tensor product#Tensor product of algebras

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 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.

Let {{mvar|V}} and {{mvar|W}} be two vector spaces over a field {{mvar|F}}, with respective bases $B\_V$ and {{tmath|1= B_W }}.

The tensor product $V\; \backslash otimes\; W$ of {{mvar|V}} and {{mvar|W}} is a vector space that has as a basis the set of all $v\backslash otimes\; w$ with $v\backslash in\; B\_V$ 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): $V\; \backslash otimes\; W$ is the set of the functions from the Cartesian product $B\_V\; \backslash times\; B\_W$ to {{mvar|F}} that have a finite number of nonzero values. The pointwise operations make $V\; \backslash otimes\; W$ a vector space. The function that maps $(v,w)$ to {{math|1}} and the other elements of $B\_V\; \backslash times\; B\_W$ to {{math|0}} is denoted {{tmath|1= v\otimes w }}.

The set $\backslash \{v\backslash otimes\; w\backslash mid\; v\backslash in\; B\_V,\; w\backslash in\; B\_W\backslash \}$ is then straightforwardly a basis of {{tmath|1= V \otimes W }}, which is called the tensor product of the bases $B\_V$ and {{tmath|1= B_W }}.

We can equivalently define $V\; \backslash otimes\; W$ to be the set of bilinear forms on $V\; \backslash times\; W$ that are nonzero at only a finite number of elements of {{tmath|1= B_V \times B_W }}. To see this, given $(x,y)\backslash in\; V\; \backslash times\; W$ and a bilinear form {{tmath|1= B : V \times W \to F }}, we can decompose $x$ and $y$ in the bases $B\_V$ and $B\_W$ as:

$$x=\backslash sum\_\{v\backslash in\; B\_V\}\; x\_v\backslash ,\; v\; \backslash quad\; \backslash text\{and\}\backslash quad\; y=\backslash sum\_\{w\backslash in\; B\_W\}\; y\_w\backslash ,\; w,$$

where only a finite number of $x\_v$'s and $y\_w$'s are nonzero, and find by the bilinearity of $B$ that:

$$B(x,\; y)\; =\backslash sum\_\{v\backslash in\; B\_V\}\backslash sum\_\{w\backslash in\; B\_W\}\; x\_v\; y\_w\backslash ,\; B(v,\; w)$$

Hence, we see that the value of $B$ for any $(x,y)\backslash in\; V\; \backslash times\; W$ 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 $v\backslash otimes\; w$ defined on $B\_V\; \backslash times\; B\_W$ as before into bilinear maps $v\; \backslash otimes\; w\; :\; V\backslash times\; W\; \backslash to\; F$ , by letting:

$$(v\; \backslash otimes\; w)(x,\; y)\; :\; =\backslash sum\_\{v\text{'}\backslash in\; B\_V\}\backslash sum\_\{w\text{'}\backslash in\; B\_W\}\; x\_\{v\text{'}\}\; y\_\{w\text{'}\}\backslash ,\; (v\; \backslash otimes\; w)(v\text{'},\; w\text{'})\; =\; x\_v\; \backslash ,\; y\_w\; .$$

Then we can express any bilinear form $B$ as a (potentially infinite) formal linear combination of the $v\backslash otimes\; w$ maps according to:

$$B\; =\; \backslash sum\_\{v\backslash in\; B\_V\}\backslash sum\_\{w\backslash in\; B\_W\}\; B(v,\; w)(v\; \backslash otimes\; w)$$

making these maps similar to a Schauder basis for the vector space $\backslash text\{Hom\}(V,\; W;\; F)$ of all bilinear forms on {{tmath|1= V \times W }}. To instead have it be a proper Hamel basis, it only remains to add the requirement that $B$ 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 $x\backslash in\; V$ and $y\; \backslash in\; W$ as before:

$$\backslash begin\{align\}$$

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}

This definition is quite clearly derived from the coefficients of $B(v,\; w)$ in the expansion by bilinearity of $B(x,\; y)$ using the bases $B\_V$ and {{tmath|1= B_W }}, as done above. It is then straightforward to verify that with this definition, the map $\{\backslash otimes\}\; :\; (x,y)\backslash mapsto\; x\backslash otimes\; y$ is a bilinear map from $V\backslash times\; W$ to $V\backslash otimes\; W$ satisfying the universal property that any construction of the tensor product satisfies (see below).

If arranged into a rectangular array, the coordinate vector of $x\backslash otimes\; y$ is the outer product of the coordinate vectors of $x$ 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.

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 spaces over a field (mathematics).

One considers first a vector space {{mvar|L}} that has the Cartesian product $V\backslash times\; W$ as a basis. That is, the basis elements of {{mvar|L}} are the pairs $(v,w)$ with $v\backslash in\; V$ and {{tmath|1= w\in W }}. To get such a vector space, one can define it as the vector space of the functions $V\backslash times\; W\; \backslash to\; F$ that have a finite number of nonzero values and identifying $(v,w)$ with the function that takes the value {{math|1}} on $(v,w)$ 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 spanned by the elements of one of the forms:

: $\backslash begin\{align\}$

(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}

where {{tmath|1= v, v_1, v_2\in V }}, $w,\; w\_1,\; w\_2\; \backslash in\; W$ and {{tmath|1= s\in F }}.

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

: $V\backslash otimes\; W=L/R,$

and the image of $(v,w)$ 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.)

File:Another universal tensor prod.svg (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 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 $\{\backslash otimes\}\; :\; (v,w)\backslash mapsto\; v\backslash otimes\; w$ from $V\backslash times\; W$ 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 $h=\backslash tilde\; h\; \backslash circ\; \{\backslash otimes\}$ (that is, $h(v,\; w)=\; \backslash tilde\; h(v\backslash otimes\; w)$ for every $v\backslash in\; V$ and {{tmath|1= w\in W }}).

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 $Z$ be complex vector spaces and let $T\; :\; X\; \backslash times\; Y\; \backslash to\; Z$ be a bilinear map. Then $(Z,\; T)$ is a tensor product of $X$ and $Y$ if and only if{{sfn|Trèves|2006|pp=403-404}} the image of $T$ spans all of $Z$ (that is, {{tmath|1= \operatorname{span} \; T(X \times Y) = Z }}), and also $X$ and $Y$ are {{em|$T$-linearly disjoint}}, which by definition means that for all positive integers $n$ and all elements $x\_1,\; \backslash ldots,\; x\_n\; \backslash in\; X$ and $y\_1,\; \backslash ldots,\; y\_n\; \backslash in\; Y$ such that {{tmath|1= \sum_{i=1}^n T\left(x_i, y_i\right) = 0 }},

1. if all $x\_1,\; \backslash ldots,\; x\_n$ are linearly independent then all $y\_i$ are {{tmath|1= 0 }}, and
2. if all $y\_1,\; \backslash ldots,\; y\_n$ are linearly independent then all $x\_i$ are {{tmath|1= 0 }}.

Equivalently, $X$ and $Y$ are $T$-linearly disjoint if and only if for all linearly independent sequences $x\_1,\; \backslash ldots,\; x\_m$ in $X$ and all linearly independent sequences $y\_1,\; \backslash ldots,\; y\_n$ in {{tmath|1= Y }}, the vectors $\backslash left\backslash \{T\backslash left(x\_i,\; y\_j\backslash right)\; :\; 1\; \backslash leq\; i\; \backslash leq\; m,\; 1\; \backslash leq\; j\; \backslash leq\; n\backslash right\backslash \}$ are linearly independent.

}}

For example, it follows immediately that if {{tmath|1=X=\C^m}} and {{tmath|1=Y=\C^n}}, where $m$ and $n$ are positive integers, then one may set $Z\; =\; \backslash Complex^\{mn\}$ and define the bilinear map as

$$\backslash begin\{align\}$$

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}

to form the tensor product of $X$ and {{tmath|1= Y }}.{{sfn|Trèves|2006|pp=407}} Often, this map $T$ is denoted by $\backslash ,\backslash otimes\backslash ,$ so that $x\; \backslash otimes\; y\; =\; T(x,\; y).$

As another example, suppose that $\backslash Complex^S$ is the vector space of all complex-valued functions on a set $S$ with addition and scalar multiplication defined pointwise (meaning that $f\; +\; g$ is the map $s\; \backslash mapsto\; f(s)\; +\; g(s)$ and $c\; f$ is the map {{tmath|1= s \mapsto c f(s) }}). Let $S$ and $T$ be any sets and for any $f\; \backslash in\; \backslash Complex^S$ and {{tmath|1= g \in \Complex^T }}, let $f\; \backslash otimes\; g\; \backslash in\; \backslash Complex^\{S\; \backslash times\; T\}$ denote the function defined by {{tmath|1= (s, t) \mapsto f(s) g(t) }}.

If $X\; \backslash subseteq\; \backslash Complex^S$ and $Y\; \backslash subseteq\; \backslash Complex^T$ are vector subspaces then the vector subspace $Z\; :=\; \backslash operatorname\{span\}\; \backslash left\backslash \{f\; \backslash otimes\; g\; :\; f\; \backslash in\; X,\; g\; \backslash in\; Y\backslash right\backslash \}$ of $\backslash Complex^\{S\; \backslash times\; T\}$ together with the bilinear map:

$$\backslash begin\{alignat\}\{4\}$$

\;&& X \times Y &&\;\to \;& Z \\[0.3ex]

&& (f, g) &&\;\mapsto\;& f \otimes g \\

\end{alignat}

form a tensor product of $X$ and {{tmath|1= Y }}.{{sfn|Trèves|2006|pp=407}}

If {{math|V}} and {{math|W}} are vector spaces of finite dimension, then $V\backslash otimes\; W$ is finite-dimensional, and its dimension is the product of the dimensions of {{math|V}} and {{math|W}}.

This results from the fact that a basis of $V\backslash otimes\; W$ is formed by taking all tensor products of a basis element of {{math|V}} and a basis element of {{math|W}}.

The tensor product is associative in the sense that, given three vector spaces {{tmath|1= U, V, W }}, there is a canonical isomorphism:

: $(U\backslash otimes\; V)\backslash otimes\; W\backslash cong\; U\backslash otimes\; (V\backslash otimes\; W),$

that maps $(u\backslash otimes\; v)\backslash otimes\; w$ to {{tmath|1= u\otimes (v \otimes w) }}.

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

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

: $V\; \backslash otimes\; W\; \backslash cong\; W\backslash otimes\; V,$

that maps $v\; \backslash otimes\; w$ to {{tmath|1= w \otimes v }}.

On the other hand, even when {{tmath|1= V=W }}, the tensor product of vectors is not commutative; that is {{tmath|1= v\otimes w \neq w \otimes v }}, in general.

{{anchor|Tensor powers and braiding}}

The map $x\backslash otimes\; y\; \backslash mapsto\; y\backslash otimes\; x$ from $V\backslash otimes\; V$ to itself induces a linear automorphism that is called a {{vanchor|braiding map}}.

More generally and as usual (see tensor algebra), let $V^\{\backslash otimes\; n\}$ denote the tensor product of {{mvar|n}} copies of the vector space {{mvar|V}}. For every permutation {{mvar|s}} of the first {{mvar|n}} positive integers, the map:

: $x\_1\backslash otimes\; \backslash cdots\; \backslash otimes\; x\_n\; \backslash mapsto\; x\_\{s(1)\}\backslash otimes\; \backslash cdots\; \backslash otimes\; x\_\{s(n)\}$

induces a linear automorphism of {{tmath|1= V^{\otimes n}\to V^{\otimes n} }}, which is called a braiding map.

{{redirect|Tensor product of linear maps|the generalization for modules|Tensor product of modules#Tensor product of linear maps and a change of base ring}}

Given a linear map {{tmath|1= f : U\to V }}, and a vector space {{mvar|W}}, the tensor product:

: $f\backslash otimes\; W\; :\; U\backslash otimes\; W\backslash to\; V\backslash otimes\; W$

is the unique linear map such that:

: $(f\backslash otimes\; W)(u\backslash otimes\; w)=f(u)\backslash otimes\; w.$

The tensor product $W\backslash otimes\; f$ is defined similarly.

Given two linear maps $f\; :\; U\backslash to\; V$ and {{tmath|1= g : W\to Z }}, their tensor product:

: $f\backslash otimes\; g\; :\; U\backslash otimes\; W\backslash to\; V\backslash otimes\; Z$

is the unique linear map that satisfies:

: $(f\backslash otimes\; g)(u\backslash otimes\; w)=f(u)\backslash otimes\; g(w).$

One has:

: $f\backslash otimes\; g=\; (f\backslash otimes\; Z)\backslash circ\; (U\backslash otimes\; g)\; =\; (V\backslash otimes\; g)\backslash circ\; (f\backslash otimes\; W).$

In terms of category theory, this means that the tensor product is a bifunctor from the category of vector spaces to itself.{{cite book| last1=Hazewinkel|first1=Michiel|last2=Gubareni|first2=Nadezhda Mikhaĭlovna| last3=Gubareni|first3=Nadiya|last4=Kirichenko|first4=Vladimir V.|title=Algebras, rings and modules|page=100|

publisher=Springer|year=2004|isbn=978-1-4020-2690-4}}

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

By choosing bases of all vector spaces involved, the linear maps {{math|f}} and {{math|g}} can be represented by matrices. Then, depending on how the tensor $v\; \backslash otimes\; w$ is vectorized, the matrix describing the tensor product $f\; \backslash otimes\; g$ is the Kronecker product of the two matrices. For example, if {{math|V, X, W}}, and {{math|U}} above are all two-dimensional and bases have been fixed for all of them, and {{math|f}} and {{math|g}} are given by the matrices:

$$A=\backslash begin\{bmatrix\}$$

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},

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

\begin{bmatrix}

a_{1,1} & a_{1,2} \\

a_{2,1} & a_{2,2} \\

\end{bmatrix}

\otimes

\begin{bmatrix}

b_{1,1} & b_{1,2} \\

b_{2,1} & b_{2,2} \\

\end{bmatrix}

=

\begin{bmatrix}

a_{1,1} \begin{bmatrix}

b_{1,1} & b_{1,2} \\

b_{2,1} & b_{2,2} \\

\end{bmatrix} & a_{1,2} \begin{bmatrix}

b_{1,1} & b_{1,2} \\

b_{2,1} & b_{2,2} \\

\end{bmatrix} \\[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}.

The resultant rank is at most 4, and thus the resultant dimension is 4. {{em|rank}} here denotes the tensor rank i.e. the number of requisite indices (while the matrix rank counts the number of degrees of freedom in the resulting array). {{tmath|1= \operatorname{Tr} A \otimes B = \operatorname{Tr} A \times \operatorname{Tr} B }}.

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

{{See also|Tensor}}

For non-negative integers {{math|r}} and {{math|s}} a type $(r,\; s)$ tensor on a vector space {{math|V}} is an element of:

$$T^r\_s(V)\; =\; \backslash underbrace\{\; V\; \backslash otimes\; \backslash cdots\; \backslash otimes\; V\}\_r\; \backslash otimes\; \backslash underbrace\{\; V^*\; \backslash otimes\; \backslash cdots\; \backslash otimes\; V^*\}\_s\; =\; V^\{\backslash otimes\; r\}\; \backslash otimes\; \backslash left(V^*\backslash right)^\{\backslash otimes\; s\}.$$

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

There is a product map, called the {{em|(tensor) product of tensors}}:{{refn|{{harvp|Bourbaki|1989|p=244}} defines the usage "tensor product of x and y", elements of the respective modules.}}

$$T^r\_s\; (V)\; \backslash otimes\_K\; T^\{r\text{'}\}\_\{s\text{'}\}\; (V)\; \backslash to\; T^\{r+r\text{'}\}\_\{s+s\text{'}\}(V).$$

It is defined by grouping all occurring "factors" {{math|V}} together: writing $v\_i$ for an element of {{math|V}} and $f\_i$ for an element of the dual space:

$$(v\_1\; \backslash otimes\; f\_1)\; \backslash otimes\; (v\text{'}\_1)\; =\; v\_1\; \backslash otimes\; v\text{'}\_1\; \backslash otimes\; f\_1.$$

If {{math|V}} is finite dimensional, then picking a basis of {{math|V}} and the corresponding dual basis of $V^*$ naturally induces a basis of $T\_s^r(V)$ (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 {{math|F}} and {{math|G}} are two covariant tensors of orders {{math|m}} and {{math|n}} respectively (i.e. $F\; \backslash in\; T\_m^0$ and {{tmath|1= G \in T_n^0 }}), then the components of their tensor product are given by: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.

$$(F\; \backslash otimes\; G)\_\{i\_1\; i\_2\; \backslash cdots\; i\_\{m+n\}\}\; =\; F\_\{i\_1\; i\_2\; \backslash cdots\; i\_m\}\; G\_\{i\_\{m+1\}\; i\_\{m+2\}\; i\_\{m+3\}\; \backslash cdots\; i\_\{m+n\}\}.$$

Thus, the components of the tensor product of two tensors are the ordinary product of the components of each tensor. Another example: let {{math|U}} be a tensor of type {{math|(1, 1)}} with components {{tmath|1= U^{\alpha}_{\beta} }}, and let {{math|V}} be a tensor of type $(1,\; 0)$ with components {{tmath|1= V^{\gamma} }}. Then:

$$\backslash left(U\; \backslash otimes\; V\backslash right)^\backslash alpha\; \{\}\_\backslash beta\; \{\}^\backslash gamma\; =\; U^\backslash alpha\; \{\}\_\backslash beta\; V^\backslash gamma$$

and:

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

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

For tensors of type {{math|(1, 1)}} there is a canonical evaluation map:

$$V\; \backslash otimes\; V^*\; \backslash to\; K$$

defined by its action on pure tensors:

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

More generally, for tensors of type {{tmath|1= (r, s) }}, with {{math|r, s > 0}}, there is a map, called tensor contraction:

$$T^r\_s\; (V)\; \backslash to\; T^\{r-1\}\_\{s-1\}(V).$$

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

On the other hand, if $V$ is {{em|finite-dimensional}}, there is a canonical map in the other direction (called the coevaluation map):

$$\backslash begin\{cases\}\; K\; \backslash to\; V\; \backslash otimes\; V^*\; \backslash \backslash \; \backslash lambda\; \backslash mapsto\; \backslash sum\_i\; \backslash lambda\; v\_i\; \backslash otimes\; v^*\_i\; \backslash end\{cases\}$$

where $v\_1,\; \backslash ldots,\; v\_n$ is any basis of {{tmath|1= V }}, and $v\_i^*$ is its dual basis. This map does not depend on the choice of basis.{{Cite web| url= https://unapologetic.wordpress.com/2008/11/13/the-coevaluation-on-vector-spaces/|title=The Coevaluation on Vector Spaces|date=2008-11-13| website=The Unapologetic Mathematician|access-date=2017-01-26| url-status=live |archive-url =https://web.archive.org/web/20170202080439/https://unapologetic.wordpress.com/2008/11/13/the-coevaluation-on-vector-spaces/| archive-date =2017-02-02}}

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

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

$$u(a\; \backslash otimes\; b)\; =\; u(a)\; \backslash otimes\; b\; -\; a\; \backslash otimes\; u^*(b),$$

where $u^*\; \backslash in\; \backslash mathrm\{End\}\backslash left(V^*\backslash right)$ is the transpose of {{math|u}}, that is, in terms of the obvious pairing on {{tmath|1= V \otimes V^* }},

$$\backslash langle\; u(a),\; b\; \backslash rangle\; =\; \backslash langle\; a,\; u^*(b)\; \backslash rangle.$$

There is a canonical isomorphism $T^1\_1(V)\; \backslash to\; \backslash mathrm\{End\}(V)$ given by:

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

Under this isomorphism, every {{math|u}} in $\backslash mathrm\{End\}(V)$ may be first viewed as an endomorphism of $T^1\_1(V)$ and then viewed as an endomorphism of {{tmath|1= \mathrm{End}(V) }}. In fact it is the adjoint representation {{math|ad(u)}} of {{tmath|1= \mathrm{End}(V) }}.

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

$$U^*\; \backslash otimes\; V\; \backslash cong\; \backslash mathrm\{Hom\}(U,\; V),$$

defined by an action of the pure tensor $f\; \backslash otimes\; v\; \backslash in\; U^*\backslash otimes\; V$ on an element of {{tmath|1= U }},

$$(f\; \backslash otimes\; v)(u)\; =\; f(u)\; v.$$

Its "inverse" can be defined using a basis $\backslash \{u\_i\backslash \}$ and its dual basis $\backslash \{u^*\_i\backslash \}$ as in the section "Evaluation map and tensor contraction" above:

$$\backslash begin\{cases\}\; \backslash mathrm\{Hom\}\; (U,V)\; \backslash to\; U^*\; \backslash otimes\; V\; \backslash \backslash \; F\; \backslash mapsto\; \backslash sum\_i\; u^*\_i\; \backslash otimes\; F(u\_i).\; \backslash end\{cases\}$$

This result implies:

$$\backslash dim(U\; \backslash otimes\; V)\; =\; \backslash dim(U)\backslash dim(V),$$

which automatically gives the important fact that $\backslash \{u\_i\backslash otimes\; v\_j\backslash \}$ forms a basis of $U\; \backslash otimes\; V$ where $\backslash \{u\_i\backslash \},\; \backslash \{v\_j\backslash \}$ are bases of {{math|U}} and {{math|V}}.

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

$$\backslash mathrm\{Hom\}\; (U\; \backslash otimes\; V,\; W)\; \backslash cong\; \backslash mathrm\{Hom\}\; (U,\; \backslash mathrm\{Hom\}(V,\; W)).$$

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

{{main|Tensor product of modules}}

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

$$A\; \backslash otimes\_R\; B\; :=\; F\; (A\; \backslash times\; B)\; /\; G\; ,$$

where now $F(A\; \backslash times\; B)$ is the Free module generated by the cartesian product and {{math|G}} is the {{math|R}}-module generated by these relations.

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

$$(ar,b)\backslash sim\; (a,rb)$$

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

The universal property also carries over, slightly modified: the map $\backslash varphi\; :\; A\; \backslash times\; B\; \backslash to\; A\; \backslash otimes\_R\; B$ defined by $(a,\; b)\; \backslash mapsto\; a\; \backslash otimes\; b$ is a middle linear map (referred to as "the canonical middle linear map"

{{cite book|last=Hungerford|first=Thomas W.|title=Algebra|

publisher=Springer|year=1974|isbn=0-387-90518-9}}); that is, it satisfies:

{{citation|last=Chen|first=Jungkai Alfred|title=Advanced Algebra II|chapter=Tensor product|chapter-url=http://www.math.ntu.edu.tw/~jkchen/S04AA/S04AAL10.pdf|type=lecture notes|date=Spring 2004|place=National Taiwan University|url-status=live|archive-url=https://web.archive.org/web/20160304040639/http://www.math.ntu.edu.tw/~jkchen/S04AA/S04AAL10.pdf|archive-date=2016-03-04}}

$$\backslash begin\{align\}$$

\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}

The first two properties make {{math|φ}} a bilinear map of the abelian group {{tmath|1= A \times B }}. For any middle linear map $\backslash psi$ of {{tmath|1= A \times B }}, a unique group homomorphism {{math|f}} of $A\; \backslash otimes\_R\; B$ satisfies {{tmath|1= \psi = f \circ \varphi }}, and this property determines $\backslash varphi$ within group isomorphism. See the main article for details.

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:

$$A\; \backslash otimes\_R\; B\; :=\; F\; (A\; \backslash times\; B)\; /\; G$$

where $F\; (A\; \backslash times\; B)$ is a free abelian group over $A\; \backslash times\; B$ and G is the subgroup of $F\; (A\; \backslash times\; B)$ generated by relations:

$$\backslash begin\{align\}$$

&\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}

The universal property can be stated as follows. Let G be an abelian group with a map $q:A\backslash times\; B\; \backslash to\; G$ that is bilinear, in the sense that:

$$\backslash begin\{align\}$$

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}

Then there is a unique map $\backslash overline\{q\}:A\backslash otimes\; B\; \backslash to\; G$ such that $\backslash overline\{q\}(a\backslash otimes\; b)\; =\; q(a,b)$ for all $a\; \backslash in\; A$ and {{tmath|1= b \in B }}.

Furthermore, we can give $A\; \backslash otimes\_R\; B$ a module structure under some extra conditions:

1. If A is a (S,R)-bimodule, then $A\; \backslash otimes\_R\; B$ is a left S-module, where {{tmath|1= s(a\otimes b):=(sa)\otimes b }}.
2. If B is a (R,S)-bimodule, then $A\; \backslash otimes\_R\; B$ is a right S-module, where {{tmath|1= (a\otimes b)s:=a\otimes (bs) }}.
3. If A is a (S,R)-bimodule and B is a (R,T)-bimodule, then $A\; \backslash otimes\_R\; B$ is a (S,T)-bimodule, where the left and right actions are defined in the same way as the previous two examples.
4. If R is a commutative ring, then A and B are (R,R)-bimodules where $ra:=ar$ and {{tmath|1= br:=rb }}. By 3), we can conclude $A\; \backslash otimes\_R\; B$ is a (R,R)-bimodule.

For vector spaces, the tensor product $V\; \backslash otimes\; W$ is quickly computed since bases of {{math|V}} of {{math|W}} immediately determine a basis of {{tmath|1= V \otimes W }}, as was mentioned above. For modules over a general (commutative) ring, not every module is free. For example, {{math|Z/nZ}} is not a free abelian group ({{math|Z}}-module). The tensor product with {{math|Z/nZ}} is given by:

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

More generally, given a presentation of some {{math|R}}-module {{math|M}}, that is, a number of generators $m\_i\; \backslash in\; M,\; i\; \backslash in\; I$ together with relations:

$$\backslash sum\_\{j\; \backslash in\; J\}\; a\_\{ji\}\; m\_i\; =\; 0,\backslash qquad\; a\_\{ij\}\; \backslash in\; R,$$

the tensor product can be computed as the following cokernel:

$$M\; \backslash otimes\_R\; N\; =\; \backslash operatorname\{coker\}\; \backslash left(N^J\; \backslash to\; N^I\backslash right)$$

Here {{tmath|1= N^J = \oplus_{j \in J} N }}, and the map $N^J\; \backslash to\; N^I$ is determined by sending some $n\; \backslash in\; N$ in the {{math|j}}th copy of $N^J$ to $a\_\{ij\}\; n$ (in {{tmath|1= N^I }}). Colloquially, this may be rephrased by saying that a presentation of {{math|M}} gives rise to a presentation of {{tmath|1= M \otimes_R N }}. This is referred to by saying that the tensor product is a right exact functor. It is not in general left exact, that is, given an injective map of {{math|R}}-modules {{tmath|1= M_1 \to M_2 }}, the tensor product:

$$M\_1\; \backslash otimes\_R\; N\; \backslash to\; M\_2\; \backslash otimes\_R\; N$$

is not usually injective. For example, tensoring the (injective) map given by multiplication with {{math|n}}, {{math|n : Z → Z}} with {{math|Z/nZ}} yields the zero map {{math|0 : Z/nZ → Z/nZ}}, which is not injective. Higher Tor functors measure the defect of the tensor product being not left exact. All higher Tor functors are assembled in the derived tensor product.

{{main|Tensor product of algebras}}

Let {{math|R}} be a commutative ring. The tensor product of {{math|R}}-modules applies, in particular, if {{math|A}} and {{math|B}} are Algebra (ring theory). In this case, the tensor product $A\; \backslash otimes\_R\; B$ is an {{math|R}}-algebra itself by putting:

$$(a\_1\; \backslash otimes\; b\_1)\; \backslash cdot\; (a\_2\; \backslash otimes\; b\_2)\; =\; (a\_1\; \backslash cdot\; a\_2)\; \backslash otimes\; (b\_1\; \backslash cdot\; b\_2).$$

For example:

$$R[x]\; \backslash otimes\_R\; R[y]\; \backslash cong\; R[x,\; y].$$

A particular example is when {{math|A}} and {{math|B}} are fields containing a common subfield {{math|R}}. The tensor product of fields is closely related to Galois theory: if, say, {{math|1=A = R[x] / f(x)}}, where {{math|f}} is some irreducible polynomial with coefficients in {{math|R}}, the tensor product can be calculated as:

$$A\; \backslash otimes\_R\; B\; \backslash cong\; B[x]\; /\; f(x)$$

where now {{math|f}} is interpreted as the same polynomial, but with its coefficients regarded as elements of {{math|B}}. In the larger field {{math|B}}, the polynomial may become reducible, which brings in Galois theory. For example, if {{math|1=A = B}} is a Galois extension of {{math|R}}, then:

$$A\; \backslash otimes\_R\; A\; \backslash cong\; A[x]\; /\; f(x)$$

is isomorphic (as an {{math|A}}-algebra) to the {{tmath|1= A^{\operatorname{deg}(f)} }}.

Square matrices $A$ with entries in a field $K$ represent linear maps of vector spaces, say {{tmath|1= K^n \to K^n }}, and thus linear maps $\backslash psi\; :\; \backslash mathbb\{P\}^\{n-1\}\; \backslash to\; \backslash mathbb\{P\}^\{n-1\}$ of projective spaces over {{tmath|1= K }}. If $A$ is nonsingular then $\backslash psi$ is well-defined everywhere, and the eigenvectors of $A$ correspond to the fixed points of {{tmath|1= \psi }}. The eigenconfiguration of $A$ consists of $n$ points in {{tmath|1= \mathbb{P}^{n-1} }}, provided $A$ is generic and $K$ is algebraically closed. The fixed points of nonlinear maps are the eigenvectors of tensors. Let $A\; =\; (a\_\{i\_1\; i\_2\; \backslash cdots\; i\_d\})$ be a $d$-dimensional tensor of format $n\; \backslash times\; n\; \backslash times\; \backslash cdots\; \backslash times\; n$ with entries $(a\_\{i\_1\; i\_2\; \backslash cdots\; i\_d\})$ lying in an algebraically closed field $K$ of characteristic zero. Such a tensor $A\; \backslash in\; (K^\{n\})^\{\backslash otimes\; d\}$ defines polynomial maps $K^n\; \backslash to\; K^n$ and $\backslash mathbb\{P\}^\{n-1\}\; \backslash to\; \backslash mathbb\{P\}^\{n-1\}$ with coordinates:

$$\backslash psi\_i(x\_1,\; \backslash ldots,\; x\_n)\; =\; \backslash sum\_\{j\_2=1\}^n\; \backslash sum\_\{j\_3=1\}^n\; \backslash cdots\; \backslash sum\_\{j\_d\; =\; 1\}^n\; a\_\{i\; j\_2\; j\_3\; \backslash cdots\; j\_d\}\; x\_\{j\_2\}\; x\_\{j\_3\}\backslash cdots\; x\_\{j\_d\}\; \backslash ;\backslash ;\; \backslash mbox\{for\; \}\; i\; =\; 1,\; \backslash ldots,\; n$$

Thus each of the $n$ coordinates of $\backslash psi$ is a homogeneous polynomial $\backslash psi\_i$ of degree $d-1$ in {{tmath|1= \mathbf{x} = \left(x_1, \ldots, x_n\right) }}. The eigenvectors of $A$ are the solutions of the constraint:

$$\backslash mbox\{rank\}\; \backslash begin\{pmatrix\}$$

x_1 & x_2 & \cdots & x_n \\

\psi_1(\mathbf{x}) & \psi_2(\mathbf{x}) & \cdots & \psi_n(\mathbf{x})

\end{pmatrix} \leq 1

and the eigenconfiguration is given by the variety of the $2\; \backslash times\; 2$ minors of this matrix.{{cite arXiv |last1=Abo |first1=H. |last2=Seigal |first2=A. |author2-link=Anna Seigal |last3=Sturmfels |first3=B. |author3-link=Bernd Sturmfels |title=Eigenconfigurations of Tensors |date=2015 |class=math.AG |eprint=1505.05729 }}

{{main|Topological tensor product|Tensor product of Hilbert spaces}}

Hilbert spaces generalize finite-dimensional vector spaces to arbitrary dimensions. There is an analogous operation, also called the "tensor product," that makes Hilbert spaces a symmetric monoidal category. It is essentially constructed as the metric space completion of the algebraic tensor product discussed above. However, it does not satisfy the obvious analogue of the universal property defining tensor products;{{cite web|url=https://www-users.cse.umn.edu/~garrett/m/v/nonexistence_tensors.pdf|date=July 22, 2010|title=Non-existence of tensor products of Hilbert spaces|first=Paul|last=Garrett}} the morphisms for that property must be restricted to Hilbert–Schmidt operators.{{Cite book | last1=Kadison | first1=Richard V. | last2=Ringrose | first2=John R. | title=Fundamentals of the theory of operator algebras | volume=I | publisher=American Mathematical Society | location=Providence, R.I. | series=Graduate Studies in Mathematics | isbn=978-0-8218-0819-1 | mr= 1468229 | year=1997 | at=Thm. 2.6.4.}}

In situations where the imposition of an inner product is inappropriate, one can still attempt to complete the algebraic tensor product, as a topological tensor product. However, such a construction is no longer uniquely specified: in many cases, there are multiple natural topologies on the algebraic tensor product.

{{main|Graded vector space#Operations on graded vector spaces}}

Some vector spaces can be decomposed into direct sums of subspaces. In such cases, the tensor product of two spaces can be decomposed into sums of products of the subspaces (in analogy to the way that multiplication distributes over addition).

{{main|Tensor product of representations}}

Vector spaces endowed with an additional multiplicative structure are called algebras. The tensor product of such algebras is described by the Littlewood–Richardson rule.

{{main|Tensor product of quadratic forms}}

Given two multilinear forms $f(x\_1,\backslash dots,x\_k)$ and $g\; (x\_1,\backslash dots,\; x\_m)$ on a vector space $V$ over the field $K$ their tensor product is the multilinear form:

$$(f\; \backslash otimes\; g)\; (x\_1,\backslash dots,x\_\{k+m\})\; =\; f(x\_1,\backslash dots,x\_k)\; g(x\_\{k+1\},\backslash dots,x\_\{k+m\}).$${{cite book |title=An Introduction to Manifolds | first=L. W. | last=Tu |series=Universitext |publisher=Springer | page=25 | isbn=978-1-4419-7399-3 | year=2010}}

This is a special case of the product of tensors if they are seen as multilinear maps (see also tensors as multilinear maps). Thus the components of the tensor product of multilinear forms can be computed by the Kronecker product.

{{main|Sheaf of modules}}

{{main|Vector bundle#Operations on vector bundles}}

{{See also|Tensor product bundle}}

{{main|Tensor product of fields}}

{{main|Tensor product of graphs}}

It should be mentioned that, though called "tensor product", this is not a tensor product of graphs in the above sense; actually it is the category-theoretic product in the category of graphs and graph homomorphisms. However it is actually the Kronecker tensor product of the adjacency matrices of the graphs. Compare also the section Tensor product of linear maps above.

The most general setting for the tensor product is the monoidal category. It captures the algebraic essence of tensoring, without making any specific reference to what is being tensored. Thus, all tensor products can be expressed as an application of the monoidal category to some particular setting, acting on some particular objects.

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

The exterior algebra is constructed from the exterior product. Given a vector space {{math|V}}, the exterior product $V\; \backslash wedge\; V$ is defined as:

$$V\; \backslash wedge\; V\; :=\; V\; \backslash otimes\; V\; \backslash big/\; \backslash \{v\backslash otimes\; v\; \backslash mid\; v\backslash in\; V\backslash \}.$$

When the underlying field of {{math|V}} does not have characteristic 2, then this definition is equivalent to:

$$V\; \backslash wedge\; V\; :=\; V\; \backslash otimes\; V\; \backslash big/\; \backslash bigl\backslash \{v\_1\; \backslash otimes\; v\_2\; +\; v\_2\; \backslash otimes\; v\_1\; \backslash mid\; (v\_1,\; v\_2)\; \backslash in\; V^2\backslash bigr\backslash \}.$$

The image of $v\_1\; \backslash otimes\; v\_2$ in the exterior product is usually denoted $v\_1\; \backslash wedge\; v\_2$ and satisfies, by construction, {{tmath|1= v_1 \wedge v_2 = - v_2 \wedge v_1}}. Similar constructions are possible for $V\; \backslash otimes\; \backslash dots\; \backslash otimes\; V$ ({{math|n}} factors), giving rise to {{tmath|1= \Lambda^n V }}, the {{math|n}}th exterior power of {{math|V}}. The latter notion is the basis of differential form.

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

$$V\; \backslash odot\; V\; :=\; V\; \backslash otimes\; V\; \backslash big/\; \backslash bigl\backslash \{\; v\_1\; \backslash otimes\; v\_2\; -\; v\_2\; \backslash otimes\; v\_1\; \backslash mid\; (v\_1,\; v\_2)\; \backslash in\; V^2\backslash bigr\backslash \}.$$

More generally:

$$\backslash operatorname\{Sym\}^n\; V\; :=\; \backslash underbrace\{V\; \backslash otimes\; \backslash dots\; \backslash otimes\; V\}\_n\; \backslash big/\; (\backslash dots\; \backslash otimes\; v\_i\; \backslash otimes\; v\_\{i+1\}\; \backslash otimes\; \backslash dots\; -\; \backslash dots\; \backslash otimes\; v\_\{i+1\}\; \backslash otimes\; v\_\{i\}\; \backslash otimes\; \backslash dots)$$

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

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

J's treatment also allows the representation of some tensor fields, as a and b may be functions instead of constants. This product of two functions is a derived function, and if a and b are differentiable, then a */ b is differentiable.

However, these kinds of notation are not universally present in array languages. Other array languages may require explicit treatment of indices (for example, MATLAB), and/or may not support higher-order functions such as the Jacobian derivative (for example, Fortran/APL).

{{wiktionary}}

• {{annotated link|Dyadics}}
• {{annotated link|Extension of scalars}}
• {{annotated link|Monoidal category}}
• {{annotated link|Tensor algebra}}
• {{annotated link|Tensor contraction}}
• {{annotated link|Topological tensor product}}

{{reflist}}

• {{cite book |first = Nicolas|last=Bourbaki|author-link=Nicolas Bourbaki | title = Elements of mathematics, Algebra I| publisher = Springer-Verlag | year = 1989|isbn=3-540-64243-9}}
• {{cite web|last=Gowers|first=Timothy|author-link=Tim Gowers|url=https://www.dpmms.cam.ac.uk/~wtg10/tensors3.html|title=How to lose your fear of tensor products|url-status=live|archive-url=https://web.archive.org/web/20210507014709/https://www.dpmms.cam.ac.uk/~wtg10/tensors3.html|archive-date=7 May 2021}}
• {{cite book |first=Pierre A.|last=Grillet|title=Abstract Algebra|year=2007|publisher=Springer Science+Business Media, LLC| isbn= 978-0387715674}}
• {{cite book |author-link=Paul Halmos|first=Paul|last=Halmos|title=Finite dimensional vector spaces| year=1974 |publisher= Springer |isbn= 0-387-90093-4}}
• {{cite book |first=Thomas W.|last=Hungerford|author-link=Thomas W. Hungerford| title=Algebra |year=2003 |publisher= Springer |isbn= 0387905189}}
• {{Lang Algebra|edition=3r}}
• {{cite book |first1=S.|last1=Mac Lane|author-link1=Saunders Mac Lane|author-link2=Garrett Birkhoff |last2= Birkhoff |first2= G. |title= Algebra|publisher=AMS Chelsea|year=1999|isbn=0-8218-1646-2}}
• {{cite book |first1=M.|last1=Aguiar|first2=S.|last2=Mahajan| title = Monoidal functors, species and Hopf algebras|publisher = CRM Monograph Series Vol 29 |year=2010|isbn=978-0-8218-4776-3}}
• {{Trèves François Topological vector spaces, distributions and kernels}}
• {{cite web |url=http://pages.bangor.ac.uk/~mas010/nonabtens.html |title=Bibliography on the nonabelian tensor product of groups }}

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