# What is the relation between $C^\infty$-linear and tensorial?

I've been self studying Riemannian Geometry through Spivak's and Lee's books, and fairly often I've seen an argument that goes somewhat like this:

We have an operator that acts on vectors in the tangent space and, in order to show that it "lives at points", we generalise it to operate on global sections in vector bundles. Then, by showing this operator is linear over $C^\infty$ functions, some argument involving bump functions implies the operator defines a tensor field.

Usually this is followed by some remark about modules, giving a strong impression that something bigger is going on.

On general relativity texts, there's an equivalent practice that involves showing the operator "transforms correctly".

This is all quite vague, since I cannot quite point my finger at what's the general phenomenon. So my question is:

What is the relation between tensor fields and $C^\infty$ linearity and what does this has to do with modules?

• Have you looked at the Tensor Characterization Lemma (Lemma 2.4 in my Riemannian Manifolds book)? If you have trouble seeing how to prove it, see Lemma 12.24 in my Introduction to Smooth Manifolds. – Jack Lee Jun 13 '15 at 14:06
• Lemma 2.4 does indeed answer the first part of my question. I believe the proof goes like a generalisation of Lemma 4.1, by picking bump functions on neighbourhoods and using them to extend the vector fields globally. I find it counterintuitive that this actually works, but I guess it's one of the niceties of differential geometry. – Felipe Jacob Jun 13 '15 at 14:40

First note that every vector bundle $E$ gives you $C^{\infty}(M)$-Module structure on global sections $\Gamma(E).$

Additionaly for vector bundles $E,F$ on maniflod $M$ there are $C^{\infty}(M)$-Module isomorphism $$\Gamma(E)\otimes_{C^{\infty}(M)}\Gamma(F)\simeq\Gamma(E\otimes_{\mathbb{R}} F)\hspace{10pt}\text{and}\hspace{10pt}\Gamma(E^{*})\simeq\Gamma(E)^{\vee}$$ where $E\otimes_{\mathbb{R}} F$ is bundle with tensor product taken fiberwise and $E^*$ is bundle with dual vector space taken fiberwise.

Now any tensor field is a section of bundle $TM\otimes\dots\otimes TM\otimes T^*M\dots\otimes T^*M.$ From previous you get that any vetor field can be treated as element of $$\Gamma(TM)\otimes\dots\otimes \Gamma(TM)\otimes \Gamma(TM)^{\vee}\dots\otimes \Gamma(TM)^{\vee}.$$ There are also isomorphism for symmetric and screw symmetric tensors. Namely $$\Gamma(E\odot E)\simeq\Gamma(E)\odot_{C^{\infty}(M)}\Gamma(E)\hspace{5pt}\text{and}\hspace{5pt}\Gamma(E\wedge E)\simeq\Gamma(E)\wedge_{C^{\infty}(M)}\Gamma(E).$$ Last, but not least due to the famous Swan's theorem you have that for any vector bundle $E$ $$\Gamma(E)^{\vee\vee}\simeq\Gamma(E).$$ It allows you to treat ${C^{\infty}(M)}$ linear functionals on $\Gamma(TM)^{\vee}$ as elements of $\Gamma(TM).$

For reference see:

Jet Nestruev, Smooth manifolds and observables, 2003

Lawrence Conlon, Differentiable Manifolds SE, 2001

• This seems exactly like what I'm looking for. However, could you clarify what the notation $^\vee$ means? – Felipe Jacob Jun 13 '15 at 14:48
• Given $R$-Module $M$ set $M^{\vee}$ is $R$-Module of $R$ linear functionals on $M,$ i.e $M^{\vee}=Hom_{R}(M,R).$ It is analogue of dual space in modules over ring. – Fallen Apart Jun 13 '15 at 14:53
• This summarises everything nicely. And thanks for the reference to Swan's Theorem, it looks like the right place to go to see how all this generalises. – Felipe Jacob Jun 13 '15 at 22:14
• This is actually a terrible answer, a good example of how not to explain mathematics. It introduces a lot of formalism and ultimately does not answer the question at all. The devil is hidden in the claim $\Gamma(E^*) = \Gamma(E)^\vee$, which sums up the whole question, and never explained in this answer. Why is $\Gamma(E^*) = \Gamma(E)^\vee$ true? The rest of the answer is then inessential formal nonsense. The real answer to the question is Jack Lee's comment. – Seub Oct 23 '19 at 13:17
• @Seub The question was about the relation not about the origin of such relation. If one wishes to dwell further into details I provided the references. – Fallen Apart Oct 29 '19 at 14:38