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I am asking the question because in the classical mechanics book by Arnold, he states that there is a distinguished $1$-form on $T^*V$. It seems that there is no such distinguished $1$-form on a general even-dimensional manifold.

So, not every even-dimensional manifold can be the phase space of a system, or not every even-dimensional manifold can be a cotangent bundle?

Is $S^2$ a cotangent bundle (intuitively it is not)? How about $S^4$?

Michael Albanese
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kaiser
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    Vector bundles are non-compact. – Michael Albanese Aug 19 '15 at 15:50
  • Great! I feel stupid to miss this simple point! – kaiser Aug 19 '15 at 15:59
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    The argument in the first paragraph is a red herring. While it's true that a manifold isn't equipped with a distinguished $1$-form, it's also true that a manifold that admits the structure of a cotangent bundle isn't equipped with a distinguished such structure. – Qiaochu Yuan Aug 19 '15 at 18:34

2 Answers2

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As I alluded to in the comments, vector bundles of positive rank are necessarily non-compact. Therefore, no compact manifold can be the total space of a vector bundle of positive rank over any topological space. In particular, neither $S^2$ nor $S^4$ can be a cotangent bundle.

On the other hand, any space can be viewed as a rank zero vector bundle over itself.

Michael Albanese
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The answer to your main question is no. Here is an argument (perhaps there are simpler arguments, but I came out with this one).

You can show that no sphere $S^{2n}$ for $n>1$ admits as symplectic structure. This was asked before here.

Let me elaborate a bit on the "distinguished" one-form to show that the cotangent bundle always admits a natural symplectic structure.

Local coordinates in the cotangent bundle are given by a pair $(x,a)$ where $p$ stands for a point on the manifold and $a$ for the coefficients of a basis in the cotangent space. In other words, $(p,a)$ stands for the point $a_idx^i$ with local coordinates centered at $p$. Consider the projection map $\pi\colon T^*M \to M$ which assings $(p,a)$ to $p$. These map induces a pullback $\pi\colon T^*M \to T^*T^*M$. You can identify $T^*T^*M$ with $T^*M$ itself. At each point $(p,a)$, define a one-form taking the value $\pi^*(p,a)$. This defines a one-form called the tautological one-form $\theta$. It is easy to prove that $d\theta$ defines a symplectic structure on $T^*M$.

This argument would prove that $S^4$ cannot be the cotangent bundle of a manifold, since it admits no symplectic structure.

I admit this is not the neatest presentation of the tautological one-form, but there is plenty about it on wikipedia.

EDIT: The other answer is definitely better. But I hope this illustrates a bit of the so-called distinguished form. The symplectic structure, after all, is import for the study of phase spaces.

Community
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Mauricio Tec
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