1
$\begingroup$

I'm having some to proof a question. I have to show that a compact manifold that admits a nowhere vanishing smooth vector field has a smooth map fixed-point free homotopic to the identity map.

I know that there exists a one-parameter group of diffeomorphisms that are smoothly homotopic to the identity, in fact the global flow of this vector field plays this role, since θ:R×M→M is a smooth left-action.

My question is: How can I show that exists t0∈R such that θt0 has no fixed points? (Here θt0(x)=θ(t0,x).

Improve this question
$\endgroup$
8
  • 1
    $\begingroup$ This isn't an appropriate forum for homework-type questions. $\endgroup$
    – Ryan Budney
    Commented Jan 14, 2015 at 18:49
  • $\begingroup$ It is not a homework, it is a little doubt of mine in my self-study. $\endgroup$
    – victor
    Commented Jan 14, 2015 at 18:53
  • $\begingroup$ You could try math.stackexchange.com, that would be more appropriate. $\endgroup$
    – Ryan Budney
    Commented Jan 14, 2015 at 18:54
  • $\begingroup$ FYI, it's relatively easy to construct a flow associated to a fixed-point free vector field where your map $\theta(t_0, \cdot)$ has fixed points for all $t_0 \in \mathbb R$. You might want to think about examples like that. A once-punctured plane does the job. You can write the vector field in polar coordinates. $\endgroup$
    – Ryan Budney
    Commented Jan 14, 2015 at 19:02
  • 1
    $\begingroup$ Crossposted: math.stackexchange.com/questions/1104186/… $\endgroup$
    – Qiaochu Yuan
    Commented Jan 14, 2015 at 19:19

1 Answer 1

Reset to default
1
$\begingroup$

A more general version of this result is proved by Brown and Fadell in:

MR0184236 (32 #1709) Reviewed Brown, Robert F.; Fadell, Edward Nonsingular path fields on compact topological manifolds. Proc. Amer. Math. Soc. 16 1965 1342–1349.

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged .