# a question about geometric quantization background

I don't understand why for geometric description of a regular system, we take always the classical phase space as a symplectic manifold?

• it seems unavoidable: points in phase space give you the states of the system, and Hamilton's equations of motion give you the symplectic structure. Without that you would not have classical mechanics as we know it. – Carlo Beenakker Oct 29 '13 at 17:31
• @Carlo: that explanation doesn't distinguish symplectic from Poisson manifolds. I can still talk about Hamilton's equations in the form $\dot{A} = \{ A, H \}$ on a Poisson manifold. – Qiaochu Yuan Oct 29 '13 at 18:56
• How should the classical phase space be taken? The initial data (or, essentially equivalently, the space of solutions) of a regular variational ODE/PDE does have symplectic structure. Do you think less of that data should be used, or more should be added? Depending on what you have in mind, there might actually be reasons for why that is not done. – Igor Khavkine Oct 29 '13 at 23:09
• Because normally mathematicians start with the easiest case? But geometric quantization for Poisson manifolds is also studied (e.g ncatlab.org/nlab/show/…) and is actually necessary with respect to quantization of for example the gauge orbit space stratification. – Tobias Diez Oct 30 '13 at 23:32

What is the classical phase space (by phase space I'll, atleast in this post, mean classical phase space)? It is simply the cotangent bundle $\mathrm{T}^{*}(\mathscr{C})$ of the configuration space $\mathscr{C}$. "... in a lecture Thurston mentioned "Any manifold can be seen as the configuration space of some physical system." (cf. MO/16460)". It can be seen as the cotangent bundle because of the Poisson bracket - see the answer to the above linked question. Any cotangent bundle of a finite-dimensional smooth manifold (the configuration space, $\mathscr{C}$, is finite-dimensional and smooth) is a symplectic manifold. For a proof, see this essay, page $7$. Hence, the classical phase space $\mathrm{T}^{*}(\mathscr{C})$ is always a symplectic manifold.