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Tensor product

From Wikipedia, the free encyclopedia

Mathematical operation on vector spaces

For generalizations of this concept, seeTensor product of modules andTensor product (disambiguation).

Inmathematics, thetensor product $V\otimes W$ of twovector spaces $V {\displaystyle V}$ and $W {\displaystyle W}$ (over the samefield) is a vector space to which is associated abilinear map $V\times W\rightarrow V\otimes W$ that maps a pair $(v,w),\ v\in V,w\in W$ to an element of $V\otimes W$ denoted⁠ $v\otimes w$ ⁠.^[1]

An element of the form $v\otimes w$ is called thetensor product of $v {\displaystyle v}$ and $w {\displaystyle w}$ . An element of $V\otimes W$ is atensor, and the tensor product of two vectors is sometimes called anelementary tensor or adecomposable tensor. The elementary tensorsspan $V\otimes W$ in the sense that every element of $V\otimes W$ is a sum of elementary tensors. Ifbases are given for $V {\displaystyle V}$ and $W {\displaystyle W}$ , a basis of $V\otimes W$ is formed by all tensor products of a basis element of $V {\displaystyle V}$ and a basis element of $W {\displaystyle W}$ .

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

Tensor products are used in many application areas, including physics and engineering. For example, ingeneral relativity, thegravitational field is described through themetric tensor, which is atensor field with one tensor at each point of thespace-time manifold, and each belonging to the tensor product of thecotangent space at the point with itself.

Definitions and constructions

Thetensor product of two vector spaces is a vector space that is definedup to anisomorphism. 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 auniversal property; see§ Universal property, below. As for every universal property, allobjects 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

LetV andW be twovector spaces over afieldF, with respectivebases $B_{V}$ and⁠ $B_{W}$ ⁠.

Thetensor product $V\otimes W$ ofV andW is a vector space that has as a basis the set of all $v\otimes w$ with $v\in B_{V}$ and⁠ $w\in B_{W}$ ⁠. This definition can be formalized in the following way (this formalization is rarely used in practice, as the preceding informal definition is generally sufficient): $V\otimes W$ is the set of thefunctions from theCartesian product $B_{V}\times B_{W}$ toF that have a finite number of nonzero values. Thepointwise operations make $V\otimes W$ a vector space. The function that maps $(v,w)$ to1 and the other elements of $B_{V}\times B_{W}$ to0 is denoted⁠ $v\otimes w$ ⁠.

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

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

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

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

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

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

If arranged into a rectangular array, thecoordinate vector of $x\otimes y$ is theouter product of the coordinate vectors of $x {\displaystyle x}$ and⁠ $y {\displaystyle y}$ ⁠. 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 acanonical 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 thetensor product of modules over aring.

As a quotient space

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

LetV andW be twovector spaces over afieldF.

One considers first a vector spaceL that has theCartesian product $V\times W$ as abasis. That is, the basis elements ofL are thepairs $(v,w)$ with $v\in V$ and⁠ $w\in W$ ⁠. To get such a vector space, one can define it as the vector space of thefunctions $V\times W\to F$ that have a finite number of nonzero values and identifying $(v,w)$ with the function that takes the value1 on $(v,w)$ and0 otherwise.

LetR be thelinear subspace ofL that is spanned by the relations that the tensor product must satisfy. More precisely,R isspanned by the elements of one of the forms:

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

Then, the tensor product is defined as thequotient space:

V\otimes W=L/R,

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

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

Universal property

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

In this section, theuniversal property satisfied by the tensor product is described. As for every universal property, two objects that satisfy the property are related by a uniqueisomorphism. 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 abilinear map is a function that isseparatelylinear in each of its arguments):

Thetensor product of two vector spacesV andW is a vector space denoted as⁠

V\otimes W

⁠, together with a bilinear map

{\varphi }:(v,w)\mapsto v\otimes w

from

V\times W

to⁠

V\otimes W

⁠, such that, for every bilinear map⁠

h:V\times W\to Z

⁠, there is aunique linear map⁠

{\tilde {h}}:V\otimes W\to Z

⁠, such that

h={\tilde {h}}\circ {\varphi }

(that is,

h(v,w)={\tilde {h}}(v\otimes w)

for every

v\in V

and⁠

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]

Theorem—Let⁠ $X, Y {\displaystyle X,Y}$ ⁠, and $Z {\displaystyle Z}$ be complex vector spaces and let $T:X\times Y\to Z$ be a bilinear map. Then $(Z,T)$ is a tensor product of $X {\displaystyle X}$ and $Y {\displaystyle Y}$ if and only if^[2] the image of $T {\displaystyle T}$ spans all of $Z {\displaystyle Z}$ (that is,⁠ $\operatorname {span} \;T(X\times Y)=Z$ ⁠), and also $X {\displaystyle X}$ and $Y {\displaystyle Y}$ are $T {\displaystyle T}$ -linearly disjoint, which by definition means that for all positive integers $n {\displaystyle n}$ and all elements $x_{1},\ldots ,x_{n}\in X$ and $y_{1},\ldots ,y_{n}\in Y$ such that⁠ $\sum _{i=1}^{n}T\left(x_{i},y_{i}\right)=0$ ⁠,

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

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

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

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

Properties

Dimension

IfV andW are vector spaces of finitedimension, then $V\otimes W$ is finite-dimensional, and its dimension is the product of the dimensions ofV andW.

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

Associativity

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

(U\otimes V)\otimes W\cong U\otimes (V\otimes W),

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

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 $V {\displaystyle V}$ and $W {\displaystyle W}$ iscommutative in the sense that there is a canonical isomorphism:

V\otimes W\cong W\otimes V,

that maps $v\otimes w$ to⁠ $w\otimes v$ ⁠.

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

The map $x\otimes y\mapsto y\otimes x$ from $V\otimes V$ to itself induces a linearautomorphism that is called abraiding map. More generally and as usual (seetensor algebra), let $V^{\otimes n}$ denote the tensor product ofn copies of the vector spaceV. For everypermutations of the firstn positive integers, the map:

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

induces a linear automorphism of⁠ $V^{\otimes n}\to V^{\otimes n}$ ⁠, which is called a braiding map.

Tensor product of linear maps

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

Given a linear map⁠ $f:U\to V$ ⁠, and a vector spaceW, thetensor product:

f\otimes W:U\otimes W\to V\otimes W

is the unique linear map such that:

(f\otimes W)(u\otimes w)=f(u)\otimes w.

The tensor product $W\otimes f$ is defined similarly.

Given two linear maps $f:U\to V$ and⁠ $g:W\to Z$ ⁠, their tensor product:

f\otimes g:U\otimes W\to V\otimes Z

is the unique linear map that satisfies:

(f\otimes g)(u\otimes w)=f(u)\otimes g(w).

One has:

f\otimes g=(f\otimes Z)\circ (U\otimes g)=(V\otimes g)\circ (f\otimes W).

In terms ofcategory theory, this means that the tensor product is abifunctor from thecategory of vector spaces to itself.^[4]

Iff andg are bothinjective orsurjective, then the same is true for all above defined linear maps. In particular, the tensor product with a vector space is anexact functor; this means that everyexact sequence is mapped to an exact sequence (tensor products of modules do not transform injections into injections, but they areright exact functors).

By choosing bases of all vector spaces involved, the linear mapsf andg can be represented bymatrices. Then, depending on how the tensor $v\otimes w$ is vectorized, the matrix describing the tensor product $f\otimes g$ is theKronecker product of the two matrices. For example, ifV,X,W, andU above are all two-dimensional and bases have been fixed for all of them, andf andg are given by the matrices: $A={\begin{bmatrix}a_{1,1}&a_{1,2}\\a_{2,1}&a_{2,2}\\\end{bmatrix}},\qquad B={\begin{bmatrix}b_{1,1}&b_{1,2}\\b_{2,1}&b_{2,2}\\\end{bmatrix}},$ respectively, then the tensor product of these two matrices is: ${\begin{aligned}{\begin{bmatrix}a_{1,1}&a_{1,2}\\a_{2,1}&a_{2,2}\\\end{bmatrix}}\otimes {\begin{bmatrix}b_{1,1}&b_{1,2}\\b_{2,1}&b_{2,2}\\\end{bmatrix}}&={\begin{bmatrix}a_{1,1}{\begin{bmatrix}b_{1,1}&b_{1,2}\\b_{2,1}&b_{2,2}\\\end{bmatrix}}&a_{1,2}{\begin{bmatrix}b_{1,1}&b_{1,2}\\b_{2,1}&b_{2,2}\\\end{bmatrix}}\\[3pt]a_{2,1}{\begin{bmatrix}b_{1,1}&b_{1,2}\\b_{2,1}&b_{2,2}\\\end{bmatrix}}&a_{2,2}{\begin{bmatrix}b_{1,1}&b_{1,2}\\b_{2,1}&b_{2,2}\\\end{bmatrix}}\\\end{bmatrix}}\\&={\begin{bmatrix}a_{1,1}b_{1,1}&a_{1,1}b_{1,2}&a_{1,2}b_{1,1}&a_{1,2}b_{1,2}\\a_{1,1}b_{2,1}&a_{1,1}b_{2,2}&a_{1,2}b_{2,1}&a_{1,2}b_{2,2}\\a_{2,1}b_{1,1}&a_{2,1}b_{1,2}&a_{2,2}b_{1,1}&a_{2,2}b_{1,2}\\a_{2,1}b_{2,1}&a_{2,1}b_{2,2}&a_{2,2}b_{2,1}&a_{2,2}b_{2,2}\\\end{bmatrix}}.\end{aligned}}$

The resultant rank is at most 4, and thus the resultant dimension is 4.rank here denotes thetensor rank i.e. the number of requisite indices (while thematrix rank counts the number of degrees of freedom in the resulting array).⁠ $\operatorname {Tr} A\otimes B=\operatorname {Tr} A\times \operatorname {Tr} B$ ⁠.

Adyadic product is the special case of the tensor product between two vectors of the same dimension.

General tensors

See also:Tensor

For non-negative integersr ands a type $(r,s)$ tensor on a vector spaceV is an element of: $T_{s}^{r}(V)=\underbrace {V\otimes \cdots \otimes V} _{r}\otimes \underbrace {V^{*}\otimes \cdots \otimes V^{*}} _{s}=V^{\otimes r}\otimes \left(V^{*}\right)^{\otimes s}.$ Here $V^{*}$ is thedual vector space (which consists of alllinear mapsf fromV to the ground fieldK).

There is a product map, called the(tensor) product of tensors:^[5] $T_{s}^{r}(V)\otimes _{K}T_{s'}^{r'}(V)\to T_{s+s'}^{r+r'}(V).$

It is defined by grouping all occurring "factors"V together: writing $v_{i}$ for an element ofV and $f_{i}$ for an element of the dual space: $(v_{1}\otimes f_{1})\otimes (v'_{1})=v_{1}\otimes v'_{1}\otimes f_{1}.$

IfV is finite dimensional, then picking a basis ofV and the correspondingdual basis of $V^{*}$ naturally induces a basis of $T_{s}^{r}(V)$ (this basis is described in thearticle on Kronecker products). In terms of these bases, thecomponents of a (tensor) product of two (or more)tensors can be computed. For example, ifF andG are twocovariant tensors of ordersm andn respectively (i.e. $F\in T_{m}^{0}$ and⁠ $G\in T_{n}^{0}$ ⁠), then the components of their tensor product are given by:^[6] $(F\otimes G)_{i_{1}i_{2}\cdots i_{m+n}}=F_{i_{1}i_{2}\cdots i_{m}}G_{i_{m+1}i_{m+2}i_{m+3}\cdots i_{m+n}}.$

Thus, the components of the tensor product of two tensors are the ordinary product of the components of each tensor. Another example: letU be a tensor of type(1, 1) with components⁠ $U_{\beta }^{\alpha }$ ⁠, and letV be a tensor of type $(1,0)$ with components⁠ $V^{\gamma }$ ⁠. Then: $\left(U\otimes V\right)^{\alpha }{}_{\beta }{}^{\gamma }=U^{\alpha }{}_{\beta }V^{\gamma }$ and: $(V\otimes U)^{\mu \nu }{}_{\sigma }=V^{\mu }U^{\nu }{}_{\sigma }.$

Tensors equipped with their product operation form analgebra, called thetensor algebra.

Evaluation map and tensor contraction

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

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

On the other hand, if $V {\displaystyle V}$ isfinite-dimensional, there is a canonical map in the other direction (called thecoevaluation map): ${\begin{cases}K\to V\otimes V^{*}\\\lambda \mapsto \sum _{i}\lambda v_{i}\otimes v_{i}^{*}\end{cases}}$ where $v_{1},\ldots ,v_{n}$ is any basis of⁠ $V {\displaystyle V}$ ⁠, and $v_{i}^{*}$ is itsdual 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 $T_{s}^{r}(V)$ may be naturally viewed as a module for theLie algebra $\mathrm {End} (V)$ by means of the diagonal action: for simplicity let us assume⁠ $r=s=1$ ⁠, then, for each⁠ $u\in \mathrm {End} (V)$ ⁠, $u(a\otimes b)=u(a)\otimes b-a\otimes u^{*}(b),$ where $u^{*}\in \mathrm {End} \left(V^{*}\right)$ is thetranspose ofu, that is, in terms of the obvious pairing on⁠ $V\otimes V^{*}$ ⁠, $\langle u(a),b\rangle =\langle a,u^{*}(b)\rangle .$

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

Under this isomorphism, everyu in $\mathrm {End} (V)$ may be first viewed as an endomorphism of $T_{1}^{1}(V)$ and then viewed as an endomorphism of⁠ $\mathrm {End} (V)$ ⁠. In fact it is theadjoint representationad(u) of⁠ $\mathrm {End} (V)$ ⁠.

Linear maps as tensors

Given two finite dimensional vector spacesU,V over the same fieldK, denote thedual space ofU asU*, and theK-vector space of all linear maps fromU toV asHom(U,V). There is an isomorphism: $U^{*}\otimes V\cong \mathrm {Hom} (U,V),$ defined by an action of the pure tensor $f\otimes v\in U^{*}\otimes V$ on an element of⁠ $U {\displaystyle U}$ ⁠, $(f\otimes v)(u)=f(u)v.$

Its "inverse" can be defined using a basis $\{u_{i}\}$ and its dual basis $\{u_{i}^{*}\}$ as in the section "Evaluation map and tensor contraction" above: ${\begin{cases}\mathrm {Hom} (U,V)\to U^{*}\otimes V\\F\mapsto \sum _{i}u_{i}^{*}\otimes F(u_{i}).\end{cases}}$

This result implies: $\dim(U\otimes V)=\dim(U)\dim(V),$ which automatically gives the important fact that $\{u_{i}\otimes v_{j}\}$ forms a basis of $U\otimes V$ where $\{u_{i}\},\{v_{j}\}$ are bases ofU andV.

Furthermore, given three vector spacesU,V,W the tensor product is linked to the vector space ofall linear maps, as follows: $\mathrm {Hom} (U\otimes V,W)\cong \mathrm {Hom} (U,\mathrm {Hom} (V,W)).$ This is an example ofadjoint functors: the tensor product is "left adjoint" to Hom.

Tensor products of modules over a ring

Main article:Tensor product of modules

The tensor product of twomodulesA andB over acommutativeringR is defined in exactly the same way as the tensor product of vector spaces over a field: $A\otimes _{R}B:=F(A\times B)/G,$ where now $F(A\times B)$ is thefreeR-module generated by the cartesian product andG is theR-module generated bythese relations.

More generally, the tensor product can be defined even if the ring isnon-commutative. In this caseA has to be a right-R-module andB is a left-R-module, and instead of the last two relations above, the relation: $(ar,b)\sim (a,rb)$ is imposed. IfR is non-commutative, this is no longer anR-module, but just anabelian group.

The universal property also carries over, slightly modified: the map $\varphi :A\times B\to A\otimes _{R}B$ defined by $(a,b)\mapsto a\otimes b$ is amiddle linear map (referred to as "the canonical middle linear map"^[9]); that is, it satisfies:^[10] ${\begin{aligned}\varphi (a+a',b)&=\varphi (a,b)+\varphi (a',b)\\\varphi (a,b+b')&=\varphi (a,b)+\varphi (a,b')\\\varphi (ar,b)&=\varphi (a,rb)\end{aligned}}$

The first two properties makeφ a bilinear map of theabelian group⁠ $A\times B$ ⁠. For any middle linear map $\psi$ of⁠ $A\times B$ ⁠, a unique group homomorphismf of $A\otimes _{R}B$ satisfies⁠ $\psi =f\circ \varphi$ ⁠, and this property determines $\varphi$ within group isomorphism. See themain article for details.

Tensor product of modules over a non-commutative ring

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

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

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

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

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

Computing the tensor product

For vector spaces, the tensor product $V\otimes W$ is quickly computed since bases ofV ofW immediately determine a basis of⁠ $V\otimes W$ ⁠, 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 withZ/nZ is given by: $M\otimes _{\mathbf {Z} }\mathbf {Z} /n\mathbf {Z} =M/nM.$

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

Here⁠ $N^{J}=\oplus _{j\in J}N$ ⁠, and the map $N^{J}\to N^{I}$ is determined by sending some $n\in N$ in thejth copy of $N^{J}$ to $a_{ij}n$ (in⁠ $N^{I}$ ⁠). Colloquially, this may be rephrased by saying that a presentation ofM gives rise to a presentation of⁠ $M\otimes _{R}N$ ⁠. This is referred to by saying that the tensor product is aright exact functor. It is not in general left exact, that is, given an injective map ofR-modules⁠ $M_{1}\to M_{2}$ ⁠, the tensor product: $M_{1}\otimes _{R}N\to M_{2}\otimes _{R}N$ is not usually injective. For example, tensoring the (injective) map given by multiplication withn,n :Z →Z withZ/nZ yields the zero map0 :Z/nZ →Z/nZ, which is not injective. HigherTor functors measure the defect of the tensor product being not left exact. All higher Tor functors are assembled in thederived tensor product.

Tensor product of algebras

Main article:Tensor product of algebras

LetR be a commutative ring. The tensor product ofR-modules applies, in particular, ifA andB areR-algebras. In this case, the tensor product $A\otimes _{R}B$ is anR-algebra itself by putting: $(a_{1}\otimes b_{1})\cdot (a_{2}\otimes b_{2})=(a_{1}\cdot a_{2})\otimes (b_{1}\cdot b_{2}).$ For example: $R[x]\otimes _{R}R[y]\cong R[x,y].$

A particular example is whenA andB are fields containing a common subfieldR. Thetensor product of fields is closely related toGalois theory: if, say,A =R[x] /f(x), wheref is someirreducible polynomial with coefficients inR, the tensor product can be calculated as: $A\otimes _{R}B\cong B[x]/f(x)$ where nowf is interpreted as the same polynomial, but with its coefficients regarded as elements ofB. In the larger fieldB, the polynomial may become reducible, which brings in Galois theory. For example, ifA =B is aGalois extension ofR, then: $A\otimes _{R}A\cong A[x]/f(x)$ is isomorphic (as anA-algebra) to the⁠ $A^{\operatorname {deg} (f)}$ ⁠.

Eigenconfigurations of tensors

Squarematrices $A {\displaystyle A}$ with entries in afield $K {\displaystyle K}$ representlinear maps ofvector spaces, say⁠ $K^{n}\to K^{n}$ ⁠, and thus linear maps $\psi :\mathbb {P} ^{n-1}\to \mathbb {P} ^{n-1}$ ofprojective spaces over⁠ $K {\displaystyle K}$ ⁠. If $A {\displaystyle A}$ isnonsingular then $\psi$ iswell-defined everywhere, and theeigenvectors of $A {\displaystyle A}$ correspond to the fixed points of⁠ $\psi$ ⁠. Theeigenconfiguration of $A {\displaystyle A}$ consists of $n {\displaystyle n}$ points in⁠ $\mathbb {P} ^{n-1}$ ⁠, provided $A {\displaystyle A}$ is generic and $K {\displaystyle K}$ isalgebraically closed. The fixed points of nonlinear maps are the eigenvectors of tensors. Let $A=(a_{i_{1}i_{2}\cdots i_{d}})$ be a $d {\displaystyle d}$ -dimensional tensor of format $n\times n\times \cdots \times n$ with entries $(a_{i_{1}i_{2}\cdots i_{d}})$ lying in an algebraically closed field $K {\displaystyle K}$ ofcharacteristic zero. Such a tensor $A\in (K^{n})^{\otimes d}$ definespolynomial maps $K^{n}\to K^{n}$ and $\mathbb {P} ^{n-1}\to \mathbb {P} ^{n-1}$ with coordinates: $\psi _{i}(x_{1},\ldots ,x_{n})=\sum _{j_{2}=1}^{n}\sum _{j_{3}=1}^{n}\cdots \sum _{j_{d}=1}^{n}a_{ij_{2}j_{3}\cdots j_{d}}x_{j_{2}}x_{j_{3}}\cdots x_{j_{d}}\;\;{\mbox{for }}i=1,\ldots ,n$

Thus each of the $n {\displaystyle n}$ coordinates of $\psi$ is ahomogeneous polynomial $\psi _{i}$ of degree $d-1$ in⁠ $\mathbf {x} =\left(x_{1},\ldots ,x_{n}\right)$ ⁠. The eigenvectors of $A {\displaystyle A}$ are the solutions of the constraint: ${\mbox{rank}}{\begin{pmatrix}x_{1}&x_{2}&\cdots &x_{n}\\\psi _{1}(\mathbf {x} )&\psi _{2}(\mathbf {x} )&\cdots &\psi _{n}(\mathbf {x} )\end{pmatrix}}\leq 1$ and the eigenconfiguration is given by thevariety of the $2\times 2$ minors of this matrix.^[11]

Other examples of tensor products

Topological tensor products

Main articles:Topological tensor product andTensor product of Hilbert spaces

Hilbert spaces generalize finite-dimensional vector spaces to arbitrary dimensions. There isan analogous operation, also called the "tensor product," that makes Hilbert spaces asymmetric monoidal category. It is essentially constructed as themetric 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 toHilbert–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 atopological 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

Main article:Graded vector space § Operations on graded vector spaces

Some vector spaces can be decomposed intodirect 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

Main article:Tensor product of representations

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

Tensor product of quadratic forms

Main article:Tensor product of quadratic forms

Tensor product of multilinear forms

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

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

Tensor product of sheaves of modules

Main article:Sheaf of modules

Tensor product of line bundles

Main article:Vector bundle § Operations on vector bundles

See also:Tensor product bundle

Tensor product of fields

Main article:Tensor product of fields

Tensor product of graphs

Main article: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 thecategory-theoretic product in the category of graphs andgraph homomorphisms. However it is actually theKronecker tensor product of theadjacency matrices of the graphs. Compare also the sectionTensor product of linear maps above.

Monoidal categories

The most general setting for the tensor product is themonoidal 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.

Quotient algebras

A number of important subspaces of thetensor algebra can be constructed asquotients: these include theexterior algebra, thesymmetric algebra, theClifford algebra, theWeyl algebra, and theuniversal enveloping algebra in general.

The exterior algebra is constructed from theexterior product. Given a vector spaceV, the exterior product $V\wedge V$ is defined as: $V\wedge V:=V\otimes V{\big /}\{v\otimes v\mid v\in V\}.$

When the underlying field ofV does not have characteristic 2, then this definition is equivalent to: $V\wedge V:=V\otimes V{\big /}{\bigl \{}v_{1}\otimes v_{2}+v_{2}\otimes v_{1}\mid (v_{1},v_{2})\in V^{2}{\bigr \}}.$

The image of $v_{1}\otimes v_{2}$ in the exterior product is usually denoted $v_{1}\wedge v_{2}$ and satisfies, by construction,⁠ $v_{1}\wedge v_{2}=-v_{2}\wedge v_{1}$ ⁠. Similar constructions are possible for $V\otimes \dots \otimes V$ (n factors), giving rise to⁠ $\Lambda ^{n}V$ ⁠, thenthexterior power ofV. The latter notion is the basis ofdifferentialn-forms.

The symmetric algebra is constructed in a similar manner, from thesymmetric product: $V\odot V:=V\otimes V{\big /}{\bigl \{}v_{1}\otimes v_{2}-v_{2}\otimes v_{1}\mid (v_{1},v_{2})\in V^{2}{\bigr \}}.$

More generally: $\operatorname {Sym} ^{n}V:=\underbrace {V\otimes \dots \otimes V} _{n}{\big /}(\dots \otimes v_{i}\otimes v_{i+1}\otimes \dots -\dots \otimes v_{i+1}\otimes v_{i}\otimes \dots )$

That is, in the symmetric algebra two adjacent vectors (and therefore all of them) can be interchanged. The resulting objects are calledsymmetric tensors.

Tensor product in programming

Array programming languages

Array programming languages may have this pattern built in. For example, inAPL the tensor product is expressed as○.× (for exampleA ○.× B orA ○.× B ○.× C). InJ the tensor product is the dyadic form of*/ (for examplea */ b ora */ b */ c).

J's treatment also allows the representation of some tensor fields, asa andb may be functions instead of constants. This product of two functions is a derived function, and ifa andb aredifferentiable, thena */ b is differentiable.

However, these kinds of notation are not universally present in array languages. Other array languages may require explicit treatment of indices (for example,MATLAB), and/or may not supporthigher-order functions such as theJacobian derivative (for example,Fortran/APL).

See also

Look uptensor product in Wiktionary, the free dictionary.

Dyadics – Second order tensor in vector algebra
Extension of scalars – Operation in algebraPages displaying short descriptions of redirect targets
Monoidal category – Category admitting tensor products
Tensor algebra – Universal construction in multilinear algebra
Tensor contraction – Operation in mathematics
Topological tensor product – Tensor product constructions for topological vector spaces

Notes

^Introduction to the Tensor Product
^^a ^bTrèves 2006, pp. 403–404.
^^a ^bTrèves 2006, pp. 407.
^Hazewinkel, Michiel; Gubareni, Nadezhda Mikhaĭlovna; Gubareni, Nadiya; Kirichenko, Vladimir V. (2004).Algebras, rings and modules. Springer. p. 100.ISBN 978-1-4020-2690-4.
^Bourbaki (1989), p. 244 defines the usage "tensor product ofx andy", elements of the respective modules.
^Analogous formulas also hold forcontravariant tensors, as well as tensors of mixed variance. Although in many cases such as when there is aninner product defined, the distinction is irrelevant.
^"The Coevaluation on Vector Spaces".The Unapologetic Mathematician. 2008-11-13.Archived from the original on 2017-02-02. Retrieved2017-01-26.
^SeeCompact closed category.
^Hungerford, Thomas W. (1974).Algebra. Springer.ISBN 0-387-90518-9.
^Chen, Jungkai Alfred (Spring 2004),"Tensor product"(PDF),Advanced Algebra II (lecture notes), National Taiwan University,archived(PDF) from the original on 2016-03-04{{citation}}: CS1 maint: location missing publisher (link)
^Abo, H.;Seigal, A.;Sturmfels, B. (2015). "Eigenconfigurations of Tensors".arXiv:1505.05729 [math.AG].
^Garrett, Paul (July 22, 2010)."Non-existence of tensor products of Hilbert spaces"(PDF).
^Kadison, Richard V.; Ringrose, John R. (1997).Fundamentals of the theory of operator algebras.Graduate Studies in Mathematics. Vol. I. Providence, R.I.:American Mathematical Society. Thm. 2.6.4.ISBN 978-0-8218-0819-1.MR 1468229.
^Tu, L. W. (2010).An Introduction to Manifolds. Universitext. Springer. p. 25.ISBN 978-1-4419-7399-3.

References

Bourbaki, Nicolas (1989).Elements of mathematics, Algebra I. Springer-Verlag.ISBN 3-540-64243-9.
Gowers, Timothy."How to lose your fear of tensor products".Archived from the original on 7 May 2021.
Grillet, Pierre A. (2007).Abstract Algebra. Springer Science+Business Media, LLC.ISBN 978-0387715674.
Halmos, Paul (1974).Finite dimensional vector spaces. Springer.ISBN 0-387-90093-4.
Hungerford, Thomas W. (2003).Algebra. Springer.ISBN 0387905189.
Lang, Serge (2002),Algebra,Graduate Texts in Mathematics, vol. 211 (Revised third ed.), New York: Springer-Verlag,ISBN 978-0-387-95385-4,MR 1878556,Zbl 0984.00001
Mac Lane, S.;Birkhoff, G. (1999).Algebra. AMS Chelsea.ISBN 0-8218-1646-2.
Aguiar, M.; Mahajan, S. (2010).Monoidal functors, species and Hopf algebras. CRM Monograph Series Vol 29.ISBN 978-0-8218-4776-3.
Trèves, François (2006) [1967].Topological Vector Spaces, Distributions and Kernels. Mineola, N.Y.: Dover Publications.ISBN 978-0-486-45352-1.OCLC 853623322.
"Bibliography on the nonabelian tensor product of groups".

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