4
$\begingroup$

Is it possible to realize every finitely presented solvable group as a fundamental group of a stably parallelizable closed n-manifold? If not, are there any known counterexamples?

$\endgroup$
4
  • 1
    $\begingroup$ I think every finitely presented group is the fundamental group of a parallelizable manifold (take a suitable finite CW complex, embed it into some sufficiently large $\mathbb{R}^n$, and thicken it to an open submanifold). Do you want to restrict attention to closed manifolds? $\endgroup$ May 15, 2015 at 22:16
  • 1
    $\begingroup$ @QiaochuYuan, For closed you can take the doubling and realize it as a hypesurface in $\mathbb R^{n+1}$, the result is stably parallelizable. $\endgroup$ May 15, 2015 at 22:19
  • $\begingroup$ @Anton Petrunin. Thank you very much for your answer. Unfortunately, I don't completely understand why doubling preserves the property of being stably parallelizable. $\endgroup$
    – user73753
    May 16, 2015 at 11:22
  • 1
    $\begingroup$ If $D$ is a compact codimension zero submanifold of $\mathbb R^n$ and $I=[0,1]$, then the double of $D$ is (after smoothing corners) the boundary of $D\times I$ which is a closed submanifold of $R^{n+1}$. The double is thus the boundary of a stably parallelizable manifold, and hence it is stably parallelizable. $\endgroup$ May 16, 2015 at 12:21

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.