Tensor product

Template:Short description Template:For

In mathematics, the tensor product <math>V \otimes W</math> of two vector spaces <math>V</math> and <math>W</math> (over the same field) is a vector space to which is associated a bilinear map <math>V\times W \rightarrow V\otimes W</math> that maps a pair <math>(v,w),\ v\in V, w\in W</math> to an element of <math>V \otimes W</math> denoted Template:Tmath.

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

The tensor product of two vector spaces captures the properties of all bilinear maps in the sense that a bilinear map from <math>V\times W</math> into another vector space <math>Z</math> factors uniquely through a linear map <math>V\otimes W\to Z</math> (see the section below titled 'Universal property'), i.e. the bilinear map is associated to a unique linear map from the tensor product <math>V \otimes W</math> to <math>Z</math>.

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.

Definitions and constructionsEdit

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 Template:Slink, 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 basesEdit

Let Template:Mvar and Template:Mvar be two vector spaces over a field Template:Mvar, with respective bases <math>B_V</math> and Template:Tmath.

The tensor product <math>V \otimes W</math> of Template:Mvar and Template:Mvar is a vector space that has as a basis the set of all <math>v\otimes w</math> with <math>v\in B_V</math> and Template:Tmath. This definition can be formalized in the following way (this formalization is rarely used in practice, as the preceding informal definition is generally sufficient): <math>V \otimes W</math> is the set of the functions from the Cartesian product <math>B_V \times B_W</math> to Template:Mvar that have a finite number of nonzero values. The pointwise operations make <math>V \otimes W</math> a vector space. The function that maps <math>(v,w)</math> to Template:Math and the other elements of <math>B_V \times B_W</math> to Template:Math is denoted Template:Tmath.

The set <math>\{v\otimes w\mid v\in B_V, w\in B_W\}</math> is then straightforwardly a basis of Template:Tmath, which is called the tensor product of the bases <math>B_V</math> and Template:Tmath.

We can equivalently define <math>V \otimes W</math> to be the set of bilinear forms on <math>V \times W</math> that are nonzero at only a finite number of elements of Template:Tmath. To see this, given <math>(x,y)\in V \times W</math> and a bilinear form Template:Tmath, we can decompose <math>x</math> and <math>y</math> in the bases <math>B_V</math> and <math>B_W</math> as: <math display="block">x=\sum_{v\in B_V} x_v\, v \quad \text{and}\quad y=\sum_{w\in B_W} y_w\, w, </math> where only a finite number of <math>x_v</math>'s and <math>y_w</math>'s are nonzero, and find by the bilinearity of <math>B</math> that: <math display="block">B(x, y) =\sum_{v\in B_V}\sum_{w\in B_W} x_v y_w\, B(v, w)</math>

Hence, we see that the value of <math>B</math> for any <math>(x,y)\in V \times W</math> is uniquely and totally determined by the values that it takes on Template:Tmath. This lets us extend the maps <math>v\otimes w</math> defined on <math>B_V \times B_W</math> as before into bilinear maps <math>v \otimes w : V\times W \to F</math> , by letting: <math display="block">(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 .</math>

Then we can express any bilinear form <math>B</math> as a (potentially infinite) formal linear combination of the <math>v\otimes w</math> maps according to: <math display="block">B = \sum_{v\in B_V}\sum_{w\in B_W} B(v, w)(v \otimes w)</math> making these maps similar to a Schauder basis for the vector space <math>\text{Hom}(V, W; F)</math> of all bilinear forms on Template:Tmath. To instead have it be a proper Hamel basis, it only remains to add the requirement that <math>B</math> is nonzero at an only a finite number of elements of Template:Tmath, 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 <math>x\in V </math> and <math>y \in W</math> as before: <math display="block">\begin{align} 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{align}</math>

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

If arranged into a rectangular array, the coordinate vector of <math>x\otimes y</math> is the outer product of the coordinate vectors of <math>x</math> and Template:Tmath. 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 spaceEdit

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

Let Template:Mvar and Template:Mvar be two vector spaces over a [[field (mathematics)|field Template:Mvar]].

One considers first a vector space Template:Mvar that has the Cartesian product <math>V\times W</math> as a basis. That is, the basis elements of Template:Mvar are the pairs <math>(v,w)</math> with <math>v\in V</math> and Template:Tmath. To get such a vector space, one can define it as the vector space of the functions <math>V\times W \to F</math> that have a finite number of nonzero values and identifying <math>(v,w)</math> with the function that takes the value Template:Math on <math>(v,w)</math> and Template:Math otherwise.

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

<math>\begin{align}

(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{align}</math> where Template:Tmath, <math>w, w_1, w_2 \in W</math> and Template:Tmath.

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

<math>V\otimes W=L/R,</math>

and the image of <math>(v,w)</math> in this quotient is denoted Template:Tmath.

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 propertyEdit

File:Another universal tensor prod.svg

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

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 Template:Mvar and Template:Mvar is a vector space denoted as Template:Tmath, together with a bilinear map <math>{\otimes} : (v,w)\mapsto v\otimes w</math> from <math>V\times W</math> to Template:Tmath, such that, for every bilinear map Template:Tmath, there is a unique linear map Template:Tmath, such that <math>h=\tilde h \circ {\otimes}</math> (that is, <math>h(v, w)= \tilde h(v\otimes w)</math> for every <math>v\in V</math> and Template:Tmath).

Linearly disjointEdit

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.Template:Sfn

Template:Math theorem

For example, it follows immediately that if Template:Tmath and Template:Tmath, where <math>m</math> and <math>n</math> are positive integers, then one may set <math>Z = \Complex^{mn}</math> and define the bilinear map as <math display=block>\begin{align} T : \Complex^m \times \Complex^n &\to \Complex^{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{align}</math> to form the tensor product of <math>X </math> and Template:Tmath.Template:Sfn Often, this map <math>T</math> is denoted by <math>\,\otimes\,</math> so that <math>x \otimes y = T(x, y).</math>

As another example, suppose that <math>\Complex^S</math> is the vector space of all complex-valued functions on a set <math>S</math> with addition and scalar multiplication defined pointwise (meaning that <math>f + g</math> is the map <math>s \mapsto f(s) + g(s)</math> and <math>c f</math> is the map Template:Tmath). Let <math>S</math> and <math>T</math> be any sets and for any <math>f \in \Complex^S</math> and Template:Tmath, let <math>f \otimes g \in \Complex^{S \times T}</math> denote the function defined by Template:Tmath. If <math>X \subseteq \Complex^S</math> and <math>Y \subseteq \Complex^T</math> are vector subspaces then the vector subspace <math>Z := \operatorname{span} \left\{f \otimes g : f \in X, g \in Y\right\}</math> of <math>\Complex^{S \times T}</math> together with the bilinear map: <math display=block>\begin{alignat}{4}

\;&& X \times Y &&\;\to    \;& Z \\[0.3ex]
    && (f, g) &&\;\mapsto\;& f \otimes g \\

\end{alignat}</math> form a tensor product of <math>X</math> and Template:Tmath.Template:Sfn

PropertiesEdit

DimensionEdit

If Template:Math and Template:Math are vector spaces of finite dimension, then <math>V\otimes W</math> is finite-dimensional, and its dimension is the product of the dimensions of Template:Math and Template:Math.

This results from the fact that a basis of <math>V\otimes W</math> is formed by taking all tensor products of a basis element of Template:Math and a basis element of Template:Math.

AssociativityEdit

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

<math>(U\otimes V)\otimes W\cong U\otimes (V\otimes W),</math>

that maps <math>(u\otimes v)\otimes w</math> to Template:Tmath.

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

Commutativity as vector space operationEdit

The tensor product of two vector spaces <math>V</math> and <math>W</math> is commutative in the sense that there is a canonical isomorphism:

<math> V \otimes W \cong W\otimes V,</math>

that maps <math>v \otimes w</math> to Template:Tmath.

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

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

<math>x_1\otimes \cdots \otimes x_n \mapsto x_{s(1)}\otimes \cdots \otimes x_{s(n)}</math>

induces a linear automorphism of Template:Tmath, which is called a braiding map.

Tensor product of linear mapsEdit

Template:Redirect Given a linear map Template:Tmath, and a vector space Template:Mvar, the tensor product:

<math>f\otimes W : U\otimes W\to V\otimes W</math>

is the unique linear map such that:

<math>(f\otimes W)(u\otimes w)=f(u)\otimes w.</math>

The tensor product <math>W\otimes f</math> is defined similarly.

Given two linear maps <math>f : U\to V</math> and Template:Tmath, their tensor product:

<math>f\otimes g : U\otimes W\to V\otimes Z</math>

is the unique linear map that satisfies:

<math>(f\otimes g)(u\otimes w)=f(u)\otimes g(w).</math>

One has:

<math>f\otimes g= (f\otimes Z)\circ (U\otimes g) = (V\otimes g)\circ (f\otimes W).</math>

In terms of category theory, this means that the tensor product is a bifunctor from the category of vector spaces to itself.<ref>Template:Cite book</ref>

If Template:Mvar and Template:Mvar 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 Template:Math and Template:Math can be represented by matrices. Then, depending on how the tensor <math>v \otimes w</math> is vectorized, the matrix describing the tensor product <math>f \otimes g</math> is the Kronecker product of the two matrices. For example, if Template:Math, and Template:Math above are all two-dimensional and bases have been fixed for all of them, and Template:Math and Template:Math are given by the matrices: <math display="block">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},</math>

respectively, then the tensor product of these two matrices is: <math>

 \begin{align}
 \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{align}

</math>

The resultant rank is at most 4, and thus the resultant dimension is 4. Template:Em 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). Template:Tmath.

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

General tensorsEdit

Template:See also For non-negative integers Template:Math and Template:Math a type <math>(r, s)</math> tensor on a vector space Template:Math is an element of: <math display="block">T^r_s(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}.</math> Here <math>V^*</math> is the dual vector space (which consists of all linear maps Template:Math from Template:Math to the ground field Template:Math).

There is a product map, called the Template:Em:Template:Refn <math display="block">T^r_s (V) \otimes_K T^{r'}_{s'} (V) \to T^{r+r'}_{s+s'}(V).</math>

It is defined by grouping all occurring "factors" Template:Math together: writing <math>v_i</math> for an element of Template:Math and <math>f_i</math> for an element of the dual space: <math display="block">(v_1 \otimes f_1) \otimes (v'_1) = v_1 \otimes v'_1 \otimes f_1.</math>

If Template:Math is finite dimensional, then picking a basis of Template:Math and the corresponding dual basis of <math>V^*</math> naturally induces a basis of <math>T_s^r(V)</math> (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 Template:Math and Template:Math are two covariant tensors of orders Template:Math and Template:Math respectively (i.e. <math>F \in T_m^0</math> and Template:Tmath), then the components of their tensor product are given by:<ref>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.</ref> <math display="block">(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}}.</math>

Thus, the components of the tensor product of two tensors are the ordinary product of the components of each tensor. Another example: let Template:Math be a tensor of type Template:Math with components Template:Tmath, and let Template:Math be a tensor of type <math>(1, 0)</math> with components Template:Tmath. Then: <math display="block">\left(U \otimes V\right)^\alpha {}_\beta {}^\gamma = U^\alpha {}_\beta V^\gamma</math> and: <math display="block">(V \otimes U)^{\mu\nu} {}_\sigma = V^\mu U^\nu {}_\sigma.</math>

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

Evaluation map and tensor contractionEdit

For tensors of type Template:Math there is a canonical evaluation map: <math display="block">V \otimes V^* \to K</math> defined by its action on pure tensors: <math display="block">v \otimes f \mapsto f(v).</math>

More generally, for tensors of type Template:Tmath, with Template:Math, there is a map, called tensor contraction: <math display="block">T^r_s (V) \to T^{r-1}_{s-1}(V).</math> (The copies of <math>V</math> and <math>V^*</math> on which this map is to be applied must be specified.)

On the other hand, if <math>V</math> is Template:Em, there is a canonical map in the other direction (called the coevaluation map): <math display="block">\begin{cases} K \to V \otimes V^* \\ \lambda \mapsto \sum_i \lambda v_i \otimes v^*_i \end{cases}</math> where <math>v_1, \ldots, v_n</math> is any basis of Template:Tmath, and <math>v_i^*</math> is its dual basis. This map does not depend on the choice of basis.<ref>{{#invoke:citation/CS1|citation |CitationClass=web }}</ref>

The interplay of evaluation and coevaluation can be used to characterize finite-dimensional vector spaces without referring to bases.<ref>See Compact closed category.</ref>

Adjoint representationEdit

The tensor product <math>T^r_s(V)</math> may be naturally viewed as a module for the Lie algebra <math>\mathrm{End}(V)</math> by means of the diagonal action: for simplicity let us assume Template:Tmath, then, for each Template:Tmath, <math display="block">u(a \otimes b) = u(a) \otimes b - a \otimes u^*(b),</math> where <math>u^* \in \mathrm{End}\left(V^*\right)</math> is the transpose of Template:Math, that is, in terms of the obvious pairing on Template:Tmath, <math display="block">\langle u(a), b \rangle = \langle a, u^*(b) \rangle.</math>

There is a canonical isomorphism <math>T^1_1(V) \to \mathrm{End}(V)</math> given by: <math display="block">(a \otimes b)(x) = \langle x, b \rangle a.</math>

Under this isomorphism, every Template:Math in <math>\mathrm{End}(V)</math> may be first viewed as an endomorphism of <math>T^1_1(V)</math> and then viewed as an endomorphism of Template:Tmath. In fact it is the adjoint representation Template:Math of Template:Tmath.

Linear maps as tensorsEdit

Given two finite dimensional vector spaces Template:Math, Template:Math over the same field Template:Math, denote the dual space of Template:Math as Template:Math, and the Template:Math-vector space of all linear maps from Template:Math to Template:Math as Template:Math. There is an isomorphism: <math display="block">U^* \otimes V \cong \mathrm{Hom}(U, V),</math> defined by an action of the pure tensor <math>f \otimes v \in U^*\otimes V</math> on an element of Template:Tmath, <math display="block">(f \otimes v)(u) = f(u) v.</math>

Its "inverse" can be defined using a basis <math>\{u_i\}</math> and its dual basis <math>\{u^*_i\}</math> as in the section "Evaluation map and tensor contraction" above: <math display="block">\begin{cases} \mathrm{Hom} (U,V) \to U^* \otimes V \\ F \mapsto \sum_i u^*_i \otimes F(u_i). \end{cases}</math>

This result implies: <math display="block">\dim(U \otimes V) = \dim(U)\dim(V),</math> which automatically gives the important fact that <math>\{u_i\otimes v_j\}</math> forms a basis of <math>U \otimes V</math> where <math>\{u_i\}, \{v_j\}</math> are bases of Template:Math and Template:Math.

Furthermore, given three vector spaces Template:Math, Template:Math, Template:Math the tensor product is linked to the vector space of all linear maps, as follows: <math display="block">\mathrm{Hom} (U \otimes V, W) \cong \mathrm{Hom} (U, \mathrm{Hom}(V, W)).</math> This is an example of adjoint functors: the tensor product is "left adjoint" to Hom.

Tensor products of modules over a ringEdit

{{#invoke:Labelled list hatnote|labelledList|Main article|Main articles|Main page|Main pages}} The tensor product of two modules Template:Math and Template:Math over a commutative ring Template:Math is defined in exactly the same way as the tensor product of vector spaces over a field: <math display="block">A \otimes_R B := F (A \times B) / G ,</math> where now <math>F(A \times B)</math> is the [[Free module|free Template:Math-module]] generated by the cartesian product and Template:Math is the Template:Math-module generated by these relations.

More generally, the tensor product can be defined even if the ring is non-commutative. In this case Template:Math has to be a right-Template:Math-module and Template:Math is a left-Template:Math-module, and instead of the last two relations above, the relation: <math display="block">(ar,b)\sim (a,rb)</math> is imposed. If Template:Math is non-commutative, this is no longer an Template:Math-module, but just an abelian group.

The universal property also carries over, slightly modified: the map <math>\varphi : A \times B \to A \otimes_R B</math> defined by <math>(a, b) \mapsto a \otimes b</math> is a middle linear map (referred to as "the canonical middle linear map"<ref> Template:Cite book</ref>); that is, it satisfies:<ref name=chen> Template:Citation</ref> <math display="block">\begin{align}

 \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{align}</math>

The first two properties make Template:Math a bilinear map of the abelian group Template:Tmath. For any middle linear map <math>\psi</math> of Template:Tmath, a unique group homomorphism Template:Math of <math>A \otimes_R B</math> satisfies Template:Tmath, and this property determines <math>\varphi</math> within group isomorphism. See the main article for details.

Tensor product of modules over a non-commutative ringEdit

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: <math display="block">A \otimes_R B := F (A \times B) / G</math> where <math>F (A \times B)</math> is a free abelian group over <math>A \times B</math> and G is the subgroup of <math>F (A \times B)</math> generated by relations: <math display="block">\begin{align}

 &\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{align}</math>

The universal property can be stated as follows. Let G be an abelian group with a map <math>q:A\times B \to G</math> that is bilinear, in the sense that: <math display="block">\begin{align}

 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{align}</math>

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

Furthermore, we can give <math>A \otimes_R B</math> a module structure under some extra conditions:

If A is a (S,R)-bimodule, then <math>A \otimes_R B</math> is a left S-module, where Template:Tmath.
If B is a (R,S)-bimodule, then <math>A \otimes_R B</math> is a right S-module, where Template:Tmath.
If A is a (S,R)-bimodule and B is a (R,T)-bimodule, then <math>A \otimes_R B</math> is a (S,T)-bimodule, where the left and right actions are defined in the same way as the previous two examples.
If R is a commutative ring, then A and B are (R,R)-bimodules where <math>ra:=ar</math> and Template:Tmath. By 3), we can conclude <math>A \otimes_R B</math> is a (R,R)-bimodule.

Computing the tensor productEdit

For vector spaces, the tensor product <math>V \otimes W</math> is quickly computed since bases of Template:Math of Template:Math immediately determine a basis of Template:Tmath, as was mentioned above. For modules over a general (commutative) ring, not every module is free. For example, Template:Math is not a free abelian group (Template:Math-module). The tensor product with Template:Math is given by: <math display="block">M \otimes_\mathbf{Z} \mathbf{Z}/n\mathbf{Z} = M/nM.</math>

More generally, given a presentation of some Template:Math-module Template:Math, that is, a number of generators <math>m_i \in M, i \in I</math> together with relations: <math display="block">\sum_{j \in J} a_{ji} m_i = 0,\qquad a_{ij} \in R,</math> the tensor product can be computed as the following cokernel: <math display="block">M \otimes_R N = \operatorname{coker} \left(N^J \to N^I\right)</math>

Here Template:Tmath, and the map <math>N^J \to N^I</math> is determined by sending some <math>n \in N</math> in the Template:Mathth copy of <math>N^J</math> to <math>a_{ij} n</math> (in Template:Tmath). Colloquially, this may be rephrased by saying that a presentation of Template:Math gives rise to a presentation of Template:Tmath. 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 Template:Math-modules Template:Tmath, the tensor product: <math display="block">M_1 \otimes_R N \to M_2 \otimes_R N</math> is not usually injective. For example, tensoring the (injective) map given by multiplication with Template:Math, Template:Math with Template:Math yields the zero map Template:Math, 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 algebrasEdit

{{#invoke:Labelled list hatnote|labelledList|Main article|Main articles|Main page|Main pages}} Let Template:Math be a commutative ring. The tensor product of Template:Math-modules applies, in particular, if Template:Math and Template:Math are [[Algebra (ring theory)|Template:Math-algebras]]. In this case, the tensor product <math>A \otimes_R B</math> is an Template:Math-algebra itself by putting: <math display="block">(a_1 \otimes b_1) \cdot (a_2 \otimes b_2) = (a_1 \cdot a_2) \otimes (b_1 \cdot b_2).</math> For example: <math display="block">R[x] \otimes_R R[y] \cong R[x, y].</math>

A particular example is when Template:Math and Template:Math are fields containing a common subfield Template:Math. The tensor product of fields is closely related to Galois theory: if, say, Template:Math, where Template:Math is some irreducible polynomial with coefficients in Template:Math, the tensor product can be calculated as: <math display="block">A \otimes_R B \cong B[x] / f(x)</math> where now Template:Math is interpreted as the same polynomial, but with its coefficients regarded as elements of Template:Math. In the larger field Template:Math, the polynomial may become reducible, which brings in Galois theory. For example, if Template:Math is a Galois extension of Template:Math, then: <math display="block">A \otimes_R A \cong A[x] / f(x)</math> is isomorphic (as an Template:Math-algebra) to the Template:Tmath.

Eigenconfigurations of tensorsEdit

Square matrices <math>A</math> with entries in a field <math>K</math> represent linear maps of vector spaces, say Template:Tmath, and thus linear maps <math>\psi : \mathbb{P}^{n-1} \to \mathbb{P}^{n-1}</math> of projective spaces over Template:Tmath. If <math>A</math> is nonsingular then <math>\psi</math> is well-defined everywhere, and the eigenvectors of <math>A</math> correspond to the fixed points of Template:Tmath. The eigenconfiguration of <math>A</math> consists of <math>n</math> points in Template:Tmath, provided <math>A</math> is generic and <math>K</math> is algebraically closed. The fixed points of nonlinear maps are the eigenvectors of tensors. Let <math>A = (a_{i_1 i_2 \cdots i_d})</math> be a <math>d</math>-dimensional tensor of format <math>n \times n \times \cdots \times n</math> with entries <math>(a_{i_1 i_2 \cdots i_d})</math> lying in an algebraically closed field <math>K</math> of characteristic zero. Such a tensor <math>A \in (K^{n})^{\otimes d}</math> defines polynomial maps <math>K^n \to K^n</math> and <math>\mathbb{P}^{n-1} \to \mathbb{P}^{n-1}</math> with coordinates: <math display="block">\psi_i(x_1, \ldots, x_n) = \sum_{j_2=1}^n \sum_{j_3=1}^n \cdots \sum_{j_d = 1}^n a_{i j_2 j_3 \cdots j_d} x_{j_2} x_{j_3}\cdots x_{j_d} \;\; \mbox{for } i = 1, \ldots, n</math>

Thus each of the <math>n</math> coordinates of <math>\psi</math> is a homogeneous polynomial <math>\psi_i</math> of degree <math>d-1</math> in Template:Tmath. The eigenvectors of <math>A</math> are the solutions of the constraint: <math display="block">\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

</math> and the eigenconfiguration is given by the variety of the <math>2 \times 2</math> minors of this matrix.<ref>Template:Cite arXiv</ref>