This is a great post that gives a great introduction to tensor prodts as a generalized form of multiplication
https://jmanton.wordpress.com/tag/intuition-behind-tensor-product/
intuition behind tensor product – Jonathan Manton's Blog
2014-07-27
None
<div><div>
<p>This note is a fresh attempt at introducing tensor products in a simple and intuitive way. The setting is a vector space on which we wish to define a multiplication. Importantly, we do not wish to limit ourselves to a multiplication producing an element of . For example, we would like to think of and as two examples of multiplying elements of . The first multiplication, , is a map from to . The second multiplication, , is a map from  to . In fact, since is well-defined even if and have different dimensions, the most general setting is the following.</p>
<p>Let and be fixed vector spaces agreed upon at the start. Assume someone has defined a multiplication rule but has not told us what it is.  Here, is a function and is a vector space, both of which are unknown to us. For example, if and , a possible choice might be where .</p>
<p>What does it mean for to be a rule for multiplication? We wish it to have certain properties characteristic of multiplication. In a field, multiplication is involved in a number of axioms, including the distributive law . Since and are vector spaces, potentially distinct from each other, not all the axioms for multiplication in a field make sense in this more general setting. (For instance, commutativity makes no sense if .) The outcome is that is declared to be a rule for multiplication if it is bilinear: , and .</p>
<p>Not knowing does not prevent us from working with it; we simply write it as and are free to manipulate it according to the aforementioned rules. We do the equivalent all the time when we prove general results about a metric; we say “let be a metric” and then make use of only the axioms of a metric, thereby ensuring our derivation is valid regardless of which metric is actually being represented by .</p>
<p>While the tensor product is more than this, in the first instance, it suffices to treat it merely as an unspecified rule for multiplication. (It is specified, but no harm comes from forgetting about this in the first instance.) Rather than write it is more convenient to write , and thinking of as multiplication reminds us that .</p>
<p>The first point is that we can treat as an unspecified multiplication, simplify expressions by manipulating it in accordance with the rules for multiplication, and at the end of the day, if we are eventually told the rule , we can evaluate its simplified expression. In computer science parlance, this can be thought of as lazy evaluation.</p>
<p>Whereas there are many metrics “incompatible” with each other, the rules for multiplication are sufficiently rigid that there exists a “most general multiplication”, and all other multiplications are special cases of it, as explained presently. The tensor product represents this most general multiplication possible.</p>
<p>To consider a specific case first, take and to be . We already know two possible multiplications are and . The claim is that the latter represents the most general multiplication possible. This means, among other things, that it must be possible to write in terms of , and indeed it is: . Note that trace is a linear operator. The precise claim is the following. No matter what is, we can pretend it is , work with this definition, then at the very end, if we are told what actually is, we can obtain the true answer by applying a particular linear transform. At the risk of belabouring the point, if we wish to simplify then we can first pretend is to obtain . Later when we are told we can deduce that the linear map required to convert from to is . Applying this linear map to yields and this is indeed the correct answer because it equals .</p>
<p>Readers wishing to think further about the above example are encouraged to consider that the most general multiplication returning a real-valued number is of the form for some matrix . How can be obtained from ? What about for multiplication rules that return a vector or a matrix?</p>
<p>The general situation works as follows. Given vector spaces and , two things can be constructed, both of which use the same symbol for brevity. We need a space for our most general multiplication, and we denote this by . We also need a rule for multiplication and we denote this by . By definition, is an element of . (Just like not all matrices can be written as , there are elements in that cannot be written as . They can, however, be written as linear combinations of such terms.)</p>
<p>At this point, I leave the reader to look up elsewhere the definition of a tensor product and relate it with the above.</p>
</div></div>||||I|||| Jonathan Manton's Blog

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 Posts Tagged ‘intuition behind tensor product’

 Tensor Products Introduced as Most General Multiplication (and as Lazy Evaluation)

 July 27, 2014 jmanton 3 comments

 This note is a fresh attempt at introducing tensor products in a simple and intuitive way. The setting is a vector space on which we wish to define a multiplication. Importantly, we do not wish to limit ourselves to a multiplication producing an element of . For example, we would like to think of and as two examples of multiplying elements of . The first multiplication, , is a map from to . The second multiplication, , is a map from to . In fact, since is well-defined even if and have different dimensions, the most general setting is the following.

 Let and be fixed vector spaces agreed upon at the start. Assume someone has defined a multiplication rule but has not told us what it is. Here, is a function and is a vector space, both of which are unknown to us. For example, if and , a possible choice might be where .

 What does it mean for to be a rule for multiplication? We wish it to have certain properties characteristic of multiplication. In a field, multiplication is involved in a number of axioms, including the distributive law . Since and are vector spaces, potentially distinct from each other, not all the axioms for multiplication in a field make sense in this more general setting. (For instance, commutativity makes no sense if .) The outcome is that is declared to be a rule for multiplication if it is bilinear: , and .

 Not knowing does not prevent us from working with it; we simply write it as and are free to manipulate it according to the aforementioned rules. We do the equivalent all the time when we prove general results about a metric; we say “let be a metric” and then make use of only the axioms of a metric, thereby ensuring our derivation is valid regardless of which metric is actually being represented by .

 While the tensor product is more than this, in the first instance, it suffices to treat it merely as an unspecified rule for multiplication. (It is specified, but no harm comes from forgetting about this in the first instance.) Rather than write it is more convenient to write , and thinking of as multiplication reminds us that .

 The first point is that we can treat as an unspecified multiplication, simplify expressions by manipulating it in accordance with the rules for multiplication, and at the end of the day, if we are eventually told the rule , we can evaluate its simplified expression. In computer science parlance, this can be thought of as lazy evaluation.

 Whereas there are many metrics “incompatible” with each other, the rules for multiplication are sufficiently rigid that there exists a “most general multiplication”, and all other multiplications are special cases of it, as explained presently. The tensor product represents this most general multiplication possible.

 To consider a specific case first, take and to be . We already know two possible multiplications are and . The claim is that the latter represents the most general multiplication possible. This means, among other things, that it must be possible to write in terms of , and indeed it is: . Note that trace is a linear operator. The precise claim is the following. No matter what is, we can pretend it is , work with this definition, then at the very end, if we are told what actually is, we can obtain the true answer by applying a particular linear transform. At the risk of belabouring the point, if we wish to simplify then we can first pretend is to obtain . Later when we are told we can deduce that the linear map required to convert from to is . Applying this linear map to yields and this is indeed the correct answer because it equals .

 Readers wishing to think further about the above example are encouraged to consider that the most general multiplication returning a real-valued number is of the form for some matrix . How can be obtained from ? What about for multiplication rules that return a vector or a matrix?

 The general situation works as follows. Given vector spaces and , two things can be constructed, both of which use the same symbol for brevity. We need a space for our most general multiplication, and we denote this by . We also need a rule for multiplication and we denote this by . By definition, is an element of . (Just like not all matrices can be written as , there are elements in that cannot be written as . They can, however, be written as linear combinations of such terms.)

 At this point, I leave the reader to look up elsewhere the definition of a tensor product and relate it with the above.

 Categories: Informal Classroom Notes Tags: introduction to tensor product , intuition behind tensor product
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This note is a fresh attempt at introducing tensor products in a simple and intuitive way. The setting is a vector space on which we wish to define a multiplication. Importantly, we do not wish to limit ourselves to a multiplication producing an element of . For example, we would like to think of and as two examples of multiplying elements of . The first multiplication, , is a map from to . The second multiplication, , is a map from  to . In fact, since is well-defined even if and have different dimensions, the most general setting is the following.
Let and be fixed vector spaces agreed upon at the start. Assume someone has defined a multiplication rule but has not told us what it is.  Here, is a function and is a vector space, both of which are unknown to us. For example, if and , a possible choice might be where .
What does it mean for to be a rule for multiplication? We wish it to have certain properties characteristic of multiplication. In a field, multiplication is involved in a number of axioms, including the distributive law . Since and are vector spaces, potentially distinct from each other, not all the axioms for multiplication in a field make sense in this more general setting. (For instance, commutativity makes no sense if .) The outcome is that is declared to be a rule for multiplication if it is bilinear: , and .
Not knowing does not prevent us from working with it; we simply write it as and are free to manipulate it according to the aforementioned rules. We do the equivalent all the time when we prove general results about a metric; we say “let be a metric” and then make use of only the axioms of a metric, thereby ensuring our derivation is valid regardless of which metric is actually being represented by .
While the tensor product is more than this, in the first instance, it suffices to treat it merely as an unspecified rule for multiplication. (It is specified, but no harm comes from forgetting about this in the first instance.) Rather than write it is more convenient to write , and thinking of as multiplication reminds us that .
The first point is that we can treat as an unspecified multiplication, simplify expressions by manipulating it in accordance with the rules for multiplication, and at the end of the day, if we are eventually told the rule , we can evaluate its simplified expression. In computer science parlance, this can be thought of as lazy evaluation.
Whereas there are many metrics “incompatible” with each other, the rules for multiplication are sufficiently rigid that there exists a “most general multiplication”, and all other multiplications are special cases of it, as explained presently. The tensor product represents this most general multiplication possible.
To consider a specific case first, take and to be . We already know two possible multiplications are and . The claim is that the latter represents the most general multiplication possible. This means, among other things, that it must be possible to write in terms of , and indeed it is: . Note that trace is a linear operator. The precise claim is the following. No matter what is, we can pretend it is , work with this definition, then at the very end, if we are told what actually is, we can obtain the true answer by applying a particular linear transform. At the risk of belabouring the point, if we wish to simplify then we can first pretend is to obtain . Later when we are told we can deduce that the linear map required to convert from to is . Applying this linear map to yields and this is indeed the correct answer because it equals .
Readers wishing to think further about the above example are encouraged to consider that the most general multiplication returning a real-valued number is of the form for some matrix . How can be obtained from ? What about for multiplication rules that return a vector or a matrix?
The general situation works as follows. Given vector spaces and , two things can be constructed, both of which use the same symbol for brevity. We need a space for our most general multiplication, and we denote this by . We also need a rule for multiplication and we denote this by . By definition, is an element of . (Just like not all matrices can be written as , there are elements in that cannot be written as . They can, however, be written as linear combinations of such terms.)
At this point, I leave the reader to look up elsewhere the definition of a tensor product and relate it with the above.
intuition behind tensor product – Jonathan Manton's Blog
https://jmanton.wordpress.com/tag/intuition-behind-tensor-product/
https://math.stackexchange.com/questions/796268/tensor-product-of-algebras-multiplication-definition
Tensor Product of Algebras: Multiplication Definition
2014-05-15
Lammey Lammey 1
<div><div>
<p>A linear map from a tensor product $A \otimes B$ is defined by first constructing a bilinear map from $A \times B$ and checking that scalars commute across the product. Specifically, a <em>bilinear</em> map $\varphi: A \times B \to C$ that also satisfies
$$
\varphi(at, b) = \varphi(a, tb)
$$
where $a \in A$, $b \in B$, and $t$ is a scalar <em>descends</em> to a map
$$
\overline{\varphi}: A \otimes B \to C,
$$
defined on pure tensors by $\overline{\varphi}(a \otimes b) = \varphi(a, b)$. The map is then extended by linearity, as pure tensors form a linear spanning set for $A \otimes B$.</p>
<p>You can think of multiplication as a bilinear map
$$
\left( A \otimes B \right) \times \left( A \otimes B \right) \to A \otimes B,
$$
or, equivalently, as a linear map
$$
\left( A \otimes B \right) \otimes \left( A \otimes B \right) \to A \otimes B.
$$
Since tensor products are associative, this is a map from the four-fold tensor product, which is most easily constructed as the composition:
$$
A \otimes B \otimes A \otimes B \xrightarrow{~~\tau~~} A \otimes A \otimes B \otimes B \xrightarrow{m_A \otimes m_B} A \otimes B,
$$
where $\tau$ is the permutation map on the second and third tensor factors and $m_A$ and $m_B$ are the multiplication maps on the respective algebras.</p>
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 Tensor Product of Algebras: Multiplication Definition

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 $\begingroup$

 I've managed to get a weak grip on what the tensor product of two vector spaces is. I'm now trying to understand the tensor product of two algebras.

 I understand that we define $(v_1\otimes w_1)(v_2\otimes w_2):=(v_1v_2)\otimes (w_1w_2)$. In order for this to be bilinear we must have relations like $x(y+z)=xy+xz$ (x,y,z are tensors here). I can't show that this is true: you can't in general simplify two tensors $y$ and $z$ into a new tensor of the form $v\otimes w$ (i.e the sum won't necessarily be a pure tensor). But then the multiplication isn't defined for a sum of pure tensors, only for pure tensors. What can I do here? Thanks for any replies!

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 * tensor-products
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 edited May 15, 2014 at 20:05
 user26857
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 asked May 15, 2014 at 19:17
 Lammey Lammey
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 $\begingroup$

 The equation $(v_1\otimes w_1)(v_2\otimes w_2)=(v_1v_1\otimes w_1w_2)$ tells us how to multiply pure tensors but does not immediately say how to multiply sums of pure tensors. But there's only one possible way to extend the multiplication operation to sums of pure tensors in a way that the distributive property holds, and that is through the distributive property itself: when multiplying sums of pure tensors, first distribute and then use the given rule. It is very often in situations like these that we only define multiplication for a set of things and then "extend linearly" to all else via distributivity.

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 answered May 15, 2014 at 19:21
 blue blue
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 $\begingroup$ Oh ok, so isn't that basically including distributivity in the definition of the multiplication? $\endgroup$
 – Lammey
 May 15, 2014 at 19:25
 * $\begingroup$ More or less. The rule still needs to be checked whenever sums of pure tensors actually do simplify to pure tensors (i.e. multiplications involving expressions like $(v_1+v_2)\otimes w$). $\endgroup$
 – blue
 May 15, 2014 at 19:27
 * $\begingroup$ Ah ok, well thanks a lot! $\endgroup$
 – Lammey
 May 15, 2014 at 19:33
 Add a comment |
 4
 $\begingroup$

 A linear map from a tensor product $A \otimes B$ is defined by first constructing a bilinear map from $A \times B$ and checking that scalars commute across the product. Specifically, a bilinear map $\varphi: A \times B \to C$ that also satisfies $$ \varphi(at, b) = \varphi(a, tb) $$ where $a \in A$, $b \in B$, and $t$ is a scalar descends to a map $$ \overline{\varphi}: A \otimes B \to C, $$ defined on pure tensors by $\overline{\varphi}(a \otimes b) = \varphi(a, b)$. The map is then extended by linearity, as pure tensors form a linear spanning set for $A \otimes B$.

 You can think of multiplication as a bilinear map $$ \left( A \otimes B \right) \times \left( A \otimes B \right) \to A \otimes B, $$ or, equivalently, as a linear map $$ \left( A \otimes B \right) \otimes \left( A \otimes B \right) \to A \otimes B. $$ Since tensor products are associative, this is a map from the four-fold tensor product, which is most easily constructed as the composition: $$ A \otimes B \otimes A \otimes B \xrightarrow{~~\tau~~} A \otimes A \otimes B \otimes B \xrightarrow{m_A \otimes m_B} A \otimes B, $$ where $\tau$ is the permutation map on the second and third tensor factors and $m_A$ and $m_B$ are the multiplication maps on the respective algebras.

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 answered May 15, 2014 at 19:34
 Sammy Black Sammy Black
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 * $\begingroup$ Thankyou very much! Does this mean that multiplication on a algebra $A$ is a linear map from $A\otimes A\rightarrow A$? It's really helping my understanding, but how does this explain the distributivity of multiplication? $\endgroup$
 – Lammey
 May 15, 2014 at 19:45
 * $\begingroup$ That's exactly right. The bilinearity of multiplication $A \times A \xrightarrow{m_A} A$ is equivalent to linearity of the map $A \otimes A \xrightarrow{\overline{m}_A} A$. $\endgroup$
 – Sammy Black
 May 15, 2014 at 19:47
 * $\begingroup$ By abuse, these two maps are often denoted by the same symbol. (I seem to have committed this abuse in my answer, where the multiplication maps ought to have been wearing overbars). $\endgroup$
 – Sammy Black
 May 15, 2014 at 19:48
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A linear map from a tensor product $A \otimes B$ is defined by first constructing a bilinear map from $A \times B$ and checking that scalars commute across the product. Specifically, a bilinear map $\varphi: A \times B \to C$ that also satisfies
$$
\varphi(at, b) = \varphi(a, tb)
$$
where $a \in A$, $b \in B$, and $t$ is a scalar descends to a map
$$
\overline{\varphi}: A \otimes B \to C,
$$
defined on pure tensors by $\overline{\varphi}(a \otimes b) = \varphi(a, b)$. The map is then extended by linearity, as pure tensors form a linear spanning set for $A \otimes B$.
You can think of multiplication as a bilinear map
$$
\left( A \otimes B \right) \times \left( A \otimes B \right) \to A \otimes B,
$$
or, equivalently, as a linear map
$$
\left( A \otimes B \right) \otimes \left( A \otimes B \right) \to A \otimes B.
$$
Since tensor products are associative, this is a map from the four-fold tensor product, which is most easily constructed as the composition:
$$
A \otimes B \otimes A \otimes B \xrightarrow{~~\tau~~} A \otimes A \otimes B \otimes B \xrightarrow{m_A \otimes m_B} A \otimes B,
$$
where $\tau$ is the permutation map on the second and third tensor factors and $m_A$ and $m_B$ are the multiplication maps on the respective algebras.
Tensor Product of Algebras: Multiplication Definition
https://math.stackexchange.com/questions/796268/tensor-product-of-algebras-multiplication-definition
https://www.johndcook.com/blog/2016/08/13/tensors-3-tensor-products/#comment-887064
Tensor products of vector spaces
2016-08-13
John
<div><div>
	<p>In the <a href="//www.johndcook.com/blog/2016/08/12/tensors-2-multilinear-operators/">previous post</a>, we defined the tensor product of two tensors, but you’ll often see tensor products of <strong>spaces</strong>. How are these tensor products defined?</p>
<h2>Tensor product splines</h2>
<p>For example, you may have seen tensor product splines. Suppose you have a function over a rectangle that you’d like to approximate by patching together polynomials so that the interpolation has the specified values at grid points, and the patches fit together smoothly. In one dimension, you do this by constructing splines. Then you can bootstrap your way from one dimension to two dimensions by using tensor product splines. A tensor product spline in <em>x</em> and <em>y</em> is a sum of terms consisting of a spline in <em>x</em> and a spline in <em>y</em>. Notice that a tensor product spline is not simply a product of two ordinary splines, but a <em>sum</em> of such products.</p>
<p>If <em>X</em> is the vector space of all splines in the <em>x</em>-direction and <em>Y</em> the space of all splines in the <em>y</em>-direction, the space of tensor product splines is the tensor product of the spaces <em>X</em> and <em>Y</em>. Suppose a set <em>s</em><sub><em>i</em></sub>, for <em>i</em> running from 1 to <em>n,</em> is a basis for <em>X</em>. Similarly, suppose <em>t</em><sub><em>j</em></sub>, for <em>j</em> running from 1 to <em>m</em>, is a basis for <em>Y</em>. Then the products <em>s</em><sub><em>i</em></sub> <em>t</em><sub><em>j</em></sub> form a basis for the tensor product <em>X</em> and <em>Y</em>, the tensor product splines over the rectangle. Notice that if <em>X</em> has dimension <em>n</em> and <em>Y</em> has dimension <em>m</em> then their tensor product has dimension <em>n</em><em>m</em>.  Notice that if we only allowed products of splines, not sums of products of splines, we’d get a much smaller space, one of dimension <em>n</em>+<em>m</em>.</p>
<p>We can use the same process to define the tensor product of any two vector spaces. A basis for the tensor product is all products of basis elements in one space and basis elements in the other. There’s a more general definition of tensor products that doesn’t involve bases sketched below.</p>
<h2>Tensor products of modules</h2>
<p>You can also define tensor products of <em>modules</em>, a generalization of vector spaces. You could think of a module as a vector space where the scalars come from a ring instead of a field. Since rings are more general than fields, modules are more general than vector spaces.</p>
<p>The tensor product of two modules over a commutative ring is defined by taking the Cartesian product and moding out by the necessary relations to make things bilinear. (This description is very hand-wavy. A detailed presentation needs its own blog post or two.)</p>
<p>Tensor products of modules hold some surprises. For example, let <em>m</em> and <em>n</em> be two relatively prime integers. You can think of the integers mod <em>m</em> or <em>n</em> as a module over the integers. The tensor product of these modules is zero because you end up moding out by everything. This kind of collapse doesn’t happen over vector spaces.</p>
<h2>Past and future</h2>
<p>The first two posts in this series:</p>
<ul>
<li><a href="//www.johndcook.com/blog/2016/08/09/what-is-a-tensor/">Tensors 1: Introduction</a></li>
<li><a href="//www.johndcook.com/blog/2016/08/12/tensors-2-multilinear-operators/">Tensors 2: Multilinear operators</a></li>
</ul>
<p>I plan to leave the algebraic perspective aside for a while, though as I mentioned above there’s more to come back to.</p>
<p>Next I plan to write about the analytic/geometric view of tensors. Here we get into things like changes of coordinates and it looks at first as if a tensor is something completely different.</p>
<p><strong>Update</strong>: <a href="//www.johndcook.com/blog/2016/08/15/tensors-4-behavior-under-change-of-coordinates/">Tensors 4: Behavior under changes of coordinates</a></p>
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 Tensors 3: Tensor products

 Posted on 13 August 2016 by John

 In the previous post, we defined the tensor product of two tensors, but you’ll often see tensor products of spaces. How are these tensor products defined?

 Tensor product splines

 For example, you may have seen tensor product splines. Suppose you have a function over a rectangle that you’d like to approximate by patching together polynomials so that the interpolation has the specified values at grid points, and the patches fit together smoothly. In one dimension, you do this by constructing splines. Then you can bootstrap your way from one dimension to two dimensions by using tensor product splines. A tensor product spline in x and y is a sum of terms consisting of a spline in x and a spline in y. Notice that a tensor product spline is not simply a product of two ordinary splines, but a sum of such products.

 If X is the vector space of all splines in the x-direction and Y the space of all splines in the y-direction, the space of tensor product splines is the tensor product of the spaces X and Y. Suppose a set si, for i running from 1 to n, is a basis for X. Similarly, suppose tj, for j running from 1 to m, is a basis for Y. Then the products si tj form a basis for the tensor product X and Y, the tensor product splines over the rectangle. Notice that if X has dimension n and Y has dimension m then their tensor product has dimension nm. Notice that if we only allowed products of splines, not sums of products of splines, we’d get a much smaller space, one of dimension n+m.

 Tensor products of vector spaces

 We can use the same process to define the tensor product of any two vector spaces. A basis for the tensor product is all products of basis elements in one space and basis elements in the other. There’s a more general definition of tensor products that doesn’t involve bases sketched below.

 Tensor products of modules

 You can also define tensor products of modules, a generalization of vector spaces. You could think of a module as a vector space where the scalars come from a ring instead of a field. Since rings are more general than fields, modules are more general than vector spaces.

 The tensor product of two modules over a commutative ring is defined by taking the Cartesian product and moding out by the necessary relations to make things bilinear. (This description is very hand-wavy. A detailed presentation needs its own blog post or two.)

 Tensor products of modules hold some surprises. For example, let m and n be two relatively prime integers. You can think of the integers mod m or n as a module over the integers. The tensor product of these modules is zero because you end up moding out by everything. This kind of collapse doesn’t happen over vector spaces.

 Past and future

 The first two posts in this series:

 * Tensors 1: Introduction
 * Tensors 2: Multilinear operators

 I plan to leave the algebraic perspective aside for a while, though as I mentioned above there’s more to come back to.

 Next I plan to write about the analytic/geometric view of tensors. Here we get into things like changes of coordinates and it looks at first as if a tensor is something completely different.

 Update: Tensors 4: Behavior under changes of coordinates

 Categories : Math
 Tags : Math Tensors
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 Previous Post Tensors 2: Multilinear operators
 Next Post Tensors 4: Behavior under change of coordinates

 4 thoughts on “ Tensors 3: Tensor products ”

 1. Eric Kernfeld
 14 August 2016 at 14:58

 You write “Notice that if we only allowed *products* of splines, not sums of products of splines, we’d get a much smaller space, one of dimension n+m.” If you don’t allow sums, it’s not a vector space; should the first *products* be *sums*?

 2. Will Brown
 15 February 2018 at 19:41

 “You could think of a module as a vector space where the scalars come from a ring ideas of a field. ” — This sentence doesn’t parse for me. Did you leave out a word?

 3. John
 16 February 2018 at 07:27

 @will: I wrote “ideas” when I meant to write “instead.” Just fixed that.

 4. Charles Peters
 11 October 2019 at 12:53

 Halmos has a nice definition. The tensor product of vector spaces X and Y is the dual of the space of bilinear functionals on XxY.

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In the previous post, we defined the tensor product of two tensors, but you’ll often see tensor products of spaces. How are these tensor products defined?
Tensor product splines
For example, you may have seen tensor product splines. Suppose you have a function over a rectangle that you’d like to approximate by patching together polynomials so that the interpolation has the specified values at grid points, and the patches fit together smoothly. In one dimension, you do this by constructing splines. Then you can bootstrap your way from one dimension to two dimensions by using tensor product splines. A tensor product spline in x and y is a sum of terms consisting of a spline in x and a spline in y. Notice that a tensor product spline is not simply a product of two ordinary splines, but a sum of such products.
If X is the vector space of all splines in the x-direction and Y the space of all splines in the y-direction, the space of tensor product splines is the tensor product of the spaces X and Y. Suppose a set s  i , for i running from 1 to n, is a basis for X. Similarly, suppose t  j , for j running from 1 to m, is a basis for Y. Then the products s  i  t  j  form a basis for the tensor product X and Y, the tensor product splines over the rectangle. Notice that if X has dimension n and Y has dimension m then their tensor product has dimension n m.  Notice that if we only allowed products of splines, not sums of products of splines, we’d get a much smaller space, one of dimension n+m.
We can use the same process to define the tensor product of any two vector spaces. A basis for the tensor product is all products of basis elements in one space and basis elements in the other. There’s a more general definition of tensor products that doesn’t involve bases sketched below.
Tensor products of modules
You can also define tensor products of modules, a generalization of vector spaces. You could think of a module as a vector space where the scalars come from a ring instead of a field. Since rings are more general than fields, modules are more general than vector spaces.
The tensor product of two modules over a commutative ring is defined by taking the Cartesian product and moding out by the necessary relations to make things bilinear. (This description is very hand-wavy. A detailed presentation needs its own blog post or two.)
Tensor products of modules hold some surprises. For example, let m and n be two relatively prime integers. You can think of the integers mod m or n as a module over the integers. The tensor product of these modules is zero because you end up moding out by everything. This kind of collapse doesn’t happen over vector spaces.
Past and future
The first two posts in this series:
 Tensors 1: Introduction 
 Tensors 2: Multilinear operators 
I plan to leave the algebraic perspective aside for a while, though as I mentioned above there’s more to come back to.
Next I plan to write about the analytic/geometric view of tensors. Here we get into things like changes of coordinates and it looks at first as if a tensor is something completely different.
 Update: Tensors 4: Behavior under changes of coordinates
Tensor products of vector spaces
https://www.johndcook.com/blog/2016/08/13/tensors-3-tensor-products/
https://en.wikipedia.org/wiki/Tensor_product
Tensor product - Wikipedia
2001-10-03
None
<div><div>
<p>In <a href="/wiki/Mathematics">mathematics</a>, the <b>tensor product</b> of two <a href="/wiki/Vector_space">vector spaces</a> <i>V</i> and <i>W</i> (over the same <a href="/wiki/Field_(mathematics)">field</a>) is a vector space to which is associated a <a href="/wiki/Bilinear_map">bilinear map</a> that maps a pair to an element of denoted 
</p><p>An element of the form is called the <b>tensor product</b> of v and w. An element of is a <a href="/wiki/Tensor">tensor</a>, and the tensor product of two vectors is sometimes called an <i>elementary tensor</i> or a <i>decomposable tensor</i>. The elementary tensors span in the sense that every element of is a sum of elementary tensors. If <a href="/wiki/Basis_(linear_algebra)">bases</a> are given for <i>V</i> and <i>W</i>, a basis of is formed by all tensor products of a basis element of V and a basis element of W.
</p><p>The tensor product of two vector spaces captures the properties of all bilinear maps in the sense that a bilinear map from into another vector space Z factors uniquely through a <a href="/wiki/Linear_map">linear map</a> (see <a href="/wiki/Universal_property">Universal property</a>).
</p><p>Tensor products are used in many application areas, including physics and engineering. For example, in <a href="/wiki/General_relativity">general relativity</a>, the <a href="/wiki/Gravitational_field">gravitational field</a> is described through the <a href="/wiki/Metric_tensor">metric tensor</a>, which is a <a href="/wiki/Vector_field">vector field</a> of tensors, one at each point of the <a href="/wiki/Space-time">space-time</a> <a href="/wiki/Manifold">manifold</a>, and each belonging to the tensor product with itself of the <a href="/wiki/Cotangent_space">cotangent space</a> at the point.
</p>
<h2>Definitions and constructions[<a href="/w/index.php?title=Tensor_product&amp;action=edit&amp;section=1">edit</a>]</h2>
<p>The <i>tensor product</i> of two vector spaces is a vector space that is defined <a href="/wiki/Up_to">up to</a> an <a href="/wiki/Isomorphism">isomorphism</a>. 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. 
</p><p>The tensor product can also be defined through a <a href="/wiki/Universal_property">universal property</a>; see <a href="#Universal_property">§ Universal property</a>, below. As for every universal property, all <a href="/wiki/Object_(category_theory)">objects</a> 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.
</p>
<h3>From bases[<a href="/w/index.php?title=Tensor_product&amp;action=edit&amp;section=2">edit</a>]</h3>
<p>Let V and W be two <a href="/wiki/Vector_space">vector spaces</a> over a <a href="/wiki/Field_(mathematics)">field</a> F, with respective <a href="/wiki/Basis_(linear_algebra)">bases</a> and 
</p><p>The <i>tensor product</i> of V and W is a vector space which has as a basis the set of all with and This definition can be formalized in the following way (this formalization is rarely used in practice, as the preceding informal definition is generally sufficient): is the set of the <a href="/wiki/Function_(mathematics)">functions</a> from the <a href="/wiki/Cartesian_product">Cartesian product</a> to F that have a finite number of nonzero values. The <a href="/wiki/Pointwise_operation">pointwise operations</a> make a vector space. The function that maps to 1 and the other elements of to 0 is denoted 
</p><p>The set is straightforwardly a basis of which is called the <i>tensor product</i> of the bases and 
</p><p>The <i>tensor product of two vectors</i> is defined from their decomposition on the bases. More precisely, if 
</p><div><p></p></div><p>
are vectors decomposed on their respective bases, then the tensor product of x and y is
</p><div><p></p></div>
<p>If arranged into a rectangular array, the <a href="/wiki/Coordinate_vector">coordinate vector</a> of is the <a href="/wiki/Outer_product">outer product</a> of the coordinate vectors of x and y. Therefore, the tensor product is a generalization of the outer product.
</p><p>It is straightforward to verify that the map is a bilinear map from to 
</p><p>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 <a href="/wiki/Canonical_isomorphism">canonical isomorphism</a> 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 <a href="/wiki/Tensor_product_of_modules">tensor product of modules</a> over a <a href="/wiki/Ring_(mathematics)">ring</a>.
</p>
<h3>As a quotient space[<a href="/w/index.php?title=Tensor_product&amp;action=edit&amp;section=3">edit</a>]</h3>
<p>A construction of the tensor product that is basis independent can be obtained in the following way.
</p><p>Let V and W be two <a href="/wiki/Vector_space">vector spaces</a> over a <a href="/wiki/Field_(mathematics)">field F</a>. 
</p><p>One considers first a vector space L that has the <a href="/wiki/Cartesian_product">Cartesian product</a> as a <a href="/wiki/Basis_(linear_algebra)">basis</a>. That is, the basis elements of L are the <a href="/wiki/Ordered_pair">pairs</a> with and To get such a vector space, one can define it as the vector space of the <a href="/wiki/Function_(mathematics)">functions</a> that have a finite number of nonzero values, and identifying with the function that takes the value 1 on and 0 otherwise.
</p><p>Let R be the <a href="/wiki/Linear_subspace">linear subspace</a> of L that is spanned by the relations that the tensor product must satisfy. More precisely R is <a href="/wiki/Linear_span">spanned by</a> the elements of one of the forms
</p>
<dl><dd></dd></dl>
<p>where and 
</p><p>Then, the tensor product is defined as the <a href="/wiki/Quotient_space_(linear_algebra)">quotient space</a>
</p>
<dl><dd></dd></dl>
<p>and the image of in this quotient is denoted 
</p><p>It is straightforward to prove that the result of this construction satisfies the <a href="/wiki/Universal_property">universal property</a> considered below. (A very similar construction can be used to define the <a href="/wiki/Tensor_product_of_modules">tensor product of modules</a>.)
</p>
<h3>Universal property[<a href="/w/index.php?title=Tensor_product&amp;action=edit&amp;section=4">edit</a>]</h3>
<div><p><a href="/wiki/File:Another_universal_tensor_prod.svg"></a></p><div><p>Universal property of tensor product: if h is bilinear, there is a unique linear map that makes the diagram <a href="/wiki/Commutative_diagram">commutative</a> (that is, ).</p></div></div>
<p>In this section, the <a href="/wiki/Universal_property">universal property</a> satisfied by the tensor product is described. As for every universal property, two objects that satisfy the property are related by a unique <a href="/wiki/Isomorphism">isomorphism</a>. 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.
</p><p>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. 
</p><p>The "universal-property definition" of the tensor product of two vector spaces is the following (recall that a <a href="/wiki/Bilinear_map">bilinear map</a> is a function that is <i>separately</i> <a href="/wiki/Linear_map">linear</a> in each of its arguments):
</p>
<dl><dd>The <i>tensor product</i> of two vector spaces V and W is a vector space denoted as together with a bilinear map from to such that, for every bilinear map there is a <i>unique</i> linear map such that (that is, for every and ).</dd></dl>
<h3>Linearly disjoint[<a href="/w/index.php?title=Tensor_product&amp;action=edit&amp;section=5">edit</a>]</h3>
<p>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.<sup><a href="#cite_note-FOOTNOTETrèves2006403–404-1">[1]</a></sup>
</p>
<p>For example, it follows immediately that if and are positive integers then and the bilinear map defined by sending to form a tensor product of and <sup><a href="#cite_note-FOOTNOTETrèves2006407-2">[2]</a></sup> Often, this map will be denoted by so that denotes this bilinear map's value at 
</p><p>As another example, suppose that is the vector space of all complex-valued functions on a set with addition and scalar multiplication defined pointwise (meaning that is the map and is the map ). Let and be any sets and for any and let denote the function defined by 
If and are vector subspaces then the vector subspace of together with the bilinear map 
</p><div><p></p></div><p>
form a tensor product of and <sup><a href="#cite_note-FOOTNOTETrèves2006407-2">[2]</a></sup></p>
<h2>Properties[<a href="/w/index.php?title=Tensor_product&amp;action=edit&amp;section=6">edit</a>]</h2>
<h3>Dimension[<a href="/w/index.php?title=Tensor_product&amp;action=edit&amp;section=7">edit</a>]</h3>
<p>If <i>V</i> and <i>W</i> are vectors spaces of finite <a href="/wiki/Dimension_(linear_algebra)">dimension</a>, then is finite-dimensional, and its dimension is the product of the dimensions of <i>V</i> and <i>W</i>.
</p><p>This results from the fact that a basis of is formed by taking all tensor products of a basis element of <i>V</i> and a basis element of <i>W</i>.
</p>
<h3>Associativity[<a href="/w/index.php?title=Tensor_product&amp;action=edit&amp;section=8">edit</a>]</h3>
<p>The tensor product is <a href="/wiki/Associative">associative</a> in the sense that, given three vector spaces there is a canonical isomorphism 
</p>
<dl><dd></dd></dl>
<p>that maps to 
</p><p>This allows omitting parentheses in the tensor product of more than two vector spaces or vectors.
</p>
<h3>Commutativity as vector space operation[<a href="/w/index.php?title=Tensor_product&amp;action=edit&amp;section=9">edit</a>]</h3>
<p>The tensor product of two vector spaces and is <a href="/wiki/Commutative">commutative </a> in the sense that there is a canonical isomorphism 
</p>
<dl><dd></dd></dl>
<p>that maps to 
</p><p>On the other hand, even when the tensor product of vectors is not commutative; that is in general.
</p><p>
The map from to itself induces a linear <a href="/wiki/Automorphism">automorphism</a> that is called a <b>braiding map. </b>
More generally and as usual (see <a href="/wiki/Tensor_algebra">tensor algebra</a>), let denote the tensor product of n copies of the vector space V. For every <a href="/wiki/Permutation">permutation</a> s of the first n positive integers, the map
</p>
<dl><dd></dd></dl>
<p>induces a linear automorphism of which is called a braiding map.
</p>
<h2>Tensor product of linear maps[<a href="/w/index.php?title=Tensor_product&amp;action=edit&amp;section=10">edit</a>]</h2>
<p>Given a linear map and a vector space W, the <i>tensor product</i> 
</p>
<dl><dd></dd></dl>
<p>is the unique linear map such that
</p>
<dl><dd></dd></dl>
<p>The tensor product is defined similarly.
</p><p>Given two linear maps and their tensor product 
</p>
<dl><dd></dd></dl>
<p>is the unique linear map that satisfies 
</p>
<dl><dd></dd></dl>
<p>One has 
</p>
<dl><dd></dd></dl>
<p>In terms of <a href="/wiki/Category_theory">category theory</a>, this means that the tensor product is a <a href="/wiki/Bifunctor">bifunctor</a> from the <a href="/wiki/Category_(mathematics)">category</a> of vector spaces to itself.<sup><a href="#cite_note-3">[3]</a></sup>
</p><p>If f and g are both <a href="/wiki/Injective">injective</a> or <a href="/wiki/Surjective">surjective</a>, then the same is true for all above defined linear maps. In particular, the tensor product with a vector space is an <a href="/wiki/Exact_functor">exact functor</a>; this means that every <a href="/wiki/Exact_sequence">exact sequence</a> is mapped to an exact sequence (<a href="/wiki/Tensor_product_of_modules">tensor products of modules</a> do not transform injections into injections, but they are <a href="/wiki/Right_exact_functor">right exact functors</a>).
</p><p>By choosing bases of all vector spaces involved, the linear maps <i>S</i> and <i>T</i> can be represented by <a href="/wiki/Matrix_(mathematics)">matrices</a>. Then, depending on how the tensor is vectorized, the matrix describing the tensor product is the <a href="/wiki/Kronecker_product">Kronecker product</a> of the two matrices. For example, if <i>V</i>, <i>X</i>, <i>W</i>, and <i>Y</i> above are all two-dimensional and bases have been fixed for all of them, and <i>S</i> and <i>T</i> are given by the matrices
</p><div><p></p></div>
<p>respectively, then the tensor product of these two matrices is
</p><div><p></p></div>
<p>The resultant rank is at most 4, and thus the resultant dimension is 4. Note that <em>rank</em> here denotes the <a href="/wiki/Tensor_rank">tensor rank</a> i.e. the number of requisite indices (while the <a href="/wiki/Matrix_rank">matrix rank</a> counts the number of degrees of freedom in the resulting array). Note 
</p><p>A <a href="/wiki/Dyadic_product">dyadic product</a> is the special case of the tensor product between two vectors of the same dimension.
</p>
<h2>General tensors[<a href="/w/index.php?title=Tensor_product&amp;action=edit&amp;section=11">edit</a>]</h2>
<p>For non-negative integers <i>r</i> and <i>s</i> a type <a href="/wiki/Tensor">tensor</a> on a vector space <i>V</i> is an element of
</p><div><p></p></div>
<p>Here is the <a href="/wiki/Dual_vector_space">dual vector space</a> (which consists of all <a href="/wiki/Linear_map">linear maps</a> <i>f</i> from <i>V</i> to the ground field <i>K</i>).
</p><p>There is a product map, called the <em>(tensor) product of tensors</em><sup><a href="#cite_note-4">[4]</a></sup>
</p><div><p></p></div>
<p>It is defined by grouping all occurring "factors" <i>V</i> together: writing for an element of <i>V</i> and for an element of the dual space,
</p><div><p></p></div>
<p>Picking a basis of <i>V</i> and the corresponding <a href="/wiki/Dual_basis">dual basis</a> of naturally induces a basis for (this basis is described in the <a href="/wiki/Kronecker_product#Abstract_properties">article on Kronecker products</a>). In terms of these bases, the <a href="/wiki/Coordinate_vector">components</a> of a (tensor) product of two (or more) <a href="/wiki/Tensor">tensors</a> can be computed. For example, if <i>F</i> and <i>G</i> are two <a href="/wiki/Covariance_and_contravariance_of_vectors">covariant</a> tensors of orders <i>m</i> and <i>n</i> respectively (i.e. and ), then the components of their tensor product are given by<sup><a href="#cite_note-5">[5]</a></sup>
</p><div><p></p></div>
<p>Thus, the components of the tensor product of two tensors are the ordinary product of the components of each tensor. Another example: let <b>U</b> be a tensor of type (1, 1) with components and let <b>V</b> be a tensor of type with components Then
</p><div><p></p></div>
<p>and
</p><div><p></p></div>
<p>Tensors equipped with their product operation form an <a href="/wiki/Algebra_over_a_field">algebra</a>, called the <a href="/wiki/Tensor_algebra">tensor algebra</a>.
</p>
<h3>Evaluation map and tensor contraction[<a href="/w/index.php?title=Tensor_product&amp;action=edit&amp;section=12">edit</a>]</h3>
<p>For tensors of type (1, 1) there is a canonical <b>evaluation map</b>
</p><div><p></p></div>
<p>defined by its action on pure tensors:
</p><div><p></p></div>
<p>More generally, for tensors of type with <i>r</i>, <i>s</i> &gt; 0, there is a map, called <a href="/wiki/Tensor_contraction">tensor contraction</a>,
</p><div><p></p></div>
<p>(The copies of and on which this map is to be applied must be specified.)
</p><p>On the other hand, if is <em>finite-dimensional</em>, there is a canonical map in the other direction (called the <b>coevaluation map</b>)
</p><div><p></p></div>
<p>where is any basis of and is its <a href="/wiki/Dual_basis">dual basis</a>. This map does not depend on the choice of basis.<sup><a href="#cite_note-6">[6]</a></sup>
</p><p>The interplay of evaluation and coevaluation can be used to characterize finite-dimensional vector spaces without referring to bases.<sup><a href="#cite_note-7">[7]</a></sup>
</p>
<h3>Adjoint representation[<a href="/w/index.php?title=Tensor_product&amp;action=edit&amp;section=13">edit</a>]</h3>
<p>The tensor product may be naturally viewed as a module for the <a href="/wiki/Lie_algebra">Lie algebra</a> by means of the diagonal action: for simplicity let us assume then, for each 
</p><div><p></p></div>
<p>where is the <a href="/wiki/Transpose">transpose</a> of <i>u</i>, that is, in terms of the obvious pairing on 
</p><div><p></p></div>
<p>There is a canonical isomorphism given by
</p><div><p></p></div>
<p>Under this isomorphism, every <i>u</i> in may be first viewed as an endomorphism of and then viewed as an endomorphism of In fact it is the <a href="/wiki/Adjoint_representation_of_a_Lie_algebra">adjoint representation</a> ad(<i>u</i>) of 
</p>
<h2>Linear maps as tensors[<a href="/w/index.php?title=Tensor_product&amp;action=edit&amp;section=14">edit</a>]</h2>
<p>Given two finite dimensional vector spaces <i>U</i>, <i>V</i> over the same field <i>K</i>, denote the <a href="/wiki/Dual_space">dual space</a> of <i>U</i> as <i>U*</i>, and the <i>K</i>-vector space of all linear maps from <i>U</i> to <i>V</i> as Hom(<i>U</i>,<i>V</i>). There is an isomorphism,
</p><div><p></p></div>
<p>defined by an action of the pure tensor on an element of 
</p><div><p></p></div>
<p>Its "inverse" can be defined using a basis and its dual basis as in the section "<a href="#Evaluation_map_and_tensor_contraction">Evaluation map and tensor contraction</a>" above:
</p><div><p></p></div>
<p>This result implies
</p><div><p></p></div>
<p>which automatically gives the important fact that forms a basis for where are bases of <i>U</i> and <i>V</i>.
</p><p>Furthermore, given three vector spaces <i>U</i>, <i>V</i>, <i>W</i> the tensor product is linked to the vector space of <i>all</i> linear maps, as follows:
</p><div><p></p></div><p>
This is an example of <a href="/wiki/Adjoint_functor">adjoint functors</a>: the tensor product is "left adjoint" to Hom.
</p>
<h2>Tensor products of modules over a ring[<a href="/w/index.php?title=Tensor_product&amp;action=edit&amp;section=15">edit</a>]</h2>
<p>The tensor product of two <a href="/wiki/Module_(mathematics)">modules</a> <i>A</i> and <i>B</i> over a <i><a href="/wiki/Commutative_ring">commutative</a></i> <a href="/wiki/Ring_(mathematics)">ring</a> <i>R</i> is defined in exactly the same way as the tensor product of vector spaces over a field:
</p><div><p></p></div><p>
where now is the <a href="/wiki/Free_module">free <i>R</i>-module</a> generated by the cartesian product and <i>G</i> is the <i>R</i>-module generated by <a href="#Putting_all_the_construction_together">the same relations as above</a>.
</p><p>More generally, the tensor product can be defined even if the ring is <a href="/wiki/Noncommutative_ring">non-commutative</a>. In this case <i>A</i> has to be a right-<i>R</i>-module and <i>B</i> is a left-<i>R</i>-module, and instead of the last two relations above, the relation
</p><div><p></p></div><p>
is imposed. If <i>R</i> is non-commutative, this is no longer an <i>R</i>-module, but just an <a href="/wiki/Abelian_group">abelian group</a>.
</p><p>The universal property also carries over, slightly modified: the map defined by is a <a href="/wiki/Tensor_product_of_modules#Balanced_product">middle linear map</a> (referred to as "the canonical middle linear map".<sup><a href="#cite_note-8">[8]</a></sup>); that is, it satisfies:<sup><a href="#cite_note-chen-9">[9]</a></sup>
</p><div><p></p></div>
<p>The first two properties make <i>φ</i> a bilinear map of the <a href="/wiki/Abelian_group">abelian group</a> For any middle linear map of a unique group homomorphism <i>f</i> of satisfies and this property determines within group isomorphism. See the <a href="/wiki/Tensor_product_of_modules">main article</a> for details.
</p>
<h3>Tensor product of modules over a non-commutative ring[<a href="/w/index.php?title=Tensor_product&amp;action=edit&amp;section=16">edit</a>]</h3>
<p>Let <i>A</i> be a right <i>R</i>-module and <i>B</i> be a left <i>R</i>-module. Then the tensor product of <i>A</i> and <i>B</i> is an abelian group defined by
</p><div><p></p></div><p>
where is a <a href="/wiki/Free_abelian_group">free abelian group</a> over and G is the subgroup of generated by relations
</p><div><p></p></div>
<p>The universal property can be stated as follows. Let <i>G</i> be an abelian group with a map that is bilinear, in the sense that
</p><div><p></p></div>
<p>Then there is a unique map such that for all and 
</p><p>Furthermore, we can give a module structure under some extra conditions:
</p>
<ol><li>If <i>A</i> is a (<i>S</i>,<i>R</i>)-bimodule, then is a left <i>S</i>-module where </li>
<li>If <i>B</i> is a (<i>R</i>,<i>S</i>)-bimodule, then is a right <i>S</i>-module where </li>
<li>If <i>A</i> is a (<i>S</i>,<i>R</i>)-bimodule and <i>B</i> is a (<i>R</i>,<i>T</i>)-bimodule, then is a (<i>S</i>,<i>T</i>)-bimodule, where the left and right actions are defined in the same way as the previous two examples.</li>
<li>If <i>R</i> is a commutative ring, then <i>A</i> and <i>B</i> are (<i>R</i>,<i>R</i>)-bimodules where and By 3), we can conclude is a (<i>R</i>,<i>R</i>)-bimodule.</li></ol>
<h3>Computing the tensor product[<a href="/w/index.php?title=Tensor_product&amp;action=edit&amp;section=17">edit</a>]</h3>
<p>For vector spaces, the tensor product is quickly computed since bases of <i>V</i> of <i>W</i> immediately determine a basis of as was mentioned above. For modules over a general (commutative) ring, not every module is free. For example, <b>Z</b>/<i>n</i><b>Z</b> is not a free abelian group (<b>Z</b>-module). The tensor product with <b>Z</b>/<i>n</i><b>Z</b> is given by
</p><div><p></p></div>
<p>More generally, given a <a href="/wiki/Presentation_of_a_module">presentation</a> of some <i>R</i>-module <i>M</i>, that is, a number of generators together with relations
</p><div><p></p></div>
<p>the tensor product can be computed as the following <a href="/wiki/Cokernel">cokernel</a>:
</p><div><p></p></div>
<p>Here and the map is determined by sending some in the <i>j</i>th copy of to (in ). Colloquially, this may be rephrased by saying that a presentation of <i>M</i> gives rise to a presentation of This is referred to by saying that the tensor product is a <a href="/wiki/Right_exact_functor">right exact functor</a>. It is not in general left exact, that is, given an injective map of <i>R</i>-modules the tensor product
</p><div><p></p></div>
<p>is not usually injective. For example, tensoring the (injective) map given by multiplication with <i>n</i>, <i>n</i> : <b>Z</b> → <b>Z</b> with <b>Z</b>/<i>n</i><b>Z</b> yields the zero map 0 : <b>Z</b>/<i>n</i><b>Z</b> → <b>Z</b>/<i>n</i><b>Z</b>, which is not injective. Higher <a href="/wiki/Tor_functor">Tor functors</a> measure the defect of the tensor product being not left exact. All higher Tor functors are assembled in the <a href="/wiki/Derived_tensor_product">derived tensor product</a>.
</p>
<h2>Tensor product of algebras[<a href="/w/index.php?title=Tensor_product&amp;action=edit&amp;section=18">edit</a>]</h2>
<p>Let <i>R</i> be a commutative ring. The tensor product of <i>R</i>-modules applies, in particular, if <i>A</i> and <i>B</i> are <a href="/wiki/Algebra_(ring_theory)"><i>R</i>-algebras</a>. In this case, the tensor product is an <i>R</i>-algebra itself by putting
</p><div><p></p></div><p>
For example,
</p><div><p></p></div>
<p>A particular example is when <i>A</i> and <i>B</i> are fields containing a common subfield <i>R</i>. The <a href="/wiki/Tensor_product_of_fields">tensor product of fields</a> is closely related to <a href="/wiki/Galois_theory">Galois theory</a>: if, say, <i>A</i> = <i>R</i>[<i>x</i>] / <i>f</i>(<i>x</i>), where <i>f</i> is some <a href="/wiki/Irreducible_polynomial">irreducible polynomial</a> with coefficients in <i>R</i>, the tensor product can be calculated as
</p><div><p></p></div><p>
where now <i>f</i> is interpreted as the same polynomial, but with its coefficients regarded as elements of <i>B</i>. In the larger field <i>B</i>, the polynomial may become reducible, which brings in Galois theory. For example, if <i>A</i> = <i>B</i> is a <a href="/wiki/Galois_extension">Galois extension</a> of <i>R</i>, then
</p><div><p></p></div><p>
is isomorphic (as an <i>A</i>-algebra) to the </p>
<h2>Eigenconfigurations of tensors[<a href="/w/index.php?title=Tensor_product&amp;action=edit&amp;section=19">edit</a>]</h2>
<p>Square <a href="/wiki/Matrix_(mathematics)">matrices</a> with entries in a <a href="/wiki/Field_(mathematics)">field</a> represent <a href="/wiki/Linear_maps">linear maps</a> of <a href="/wiki/Vector_spaces">vector spaces</a>, say and thus linear maps of <a href="/wiki/Projective_spaces">projective spaces</a> over If is <a href="/wiki/Invertible_matrix">nonsingular</a> then is <a href="/wiki/Well-defined">well-defined</a> everywhere, and the <a href="/wiki/Eigenvalues_and_eigenvectors">eigenvectors</a> of correspond to the fixed points of The <i>eigenconfiguration</i> of consists of points in provided is generic and is <a href="/wiki/Algebraically_closed_field">algebraically closed</a>. The fixed points of nonlinear maps are the eigenvectors of tensors. Let be a -dimensional tensor of format with entries lying in an algebraically closed field of <a href="/wiki/Characteristic_(algebra)">characteristic</a> zero. Such a tensor defines <a href="/wiki/Morphism_of_algebraic_varieties">polynomial maps</a> and with coordinates
</p><div><p></p></div>
<p>Thus each of the coordinates of is a <a href="/wiki/Homogeneous_polynomial">homogeneous polynomial</a> of degree in The eigenvectors of are the solutions of the constraint
</p><div><p></p></div>
<p>and the eigenconfiguration is given by the <a href="/wiki/Algebraic_variety">variety</a> of the <a href="/wiki/Minor_(linear_algebra)">minors</a> of this matrix.<sup><a href="#cite_note-10">[10]</a></sup>
</p>
<h2>Other examples of tensor products[<a href="/w/index.php?title=Tensor_product&amp;action=edit&amp;section=20">edit</a>]</h2>
<h3>Tensor product of Hilbert spaces[<a href="/w/index.php?title=Tensor_product&amp;action=edit&amp;section=21">edit</a>]</h3>
<p><a href="/wiki/Hilbert_space">Hilbert spaces</a> generalize finite-dimensional vector spaces to <a href="/wiki/Countable">countably-infinite</a> dimensions. The tensor product is still defined; it is the <a href="/wiki/Tensor_product_of_Hilbert_spaces">tensor product of Hilbert spaces</a>.
</p>
<h3>Topological tensor product[<a href="/w/index.php?title=Tensor_product&amp;action=edit&amp;section=22">edit</a>]</h3>
<p>When the basis for a vector space is no longer countable, then the appropriate axiomatic formalization for the vector space is that of a <a href="/wiki/Topological_vector_space">topological vector space</a>. The tensor product is still defined, it is the <a href="/wiki/Topological_tensor_product">topological tensor product</a>.
</p>
<h3>Tensor product of graded vector spaces[<a href="/w/index.php?title=Tensor_product&amp;action=edit&amp;section=23">edit</a>]</h3>
<p>Some vector spaces can be decomposed into <a href="/wiki/Direct_sum">direct sums</a> 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).
</p>
<h3>Tensor product of representations[<a href="/w/index.php?title=Tensor_product&amp;action=edit&amp;section=24">edit</a>]</h3>
<p>Vector spaces endowed with an additional multiplicative structure are called <a href="/wiki/Algebra_over_a_field">algebras</a>. The tensor product of such algebras is described by the <a href="/wiki/Littlewood%E2%80%93Richardson_rule">Littlewood–Richardson rule</a>.
</p>
<h3>Tensor product of quadratic forms[<a href="/w/index.php?title=Tensor_product&amp;action=edit&amp;section=25">edit</a>]</h3>
<h3>Tensor product of multilinear forms[<a href="/w/index.php?title=Tensor_product&amp;action=edit&amp;section=26">edit</a>]</h3>
<p>Given two <a href="/wiki/Multilinear_form">multilinear forms</a> and on a vector space over the field their tensor product is the multilinear form
</p><div><p></p></div><p><sup><a href="#cite_note-An_Introduction_to_Manifolds-11">[11]</a></sup></p><p>This is a special case of the <a href="#Product_of_tensors">product of tensors</a> if they are seen as multilinear maps (see also <a href="/wiki/Tensor#As_multilinear_maps">tensors as multilinear maps</a>). Thus the components of the tensor product of multilinear forms can be computed by the <a href="/wiki/Kronecker_product">Kronecker product</a>.
</p>
<h3>Tensor product of sheaves of modules[<a href="/w/index.php?title=Tensor_product&amp;action=edit&amp;section=27">edit</a>]</h3>
<h3>Tensor product of line bundles[<a href="/w/index.php?title=Tensor_product&amp;action=edit&amp;section=28">edit</a>]</h3>
<h3>Tensor product of fields[<a href="/w/index.php?title=Tensor_product&amp;action=edit&amp;section=29">edit</a>]</h3>
<h3>Tensor product of graphs[<a href="/w/index.php?title=Tensor_product&amp;action=edit&amp;section=30">edit</a>]</h3>
<p>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 <a href="/wiki/Product_(category_theory)">category-theoretic product</a> in the category of graphs and <a href="/wiki/Graph_homomorphism">graph homomorphisms</a>. However it is actually the <a href="/wiki/Kronecker_product">Kronecker tensor product</a> of the <a href="/wiki/Adjacency_matrix">adjacency matrices</a> of the graphs. Compare also the section <a href="#Tensor_product_of_linear_maps">Tensor product of linear maps</a> above.
</p>
<h3>Monoidal categories[<a href="/w/index.php?title=Tensor_product&amp;action=edit&amp;section=31">edit</a>]</h3>
<p>The most general setting for the tensor product is the <a href="/wiki/Monoidal_category">monoidal category</a>. 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.
</p>
<h2>Quotient algebras[<a href="/w/index.php?title=Tensor_product&amp;action=edit&amp;section=32">edit</a>]</h2>
<p>A number of important subspaces of the <a href="/wiki/Tensor_algebra">tensor algebra</a> can be constructed as <a href="/wiki/Quotient_space_(linear_algebra)">quotients</a>: these include the <a href="/wiki/Exterior_algebra">exterior algebra</a>, the <a href="/wiki/Symmetric_algebra">symmetric algebra</a>, the <a href="/wiki/Clifford_algebra">Clifford algebra</a>, the <a href="/wiki/Weyl_algebra">Weyl algebra</a>, and the <a href="/wiki/Universal_enveloping_algebra">universal enveloping algebra</a> in general.
</p><p>The exterior algebra is constructed from the <a href="/wiki/Exterior_product">exterior product</a>. Given a vector space <i>V</i>, the exterior product is defined as
</p><div><p></p></div><p>
Note that when the underlying field of <i>V</i> does not have characteristic 2, then this definition is equivalent to
</p><div><p></p></div><p>
The image of in the exterior product is usually denoted and satisfies, by construction, Similar constructions are possible for (<i>n</i> factors), giving rise to the <i>n</i>th <a href="/wiki/Exterior_power">exterior power</a> of <i>V</i>. The latter notion is the basis of <a href="/wiki/Differential_form">differential <i>n</i>-forms</a>.
</p><p>The symmetric algebra is constructed in a similar manner, from the <a href="/wiki/Symmetric_tensor#symmetric_product">symmetric product</a>
</p><div><p></p></div><p>
More generally
</p><div><p></p></div><p>
That is, in the symmetric algebra two adjacent vectors (and therefore all of them) can be interchanged. The resulting objects are called <a href="/wiki/Symmetric_tensor">symmetric tensors</a>.
</p>
<h2>Tensor product in programming[<a href="/w/index.php?title=Tensor_product&amp;action=edit&amp;section=33">edit</a>]</h2>
<h3>Array programming languages[<a href="/w/index.php?title=Tensor_product&amp;action=edit&amp;section=34">edit</a>]</h3>
<p><a href="/wiki/Array_programming_languages">Array programming languages</a> may have this pattern built in. For example, in <a href="/wiki/APL_programming_language">APL</a> the tensor product is expressed as <code>○.×</code> (for example <code>A ○.× B</code> or <code>A ○.× B ○.× C</code>). In <a href="/wiki/J_programming_language">J</a> the tensor product is the dyadic form of <code>*/</code> (for example <code>a */ b</code> or <code>a */ b */ c</code>).
</p><p>Note that J's treatment also allows the representation of some tensor fields, as <code>a</code> and <code>b</code> may be functions instead of constants. This product of two functions is a derived function, and if <code>a</code> and <code>b</code> are <a href="/wiki/Differentiable_function">differentiable</a>, then <code>a */ b</code> is differentiable.
</p><p>However, these kinds of notation are not universally present in array languages. Other array languages may require explicit treatment of indices (for example, <a href="/wiki/MATLAB">MATLAB</a>), and/or may not support <a href="/wiki/Higher-order_function">higher-order functions</a> such as the <a href="/wiki/Jacobian_matrix_and_determinant">Jacobian derivative</a> (for example, <a href="/wiki/Fortran">Fortran</a>/APL).
</p>
<h2>See also[<a href="/w/index.php?title=Tensor_product&amp;action=edit&amp;section=35">edit</a>]</h2>
<ul><li><a href="/wiki/Dyadics">Dyadics</a> – second order tensor, written in a notation that fits in with vector algebra</li>
<li><a href="/wiki/Extension_of_scalars">Extension of scalars</a></li>
<li><a href="/wiki/Monoidal_category">Monoidal category</a> – Category admitting tensor products</li>
<li><a href="/wiki/Tensor_algebra">Tensor algebra</a> – Universal construction in multilinear algebra</li>
<li><a href="/wiki/Tensor_contraction">Tensor contraction</a> – in mathematics and physics, an operation on tensors</li>
<li><a href="/wiki/Topological_tensor_product">Topological tensor product</a> – Tensor product constructions for topological vector spaces</li></ul>
<h2>Notes[<a href="/w/index.php?title=Tensor_product&amp;action=edit&amp;section=36">edit</a>]</h2>
<div><ol>
<li>^ <a href="#cite_ref-FOOTNOTETrèves2006403–404_1-0"><sup><i><b>a</b></i></sup></a> <a href="#cite_ref-FOOTNOTETrèves2006403–404_1-1"><sup><i><b>b</b></i></sup></a> <a href="#CITEREFTrèves2006">Trèves 2006</a>, pp. 403–404.
</li>
<li>^ <a href="#cite_ref-FOOTNOTETrèves2006407_2-0"><sup><i><b>a</b></i></sup></a> <a href="#cite_ref-FOOTNOTETrèves2006407_2-1"><sup><i><b>b</b></i></sup></a> <a href="#CITEREFTrèves2006">Trèves 2006</a>, pp. 407.
</li>
<li><b><a href="#cite_ref-3">^</a></b> <cite>Hazewinkel, Michiel; Gubareni, Nadezhda Mikhaĭlovna; Gubareni, Nadiya; Kirichenko, Vladimir V. (2004). <i>Algebras, rings and modules</i>. Springer. p. 100. <a href="/wiki/ISBN_(identifier)">ISBN</a> <a href="/wiki/Special:BookSources/978-1-4020-2690-4"><bdi>978-1-4020-2690-4</bdi></a>.</cite>
</li>
<li><b><a href="#cite_ref-4">^</a></b> <a href="#CITEREFBourbaki1989">Bourbaki (1989)</a>, p. 244 defines the usage "tensor product of <i>x</i> and <i>y</i>", elements of the respective modules.
</li>
<li><b><a href="#cite_ref-5">^</a></b> Analogous formulas also hold for <a href="/wiki/Covariance_and_contravariance_of_vectors">contravariant</a> tensors, as well as tensors of mixed variance. Although in many cases such as when there is an <a href="/wiki/Inner_product">inner product</a> defined, the distinction is irrelevant.
</li>
<li><b><a href="#cite_ref-6">^</a></b> <cite><a href="https://unapologetic.wordpress.com/2008/11/13/the-coevaluation-on-vector-spaces/">"The Coevaluation on Vector Spaces"</a>. <i>The Unapologetic Mathematician</i>. 2008-11-13. <a href="https://web.archive.org/web/20170202080439/https://unapologetic.wordpress.com/2008/11/13/the-coevaluation-on-vector-spaces/">Archived</a> from the original on 2017-02-02. Retrieved 2017-01-26.</cite>
</li>
<li><b><a href="#cite_ref-7">^</a></b> See <a href="/wiki/Compact_closed_category">Compact closed category</a>.
</li>
<li><b><a href="#cite_ref-8">^</a></b> 
<cite>Hungerford, Thomas W. (1974). <i>Algebra</i>. Springer. <a href="/wiki/ISBN_(identifier)">ISBN</a> <a href="/wiki/Special:BookSources/0-387-90518-9"><bdi>0-387-90518-9</bdi></a>.</cite>
</li>
<li><b><a href="#cite_ref-chen_9-0">^</a></b> 
<cite>Chen, Jungkai Alfred (Spring 2004), <a href="http://www.math.ntu.edu.tw/~jkchen/S04AA/S04AAL10.pdf">"Tensor product"</a> (PDF), <i>Advanced Algebra II</i> (lecture notes), National Taiwan University, <a href="https://web.archive.org/web/20160304040639/http://www.math.ntu.edu.tw/~jkchen/S04AA/S04AAL10.pdf">archived</a> (PDF) from the original on 2016-03-04</cite>
</li>
<li><b><a href="#cite_ref-10">^</a></b> <cite>Abo, H.; Seigal, A.; Sturmfels, B. (2015). "Eigenconfigurations of Tensors". <a href="/wiki/ArXiv_(identifier)">arXiv</a>:<a href="//arxiv.org/abs/1505.05729">1505.05729</a> [<a href="//arxiv.org/archive/math.AG">math.AG</a>].</cite>
</li>
<li><b><a href="#cite_ref-An_Introduction_to_Manifolds_11-0">^</a></b> <cite>Tu, L. W. (2010). <i>An Introduction to Manifolds</i>. Universitext. Springer. p. 25. <a href="/wiki/ISBN_(identifier)">ISBN</a> <a href="/wiki/Special:BookSources/978-1-4419-7399-3"><bdi>978-1-4419-7399-3</bdi></a>.</cite>
</li>
</ol></div>
<h2>References[<a href="/w/index.php?title=Tensor_product&amp;action=edit&amp;section=37">edit</a>]</h2>
<ul><li><cite><a href="/wiki/Nicolas_Bourbaki">Bourbaki, Nicolas</a> (1989). <i>Elements of mathematics, Algebra I</i>. Springer-Verlag. <a href="/wiki/ISBN_(identifier)">ISBN</a> <a href="/wiki/Special:BookSources/3-540-64243-9"><bdi>3-540-64243-9</bdi></a>.</cite></li>
<li><cite><a href="/wiki/Tim_Gowers">Gowers, Timothy</a>. <a href="https://web.archive.org/web/20210507014709/https://www.dpmms.cam.ac.uk/~wtg10/tensors3.html">"How to lose your fear of tensor products"</a>. Archived from <a href="https://www.dpmms.cam.ac.uk/~wtg10/tensors3.html">the original</a> on 23 July 2021.</cite></li>
<li><cite>Grillet, Pierre A. (2007). <i>Abstract Algebra</i>. Springer Science+Business Media, LLC. <a href="/wiki/ISBN_(identifier)">ISBN</a> <a href="/wiki/Special:BookSources/978-0387715674"><bdi>978-0387715674</bdi></a>.</cite></li>
<li><cite><a href="/wiki/Paul_Halmos">Halmos, Paul</a> (1974). <i>Finite dimensional vector spaces</i>. Springer. <a href="/wiki/ISBN_(identifier)">ISBN</a> <a href="/wiki/Special:BookSources/0-387-90093-4"><bdi>0-387-90093-4</bdi></a>.</cite></li>
<li><cite><a href="/wiki/Thomas_W._Hungerford">Hungerford, Thomas W.</a> (2003). <i>Algebra</i>. Springer. <a href="/wiki/ISBN_(identifier)">ISBN</a> <a href="/wiki/Special:BookSources/0387905189"><bdi>0387905189</bdi></a>.</cite></li>
<li><cite><a href="/wiki/Serge_Lang">Lang, Serge</a> (2002), <i>Algebra</i>, <a href="/wiki/Graduate_Texts_in_Mathematics">Graduate Texts in Mathematics</a>, vol. 211 (Revised third ed.), New York: Springer-Verlag, <a href="/wiki/ISBN_(identifier)">ISBN</a> <a href="/wiki/Special:BookSources/978-0-387-95385-4"><bdi>978-0-387-95385-4</bdi></a>, <a href="/wiki/MR_(identifier)">MR</a> <a href="//mathscinet.ams.org/mathscinet-getitem?mr=1878556">1878556</a>, <a href="/wiki/Zbl_(identifier)">Zbl</a> <a href="//zbmath.org/?format=complete&amp;q=an:0984.00001">0984.00001</a></cite></li>
<li><cite><a href="/wiki/Saunders_Mac_Lane">Mac Lane, S.</a>; <a href="/wiki/Garrett_Birkhoff">Birkhoff, G.</a> (1999). <i>Algebra</i>. AMS Chelsea. <a href="/wiki/ISBN_(identifier)">ISBN</a> <a href="/wiki/Special:BookSources/0-8218-1646-2"><bdi>0-8218-1646-2</bdi></a>.</cite></li>
<li><cite>Aguiar, M.; Mahajan, S. (2010). <i>Monoidal functors, species and Hopf algebras</i>. CRM Monograph Series Vol 29. <a href="/wiki/ISBN_(identifier)">ISBN</a> <a href="/wiki/Special:BookSources/978-0-8218-4776-3"><bdi>978-0-8218-4776-3</bdi></a>.</cite></li>
<li><cite><a href="/wiki/Fran%C3%A7ois_Tr%C3%A8ves">Trèves, François</a> (2006) [1967]. <i>Topological Vector Spaces, Distributions and Kernels</i>. Mineola, N.Y.: Dover Publications. <a href="/wiki/ISBN_(identifier)">ISBN</a> <a href="/wiki/Special:BookSources/978-0-486-45352-1"><bdi>978-0-486-45352-1</bdi></a>. <a href="/wiki/OCLC_(identifier)">OCLC</a> <a href="//www.worldcat.org/oclc/853623322">853623322</a>.</cite></li>
<li><cite><a href="http://pages.bangor.ac.uk/~mas010/nonabtens.html">"Bibliography on the nonabelian tensor product of groups"</a>.</cite></li></ul>
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 Contents

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 * (Top)
 * 1 Definitions and constructions
 + 1.1 From bases
 + 1.2 As a quotient space
 + 1.3 Universal property
 + 1.4 Linearly disjoint
 * 2 Properties
 + 2.1 Dimension
 + 2.2 Associativity
 + 2.3 Commutativity as vector space operation
 * 3 Tensor product of linear maps
 * 4 General tensors
 + 4.1 Evaluation map and tensor contraction
 + 4.2 Adjoint representation
 * 5 Linear maps as tensors
 * 6 Tensor products of modules over a ring
 + 6.1 Tensor product of modules over a non-commutative ring
 + 6.2 Computing the tensor product
 * 7 Tensor product of algebras
 * 8 Eigenconfigurations of tensors
 * 9 Other examples of tensor products
 + 9.1 Tensor product of Hilbert spaces
 + 9.2 Topological tensor product
 + 9.3 Tensor product of graded vector spaces
 + 9.4 Tensor product of representations
 + 9.5 Tensor product of quadratic forms
 + 9.6 Tensor product of multilinear forms
 + 9.7 Tensor product of sheaves of modules
 + 9.8 Tensor product of line bundles
 + 9.9 Tensor product of fields
 + 9.10 Tensor product of graphs
 + 9.11 Monoidal categories
 * 10 Quotient algebras
 * 11 Tensor product in programming
 + 11.1 Array programming languages
 * 12 See also
 * 13 Notes
 * 14 References
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 Tensor product

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 From Wikipedia, the free encyclopedia
 For generalizations of this concept, see Tensor product of modules and Tensor product (disambiguation).

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

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

 The tensor product of two vector spaces captures the properties of all bilinear maps in the sense that a bilinear map from into another vector space Z factors uniquely through a linear map (see Universal property).

 Tensor products are used in many application areas, including physics and engineering. For example, in general relativity, the gravitational field is described through the metric tensor, which is a vector field of tensors, one at each point of the space-time manifold, and each belonging to the tensor product with itself of the cotangent space at the point.

 Definitions and constructions [ edit ]

 The tensor product of two vector spaces is a vector space that is defined up to an isomorphism. There are several equivalent ways to define it. Most consist of defining explicitly a vector space that is called a tensor product, and, generally, the equivalence proof results almost immediately from the basic properties of the vector spaces that are so defined.

 The tensor product can also be defined through a universal property; see § Universal property, below. As for every universal property, all objects that satisfy the property are isomorphic through a unique isomorphism that is compatible with the universal property. When this definition is used, the other definitions may be viewed as constructions of objects satisfying the universal property and as proofs that there are objects satisfying the universal property, that is that tensor products exist.

 From bases [ edit ]

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

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

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

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

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

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

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

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

 As a quotient space [ edit ]

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

 Let V and W be two vector spaces over a field F .

 One considers first a vector space L that has the Cartesian product as a basis. That is, the basis elements of L are the pairs with and To get such a vector space, one can define it as the vector space of the functions that have a finite number of nonzero values, and identifying with the function that takes the value 1 on and 0 otherwise.

 Let R be the linear subspace of L that is spanned by the relations that the tensor product must satisfy. More precisely R is spanned by the elements of one of the forms

 where and

 Then, the tensor product is defined as the quotient space

 and the image of in this quotient is denoted

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

 Universal property [ edit ]

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

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

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

 The "universal-property definition" of the tensor product of two vector spaces is the following (recall that a bilinear map is a function that is separately linear in each of its arguments):

 The tensor product of two vector spaces V and W is a vector space denoted as together with a bilinear map from to such that, for every bilinear map there is a unique linear map such that (that is, for every and ).

 Linearly disjoint [ edit ]

 Like the universal property above, the following characterization may also be used to determine whether or not a given vector space and given bilinear map form a tensor product.[1]

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

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

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

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

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

 form a tensor product of and [2]

 Properties [ edit ]

 Dimension [ edit ]

 If V and W are vectors spaces of finite dimension, then is finite-dimensional, and its dimension is the product of the dimensions of V and W.

 This results from the fact that a basis of is formed by taking all tensor products of a basis element of V and a basis element of W.

 Associativity [ edit ]

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

 that maps to

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

 Commutativity as vector space operation [ edit ]

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

 that maps to

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

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

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

 Tensor product of linear maps [ edit ]

 "Tensor product of linear maps" redirects here. For the generalization for modules, see Tensor product of modules § Tensor product of linear maps and a change of base ring.

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

 is the unique linear map such that

 The tensor product is defined similarly.

 Given two linear maps and their tensor product

 is the unique linear map that satisfies

 One has

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

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

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

 respectively, then the tensor product of these two matrices is

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

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

 General tensors [ edit ]

 See also: Tensor

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

 Here is the dual vector space (which consists of all linear maps f from V to the ground field K).

 There is a product map, called the (tensor) product of tensors[4]

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

 Picking a basis of V and the corresponding dual basis of naturally induces a basis for (this basis is described in the article on Kronecker products). In terms of these bases, the components of a (tensor) product of two (or more) tensors can be computed. For example, if F and G are two covariant tensors of orders m and n respectively (i.e. and ), then the components of their tensor product are given by[5]

 Thus, the components of the tensor product of two tensors are the ordinary product of the components of each tensor. Another example: let U be a tensor of type (1, 1) with components and let V be a tensor of type with components Then

 and

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

 Evaluation map and tensor contraction [ edit ]

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

 defined by its action on pure tensors:

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

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

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

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

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

 Adjoint representation [ edit ]

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

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

 There is a canonical isomorphism given by

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

 Linear maps as tensors [ edit ]

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

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

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

 This result implies

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

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

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

 Tensor products of modules over a ring [ edit ]

 Main article: Tensor product of modules

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

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

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

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

 The universal property also carries over, slightly modified: the map defined by is a middle linear map (referred to as "the canonical middle linear map".[8]); that is, it satisfies:[9]

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

 Tensor product of modules over a non-commutative ring [ edit ]

 Let A be a right R-module and B be a left R-module. Then the tensor product of A and B is an abelian group defined by

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

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

 Then there is a unique map such that for all and

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

 1. If A is a (S,R)-bimodule, then is a left S-module where
 2. If B is a (R,S)-bimodule, then is a right S-module where
 3. If A is a (S,R)-bimodule and B is a (R,T)-bimodule, then is a (S,T)-bimodule, where the left and right actions are defined in the same way as the previous two examples.
 4. If R is a commutative ring, then A and B are (R,R)-bimodules where and By 3), we can conclude is a (R,R)-bimodule.

 Computing the tensor product [ edit ]

 For vector spaces, the tensor product is quickly computed since bases of V of W immediately determine a basis of as was mentioned above. For modules over a general (commutative) ring, not every module is free. For example, Z / nZ is not a free abelian group (Z-module). The tensor product with Z / nZ is given by

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

 the tensor product can be computed as the following cokernel:

 Here and the map is determined by sending some in the jth copy of to (in ). Colloquially, this may be rephrased by saying that a presentation of M gives rise to a presentation of This is referred to by saying that the tensor product is a right exact functor. It is not in general left exact, that is, given an injective map of R-modules the tensor product

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

 Tensor product of algebras [ edit ]

 Main article: Tensor product of algebras

 Let R be a commutative ring. The tensor product of R-modules applies, in particular, if A and B are R-algebras. In this case, the tensor product is an R-algebra itself by putting

 For example,

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

 where now f is interpreted as the same polynomial, but with its coefficients regarded as elements of B. In the larger field B, the polynomial may become reducible, which brings in Galois theory. For example, if A = B is a Galois extension of R, then
 is isomorphic (as an A-algebra) to the

 Eigenconfigurations of tensors [ edit ]

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

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

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

 Other examples of tensor products [ edit ]

 Tensor product of Hilbert spaces [ edit ]

 Main article: Tensor product of Hilbert spaces

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

 Topological tensor product [ edit ]

 Main article: Topological tensor product

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

 Tensor product of graded vector spaces [ edit ]

 Main article: 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).

 Tensor product of representations [ edit ]

 Main article: 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.

 Tensor product of quadratic forms [ edit ]

 Main article: Tensor product of quadratic forms

 Tensor product of multilinear forms [ edit ]

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

 [11]

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

 Tensor product of sheaves of modules [ edit ]

 Main article: Sheaf of modules

 Tensor product of line bundles [ edit ]

 Main article: Vector bundle § Operations on vector bundles
 See also: tensor product bundle

 Tensor product of fields [ edit ]

 Main article: Tensor product of fields

 Tensor product of graphs [ edit ]

 Main article: 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.

 Monoidal categories [ edit ]

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

 Quotient algebras [ edit ]

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

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

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

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

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

 Tensor product in programming [ edit ]

 Array programming languages [ edit ]

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

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

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

 See also [ edit ]

 Look up tensor product in Wiktionary, the free dictionary.
 * Dyadics – second order tensor, written in a notation that fits in with vector algebra
 * Extension of scalars
 * Monoidal category – Category admitting tensor products
 * Tensor algebra – Universal construction in multilinear algebra
 * Tensor contraction – in mathematics and physics, an operation on tensors
 * Topological tensor product – Tensor product constructions for topological vector spaces

 Notes [ edit ]

 1. ^ a b Trèves 2006 , pp. 403–404.
 2. ^ a b Trèves 2006 , pp. 407.
 3. ^ Hazewinkel, Michiel; Gubareni, Nadezhda Mikhaĭlovna; Gubareni, Nadiya; Kirichenko, Vladimir V. (2004). Algebras, rings and modules. Springer. p. 100. ISBN 978-1-4020-2690-4.
 4. ^ Bourbaki (1989) , p. 244 defines the usage "tensor product of x and y ", elements of the respective modules.
 5. ^ Analogous formulas also hold for contravariant tensors, as well as tensors of mixed variance. Although in many cases such as when there is an inner product defined, the distinction is irrelevant.
 6. ^ "The Coevaluation on Vector Spaces". The Unapologetic Mathematician. 2008-11-13. Archived from the original on 2017-02-02 . Retrieved 2017-01-26 .
 7. ^ See Compact closed category .
 8. ^ Hungerford, Thomas W. (1974). Algebra. Springer. ISBN 0-387-90518-9.
 9. ^ Chen, Jungkai Alfred (Spring 2004), "Tensor product" (PDF) , Advanced Algebra II (lecture notes), National Taiwan University, archived (PDF) from the original on 2016-03-04
 10. ^ Abo, H.; Seigal, A.; Sturmfels, B. (2015). "Eigenconfigurations of Tensors". arXiv:1505.05729 [math.AG].
 11. ^ Tu, L. W. (2010). An Introduction to Manifolds. Universitext. Springer. p. 25. ISBN 978-1-4419-7399-3.

 References [ edit ]

 * Bourbaki, Nicolas (1989). Elements of mathematics, Algebra I. Springer-Verlag. ISBN 3-540-64243-9.
 * Gowers, Timothy. "How to lose your fear of tensor products". Archived from the original on 23 July 2021.
 * Grillet, Pierre A. (2007). Abstract Algebra. Springer Science+Business Media, LLC. ISBN 978-0387715674.
 * Halmos, Paul (1974). Finite dimensional vector spaces. Springer. ISBN 0-387-90093-4.
 * Hungerford, Thomas W. (2003). Algebra. Springer. ISBN 0387905189.
 * Lang, Serge (2002), Algebra, Graduate Texts in Mathematics, vol. 211 (Revised third ed.), New York: Springer-Verlag, ISBN 978-0-387-95385-4, MR 1878556, Zbl 0984.00001
 * Mac Lane, S.; Birkhoff, G. (1999). Algebra. AMS Chelsea. ISBN 0-8218-1646-2.
 * Aguiar, M.; Mahajan, S. (2010). Monoidal functors, species and Hopf algebras. CRM Monograph Series Vol 29. ISBN 978-0-8218-4776-3.
 * Trèves, François (2006) [1967]. Topological Vector Spaces, Distributions and Kernels. Mineola, N.Y.: Dover Publications. ISBN 978-0-486-45352-1. OCLC 853623322.
 * "Bibliography on the nonabelian tensor product of groups".
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In mathematics, the tensor product of two vector spaces V and W (over the same field) is a vector space to which is associated a bilinear map that maps a pair to an element of denoted 
 An element of the form is called the tensor product of v and w. An element of is a tensor, and the tensor product of two vectors is sometimes called an elementary tensor or a decomposable tensor. The elementary tensors span in the sense that every element of is a sum of elementary tensors. If bases are given for V and W, a basis of is formed by all tensor products of a basis element of V and a basis element of W.
 The tensor product of two vector spaces captures the properties of all bilinear maps in the sense that a bilinear map from into another vector space Z factors uniquely through a linear map (see Universal property).
 Tensor products are used in many application areas, including physics and engineering. For example, in general relativity, the gravitational field is described through the metric tensor, which is a vector field of tensors, one at each point of the space-time manifold, and each belonging to the tensor product with itself of the cotangent space at the point.
Definitions and constructions[edit]
The tensor product of two vector spaces is a vector space that is defined up to an isomorphism. There are several equivalent ways to define it. Most consist of defining explicitly a vector space that is called a tensor product, and, generally, the equivalence proof results almost immediately from the basic properties of the vector spaces that are so defined. 
 The tensor product can also be defined through a universal property; see § Universal property, below. As for every universal property, all objects that satisfy the property are isomorphic through a unique isomorphism that is compatible with the universal property. When this definition is used, the other definitions may be viewed as constructions of objects satisfying the universal property and as proofs that there are objects satisfying the universal property, that is that tensor products exist.
From bases[edit]
Let V and W be two vector spaces over a field F, with respective bases and 
 The tensor product of V and W is a vector space which has as a basis the set of all with and This definition can be formalized in the following way (this formalization is rarely used in practice, as the preceding informal definition is generally sufficient): is the set of the functions from the Cartesian product to F that have a finite number of nonzero values. The pointwise operations make a vector space. The function that maps to 1 and the other elements of to 0 is denoted 
 The set is straightforwardly a basis of which is called the tensor product of the bases and 
 The tensor product of two vectors is defined from their decomposition on the bases. More precisely, if 
     
are vectors decomposed on their respective bases, then the tensor product of x and y is
    
If arranged into a rectangular array, the coordinate vector of is the outer product of the coordinate vectors of x and y. Therefore, the tensor product is a generalization of the outer product.
 It is straightforward to verify that the map is a bilinear map from to 
 A limitation of this definition of the tensor product is that, if one changes bases, a different tensor product is defined. However, the decomposition on one basis of the elements of the other basis defines a canonical isomorphism between the two tensor products of vector spaces, which allows identifying them. Also, contrarily to the two following alternative definitions, this definition cannot be extended into a definition of the tensor product of modules over a ring.
As a quotient space[edit]
A construction of the tensor product that is basis independent can be obtained in the following way.
 Let V and W be two vector spaces over a field F. 
 One considers first a vector space L that has the Cartesian product as a basis. That is, the basis elements of L are the pairs with and To get such a vector space, one can define it as the vector space of the functions that have a finite number of nonzero values, and identifying with the function that takes the value 1 on and 0 otherwise.
 Let R be the linear subspace of L that is spanned by the relations that the tensor product must satisfy. More precisely R is spanned by the elements of one of the forms
   
where and 
 Then, the tensor product is defined as the quotient space
   
and the image of in this quotient is denoted 
 It is straightforward to prove that the result of this construction satisfies the universal property considered below. (A very similar construction can be used to define the tensor product of modules.)
Universal property[edit]
      Universal property of tensor product: if h is bilinear, there is a unique linear map that makes the diagram commutative (that is, ).  
In this section, the universal property satisfied by the tensor product is described. As for every universal property, two objects that satisfy the property are related by a unique isomorphism. It follows that this is a (non-constructive) way to define the tensor product of two vector spaces. In this context, the preceding constructions of tensor products may be viewed as proofs of existence of the tensor product so defined.
 A consequence of this approach is that every property of the tensor product can be deduced from the universal property, and that, in practice, one may forget the method that has been used to prove its existence. 
 The "universal-property definition" of the tensor product of two vector spaces is the following (recall that a bilinear map is a function that is separately linear in each of its arguments):
 The tensor product of two vector spaces V and W is a vector space denoted as together with a bilinear map from to such that, for every bilinear map there is a unique linear map such that (that is, for every and ). 
Linearly disjoint[edit]
Like the universal property above, the following characterization may also be used to determine whether or not a given vector space and given bilinear map form a tensor product. [1] 
For example, it follows immediately that if and are positive integers then and the bilinear map defined by sending to form a tensor product of and  [2]  Often, this map will be denoted by so that denotes this bilinear map's value at 
 As another example, suppose that is the vector space of all complex-valued functions on a set with addition and scalar multiplication defined pointwise (meaning that is the map and is the map ). Let and be any sets and for any and let denote the function defined by 
If and are vector subspaces then the vector subspace of together with the bilinear map 
     
form a tensor product of and  [2]  
Properties[edit]
Dimension[edit]
If V and W are vectors spaces of finite dimension, then is finite-dimensional, and its dimension is the product of the dimensions of V and W.
 This results from the fact that a basis of is formed by taking all tensor products of a basis element of V and a basis element of W.
Associativity[edit]
The tensor product is associative in the sense that, given three vector spaces there is a canonical isomorphism 
   
that maps to 
 This allows omitting parentheses in the tensor product of more than two vector spaces or vectors.
Commutativity as vector space operation[edit]
The tensor product of two vector spaces and is commutative  in the sense that there is a canonical isomorphism 
   
that maps to 
 On the other hand, even when the tensor product of vectors is not commutative; that is in general.
 
The map from to itself induces a linear automorphism that is called a braiding map. 
More generally and as usual (see tensor algebra), let denote the tensor product of n copies of the vector space V. For every permutation s of the first n positive integers, the map
   
induces a linear automorphism of which is called a braiding map.
Tensor product of linear maps[edit]
Given a linear map and a vector space W, the tensor product 
   
is the unique linear map such that
   
The tensor product is defined similarly.
 Given two linear maps and their tensor product 
   
is the unique linear map that satisfies 
   
One has 
   
In terms of category theory, this means that the tensor product is a bifunctor from the category of vector spaces to itself. [3] 
 If f and g are both injective or surjective, then the same is true for all above defined linear maps. In particular, the tensor product with a vector space is an exact functor; this means that every exact sequence is mapped to an exact sequence (tensor products of modules do not transform injections into injections, but they are right exact functors).
 By choosing bases of all vector spaces involved, the linear maps S and T can be represented by matrices. Then, depending on how the tensor is vectorized, the matrix describing the tensor product is the Kronecker product of the two matrices. For example, if V, X, W, and Y above are all two-dimensional and bases have been fixed for all of them, and S and T are given by the matrices
    
respectively, then the tensor product of these two matrices is
    
The resultant rank is at most 4, and thus the resultant dimension is 4. Note that rank here denotes the tensor rank i.e. the number of requisite indices (while the matrix rank counts the number of degrees of freedom in the resulting array). Note 
 A dyadic product is the special case of the tensor product between two vectors of the same dimension.
General tensors[edit]
For non-negative integers r and s a type tensor on a vector space V is an element of
    
Here is the dual vector space (which consists of all linear maps f from V to the ground field K).
 There is a product map, called the (tensor) product of tensors  [4] 
    
It is defined by grouping all occurring "factors" V together: writing for an element of V and for an element of the dual space,
    
Picking a basis of V and the corresponding dual basis of naturally induces a basis for (this basis is described in the article on Kronecker products). In terms of these bases, the components of a (tensor) product of two (or more) tensors can be computed. For example, if F and G are two covariant tensors of orders m and n respectively (i.e. and ), then the components of their tensor product are given by [5] 
    
Thus, the components of the tensor product of two tensors are the ordinary product of the components of each tensor. Another example: let U be a tensor of type (1, 1) with components and let V be a tensor of type with components Then
    
and
    
Tensors equipped with their product operation form an algebra, called the tensor algebra.
Evaluation map and tensor contraction[edit]
For tensors of type (1, 1) there is a canonical evaluation map
    
defined by its action on pure tensors:
    
More generally, for tensors of type with r, s &gt; 0, there is a map, called tensor contraction,
    
(The copies of and on which this map is to be applied must be specified.)
 On the other hand, if is finite-dimensional, there is a canonical map in the other direction (called the coevaluation map)
    
where is any basis of and is its dual basis. This map does not depend on the choice of basis. [6] 
 The interplay of evaluation and coevaluation can be used to characterize finite-dimensional vector spaces without referring to bases. [7] 
Adjoint representation[edit]
The tensor product may be naturally viewed as a module for the Lie algebra by means of the diagonal action: for simplicity let us assume then, for each 
    
where is the transpose of u, that is, in terms of the obvious pairing on 
    
There is a canonical isomorphism given by
    
Under this isomorphism, every u in may be first viewed as an endomorphism of and then viewed as an endomorphism of In fact it is the adjoint representation ad(u) of 
Linear maps as tensors[edit]
Given two finite dimensional vector spaces U, V over the same field K, denote the dual space of U as U*, and the K-vector space of all linear maps from U to V as Hom(U,V). There is an isomorphism,
    
defined by an action of the pure tensor on an element of 
    
Its "inverse" can be defined using a basis and its dual basis as in the section "Evaluation map and tensor contraction" above:
    
This result implies
    
which automatically gives the important fact that forms a basis for where are bases of U and V.
 Furthermore, given three vector spaces U, V, W the tensor product is linked to the vector space of all linear maps, as follows:
     
This is an example of adjoint functors: the tensor product is "left adjoint" to Hom.
Tensor products of modules over a ring[edit]
The tensor product of two modules A and B over a  commutative  ring R is defined in exactly the same way as the tensor product of vector spaces over a field:
     
where now is the free R-module generated by the cartesian product and G is the R-module generated by the same relations as above.
 More generally, the tensor product can be defined even if the ring is non-commutative. In this case A has to be a right-R-module and B is a left-R-module, and instead of the last two relations above, the relation
     
is imposed. If R is non-commutative, this is no longer an R-module, but just an abelian group.
 The universal property also carries over, slightly modified: the map defined by is a middle linear map (referred to as "the canonical middle linear map". [8] ); that is, it satisfies: [9] 
    
The first two properties make φ a bilinear map of the abelian group For any middle linear map of a unique group homomorphism f of satisfies and this property determines within group isomorphism. See the main article for details.
Tensor product of modules over a non-commutative ring[edit]
Let A be a right R-module and B be a left R-module. Then the tensor product of A and B is an abelian group defined by
     
where is a free abelian group over and G is the subgroup of generated by relations
    
The universal property can be stated as follows. Let G be an abelian group with a map that is bilinear, in the sense that
    
Then there is a unique map such that for all and 
 Furthermore, we can give a module structure under some extra conditions:
 If A is a (S,R)-bimodule, then is a left S-module where 
If B is a (R,S)-bimodule, then is a right S-module where 
If A is a (S,R)-bimodule and B is a (R,T)-bimodule, then is a (S,T)-bimodule, where the left and right actions are defined in the same way as the previous two examples.
If R is a commutative ring, then A and B are (R,R)-bimodules where and By 3), we can conclude is a (R,R)-bimodule. 
Computing the tensor product[edit]
For vector spaces, the tensor product is quickly computed since bases of V of W immediately determine a basis of as was mentioned above. For modules over a general (commutative) ring, not every module is free. For example, Z/n Z is not a free abelian group (Z-module). The tensor product with Z/n Z is given by
    
More generally, given a presentation of some R-module M, that is, a number of generators together with relations
    
the tensor product can be computed as the following cokernel:
    
Here and the map is determined by sending some in the jth copy of to (in ). Colloquially, this may be rephrased by saying that a presentation of M gives rise to a presentation of This is referred to by saying that the tensor product is a right exact functor. It is not in general left exact, that is, given an injective map of R-modules the tensor product
    
is not usually injective. For example, tensoring the (injective) map given by multiplication with n, n : Z → Z with Z/n Z yields the zero map 0 : Z/n Z → Z/n Z, which is not injective. Higher Tor functors measure the defect of the tensor product being not left exact. All higher Tor functors are assembled in the derived tensor product.
Tensor product of algebras[edit]
Let R be a commutative ring. The tensor product of R-modules applies, in particular, if A and B are  R-algebras. In this case, the tensor product is an R-algebra itself by putting
     
For example,
    
A particular example is when A and B are fields containing a common subfield R. The tensor product of fields is closely related to Galois theory: if, say, A = R[x] / f(x), where f is some irreducible polynomial with coefficients in R, the tensor product can be calculated as
     
where now f is interpreted as the same polynomial, but with its coefficients regarded as elements of B. In the larger field B, the polynomial may become reducible, which brings in Galois theory. For example, if A = B is a Galois extension of R, then
     
is isomorphic (as an A-algebra) to the 
Eigenconfigurations of tensors[edit]
Square matrices with entries in a field represent linear maps of vector spaces, say and thus linear maps of projective spaces over If is nonsingular then is well-defined everywhere, and the eigenvectors of correspond to the fixed points of The eigenconfiguration of consists of points in provided is generic and is algebraically closed. The fixed points of nonlinear maps are the eigenvectors of tensors. Let be a -dimensional tensor of format with entries lying in an algebraically closed field of characteristic zero. Such a tensor defines polynomial maps and with coordinates
    
Thus each of the coordinates of is a homogeneous polynomial of degree in The eigenvectors of are the solutions of the constraint
    
and the eigenconfiguration is given by the variety of the minors of this matrix. [10] 
Other examples of tensor products[edit]
Tensor product of Hilbert spaces[edit]
 Hilbert spaces generalize finite-dimensional vector spaces to countably-infinite dimensions. The tensor product is still defined; it is the tensor product of Hilbert spaces.
Topological tensor product[edit]
When the basis for a vector space is no longer countable, then the appropriate axiomatic formalization for the vector space is that of a topological vector space. The tensor product is still defined, it is the topological tensor product.
Tensor product of graded vector spaces[edit]
Some vector spaces can be decomposed into direct sums of subspaces. In such cases, the tensor product of two spaces can be decomposed into sums of products of the subspaces (in analogy to the way that multiplication distributes over addition).
Tensor product of representations[edit]
Vector spaces endowed with an additional multiplicative structure are called algebras. The tensor product of such algebras is described by the Littlewood–Richardson rule.
Tensor product of quadratic forms[edit]
Tensor product of multilinear forms[edit]
Given two multilinear forms and on a vector space over the field their tensor product is the multilinear form
       [11]   This is a special case of the product of tensors if they are seen as multilinear maps (see also tensors as multilinear maps). Thus the components of the tensor product of multilinear forms can be computed by the Kronecker product.
Tensor product of sheaves of modules[edit]
Tensor product of line bundles[edit]
Tensor product of fields[edit]
Tensor product of graphs[edit]
It should be mentioned that, though called "tensor product", this is not a tensor product of graphs in the above sense; actually it is the category-theoretic product in the category of graphs and graph homomorphisms. However it is actually the Kronecker tensor product of the adjacency matrices of the graphs. Compare also the section Tensor product of linear maps above.
Monoidal categories[edit]
The most general setting for the tensor product is the monoidal category. It captures the algebraic essence of tensoring, without making any specific reference to what is being tensored. Thus, all tensor products can be expressed as an application of the monoidal category to some particular setting, acting on some particular objects.
Quotient algebras[edit]
A number of important subspaces of the tensor algebra can be constructed as quotients: these include the exterior algebra, the symmetric algebra, the Clifford algebra, the Weyl algebra, and the universal enveloping algebra in general.
 The exterior algebra is constructed from the exterior product. Given a vector space V, the exterior product is defined as
     
Note that when the underlying field of V does not have characteristic 2, then this definition is equivalent to
     
The image of in the exterior product is usually denoted and satisfies, by construction, Similar constructions are possible for (n factors), giving rise to the nth exterior power of V. The latter notion is the basis of differential n-forms.
 The symmetric algebra is constructed in a similar manner, from the symmetric product
     
More generally
     
That is, in the symmetric algebra two adjacent vectors (and therefore all of them) can be interchanged. The resulting objects are called symmetric tensors.
Tensor product in programming[edit]
Array programming languages[edit]
 Array programming languages may have this pattern built in. For example, in APL the tensor product is expressed as ○.× (for example A ○.× B or A ○.× B ○.× C). In J the tensor product is the dyadic form of */ (for example a */ b or a */ b */ c).
 Note that J's treatment also allows the representation of some tensor fields, as a and b may be functions instead of constants. This product of two functions is a derived function, and if a and b are differentiable, then a */ b is differentiable.
 However, these kinds of notation are not universally present in array languages. Other array languages may require explicit treatment of indices (for example, MATLAB), and/or may not support higher-order functions such as the Jacobian derivative (for example, Fortran/APL).
See also[edit]
  Dyadics – second order tensor, written in a notation that fits in with vector algebra
 Extension of scalars 
 Monoidal category – Category admitting tensor products
 Tensor algebra – Universal construction in multilinear algebra
 Tensor contraction – in mathematics and physics, an operation on tensors
 Topological tensor product – Tensor product constructions for topological vector spaces 
Notes[edit]
 
^    a       b    Trèves 2006, pp. 403–404.
^    a       b    Trèves 2006, pp. 407.
  ^  Hazewinkel, Michiel; Gubareni, Nadezhda Mikhaĭlovna; Gubareni, Nadiya; Kirichenko, Vladimir V. (2004). Algebras, rings and modules. Springer. p. 100. ISBN  978-1-4020-2690-4 .
  ^  Bourbaki (1989), p. 244 defines the usage "tensor product of x and y", elements of the respective modules.
  ^  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.
  ^   "The Coevaluation on Vector Spaces". The Unapologetic Mathematician. 2008-11-13. Archived from the original on 2017-02-02. Retrieved 2017-01-26.
  ^  See Compact closed category.
  ^  
Hungerford, Thomas W. (1974). Algebra. Springer. ISBN  0-387-90518-9 .
  ^  
Chen, Jungkai Alfred (Spring 2004), "Tensor product" (PDF), Advanced Algebra II (lecture notes), National Taiwan University, archived (PDF) from the original on 2016-03-04
  ^  Abo, H.; Seigal, A.; Sturmfels, B. (2015). "Eigenconfigurations of Tensors". arXiv:1505.05729 [math.AG].
  ^  Tu, L. W. (2010). An Introduction to Manifolds. Universitext. Springer. p. 25. ISBN  978-1-4419-7399-3 .
 
References[edit]
   Bourbaki, Nicolas (1989). Elements of mathematics, Algebra I. Springer-Verlag. ISBN  3-540-64243-9 . 
  Gowers, Timothy. "How to lose your fear of tensor products". Archived from the original on 23 July 2021. 
 Grillet, Pierre A. (2007). Abstract Algebra. Springer Science+Business Media, LLC. ISBN  978-0387715674 . 
  Halmos, Paul (1974). Finite dimensional vector spaces. Springer. ISBN  0-387-90093-4 . 
  Hungerford, Thomas W. (2003). Algebra. Springer. ISBN  0387905189 . 
  Lang, Serge (2002), Algebra, Graduate Texts in Mathematics, vol. 211 (Revised third ed.), New York: Springer-Verlag, ISBN  978-0-387-95385-4 , MR 1878556, Zbl 0984.00001  
  Mac Lane, S.; Birkhoff, G. (1999). Algebra. AMS Chelsea. ISBN  0-8218-1646-2 . 
 Aguiar, M.; Mahajan, S. (2010). Monoidal functors, species and Hopf algebras. CRM Monograph Series Vol 29. ISBN  978-0-8218-4776-3 . 
  Trèves, François (2006) [1967]. Topological Vector Spaces, Distributions and Kernels. Mineola, N.Y.: Dover Publications. ISBN  978-0-486-45352-1 . OCLC 853623322. 
  "Bibliography on the nonabelian tensor product of groups".
Tensor product - Wikipedia
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