We can define vector fields on manifolds in two ways. The way I first saw was that a vector field was a linear map $C^\infty(M) \to C^\infty(M)$ satisfying the Leibniz rule (aka product rule). We can also define a vector field to be a smooth section of $TM \to M$.

I get that given a section $s$ of $TM \to M$, we can define $\hat{s}(f)(p) = s(p)(f)$, but I don't understand why $\hat{}:Sec(TM \to M) \to Vect(M)$ is an isomorphism. It's clearly linear, but I don't see why it's injective or surjective.

$s \in Sec(TM \to M)$ has a left inverse (the inverse being the projection). Maybe that is used to show $\hat{} :Sec(TM \to M) \to Vect(M)$ is injective?

  • $\begingroup$ So you're thinking about $Vect(M)$ as the vector space of vector fields and you already know the map is linear? Why is it not trivial to see that this is injective? If some section maps to $0$ (meaning the zero vector field!), then by your formula it had to be the zero section. Maybe I'm misunderstanding something ... $\endgroup$ – Matt Apr 6 '13 at 19:38
  • $\begingroup$ My question was indeed silly. Thanks for the reply. I must admit that I was somewhat confused with the concepts. $\endgroup$ – nigel Apr 7 '13 at 3:10
  • $\begingroup$ What is your definition of tangent bundle? By linear functional satisfying Leibniz rule or by equivalent classes of curves passing a given point? $\endgroup$ – Yuchen Liu Apr 7 '13 at 3:33
  • $\begingroup$ The tangent bundle (for me) is defined as $\bigcup\limits_{x \in M} T_x M$ $\endgroup$ – nigel Apr 7 '13 at 15:37
  • $\begingroup$ and what is $T_xM$ for you? I think this is what jerrysciencemath wanted to know. $\endgroup$ – Sam Lisi Apr 8 '13 at 9:52

tl;dr: The injectivity of the section boils down to TM being the union of the tangent space at every point. If you now map from or to $TM$, this domain/range contains all of $M$!

You are talking about a vector field being a section $TM \to M$,..

If I'm not mistaken, this way round goes the surjective projection $\pi: TM \to M$, because taking a an element from $TM$ (vector with basepoint) it maps them to a point in the manifold. The projection is surjective, because you took $T_pM$ for $\forall p \in M$!

Now the section $s: M \to TM$ can at most be injective. This is obvious, since TM contains all the vectors at each point, but with your field you just want to show the "user" one distinct vector at each point. This "at each point" again tells you, that the section is injective (takes every point to some element in $TM$).

Furthermore, by tedious execution of the vector space rules you'll find, that nearly all of them are fulfilled for the tangent bundle. However, there is no inverse element for multiplication by a smooth function because there are functions with local zeros. The conclusion is, that every element in TM, which I call $\Gamma(TM)$ is a $\mathcal{C}^{\infty}$-module (not quite a vector space). Since you map into this (not really) vector space, the section is a structure preserving map.

Further details, like the basis for $TM$ and continuity/smoothness of the projection are discussed in this excellent (long) video.

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