Editing Tensor field (section)

== Tensor fields as multilinear forms ==
There is another more abstract (but often useful) way of characterizing tensor fields on a manifold ''M'', which makes tensor fields into honest tensors (i.e. ''single'' multilinear mappings), though of a different type (although this is ''not'' usually why one often says "tensor" when one really means "tensor field"). First, we may consider the set of all smooth (''C''<sup>∞</sup>) vector fields on ''M'', <math>\mathfrak{X}(M):=\mathcal T^1_0(M)</math> (see the section on notation above) as a single space – a [[module (mathematics)|module]] over the [[ring (mathematics)|ring]] of smooth functions, ''C''<sup>∞</sup>(''M''), by pointwise scalar multiplication. The notions of multilinearity and tensor products extend easily to the case of modules over any [[commutative ring]].

As a motivating example, consider the space <math>\Omega^1(M)=\mathcal{T}^0_1(M)</math> of smooth covector fields ([[differential form|1-forms]]), also a module over the smooth functions. These act on smooth vector fields to yield smooth functions by pointwise evaluation, namely, given a covector field ''ω'' and a vector field ''X'', we define
: <math>\tilde{\omega}(X)(p):=\omega(p)(X(p)).</math>

Because of the pointwise nature of everything involved, the action of <math>\tilde \omega </math> on ''X'' is a ''C''<sup>∞</sup>(''M'')-linear map, that is,
: <math>\tilde \omega(fX)(p)=\omega(p)((fX)(p))=\omega(p)(f(p)X(p))=f(p)\omega(p)(X(p))=(f\omega)(p)(X(p))=(f\tilde \omega)(X)(p)</math>
for any ''p'' in ''M'' and smooth function ''f''.  Thus we can regard covector fields not just as sections of the cotangent bundle, but also linear mappings of vector fields into functions. By the double-dual construction, vector fields can similarly be expressed as mappings of covector fields into functions (namely, we could start "natively" with covector fields and work up from there).

In a complete parallel to the construction of ordinary single tensors (not tensor fields!) on ''M''  as multilinear maps on vectors and covectors, we can regard general (''k'',''l'') tensor fields on ''M'' as ''C''<sup>∞</sup>(''M'')-multilinear maps defined on ''k'' copies of <math>\mathfrak{X}(M)</math> and ''l'' copies of <math>\Omega^1(M)</math> into ''C''<sup>∞</sup>(''M'').

Now, given any arbitrary mapping ''T'' from a product of ''k'' copies of <math>\mathfrak{X}(M)</math> and  ''l'' copies of <math>\Omega^1(M)</math> into ''C''<sup>∞</sup>(''M''), it turns out that it arises from a tensor field on ''M'' if and only if it is multilinear over ''C''<sup>∞</sup>(''M''). Namely ''C''<sup>∞</sup>(''M'')-module of tensor fields of type <math>(k,l)</math> over ''M'' is canonically isomorphic to ''C''<sup>∞</sup>(''M'')-module of ''C''<sup>∞</sup>(''M'')-[[multilinear form]]s 
: <math>\underbrace{\Omega^1(M) \times \ldots \times \Omega^1(M)}_{l\ \mathrm{times}} \times \underbrace{ \mathfrak X(M)\times \ldots \times \mathfrak X(M)}_{k\ \mathrm{times}} \to C^ \infty (M).</math><ref>{{Cite web|title=Notes on Smooth Manifolds|url = https://www.ime.usp.br/~gorodski/teaching/mat5799-2015/gorodski-smooth-manifolds-2013.pdf|quote = |access-date = 2024-06-24|author = Claudio Gorodski}} </ref> 

This kind of multilinearity implicitly expresses the fact that we're really dealing with a pointwise-defined object, i.e. a tensor field, as opposed to a function which, even when evaluated at a single point, depends on all the values of vector fields and 1-forms simultaneously.

A frequent example application of this general rule is showing that the [[Levi-Civita connection]], which is a mapping of smooth vector fields <math>(X,Y) \mapsto \nabla_{X} Y</math> taking a pair of vector fields to a vector field, does not define a tensor field on ''M''. This is because it is only <math>\mathbb R</math>-linear in ''Y'' (in place of full ''C''<sup>∞</sup>(''M'')-linearity, it satisfies the ''Leibniz rule,'' <math>\nabla_{X}(fY) = (Xf) Y +f \nabla_X Y</math>)). Nevertheless, it must be stressed that even though it is not a tensor field, it still qualifies as a geometric object with a component-free interpretation.