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Hamiltonian mechanics occurs in a sympletic manifold called phase space. Lagrangian mechanics take place in the tangent bundle of the configuration manifold.

Using Legendre transform makes possible to pass to Hamiltonian formulation, because this transform allows the construction of the cotangent bundle which is a sympletic manifold.

Lagrangian formulation is a sub-set of Hamiltonian formulation, because not all the phase spaces are cotangent bundles, there are indeed phase spaces which are compact, a property not present in cotangent bundles.

My question is how to show that cotangent bundles are not compact. My first approach to this demonstration was to use Whitney embedding theorem, but I failed since this theorem holds only for compact manifolds... Do you know another approach to show this fact?

Ruslan
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Edgar Ortiz
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2 Answers2

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No $\mathbb{R}$-vector bundle (of dimension $\geq 1$) is ever compact.

Proof: Let $S$ be an arbitrary topological space and let $T$ be a vector bundle of dimension $n\geq 1$, and let $p:T\rightarrow S$ be the bundle projection.

If $S$ a manifold, or is even $T_1$, or even more generally if any point of $S$ is closed, then the proof is particularly easy: let $s\in S$ be such that $\{s\}$ is a closed set. Then $p^{-1}(s) \cong \mathbb{R}^n$ is a closed set in $T$ because it is the preimage of a closed set under a continuous map. This set is obviously not compact. But if $T$ were compact, it would have to be: closed subsets of compact spaces are compact. This shows $T$ is not compact.

But there is a similar proof that works for any topological space, even one with no closed points. Let $S$ be such a space and let $T,p,n$ be as above. Let $\{U_i\}$ be a local trivialization of $T$, i.e. an open cover of $S$ such that $p^{-1}(U_i) \cong U_i\times \mathbb{R}^n$ for each $i$. This exists by definition of a vector bundle. There is no hope for $T$ to be compact unless $S$ is compact, since it is $T$'s continuous image under $p$, so we can restrict to this case and therefore can take $\{U_i\}$ to be a finite cover. Without loss of generality we can then take $\{U_i\}$ to be minimal in the sense that for every $U_i$, there exist points of $S$ that are only in $U_i$ and not in $U_j$ for any $j\neq i$. (I.e. no open set of the cover can be dropped. We achieve this by dropping redundant $U_i$'s until none are redundant.) Then, fixing some $i$, $V = \bigcap_{j\neq i} U_j^c$, the set of points that are only in $U_i$, is a nonempty closed set contained in $U_i$. Then $p^{-1}(V)\cong V\times \mathbb{R}^n$ since $V\subset U_i$, and this set is not compact since $\mathbb{R}^n$ is not. On the other hand, as above, $p^{-1}(V)$ is closed in $T$, so if $T$ were compact it would be too.

This argument shows in general that a fiber bundle can't be compact unless the fiber is compact. (Also, since the bundle projection is continuous, the bundle can't be compact unless the base is compact.)

Ben Blum-Smith
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  • [Trivial bundles over compact manifolds] whose vector spaces are compact give counterexamples to $\hspace{.3 in}$ your claim. $:$ (For example, consider a finite field with the discrete topology as a vector space over itself.) $;;;;$ –  Apr 26 '15 at 08:35
  • @RickyDemer The claim is only for rank at least 1 vector bundles over $\mathbb{R}$. – Ben Blum-Smith Apr 26 '15 at 16:36
  • The real claim is the final paragraph: a fiber bundle isn't compact if the fiber isn't compact. – Ben Blum-Smith Apr 26 '15 at 16:37
  • I added language to the answer to indicate that I mean $\mathbb{R}$-bundles. – Ben Blum-Smith Apr 26 '15 at 17:05
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Hints: It suffices to find a closed, non-compact subset of the total space. In an arbitrary vector bundle (of positive rank), there's a particularly natural way to do this.

Andrew D. Hwang
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