Tensor product

In mathematics, the tensor product, denoted by ⊗, may be applied in different contexts to vectors, matrices, tensors, vector spaces, algebras, topological vector spaces, and modules, among many other structures or objects. In each case the significance of the symbol is the same: the most general bilinear operation. In some contexts, this product is also referred to as outer product. The term "tensor product" is also used in relation to monoidal categories. The $\boxtimes$ variant of ⊗ is used in control theory that expresses that the elements in the tensor product are vectors, matrices or tensors which define the vertexes of a given polytopic model, as in TP model transformation.

Tensor product of vector spaces

The tensor product of two vector spaces V and W over a field K is another vector space over K. It is denoted V ⊗_K W or V ⊗ W when the underlying field K is understood.

Prerequisite: the free vector space

The construction of V ⊗ W requires the notion of the free vector space F(S) on some set S. The elements of the vector space F(S) are expressions of the form

$a_{1}\cdot s_{1}+a_{2}\cdot s_{2}+\dots +a_{n}\cdot s_{n}.$

Here the coefficients $a_{1},\dots ,a_{n}$ are elements of the ground field K and the $s_{1},\dots ,s_{n}$ are arbitrary elements of S. The plus symbol and the dots are purely formal notation. Addition of such formal linear sums does not mean that elements of S are added, nor is any element of K actually multiplied with one of S. Instead, for example

$(a_{1}\cdot s_{1}+\dots +a_{n}\cdot s_{n})+(a'_{1}\cdot s_{1}+b'\cdot s')=(a_{1}+a'_{1})\cdot s_{1}+a_{2}\cdot s_{2}+\dots a_{n}\cdot s_{n}+b'\cdot s',$

if s' is different from all the elements appearing in the first summand. Moreover, the product (scalar multiplication) of the above formal linear sum with some element x in K is defined as

$(xa_{1})\cdot s_{1}+(xa_{2})\cdot s_{2}+\dots +(xa_{n})\cdot s_{n}.$

This concludes the definition of the vector space F(S). For example, if S has just 3 elements, then F(S) is a 3-dimensional vector space.

Definition

Given two vector spaces V and W, the Cartesian product V × W is the set consisting of pairs (v, w) with v in V and w in W. (This Cartesian product is also a vector space in its own right, but is regarded as a set, only, at this point.) The tensor product is defined as a certain quotient vector space of F(V × W), the free K-vector space on the Cartesian product.

The free vector space is called like this since different elements of the set V x W are not at all related in F(V × W). For example, given two different elements $v_{1},v_{2}\in V$, in the free vector space F(V x W), the equation

$(v_{1},w_{1})+(v_{2},w_{1})=(v_{1}+v_{2},w_{1})$

does not hold. Likewise, for x in K, the equation

$x(v_{1},w_{1})=(xv_{1},w_{1})$

does not hold in the free vector space. In this sense, even though V × W is a vector space itself, that structure is not reflected in F(V × W). The idea of the tensor product is to enforce these two relations and similar ones for the second variable. To do so, consider the subspace R of F(V × W) generated by the following elements. To simplify notation, it is customary to drop the coefficient if it is one, i.e., $(v,w)$ stands for $1\cdot (v,w)\in F(V\times W)$.

${\begin{aligned}(v_{1},w)+(v_{2},w)-(v_{1}+v_{2},w)\\(v,w_{1})+(v,w_{2})-(v,w_{1}+w_{2})\\c\cdot (v,w)-(cv,w)\\c\cdot (v,w)-(v,cw),\end{aligned}}$

where v, v₁ and v₂ are arbitrary elements of V, while w, w₁, and w₂ are vectors from W, and c is from the underlying field K.

The tensor product is defined as the vector space

$V\otimes W:=F(V\times W)/R.$

The tensor product of two vectors v and w is the equivalence class ((v,w) + R) of (v,w) in V ⊗ W. It is denoted v ⊗ w. The effect of dividing out R in the free vector space is the following equations hold in V ⊗ W:

${\begin{aligned}(v_{1}+v_{2})\otimes w&=v_{1}\otimes w+v_{2}\otimes w;\\v\otimes (w_{1}+w_{2})&=v\otimes w_{1}+v\otimes w_{2};\\cv\otimes w&=v\otimes cw=c(v\otimes w).\end{aligned}}$

Notation and examples

Given bases {v_i} and {w_i} for V and W respectively, the tensors {v_i ⊗ w_j} form a basis for V ⊗ W. The dimension of the tensor product therefore is the product of dimensions of the original spaces; for instance R^m ⊗ Rⁿ will have dimension mn.

Elements of V ⊗ W are sometimes referred to as tensors, although this term refers to many other related concepts as well. An element of V ⊗ W of the form v ⊗ w is called a pure or simple tensor. In general, an element of the tensor product space is not a pure tensor, but rather a finite linear combination of pure tensors. That is to say, if v₁ and v₂ are linearly independent, and w₁ and w₂ are also linearly independent, then v₁ ⊗ w₁ + v₂ ⊗ w₂ cannot be written as a pure tensor. The number of simple tensors required to express an element of a tensor product is called the tensor rank (not to be confused with tensor order, which is the number of spaces one has taken the product of, in this case 2; in notation, the number of indices), and for linear operators or matrices, thought of as (1,1) tensors (elements of the space V ⊗ V*), it agrees with matrix rank.

Tensor product of linear maps

The tensor product also operates on linear maps between vector spaces. Specifically, given two linear maps S : V → X and T : W → Y between vector spaces, the tensor product of the two linear maps S and T is a linear map

$S\otimes T:V\otimes W\rightarrow X\otimes Y$

defined by

$(S\otimes T)(v\otimes w)=S(v)\otimes T(w).$

In this way, the tensor product becomes a bifunctor from the category of vector spaces to itself, covariant in both arguments.

By choosing bases of all vector spaces involved, the linear maps S and T can be represented by matrices. Then, the matrix describing the tensor product $S\otimes T$ is the Kronecker product of the two matrices. For example, if V, X, W, and Y are all two-dimensional above and bases have been fixed for all of them, and S and T are given by the matrices ${\begin{bmatrix}a_{1,1}&a_{1,2}\\a_{2,1}&a_{2,2}\\\end{bmatrix}}$ and ${\begin{bmatrix}b_{1,1}&b_{1,2}\\b_{2,1}&b_{2,2}\\\end{bmatrix}}$, respectively, then the tensor product of these two matrices is

${\begin{bmatrix}a_{1,1}&a_{1,2}\\a_{2,1}&a_{2,2}\\\end{bmatrix}}\otimes {\begin{bmatrix}b_{1,1}&b_{1,2}\\b_{2,1}&b_{2,2}\\\end{bmatrix}}={\begin{bmatrix}a_{1,1}{\begin{bmatrix}b_{1,1}&b_{1,2}\\b_{2,1}&b_{2,2}\\\end{bmatrix}}&a_{1,2}{\begin{bmatrix}b_{1,1}&b_{1,2}\\b_{2,1}&b_{2,2}\\\end{bmatrix}}\\&\\a_{2,1}{\begin{bmatrix}b_{1,1}&b_{1,2}\\b_{2,1}&b_{2,2}\\\end{bmatrix}}&a_{2,2}{\begin{bmatrix}b_{1,1}&b_{1,2}\\b_{2,1}&b_{2,2}\\\end{bmatrix}}\\\end{bmatrix}}={\begin{bmatrix}a_{1,1}b_{1,1}&a_{1,1}b_{1,2}&a_{1,2}b_{1,1}&a_{1,2}b_{1,2}\\a_{1,1}b_{2,1}&a_{1,1}b_{2,2}&a_{1,2}b_{2,1}&a_{1,2}b_{2,2}\\a_{2,1}b_{1,1}&a_{2,1}b_{1,2}&a_{2,2}b_{1,1}&a_{2,2}b_{1,2}\\a_{2,1}b_{2,1}&a_{2,1}b_{2,2}&a_{2,2}b_{2,1}&a_{2,2}b_{2,2}\\\end{bmatrix}}.$

The resultant rank is at most 4, and the resultant dimension 16. Here rank denotes the tensor rank (number of requisite indices), while the matrix rank counts the number of degrees of freedom in the resulting array.

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

Universal property

The tensor product as defined above has a universal property. In general, a universal property means that some mathematical object is characterized by the maps with target (or, alternatively, domain) this object. In the context of linear algebra and vector spaces, the maps in question are required to be linear maps. The tensor product of vector spaces, as defined above, satisfies the following universal property: there is a bilinear (i.e., linear in each variable v and w) map $\varphi :V\times W\to V\otimes W$ such that given any other vector space Z together with a bilinear map $h:V\times W\to Z$, there is a unique linear map ${\tilde {h}}:V\otimes W\to Z$ verifying $h={\tilde {h}}\circ \varphi$. In this sense, $\varphi$ is the most general bilinear map that can be built from $V\times W$.

This characterization can simplify proving statements about the tensor product. For example, the tensor product is symmetric: that is, there is a canonical isomorphism:

$V\otimes W\cong W\otimes V.$

To construct, say, a map from left to right, it suffices, by the universal property, to give a bilinear map $V\times W\to W\otimes V.$ This is done by mapping (v, w) to $w\otimes v$. Constructing a map in the opposite direction is done similarly, as is checking that the two linear maps $V\otimes W\to W\otimes V$ and $W\otimes V\to V\otimes W$ are inverse to one another.

A similar reasoning can be used to show that the tensor product is associative, that is, there are natural isomorphisms

$V_{1}\otimes (V_{2}\otimes V_{3})\cong (V_{1}\otimes V_{2})\otimes V_{3}.$

Therefore, it is customary to omit the parentheses and write $V_{1}\otimes V_{2}\otimes V_{3}$.

Tensor powers and braiding

Let n be a non-negative integer. The nth tensor power of the vector space V is the n-fold tensor product of V with itself. That is

$V^{\otimes n}\;{\overset {\mathrm {def} }{=}}\;\underbrace {V\otimes \cdots \otimes V} _{n}.$

A permutation σ of the set {1, 2, ..., n} determines a mapping of the nth Cartesian power of V

$\sigma :V^{n}\to V^{n}$

defined by

$\sigma (v_{1},v_{2},\dots ,v_{n})=(v_{\sigma 1},v_{\sigma 2},\dots ,v_{\sigma n}).$

Let

$\varphi :V^{n}\to V^{\otimes n}$

be the natural multilinear embedding of the Cartesian power of V into the tensor power of V. Then, by the universal property, there is a unique isomorphism

$\tau _{\sigma }:V^{\otimes n}\to V^{\otimes n}$

such that

$\varphi \circ \sigma =\tau _{\sigma }\circ \varphi .$

The isomorphism τ_σ is called the braiding map associated to the permutation σ.

Product of tensors

See also: Classical treatment of tensors

For non-negative integers r and s a r,s-tensor on a vector space V is an element of

$T_{s}^{r}(V)=\underbrace {V\otimes \dots \otimes V} _{r}\otimes \underbrace {V^{*}\otimes \dots \otimes V^{*}} _{s}=V^{\otimes r}\otimes V^{*\otimes s}.$

Here $V^{*}$ is the dual vector space (which consists of all linear maps f from V to the ground field K).

There is a product map, called the (tensor) product of tensors

$T_{s}^{r}(V)\otimes _{K}T_{s'}^{r'}(V)\to T_{s+s'}^{r+r'}(V).$

It is defined by grouping all occuring "factors" V together: writing $v_{i}$ for an element of V and $f_{i}$ for elements of the dual space,

$(v_{1}\otimes f_{1})\otimes (v'_{1})=v_{1}\otimes v'_{1}\otimes f_{1}.$

Picking a basis of V and the corresponding dual basis of $V^{*}$, $T_{s}^{r}(V)$ is endowed with a natural basis (this basis is described in the article on Kronecker products). In terms of these bases, the components of a (tensor) product of two (or more) tensors can be computed. For example, if F and G are two covariant tensors of rank m and n (respectively) (i.e. F ∈ T_m⁰, and G ∈ T_n⁰), then the components of their tensor product are given by

$(F\otimes G)_{i_{1}i_{2}...i_{m+n}}=F_{i_{1}i_{2}...i_{m}}G_{i_{m+1}i_{m+2}i_{m+3}...i_{m+n}}.$

Thus, the components of the tensor product of two tensors are the ordinary product of the components of each tensor. Another example: let U be a tensor of type (1,1) with components U^α_β, and let V be a tensor of type (1,0) with components V^γ. Then

$U^{\alpha }{}_{\beta }V^{\gamma }=(U\otimes V)^{\alpha }{}_{\beta }{}^{\gamma }$

and

$V^{\mu }U^{\nu }{}_{\sigma }=(V\otimes U)^{\mu \nu }{}_{\sigma }.$

Relation to dual space

A particular example is the tensor product of some vector space V with its dual vector space $V^{*}$ (which consists of all linear maps f from V to the ground field K). In this case, there is a natural "evaluation" map

$V\otimes V^{*}\to K$

which on elementary tensors is defined by

$v\otimes f\mapsto f(v).$

The resulting map

$T_{s}^{r}(V)\to T_{s-1}^{r-1}(V)$

is called tensor contraction (for r, s > 0).

On the other hand, if V is finite-dimensional, there is a map in the other direction (called coevaluation)

$K\to V\otimes V^{*},\lambda \mapsto \sum _{i}\lambda v_{i}\otimes v_{i}^{*}.$

where $v_{1},\dots ,v_{n}$ is a basis of V, and $v_{i}^{*}$ is its dual basis. The interplay of evaluation and coevaluation map can be used to characterize finite-dimensional vector spaces without referring to bases.

Tensor product vs. Hom

Given three vector spaces U, V, W the tensor product is linked to the vector space of all linear maps, as follows:

$Hom(U\otimes V,W)\cong Hom(U,Hom(V,W)).$

Here $Hom(-,-)$ denotes the K-vector space of all linear maps. This is an example of adjoint functors: the tensor product is "left adjoint" to Hom.

Adjoint representation

The tensor $\scriptstyle T_{s}^{r}(V)$ may be naturally viewed as a module for the Lie algebra End(V) by means of the diagonal action: for simplicity let us assume r = s = 1, then, for each $\scriptstyle u\in \mathrm {End} (V)$,

$u(a\otimes b)=u(a)\otimes b-a\otimes u^{*}(b),$

where u* in End(V*) is the transpose of u, that is, in terms of the obvious pairing on V ⊗ V*,

$\langle u(a),b\rangle =\langle a,u^{*}(b)\rangle$.

There is a canonical isomorphism $\scriptstyle T_{1}^{1}(V)\rightarrow \mathrm {End} (V)$ given by

$(a\otimes b)(x)=\langle x,b\rangle a.$

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

Tensor products of modules over a ring

Main article: Tensor product of modules

The tensor product of two modules A and B over a commutative ring $R$ is defined in the exact same way as the tensor product of vector spaces over a field:

$M\otimes _{R}N:=F(A\times B)/G$

where now $F(A\times B)$ is the free R-module generated by the cartesian product and G is the R-module generated by the same relations as above.

More generally, the tensor product can be defined even if the ring is non-commutative (ab ≠ ba). In this case M has to be a right-R-module and N is a left-R-module, and instead of the last two relations above, the relation

$(ar,b)-(a,rb)$

is imposed. If R is non-commutative, this is no longer an R-module, but just an abelian group.

The universal property also carries over, slightly modified: the map $\phi :A\times B\to A\otimes _{R}B$ defined by $(a,b)\mapsto a\otimes b$ is a middle linear map (referred to as "canonical Middle Linear Map".); that is, it satisfies:

${\begin{aligned}\phi (a+a',b)=\phi (a,b)+\phi (a',b)\\\phi (a,b+b')=\phi (a,b)+\phi (a,b')\\\phi (ar,b)=\phi (a,rb)\end{aligned}}$

The first two properties make $\phi$ a homomorphism of the abelian group $A\times B$. For any middle linear map $\psi$ of $A\times B$, a unique group homomorphism $f$ of $A\otimes _{R}B$ satisfies $\psi =f\circ \phi$, and this property determines $\phi$ within group isomorphism. See the main article for details.

Computing the tensor product

For vector spaces, the tensor product $V\otimes W$ is quickly computed since bases of V of W 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/n is not a free abelian group (=Z-module). The tensor product with Z/n is given by

$M\otimes _{\mathbf {Z} }\mathbf {Z} /n=M/n.$

More generally, given a presentation of some R-module M, that is, a number of generators $m_{i}\in M,i\in I$ together with relations $\sum _{j\in J}a_{ji}m_{i}=0$, with $a_{ji}\in R$, the tensor product can be computed as the following cokernel:

$M\otimes _{R}N=\operatorname {coker} (N^{J}\rightarrow N^{I})$

Here $N^{J}:=\oplus _{j\in J}N$ and the map is determined by sending some $n\in N$ in the j-th copy of $N^{J}$ to $a_{ji}n$ (in $N^{I}$). Colloquially, this may be rephrased by saying that a presentation of M gives rise to a presentation of $M\otimes _{R}N$. 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 $M_{1}\to M_{2}$, 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 with n, $n:\mathbf {Z} \to \mathbf {Z}$ with $\mathbf {Z} /n$ yields the 0 map $0:\mathbf {Z} /n\to \mathbf {Z} /n$ which is not injective. Higher Tor functors measure the defect of the tensor product being not left exact.

Tensor product of algebras

Main article: Tensor product of algebras

Let R be a commutative ring. The tensor product of R-modules applies, in particular, if A and B are R-algebras. In this case, the tensor product $A\otimes _{R}B$ is an R-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,

$R[x]\otimes _{R}R[y]=R[x,y].$

A particular example is when A and B are fields containing a common subfield R. The tensor product of fields is closely related to Galois theory: if, say, $A=R[x]/f(x)$, where f is some irreducible polynomial with coefficients in R, the tensor product can be calculated as

$A\otimes _{R}B=B[x]/f(x)$

where now f is interpreted as the same polynomial, but with its coefficients regarded as elements of B. In the larger field B, the polynomial may become reducible, which brings in Galois theory. For example, if A = B is a Galois extension of R, then

$A\otimes _{R}A=A[x]/f(x)$

is isomorphic (as an A-algebra) to the $A^{deg(f)}$.

Other examples of tensor products

Tensor product of Hilbert spaces

Main article: Tensor product of Hilbert spaces

Topological tensor product

Main article: Topological tensor product

Tensor product of graded vector spaces

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

Tensor product of quadratic forms

Main article: Tensor product of quadratic forms

Tensor product of multilinear maps

Given multilinear maps $\scriptstyle f(x_{1},\dots ,x_{k})$ and $\scriptstyle g(x_{1},\dots ,x_{m})$ their tensor product is the multilinear function

$(f\otimes g)(x_{1},\dots ,x_{k+m})=f(x_{1},\dots ,x_{k})g(x_{k+1},\dots ,x_{k+m}).$

Tensor product of graphs

Main article: Tensor product of graphs

Applications

Exterior and symmetric algebra

Two notable constructions in linear algebra can be constructed as quotients of the tensor product: the exterior algebra and the symmetric algebra. For example, given a vector space V, the exterior product

$V\wedge V$

is defined as

$V\otimes V/(v_{1}\otimes v_{2}+v_{2}\otimes v_{1}{\text{ for all }}v_{1},v_{2}\in V).$

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 $\Lambda ^{n}V$, the n-th exterior power of V. The latter notion is the basis of differential n-forms.

The symmetric algebra is constructed in a similar manner:

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

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

Tensor product of line bundles

Main article: Vector bundle § Operations on vector bundles

Tensor product for computer programmers

Array programming languages

Array programming languages may have this pattern built in. For example, in APL the tensor product is expressed as $\scriptstyle \circ .\times$ (for example $\scriptstyle A\circ .\times B$ or $\scriptstyle A\circ .\times B\circ .\times C$). In J the tensor product is the dyadic form of */ (for example a */ b or a */ b */ c).

Note that J's treatment also allows the representation of some tensor fields, as a and b may be functions instead of constants. This product of two functions is a derived function, and if a and b are differentiable, then a*/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 support higher-order functions such as the Jacobian derivative (for example, Fortran/APL).

See also

• Dyadic product
• Extension of scalars
• Multilinear subspace learning
• Tensor algebra
• Tensor contraction
• Topological tensor product

References

• Bourbaki, Nicolas (1989), Elements of mathematics, Algebra I, Springer-Verlag, ISBN 3-540-64243-9.
• Halmos, Paul (1974), Finite dimensional vector spaces, Springer, ISBN 0-387-90093-4.
• 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.