Editing Tensor product (section)

== General tensors ==
{{See also|Tensor}}
For non-negative integers {{math|''r''}} and {{math|''s''}} a type <math>(r, s)</math> [[tensor]] on a vector space {{math|''V''}} is an element of:
<math display="block">T^r_s(V) = \underbrace{ V \otimes \cdots \otimes V}_r \otimes \underbrace{ V^* \otimes \cdots \otimes V^*}_s = V^{\otimes r} \otimes \left(V^*\right)^{\otimes s}.</math>
Here <math>V^*</math> is the [[dual vector space]] (which consists of all [[linear map]]s {{math|''f''}} from {{math|''V''}} to the ground field {{math|''K''}}).

There is a product map, called the {{em|(tensor) product of tensors}}:{{refn|{{harvp|Bourbaki|1989|p=244}} defines the usage "tensor product of ''x'' and ''y''", elements of the respective modules.}}
<math display="block">T^r_s (V) \otimes_K T^{r'}_{s'} (V) \to T^{r+r'}_{s+s'}(V).</math>

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

If {{math|''V''}} is finite dimensional, then picking a basis of {{math|''V''}} and the corresponding [[dual basis]] of <math>V^*</math> naturally induces a basis of <math>T_s^r(V)</math> (this basis is described in the [[Kronecker product#Abstract properties|article on Kronecker products]]). In terms of these bases, the [[Coordinate vector|components]] of a (tensor) product of two (or more) [[tensor]]s can be computed. For example, if {{math|''F''}} and {{math|''G''}} are two [[covariance and contravariance of vectors|covariant]] tensors of orders {{math|''m''}} and {{math|''n''}} respectively (i.e. <math>F \in T_m^0</math> and {{tmath|1= G \in T_n^0 }}), then the components of their tensor product are given by:<ref>Analogous formulas also hold for [[covariance and contravariance of vectors|contravariant]] tensors, as well as tensors of mixed variance. Although in many cases such as when there is an [[inner product]] defined, the distinction is irrelevant.</ref>
<math display="block">(F \otimes G)_{i_1 i_2 \cdots i_{m+n}} = F_{i_1 i_2 \cdots i_m} G_{i_{m+1} i_{m+2} i_{m+3} \cdots i_{m+n}}.</math>

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

Tensors equipped with their product operation form an [[algebra over a field|algebra]], called the [[tensor algebra]].

=== Evaluation map and tensor contraction ===
For tensors of type {{math|(1, 1)}} there is a canonical '''evaluation map:'''
<math display="block">V \otimes V^* \to K</math>
defined by its action on pure tensors:
<math display="block">v \otimes f \mapsto f(v).</math>

More generally, for tensors of type {{tmath|1= (r, s) }}, with {{math|''r'', ''s'' > 0}}, there is a map, called [[tensor contraction]]:
<math display="block">T^r_s (V) \to T^{r-1}_{s-1}(V).</math>
(The copies of <math>V</math> and <math>V^*</math> on which this map is to be applied must be specified.)

On the other hand, if <math>V</math> is {{em|finite-dimensional}}, there is a canonical map in the other direction (called the '''coevaluation map'''):
<math display="block">\begin{cases} K \to V \otimes V^* \\ \lambda \mapsto \sum_i \lambda v_i \otimes v^*_i \end{cases}</math>
where <math>v_1, \ldots, v_n</math> is any basis of {{tmath|1= V }}, and <math>v_i^*</math> is its [[dual basis]]. This map does not depend on the choice of basis.<ref>{{Cite web| url= https://unapologetic.wordpress.com/2008/11/13/the-coevaluation-on-vector-spaces/|title=The Coevaluation on Vector Spaces|date=2008-11-13| website=The Unapologetic Mathematician|access-date=2017-01-26| url-status=live |archive-url =https://web.archive.org/web/20170202080439/https://unapologetic.wordpress.com/2008/11/13/the-coevaluation-on-vector-spaces/| archive-date =2017-02-02}}</ref>

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

=== Adjoint representation ===
The tensor product <math>T^r_s(V)</math> may be naturally viewed as a module for the [[Lie algebra]] <math>\mathrm{End}(V)</math> by means of the diagonal action: for simplicity let us assume {{tmath|1= r = s = 1 }}, then, for each {{tmath|1= u \in \mathrm{End}(V) }},
<math display="block">u(a \otimes b) = u(a) \otimes b - a \otimes u^*(b),</math>
where <math>u^* \in \mathrm{End}\left(V^*\right)</math> is the [[transpose]] of {{math|''u''}}, that is, in terms of the obvious pairing on {{tmath|1= V \otimes V^* }},
<math display="block">\langle u(a), b \rangle = \langle a, u^*(b) \rangle.</math>

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

Under this isomorphism, every {{math|''u''}} in <math>\mathrm{End}(V)</math> may be first viewed as an endomorphism of <math>T^1_1(V)</math> and then viewed as an endomorphism of {{tmath|1= \mathrm{End}(V) }}. In fact it is the [[Adjoint representation of a Lie algebra|adjoint representation]] {{math|ad(''u'')}} of {{tmath|1= \mathrm{End}(V) }}.