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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 Lagrange equations), or using canonical variables, $(\pmb{q},\pmb{p})\in\mathbb{R}^{2n}$ (satisfying Hamilton'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 $\pmb{z}=(a,e,i,\Omega,\omega,\nu)\in\mathbb{R}^6$). 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. But we could also transform to the classic Keplerian elements, $\pmb{z}=(a,e,i,\Omega,\omega,\nu)$ — semi-major axis, eccentricity, inclination, right ascension, argument of perigee, true anomaly — which may be mapped back and forth between $(\pmb{q},\dot{\pmb{q}})$ and $(\pmb{q},\pmb{p})$. Each $z^i$ is a function of both position and velocity (or momenta) coordinates. These $\pmb{z}$ are general 'state space' coordinates but exactly what are they coordinates for? A point on $\mathbf{T}\mathbb{Q}$? a point on $\mathbf{T}^*\mathbb{Q}$? some other manifold?

Edit: Essentially, what I called $(\pmb{q},\dot{\pmb{q}})$ and $(\pmb{q},\pmb{p})$ seem to be two particular "classes" of state space coordinates which have a certain geometric meaning behind them. In basic dynamical system theory, we often have a 'state vector' $\pmb{z}\in\mathbb{R}^m$, with dynamics given by some m first order ODEs, $\dot{\pmb{z}}=\pmb{f}(\pmb{z},t)$. The cases that $\pmb{z}=(\pmb{q},\dot{\pmb{q}})$ and $\pmb{z}=(\pmb{q},\pmb{p})$ seem to be two classes of state space coordinates for which there is a certain geometrical meaning behind them (tangent bundle and cotangent bundle coordinates, respectively). For a mechanical system, can $\pmb{z}$ always be classified into one of these two classes? What is the geometric meaning behind some general 'state vector'?

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    $\begingroup$ Are you looking for the word "polarization" of a symplectic manifold? Also if your manifold is Kahler then there is some notion of holomorphic/antiholomorphic which also cuts the phase space in hald $\endgroup$
    – user1379857
    Commented Nov 7, 2022 at 18:55
  • $\begingroup$ @user1379857 I'm not familiar with the polarization of a symplectic manifold or Kahler manifolds. But, after a quick search, I do think those concepts are related to my question, albeit perhaps in a more rigorous way than I am currently looking for. Or perhaps I am being lazy; I'll read a bit more and see. $\endgroup$
    – J Peterson
    Commented Nov 7, 2022 at 19:23
  • $\begingroup$ I am not sure about the geometrical meaning, but physically, points in phase space correspond to solutions of the equations of motion. This is a coordinate-independent definition of phase space. If you choose different coordinates, you are still describing points on the same phase space manifold. So in particular if your phase space is a (co)tangent bundle, it is the same bundle regardless of the coordinates you choose. $\endgroup$
    – d_b
    Commented Nov 7, 2022 at 19:49
  • $\begingroup$ @d_b Ok yeah I understand that but I only know how to think of state space, in general, as coordinates for $\mathbb{R}^{m}$ (m=2n). Or, in your words, as solutions of ODEs, where the ODEs are on $\mathbb{R}^{m}$. But the system we are interested in is usually not $\mathbb{R}^m$; it is some smooth manifold diffeomorphic to $\mathbb{R}^n$ such that its tangent and cotangent bundles are diffeomorphic to $\mathbb{R}^{2n}$. $\endgroup$
    – J Peterson
    Commented Nov 7, 2022 at 21:42
  • $\begingroup$ Further, what confuses me is that we can often easily come up with a coordinated transformation $(\pmb{q},\dot{\pmb{q}})\mapsto (\pmb{q},\pmb{p})$ on $\mathbb{R}^{m}$. But we've now switched from the tangent to the cotangent bundle, two distinctly different spaces. So are $(\pmb{q},\pmb{p})$ then also somehow coordinates for the tangent bundle, just in some chart that happens to coincide with coordinates for the cotangent bundle? $\endgroup$
    – J Peterson
    Commented Nov 7, 2022 at 21:44

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