Put in other words, given an even-dimensional sphere $S^{2k}$: is there a manifold $M$ such that $T^* M$ is diffeomorphic to $S^{2k}$?
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8$\begingroup$ The sphere is compact and $T^*M$ is compact only for zero-dimensional $M$. $\endgroup$– HenrikRüpingCommented Nov 27, 2012 at 9:46
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$\begingroup$ $T^*M$ is not compact. $\endgroup$– ThiKuCommented Nov 27, 2012 at 9:47
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$\begingroup$ One may still ask whether a sphere can arise as the unit tangent bundle of some manifold. But then of course ist should have odd dimension. $\endgroup$– ThiKuCommented Nov 27, 2012 at 9:49
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4$\begingroup$ In physics, a phase space is no more and no less than a symplectic manifold. It only happens that the most common examples are cotangent bundles. In this sense, to make $S^{2k}$ a phase space, you need only find a symplectic form on it. I think that considering it as a coadjoint orbit of $SO(2k+1)$ might do the trick. $\endgroup$– Igor KhavkineCommented Nov 27, 2012 at 9:55
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3$\begingroup$ @Igor, I don't think so. For $k > 1$, $H^2(S^{2k}) = 0$. $\endgroup$– Oliver NashCommented Nov 27, 2012 at 10:02
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2 Answers
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Of course, the spheres are compact while cotangent bundles are noncompact (unless in dimension 0). Nevertheless, a bit more interesting is the question whether the even dimensional spheres can be phase spaces in the sense of symplectic manifolds. There the $\mathbb{S}^2$ is an example: the volume form is non-degenerate and a two-form. Closedness is for free in 2 dimensions. The higher dimensional spheres $\mathbb{S}^{2n}$ are never symplectic as on a compact symplectic manifold, the deRham cohomology has to be sufficiently non-trivial: the class of the symplectic form and all its $\wedge$-powers up to $n$ are non-trivial. For $\mathbb{S}^{2n}$ and $n \ge 2$ this is known to be not true: all cohomologies vanish except for the zeroth and the $2n$-th, which are both one-dimensional.
answered Nov 27, 2012 at 9:54
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$\begingroup$ Thanks for the quick reply. And please forgive my sloppiness in writing the question, which made it sound trivial (I put the cotangent thing thinking in the physics context). This answers exactly what I had in mind! $\endgroup$ Commented Nov 27, 2012 at 11:04
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To answer the question in the title: if by phase space we mean a symplectic manifold, then only for $k=1$ is there a symplectic structure. This is the phase space of a classical spin.
It is not necessary for a manifold to be identified with $T^*M$ for some $M$ to qualify as a phase space. This is the first place we encounter the idea, with $M$ being the configuration space of a system, but the concept is more general
