2
$\begingroup$

Is there a simple proof that there is no Anosov flow on $S^2$? Where can I find it?

$\endgroup$

1 Answer 1

7
$\begingroup$

The usual definition of Anosov flow requires three invariant sub-bundles, so I guess you are actually asking about the 3-sphere?

Plante and Thurston have proved in

Plante, J. F.; Thurston, W. P., Anosov flows and the fundamental group, Topology 11, 147-150 (1972). ZBL0246.58014.

that if a manifold admits a codimension 1 Anosov flow, then its fundamental group has exponential growth.

$\endgroup$
3
  • 2
    $\begingroup$ You're right! What about Anosov diffeomorphisms? $\endgroup$
    – Uagi
    Nov 11, 2022 at 11:52
  • 3
    $\begingroup$ The only (orientable) surface admitting an Anosov diffeomorphism is the torus. One way to see this is using the Euler characteristic: the unstable foliation has no closed leaf (otherwise the diffeomorphism would be expanding on the closed leaf), and the only surface admitting such a foliation is the torus. $\endgroup$ Nov 11, 2022 at 12:20
  • $\begingroup$ @Uagi - If you admit "poles" (once-pronged singularities of the foliations) then you can obtain pseudo-Anosov homoemorphisms. This is why "pseudo-Anosov braids" can exist. $\endgroup$
    – Sam Nead
    Nov 23, 2022 at 11:33

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.