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There are two definitions of a vector field on a smooth manifold $M$.

  1. A smooth map $V:M \rightarrow TM, \forall p \in M:V(p) \in T_p M$.

  2. A linear map $V:C^{\infty}(M) \rightarrow C^{\infty}(M), \forall f,g:V(fg)=fV(g)+gV(f)$

I can't undestand why they are equivalent. We must somehow build $2$ maps and show that their composition is $id$, but i don't have any ideas how. Please help.

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This becomes easier if you think of a vector as a directional derivative on the vector's direction. With this in mind, you can define a map that sends a vector in $T_p M$ to a differential operator (namely, the directional derivative), and vice-versa. – Marra May 11 '14 at 18:22
Marra, how does it work? We take a function $f$ and define $V(f)(p)$ as a derivative of $f$ in the direction of $V(p)$ in the point $p$ (last $V$ is from $(1)$? – soporhs May 11 '14 at 18:38
I'll try to write it down as an answer, gimme one minute :) – Marra May 11 '14 at 18:46
What is your definition of $TM$? – T. Eskin May 11 '14 at 19:05
up vote 4 down vote accepted

This depends heavily on your definition of the tangent space $T_{p}M$, and thus the tangent bundle $TM$. There are several equivalent ways of defining it. Which book are you following?

If your definition of the tangent space $T_{p}M$ is a vector space of linear maps $V : C^{\infty}(p)\to\mathbb{R}$ that satisfy the Leibniz rule, i.e. $$V(fg)=f(p)V(g)+g(p)V(f),$$ where $C^{\infty}(p)$ is defined as \begin{align*} C^{\infty}(p)=\{f:U\to\mathbb{R}\,\,|\,\,f\,\,\mathrm{is}\,\,\mathrm{smooth}\,\,\mathrm{at}\,\,p\in U\,\,\mathrm{and}\,\,U\subseteq M\,\,\mathrm{is}\,\,\mathrm{open}\}, \end{align*} as is usually done, then this exercise is pretty straight forward.

Get a map $\Psi:(1)\to (2)$ as follows. For each $V$ satisfying $(1)$ assign a linear map $\Psi(V)$ satisfying $(2)$ by taking $\Psi(V)(f)(p)=V(p)(f)$ for all $f\in C^{\infty}(p)$ and $p\in M$. Show that this is one-to-one and onto, or alternatively define an inverse $\Phi:(2)\to (1)$ by assigning for each $V$ satisfying $(2)$ a smooth map $\Phi(V)$ satisfying $(1)$ by taking $\Phi(V)(p)(f)=V(f)(p)$ for all $p\in M$ and $f\in C^{\infty}(p)$. So you get that the two definitions are equivalent.

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Yes, this is our definition of $T_p M$. Thanks. – soporhs May 11 '14 at 19:35
@soporhs. You're welcome. I'll gladly answer any questions you might have, but I advice that you try to work the details on your own at first. – T. Eskin May 11 '14 at 19:44

I'll show that for every $X:M\rightarrow TM$ vector field, we can associate a differential operator $X:C^{\infty}(M)\rightarrow C^{\infty}(M)$.

By now you probably know that we can work in coordinates $(U,\mathcal{X})$, $U\underset{open}\subset M$, $\mathcal{X}:U\rightarrow \mathbb{R}^n$, and so we can work with the vector field $X$ written locally as $X=\sum_{i=1}^nX_i\partial x_i$. That is, we are writting $\dfrac{\partial}{\partial x_i}(p)=\partial x_i(p)$ for convenience and erasing the $p$ to avoid heavy notation. Also remember that $\{\partial x_1(p),...,\partial x_n (p)\}$ is a basis for $T_p M$.

Now, given this vector field, define a mapping $G:\mathfrak{X}^\infty(M)\rightarrow \mathcal{D}^\infty(M)$ from the space of vector fields $\mathfrak{X}^\infty(M)$ over $M$ to the set of 'directional differential operators' $\mathcal{D}^\infty(M)$ (I just made this up) of mappings $\mathcal{C}^\infty (M)\rightarrow \mathcal{C}^\infty (M)$, such that $$ G(X)(f)=\sum_{i-1}^n X_i\dfrac{\partial f}{\partial x_i} $$ which is in $C^\infty (M)$ (remember that every $X_i$ is in $C^\infty (M)$). Basically, $G$ is just the 'change of point of view' that allows you to see a vector field $X$ as a directional derivative $X$.

Now, given a directional differential operator, you can define a vector field that is associated to it. I hope you have already realised now that the notation for the basis of $T_p M$ is not accidental and is motivated by this association between vector fields and directional differentiation.

If you need a stronger reference, try looking for Do Carmo's "Riemannian Geometry". I'm also taking this course this semester and this book is being of great help. If you have any question, don't hesitate to ask.

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What definition of $TM$ are you working with? It seems that this answer only opens the usual definition of the tangent space but I can't see how you identify those two definitions of a vector field. – T. Eskin May 11 '14 at 19:32
Thanks. After our course of Riemannian geometry, it seems for me like a large composition of strange definitions and for some strange reasons it all works exactly as we want it to work. – soporhs May 11 '14 at 19:36
@ThomasE. I'm using it as stated in Do Carmo's. – Marra May 11 '14 at 19:51

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