Editing Tensor product (section)

== Definitions and constructions ==

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 {{slink||Universal_property}}, below. As for every universal property, all [[object (category theory)|objects]] that satisfy the property are isomorphic through a unique isomorphism that is compatible with the universal property. When this definition is used, the other definitions may be viewed as constructions of objects satisfying the universal property and as proofs that there are objects satisfying the universal property, that is that tensor products exist.

=== From bases ===
Let {{mvar|V}} and {{mvar|W}} be two [[vector space]]s over a [[field (mathematics)|field]] {{mvar|F}}, with respective [[basis (linear algebra)|bases]] <math>B_V</math> and {{tmath|1= B_W }}.

The ''tensor product'' <math>V \otimes W</math> of {{mvar|V}} and {{mvar|W}} 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 {{tmath|1= 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): <math>V \otimes W</math> is the set of the [[function (mathematics)|functions]] from the [[Cartesian product]] <math>B_V \times B_W</math> to {{mvar|F}} that have a finite number of nonzero values. The [[pointwise operation]]s make <math>V \otimes W</math> a vector space. The function that maps <math>(v,w)</math> to {{math|1}} and the other elements of <math>B_V \times B_W</math> to {{math|0}} is denoted {{tmath|1= v\otimes w }}.

The set <math>\{v\otimes w\mid v\in B_V, w\in B_W\}</math> is then straightforwardly a basis of {{tmath|1= V \otimes W }}, which is called the ''tensor product'' of the bases <math>B_V</math> and {{tmath|1= B_W }}.

We can equivalently define <math>V \otimes W</math> to be the set of [[Bilinear form|bilinear forms]] on <math>V \times W</math> that are nonzero at only a finite number of elements of {{tmath|1= B_V \times B_W }}. To see this, given <math>(x,y)\in V \times W</math>  and a bilinear form {{tmath|1= B : V \times W \to F }},  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 {{tmath|1= B_V \times B_W }}. 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 {{tmath|1= V \times W }}. To instead have it be a proper Hamel [[Basis (linear algebra)|basis]], it only remains to add the requirement that <math>B</math> is nonzero at an only a finite number of elements of {{tmath|1= B_V \times B_W }}, and consider the subspace of such maps instead. 

In either construction, the ''tensor product of two vectors'' is defined from their decomposition on the bases. More precisely, taking the basis decompositions of <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 {{tmath|1= B_W }}, 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 {{tmath|1= y }}. Therefore, the tensor product is a generalization of the outer product, that is, an abstraction of it beyond coordinate vectors. 

A limitation of this definition of the tensor product is that, if one changes bases, a different tensor product is defined. However, the decomposition on one basis of the elements of the other basis defines a [[canonical isomorphism]] between the two tensor products of vector spaces, which allows identifying them. Also, contrarily to the two following alternative definitions, this definition cannot be extended into a definition of the [[tensor product of modules]] over a [[ring (mathematics)|ring]].

=== As a quotient space ===
A construction of the tensor product that is basis independent can be obtained in the following way.

Let {{mvar|V}} and {{mvar|W}} be two [[vector space]]s over a [[field (mathematics)|field {{mvar|F}}]]. 

One considers first a vector space {{mvar|L}} that has the [[Cartesian product]] <math>V\times W</math> as a [[basis (linear algebra)|basis]]. That is, the basis elements of {{mvar|L}} are the [[ordered pair|pairs]] <math>(v,w)</math> with <math>v\in V</math> and {{tmath|1= w\in W }}. To get such a vector space, one can define it as the vector space of the [[function (mathematics)|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 {{math|1}} on <math>(v,w)</math> and {{math|0}} otherwise.

Let {{mvar|R}} be the [[linear subspace]] of {{mvar|L}} that is spanned by the relations that the tensor product must satisfy. More precisely, {{mvar|R}} is [[Linear span|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 {{tmath|1= v, v_1, v_2\in V }}, <math>w, w_1, w_2 \in W</math> and {{tmath|1= s\in F }}.

Then, the tensor product is defined as the [[quotient space (linear algebra)|quotient space]]:
: <math>V\otimes W=L/R,</math>
and the image of <math>(v,w)</math> in this quotient is denoted {{tmath|1= v\otimes w }}.

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

=== Universal property ===
[[File:Another universal tensor prod.svg|class=skin-invert-image|right|thumb|200px|Universal property of tensor product: if {{math|''h''}} is bilinear, there is a unique linear map {{math|{{overset|lh=0.7|~|''h''}}}} that makes the diagram [[commutative diagram|commutative]] (that is, {{math|1=''h'' = {{overset|lh=0.7|~|''h''}} ∘ ''φ''}}).]]

In this section, the [[universal property]] satisfied by the tensor product is described. As for every universal property, two objects that satisfy the property are related by a unique [[isomorphism]]. It follows that this is a (non-constructive) way to define the tensor product of two vector spaces. In this context, the preceding constructions of tensor products may be viewed as proofs of existence of the tensor product so defined.

A consequence of this approach is that every property of the tensor product can be deduced from the universal property, and that, in practice, one may forget the method that has been used to prove its existence. 

The "universal-property definition" of the tensor product of two vector spaces is the following (recall that a [[bilinear map]] is a function that is ''separately'' [[linear map|linear]] in each of its arguments):
:The ''tensor product'' of two vector spaces {{mvar|V}} and {{mvar|W}} is a vector space denoted as {{tmath|1= V\otimes W }}, together with a bilinear map <math>{\otimes} : (v,w)\mapsto v\otimes w</math> from <math>V\times W</math> to {{tmath|1= V\otimes W }}, such that, for every bilinear map {{tmath|1= h : V\times W\to Z }}, there is a ''unique'' linear map {{tmath|1= \tilde h : V\otimes W\to Z }}, 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 {{tmath|1= w\in W }}).<!-- Commenting out considerations that are misplaced here; kept for a possible use elsewhere.

This characterization can simplify proofs about the tensor product. For example, the tensor product is symmetric, meaning there is a [[canonical isomorphism]]:
<math display="block">V \otimes W \cong W \otimes V.</math>

To construct, say, a map from <math>V \otimes W</math> to {{tmath|1= W \otimes V }}, it suffices to give a bilinear map <math>h: V \times W \to W \otimes V</math> that maps <math>(v,w)</math> to {{tmath|1= w \otimes v }}. Then the universal property of <math>V \otimes W</math> means <math>h</math> factors into a map {{tmath|1= \tilde{h}:V \otimes W \to W \otimes V }}.
A map <math>\tilde{g}:W \otimes V \to V \otimes W</math> in the opposite direction is similarly defined, and one checks that the two linear maps <math>\tilde{h}</math> and <math>\tilde{g}</math> are [[Inverse function|inverse]] to one another by again using their universal properties.

The universal property is extremely useful in showing that a map to a tensor product is injective. For example, suppose we want to show that <math>\R \otimes \R</math> is isomorphic to {{tmath|1= \R }}. Since all simple tensors are of the form {{tmath|1= a \otimes b = (ab) \otimes 1 }}, and hence all elements of the tensor product are of the form <math>x \otimes 1</math> by additivity in the first coordinate, we have a natural candidate for an isomorphism <math>\R \rightarrow \R \otimes \R</math> given by mapping <math>x</math> to {{tmath|1= x \otimes 1 }}, and this map is trivially surjective.

Showing injectivity directly would involve somehow showing that there are no non-trivial relationships between <math>x \otimes 1</math> and <math>y \otimes 1</math> for {{tmath|1= x \neq y }}, which seems daunting. However, we know that there is a bilinear map <math>\R \times \R \rightarrow \R</math> given by multiplying the coordinates together, and the universal property of the tensor product then furnishes a map of vector spaces <math>\R \otimes \R \rightarrow \R</math> which maps <math>x \otimes 1</math> to {{tmath|1= x }}, and hence is an inverse of the previously constructed homomorphism, immediately implying the desired result. A priori, it is not even clear that this inverse map is well-defined, but the universal property and associated bilinear map together imply this is the case.

Similar reasoning can be used to show that the tensor product is associative, that is, there are natural isomorphisms
<math display="block">V_1 \otimes \left(V_2\otimes V_3\right) \cong \left(V_1\otimes V_2\right) \otimes V_3.</math>
Therefore, it is customary to omit the parentheses and write {{tmath|1= V_1 \otimes V_2 \otimes V_3 }}, so the ''ijk-th'' component of <math> \mathbf{u} \otimes \mathbf{v} \otimes \mathbf{w} </math> is 
<math display="block">(\mathbf{u} \otimes \mathbf{v}\otimes \mathbf{w})_{ijk} = u_i v_j w_k,</math>
similar to the first [[#Tensors in finite dimensions, and the outer product|example]] on this page.

The category of vector spaces with tensor product is an example of a [[symmetric monoidal category]].
 -->

=== 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.{{sfn|Trèves|2006|pp=403-404}}

{{math theorem|math_statement=
Let {{tmath|1= X, Y }}, and <math>Z</math> be complex vector spaces and let <math>T : X \times Y \to Z</math> be a bilinear map. Then <math>(Z, T)</math> is a tensor product of <math>X</math> and <math>Y</math> if and only if{{sfn|Trèves|2006|pp=403-404}} the image of <math>T</math> spans all of <math>Z</math> (that is, {{tmath|1= \operatorname{span} \; T(X \times Y) = Z }}), and also <math>X</math> and <math>Y</math> are {{em|<math>T</math>-linearly disjoint}}, which by definition means that for all positive integers <math>n</math> and all elements <math>x_1, \ldots, x_n \in X</math> and <math>y_1, \ldots, y_n \in Y</math> such that {{tmath|1= \sum_{i=1}^n T\left(x_i, y_i\right) = 0 }}, 
# if all <math>x_1, \ldots, x_n</math> are [[linearly independent]] then all <math>y_i</math> are {{tmath|1= 0 }}, and 
# if all <math>y_1, \ldots, y_n</math> are linearly independent then all <math>x_i</math> are {{tmath|1= 0 }}.

Equivalently, <math>X</math> and <math>Y</math> are <math>T</math>-linearly disjoint if and only if for all linearly independent sequences <math>x_1, \ldots, x_m</math> in <math>X</math> and all linearly independent sequences <math>y_1, \ldots, y_n</math> in {{tmath|1= Y }}, the vectors <math>\left\{T\left(x_i, y_j\right) : 1 \leq i \leq m, 1 \leq j \leq n\right\}</math> are linearly independent.
}}

For example, it follows immediately that if {{tmath|1=X=\C^m}} and {{tmath|1=Y=\C^n}}, 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 {{tmath|1= Y }}.{{sfn|Trèves|2006|pp=407}} 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 {{tmath|1= s \mapsto c f(s) }}). Let <math>S</math> and <math>T</math> be any sets and for any <math>f \in \Complex^S</math> and {{tmath|1= g \in \Complex^T }}, let <math>f \otimes g \in \Complex^{S \times T}</math> denote the function defined by {{tmath|1= (s, t) \mapsto f(s) g(t) }}. 
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 {{tmath|1= Y }}.{{sfn|Trèves|2006|pp=407}}