I am a physicist who is trying to understand more formal differential geometry in the context of classical mechanics. I came across three ways of computing the Lie derivative of differential one-forms. The first two are consistent, but the third is not. I would like you to help me figure out if there is a legitimate interpretation for that case, or if it's just a mistake.
=== INTRODUCTION AND DEFINITIONS ===
For simplicity, let $\mathbb{Q}$ be a 1-dimensional manifold. The manifold $\mathbb{Q}$ has some local coordinate $q$.
We can then define the 2-dimensional tangent bundle $T\mathbb{Q}$. The tangent bundle has local coordinates $(q,\dot{q})$, with $\dot{q}$ a tangent vector at $q$.
We can also define the 2-dimensional cotangent bundle $T^* \mathbb{Q}$, i.e., the phase space. The phase space has local coordinates $(q,p)$, with $p$ a cotangent vector (covector) at $q$.
We can also define the 4-dimensional tangent bundle $T(T^*\mathbb{Q})$ associated with the phase space. The local coordinates of $T(T^*\mathbb{Q})$ are $(q,p,\dot{q},\dot{p})$, where $(\dot{q},\dot{p})$ is a vector tangent to $T^*\mathbb{Q}$ at $(q,p)$.
=== FLOW ===
Also for simplicity, consider the flow generated by the Hamiltonian function $H(q,p)$ of the simple harmonic oscillator on $T^* \mathbb{Q}$:
$$H(q,p) = \frac{q^2}{2} + \frac{p^2}{2}.$$
The integral curves in local coordinates are the solutions of the Hamilton differential equations
$$\dot{q} \frac{\partial}{\partial q} + \dot{p}\frac{\partial}{\partial p} = p \frac{\partial}{\partial q} - q \frac{\partial}{\partial p}.$$
Note that the left-hand side is necessarily a vector field defined on $T(T^*\mathbb{Q})$. By contrast, the right-hand side can be seen as a vector field restricted to $T^* \mathbb{Q}$. Let us define these vector fields separately:
$$\Delta_H = p \frac{\partial}{\partial q} - q \frac{\partial}{\partial p},\;\; \Delta_T = \dot{q} \frac{\partial}{\partial q} + \dot{p}\frac{\partial}{\partial p},$$
so that the Hamilton equations are given by the constraint $\Delta_H = \Delta_T$. Note that we denote the contraction with respect to a field $\Delta$ as $i_\Delta$.
Now I am interested in calculating the Lie derivative of the fundamental one-form
$$\theta = p\,dq.$$
We can conceptualize this either as a one-form defined on $T^* \mathbb{Q}$,
$$\theta = p\,dq + 0\,dp,$$
or as a one-form defined on $T(T^* \mathbb{Q})$,
$$\theta = p\,dq + 0\,dp + 0\,d\dot{q} + 0\,d\dot{p}.$$
=== #1: LIE DERIVATIVE ON $T^*\mathbb{Q}$ ===
First let's interpret $\theta$ as a one-form on $T^*\mathbb{Q}$ and take the Lie derivative of $\theta$ with respect to the vector field $\Delta_H$ on $T^*\mathbb{Q}$. We get:
$$\begin{align} L_{\Delta_H}(\theta) &= d(i_{\Delta_H}\theta) + i_{\Delta_H}d\theta\\ &= d(p^2) + i_{\Delta_H} dp \wedge dq\\ &= \left(\frac{\partial (p^2)}{\partial q}dq + \frac{\partial (p^2)}{\partial p} dp\right) + (-q\,dq - p\,dp)\\ &= 2p\,dp + (-q\,dq - p\,dp)\\ L_{\Delta_H}(\theta)&= p\,dp - q\,dq = d\left(\frac{p^2}{2}-\frac{q^2}{2}\right). \end{align}$$
This is consistent with the Euler-Lagrange equation usually defined on the tangent bundle $T\mathbb{Q}$ for the Lagrangian $\mathcal{L}(q,\dot{q}) = \dot{q}^2/2 - q^2/2$. This one makes sense to me.
=== #2: LIE DERIVATIVE ON $T(T^*\mathbb{Q})$ ===
However, I could also interpret $\theta$ as a one-form on $T(T^*Q)$ and study its flow with respect to the field $\Delta_T$. To be precise, I could calculate the Lie derivative on $T(T^*\mathbb{Q})$ in the following way:
$$\begin{align} L_{\Delta_T}(\theta) &= d(i_{\Delta_T}\theta) + i_{\Delta_T}d\theta, \\ &= d(p\dot{q}) + i_{\Delta_T}(dp\wedge dq), \\ &= \left[\frac{(p\dot{q})}{\partial q}dq + \frac{(p\dot{q})}{\partial p}dp + \frac{(p\dot{q})}{\partial \dot{q}}d\dot{q} + \frac{(p\dot{q})}{\partial \dot{p}}d\dot{p}\right] + (\dot{p}\,dq - \dot{q}\,dp), \\ &= (\dot{q}\,dp + p\,d\dot{q}) + (\dot{p}\,dq - \dot{q}\,dp), \\ L_{\Delta_T}(\theta) &= p\,d\dot{q} + \dot{p}\,dq. \end{align}$$
If we enforce the constraint $\dot{q} = p$ and $\dot{p} = -q$ from Hamilton's equations, we get the same result as in #1,
$$L_{\Delta_T}(\theta) = p\,dp - q\,dq.$$
So the first two calculations look consistent.
=== #3: SEEMINGLY MIXED APPROACH ===
Here we can again consider $\theta$ as a one-form on $T^* \mathbb{Q}$, but we try to calculate the Lie derivative with respect to the field $\Delta_T$ defined on $T(T^*\mathbb{Q})$. Naively,
$$ \begin{align} \begin{aligned} L_{\Delta_T}(\theta) &= d(i_{\Delta_T}\theta) + i_{\Delta_T}(d\theta) \\ &= d(p\dot{q}) + i_{\Delta_T}(dp \wedge dq) \\ &= \left(\frac{\partial (p \dot{q})}{\partial q}dq + \frac{\partial (p \dot{q})}{\partial p} dp \right) + \left(\dot{p}\,dq - \dot{q}\,dp \right) \\ &= \left[ \left(\frac{\partial p}{\partial q} \dot{q} + p \frac{\partial \dot{q}}{\partial q}\right) dq + \left(\frac{\partial p}{\partial p} \dot{q} + p \frac{\partial \dot{q}}{\partial p} \right) dp \right] + \left(\dot{p}\,dq - \dot{q}\,dp \right) \\ &= \dot{q} dp + \left(\dot{p}\,dq - \dot{q}\,dp \right) \\ L_{\Delta_T}(\theta) &= \dot{p}\,dq. \end{aligned} \end{align} $$
Here, the difference is that terms like $d\dot{q}$ and $d\dot{p}$ are not included in the differential because we work on $T^*\mathbb{Q}$. Moreover, the derivatives of $\dot{q}$ and $\dot{p}$ are interpreted as:
$$\frac{\partial \dot{q}}{\partial q} = \frac{d}{dt}\left(\frac{\partial{q}}{\partial{q}}\right) = \frac{d}{dt} \left( 1 \right) = 0,\;\;\textrm{and}\;\;\frac{\partial \dot{p}}{\partial q} = \frac{d}{dt}\left(\frac{\partial{p}}{\partial{q}}\right) = \frac{d}{dt} \left( 0 \right) = 0$$
where $d/dt$ is the total time derivative along the flow. Note that if I set the constraint $\dot{p} = -q$, I get $L_{\Delta_T}(\theta) = -q\,dq$. This is not the same as in #1 and #2.
I saw someone more knowledgeable than me perform this last calculation. But it seems very fishy to me and I can't find a good interpretation for it. Can anyone make sense of it in some context? Or is it just a plain mistake on their part?
Thank you in advance.