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?
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.
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.
