I am doing the following exercise from Frankel's The geometry of physics and I am having some doubts:
2.3(2) Consider the tangent bundle to a manifold $M$.
(i) Show that under a change of coordinates in $M$, $\partial/\partial q$ depends on both $\partial/\partial q'$ and $\partial/{\partial \dot q'}$.
(ii) Is the locally defined vector field $\sum_j\dot q^j \partial/\partial q^j$ well defined on all of $TM$?
(iii) Is $\sum_j\dot q^j \partial/\partial \dot q^j$ well defined?
(iv) If any of the above in (ii),(iii) is well defined, can you produce an intrinsic definition?
For (i) first of all, I will assume, as in here, that by $\frac{\partial}{\partial \dot q'}$ we mean a tangent vector of a point of $TM$, the tangent bundle of $M$.
If we indicate the transition functions for a coordinate patch of $M$ as $q^i=q^i(q'^1,...,q'^n)$ where $n$ is the dimension of $M$, the coordinates for the corresponding patch of the tangent space $TM$ are $$q^1,...,q^n, \dot q^1,...,\dot q^n $$ where $q^i=q^i(q'^1,...,q'^n)$ and $\dot q^i=\frac{\partial q^i}{\partial q'^j}\dot q'^j$.
To see how the vector $\frac{\partial}{\partial q}$ is represented in the "$q'$" system, I worked as follows: I took a function on TM and differentiated it by $\frac{\partial}{\partial q}$. Then I applied the chain rule (Einstein's notation is used below):
$$\frac{\partial f(q'^1,...,q'^n,\dot q'^1,...,\dot q'^n) }{\partial q^i}=\frac{\partial f}{\partial q'^j} \frac{\partial q'^j}{\partial q^i}+\frac{\partial f}{\partial \dot q'^j} \frac{\partial \dot q'^j}{\partial q^i}$$ so $$\frac{\partial }{\partial q^i}=\frac{\partial q'^j}{\partial q^i}\frac{\partial }{\partial q'^j} +\frac{\partial \dot q'^j}{\partial q^i}\frac{\partial }{\partial \dot q'^j}$$
then I further expanded $\frac{\partial \dot q'^j}{\partial q^i}$ as $\frac{\partial}{\partial q^i}(\frac{\partial q'^j}{\partial q^k}\dot q^k)=\frac{\partial}{\partial q^i}(\frac{\partial q'^j}{\partial q^k})\dot q^k+\frac{\partial q'^j}{\partial q^k}\frac{\partial}{\partial q^i}(\dot q^k)$ getting, as $q^k$ and $\dot q^k$ are two independent variables, $$\frac{\partial }{\partial q^i}=\frac{\partial q'^j}{\partial q^i}\frac{\partial }{\partial q'^j} +\{\frac{\partial}{\partial q^i}(\frac{\partial q'^j}{\partial q^k})\dot q^k\}\frac{\partial }{\partial \dot q'^j}$$ Equivalently, I got the same result considering a curve $c(t)$ with values in $TM$ and calculating the "differential" of the transition function (of the full coordinates of $TM$), then applying such differential to the vector $\frac{\partial}{\partial q^k}$.
I wonder if this result is correct and if it was required by the exercise to further expand $\frac{\partial \dot q'^j}{\partial q^i}$ as I didn't know anything a priori about this expression.
As for (ii) and (iii) I believe that "well defined" means "invariant in form to coordinate transformation"?
Let's start from (iii). I wanted to evaluate how $\sum_j\dot q^j \partial/\partial \dot q^j$ is represented in $q'$ system so I applied the same logic as in (i) getting
$$\frac{\partial f(q'^1,...,q'^n,\dot q'^1,...,\dot q'^n) }{\partial \dot q^i}=\frac{\partial f}{\partial q'^j} \frac{\partial q'^j}{\partial \dot q^i}+\frac{\partial f}{\partial \dot q'^j} \frac{\partial \dot q'^j}{\partial \dot q^i}$$
I think that the first term of the sum is zero, because if I take a curve in $TM$ which stays on the same fibre of a point $p \in M$, the $M$ coordinates will stay constant for any coordinate patch. A curve with velocity $\partial / \partial \dot q$ is such a curve, so we have that $\frac{\partial q'^j}{\partial \dot q^i}=0$. Then, as we know that $\dot q^j=\sum_i \frac{\partial q^j}{\partial q'^i}\dot q'^i$, as this is a linear relationship, all the $\dot q^i$ are independent and, again, the $\frac{\partial q^j}{\partial \dot q'^i}=0$, we can write that $\frac{\partial \dot q'^j}{\partial \dot q^i}=\frac{\partial q'^j}{\partial q^i}$. Summing all together we get
$$ \sum_j\dot q^j \partial/\partial \dot q^j = \sum_j\{(\sum_k \frac{\partial q^j}{\partial q'^k}\dot q'^k)(\sum_i \frac{\partial q'^i}{\partial q^j}\frac{\partial }{\partial \dot q'^i})\}= \sum_k \sum_i \{ \sum_j( \frac{\partial q^j}{\partial q'^k}\dot q'^k)(\frac{\partial q'^i}{\partial q^j}\frac{\partial }{\partial \dot q'^i})\}= \sum_k\dot q^k \partial/\partial \dot q^k$$ as in the last term I have the product of the Jacobian by its inverse.
On the other hand, for (ii) I write $\sum_j\dot q^j\frac{\partial}{\partial q^j}$ in the $q'$ coordinates using the result in (i). Is it enough to note that the terms in $\partial/\partial \dot q'^j$ do not vanish, to prove that $\sum_j\dot q^j\frac{\partial}{\partial q^j}$ is not invariant in form? As I don't see this time how this could simplify.
Last part of the exercise is (iv), which I assume should be to find a coordinate independent version for $\sum_j\dot q^j \partial/\partial \dot q^j$. I feel this should be a similar procedure to the definition of the Poincare' 1-form, but I cannot formalize anything useful. Any hint?
thanks