In mathematics, the tensor product of two vector spaces V and W (over the same field) is a vector space to which is associated a bilinear map that maps a pair to an element of denoted

An element of the form is called the tensor product of v and w. An element of is a tensor, and the tensor product of two vectors is sometimes called an elementary tensor or a decomposable tensor. The elementary tensors span in the sense that every element of is a sum of elementary tensors. If bases are given for V and W, a basis of 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 into another vector space Z factors uniquely through a linear map (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 vector field of tensors, one at each point of the space-time manifold, and each belonging to the tensor product with itself of the cotangent space at the point.

Definitions and constructions Edit

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 Edit

Let V and W be two vector spaces over a field F, with respective bases   and  

The tensor product   of V and W is a vector space which has as a basis the set of all   with   and   This definition can be formalized in the following way (this formalization is rarely used in practice, as the preceding informal definition is generally sufficient):   is the set of the functions from the Cartesian product   to F that have a finite number of nonzero values. The pointwise operations make   a vector space. The function that maps   to 1 and the other elements of   to 0 is denoted  

The set   is straightforwardly a basis of   which is called the tensor product of the bases   and  

The tensor product of two vectors is defined from their decomposition on the bases. More precisely, if

are vectors decomposed on their respective bases, then the tensor product of x and y is

If arranged into a rectangular array, the coordinate vector of   is the outer product of the coordinate vectors of x and y. Therefore, the tensor product is a generalization of the outer product.

It is straightforward to verify that the map   is a bilinear map from   to  

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 Edit

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   as a basis. That is, the basis elements of L are the pairs   with   and   To get such a vector space, one can define it as the vector space of the functions   that have a finite number of nonzero values, and identifying   with the function that takes the value 1 on   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


where     and  

Then, the tensor product is defined as the quotient space


and the image of   in this quotient is denoted  

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 Edit

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

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   together with a bilinear map   from   to   such that, for every bilinear map   there is a unique linear map   such that   (that is,   for every   and  ).

Linearly disjoint Edit

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   and   be complex vector spaces and let   be a bilinear map. Then   is a tensor product of   and   if and only if[1] the image of   spans all of   (that is,  ), and also   and   are  -linearly disjoint, which by definition means that for all positive integers   and all elements   and   such that  

  1. if all   are linearly independent then all   are   and
  2. if all   are linearly independent then all   are  

Equivalently,   and   are  -linearly disjoint if and only if for all linearly independent sequences   in   and all linearly independent sequences   in   the vectors   are linearly independent.

For example, it follows immediately that if   and   are positive integers then   and the bilinear map   defined by sending   to   form a tensor product of   and  [2] Often, this map   will be denoted by   so that   denotes this bilinear map's value at  

As another example, suppose that   is the vector space of all complex-valued functions on a set   with addition and scalar multiplication defined pointwise (meaning that   is the map   and   is the map  ). Let   and   be any sets and for any   and   let   denote the function defined by   If   and   are vector subspaces then the vector subspace   of   together with the bilinear map

form a tensor product of   and  [2]

Properties Edit

Dimension Edit

If V and W are vectors spaces of finite dimension, then   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   is formed by taking all tensor products of a basis element of V and a basis element of W.

Associativity Edit

The tensor product is associative in the sense that, given three vector spaces   there is a canonical isomorphism


that maps   to  

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

Commutativity as vector space operation Edit

The tensor product of two vector spaces   and   is commutative in the sense that there is a canonical isomorphism


that maps   to  

On the other hand, even when   the tensor product of vectors is not commutative; that is   in general.

The map   from   to itself induces a linear automorphism that is called a braiding map. More generally and as usual (see tensor algebra), let denote   the tensor product of n copies of the vector space V. For every permutation s of the first n positive integers, the map


induces a linear automorphism of   which is called a braiding map.

Tensor product of linear maps Edit

Given a linear map   and a vector space W, the tensor product


is the unique linear map such that


The tensor product   is defined similarly.

Given two linear maps   and   their tensor product


is the unique linear map that satisfies


One has


In terms of category theory, this means that the tensor product is a bifunctor from the category of vector spaces to itself.[3]

If f and g are both injective or surjective, then the same is true for all above defined linear maps. In particular, the tensor product with a vector space is an exact functor; this means that every exact sequence is mapped to an exact sequence (tensor products of modules do not transform injections into injections, but they are right exact functors).

By choosing bases of all vector spaces involved, the linear maps S and T can be represented by matrices. Then, depending on how the tensor   is vectorized, the matrix describing the tensor product   is the Kronecker product of the two matrices. For example, if V, X, W, and Y above are all two-dimensional and bases have been fixed for all of them, and S and T are given by the matrices


respectively, then the tensor product of these two matrices is


The resultant rank is at most 4, and thus the resultant dimension is 4. Note that rank here denotes the tensor rank i.e. the number of requisite indices (while the matrix rank counts the number of degrees of freedom in the resulting array). Note  

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

General tensors Edit

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


Here   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[4]


It is defined by grouping all occurring "factors" V together: writing   for an element of V and   for an element of the dual space,


Picking a basis of V and the corresponding dual basis of   naturally induces a basis for   (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 orders m and n respectively (i.e.   and  ), then the components of their tensor product are given by[5]


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   and let V be a tensor of type   with components   Then




Tensors equipped with their product operation form an algebra, called the tensor algebra.

Evaluation map and tensor contraction Edit

For tensors of type (1, 1) there is a canonical evaluation map


defined by its action on pure tensors:


More generally, for tensors of type   with r, s > 0, there is a map, called tensor contraction,


(The copies of   and   on which this map is to be applied must be specified.)

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


where   is any basis of   and   is its dual basis. This map does not depend on the choice of basis.[6]

The interplay of evaluation and coevaluation can be used to characterize finite-dimensional vector spaces without referring to bases.[7]

Adjoint representation Edit

The tensor product   may be naturally viewed as a module for the Lie algebra   by means of the diagonal action: for simplicity let us assume   then, for each  


where   is the transpose of u, that is, in terms of the obvious pairing on  


There is a canonical isomorphism   given by


Under this isomorphism, every u in   may be first viewed as an endomorphism of   and then viewed as an endomorphism of   In fact it is the adjoint representation ad(u) of  

Linear maps as tensors Edit

Given two finite dimensional vector spaces U, V over the same field K, denote the dual space of U as U*, and the K-vector space of all linear maps from U to V as Hom(U,V). There is an isomorphism,


defined by an action of the pure tensor   on an element of  


Its "inverse" can be defined using a basis   and its dual basis   as in the section "Evaluation map and tensor contraction" above:


This result implies


which automatically gives the important fact that   forms a basis for   where   are bases of U and V.

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

This is an example of adjoint functors: the tensor product is "left adjoint" to Hom.

Tensor products of modules over a ring Edit

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

where now   is the free R-module generated by the cartesian product and G is the R-module generated by these relations.

More generally, the tensor product can be defined even if the ring is non-commutative. In this case A has to be a right-R-module and B is a left-R-module, and instead of the last two relations above, the relation

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   defined by   is a middle linear map (referred to as "the canonical middle linear map".[8]); that is, it satisfies:[9]


The first two properties make φ a bilinear map of the abelian group   For any middle linear map   of   a unique group homomorphism f of   satisfies   and this property determines   within group isomorphism. See the main article for details.

Tensor product of modules over a non-commutative ring Edit

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

where   is a free abelian group over   and G is the subgroup of   generated by relations

The universal property can be stated as follows. Let G be an abelian group with a map   that is bilinear, in the sense that


Then there is a unique map   such that   for all   and  

Furthermore, we can give   a module structure under some extra conditions:

  1. If A is a (S,R)-bimodule, then   is a left S-module where  
  2. If B is a (R,S)-bimodule, then   is a right S-module where  
  3. If A is a (S,R)-bimodule and B is a (R,T)-bimodule, then   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   and   By 3), we can conclude   is a (R,R)-bimodule.

Computing the tensor product Edit

For vector spaces, the tensor product   is quickly computed since bases of V of W immediately determine a basis of   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


More generally, given a presentation of some R-module M, that is, a number of generators   together with relations


the tensor product can be computed as the following cokernel:


Here   and the map   is determined by sending some   in the jth copy of   to   (in  ). Colloquially, this may be rephrased by saying that a presentation of M gives rise to a presentation of   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   the tensor product


is not usually injective. For example, tensoring the (injective) map given by multiplication with n, n : ZZ with Z/nZ yields the zero map 0 : Z/nZZ/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.

Tensor product of algebras Edit

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   is an R-algebra itself by putting

For example,

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

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
is isomorphic (as an A-algebra) to the  

Eigenconfigurations of tensors Edit

Square matrices   with entries in a field   represent linear maps of vector spaces, say   and thus linear maps   of projective spaces over   If   is nonsingular then   is well-defined everywhere, and the eigenvectors of   correspond to the fixed points of   The eigenconfiguration of   consists of   points in   provided   is generic and   is algebraically closed. The fixed points of nonlinear maps are the eigenvectors of tensors. Let   be a  -dimensional tensor of format   with entries   lying in an algebraically closed field   of characteristic zero. Such a tensor   defines polynomial maps   and   with coordinates


Thus each of the   coordinates of   is a homogeneous polynomial   of degree   in   The eigenvectors of   are the solutions of the constraint


and the eigenconfiguration is given by the variety of the   minors of this matrix.[10]

Other examples of tensor products Edit

Tensor product of Hilbert spaces Edit

Hilbert spaces generalize finite-dimensional vector spaces to countably-infinite dimensions. The tensor product is still defined; it is the tensor product of Hilbert spaces.

Topological tensor product Edit

When the basis for a vector space is no longer countable, then the appropriate axiomatic formalization for the vector space is that of a topological vector space. The tensor product is still defined, it is the topological tensor product.

Tensor product of graded vector spaces Edit

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 Edit

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 quadratic forms Edit

Tensor product of multilinear forms Edit

Given two multilinear forms   and   on a vector space   over the field   their tensor product is the multilinear form


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 Edit

Tensor product of line bundles Edit

Tensor product of fields Edit

Tensor product of graphs Edit

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 Edit

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.

Quotient algebras Edit

A number of important subspaces of the tensor algebra can be constructed as quotients: these include the exterior algebra, the symmetric algebra, the Clifford algebra, the Weyl algebra, and the universal enveloping algebra in general.

The exterior algebra is constructed from the exterior product. Given a vector space V, the exterior product   is defined as

Note that when the underlying field of V does not have characteristic 2, then this definition is equivalent to
The image of   in the exterior product is usually denoted   and satisfies, by construction,   Similar constructions are possible for   (n factors), giving rise to   the nth exterior power of V. The latter notion is the basis of differential n-forms.

The symmetric algebra is constructed in a similar manner, from the symmetric product

More generally
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 in programming Edit

Array programming languages Edit

Array programming languages may have this pattern built in. For example, in APL the tensor product is expressed as ○.× (for example A ○.× B or A ○.× B ○.× 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 Edit

Notes Edit

  1. ^ a b Trèves 2006, pp. 403–404.
  2. ^ a b Trèves 2006, pp. 407.
  3. ^ Hazewinkel, Michiel; Gubareni, Nadezhda Mikhaĭlovna; Gubareni, Nadiya; Kirichenko, Vladimir V. (2004). Algebras, rings and modules. Springer. p. 100. ISBN 978-1-4020-2690-4.
  4. ^ Bourbaki (1989), p. 244 defines the usage "tensor product of x and y", elements of the respective modules.
  5. ^ Analogous formulas also hold for contravariant tensors, as well as tensors of mixed variance. Although in many cases such as when there is an inner product defined, the distinction is irrelevant.
  6. ^ "The Coevaluation on Vector Spaces". The Unapologetic Mathematician. 2008-11-13. Archived from the original on 2017-02-02. Retrieved 2017-01-26.
  7. ^ See Compact closed category.
  8. ^ Hungerford, Thomas W. (1974). Algebra. Springer. ISBN 0-387-90518-9.
  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)
  10. ^ Abo, H.; Seigal, A.; Sturmfels, B. (2015). "Eigenconfigurations of Tensors". arXiv:1505.05729 [math.AG].
  11. ^ Tu, L. W. (2010). An Introduction to Manifolds. Universitext. Springer. p. 25. ISBN 978-1-4419-7399-3.

References Edit