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Is there a manifold $M$ such that for every $x\in M$, $M-\{x\}$ is not parallelizable but there is a finite set $S\subset M$, with $\# S>1$, such that $M-S$ is parallelizable?

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    $\begingroup$ The question presupposes that $M$ is connected; otherwise, the answer would trivially be yes. $\endgroup$ Jul 26, 2014 at 17:45
  • $\begingroup$ @AndreasBlass Thanks for the comment. I think you consider the disjoint union of two sphere? $\endgroup$ Jul 26, 2014 at 18:24

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No: any finite set $S \subset M$ can be contained in the interior of an embedded closed disc in $M$, and cutting this out gives a manifold diffeomorphic to $M - \{x\}$. So if $M - S$ were parallelisable, $M-\{x\}$ would be too.

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  • $\begingroup$ Thank you for your answer. Why a manifold is smoothly n -homogeneous? $\endgroup$ Jul 26, 2014 at 10:39
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    $\begingroup$ mathoverflow.net/questions/91591/… : in the answers to this question you will find arguments that show that the diffeomorphism group acts $n$-transitively on the manifold. $\endgroup$ Jul 26, 2014 at 11:13

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