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{{Short description|Coordinate-free definition of a tensor}}
{{hatnote|This article assumes an understanding of the [[tensor product]] of [[vector space]]s without chosen [[Basis (linear algebra)|bases]]. An introduction to the nature and significance of tensors in a broad context can be found in the main [[Tensor]] article.}}
{{More footnotes needed|date=November 2024}}

In [[mathematics]], the modern [[component-free]] approach to the theory of a '''tensor''' views a tensor as an [[abstract object]], expressing some definite type of [[Multilinear_map|multilinear]] concept. Their  properties can be derived from their definitions, as [[linear map]]s or more generally; and the rules for manipulations of tensors arise as an extension of [[linear algebra]] to [[multilinear algebra]].

In [[differential geometry]], an intrinsic{{Definition needed|date=May 2020}} geometric statement may be described by a [[tensor field]] on a [[manifold]], and then doesn't need to make reference to coordinates at all. The same is true in [[general relativity]], of tensor fields describing a [[physical property]]. The component-free approach is also used extensively in [[abstract algebra]] and [[homological algebra]], where tensors arise naturally.

==Definition via tensor products of vector spaces==
Given a finite set {{math|{{brace|''V''<sub>1</sub>, ..., ''V''<sub>''n''</sub>}}}} of [[vector space]]s over a common [[Field (mathematics)|field]] {{mvar|F}}, one may form their [[Tensor product#Tensor product of vector spaces|tensor product]] {{math|''V''<sub>1</sub> ⊗ ... ⊗ ''V''<sub>''n''</sub>}}, an element of which is termed a '''tensor'''.

A '''tensor on the vector space''' {{mvar|V}} is then defined to be an element of (i.e., a vector in) a vector space of the form:
<math display="block">V \otimes \cdots \otimes V \otimes V^* \otimes \cdots \otimes V^*</math>
where {{mvar|V{{sup|∗}}}} is the [[dual space]] of {{mvar|V}}.

If there are {{mvar|m}} copies of {{mvar|V}} and {{mvar|n}} copies of {{mvar|V{{sup|∗}}}} in our product, the tensor is said to be of {{nowrap|'''type ({{mvar|m}}, {{mvar|n}})'''}} and [[Covariance and contravariance of vectors|contravariant]] of order {{mvar|m}} and covariant of order {{mvar|n}} and of total [[tensor order|order]] {{math|''m'' + ''n''}}. The tensors of order zero are just the scalars (elements of the field {{mvar|F}}), those of contravariant order 1 are the vectors in {{mvar|V}}, and those of covariant order 1 are the [[linear functional|one-forms]] in {{mvar|V{{sup|∗}}}} (for this reason, the elements of the last two spaces are often called the contravariant and covariant vectors). The space of all tensors of type {{math|(''m'', ''n'')}} is denoted
<math display="block"> T^m_n(V) = \underbrace{ V\otimes \dots \otimes V}_{m} \otimes \underbrace{ V^*\otimes \dots \otimes V^*}_{n}.</math>

'''Example 1.''' The space of type {{math|(1, 1)}} tensors, <math>T^1_1(V) = V \otimes V^*,</math> is [[isomorphic]] in a natural way to the space of [[linear transformations]] from {{mvar|V}} to {{mvar|V}}. 

'''Example 2.''' A [[bilinear form]] on a real vector space {{mvar|V}}, <math>V\times V \to F,</math> corresponds in a natural way to a type {{math|(0, 2)}} tensor in <math>T^0_2 (V) = V^* \otimes V^*.</math> An example of such a bilinear form may be defined,{{clarify|date=October 2023}} termed the associated ''[[metric tensor]]'', and is usually denoted {{mvar|g}}.

==Tensor rank==
{{Main|Tensor rank decomposition}}

A '''simple tensor''' (also called a tensor of rank one, elementary tensor or decomposable tensor{{sfnp|Hackbusch|2012|pp=4}}) is a tensor that can be written as a product of tensors of the form
<math display="block">T=a\otimes b\otimes\cdots\otimes d</math>
where {{math|''a'', ''b'', ..., ''d''}} are nonzero and in {{mvar|V}} or {{mvar|V{{sup|∗}}}} – that is, if the tensor is nonzero and completely [[factorization|factorizable]]. Every tensor can be expressed as a sum of simple tensors. The '''rank of a tensor''' {{mvar|T}} is the minimum number of simple tensors that sum to {{mvar|T}}.{{sfnp|Bourbaki|1989|loc=II, §7, no. 8}}

The [[zero tensor]] has rank zero. A nonzero order 0 or 1 tensor always has rank 1. The rank of a non-zero order 2 or higher tensor is less than or equal to the product of the dimensions of all but the highest-dimensioned vectors in (a sum of products of) which the tensor can be expressed, which is {{math|''d''{{i sup|''n''−1}}}} when each product is of {{mvar|n}} vectors from a finite-dimensional vector space of dimension {{mvar|d}}.

The term ''rank of a tensor'' extends the notion of the [[rank of a matrix]] in linear algebra, although the term is also often used to mean the order (or degree) of a tensor. The rank of a matrix is the minimum number of column vectors needed to span the [[Row and column spaces|range of the matrix]]. A matrix thus has rank one if it can be written as an [[outer product]] of two nonzero vectors:
<math display="block">A = v w^{\mathrm{T}}.</math>

The rank of a matrix {{mvar|A}} is the smallest number of such outer products that can be summed to produce it:
<math display="block">A = v_1w_1^\mathrm{T} + \cdots + v_k w_k^\mathrm{T}.</math>

In indices, a tensor of rank 1 is a tensor of the form
<math display="block">T_{ij\dots}^{k\ell\dots}=a_i b_j \cdots c^k d^\ell\cdots.</math>

The rank of a tensor of order 2 agrees with the rank when the tensor is regarded as a [[Matrix (mathematics)|matrix]],{{sfnp|Halmos|1974|loc=§51}} and can be determined from [[Gaussian elimination]] for instance. The rank of an order 3 or higher tensor is however often {{em|very difficult}} to determine, and low rank decompositions of tensors are sometimes of great practical interest.{{sfnp|de Groote|1987}} In fact, the problem of finding the rank of an order 3 tensor over any [[finite field]] is [[NP-completeness|NP-Complete]], and over the rationals, is [[NP-hardness|NP-Hard]].{{sfnp|Håstad|1989}} Computational tasks such as the efficient multiplication of matrices and the efficient evaluation of [[polynomial]]s can be recast as the problem of simultaneously evaluating a set of [[bilinear form]]s
<math display="block">z_k = \sum_{ij} T_{ijk}x_iy_j</math>
for given inputs {{mvar|x{{sub|i}}}} and {{mvar|y{{sub|j}}}}. If a low-rank decomposition of the tensor {{mvar|T}} is known, then an efficient [[evaluation strategy]] is known.{{sfnp|Knuth|1998|pp=506–508}}

==Universal property==
The space <math>T^m_n(V)</math> can be characterized by a [[universal property]] in terms of [[multilinear map]]pings. Amongst the advantages of this approach are that it gives a way to show that many linear mappings are "natural" or "geometric" (in other words are independent of any choice of basis). Explicit computational information can then be written down using bases, and this order of priorities can be more convenient than proving a formula gives rise to a natural mapping. Another aspect is that tensor products are not used only for [[free module]]s, and the "universal" approach carries over more easily to more general situations.

A scalar-valued function on a [[Cartesian product]] (or [[Direct sum of modules|direct sum]]) of vector spaces
<math display="block">f : V_1\times\cdots\times V_N \to F</math>
is multilinear if it is linear in each argument. The space of all multilinear mappings from {{math|''V''<sub>1</sub> × ... × ''V<sub>N</sub>''}} to {{mvar|W}} is denoted {{math|''L<sup>N</sup>''(''V''<sub>1</sub>, ..., ''V<sub>N</sub>'';&nbsp;''W'')}}. When {{math|1= ''N'' = 1}}, a multilinear mapping is just an ordinary linear mapping, and the space of all linear mappings from {{mvar|V}} to {{mvar|W}} is denoted {{math|''L''(''V''; ''W'')}}.

The [[tensor product#Universal property|universal characterization of the tensor product]] implies that, for each multilinear function
<math display="block">f\in L^{m+n}(\underbrace{V^*,\ldots,V^*}_m,\underbrace{V,\ldots,V}_n;W)</math>
(where {{mvar|W}} can represent the field of scalars, a vector space, or a tensor space) there exists a unique linear function
<math display="block">T_f \in L(\underbrace{V^*\otimes\cdots\otimes V^*}_m \otimes \underbrace{V\otimes\cdots\otimes V}_n; W)</math>
such that
<math display="block">f(\alpha_1,\ldots,\alpha_m, v_1,\ldots,v_n) = T_f(\alpha_1\otimes\cdots\otimes\alpha_m \otimes v_1\otimes\cdots\otimes v_n)</math>
for all {{mvar|v{{sub|i}}}} in {{mvar|V}} and {{mvar|α{{sub|i}}}} in {{mvar|V{{sup|∗}}}}.

Using the universal property, it follows, when {{mvar|V}} is [[Dimension_(vector_space)|finite dimensional]], that the space of {{math|(''m'', ''n'')}}-tensors admits a [[natural isomorphism]]
<math display="block">T^m_n(V) \cong L(\underbrace{V^* \otimes \cdots \otimes V^*}_m \otimes \underbrace{V \otimes \cdots \otimes V}_n; F) \cong L^{m+n}(\underbrace{V^*, \ldots,V^*}_m,\underbrace{V,\ldots,V}_n; F).</math>

Each {{mvar|V}} in the definition of the tensor corresponds to a {{mvar|V{{sup|∗}}}} inside the argument of the linear maps, and vice versa. (Note that in the former case, there are {{mvar|m}} copies of {{mvar|V}} and {{mvar|n}} copies of {{mvar|V{{sup|∗}}}}, and in the latter case vice versa). In particular, one has
<math display="block">\begin{align}
T^1_0(V) &\cong L(V^*;F) \cong V,\\
T^0_1(V) &\cong L(V;F) = V^*,\\
T^1_1(V) &\cong L(V;V).
\end{align}</math>

==Tensor fields==
{{Main|tensor field}}

[[Differential geometry]], [[physics]] and [[engineering]] must often deal with [[tensor field]]s on [[smooth manifold]]s. The term ''tensor'' is sometimes used as a shorthand for ''tensor field''. A tensor field expresses the concept of a tensor that varies from point to point on the manifold.

==References==
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[[Category:Tensors]]