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I want to understand if it is always possible to find the following on a simply connected manifold $M$:

A countable family $\{U_{n}\}_{n=1}^{\infty}$ of simply connected open subsets with compact closure such that

  • $U_{n}\subset U_{n+1}$ $\forall n\in\mathbb{N}$,
  • $\bigcup_{n=1}^{\infty}U_{n} = M$.

Maybe this can be done by finding an adequate Moser function $f:M\rightarrow\mathbb{R}$.

In particular, I am interested in doing this for any symplectic manifold $(M,\omega)$.

Examples of manifolds admitting such exhaustions are $\mathbb{R}^{n}$ and compact simply connected manifolds (which no need to be symplectic).

Do you have any counterexample? Or, do you know a larger family of (symplectic) manifolds for which this construction holds?

Thank you.

Pd: Just to be precise, my definition of simply connected implies connected.

LaloVelasco
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  • I am interested in the case of symplectic manifolds. On a first moment I thought that this would hold for any simply connected manifold (not necessarily symplectic), but the Whitehead manifold is a counter example: http://math.stackexchange.com/questions/1081994/do-simply-connected-open-sets-admit-simply-connected-compact-exhaustions – LaloVelasco Sep 16 '16 at 09:44
  • The Whitehand manifold is not (yet) a counterexample, the answer further assume that the open sets are manifold with boundary. –  Sep 16 '16 at 19:21
  • Take the Whitehead manifold times $\Bbb R$ if you want a symplectic example.. – PVAL-inactive Sep 17 '16 at 02:50
  • @PVAL: Whitehead manifold times ${\mathbb R}$ is diffeomorphic to ${\mathbb R}^4$. However, there should be other 4-dimensional examples on non-tame simply-connected manifolds. – Moishe Kohan Sep 18 '16 at 22:31

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