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?
$\begingroup$
$\endgroup$
2
asked Jul 26, 2014 at 10:03
-
1$\begingroup$ The question presupposes that $M$ is connected; otherwise, the answer would trivially be yes. $\endgroup$– Andreas BlassCommented Jul 26, 2014 at 17:45
-
$\begingroup$ @AndreasBlass Thanks for the comment. I think you consider the disjoint union of two sphere? $\endgroup$– Ali TaghaviCommented Jul 26, 2014 at 18:24
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
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.
answered Jul 26, 2014 at 10:28
-
$\begingroup$ Thank you for your answer. Why a manifold is smoothly n -homogeneous? $\endgroup$ Commented Jul 26, 2014 at 10:39
-
1$\begingroup$ 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. $\endgroup$ Commented Jul 26, 2014 at 11:13