Is there a simple proof that there is no Anosov flow on $S^2$? Where can I find it?
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
3
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.
-
2$\begingroup$ You're right! What about Anosov diffeomorphisms? $\endgroup$– UagiCommented 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. $\endgroup$ Commented 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 NeadCommented Nov 23, 2022 at 11:33