User:Tim Ocean/Tensor field

Notation
The notation for tensor fields can sometimes be confusingly similar to the notation for tensor spaces. Thus, the tangent bundle TM = T(M) might sometimes be written as
 * $$T_0^1(M)=T(M) =TM $$

to emphasize that the tangent bundle is the range space of the (1,0) tensor fields (i.e., vector fields) on the manifold M. This should not be confused with the very similar looking notation


 * $$T_0^1(V)$$;

in the latter case, we just have one tensor space, whereas in the former, we have a tensor space defined for each point in the manifold M.

Curly (script) letters are sometimes used to denote the set of infinitely-differentiable tensor fields on M. Thus,
 * $$\mathcal{T}^m_n(M)$$

are the sections of the (m,n) tensor bundle on M that are infinitely-differentiable. A tensor field is an element of this set.

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∞) vector fields on M, $$\mathfrak{X}(M):=\mathcal T^1_0(M)$$ (see the section on notation above) as a single space — a module over the ring of smooth functions, C∞(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 $$\Omega^1(M)=\mathcal{T}^0_1(M)$$ of smooth covector fields (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


 * $$\tilde{\omega}(X)(p):=\omega(p)(X(p)).$$

Because of the pointwise nature of everything involved, the action of $$\tilde \omega $$ on X is a C∞(M)-linear map, that is,


 * $$\tilde \omega(fX)(p)=\omega(p)((fX)(p))=\omega(p)(f(p)X(p))=f(p)\omega(p)(X(p))=(f\omega)(p)(X(p))=(f\tilde \omega)(X)(p)$$

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∞(M)-multilinear maps defined on k copies of $$\mathfrak{X}(M)$$ and l copies of $$\Omega^1(M)$$ into C∞(M).

Now, given any arbitrary mapping T from a product of k copies of $$\mathfrak{X}(M)$$ and l copies of $$\Omega^1(M)$$ into C∞(M), it turns out that it arises from a tensor field on M if and only if it is multilinear over C∞(M). Namely $$C^\infty (M)$$-module of tensor fields of type $$(k,l)$$ over M is canonically isomorphic to $$C^\infty (M)$$-module of $$C^\infty (M)$$-multilinear forms
 * $$\underbrace{\Omega^1(M) \times \ldots \times \Omega^1(M)}_{l\ \mathrm{times}} \times \underbrace{ \mathfrak X(M)\times \ldots \times \mathfrak X(M)}_{k\ \mathrm{times}} \to C^ \infty (M).$$

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 $$(X,Y) \mapsto \nabla_{X} Y$$ taking a pair of vector fields to a vector field, does not define a tensor field on M. This is because it is only $$\mathbb R$$-linear in Y (in place of full C∞(M)-linearity, it satisfies the Leibniz rule, $$\nabla_{X}(fY) = (Xf) Y +f \nabla_X Y$$)). 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.