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The tensor product of two unitary modules <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t0924101.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t0924102.png" /> over an associative commutative ring <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t0924103.png" /> with a unit is the <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t0924104.png" />-module <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t0924105.png" /> together with an <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t0924106.png" />-bilinear mapping
+
{{TEX|done}}
 +
$\newcommand{\tensor}{\otimes}$
 +
$\newcommand{\lieg}{\mathfrak{g}}$
 +
$\newcommand{\iso}{\cong}$
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t0924107.png" /></td> </tr></table>
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====Tensor product of two unitary modules====
  
which is universal in the following sense: For any <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t0924108.png" />-bilinear mapping <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t0924109.png" />, where <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241010.png" /> is an arbitrary <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241011.png" />-module, there is a unique <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241012.png" />-linear mapping <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241013.png" /> such that
+
The tensor product of two unitary modules $V_1$ and $V_2$ over an associative commutative ring $A$ with a unit is the $A$-module $V_1 \tensor_A V_2$ together with an $A$-bilinear mapping
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241014.png" /></td> </tr></table>
+
$$(x_1, x_2) \mapsto x_1 \tensor x_2 \in V_1 \tensor_A V_2$$
 +
which is universal in the following sense: For any $A$-bilinear mapping $\beta: V_1 \times V_2 \to W$, where $W$ is an arbitrary $A$-module, there is a unique $A$-linear mapping $b : V_1 \tensor_A V_2 \to W$ such that
  
The tensor product is uniquely defined up to a natural isomorphism. It always exists and can be constructed as the quotient module of the free <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241015.png" />-module <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241016.png" /> generated by the set <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241017.png" /> modulo the <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241018.png" />-submodule <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241019.png" /> generated by the elements of the form
+
$$\beta(x_1, x_2) = b(x_1 \tensor x_2), \qquad x_1 \in V_1, \qquad x_2 \in V_2.$$
 +
The tensor product is uniquely defined up to a natural isomorphism. It always exists and can be constructed as the quotient module of the free $A$-module $F$ generated by the set $V_1 \times V_2$ modulo the $A$-submodule $R$ generated by the elements of the form
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241020.png" /></td> </tr></table>
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$$(x_1 + y, x_2) - (x_1, x_2) - (y, x_2),$$
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241021.png" /></td> </tr></table>
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$$(x_1, x_2 + z) - (x_1, x_2) - (x_1, z),$$
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241022.png" /></td> </tr></table>
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$$(cx_1, x_2) - c(x_1, x_2),$$
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241023.png" /></td> </tr></table>
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$$(x_1, cx_2) - c(x_1, x_2),$$
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241024.png" /></td> </tr></table>
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$$x_1, y \in V_1, \qquad x_2, z \in V_2, \qquad c \in A;$$
 
+
then $x_1 \tensor x_2 = (x_1, x_2) + R$. If one gives up the requirement of commutativity of $A$, a construction close to the one described above allows one to form from a right $A$-module $V_1$ and a left $A$-module $V_2$ an Abelian group $V_1 \tensor_A V_2$, also called the tensor product of these modules
then <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241025.png" />. If one gives up the requirement of commutativity of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241026.png" />, a construction close to the one described above allows one to form from a right <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241027.png" />-module <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241028.png" /> and a left <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241029.png" />-module <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241030.png" /> an Abelian group <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241031.png" />, also called the tensor product of these modules [[#References|[1]]]. In what follows <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241032.png" /> will be assumed to be commutative.
+
[[#References|[1]]]. In what follows $A$ will be assumed to be commutative.
  
 
The tensor product has the following properties:
 
The tensor product has the following properties:
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241033.png" /></td> </tr></table>
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$$A \tensor_A V \iso V,$$
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241034.png" /></td> </tr></table>
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$$V_1 \tensor_A V_2 \iso V_2 \tensor_A V_1,$$
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241035.png" /></td> </tr></table>
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$$(V_1 \tensor_A V_2) \tensor V_3 \iso V_1 \tensor_A (V_2 \tensor_A V_3),$$
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241036.png" /></td> </tr></table>
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$$\left( \bigoplus_{i \in I} V_i \right) \tensor_A W \iso \bigoplus_{i \in I} (V_i \tensor_A W)$$
 +
for any $A$-modules $V$, $V_i$ and $W$.
  
for any <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241037.png" />-modules <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241038.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241039.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241040.png" />.
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If $(x_i)_{i \in I}$ and $(y_j)_{j \in J}$ are bases of the free $A$-modules $V_1$ and $V_2$, then $(x_i \tensor y_j)_{(i,j) \in I\times J}$ is a basis of the module $V_1 \tensor_A V_2$. In particular,
  
If <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241041.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241042.png" /> are bases of the free <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241043.png" />-modules <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241044.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241045.png" />, then <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241046.png" /> is a basis of the module <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241047.png" />. In particular,
+
$$\dim(V_1 \tensor_A V_2) = \dim V_1 \cdot \dim V_2$$
 +
if the $V_i$ are free finitely-generated modules (for instance, finite-dimensional vector spaces over a field $A$). The tensor product of cyclic $A$-modules is computed by the formula
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241048.png" /></td> </tr></table>
+
$$(A/I) \tensor_A (A/J) \iso A/(I+J)$$
 +
where $I$ and $J$ are ideals in $A$.
  
if the <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241049.png" /> are free finitely-generated modules (for instance, finite-dimensional vector spaces over a field <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241050.png" />). The tensor product of cyclic <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241051.png" />-modules is computed by the formula
+
One also defines the tensor product of arbitrary (not necessarily finite) families of $A$-modules. The tensor product
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241052.png" /></td> </tr></table>
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$$\bigotimes^p V = V \tensor_A \cdots \tensor_A V \qquad (p \text{ factors})$$
 +
is called the $p$-th tensor power of the $A$-module $V$; its elements are the contravariant tensors (cf.
 +
[[Tensor on a vector space|Tensor on a vector space]]) of degree $p$ on $V$.
  
where <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241053.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241054.png" /> are ideals in <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241055.png" />.
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To any pair of homomorphisms of $A$-modules $\alpha_i : V_i \to W_i$, $i=1,2$, corresponds their tensor product $\alpha_1 \tensor \alpha_2$, which is a homomorphism of $A$-modules $V_1 \tensor_A V_2 \to W_1 \tensor_A W_2$ and is defined by the formula
  
One also defines the tensor product of arbitrary (not necessarily finite) families of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241056.png" />-modules. The tensor product
+
$$(\alpha_1 \tensor \alpha_2) (x_1 \tensor x_2) = \alpha(x_1)\tensor \alpha_2(x_2), \qquad x_i \in V_i.$$
 +
This operation can also be extended to arbitrary families of homomorphisms and has functorial properties (see
 +
[[Module|Module]]). It defines a homomorphism of $A$-modules
 +
$$\Hom_A(V_1, W_1) \tensor_A \Hom_A(V_2, W_2) \to \Hom_A(V_1 \tensor V_2, W_1 \tensor W_2),$$
 +
which is an isomorphism if all the $V_i$ and $W_i$ are free and finitely generated.
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241057.png" /></td> </tr></table>
 
  
is called the <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241059.png" />-th tensor power of the <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241060.png" />-module <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241061.png" />; its elements are the contravariant tensors (cf. [[Tensor on a vector space|Tensor on a vector space]]) of degree <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241062.png" /> on <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241063.png" />.
 
  
To any pair of homomorphisms of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241064.png" />-modules <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241065.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241066.png" />, corresponds their tensor product <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241067.png" />, which is a homomorphism of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241068.png" />-modules <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241069.png" /> and is defined by the formula
+
=====Comments=====
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241070.png" /></td> </tr></table>
+
An important interpretation of the tensor product in (theoretical) physics is as follows. Often the states of an object, say, a particle, are defined as the vector space $V$ over $\C$ of all complex linear combinations of a set of pure states $e_i$, $i \in I$. Let the pure states of a second similar object be $f_j$, $j \in J$, yielding a second vector space $W$. Then the pure states of the ordered pair of objects are all pairs $(e_i, f_j)$ and the space of states of this ordered pair is the tensor product $V\tensor_\C W$.
  
This operation can also be extended to arbitrary families of homomorphisms and has functorial properties (see [[Module|Module]]). It defines a homomorphism of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241071.png" />-modules
+
====Tensor product of two algebras====
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241072.png" /></td> </tr></table>
+
The tensor product of two algebras $C_1$ and $C_2$ over an associative commutative ring $A$ with a unit is the algebra $C_1 \tensor_A C_2$ over $A$ which is obtained by introducing on the tensor product $C_1 \tensor_A C_2$ of $A$-modules a multiplication according to the formula
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241073.png" /></td> </tr></table>
+
$$(x_1 \tensor x_2)(y_1 \tensor y_2) = (x_1 y_1) \tensor (x_2 y_2), \qquad x_i, y_i \in C_i.$$
 +
This definition can be extended to the case of an arbitrary family of factors. The tensor product $C_1 \tensor_A C_2$ is associative and commutative and contains a unit if both algebras $C_i$ have a unit. If $C_1$ and $C_2$ are algebras with a unit over the field $A$, then $\widetilde C_1 = C_1 \tensor \mathbf{1}$ and $\widetilde C_2 = \mathbf{1} \tensor C_2$ are subalgebras of $C_1 \tensor_A C_2$ which are isomorphic to $C_1$ and $C_2$ and commute elementwise. Conversely, let $C$ be an algebra with a unit over the field $A$, and let $C_1$ and $C_2$ be subalgebras of it containing its unit and such that $x_1 x_2 = x_2 x_1$ for any $x_i \in C_i$. Then there is a homomorphism of $A$-algebras $\phi : C_1 \tensor_A C_2 \to C$ such that $\phi(x_1 \tensor x_2) = x_1 x_2$, $x_i \in C_i$. For $\phi$ to be an isomorphism it is necessary and sufficient that there is in $C_1$ a basis over $A$ which is also a basis of the right $C_2$-module $C$.
  
which is an isomorphism if all the <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241074.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241075.png" /> are free and finitely generated.
+
====Tensor product of two matrices (by D.A. Suprunenko)====
  
====References====
+
The tensor product, or
<table><TR><TD valign="top">[1]</TD> <TD valign="top">  N. Bourbaki,  "Elements of mathematics. Algebra: Algebraic structures. Linear algebra" , '''1''' , Addison-Wesley  (1974)  pp. Chapt.1;2  (Translated from French)</TD></TR><TR><TD valign="top">[2]</TD> <TD valign="top">  F. Kasch,   "Modules and rings" , Acad. Press  (1982)  (Translated from German)</TD></TR><TR><TD valign="top">[3]</TD> <TD valign="top">  A.I. Kostrikin,  Yu.I. Manin,  "Linear algebra and geometry" , Gordon &amp; Breach  (1989)  (Translated from Russian)</TD></TR></table>
+
[[Kronecker product]] (cf.
 +
[[Matrix multiplication]]), of two matrices $A = [ \alpha_{ij} ]$ and $B$ is the matrix
  
 +
$$A \tensor B = \begin{bmatrix} \alpha_{11} B & \cdots & \alpha_{1n} B \\ \vdots & \ddots & \vdots \\ \alpha_{m1} B & \cdots & \alpha_{mn} B \end{bmatrix}.$$
 +
Here, $A$ is an $(m\times n)$-matrix, $B$ is a $(p \times q)$-matrix and $A \tensor B$ is an $(mp \times nq)$-matrix over an associative commutative ring $k$ with a unit.
  
 +
Properties of the tensor product of matrices are:
  
====Comments====
+
$$(A_1 + A_2) \tensor B = A_1 \tensor B + A_2 \tensor B,$$
An important interpretation of the tensor product in (theoretical) physics is as follows. Often the states of an object, say, a particle, are defined as the vector space <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241076.png" /> over <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241077.png" /> of all complex linear combinations of a set of pure states <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241078.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241079.png" />. Let the pure states of a second similar object be <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241080.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241081.png" />, yielding a second vector space <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241082.png" />. Then the pure states of the ordered pair of objects are all pairs <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241083.png" /> and the space of states of this ordered pair is the tensor product <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241084.png" />.
 
  
The tensor product of two algebras <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241085.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241086.png" /> over an associative commutative ring <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241087.png" /> with a unit is the algebra <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241088.png" /> over <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241089.png" /> which is obtained by introducing on the tensor product <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241090.png" /> of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241091.png" />-modules a multiplication according to the formula
+
$$A \tensor (B_1 + B_2) = A \tensor B_1 + A\tensor B_2,$$
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241092.png" /></td> </tr></table>
+
$$\alpha(A \tensor B) = \alpha A \tensor B = A \tensor \alpha B,$$
 +
where $\alpha \in k$,
  
This definition can be extended to the case of an arbitrary family of factors. The tensor product <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241093.png" /> is associative and commutative and contains a unit if both algebras <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241094.png" /> have a unit. If <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241095.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241096.png" /> are algebras with a unit over the field <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241097.png" />, then <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241098.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t09241099.png" /> are subalgebras of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410100.png" /> which are isomorphic to <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410101.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410102.png" /> and commute elementwise. Conversely, let <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410103.png" /> be an algebra with a unit over the field <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410104.png" />, and let <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410105.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410106.png" /> be subalgebras of it containing its unit and such that <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410107.png" /> for any <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410108.png" />. Then there is a homomorphism of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410109.png" />-algebras <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410110.png" /> such that <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410111.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410112.png" />. For <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410113.png" /> to be an isomorphism it is necessary and sufficient that there is in <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410114.png" /> a basis over <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410115.png" /> which is also a basis of the right <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410116.png" />-module <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410117.png" />.
+
$$(A \tensor B)(C \tensor D) = AC \tensor BD.$$
 +
If $m=n$ and $p=q$, then
  
====References====
+
$$\det(A \tensor B) = (\det A)^p (\det B)^n.$$
<table><TR><TD valign="top">[1]</TD> <TD valign="top">  N. Bourbaki,   "Elements of mathematics. Algebra: Algebraic structures. Linear algebra" , '''1''' , Addison-Wesley  (1974)  pp. Chapt.1;2  (Translated from French)</TD></TR></table>
+
Let $k$ be a field, $m=n$ and $p=q$. Then $A\tensor B$ is similar to $B \tensor A$, and $\det(A \tensor E_p - E_n \tensor B)$, where $E_s$ is the unit matrix, coincides with the resultant of the characteristic polynomials of $A$ and $B$.
  
 +
If $\alpha : V \to V'$ and $\beta : W \to W'$ are homomorphisms of unitary free finitely-generated $k$-modules and $A$ and $B$ are their matrices in certain bases, then $A \tensor B$ is the matrix of the homomorphism $\alpha \tensor \beta : V \tensor W \to V' \tensor W'$ in the basis consisting of the tensor products of the basis vectors.
  
 +
====Tensor product of two representations (by A.I. Shtern)====
  
The tensor product, or Kronecker product, of two matrices <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410118.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410119.png" /> is the matrix
+
The tensor product of two representations $\pi_1$ and $\pi_2$ of a group $G$ in vector spaces $E_1$ and $E_2$, respectively, is the representation $\pi_1 \tensor \pi_2$ of $G$ in $E_1 \tensor E_2$ uniquely defined by the condition
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410120.png" /></td> </tr></table>
+
$$(\pi_1 \tensor \pi_2) (g) (\xi_1 \tensor \xi_2) = \pi_1(g) \xi_1 \tensor \pi_2(g) \xi_2 \tag{*}$$
 +
for all $\xi_1 \in E_1$, $\xi_2 \in E_2$ and $g \in G$. If $\pi_1$ and $\pi_2$ are continuous unitary representations of a topological group $G$ in Hilbert spaces $E_1$ and $E_2$, respectively, then the operators $(\pi_1 \tensor \pi_2)(g)$, $g \in G$, in the vector space $E_1 \tensor E_2$ admit a unique extension by continuity to continuous linear operators $(\pi_1 \tensor -\pi_2)g$, $g\in G$, in the Hilbert space $E_1 \tensor -E_2$ (being the completion of the space $E_1 \tensor E_2$ with respect to the scalar product defined by the formula $(\xi_1 \tensor \xi_2, \eta_1 \tensor \eta_2) = (\xi_1, \eta_1)(\xi_2, \eta_2)$) and the mapping $\pi_1 \tensor \pi_2 : g \to (\pi_1 \tensor -\pi_2)g$, $g \in G$, is a continuous
 +
[[Unitary representation|unitary representation]] of the group $G$ in the Hilbert space $E_1 \tensor -E_2$, called the tensor product of the unitary representations $\pi_1$ and $\pi_2$. The representations $\pi_1 \tensor \pi_2$ and $\pi_2 \tensor \pi_1$ are equivalent (unitarily equivalent if $\pi_1$ and $\pi_2$ are unitary). The operation of tensor multiplication can be defined also for continuous representations of a topological group in topological vector spaces of a general form.
  
Here, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410121.png" /> is an <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410122.png" />-matrix, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410123.png" /> is a <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410124.png" />-matrix and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410125.png" /> is an <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410126.png" />-matrix over an associative commutative ring <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410127.png" /> with a unit.
 
  
Properties of the tensor product of matrices are:
 
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410128.png" /></td> </tr></table>
+
=====Comments=====
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410129.png" /></td> </tr></table>
+
If $\pi_i$ is a representation of an algebra $A_i$ in a vector space $E_i$, $i=1,2$, one defines the tensor product $\pi_1 \tensor \pi_2$, which is a representation of $A_1\tensor A_2$ in $E_1\tensor E_2$, by
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410130.png" /></td> </tr></table>
+
$$(\pi_1 \tensor \pi) (a_1 \tensor a_2) = \pi_1(a_1) \tensor \pi_2(a_2).$$
 +
In case $A = A_1 = A_2$ is a bi-algebra (cf.
 +
[[Hopf algebra|Hopf algebra]]), composition of this representation with the comultiplication $A \to A \tensor A$ (which is an algebra homomorphism) yields a new representation of $A$, (also) called the tensor product.
  
where <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410131.png" />,
+
In case $G$ is a group, a representation of $G$ is the same as a representation of the
 +
[[Group algebra|group algebra]] $k[G]$ of $G$, which is a bi-algebra, so that the previous construction applies, giving the same definition as (*) above. (The comultiplication on $k[G]$ is given by $g\mapsto g \tensor g$.)
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410132.png" /></td> </tr></table>
+
In case $\lieg$ is a Lie algebra, a representation of $\lieg$ is the same as a representation of its
 +
[[Universal enveloping algebra|universal enveloping algebra]], $U_\lieg$, which is also a bi-algebra (with comultiplication defined by $x\mapsto 1 \tensor x + x \tensor 1$, $x \in \lieg$). This permits one to define the tensor product of two representations of a Lie algebra:
  
If <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410133.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410134.png" />, then
+
$$(\pi_1 \tensor \pi_2)(x) = 1 \tensor \pi_2(x) + \pi_1(x) \tensor 1.$$
 +
====Tensor product of two vector bundles====
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410135.png" /></td> </tr></table>
+
The tensor product of two vector bundles $E$ and $F$ over a topological space $X$ is the vector bundle $E\tensor F$ over $X$ whose fibre at a point $x \in X$ is the tensor product of the fibres $E_x \tensor F_x$. The tensor product can be defined as the bundle whose transfer function is the tensor product of the transfer functions of the bundles $E$ and $F$ in the same trivializing covering (see Tensor product of matrices, above).
  
Let <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410136.png" /> be a field, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410137.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410138.png" />. Then <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410139.png" /> is similar to <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410140.png" />, and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410141.png" />, where <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410142.png" /> is the unit matrix, coincides with the resultant of the characteristic polynomials of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410143.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410144.png" />.
 
  
If <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410145.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410146.png" /> are homomorphisms of unitary free finitely-generated <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410147.png" />-modules and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410148.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410149.png" /> are their matrices in certain bases, then <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410150.png" /> is the matrix of the homomorphism <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410151.png" /> in the basis consisting of the tensor products of the basis vectors.
 
  
====References====
+
=====Comments=====
<table><TR><TD valign="top">[1]</TD> <TD valign="top">  P.R. Halmos,  "Finite-dimensional vector spaces" , v. Nostrand  (1958)</TD></TR><TR><TD valign="top">[2]</TD> <TD valign="top">  N. Bourbaki,  "Elements of mathematics. Algebra: Algebraic structures. Linear algebra" , '''1''' , Addison-Wesley  (1974)  pp. Chapt.1;2  (Translated from French)</TD></TR></table>
 
 
 
''D.A. Suprunenko''
 
 
 
The tensor product of two representations <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410152.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410153.png" /> of a group <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410154.png" /> in vector spaces <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410155.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410156.png" />, respectively, is the representation <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410157.png" /> of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410158.png" /> in <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410159.png" /> uniquely defined by the condition
 
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410160.png" /></td> <td valign="top" style="width:5%;text-align:right;">(*)</td></tr></table>
+
For a vector bundle $E$ over a space $X$ and a vector bundle $F$ over a space $Y$ one defines the vector bundle $E \times F$ over $X \times Y$ (sometimes written $E \tensor F$) as the vector bundle over $X \times Y$ with fibre $E_x \tensor F_y$ over $(x, y)$. Pulling back this bundle by the diagonal mapping $x \mapsto (x, x)$ defines the tensor product defined above.
  
for all <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410161.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410162.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410163.png" />. If <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410164.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410165.png" /> are continuous unitary representations of a topological group <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410166.png" /> in Hilbert spaces <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410167.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410168.png" />, respectively, then the operators <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410169.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410170.png" />, in the vector space <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410171.png" /> admit a unique extension by continuity to continuous linear operators <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410172.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410173.png" />, in the Hilbert space <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410174.png" /> (being the completion of the space <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410175.png" /> with respect to the scalar product defined by the formula <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410176.png" />) and the mapping <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410177.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410178.png" />, is a continuous [[Unitary representation|unitary representation]] of the group <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410179.png" /> in the Hilbert space <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410180.png" />, called the tensor product of the unitary representations <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410181.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410182.png" />. The representations <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410183.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410184.png" /> are equivalent (unitarily equivalent if <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410185.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410186.png" /> are unitary). The operation of tensor multiplication can be defined also for continuous representations of a topological group in topological vector spaces of a general form.
 
  
''A.I. Shtern''
 
 
====Comments====
 
If <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410187.png" /> is a representation of an algebra <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410188.png" /> in a vector space <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410189.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410190.png" />, one defines the tensor product <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410191.png" />, which is a representation of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410192.png" /> in <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410193.png" />, by
 
 
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410194.png" /></td> </tr></table>
 
 
In case <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410195.png" /> is a bi-algebra (cf. [[Hopf algebra|Hopf algebra]]), composition of this representation with the comultiplication <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410196.png" /> (which is an algebra homomorphism) yields a new representation of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410197.png" />, (also) called the tensor product.
 
 
In case <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410198.png" /> is a group, a representation of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410199.png" /> is the same as a representation of the [[Group algebra|group algebra]] <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410200.png" /> of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410201.png" />, which is a bi-algebra, so that the previous construction applies, giving the same definition as (*) above. (The comultiplication on <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410202.png" /> is given by <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410203.png" />.)
 
 
In case <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410204.png" /> is a Lie algebra, a representation of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410205.png" /> is the same as a representation of its [[Universal enveloping algebra|universal enveloping algebra]], <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410206.png" />, which is also a bi-algebra (with comultiplication defined by <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410207.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410208.png" />). This permits one to define the tensor product of two representations of a Lie algebra:
 
 
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410209.png" /></td> </tr></table>
 
 
The tensor product of two vector bundles <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410210.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410211.png" /> over a topological space <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410212.png" /> is the vector bundle <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410213.png" /> over <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410214.png" /> whose fibre at a point <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410215.png" /> is the tensor product of the fibres <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410216.png" />. The tensor product can be defined as the bundle whose transfer function is the tensor product of the transfer functions of the bundles <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410217.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410218.png" /> in the same trivializing covering (see Tensor product of matrices, above).
 
  
 
====References====
 
====References====
<table><TR><TD valign="top">[1]</TD> <TD valign="top">  M.F. Atiyah,  "<img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410219.png" />-theory: lectures" , Benjamin  (1967)</TD></TR></table>
 
 
 
  
====Comments====
+
<table> <TR><TD valign="top">[1]</TD>
For a vector bundle <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410220.png" /> over a space <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410221.png" /> and a vector bundle <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410222.png" /> over a space <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410223.png" /> one defines the vector bundle <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410224.png" /> over <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410225.png" /> (sometimes written <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410226.png" />) as the vector bundle over <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410227.png" /> with fibre <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410228.png" /> over <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410229.png" />. Pulling back this bundle by the diagonal mapping <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/t/t092/t092410/t092410230.png" /> defines the tensor product defined above.
+
<TD valign="top">  N. Bourbaki,  "Elements of mathematics. Algebra: Algebraic structures. Linear algebra" , '''1''' , Addison-Wesley  (1974)  pp. Chapt.1;2  (Translated from French)</TD>
 +
</TR> <TR><TD valign="top">[2]</TD>
 +
<TD valign="top">  F. Kasch,  "Modules and rings" , Acad. Press  (1982)  (Translated from German)</TD>
 +
</TR> <TR><TD valign="top">[3]</TD>
 +
<TD valign="top">  A.I. Kostrikin,  Yu.I. Manin,  "Linear algebra and geometry" , Gordon &amp; Breach  (1989)  (Translated from Russian)</TD>
 +
</TR> <TR><TD valign="top">[4]</TD>
 +
<TD valign="top">  P.R. Halmos,  "Finite-dimensional vector spaces" , v. Nostrand  (1958)</TD>
 +
</TR> <TR><TD valign="top">[5]</TD>
 +
<TD valign="top">  M.F. Atiyah,  "$K$-theory: lectures" , Benjamin  (1967)</TD>
 +
</TR> </table>

Latest revision as of 03:52, 23 July 2018

$\newcommand{\tensor}{\otimes}$ $\newcommand{\lieg}{\mathfrak{g}}$ $\newcommand{\iso}{\cong}$

Tensor product of two unitary modules

The tensor product of two unitary modules $V_1$ and $V_2$ over an associative commutative ring $A$ with a unit is the $A$-module $V_1 \tensor_A V_2$ together with an $A$-bilinear mapping

$$(x_1, x_2) \mapsto x_1 \tensor x_2 \in V_1 \tensor_A V_2$$ which is universal in the following sense: For any $A$-bilinear mapping $\beta: V_1 \times V_2 \to W$, where $W$ is an arbitrary $A$-module, there is a unique $A$-linear mapping $b : V_1 \tensor_A V_2 \to W$ such that

$$\beta(x_1, x_2) = b(x_1 \tensor x_2), \qquad x_1 \in V_1, \qquad x_2 \in V_2.$$ The tensor product is uniquely defined up to a natural isomorphism. It always exists and can be constructed as the quotient module of the free $A$-module $F$ generated by the set $V_1 \times V_2$ modulo the $A$-submodule $R$ generated by the elements of the form

$$(x_1 + y, x_2) - (x_1, x_2) - (y, x_2),$$

$$(x_1, x_2 + z) - (x_1, x_2) - (x_1, z),$$

$$(cx_1, x_2) - c(x_1, x_2),$$

$$(x_1, cx_2) - c(x_1, x_2),$$

$$x_1, y \in V_1, \qquad x_2, z \in V_2, \qquad c \in A;$$ then $x_1 \tensor x_2 = (x_1, x_2) + R$. If one gives up the requirement of commutativity of $A$, a construction close to the one described above allows one to form from a right $A$-module $V_1$ and a left $A$-module $V_2$ an Abelian group $V_1 \tensor_A V_2$, also called the tensor product of these modules [1]. In what follows $A$ will be assumed to be commutative.

The tensor product has the following properties:

$$A \tensor_A V \iso V,$$

$$V_1 \tensor_A V_2 \iso V_2 \tensor_A V_1,$$

$$(V_1 \tensor_A V_2) \tensor V_3 \iso V_1 \tensor_A (V_2 \tensor_A V_3),$$

$$\left( \bigoplus_{i \in I} V_i \right) \tensor_A W \iso \bigoplus_{i \in I} (V_i \tensor_A W)$$ for any $A$-modules $V$, $V_i$ and $W$.

If $(x_i)_{i \in I}$ and $(y_j)_{j \in J}$ are bases of the free $A$-modules $V_1$ and $V_2$, then $(x_i \tensor y_j)_{(i,j) \in I\times J}$ is a basis of the module $V_1 \tensor_A V_2$. In particular,

$$\dim(V_1 \tensor_A V_2) = \dim V_1 \cdot \dim V_2$$ if the $V_i$ are free finitely-generated modules (for instance, finite-dimensional vector spaces over a field $A$). The tensor product of cyclic $A$-modules is computed by the formula

$$(A/I) \tensor_A (A/J) \iso A/(I+J)$$ where $I$ and $J$ are ideals in $A$.

One also defines the tensor product of arbitrary (not necessarily finite) families of $A$-modules. The tensor product

$$\bigotimes^p V = V \tensor_A \cdots \tensor_A V \qquad (p \text{ factors})$$ is called the $p$-th tensor power of the $A$-module $V$; its elements are the contravariant tensors (cf. Tensor on a vector space) of degree $p$ on $V$.

To any pair of homomorphisms of $A$-modules $\alpha_i : V_i \to W_i$, $i=1,2$, corresponds their tensor product $\alpha_1 \tensor \alpha_2$, which is a homomorphism of $A$-modules $V_1 \tensor_A V_2 \to W_1 \tensor_A W_2$ and is defined by the formula

$$(\alpha_1 \tensor \alpha_2) (x_1 \tensor x_2) = \alpha(x_1)\tensor \alpha_2(x_2), \qquad x_i \in V_i.$$ This operation can also be extended to arbitrary families of homomorphisms and has functorial properties (see Module). It defines a homomorphism of $A$-modules $$\Hom_A(V_1, W_1) \tensor_A \Hom_A(V_2, W_2) \to \Hom_A(V_1 \tensor V_2, W_1 \tensor W_2),$$ which is an isomorphism if all the $V_i$ and $W_i$ are free and finitely generated.


Comments

An important interpretation of the tensor product in (theoretical) physics is as follows. Often the states of an object, say, a particle, are defined as the vector space $V$ over $\C$ of all complex linear combinations of a set of pure states $e_i$, $i \in I$. Let the pure states of a second similar object be $f_j$, $j \in J$, yielding a second vector space $W$. Then the pure states of the ordered pair of objects are all pairs $(e_i, f_j)$ and the space of states of this ordered pair is the tensor product $V\tensor_\C W$.

Tensor product of two algebras

The tensor product of two algebras $C_1$ and $C_2$ over an associative commutative ring $A$ with a unit is the algebra $C_1 \tensor_A C_2$ over $A$ which is obtained by introducing on the tensor product $C_1 \tensor_A C_2$ of $A$-modules a multiplication according to the formula

$$(x_1 \tensor x_2)(y_1 \tensor y_2) = (x_1 y_1) \tensor (x_2 y_2), \qquad x_i, y_i \in C_i.$$ This definition can be extended to the case of an arbitrary family of factors. The tensor product $C_1 \tensor_A C_2$ is associative and commutative and contains a unit if both algebras $C_i$ have a unit. If $C_1$ and $C_2$ are algebras with a unit over the field $A$, then $\widetilde C_1 = C_1 \tensor \mathbf{1}$ and $\widetilde C_2 = \mathbf{1} \tensor C_2$ are subalgebras of $C_1 \tensor_A C_2$ which are isomorphic to $C_1$ and $C_2$ and commute elementwise. Conversely, let $C$ be an algebra with a unit over the field $A$, and let $C_1$ and $C_2$ be subalgebras of it containing its unit and such that $x_1 x_2 = x_2 x_1$ for any $x_i \in C_i$. Then there is a homomorphism of $A$-algebras $\phi : C_1 \tensor_A C_2 \to C$ such that $\phi(x_1 \tensor x_2) = x_1 x_2$, $x_i \in C_i$. For $\phi$ to be an isomorphism it is necessary and sufficient that there is in $C_1$ a basis over $A$ which is also a basis of the right $C_2$-module $C$.

Tensor product of two matrices (by D.A. Suprunenko)

The tensor product, or Kronecker product (cf. Matrix multiplication), of two matrices $A = [ \alpha_{ij} ]$ and $B$ is the matrix

$$A \tensor B = \begin{bmatrix} \alpha_{11} B & \cdots & \alpha_{1n} B \\ \vdots & \ddots & \vdots \\ \alpha_{m1} B & \cdots & \alpha_{mn} B \end{bmatrix}.$$ Here, $A$ is an $(m\times n)$-matrix, $B$ is a $(p \times q)$-matrix and $A \tensor B$ is an $(mp \times nq)$-matrix over an associative commutative ring $k$ with a unit.

Properties of the tensor product of matrices are:

$$(A_1 + A_2) \tensor B = A_1 \tensor B + A_2 \tensor B,$$

$$A \tensor (B_1 + B_2) = A \tensor B_1 + A\tensor B_2,$$

$$\alpha(A \tensor B) = \alpha A \tensor B = A \tensor \alpha B,$$ where $\alpha \in k$,

$$(A \tensor B)(C \tensor D) = AC \tensor BD.$$ If $m=n$ and $p=q$, then

$$\det(A \tensor B) = (\det A)^p (\det B)^n.$$ Let $k$ be a field, $m=n$ and $p=q$. Then $A\tensor B$ is similar to $B \tensor A$, and $\det(A \tensor E_p - E_n \tensor B)$, where $E_s$ is the unit matrix, coincides with the resultant of the characteristic polynomials of $A$ and $B$.

If $\alpha : V \to V'$ and $\beta : W \to W'$ are homomorphisms of unitary free finitely-generated $k$-modules and $A$ and $B$ are their matrices in certain bases, then $A \tensor B$ is the matrix of the homomorphism $\alpha \tensor \beta : V \tensor W \to V' \tensor W'$ in the basis consisting of the tensor products of the basis vectors.

Tensor product of two representations (by A.I. Shtern)

The tensor product of two representations $\pi_1$ and $\pi_2$ of a group $G$ in vector spaces $E_1$ and $E_2$, respectively, is the representation $\pi_1 \tensor \pi_2$ of $G$ in $E_1 \tensor E_2$ uniquely defined by the condition

$$(\pi_1 \tensor \pi_2) (g) (\xi_1 \tensor \xi_2) = \pi_1(g) \xi_1 \tensor \pi_2(g) \xi_2 \tag{*}$$ for all $\xi_1 \in E_1$, $\xi_2 \in E_2$ and $g \in G$. If $\pi_1$ and $\pi_2$ are continuous unitary representations of a topological group $G$ in Hilbert spaces $E_1$ and $E_2$, respectively, then the operators $(\pi_1 \tensor \pi_2)(g)$, $g \in G$, in the vector space $E_1 \tensor E_2$ admit a unique extension by continuity to continuous linear operators $(\pi_1 \tensor -\pi_2)g$, $g\in G$, in the Hilbert space $E_1 \tensor -E_2$ (being the completion of the space $E_1 \tensor E_2$ with respect to the scalar product defined by the formula $(\xi_1 \tensor \xi_2, \eta_1 \tensor \eta_2) = (\xi_1, \eta_1)(\xi_2, \eta_2)$) and the mapping $\pi_1 \tensor \pi_2 : g \to (\pi_1 \tensor -\pi_2)g$, $g \in G$, is a continuous unitary representation of the group $G$ in the Hilbert space $E_1 \tensor -E_2$, called the tensor product of the unitary representations $\pi_1$ and $\pi_2$. The representations $\pi_1 \tensor \pi_2$ and $\pi_2 \tensor \pi_1$ are equivalent (unitarily equivalent if $\pi_1$ and $\pi_2$ are unitary). The operation of tensor multiplication can be defined also for continuous representations of a topological group in topological vector spaces of a general form.


Comments

If $\pi_i$ is a representation of an algebra $A_i$ in a vector space $E_i$, $i=1,2$, one defines the tensor product $\pi_1 \tensor \pi_2$, which is a representation of $A_1\tensor A_2$ in $E_1\tensor E_2$, by

$$(\pi_1 \tensor \pi) (a_1 \tensor a_2) = \pi_1(a_1) \tensor \pi_2(a_2).$$ In case $A = A_1 = A_2$ is a bi-algebra (cf. Hopf algebra), composition of this representation with the comultiplication $A \to A \tensor A$ (which is an algebra homomorphism) yields a new representation of $A$, (also) called the tensor product.

In case $G$ is a group, a representation of $G$ is the same as a representation of the group algebra $k[G]$ of $G$, which is a bi-algebra, so that the previous construction applies, giving the same definition as (*) above. (The comultiplication on $k[G]$ is given by $g\mapsto g \tensor g$.)

In case $\lieg$ is a Lie algebra, a representation of $\lieg$ is the same as a representation of its universal enveloping algebra, $U_\lieg$, which is also a bi-algebra (with comultiplication defined by $x\mapsto 1 \tensor x + x \tensor 1$, $x \in \lieg$). This permits one to define the tensor product of two representations of a Lie algebra:

$$(\pi_1 \tensor \pi_2)(x) = 1 \tensor \pi_2(x) + \pi_1(x) \tensor 1.$$

Tensor product of two vector bundles

The tensor product of two vector bundles $E$ and $F$ over a topological space $X$ is the vector bundle $E\tensor F$ over $X$ whose fibre at a point $x \in X$ is the tensor product of the fibres $E_x \tensor F_x$. The tensor product can be defined as the bundle whose transfer function is the tensor product of the transfer functions of the bundles $E$ and $F$ in the same trivializing covering (see Tensor product of matrices, above).


Comments

For a vector bundle $E$ over a space $X$ and a vector bundle $F$ over a space $Y$ one defines the vector bundle $E \times F$ over $X \times Y$ (sometimes written $E \tensor F$) as the vector bundle over $X \times Y$ with fibre $E_x \tensor F_y$ over $(x, y)$. Pulling back this bundle by the diagonal mapping $x \mapsto (x, x)$ defines the tensor product defined above.


References

[1] N. Bourbaki, "Elements of mathematics. Algebra: Algebraic structures. Linear algebra" , 1 , Addison-Wesley (1974) pp. Chapt.1;2 (Translated from French)
[2] F. Kasch, "Modules and rings" , Acad. Press (1982) (Translated from German)
[3] A.I. Kostrikin, Yu.I. Manin, "Linear algebra and geometry" , Gordon & Breach (1989) (Translated from Russian)
[4] P.R. Halmos, "Finite-dimensional vector spaces" , v. Nostrand (1958)
[5] M.F. Atiyah, "$K$-theory: lectures" , Benjamin (1967)
How to Cite This Entry:
Tensor product. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Tensor_product&oldid=12437
This article was adapted from an original article by A.L. Onishchik (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article