Intrinsic definition of canonical vector field on tangent bundle I have been trying to solve the following question's last part (iv):



on giving an intrinsic definition to the vector field whose local definition is that in part (iii). I believe this is called the canonical vector field or the Liouville vector. This question is from page 58 of Frankel's book called 'Geometry of Physics'. Some discussion on the question can be found here: https://www.physicsforums.com/threads/transformation-of-coordinate-basis.998435/.
The meaning of intrinsic as far as I understand it is that the vector field should be defined in a coordinate free way so that it does not use any local coordinates. Given this is at the start of the book before the Lie derivative and all that are introduced, I believe the only objects we have are the fiber projection and the section, as well as the basis vectors. The way the question is phrased makes it seem that there is such an intrinsic definition but at first seems a bit strange given the vector field has 0 components in the directions described by the basis vectors coming from the coordinates of $M$.
 A: You're right to be concerned because at first sight, it is indeed not obvious that one can extend the definition of $\xi$ to yield a smooth vector field on $TM$ which has a similar expression in every adapted coordinate chart. However it is precisely because we're only taking a specific combination of the basis vector fields that this induces a global definition for a vector field on $TM$. Proving that this is true is the content of part (iii).
Usually, coming up with an "intrinsic definition" (i.e one which makes no reference to coordinate charts) isn't easy, but here it isn't too bad. One needs to stay very clear on all the spaces and all the definitions. First I would suggest you take a look at (the first part of) my MSE answer Second derivatives, Hamilton and tangent bundle of tangent bundle TTM in order to understand the second tangent bundle and the associated charts. The definition of tangent space which I find most intuitive is as an equivalence class of smooth curves. Recall that a (smooth) vector field on $TM$ is a (smooth) mapping $\xi:TM\to TTM$. So, to each equivalence class of smooth curves $[s\mapsto \gamma(s)]$ in $M$, we must assign another equivalence class of smooth curves in $TM$. So, consider the mapping $\xi:TM\to TTM$, defined as
\begin{align}
\xi([\gamma]):= [t\mapsto [s\mapsto \gamma(s+ts)]]
\end{align}
In words, what's going on is that we're taking a smooth curve $\gamma:I\subset\Bbb{R}\to M$, then for each (small enough) $t$, we consider the curve $s\mapsto \gamma(s+ts)$ in $M$. The equivalence class of this curve is a vector tangent to $M$, i.e $[s\mapsto \gamma(s+ts)]\in T_{\gamma(0+t\cdot 0)}M=T_{\gamma(0)}M$. So, if we call $\Gamma(t)=[s\mapsto \gamma(s+ts)]$, then we have just defined a mapping $\Gamma:\Bbb{R}\to TM$, such that $\Gamma(0)=[\gamma]$. Therefore, we have a curve $\Gamma$ in $TM$ whose base point is $[\gamma]$. So, the equivalence class $[\Gamma]$ is thus an element of $T_{\Gamma(0)}(TM)=T_{[\gamma]}(TM)$. One can unwind the definitions to check that the mapping $\xi$ doesn't depend on the particular representative $\gamma$ of the equivalence class. Hence, $\xi:TM\to T(TM)$ is a well-defined mapping.
It is a good exercise for you to make sure you understand all the definitions, and verify for yourself that given a chart $(U,\alpha)$ on $M$, the local representation of $\xi$ is $(TT\alpha)\circ \xi\circ (T\alpha)^{-1}: \alpha[U]\times\Bbb{R}^m\to (\alpha[U]\times\Bbb{R}^m)\times (\Bbb{R}^m\times\Bbb{R}^m)$, given as
\begin{align}
(x,v)\mapsto ((x,v),(0,v)) = (x,v,0,v).
\end{align}
(here the first two entries being $(x,v)$ tell us that $\xi$ is a vector field; i.e that it has the correct base point of $(x,v)$. The last two entries being $(0,v)$ tell us that $\xi$ is pointing "along the fiber direction" and not "in the base direction")
In other words, $\xi = \dot{q}^i\frac{\partial}{\partial \dot{q}^i}$.
The way I came up with this intrinsic definition is I took the coordinate expression $\dot{q}^i\frac{\partial}{\partial \dot{q}^i}$, from which I knew that the lcoal expression for $\xi$ had to be $(x,v)\mapsto ((x,v), (0,v))$. From here, I just worked backward (from the definition of $TT\alpha$ and $(T\alpha)^{-1}$, as presented in my link above) to figure
the abstract definition of $\xi$.

As an aside, one has a very similar (perhaps more well-known?) situation in the cotangent bundle, where one has the tautological/Liouville one-form given in coordinates as $\theta=\pm p_i\,dq^i$ (sign conventions are notorious). Proving this definitions extends to a global one on $T^*M$ is also a standard exercise which I explain in detail here (so if you have trouble with the coordinate proof of part (iii), I would suggest you take a look at my answer here, and try to mimic the proof accordingly). Also, the tautological form has a not-too-difficult intrinsic definition which can be easily reverse-engineered/googled.
