ → Z/nZ, which is not injective. Higher Tor functors measure the defect of the tensor product being not left exact. All higher Tor functors are assembled in the derived tensor product.

Topological tensor products

Hilbert spaces generalize finite-dimensional vector spaces to arbitrary dimensions. There is an analogous operation, also called the "tensor product," that makes Hilbert spaces a symmetric monoidal category. It is essentially constructed as the metric space completion of the algebraic tensor product discussed above. However, it does not satisfy the obvious analogue of the universal property defining tensor products;^[12] the morphisms for that property must be restricted to Hilbert–Schmidt operators.^[13]

In situations where the imposition of an inner product is inappropriate, one can still attempt to complete the algebraic tensor product, as a topological tensor product. However, such a construction is no longer uniquely specified: in many cases, there are multiple natural topologies on the algebraic tensor product.

Tensor product of graded vector spaces

Some vector spaces can be decomposed into direct sums of subspaces. In such cases, the tensor product of two spaces can be decomposed into sums of products of the subspaces (in analogy to the way that multiplication distributes over addition).

Tensor product of representations

Vector spaces endowed with an additional multiplicative structure are called algebras. The tensor product of such algebras is described by the Littlewood–Richardson rule.

Tensor product of algebraic fields

Given two fields that are algebraic extensions of a common base field, their tensor product over the base field is again algebraic over the base field. Specifically, it is the algebraic extension generated by the products of the generators of the original two fields.^{[citation needed]} For example, the tensor product over the rational numbers of ${\displaystyle \mathbb {Q} [{\sqrt {2}}]}$ and ${\displaystyle \mathbb {Q} [{\sqrt {3}}]}$ is generated by ${\displaystyle 1\cdot 1=1}$, ${\displaystyle 1\cdot {\sqrt {2}}={\sqrt {2}}}$, ${\displaystyle 1\cdot {\sqrt {3}}={\sqrt {3}}}$, and ${\displaystyle {\sqrt {2}}\cdot {\sqrt {3}}={\sqrt {6}}}$, and can be denoted ${\displaystyle \mathbb {Q} [{\sqrt {2}},{\sqrt {3}}]}$.

Tensor product of quadratic forms

Tensor product of multilinear forms

Given two multilinear forms ${\displaystyle f(x_{1},\dots ,x_{k})}$ and ${\displaystyle g(x_{1},\dots ,x_{m})}$ on a vector space ${\displaystyle V}$ over the field ${\displaystyle K}$ their tensor product is the multilinear form:^[14] $${\displaystyle (f\otimes g)(x_{1},\dots ,x_{k+m})=f(x_{1},\dots ,x_{k})g(x_{k+1},\dots ,x_{k+m}).}$$

This is a special case of the product of tensors if they are seen as multilinear maps (see also tensors as multilinear maps). Thus the components of the tensor product of multilinear forms can be computed by the Kronecker product.

Tensor product of sheaves of modules

Tensor product of line bundles

Tensor product of fields

Tensor product of graphs

It should be mentioned that, though called "tensor product", this is not a tensor product of graphs in the above sense; actually it is the category-theoretic product in the category of graphs and graph homomorphisms. However it is actually the Kronecker tensor product of the adjacency matrices of the graphs. Compare also the section Tensor product of linear maps above.

Monoidal categories

The most general setting for the tensor product is the monoidal category. It captures the algebraic essence of tensoring, without making any specific reference to what is being tensored. Thus, all tensor products can be expressed as an application of the monoidal category to some particular setting, acting on some particular objects.