what is the geometric interpretation of a general 'state space' in classical mechanics?
Let $\pmb{q}\in\mathbb{R}^n$ be some n generalized coordinates for the system (say, a double pendulum). Then the 'state space' is often examined using either the 'Lagrangian variables', $(\pmb{q},\dot{\pmb{q}})\in\mathbb{R}^{2n}$ (satisfying the Euler-Lagrange equations), or using canonical variables, $(\pmb{q},\pmb{p})\in\mathbb{R}^{2n}$ (satisfying <PERSON>'s canonical equations). But these are just coordinate representations (in some chart) in $\mathbb{R}^{2n}$; the former set, $(\pmb{q},\dot{\pmb{q}})$, are coordinates for a point on the tangent bundle, $\mathbf{T}\mathbb{Q}$, and the latter set, $(\pmb{q},\pmb{p})$, are coordinates for a point on the cotangent bundle, $\mathbf{T}^\mathbb{Q}$ (where $\mathbb{Q}$ is the n-dimensional configuration manifold). We could use some other coordinate chart, $(\pmb{q}',\dot{\pmb{q}}')$ or $(\pmb{q}',\pmb{p}')$, but these would still be coordinates for the same point on either $\mathbf{T}\mathbb{Q}$ or $\mathbf{T}^\mathbb{Q}$, respectively. Further, in these two types of state space coordinates, the first n coordinates are always 'position level' coordinates, while the last n coordinates are always 'velocity/momenta level' coordinates.
My question: We can also transform some other state space coordinates, $\pmb{z}\in\mathbb{R}^{2n}$, where the separation between 'position coordinates' and 'velocity/momenta coordinates' is lost (for example, the classic Keplerian orbital elements for the two body problem $\pmb{z}=(a,e,i,\Omega,\omega,\nu)\in\mathbb{R}^6$). 101
In general, what is the geometric meaning of some general state space coordinates, $\pmb{z}\in\mathbb{R}^{2n}$, which are simply any 2n independent coordinates which fully define the state of the system? This is the coordinate representation of a point on...what?
I suppose we could say that $\pmb{z}\mapsto (\pmb{q},\dot{\pmb{q}})$ is just a coordinate transformation (transition function) for a point $(\text{x},\mathbf{v})\in\mathbf{T}_{\text{x}}\mathbb{Q}$ in which case $\pmb{z}$ is just some alternative coordinate chart for $\mathbf{T}\mathbb{Q}$. But we could also come up with a coordinate transformation $\pmb{z}\mapsto (\pmb{q},\pmb{p})$ for some point $(\text{x},\mathbf{p})\in\mathbf{T}^_{\text{x}}\mathbb{Q}$ in which case $\pmb{z}$ would be some coordinate chart for $\mathbf{T}^\mathbb{Q}$. But $\mathbf{T}^*\mathbb{Q}\neq \mathbf{T}\mathbb{Q}$.
Example: The classic two-body problem. We could use 'lagrangian coordinates', $(\pmb{q},\dot{\pmb{q}})=(r,\phi,\theta,\dot{r},\dot{\phi},\dot{\theta})$, or canonical coordinates $(\pmb{q},\pmb{p})=(r,\phi,\theta,p_r,p_\phi , p_\theta )$, which are coordinates (in the spherical coordinate chart) for $\mathbf{T}\mathbb{Q}$ and $\mathbf{T}^\mathbb{Q}$, respectively. We could transform to their cartesian counterparts but they are just different coordinates for the same points. 101