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- TL;DR Summary
- Confusion about coordinate transformation on a tangent bundle

So while reading T. Frankel's "The Geometry of Physics", I was going through the part on cotangent bundles which ended with the definition of Poincare 1-form. The author argued that cotangent bundles are better suited than tangent bundles for some problems in physics and that there is no natural isomorphism between them, that is, one has to include a metric as an additional structure to relate vectors to covectors. He said that some objects naturally live in the cotangent bundle(for example canonical momenta), etc.

So at the end of that discussion there is an exercise that is(I presume) supposed to show that an equivalent of Poincare 1-form doesn't exist on a tangent bundle. The precise statement of the exercise is given in the screenshot below.

The exercise doesn't look hard, but still, when I look at the (i) I don't see why ##\bf{\partial/\partial q}## would depend on both ##\bf{\partial/\partial q}## and ##\bf{\partial/\partial \dot{q}}##. ##q## is a letter reserved for the coordinate basis. ##\dot{q}## is a generalized velocity(that is, the on tangent bundle the coordinates are of form ##(q,\dot{q})##).

Doesn't the coordinate change ##q'^i = q'^i(q^1,\dots , q^n)## imply the transformation:

$$\frac{\partial}{\partial q^i} = \frac{\partial q'^j}{\partial q^i}\frac{\partial}{\partial q'^j}$$

P.S. This is not a homework, just something I'm doing in my free time.

So at the end of that discussion there is an exercise that is(I presume) supposed to show that an equivalent of Poincare 1-form doesn't exist on a tangent bundle. The precise statement of the exercise is given in the screenshot below.

The exercise doesn't look hard, but still, when I look at the (i) I don't see why ##\bf{\partial/\partial q}## would depend on both ##\bf{\partial/\partial q}## and ##\bf{\partial/\partial \dot{q}}##. ##q## is a letter reserved for the coordinate basis. ##\dot{q}## is a generalized velocity(that is, the on tangent bundle the coordinates are of form ##(q,\dot{q})##).

Doesn't the coordinate change ##q'^i = q'^i(q^1,\dots , q^n)## imply the transformation:

$$\frac{\partial}{\partial q^i} = \frac{\partial q'^j}{\partial q^i}\frac{\partial}{\partial q'^j}$$

P.S. This is not a homework, just something I'm doing in my free time.