Tensor product

For generalizations of this concept, see Tensor product of modules and Tensor product (disambiguation).

In mathematics, the tensor product ${\displaystyle V\otimes W}$ of two vector spaces V and W (over the same field) is a vector space to which is associated a bilinear map ${\displaystyle V\times W\rightarrow V\otimes W}$ that maps a pair ${\displaystyle (v,w),\ v\in V,w\in W}$ to an element of ${\displaystyle V\otimes W}$ denoted ⁠${\displaystyle v\otimes w}$⁠.

An element of the form ${\displaystyle v\otimes w}$ is called the tensor product of v and w. An element of ${\displaystyle V\otimes W}$ is a tensor, and the tensor product of two vectors is sometimes called an elementary tensor or a decomposable tensor. The elementary tensors span ${\displaystyle V\otimes W}$ in the sense that every element of ${\displaystyle V\otimes W}$ is a sum of elementary tensors. If bases are given for V and W, a basis of ${\displaystyle V\otimes W}$ is formed by all tensor products of a basis element of V and a basis element of W.

The tensor product of two vector spaces captures the properties of all bilinear maps in the sense that a bilinear map from ${\displaystyle V\times W}$ into another vector space Z factors uniquely through a linear map ${\displaystyle V\otimes W\to Z}$ (see Universal property).

Tensor products are used in many application areas, including physics and engineering. For example, in general relativity, the gravitational field is described through the metric tensor, which is a tensor field with one tensor at each point of the space-time manifold, and each belonging to the tensor product of the cotangent space at the point with itself.

The tensor product of two vector spaces is a vector space that is defined up to an isomorphism. There are several equivalent ways to define it. Most consist of defining explicitly a vector space that is called a tensor product, and, generally, the equivalence proof results almost immediately from the basic properties of the vector spaces that are so defined.

The tensor product can also be defined through a universal property; see § Universal property, below. As for every universal property, all objects that satisfy the property are isomorphic through a unique isomorphism that is compatible with the universal property. When this definition is used, the other definitions may be viewed as constructions of objects satisfying the universal property and as proofs that there are objects satisfying the universal property, that is that tensor products exist.

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

We can equivalently 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

Universal property of tensor product: if h is bilinear, there is a unique linear map ~h that makes the diagram commutative (that is, h = ~h ∘ φ).

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 ${\displaystyle {\otimes }:(v,w)\mapsto v\otimes w}$ 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 ${\displaystyle h={\tilde {h}}\circ {\otimes }}$ (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.^[1]

Theorem — Let ⁠${\displaystyle X,Y}$⁠, and ${\displaystyle Z}$ be complex vector spaces and let ${\displaystyle T:X\times Y\to Z}$ be a bilinear map. Then ${\displaystyle (Z,T)}$ is a tensor product of ${\displaystyle X}$ and ${\displaystyle Y}$ if and only if^[1] the image of ${\displaystyle T}$ spans all of ${\displaystyle Z}$ (that is, ⁠${\displaystyle \operatorname {span} \;T(X\times Y)=Z}$⁠), and also ${\displaystyle X}$ and ${\displaystyle Y}$ are ${\displaystyle T}$-linearly disjoint, which by definition means that for all positive integers ${\displaystyle n}$ and all elements ${\displaystyle x_{1},\ldots ,x_{n}\in X}$ and ${\displaystyle y_{1},\ldots ,y_{n}\in Y}$ such that ⁠${\displaystyle \sum _{i=1}^{n}T\left(x_{i},y_{i}\right)=0}$⁠,

1. if all ${\displaystyle x_{1},\ldots ,x_{n}}$ are linearly independent then all ${\displaystyle y_{i}}$ are ⁠${\displaystyle 0}$⁠, and
2. if all ${\displaystyle y_{1},\ldots ,y_{n}}$ are linearly independent then all ${\displaystyle x_{i}}$ are ⁠${\displaystyle 0}$⁠.

Equivalently, ${\displaystyle X}$ and ${\displaystyle Y}$ are ${\displaystyle T}$-linearly disjoint if and only if for all linearly independent sequences ${\displaystyle x_{1},\ldots ,x_{m}}$ in ${\displaystyle X}$ and all linearly independent sequences ${\displaystyle y_{1},\ldots ,y_{n}}$ in ⁠${\displaystyle Y}$⁠, the vectors ${\displaystyle \left\{T\left(x_{i},y_{j}\right):1\leq i\leq m,1\leq j\leq n\right\}}$ are linearly independent.

For example, it follows immediately that if ${\displaystyle m}$ and ${\displaystyle n}$ are positive integers then ${\displaystyle Z:=\mathbb {C} ^{mn}}$ and the bilinear map ${\displaystyle T:\mathbb {C} ^{m}\times \mathbb {C} ^{n}\to \mathbb {C} ^{mn}}$ defined by sending ${\displaystyle (x,y)=\left(\left(x_{1},\ldots ,x_{m}\right),\left(y_{1},\ldots ,y_{n}\right)\right)}$ to ${\displaystyle \left(x_{i}y_{j}\right)_{\stackrel {i=1,\ldots ,m}{j=1,\ldots ,n}}}$ form a tensor product of ${\displaystyle X:=\mathbb {C} ^{m}}$ and ⁠${\displaystyle Y:=\mathbb {C} ^{n}}$⁠.^[2] Often, this map ${\displaystyle T}$ will be denoted by ${\displaystyle \,\otimes \,}$ so that ${\displaystyle x\otimes y\;:=\;T(x,y)}$ denotes this bilinear map's value at ⁠${\displaystyle (x,y)\in X\times Y}$⁠.

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 ⁠${\displaystyle Y}$⁠.^[2]

Dimension

If V and W are vectors 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.

"Tensor product of linear maps" redirects here. For the generalization for modules, see Tensor product of modules § Tensor product of linear maps and a change of base ring.

Given a linear map ⁠${\displaystyle f:U\to V}$⁠, and a vector space W, the tensor product:

${\displaystyle f\otimes W:U\otimes W\to V\otimes W}$

is the unique linear map such that:

${\displaystyle (f\otimes W)(}$