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Generally we "draw" phase space as typical coordinate system, where $q$s and $p$s are treated like perpendicular axes. Why do we then regard phase space as generall differential manifold while it seems to be typical euclidean space? There is no curvature for example of phase space. I know that there is something special (symplecticity) about phase space, but it's not my question.

Why do we call phase space as differential manifold, while we don't consider curvature, only flat Euclidean space?

Qmechanic
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Vicolls
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  • I don't understand your question. "Curvature" is a property of (pseudo-)Riemannian manifolds, i.e. differential manifold with a metric. A "pure" differential manifold has no metric and no curvature. As you also say, phase space is more properly a symplectic manifold, which is a differential manifold with a symplectic form. What, exactly, do you want to know here? – ACuriousMind Dec 26 '18 at 16:21
  • Although we don't have curvature we can create differential manifold witch is for example cylinder-shape 2D differential manniforld in 3D space. Can't we? We don't have to recall curvature or metric to "see" that something is cylinder. Yet phase space is always drawn as perpendicular axes for qs and ps. – Vicolls Dec 26 '18 at 16:51

1 Answers1

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  1. The geometric data in a Hamiltonian formulation is typically a symplectic manifold $(M,\omega)$. E.g. the manifold $M$ does not need to be an affine space, see e.g. this & this Phys.SE posts.

  2. One can (under mild technical assumptions) endow a symplectic manifold with a torsionfree symplectic tangent space connection $\nabla$, see e.g. this Phys.SE post for details. Although the corresponding scalar$^1$ curvature indeed vanishes, this is generically not so for the corresponding Riemann curvature tensor.

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$^1$ In symplectic geometry the scalar curvature is naturally defined by contracting the Riemann curvature tensor with the symplectic bi-vector.

Qmechanic
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